1The Foundations
of Chemistry
OUTLINE
1-1 Matter and Energy
1-2 States of Matter
1-3 Chemical and Physical Properties
1-4 Chemical and Physical Changes
1-5 Mixtures, Substances,
Compounds, and Elements
1-6 Measurements in Chemistry
1-7 Units of Measurement
1-8 Use of Numbers
1-9 The Unit Factor Method
(Dimensional Analysis)
1-10 Percentage
1-11 Density and Specific Gravity
1-12 Heat and Temperature
1-13 Heat Transfer and the
Measurement of Heat
T
OBJECTIVES
After you have studied this chapter, you should be able to
• Use the basic vocabulary of matter and energy
• Distinguish between chemical and physical properties and between chemical and
physical changes
• Recognize various forms of matter: homogeneous and heterogeneous mixtures,
substances, compounds, and elements
• Apply the concept of significant figures
• Apply appropriate units to describe the results of measurement
• Use the unit factor method to carry out conversions among units
• Describe temperature measurements on various common scales, and convert between
these scales
• Carry out calculations relating temperature change to heat absorbed or liberated
T
housands of practical questions are studied by chemists. A few of them are
How can we modify a useful drug so as to improve its effectiveness while minimizing
any harmful or unpleasant side effects?
How can we develop better materials to be used as synthetic bone for replacement
surgery?
Which substances could help to avoid rejection of foreign tissue in organ transplants?
What improvements in fertilizers or pesticides can increase agricultural yields? How
can this be done with minimal environmental danger?
How can we get the maximum work from a fuel while producing the least harmful emissions
possible?
The earth is a huge chemical
system, including innumerable
reactions taking place constantly,
with some energy input from
sunlight. The earth serves as the
source of raw materials for all
human activities as well as the
depository for the products of these
activities. Maintaining life on the
planet requires understanding and
intelligent use of these resources.
Scientists can provide important
information about the processes, but
each of us must share in the
responsibility for our environment.
Which really poses the greater environmental threat—the burning of fossil fuels and
its contribution to the greenhouse effect and climatic change, or the use of nuclear
power and the related radiation and disposal problems?
How can we develop suitable materials for the semiconductor and microelectronics industry?
Can we develop a battery that is cheaper, lighter, and more powerful?
What changes in structural materials could help to make aircraft lighter and more economical,
yet at the same time stronger and safer?
What relationship is there between the substances we eat, drink, or breathe and the
possibility of developing cancer? How can we develop substances that are effective in
killing cancer cells preferentially over normal cells?
Can we economically produce fresh water from sea water for irrigation or consump-
tion?
How can we slow down unfavorable reactions, such as the corrosion of metals, while
speeding up favorable ones, such as the growth of foodstuffs?
Chemistry touches almost every aspect of our lives, our culture, and our environment. Its
scope encompasses the air we breathe, the food we eat, the fluids we drink, our clothing,
dwellings, transportation and fuel supplies, and our fellow creatures.
Chemistry is the science that describes matter—its properties, the changes it undergoes,
and the energy changes that accompany those processes.
Matter includes everything that is tangible, from our bodies and the stuff of our everyday
lives to the grandest objects in the universe. Some call chemistry the central science.
It rests on the foundation of mathematics and physics and in turn underlies the life
sciences—biology and medicine. To understand living systems fully, we must first
understand the chemical reactions and chemical influences that operate within them. The
chemicals of our bodies profoundly affect even the personal world of our thoughts and
emotions.
No one can be expert in all aspects of such a broad science as chemistry. Sometimes
we arbitrarily divide the study of chemistry into various branches. Carbon is very versatile
in its bonding and behavior and is a key element in many substances that are essential
to life. All living matter contains carbon combined with hydrogen. The chemistry of
compounds of carbon and hydrogen is called organic chemistry. (In the early days of
chemistry, living matter and inanimate matter were believed to be entirely different. We
now know that many of the compounds found in living matter can be made from nonliving,
or “inorganic,” sources. Thus, the terms “organic” and “inorganic” have different
meanings than they did originally.) The study of substances that do not contain carbon
combined with hydrogen is called inorganic chemistry. The branch of chemistry that is
concerned with the detection or identification of substances present in a sample (qualitative
analysis) or with the amount of each that is present (quantitative analysis) is called
analytical chemistry. Physical chemistry applies the mathematical theories and
methods of physics to the properties of matter and to the study of chemical processes and
the accompanying energy changes. As its name suggests, biochemistry is the study of
the chemistry of processes in living organisms. Such divisions are arbitrary, and most
chemical studies involve more than one of these traditional areas of chemistry. The
principles you will learn in a general chemistry course are the foundation of all branches
of chemistry.
Enormous numbers of chemical
reactions are necessary to produce
a human embryo (here at 10 weeks,
6 cm long).
CHAPTER 1: The Foundations of Chemistry 3
We understand simple chemical systems well; they lie near chemistry’s fuzzy boundary
with physics. They can often be described exactly by mathematical equations. We fare less
well with more complicated systems. Even where our understanding is fairly thorough,
we must make approximations, and often our knowledge is far from complete. Each year
researchers provide new insights into the nature of matter and its interactions. As chemists
find answers to old questions, they learn to ask new ones. Our scientific knowledge has
been described as an expanding sphere that, as it grows, encounters an ever-enlarging
frontier.
In our search for understanding, we eventually must ask fundamental questions, such
as the following:
How do substances combine to form other substances? How much energy is involved
in changes that we observe?
How is matter constructed in its intimate detail? How are atoms and the ways that they
combine related to the properties of the matter that we can measure, such as color,
hardness, chemical reactivity, and electrical conductivity?
What fundamental factors influence the stability of a substance? How can we force a
desired (but energetically unfavorable) change to take place? What factors control the
rate at which a chemical change takes place?
In your study of chemistry, you will learn about these and many other basic ideas that
chemists have developed to help them describe and understand the behavior of matter.
Along the way, we hope that you come to appreciate the development of this science, one
of the grandest intellectual achievements of human endeavor. You will also learn how to
apply these fundamental principles to solve real problems. One of your major goals in the
study of chemistry should be to develop your ability to think critically and to solve problems
(not just do numerical calculations!). In other words, you need to learn to manipulate
not only numbers, but also quantitative ideas, words, and concepts.
In the first chapter, our main goals are (1) to begin to get an idea of what chemistry is
about and the ways in which chemists view and describe the material world and (2) to
acquire some skills that are useful and necessary in the understanding of chemistry, its
contribution to science and engineering, and its role in our daily lives.
MATTER AND ENERGY
Matter is anything that has mass and occupies space. Mass is a measure of the quantity
of matter in a sample of any material. The more massive an object is, the more force is
required to put it in motion. All bodies consist of matter. Our senses of sight and touch
usually tell us that an object occupies space. In the case of colorless, odorless, tasteless
gases (such as air), our senses may fail us.
Energy is defined as the capacity to do work or to transfer heat. We are familiar with
many forms of energy, including mechanical energy, light energy, electrical energy, and
heat energy. Light energy from the sun is used by plants as they grow, electrical energy
allows us to light a room by flicking a switch, and heat energy cooks our food and warms
our homes. Energy can be classified into two principal types: kinetic energy and potential
energy.
A body in motion, such as a rolling boulder, possesses energy because of its motion.
Such energy is called kinetic energy. Kinetic energy represents the capacity for doing
work directly. It is easily transferred between objects. Potential energy is the energy an
1-1
We might say that we can “touch” air
when it blows in our faces, but we
depend on other evidence to show that
a still body of air fits our definition of
matter.
The term comes from the Greek word
kinein, meaning “to move.” The word
“cinema” is derived from the same
Greek word.
4 CHAPTER 1: The Foundations of Chemistry
object possesses because of its position, condition, or composition. Coal, for example,
possesses chemical energy, a form of potential energy, because of its composition. Many
electrical generating plants burn coal, producing heat, which is converted to electrical
energy. A boulder located atop a mountain possesses potential energy because of its height.
It can roll down the mountainside and convert its potential energy into kinetic energy.
We discuss energy because all chemical processes are accompanied by energy changes. As
some processes occur, energy is released to the surroundings, usually as heat energy. We
call such processes exothermic. Any combustion (burning) reaction is exothermic. Some
chemical reactions and physical changes, however, are endothermic; that is, they absorb
energy from their surroundings. An example of a physical change that is endothermic is
the melting of ice.
The Law of Conservation of Matter
When we burn a sample of metallic magnesium in the air, the magnesium combines with
oxygen from the air (Figure 1-1) to form magnesium oxide, a white powder. This chemical
reaction is accompanied by the release of large amounts of heat energy and light
energy. When we weigh the product of the reaction, magnesium oxide, we find that it is
heavier than the original piece of magnesium. The increase in the mass of a solid is due
to the combination of oxygen from the air with magnesium to form magnesium oxide.
Many experiments have shown that the mass of the magnesium oxide is exactly the sum
of the masses of magnesium and oxygen that combined to form it. Similar statements can
be made for all chemical reactions. These observations are summarized in the Law of
Conservation of Matter:
There is no observable change in the quantity of matter during a chemical reaction
or during a physical change.
This statement is an example of a scientific (natural) law, a general statement based on
the observed behavior of matter to which no exceptions are known. A nuclear reaction is
not a chemical reaction.
The Law of Conservation of Energy
In exothermic chemical reactions, chemical energy is usually converted into heat energy.
Some exothermic processes involve other kinds of energy changes. For example, some liberate
light energy without heat, and others produce electrical energy without heat or light.
In endothermic reactions, heat energy, light energy, or electrical energy is converted into
chemical energy. Although chemical changes always involve energy changes, some energy
transformations do not involve chemical changes at all. For example, heat energy may be
converted into electrical energy or into mechanical energy without any simultaneous
chemical changes. Many experiments have demonstrated that all of the energy involved
in any chemical or physical change appears in some form after the change. These observations
are summarized in the Law of Conservation of Energy:
Energy cannot be created or destroyed in a chemical reaction or in a physical change.
It can only be converted from one form to another.
Nuclear energy is an important kind of
potential energy.
1-1 Matter and Energy 5
Electricity is produced in hydroelectric
plants by the conversion of mechanical
energy (from flowing water) into
electrical energy.
Figure 1-1 Magnesium burns in
the oxygen of the air to form magnesium
oxide, a white solid. The mass
of magnesium oxide formed is equal
to the sum of the masses of oxygen
and magnesium that formed it.
The Law of Conservation of Matter and Energy
With the dawn of the nuclear age in the 1940s, scientists, and then the world, became
aware that matter can be converted into energy. In nuclear reactions (Chapter 26),
matter is transformed into energy. The relationship between matter and energy is given
by Albert Einstein’s now famous equation
E ϭ mc2
This equation tells us that the amount of energy released when matter is transformed into
energy is the product of the mass of matter transformed and the speed of light squared.
At the present time, we have not (knowingly) observed the transformation of energy into
matter on a large scale. It does, however, happen on an extremely small scale in “atom
smashers,” or particle accelerators, used to induce nuclear reactions. Now that the equivalence
of matter and energy is recognized, the Law of Conservation of Matter and
Energy can be stated in a single sentence:
The combined amount of matter and energy in the universe is fixed.
STATES OF MATTER
Matter can be classified into three states (Figure 1-2), although most of us can think of
examples that do not fit neatly into any of the three categories. In the solid state, substances
are rigid and have definite shapes. Volumes of solids do not vary much with changes
in temperature and pressure. In many solids, called crystalline solids, the individual particles
that make up the solid occupy definite positions in the crystal structure. The strengths
of interaction between the individual particles determine how hard and how strong the
crystals are. In the liquid state, the individual particles are confined to a given volume. A
liquid flows and assumes the shape of its container up to the volume of the liquid. Liquids
are very hard to compress. Gases are much less dense than liquids and solids. They
occupy all parts of any vessel in which they are confined. Gases are capable of infinite
expansion and are compressed easily. We conclude that they consist primarily of empty
space, meaning that the individual particles are quite far apart.
CHEMICAL AND PHYSICAL PROPERTIES
To distinguish among samples of different kinds of matter, we determine and compare
their properties. We recognize different kinds of matter by their properties, which are
broadly classified into chemical properties and physical properties.
Chemical properties are exhibited by matter as it undergoes changes in composition.
These properties of substances are related to the kinds of chemical changes that the
substances undergo. For instance, we have already described the combination of metallic
magnesium with gaseous oxygen to form magnesium oxide, a white powder. A chemical
property of magnesium is that it can combine with oxygen, releasing energy in the process.
A chemical property of oxygen is that it can combine with magnesium.
All substances also exhibit physical properties that can be observed in the absence of
any change in composition. Color, density, hardness, melting point, boiling point, and electrical
and thermal conductivities are physical properties. Some physical properties of a
1-3
1-2
Einstein formulated this equation
in 1905 as a part of his theory of
relativity. Its validity was demonstrated
in 1939 with the first controlled
nuclear reaction.
6 CHAPTER 1: The Foundations of Chemistry
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 1.3, States of Mattter.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 1.2, Physical Properties of
Matter.
The properties of a person include
height, weight, sex, skin and hair color,
and the many subtle features that
constitute that person’s general
appearance.
substance depend on the conditions, such as temperature and pressure, under which they
are measured. For instance, water is a solid (ice) at low temperatures but is a liquid at
higher temperatures. At still higher temperatures, it is a gas (steam). As water is converted
from one state to another, its composition is constant. Its chemical properties change very
little. On the other hand, the physical properties of ice, liquid water, and steam are different
(Figure 1-3).
Properties of matter can be further classified according to whether or not they depend
on the amount of substance present. The volume and the mass of a sample depend on,
and are directly proportional to, the amount of matter in that sample. Such properties,
which depend on the amount of material examined, are called extensive properties. By
contrast, the color and the melting point of a substance are the same for a small sample
and for a large one. Properties such as these, which are independent of the amount of
material examined, are called intensive properties. All chemical properties are intensive
properties.
Figure 1-2 A comparison of some physical properties of the three states of matter.
(Left) Iodine, a solid element. (Center) Bromine, a liquid element. (Right) Chlorine, a
gaseous element.
1-3 Chemical and Physical Properties 7
Property Solid
Rigidity Rigid
Expansion Slight
on heating
Compressibility Slight
Liquid
Flows and assumes
shape of container
Slight
Slight
Gas
Fills any container
completely
Expands infinitely
Easily compressed
Many compilations of chemical
and physical properties of matter can
be found on the World Wide Web. One
site is maintained by the U. S. National
Institute of Standards and Technology
(NIST) at http://webbook.nist.gov
Perhaps you can find other sites.
Figure 1-3 Physical changes that
occur among the three states of
matter. Sublimation is the conversion
of a solid directly to a gas without
passing through the liquid state;
the reverse of that process is called
deposition. The changes shown in
blue are endothermic (absorb heat);
those shown in red are exothermic
(release heat). Water is a substance
that is familiar to us in all three
physical states. The molecules are
close together in the solid and the
liquid but far apart in the gas. The
molecules in the solid are relatively
fixed in position, but those in the
liquid and gas can flow around each
other.
8 CHAPTER 1: The Foundations of Chemistry
Figure 1-4 Some physical and chemical properties of water. Physical: (a) It melts at 0°C;
(b) it boils at 100°C (at normal atmospheric pressure); (c) it dissolves a wide range of
substances, including copper(II) sulfate, a blue solid. Chemical: (d) It reacts with sodium
to form hydrogen gas and a solution of sodium hydroxide. The solution contains a little
phenolphthalein, which is pink in the presence of sodium hydroxide.
Gas
LiquidSolid
Melting
Freezing
Condensation
Evaporation
DepositionSublimation
(c)(a) (b) (d)
Because no two different substances have identical sets of chemical and physical properties
under the same conditions, we are able to identify and distinguish among different
substances. For instance, water is the only clear, colorless liquid that freezes at 0°C, boils
at 100°C at one atmosphere of pressure, dissolves a wide variety of substances (e.g.,
copper(II) sulfate), and reacts violently with sodium (Figure 1-4). Table 1-1 compares several
physical properties of a few substances. A sample of any of these substances can be
distinguished from the others by observing their properties.
CHEMICAL AND PHYSICAL CHANGES
We described the reaction of magnesium as it burns in the oxygen of the air (see Figure
1-1). This reaction is a chemical change. In any chemical change, (1) one or more substances
are used up (at least partially), (2) one or more new substances are formed, and
(3) energy is absorbed or released. As substances undergo chemical changes, they demonstrate
their chemical properties. A physical change, on the other hand, occurs with no
change in chemical composition. Physical properties are usually altered significantly as
matter undergoes physical changes (Figure 1-3). In addition, a physical change may
suggest that a chemical change has also taken place. For instance, a color change, a
warming, or the formation of a solid when two solutions are mixed could indicate a
chemical change.
Energy is always released or absorbed when chemical or physical changes occur. Energy
is required to melt ice, and energy is required to boil water. Conversely, the condensation
of steam to form liquid water always liberates energy, as does the freezing of liquid
1-4
One atmosphere of pressure is the
average atmospheric pressure at
sea level.
1-4 Chemical and Physical Changes 9
TABLE 1-1 Physical Properties of a Few Common Substances
(at 1 atm pressure)
Solubility at 25°C
(g/100 g)
Melting Boiling In In ethyl Density
Substance Point (°C) Point (°C) water alcohol (g/cm3)
acetic acid 16.6 118.1 infinite infinite 1.05
benzene 5.5 80.1 0.07 infinite 0.879
bromine Ϫ7.1 58.8 3.51 infinite 3.12
iron 1530 3000 insoluble insoluble 7.86
methane Ϫ182.5 Ϫ161.5 0.0022 0.033 6.67 ϫ 10Ϫ4
oxygen Ϫ218.8 Ϫ183.0 0.0040 0.037 1.33 ϫ 10Ϫ3
sodium chloride 801 1473 36.5 0.065 2.16
water 0 100 — infinite 1.00
See the Saunders Interactive
General Chemistry CD-ROM,
Screens 1.11, Chemical Change,
and 1.12, Chemical Change on the
Molecular Scale.
water to form ice. The changes in energy that accompany these physical changes for water
are shown in Figure 1-5. At a pressure of one atmosphere, ice always melts at the same
temperature (0°C), and pure water always boils at the same temperature (100°C).
MIXTURES, SUBSTANCES, COMPOUNDS, AND ELEMENTS
Mixtures are combinations of two or more pure substances in which each substance retains
its own composition and properties. Almost every sample of matter that we ordinarily
encounter is a mixture. The most easily recognized type of mixture is one in which
different portions of the sample have recognizably different properties. Such a mixture,
which is not uniform throughout, is called heterogeneous. Examples include mixtures of
salt and charcoal (in which two components with different colors can be distinguished
readily from each other by sight), foggy air (which includes a suspended mist of water
droplets), and vegetable soup. Another kind of mixture has uniform properties throughout;
such a mixture is described as a homogeneous mixture and is also called a
solution. Examples include salt water; some alloys, which are homogeneous mixtures of
metals in the solid state; and air (free of particulate matter or mists). Air is a mixture of
gases. It is mainly nitrogen, oxygen, argon, carbon dioxide, and water vapor. There are
only trace amounts of other substances in the atmosphere.
An important characteristic of all mixtures is that they can have variable composition.
(For instance, we can make an infinite number of different mixtures of salt and sugar by
varying the relative amounts of the two components used.) Consequently, repeating the
same experiment on mixtures from different sources may give different results, whereas
the same treatment of a pure sample will always give the same results. When the distinction
between homogeneous mixtures and pure substances was realized and methods were
developed (in the late 1700s) for separating mixtures and studying pure substances, consistent
results could be obtained. This resulted in reproducible chemical properties, which
formed the basis of real progress in the development of chemical theory.
1-5
Figure 1-5 Changes in energy that accompany some physical changes for water. The
energy unit joules (J) is defined in Section 1-13.
10 CHAPTER 1: The Foundations of Chemistry
1.00 g ice at 0°C 1.00 g liq H2O
at 0°C
1.00 g liq H2O
at 100°C
1.00 g steam
at 100°C
334 J absorbed
334 J released
418 J absorbed
–418 J released
+2260 J absorbed
–2260 J released
A heterogeneous mixture of two minerals: galena (black) and quartz (white).
By “composition of a mixture,” we
mean both the identities of the
substances present and their relative
amounts in the mixture.
The blue copper(II) sulfate solution in
Figure 1-4c is a homogeneous mixture.
The Development of ScienceCCHEMISTRY IN USE
The Resources of the Ocean
As is apparent to anyone who has swum in the ocean, sea
water is not pure water but contains a large amount of dissolved
solids. In fact, each cubic kilometer of seawater
contains about 3.6 ϫ 1010 kilograms of dissolved solids.
Nearly 71% of the earth’s surface is covered with water. The
oceans cover an area of 361 million square kilometers at an
average depth of 3729 meters, and hold approximately 1.35
billion cubic kilometers of water. This means that the oceans
contain a total of more than 4.8 ϫ 1021 kilograms of dissolved
material (or more than 100,000,000,000,000,000,000
pounds). Rivers flowing into the oceans and submarine volcanoes
constantly add to this storehouse of minerals. The
formation of sediment and the biological demands of organisms
constantly remove a similar amount.
Sea water is a very complicated solution of many
substances. The main dissolved component of sea water is
sodium chloride, common salt. Besides sodium and chlorine,
the main elements in sea water are magnesium, sulfur,
calcium, potassium, bromine, carbon, nitrogen, and strontium.
Together these 10 elements make up more than 99%
of the dissolved materials in the oceans. In addition to sodium
chloride, they combine to form such compounds as magnesium
chloride, potassium sulfate, and calcium carbonate
(lime). Animals absorb the latter from the sea and build it
into bones and shells.
Many other substances exist in smaller amounts in sea
water. In fact, most of the 92 naturally occurring elements
have been measured or detected in sea water, and the remainder
will probably be found as more sensitive analytical
techniques become available. There are staggering amounts
of valuable metals in sea water, including approximately 1.3 ϫ
1011 kilograms of copper, 4.2 ϫ 1012 kilograms of uranium,
5.3 ϫ 109 kilograms of gold, 2.6 ϫ 109 kilograms of silver,
and 6.6 ϫ 108 kilograms of lead. Other elements of economic
importance include 2.6 ϫ 1012 kilograms of aluminum, 1.3 ϫ
1010 kilograms of tin, 26 ϫ 1011 kilograms of manganese, and
4.0 ϫ 1010 kilograms of mercury.
One would think that with such a large reservoir of dissolved
solids, considerabe “chemical mining” of the ocean
would occur. At present, only four elements are commercially
extracted in large quantities. They are sodium and chlorine,
which are produced from the sea by solar evaporation; magnesium;
and bromine. In fact, most of the U. S. production
of magnesium is derived from sea water, and the ocean is one
of the principal sources of bromine. Most of the other elements
are so thinly scattered through the ocean that the cost
of their recovery would be much higher than their economic
value. However, it is probable that as resources become more
and more depleted from the continents, and as recovery techniques
become more efficient, mining of sea water will
become a much more desirable and feasible prospect.
One promising method of extracting elements from sea
water uses marine organisms. Many marine animals concentrate
certain elements in their bodies at levels many times
higher than the levels in sea water. Vanadium, for example,
is taken up by the mucus of certain tunicates and can be concentrated
in these animals to more than 280,000 times its
concentration in sea water. Other marine organisms can concentrate
copper and zinc by a factor of about 1 million. If
these animals could be cultivated in large quantities without
endangering the ocean ecosystem, they could become a valuable
source of trace metals.
In addition to dissolved materials, sea water holds a great
store of suspended particulate matter that floats through the
water. Some 15% of the manganese in sea water is present
in particulate form, as are appreciable amounts of lead and
iron. Similarly, most of the gold in sea water is thought to
adhere to the surfaces of clay minerals in suspension. As in
the case of dissolved solids, the economics of filtering these
very fine particles from sea water are not favorable at present.
However, because many of the particles suspended in
sea water carry an electric charge, ion exchange techniques
and modifications of electrostatic processes may someday
provide important methods for the recovery of trace metals.
Mixtures can be separated by physical means because each component retains its properties
(Figures 1-6 and 1-7). For example, a mixture of salt and water can be separated by
evaporating the water and leaving the solid salt behind. To separate a mixture of sand and
salt, we could treat it with water to dissolve the salt, collect the sand by filtration, and
then evaporate the water to reclaim the solid salt. Very fine iron powder can be mixed
with powdered sulfur to give what appears to the naked eye to be a homogeneous mixture
of the two. Separation of the components of this mixture is easy, however. The iron
may be removed by a magnet, or the sulfur may be dissolved in carbon disulfide, which
does not dissolve iron (Figure 1-6).
12 CHAPTER 1: The Foundations of Chemistry
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 1.13, Mixtures and Pure
Substances.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 1.14, Separation of Mixtures.
(a) (b)
MATTER
Everything that has mass
MIXTURES
Variable composition
HOMOGENEOUS MIXTURES
Have same composition throughout
COMPOUNDS
Chemical
changes
Physical
changes
Can be decomposed into
simpler substances by
chemical changes, always
in a definite ratio
ELEMENTS
Cannot be
decomposed into
simpler substances by
chemical changes
Components are indistinguishable
HETEROGENEOUS
MIXTURES
Do not have same composition
throughout
Components are distinguishable
Components retain their characteristic properties
PURE SUBSTANCES
Fixed composition
Properties do not vary
Cannot be separated into simpler substances by
physical methods
Can only be changed in identity and properties by
chemical methods
May be separated into pure substances by physical methods
Mixtures of different compositions may have widely
different properties
Figure 1-7 One scheme for classification of matter. Arrows indicate the general means by
which matter can be separated.
Figure 1-6 (a) A mixture of iron and sulfur is a heterogeneous mixture. (b) Like any mixture,
it can be separated by physical means, such as removing the iron with a magnet.
In any mixture, (1) the composition can be varied and (2) each component of the
mixture retains its own properties.
Imagine that we have a sample of muddy river water (a heterogeneous mixture). We
might first separate the suspended dirt from the liquid by filtration. Then we could remove
dissolved air by warming the water. Dissolved solids might be removed by cooling
the sample until some of it freezes, pouring off the liquid, and then melting the ice. Other
dissolved components might be separated by distillation or other methods. Eventually we
would obtain a sample of pure water that could not be further separated by any physical
separation methods. No matter what the original source of the impure water—the ocean,
the Mississippi River, a can of tomato juice, and so on—water samples obtained by purification
all have identical composition, and, under identical conditions, they all have
identical properties. Any such sample is called a substance, or sometimes a pure substance.
A substance cannot be further broken down or purified by physical means. A substance
is matter of a particular kind. Each substance has its own characteristic properties
that are different from the set of properties of any other substance.
Now suppose we decompose some water by passing electricity through it (Figure
1-8). (An electrolysis process is a chemical reaction.) We find that the water is converted
into two simpler substances, hydrogen and oxygen; more significantly, hydrogen and
Figure 1-8 Electrolysis
apparatus for small-scale chemical
decomposition of water by electrical
energy. The volume of hydrogen
produced (right) is twice that of
oxygen (left). Some dilute sulfuric
acid is added to increase the
conductivity.
1-5 Mixtures, Substances, Compounds, and Elements 13
The first ice that forms is quite pure.
The dissolved solids tend to stay
behind in the remaining liquid.
If we use the definition given here of a
substance, the phrase pure substance may
appear to be redundant.
oxygen
hydrogen
water
oxygen are always present in the same ratio by mass, 11.1% to 88.9%. These observations
allow us to identify water as a compound.
A compound is a substance that can be decomposed by chemical means into simpler
substances, always in the same ratio by mass.
As we continue this process, starting with any substance, we eventually reach a stage
at which the new substances formed cannot be further broken down by chemical means.
The substances at the end of this chain are called elements.
An element is a substance that cannot be decomposed into simpler substances by
chemical changes.
For instance, neither of the two gases obtained by the electrolysis of water—hydrogen
and oxygen—can be further decomposed, so we know that they are elements.
As another illustration (Figure 1-9), pure calcium carbonate (a white solid present in
limestone and seashells) can be broken down by heating to give another white solid (call
it A) and a gas (call it B) in the mass ratio 56.0Ϻ44.0. This observation tells us that calcium
carbonate is a compound. The white solid A obtained from calcium carbonate can
be further broken down into a solid and a gas in a definite ratio by mass, 71.5Ϻ28.5. But
neither of these can be further decomposed, so they must be elements. The gas is identical
to the oxygen obtained from the electrolysis of water; the solid is a metallic element
called calcium. Similarly, the gas B, originally obtained from calcium carbonate, can be
decomposed into two elements, carbon and oxygen, in a fixed mass ratio, 27.3Ϻ72.7. This
sequence illustrates that a compound can be broken apart into simpler substances at a
fixed mass ratio; those simpler substances may be either elements or simpler compounds.
Figure 1-9 Diagram of the decomposition of calcium carbonate to give a white solid A
(56.0% by mass) and a gas B (44.0% by mass). This decomposition into simpler substances
at a fixed ratio proves that calcium carbonate is a compound. The white solid A further
decomposes to give the elements calcium (71.5% by mass) and oxygen (28.5% by mass).
This proves that the white solid A is a compound; it is known as calcium oxide. The gas B
also can be broken down to give the elements carbon (27.3% by mass) and oxygen (72.7%
by mass). This establishes that gas B is a compound; it is known as carbon dioxide.
14 CHAPTER 1: The Foundations of Chemistry
pure calcium carbonate
white solid A
56.0% by mass 44.0% by mass
72.7%
by mass
27.3%
by mass
28.5%
by mass
71.5%
by mass
gas B
calcium oxygen carbon oxygen
Furthermore, we may say that a compound is a pure substance consisting of two or more different
elements in a fixed ratio. Water is 11.1% hydrogen and 88.9% oxygen by mass.
Similarly, carbon dioxide is 27.3% carbon and 72.7% oxygen by mass, and calcium oxide
(the white solid A in the previous discussion) is 71.5% calcium and 28.5% oxygen by mass.
We could also combine the numbers in the previous paragraph to show that calcium carbonate
is 40.1% calcium, 12.0% carbon, and 47.9% oxygen by mass. Observations such
as these on innumerable pure compounds led to the statement of the Law of Definite
Proportions (also known as the Law of Constant Composition):
Different samples of any pure compound contain the same elements in the same
proportions by mass.
The physical and chemical properties of a compound are different from the properties
of its constituent elements. Sodium chloride is a white solid that we ordinarily use as table
salt (Figure 1-10). This compound is formed by the combination of the element sodium
(a soft, silvery white metal that reacts violently with water; see Figure 1-4d) and the element
chlorine (a pale green, corrosive, poisonous gas; see Figure 1-2c).
Recall that elements are substances that cannot be decomposed into simpler substances
by chemical changes. Nitrogen, silver, aluminum, copper, gold, and sulfur are other examples
of elements.
We use a set of symbols to represent the elements. These symbols can be written more
quickly than names, and they occupy less space. The symbols for the first 109 elements
consist of either a capital letter or a capital letter and a lowercase letter, such as C (carbon)
or Ca (calcium). A list of the known elements and their symbols is given inside the front
cover.
In the past, the discoverers of elements claimed the right to name them (see the essay
“The Names of the Elements” on page 68), although the question of who had actually
discovered the elements first was sometimes disputed. In modern times, new elements are
given temporary names and three-letter symbols based on a numerical system. These
designations are used until the question of the right to name the newly discovered
elements is resolved. Decisions resolving the names of elements 104 through 109 were
announced in 1997 by the International Union of Pure and Applied Chemistry (IUPAC),
an international organization that represents chemical societies from 40 countries.
IUPAC makes recommendations regarding many matters of convention and terminology
in chemistry. These recommendations carry no legal force, but they are normally viewed
as authoritative throughout the world.
A short list of symbols of common elements is given in Table 1-2. Learning this list
will be helpful. Many symbols consist of the first one or two letters of the element’s English
name. Some are derived from the element’s Latin name (indicated in parentheses in Table
1-2) and one, W for tungsten, is from the German Wolfram. Names and symbols for
additional elements should be learned as they are encountered.
Most of the earth’s crust is made up of a relatively small number of elements. Only 10
of the 88 naturally occurring elements make up more than 99% by mass of the earth’s
crust, oceans, and atmosphere (Table 1-3). Oxygen accounts for roughly half. Relatively
few elements, approximately one fourth of the naturally occurring ones, occur in nature
as free elements. The rest are always found chemically combined with other elements.
A very small amount of the matter in the earth’s crust, oceans, and atmosphere is
involved in living matter. The main element in living matter is carbon, but only a tiny
Figure 1-10 The reaction of
sodium, a solid element, and
chlorine, a gaseous element, to
produce sodium chloride (table salt).
This reaction gives off considerable
energy in the form of heat and light.
1-5 Mixtures, Substances, Compounds, and Elements 15
The other known elements have been
made artificially in laboratories, as
described in Chapter 26.
fraction of the carbon in the environment occurs in living organisms. More than one
quarter of the total mass of the earth’s crust, oceans, and atmosphere is made up of
silicon, yet it has almost no biological role.
MEASUREMENTS IN CHEMISTRY
In the next section, we introduce the standards for basic units of measurement. These
standards were selected because they allow us to make precise measurements and because
they are reproducible and unchanging. The values of fundamental units are arbitrary.1 In
1-6
Mercury is the only metal that is a
liquid at room temperature.
1Prior to the establishment of the National Bureau of Standards in 1901, at least 50 different distances had been
used as “1 foot” in measuring land within New York City. Thus the size of a 100-ft by 200-ft lot in New York City
depended on the generosity of the seller and did not necessarily represent the expected dimensions.
TABLE 1-2 Some Common Elements and Their Symbols
Symbol Element Symbol Element Symbol Element
Ag silver (argentum) F fluorine Ni nickel
Al aluminum Fe iron (ferrum) O oxygen
Au gold (aurum) H hydrogen P phosphorus
B boron He helium Pb lead (plumbum)
Ba barium Hg mercury (hydrargyrum) Pt platinum
Bi bismuth I iodine S sulfur
Br bromine K potassium (kalium) Sb antimony (stibium)
C carbon Kr krypton Si silicon
Ca calcium Li lithium Sn tin (stannum)
Cd cadmium Mg magnesium Sr strontium
Cl chlorine Mn manganese Ti titanium
Co cobalt N nitrogen U uranium
Cr chromium Na sodium (natrium) W tungsten (Wolfram)
Cu copper (cuprum) Ne neon Zn zinc
TABLE 1-3 Abundance of Elements in the Earth’s Crust, Oceans,
and Atmosphere
Element Symbol % by Mass Element Symbol % by Mass
oxygen O 49.5% chlorine Cl 0.19%
silicon Si 25.7 phosphorus P 0.12
aluminum Al 7.5 manganese Mn 0.09
iron Fe 4.7 carbon C 0.08
calcium Ca 3.4
99.2%
sulfur S 0.06
0.7%
sodium Na 2.6 barium Ba 0.04
potassium K 2.4 chromium Cr 0.033
magnesium Mg 1.9 nitrogen N 0.030
hydrogen H 0.87 fluorine F 0.027
titanium Ti 0.58 zirconium Zr 0.023
All others combined Ϸ0.1%






























The stable form of sulfur at room
temperature is a solid.
the United States, all units of measure are set by the National Institute of Standards and
Technology, NIST (formerly the National Bureau of Standards, NBS). Measurements in
the scientific world are usually expressed in the units of the metric system or its modernized
successor, the International System of Units (SI). The SI, adopted by the National
Bureau of Standards in 1964, is based on the seven fundamental units listed in Table 1-4.
All other units of measurement are derived from them.
In this text we shall use both metric units and SI units. Conversions between non-SI
and SI units are usually straightforward. Appendix C lists some important units of
measurement and their relationships to one another. Appendix D lists several useful
physical constants. The most frequently used of these appear on the inside back
cover.
The metric and SI systems are decimal systems, in which prefixes are used to indicate fractions
and multiples of ten. The same prefixes are used with all units of measurement. The
distances and masses in Table 1-5 illustrate the use of some common prefixes and the
relationships among them.
The abbreviation SI comes from the
French le Système International.
1-6 Measurements in Chemistry 17
The prefixes used in the SI and
metric systems may be thought of as
multipliers. For example, the prefix kiloindicates
multiplication by 1000 or
103, and milli- indicates multiplication
by 0.001 or 10Ϫ3.
TABLE 1-4 The Seven Fundamental
Units of Measurement (SI)
Physical Property Name of Unit Symbol
length meter m
mass kilogram kg
time second s
electric current ampere A
temperature kelvin K
luminous intensity candela cd
amount of substance mole mol
TABLE 1-5 Common Prefixes Used in the SI and Metric Systems
Prefix Abbreviation Meaning Example
mega- M 106 1 megameter (Mm) ϭ 1 ϫ 106 m
kilo-* k 103 1 kilometer (km) ϭ 1 ϫ 103 m
deci- d 10Ϫ1 1 decimeter (dm) ϭ 1 ϫ 10Ϫ1 m
centi-* c 10Ϫ2 1 centimeter (cm) ϭ 1 ϫ 10Ϫ2 m
milli-* m 10Ϫ3 1 milligram (mg) ϭ 1 ϫ 10Ϫ3 g
micro-* ␮† 10Ϫ6 1 microgram (␮g) ϭ 1 ϫ 10Ϫ6 g
nano-* n 10Ϫ9 1 nanogram (ng) ϭ 1 ϫ 10Ϫ9 g
pico- p 10Ϫ12 1 picogram (pg) ϭ 1 ϫ 10Ϫ12 g
*These prefixes are commonly used in chemistry.
†This is the Greek letter ␮ (pronounced “mew”).
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 1.16, The Metric System.
UNITS OF MEASUREMENT
Mass and Weight
We distinguish between mass and weight. Mass is the measure of the quantity of matter
a body contains (see Section 1-1). The mass of a body does not vary as its position changes.
On the other hand, the weight of a body is a measure of the gravitational attraction of
the earth for the body, and this varies with distance from the center of the earth. An object
weighs very slightly less high up on a mountain than at the bottom of a deep valley.
Because the mass of a body does not vary with its position, the mass of a body is a more
fundamental property than its weight. We have become accustomed, however, to using
the term “weight” when we mean mass, because weighing is one way of measuring mass
(Figure 1-11). Because we usually discuss chemical reactions at constant gravity, weight
relationships are just as valid as mass relationships. Nevertheless, we should keep in mind
that the two are not identical.
The basic unit of mass in the SI system is the kilogram (Table 1-6). The kilogram is
defined as the mass of a platinum–iridium cylinder stored in a vault in Sèvres, near Paris,
France. A 1-lb object has a mass of 0.4536 kg. The basic mass unit in the earlier metric
system was the gram. A U.S. five-cent coin (a “nickel”) has a mass of about 5 g.
Length
The meter is the standard unit of length (distance) in both SI and metric systems. The
meter is defined as the distance light travels in a vacuum in 1/299,792,468 second. It is
approximately 39.37 inches. In situations in which the English system would use inches,
the metric centimeter (1/100 meter) is convenient. The relationship between inches and
centimeters is shown in Figure 1-12.
1-7
The meter was originally defined
(1791) as one ten-millionth of the
distance between the North Pole
and the equator.
18 CHAPTER 1: The Foundations of Chemistry
Figure 1-11 Three types of laboratory balances. (a) A triple-beam balance used for
determining mass to about Ϯ0.01 g. (b) A modern electronic top-loading balance that gives
a direct readout of mass to Ϯ0.001 g. (c) A modern analytical balance that can be used to
determine mass to Ϯ0.0001 g. Analytical balances are used when masses must be determined
as precisely as possible.
TABLE 1-6 Some SI Units
of Mass
kilogram, kg base unit
gram, g 1,000 g ϭ 1 kg
milligram, mg 1,000 mg ϭ 1 g
microgram, ␮g 1,000,000 ␮g ϭ 1 g
(a) (b)
(c)
Volume
Volumes are often measured in liters or milliliters in the metric system. One liter (1 L)
is one cubic decimeter (1 dm3), or 1000 cubic centimeters (1000 cm3). One milliliter
(1 mL) is 1 cm3. In medical laboratories, the cubic centimeter (cm3) is often abbreviated
cc. In the SI, the cubic meter is the basic volume unit and the cubic decimeter replaces
the metric unit, liter. Different kinds of glassware are used to measure the volume of liquids.
The one we choose depends on the accuracy we desire. For example, the volume of
a liquid dispensed can be measured more accurately with a buret than with a small graduated
cylinder (Figure 1-13). Equivalences between common English units and metric
units are summarized in Table 1-7.
Sometimes we must combine two or more units to describe a quantity. For instance,
we might express the speed of a car as 60 mi/h (also mph). Recall that the algebraic
notation xϪ1 means 1/x; applying this notation to units, we see that hϪ1 means 1/h, or
“per hour.” So the unit of speed could also be expressed as miиhϪ1.
Figure 1-13 Some laboratory
apparatus used to measure volumes
of liquids: 150-mL beaker (bottom
left, green liquid); 25-mL buret (top
left, red); 1000-mL volumetric flask
(center, yellow); 100-mL graduated
cylinder (right front, blue); and 10mL
volumetric pipet (right rear,
green).
1-7 Units of Measurement 19
Figure 1-12 The relationship between inches and centimeters: 1 in. ϭ 2.54 cm (exactly).
TABLE 1-7 Conversion Factors Relating Length, Volume, and Mass (weight) Units
Metric English Metric–English Equivalents
Length 1 km ϭ 103 m 1 ft ϭ 12 in. 2.54 cm ϭ 1 in.
1 cm ϭ 10Ϫ2 m 1 yd ϭ 3 ft 39.37 in.* ϭ 1 m
1 mm ϭ 10Ϫ3 m 1 mile ϭ 5280 ft 1.609 km* ϭ 1 mile
1 nm ϭ 10Ϫ9 m
1 Å ϭ 10Ϫ10 m
Volume 1 mL ϭ 1 cm3 ϭ 10Ϫ3 L 1 gal ϭ 4 qt ϭ 8 pt 1L ϭ 1.057 qt*
1 m3 ϭ 106 cm3 ϭ 103 L 1 qt ϭ 57.75 in.3* 28.32 L ϭ 1 ft3
Mass 1 kg ϭ 103 g 1 lb ϭ 16 oz 453.6 g* ϭ 1 lb
1 mg ϭ 10Ϫ3 g 1 g ϭ 0.03527 oz*
1 metric tonne ϭ 103 kg 1 short ton ϭ 2000 lb 1 metric tonne ϭ 1.102 short ton*
*These conversion factors, unlike the others listed, are inexact. They are quoted to four significant figures, which is ordinarily more than sufficient.
USE OF NUMBERS
In chemistry, we measure and calculate many things, so we must be sure we understand
how to use numbers. In this section we discuss two aspects of the use of numbers: (1) the
notation of very large and very small numbers and (2) an indication of how well we
actually know the numbers we are using. You will carry out many calculations with calculators.
Please refer to Appendix A for some instructions about the use of electronic
calculators.
Scientific Notation
We use scientific notation when we deal with very large and very small numbers. For
example, 197 grams of gold contains approximately
602,000,000,000,000,000,000,000 gold atoms
The mass of one gold atom is approximately
0.000 000 000 000 000 000 000 327 gram
In using such large and small numbers, it is inconvenient to write down all the zeroes. In
scientific (exponential) notation, we place one nonzero digit to the left of the decimal.
4,300,000. ϭ 4.3 ϫ 106
6 places to the left, І exponent of 10 is 6
0.000348 ϭ 3.48 ϫ 10Ϫ4
4 places to the right, І exponent of 10 is Ϫ4
The reverse process converts numbers from exponential to decimal form. See Appendix
A for more detail, if necessary.
1-8
In exponential form, these numbers
are
6.02 ϫ 1023 gold atoms
3.27 ϫ 10Ϫ22 gram
20 CHAPTER 1: The Foundations of Chemistry
Problem-Solving Tip: Know How to Use Your Calculator
Students sometimes make mistakes when they try to enter numbers into their calculators
in scientific notation. Suppose you want to enter the number 4.36 ϫ 10Ϫ2. On most
calculators, you would
(1) press 4.36
(2) press EE or EXP, which stands for “times ten to the”
(3) press 2 (the magnitude of the exponent) and then Ϯ or CHS (to change its sign)
The calculator display might show the value as 4.36 Ϫ02 or as 0.0436 . Different
calculators show different numbers of digits, which can sometimes be adjusted.
If you wished to enter Ϫ4.36 ϫ 102, you would
(1) press 4.36, then press Ϯ or CHS to change its sign,
(2) press EE or EXP, and then press 2
The calculator would then show Ϫ4.36 02 or Ϫ436.0 .
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 1.17, Using Numerical
Information.
Significant Figures
There are two kinds of numbers. Numbers obtained by counting or from definitions are exact
numbers. They are known to be absolutely accurate. For example, the exact number of
people in a closed room can be counted, and there is no doubt about the number of people.
A dozen eggs is defined as exactly 12 eggs, no more, no fewer (Figure 1-14).
An exact number may be thought of
as containing an infinite number of
significant figures.
1-8 Use of Numbers 21
Caution: Be sure you remember that the EE or EXP button includes the “times 10”
operation. An error that beginners often make is to enter “ ϫ 10” explicitly when trying
to enter a number in scientific notation. Suppose you mistakenly enter 3.7 ϫ 102 as fol-
lows:
(1) enter 3.7
(2) press ϫ and then enter 10
(3) press EXP or EE and then enter 2
The calculator then shows the result as 3.7 ϫ 103 or 3700—why? This sequence is
processed by the calculator as follows: Step (1) enters the number 3.7; step (2) multiplies
by 10, to give 37; step (3) multiplies this by 102, to give 37 ϫ 102 or 3.7 ϫ 103.
Other common errors include changing the sign of the exponent when the intent was
to change the sign of the entire number (e.g., Ϫ3.48 ϫ 104 entered as 3.48 ϫ 10Ϫ4).
When in doubt, carry out a trial calculation for which you already know the answer.
For instance, multiply 300 by 2 by entering the first value as 3.00 ϫ 102 and then multiplying
by 2; you know the answer should be 600, and if you get any other answer, you
know you have done something wrong. If you cannot find (or understand) the printed
instructions for your calculator, your instructor or a classmate might be able to help.
Figure 1-14 (a) A dozen eggs is exactly 12 eggs. (b) A specific swarm of honeybees
contains an exact number of live bees (but it would be difficult to count them, and any
two swarms would be unlikely to contain the same exact number of bees).
(a) (b)
Numbers obtained from measurements are not exact. Every measurement involves an estimate.
For example, suppose you are asked to measure the length of this page to the nearest
0.1 mm. How do you do it? The smallest divisions (calibration lines) on a meter stick
are 1 mm apart (see Figure 1-12). An attempt to measure to 0.1 mm requires estimation.
If three different people measure the length of the page to 0.1 mm, will they get the same
answer? Probably not. We deal with this problem by using significant figures.
Significant figures are digits believed to be correct by the person who makes a measurement.
We assume that the person is competent to use the measuring device. Suppose
one measures a distance with a meter stick and reports the distance as 343.5 mm. What
does this number mean? In this person’s judgment, the distance is greater than 343.4 mm
but less than 343.6 mm, and the best estimate is 343.5 mm. The number 343.5 mm contains
four significant figures. The last digit, 5, is a best estimate and is therefore doubtful,
but it is considered to be a significant figure. In reporting numbers obtained from
measurements, we report one estimated digit, and no more. Because the person making the
measurement is not certain that the 5 is correct, it would be meaningless to report the
distance as 343.53 mm.
To see more clearly the part significant figures play in reporting the results of measurements,
consider Figure 1-15a. Graduated cylinders are used to measure volumes of
liquids when a high degree of accuracy is not necessary. The calibration lines on a 50-mL
graduated cylinder represent 1-mL increments. Estimation of the volume of liquid in a
50-mL cylinder to within 0.2 mL (ᎏ
1
5
ᎏ of one calibration increment) with reasonable certainty
is possible. We might measure a volume of liquid in such a cylinder and report the
volume as 38.6 mL, that is, to three significant figures.
There is some uncertainty in all
measurements.
22 CHAPTER 1: The Foundations of Chemistry
Figure 1-15 Measurement of the volume of water using two types of volumetric glassware.
For consistency, we always read the bottom of the meniscus (the curved surface of the
water). (a) A graduated cylinder is used to measure the amount of liquid contained in the
glassware, so the scale increases from bottom to top. The level in a 50-mL graduated
cylinder can usually be estimated to within 0.2 mL. The level here is 38.6 mL (three
significant figures). (b) We use a buret to measure the amount of liquid delivered from
the glassware, by taking the difference between an initial and a final volume reading. The
level in a 50-mL buret can be read to within 0.02 mL. The level here is 38.57 mL (four
significant figures).
50
40
30
Read as
38.6 mL
Graduated
cylinder
38
39
40
Read as
38.57 mL
Buret
(a) (b)
Significant figures indicate the
uncertainty in measurements.
Burets are used to measure volumes of liquids when higher accuracy is required. The
calibration lines on a 50-mL buret represent 0.1-mL increments, allowing us to make
estimates to within 0.02 mL (ᎏ
1
5
ᎏ of one calibration increment) with reasonable certainty
(Figure 1-15b). Experienced individuals estimate volumes in 50-mL burets to 0.01 mL
with considerable reproducibility. For example, using a 50-mL buret, we can measure out
38.57 mL (four significant figures) of liquid with reasonable accuracy.
Accuracy refers to how closely a measured value agrees with the correct value.
Precision refers to how closely individual measurements agree with one another. Ideally,
all measurements should be both accurate and precise. Measurements may be quite precise
yet quite inaccurate because of some systematic error, which is an error repeated in
each measurement. (A faulty balance, for example, might produce a systematic error.) Very
accurate measurements are seldom imprecise.
Measurements are frequently repeated to improve accuracy and precision. Average
values obtained from several measurements are usually more reliable than individual
measurements. Significant figures indicate how precisely measurements have been made
(assuming the person who made the measurements was competent).
Some simple rules govern the use of significant figures.
1. Nonzero digits are always significant.
For example, 38.57 mL has four significant figures; 288 g has three significant figures.
2. Zeroes are sometimes significant, and sometimes they are not.
a. Zeroes at the beginning of a number (used just to position the decimal point)
are never significant.
For example, 0.052 g has two significant figures; 0.00364 m has three significant figures.
These could also be reported in scientific notation (Appendix A) as 5.2 ϫ 10Ϫ2 g and
3.64 ϫ 10Ϫ3 m, respectively.
b. Zeroes between nonzero digits are always significant.
For example, 2007 g has four significant figures; 6.08 km has three significant figures.
c. Zeroes at the end of a number that contains a decimal point are always sig-
nificant.
For example, 38.0 cm has three significant figures; 440.0 m has four significant figures.
These could also be reported as 3.80 ϫ 101 cm and 4.400 ϫ 102 m, respectively.
d. Zeroes at the end of a number that does not contain a decimal point may or
may not be significant.
1-8 Use of Numbers 23
For example, the quantity 24,300 km could represent three, four, or five significant figures.
We are given insufficient information to answer the question. If both of the zeroes
are used just to place the decimal point, the number should appear as 2.43 ϫ 104 km (three
significant figures). If only one of the zeroes is used to place the decimal point (i.e., the
number was measured Ϯ10), the number is 2.430 ϫ 104 km (four significant figures). If
the number is actually known to be 24,300 Ϯ 1, it should be written as 2.4300 ϫ 104 km
(five significant figures).
3. Exact numbers can be considered as having an unlimited number of significant
figures. This applies to defined quantities.
For example, in the equivalence 1 yard ϭ 3 feet, the numbers 1 and 3 are exact, and
we do not apply the rules of significant figures to them. The equivalence 1 inch ϭ 2.54
centimeters is an exact one.
A calculated number can never be more precise than the numbers used to calculate it.
The following rules show how to get the number of significant figures in a calculated
number.
4. In addition and subtraction, the last digit retained in the sum or difference is determined
by the position of the first doubtful digit.
EXAMPLE 1-1 Significant Figures (Addition and Subtraction)
(a) Add 37.24 mL and 10.3 mL. (b) Subtract 21.2342 g from 27.87 g.
Plan
We first check to see that the quantities to be added or subtracted are expressed in the same
units. We carry out the addition or subtraction. Then we follow Rule 4 for significant figures
to express the answer to the correct number of significant figures.
Solution
(a) 37.24 mL
ϩ10.30 mL
47.54 mL is reported as 47.5 mL (calculator gives 47.54)
(b) 27.8700g
Ϫ21.2342g
6.6358g is reported as 6.64 g (calculator gives 6.6358)
5. In multiplication and division, an answer contains no more significant figures
than the least number of significant figures used in the operation.
When we wish to specify that all of
the zeroes in such a number are
significant, we may indicate this by
placing a decimal point after the
number. For instance, 130. grams
can represent a mass known to three
significant figures, that is, 130 Ϯ 1
gram.
24 CHAPTER 1: The Foundations of Chemistry
Doubtful digits are underlined in this
example.
EXAMPLE 1-2 Significant Figures (Multiplication)
What is the area of a rectangle 1.23 cm wide and 12.34 cm long?
Plan
The area of a rectangle is its length times its width. We must first check to see that the width
and length are expressed in the same units. (They are, but if they were not, we must first convert
one to the units of the other.) Then we multiply the width by the length. We then follow
Rule 5 for significant figures to find the correct number of significant figures. The units for
the result are equal to the product of the units for the individual terms in the multiplication.
Solution
A ϭ ᐉ ϫ w ϭ (12.34 cm)(1.23 cm) ϭ 15.2 cm2
(calculator result ϭ 15.1782)
Because three is the smallest number of significant figures used, the answer should contain only
three significant figures. The number generated by an electronic calculator (15.1782) implies
more accuracy than is justified; the result cannot be more accurate than the information that
led to it. Calculators have no judgment, so you must exercise yours.
You should now work Exercise 27.
The step-by-step calculation in the margin demonstrates why the area is reported as
15.2 cm2 rather than 15.1782 cm2. The length, 12.34 cm, contains four significant figures,
whereas the width, 1.23 cm, contains only three. If we underline each uncertain figure,
as well as each figure obtained from an uncertain figure, the step-by-step multiplication
gives the result reported in Example 1-2. We see that there are only two certain
figures (15) in the result. We report the first doubtful figure (.2), but no more. Division
is just the reverse of multiplication, and the same rules apply.
In the three simple arithmetic operations we have performed, the number combination
generated by an electronic calculator is not the “answer” in a single case! The correct
result of each calculation, however, can be obtained by “rounding off.” The rules of
significant figures tell us where to round off.
In rounding off, certain conventions have been adopted. When the number to be
dropped is less than 5, the preceding number is left unchanged (e.g., 7.34 rounds off to
7.3). When it is more than 5, the preceding number is increased by 1 (e.g., 7.37 rounds
off to 7.4). When the number to be dropped is 5, the preceding number is set to the nearest
even number (e.g., 7.45 rounds off to 7.4, and 7.35 rounds off to 7.4).
With many examples we suggest
selected exercises from the end of the
chapter. These exercises use the skills
or concepts from that example. Now
you should work Exercise 27 from the
end of this chapter.
12.34 cm
ϫ 1.23 cm
3702
2468
1234
15.1782 cm2 ϭ 15.2 cm2
Rounding off to an even number is
intended to reduce the accumulation of
errors in chains of calculations.
1-8 Use of Numbers 25
Problem-Solving Tip: When Do We Round?
When a calculation involves several steps, we often show the answer to each step to the
correct number of significant figures. We carry all digits in the calculator to the end of
the calculation, however. Then we round the final answer to the appropriate number of
significant figures. When carrying out such a calculation, it is safest to carry extra figures
through all steps and then to round the final answer appropriately.
THE UNIT FACTOR METHOD
(DIMENSIONAL ANALYSIS)
Many chemical and physical processes can be described by numerical relationships. In fact,
many of the most useful ideas in science must be treated mathematically. In this section,
we review some problem-solving skills.
The units must always accompany the numeric value of a measurement, whether we
are writing about the quantity, talking about it, or using it in calculations.
Multiplication by unity (by one) does not change the value of an expression. If we represent
“one” in a useful way, we can do many conversions by just “multiplying by one.”
This method of performing calculations is known as dimensional analysis, the factorlabel
method, or the unit factor method. Regardless of the name chosen, it is a powerful
mathematical tool that is almost foolproof.
Unit factors may be constructed from any two terms that describe the same or equivalent
“amounts” of whatever we may consider. For example, 1 foot is equal to exactly 12
inches, by definition. We may write an equation to describe this equality:
1 ft ϭ 12 in.
Dividing both sides of the equation by 1 ft gives
ϭ or 1 ϭ
The factor (fraction) 12 in./1 ft is a unit factor because the numerator and denominator
describe the same distance. Dividing both sides of the original equation by 12 in. gives
1 ϭ 1 ft/12 in., a second unit factor that is the reciprocal of the first. The reciprocal of any
unit factor is also a unit factor. Stated differently, division of an amount by the same amount
always yields one!
In the English system we can write many unit factors, such as
, , , ,
The reciprocal of each of these is also a unit factor. Items in retail stores are frequently
priced with unit factors, such as 39¢/lb and $3.98/gal. When all the quantities in a unit
factor come from definitions, the unit is known to an unlimited (infinite) number of significant
figures. For instance, if you bought eight 1-gallon jugs of something priced at
$3.98/gal, the total cost would be 8 ϫ $3.98, or $31.84; the merchant would not round
this to $31.80, let alone to $30.
In science, nearly all numbers have units. What does 12 mean? Usually we must supply
appropriate units, such as 12 eggs or 12 people. In the unit factor method, the units guide
us through calculations in a step-by-step process, because all units except those in the desired
result cancel.
2000 lb
ᎏ
1 ton
4 qt
ᎏ
1 gal
1 mi
ᎏ
5280 ft
1 yd
ᎏ
36 in.
1 yd
ᎏ
3 ft
12 in.
ᎏ
1 ft
12 in.
ᎏ
1 ft
1 ft
ᎏ
1 ft
1-9
It would be nonsense to say that the
length of a piece of cloth is 4.7. We
must specify units with the number—
4.7 inches, 4.7 feet, or 4.7 meters, for
instance.
26 CHAPTER 1: The Foundations of Chemistry
Unless otherwise indicated, a “ton”
refers to a “short ton,” 2000 lb. There
are also the “long ton,” which is 2240
lb, and the metric tonne, which is
1000 kg.
EXAMPLE 1-3 Unit Factors
Express 1.47 mi in inches.
Plan
First we write down the units of what we wish to know, preceded by a question mark. Then
we set it equal to whatever we are given:
__? inches ϭ 1.47 miles
Then we choose unit factors to convert the given units (miles) to the desired units (inches):
miles 88n feet 88n inches
Solution
__? in. ϭ 1.47 mi ϫ ϫ ϭ 9.31 ϫ 104 in. (calculator gives 93139.2)
Note that both miles and feet cancel, leaving only inches, the desired unit. Thus, there is no
ambiguity as to how the unit factors should be written. The answer contains three significant
figures because there are three significant figures in 1.47 miles.
12 in.
ᎏ
1 ft
5280 ft
ᎏ
1 mi
We relate (a) miles to feet and then
(b) feet to inches.
In the interest of clarity, cancellation
of units will be omitted in the
remainder of this book. You may find
it useful to continue the cancellation
of units.
1-9 The Unit Factor Method (Dimensional Analysis) 27
Within the SI and metric systems, many measurements are related to one another by
powers of ten.
EXAMPLE 1-4 Unit Conversions
The Ångstrom (Å) is a unit of length, 1 ϫ 10Ϫ10 m, that provides a convenient scale on which
to express the radii of atoms. Radii of atoms are often expressed in nanometers. The radius of
a phosphorus atom is 1.10 Å. What is the distance expressed in centimeters and nanometers?
Problem-Solving Tip: Significant Figures
“How do defined quantities affect significant figures?” Any quantity that comes from a
definition is exact, that is, it is known to an unlimited number of significant figures. In
Example 1-3, the quantities 5280 ft, 1 mile, 12 in., and 1 ft all come from definitions,
so they do not limit the significant figures in the answer.
Problem-Solving Tip: Think About Your Answer!
It is often helpful to ask yourself, “Does the answer make sense?” In Example 1-3, the
distance involved is more than a mile. We expect this distance to be many inches, so a
large answer is not surprising. Suppose we had mistakenly multiplied by the unit factor
(and not noticed that the units did not cancel properly); we would have
gotten the answer 3.34 ϫ 10Ϫ3 in. (0.00334 in.), which we should have immediately recognized
as nonsense!
1 mile
ᎏᎏ
5280 feet
Plan
We use the equalities 1 Å ϭ 1 ϫ 10Ϫ10 m, 1 cm ϭ 1 ϫ 10Ϫ2 m, and 1 nm ϭ 1 ϫ 10Ϫ9 m to
construct the unit factors that convert 1.10 Å to the desired units.
Solution
__? cm ϭ 1.10 Å ϫ ϫ ϭ 1.10 ϫ 10Ϫ8 cm
__? nm ϭ 1.10 Å ϫ ϫ ϭ 1.10 ϫ 10Ϫ1 nm
You should now work Exercise 30.
EXAMPLE 1-5 Volume Calculation
Assuming a phosphorus atom is spherical, calculate its volume in Å3, cm3, and nm3. The formula
for the volume of a sphere is V ϭ (ᎏ4
3
ᎏ)␲r3. Refer to Example 1-4.
Plan
We use the results of Example 1-4 to calculate the volume in each of the desired units.
Solution
__? Å3 ϭ (ᎏ4
3
ᎏ)␲(1.10 Å)3 ϭ 5.58 Å3
__? cm3 ϭ (ᎏ4
3
ᎏ)␲(1.10 ϫ 10Ϫ8 cm)3 ϭ 5.58 ϫ 10Ϫ24 cm3
__? nm3 ϭ (ᎏ4
3
ᎏ)␲(1.10 ϫ 10Ϫ1 nm)3 ϭ 5.58 ϫ 10Ϫ3 nm3
You should now work Exercise 34.
EXAMPLE 1-6 Mass Conversion
A sample of gold has a mass of 0.234 mg. What is its mass in g? in kg?
Plan
We use the relationships 1 g ϭ 1000 mg and 1 kg ϭ 1000 g to write the required unit factors.
Solution
__? g ϭ 0.234 mg ϫ ϭ 2.34 ϫ 10Ϫ4 g
__? kg ϭ 2.34 ϫ 10Ϫ4 g ϫ ϭ 2.34 ϫ 10Ϫ7 kg
Again, this example includes unit factors that contain only exact numbers.
1 kg
ᎏ
1000 g
1 g
ᎏᎏ
1000 mg
1 nm
ᎏᎏ
1 ϫ 10Ϫ9 m
1.0 ϫ 10Ϫ10 m
ᎏᎏ
1 Å
1 cm
ᎏᎏ
1 ϫ 10Ϫ2 m
1 ϫ 10Ϫ10 m
ᎏᎏ
1 Å
All the unit factors used in this
example contain only exact numbers.
28 CHAPTER 1: The Foundations of Chemistry
Problem-Solving Tip: Conversions Within the Metric or SI System
The SI and metric systems of units are based on powers of ten. This means that many
unit conversions within these systems can be carried out just by shifting the decimal point.
For instance, the conversion from milligrams to grams in Example 1-6 just
Å n m n cm
Å n m n nm
1 Å ϭ 10Ϫ10 m ϭ 10Ϫ8 cm
Suppose we start with the equality
1 in. ϭ 2.54 cm
We can perform the same operation on
both sides of the equation. Let’s cube
both sides:
(1 in.)3 ϭ (2.54 cm)3 ϭ 16.4 cm3
so the quantity
ϭ 1
is a unit factor.
(1 in.)3
ᎏᎏ
(2.54 cm)3
1-9 The Unit Factor Method (Dimensional Analysis) 29
involves shifting the decimal point to the left by three places. How do we know to move
it to the left? We know that the gram is a larger unit of mass than the milligram, so the
number of grams in a given mass must be a smaller number than the number of
milligrams. After you carry out many such conversions using unit factors, you will
probably begin to take such shortcuts. Always think about the answer, to see whether it
should be larger or smaller than the quantity was before conversion.
Unity raised to any power is 1. Any unit factor raised to a power is still a unit factor,
as the next example shows.
EXAMPLE 1-7 Volume Conversion
One liter is exactly 1000 cm3. How many cubic inches are there in 1000 cm3?
Plan
We would multiply by the unit factor ᎏ
2.
1
54
in
c
.
m
ᎏ to convert cm to in. Here we require the cube
of this unit factor.
Solution
__? in.3 ϭ 1000 cm3 ϫ
΂ ΃
3
ϭ 1000 cm3 ϫ ϭ 61.0 in.3
Example 1-7 shows that a unit factor cubed is still a unit factor.
EXAMPLE 1-8 Energy Conversion
A common unit of energy is the erg. Convert 3.74 ϫ 10Ϫ2 erg to the SI units of energy, joules,
and kilojoules. One erg is exactly 1 ϫ 10Ϫ7 joule (J).
Plan
The definition that relates ergs and joules is used to generate the needed unit factor. The second
conversion uses a unit factor that is based on the definition of the prefix kilo-.
Solution
__? J ϭ 3.74 ϫ 10Ϫ2 erg ϫ ϭ 3.74 ϫ 10Ϫ9 J
__? kJ ϭ 3.74 ϫ 10Ϫ9 J ϫ ϭ 3.74 ϫ 10Ϫ12 kJ
Conversions between the English and SI (metric) systems are conveniently made by
the unit factor method. Several conversion factors are listed in Table 1-7. It may be helpful
to remember one each for
length 1 in. ϭ 2.54 cm (exact)
mass and weight 1 lb ϭ 454 g (near sea level)
volume 1 qt ϭ 0.946 L or 1 L ϭ 1.06 qt
1 kJ
ᎏ
1000 J
1 ϫ 10Ϫ7 J
ᎏᎏ
1 erg
1 in.
ᎏᎏ
16.4 cm3
1 in.
ᎏ
2.54 cm
We relate
(a) gallons to quarts, then
(b) quarts to liters, and then
(c) liters to milliliters.
30 CHAPTER 1: The Foundations of Chemistry
EXAMPLE 1-9 English–Metric Conversion
Express 1.0 gallon in milliliters.
Plan
We ask __? mL ϭ 1.0 gal and multiply by the appropriate factors.
gallons 88n quarts 88n liters 88n milliliters
Solution
__? mL ϭ 1.0 gal ϫ ϫ ϫ ϭ 3.8 ϫ 103 mL
You should now work Exercise 32.
The fact that all other units cancel to give the desired unit, milliliters, shows that we used
the correct unit factors. The factors 4 qt/gal and 1000 mL/L contain only exact numbers.
The factor 1 L/1.06 qt contains three significant figures. Because 1.0 gal contains only
two, the answer contains only two significant figures.
Examples 1-1 through 1-9 show that multiplication by one or more unit factors
changes the units and the number of units, but not the amount of whatever we are
calculating.
PERCENTAGE
We often use percentages to describe quantitatively how a total is made up of its parts.
In Table 1-3, we described the amounts of elements present in terms of the percentage
of each element.
Percentages can be treated as unit factors. For any mixture containing substance A,
ϭ
Mass A m88888888888888888888888888888888888888888n Mass mixture
If we say that a sample is 24.4% carbon by mass, we mean that out of every 100 parts (exactly)
by mass of sample, 24.4 parts by mass are carbon. This relationship can be represented
by whichever of the two unit factors we find useful:
or
This ratio can be expressed in terms of grams of carbon for every 100 grams of sample,
pounds of carbon for every 100 pounds of sample, or any other mass or weight unit. The
next example illustrates the use of dimensional analysis involving percentage.
100 parts sample
ᎏᎏ
24.4 parts carbon
24.4 parts carbon
ᎏᎏ
100 parts sample
Parts A (by mass)
ᎏᎏᎏᎏ
100 parts mixture (by mass)
% A
ᎏᎏ
(by mass)
1-10
1000 mL
ᎏᎏ
1 L
1 L
ᎏ
1.06 qt
4 qt
ᎏ
1 gal
EXAMPLE 1-10 Percentage
U.S. pennies made since 1982 consist of 97.6% zinc and 2.4% copper. The mass of a particular
penny is measured to be 1.494 grams. How many grams of zinc does this penny contain?
Plan
From the percentage information given, we may write the required unit factor
Solution
__? g zinc ϭ 1.494 g sample ϫ ϭ 1.46 g zinc
The number of significant figures in the result is limited by the three significant figures in
97.6%. Because the definition of percentage involves exactly 100 parts, the number 100 is known
to an infinite number of significant figures.
You should now work Exercises 59 and 60.
DENSITY AND SPECIFIC GRAVITY
In science, we use many terms that involve combinations of different units. Such quantities
may be thought of as unit factors that can be used to convert among these units. The
density of a sample of matter is defined as the mass per unit volume:
density ϭ or D ϭ
Densities may be used to distinguish between two substances or to assist in identifying a
particular substance. They are usually expressed as g/cm3 or g/mL for liquids and solids
and as g/L for gases. These units can also be expressed as gиcmϪ3, gиmLϪ1, and
gиLϪ1, respectively. Densities of several substances are listed in Table 1-8.
EXAMPLE 1-11 Density, Mass, Volume
A 47.3-mL sample of ethyl alcohol (ethanol) has a mass of 37.32 g. What is its density?
Plan
We use the definition of density.
Solution
D ϭ ϭ ϭ 0.789 g/mL
You should now work Exercise 36.
37.32 g
ᎏ
47.3 mL
m
ᎏ
V
m
ᎏ
V
mass
ᎏ
volume
1-11
97.6 g zinc
ᎏᎏ
100 g sample
97.6 g zinc
ᎏᎏ
100 g sample
The intensive property density relates
the two extensive properties: mass and
volume.
1-11 Density and Specific Gravity 31
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 1.8, Density.
Mass
Density ϭ
Volume
Mass
ᎏ
Volume
h
666666666
g
Six materials with different densities.
The liquid layers are gasoline (top),
water (middle), and mercury (bottom).
A cork floats on gasoline. A piece of
oak wood sinks in gasoline but floats
on water. Brass sinks in water but
floats on mercury.
EXAMPLE 1-12 Density, Mass, Volume
If 116 g of ethanol is needed for a chemical reaction, what volume of liquid would you use?
Plan
We determined the density of ethanol in Example 1-11. Here we are given the mass, m, of a
sample of ethanol. So we know values for D and m in the relationship
D ϭ
We rearrange this relationship to solve for V, put in the known values, and carry out the
calculation. Alternatively, we can use the unit factor method to solve the problem.
Solution
The density of ethanol is 0.789 g/mL (Table 1-8).
D ϭ , so V ϭ ϭ ϭ 147 mL
Alternatively,
__? mL ϭ 116 g ϫ ϭ 147 mL
You should now work Exercise 39.
EXAMPLE 1-13 Unit Conversion
Express the density of mercury in lb/ft3.
Plan
The density of mercury is 13.59 g/cm3 (see Table 1-8). To convert this value to the desired
units, we can use unit factors constructed from the conversion factors in Table 1-7.
1 mL
ᎏ
0.789 g
116 g
ᎏᎏ
0.789 g/mL
m
ᎏ
D
m
ᎏ
V
m
ᎏ
V
These densities are given at room
temperature and one atmosphere
pressure, the average atmospheric
pressure at sea level. Densities of solids
and liquids change only slightly, but
densities of gases change great]y, with
changes in temperature and pressure.
32 CHAPTER 1: The Foundations of Chemistry
Observe that density gives two unit
factors. In this case, they are ᎏ
0
1
.7
m
89
L
g
ᎏ
and ᎏ
0
1
.7
m
89
L
g
ᎏ
TABLE 1-8 Densities of Common Substances*
Substance Density (g/cm3) Substance Density (g/cm3)
hydrogen (gas) 0.000089 sand* 2.32
carbon dioxide (gas) 0.0019 aluminum 2.70
cork* 0.21 iron 7.86
oak wood* 0.71 copper 8.92
ethyl alcohol 0.789 silver 10.50
water 1.00 lead 11.34
magnesium 1.74 mercury 13.59
table salt 2.16 gold 19.30
*Cork, oak wood, and sand are common materials that have been included to provide familiar reference points. They
are not pure elements or compounds as are the other substances listed.
Solution
__? ϭ 13.59 ϫ ϫ ΂ ΃
3
ϫ
΂ ΃
3
ϭ 848.4 lb/ft3
It would take a very strong person to lift a cubic foot of mercury!
The specific gravity (Sp. Gr.) of a substance is the ratio of its density to the density
of water, both at the same temperature.
Sp. Gr. ϭ
The density of water is 1.000 g/mL at 3.98°C, the temperature at which the density of
water is greatest. Variations in the density of water with changes in temperature, however,
are small enough that we can use 1.00 g/mL up to 25°C without introducing significant
errors into our calculations.
EXAMPLE 1-14 Density, Specific Gravity
The density of table salt is 2.16 g/mL at 20°C. What is its specific gravity?
Plan
We use the preceding definition of specific gravity. The numerator and denominator have the
same units, so the result is dimensionless.
Solution
Sp. Gr. ϭ ϭ ϭ 2.16
2.16 g/mL
ᎏᎏ
1.00 g/mL
Dsalt
ᎏ
Dwater
Dsubstance
ᎏᎏ
Dwater
12 in.
ᎏ
1 ft
2.54 cm
ᎏ
1 in.
1 lb
ᎏ
453.6 g
g
ᎏ
cm3
lb
ᎏ
ft3
Density and specific gravity are both
intensive properties; that is, they do
not depend on the size of the sample.
Specific gravities are dimensionless
numbers.
1-11 Density and Specific Gravity 33
Ice is slightly less dense than liquid
water, so ice floats in water.
Solid ethyl alcohol is more
dense than liquid ethyl alcohol.
This is true of nearly
every known substance.
This example also demonstrates that the density and specific gravity of a substance are
numerically equal near room temperature if density is expressed in g/mL (g/cm3).
Labels on commercial solutions of acids give specific gravities and the percentage by
mass of the acid present in the solution. From this information, the amount of acid present
in a given volume of solution can be calculated.
EXAMPLE 1-15 Specific Gravity, Volume, Percentage by Mass
Battery acid is 40.0% sulfuric acid, H2SO4, and 60.0% water by mass. Its specific gravity is
1.31. Calculate the mass of pure H2SO4 in 100.0 mL of battery acid.
Plan
The percentages are given on a mass basis, so we must first convert the 100.0 mL of acid
solution (soln) to mass. To do this, we need a value for the density. We have demonstrated
that density and specific gravity are numerically equal at 20°C because the density of water
is 1.00 g/mL. We can use the density as a unit factor to convert the given volume of the
solution to mass of the solution. Then we use the percentage by mass to convert the mass of
the solution to the mass of the acid.
Solution
From the given value for specific gravity, we may write
Density ϭ 1.31 g/mL
The solution is 40.0% H2SO4 and 60.0% H2O by mass. From this information we may construct
the desired unit factor:
88n
We can now solve the problem:
__? H2SO4 ϭ 100.0 mL soln ϫ ϫ ϭ 52.4 g H2SO4
You should now work Exercise 43.
HEAT AND TEMPERATURE
In Section 1-1 you learned that heat is one form of energy. You also learned that the many
forms of energy can be interconverted and that in chemical processes, chemical energy is
converted to heat energy or vice versa. The amount of heat a process uses (endothermic)
or gives off (exothermic) can tell us a great deal about that process. For this reason it is
important for us to be able to measure the intensity of heat.
Temperature measures the intensity of heat, the “hotness” or “coldness” of a body. A
piece of metal at 100°C feels hot to the touch, whereas an ice cube at 0°C feels cold.
Why? Because the temperature of the metal is higher, and that of the ice cube lower, than
body temperature. Heat is a form of energy that always flows spontaneously from a hotter
body to a colder body—never in the reverse direction.
Temperatures can be measured with mercury-in-glass thermometers. A mercury thermometer
consists of a reservoir of mercury at the base of a glass tube, open to a very thin
1-12
40.0 g H2SO4
ᎏᎏ
100 g soln
1.31 g soln
ᎏᎏ
1 mL soln
because 100 g of solution
ᎏᎏᎏ
contains 40.0 g of H2SO4
40.0 g H2SO4
ᎏᎏ
100 g soln
34 CHAPTER 1: The Foundations of Chemistry
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 1.10, Temperature.
(capillary) column extending upward. Mercury expands more than most other liquids as
its temperature rises. As it expands, its movement up into the evacuated column can be
seen.
Anders Celsius (1701–1744), a Swedish astronomer, developed the Celsius temperature
scale, formerly called the centigrade temperature scale. When we place a Celsius
thermometer in a beaker of crushed ice and water, the mercury level stands at exactly 0°C,
the lower reference point. In a beaker of water boiling at one atmosphere pressure, the
mercury level stands at exactly 100°C, the higher reference point. There are 100 equal
steps between these two mercury levels. They correspond to an interval of 100 degrees
between the melting point of ice and the boiling point of water at one atmosphere. Figure
1-16 shows how temperature marks between the reference points are established.
In the United States, temperatures are frequently measured on the temperature scale
devised by Gabriel Fahrenheit (1686–1736), a German instrument maker. On this scale
the freezing and boiling points of water are defined as 32°F and 212°F, respectively. In
scientific work, temperatures are often expressed on the Kelvin (absolute) temperature
scale. As we shall see in Section 12-5, the zero point of the Kelvin temperature scale is
derived from the observed behavior of all matter.
Relationships among the three temperature scales are illustrated in Figure 1-17.
Between the freezing point of water and the boiling point of water, there are 100 steps
(°C or kelvins, respectively) on the Celsius and Kelvin scales. Thus the “degree” is the
same size on the Celsius and Kelvin scales. But every Kelvin temperature is 273.15 units
above the corresponding Celsius temperature. The relationship between these two scales
is as follows:
__? K ϭ °C ϩ 273.15° or __? °C ϭ K Ϫ 273.15°
Figure 1-16 At 45°C, as read on
a mercury-in-glass thermometer,
d equals 0.45d0 where d0 is the
distance from the mercury level
at 0°C to the level at 100°C.
1-12 Heat and Temperature 35
We shall usually round 273.15 to 273.
Figure 1-17 The relationships among the Kelvin,
Celsius (centigrade), and Fahrenheit temperature
scales.
100ЊC
0ЊC
d
d0
273
283
293
303
313
323
333
343
353
363
373
0Њ
10Њ
20Њ
30Њ
40Њ
50Њ
60Њ
70Њ
80Њ
90Њ
100Њ 212Њ
192Њ
172Њ
152Њ
132Њ
112Њ
92Њ
72Њ
52Њ
32Њ
42Њ
62Њ
Boiling
point
of water
Freezing
point
of water
180Њ100Њ
K C FBoiling
point
of water
Freezing
point
of water
In the SI system, “degrees Kelvin” are abbreviated simply as K rather than °K and are
called kelvins.
Any temperature change has the same numerical value whether expressed on the Celsius
scale or on the Kelvin scale. For example, a change from 25°C to 59°C represents a change
of 34 Celsius degrees. Converting these to the Kelvin scale, the same change is expressed
as (273 ϩ 25) ϭ 298 K to (59 ϩ 273) ϭ 332 K, or a change of 34 kelvins.
Comparing the Fahrenheit and Celsius scales, we find that the intervals between the
same reference points are 180 Fahrenheit degrees and 100 Celsius degrees, respectively.
Thus a Fahrenheit degree must be smaller than a Celsius degree. It takes 180 Fahrenheit
degrees to cover the same temperature interval as 100 Celsius degrees. From this information,
we can construct the unit factors for temperature changes:
or and or
But the starting points of the two scales are different, so we cannot convert a temperature
on one scale to a temperature on the other just by multiplying by the unit factor. In converting
from °F to °C, we must subtract 32 Fahrenheit degrees to reach the zero point
on the Celsius scale (Figure 1-17).
__? °F ϭ ΂x°C ϫ ΃ϩ 32°F and __? °C ϭ (x°F Ϫ 32°F)
EXAMPLE 1-16 Temperature Conversion
When the temperature reaches “100.°F in the shade,” it’s hot. What is this temperature on the
Celsius scale?
Plan
We use the relationship __? °C ϭ ᎏ
1
1
.
.
0
8
°
°
C
F
ᎏ (x°F Ϫ 32°F) to carry out the desired conversion.
Solution
__? °C ϭ (100.°F Ϫ 32°F) ϭ (68°F) ϭ 38°C
EXAMPLE 1-17 Temperature Conversion
When the absolute temperature is 400 K, what is the Fahrenheit temperature?
Plan
We first use the relationship __? °C ϭ K Ϫ 273° to convert from kelvins to degrees Celsius, then
we carry out the further conversion from degrees Celsius to degrees Fahrenheit.
Solution
__? °C ϭ (400 K Ϫ 273 K)ᎏ
1
1
.
.
0
0
°
K
C
ᎏ ϭ 127°C
__? °F ϭ
΂127°C ϫ ᎏ
1
1
.
.
0
8
°
°
C
F
ᎏ΃ϩ 32°F ϭ 261°F
You should now work Exercise 46.
1.0°C
ᎏ
1.8°F
1.0°C
ᎏ
1.8°F
1.0°C
ᎏ
1.8°F
1.8°F
ᎏ
1.0°C
1.0°C
ᎏ
1.8°F
100°C
ᎏ
180°F
1.8°F
ᎏ
1.0°C
180°F
ᎏ
100°C
The numbers in these ratios are exact
numbers, so they do not affect the
number of significant figures in the
calculated result.
These are often remembered in
abbreviated form:
°F ϭ 1.8°C ϩ 32°
°C ϭ
Either of these equations can be
rearranged to obtain the other one, so
you need to learn only one of them.
A temperature of 100.°F is 38°C.
(°F Ϫ 32°)
ᎏᎏ
1.8
36 CHAPTER 1: The Foundations of Chemistry
HEAT TRANSFER AND THE MEASUREMENT OF HEAT
Chemical reactions and physical changes occur with either the simultaneous evolution of
heat (exothermic processes) or the absorption of heat (endothermic processes). The
amount of heat transferred in a process is usually expressed in joules or in calories.
The SI unit of energy and work is the joule (J), which is defined as 1 kgиm2/s2. The
kinetic energy (KE) of a body of mass m moving at speed v is given by ᎏ1
2
ᎏmv2. A 2-kg
object moving at one meter per second has KE ϭ ᎏ1
2
ᎏ(2 kg)(1 m/s)2 ϭ 1 kgиm2/s2 ϭ 1 J. You
may find it more convenient to think in terms of the amount of heat required to raise the
temperature of one gram of water from 14.5°C to 15.5°C, which is 4.184 J.
One calorie is defined as exactly 4.184 J. The so-called “large calorie,” used to indicate
the energy content of foods, is really one kilocalorie, that is, 1000 calories. We shall
do most calculations in joules.
The specific heat of a substance is the amount of heat required to raise the temperature
of one gram of the substance one degree Celsius (also one kelvin) with no change
in phase. Changes in phase (physical state) absorb or liberate relatively large amounts of
energy (see Figure 1-5). The specific heat of each substance, a physical property, is different
for the solid, liquid, and gaseous phases of the substance. For example, the specific
heat of ice is 2.09 J/gи°C near 0°C; for liquid water it is 4.18 J/gи°C; and for steam it is
2.03 J/gи°C near 100°C. The specific heat for water is quite high. A table of specific heats
is provided in Appendix E.
Specific heat ϭ
The units of specific heat are or JиgϪ1и°CϪ1.
The heat capacity of a body is the amount of heat required to raise its temperature
1°C. The heat capacity of a body is its mass in grams times its specific heat. The heat capacity
refers to the mass of that particular body, so its units do not include mass. The
units are J/°C or Jи°CϪ1.
EXAMPLE 1-18 Specific Heat
How much heat, in joules, is required to raise the temperature of 205 g of water from 21.2°C
to 91.4°C?
Plan
The specific heat of a substance is the amount of heat required to raise the temperature of
1 g of substance 1°C:
Specific heat ϭ
We can rearrange the equation so that
(Amount of heat) ϭ (mass of substance) (specific heat) (temperature change)
Alternatively, we can use the unit factor approach.
Solution
Amount of heat ϭ (205 g) (4.18 J/gи°C) (70.2°C) ϭ 6.02 ϫ 104 J
(amount of heat in J)
ᎏᎏᎏᎏᎏᎏ
(mass of substance in g)(temperature change in °C)
J
ᎏ
gи°C
(amount of heat in J)
ᎏᎏᎏᎏᎏᎏ
(mass of substance in g)(temperature change in °C)
1-13
In English units this corresponds to a
4.4-pound object moving at 197 feet
per minute, or 2.2 miles per hour. In
terms of electrical energy, one joule is
equal to one watt и second. Thus, one
joule is enough energy to operate a 10watt
light bulb for ᎏ
1
1
0
ᎏ second.
The calorie was originally defined as
the amount of heat necessary to raise
the temperature of one gram of water
at one atmosphere from 14.5°C to
15.5°C.
1-13 Heat Transfer and the Measurement of Heat 37
The specific heat of a substance varies
slightly with temperature and pressure.
These variations can be ignored for
calculations in this text.
In this example, we calculate the
amount of heat needed to prepare a
cup of hot tea.
By the unit factor approach,
Amount of heat ϭ (205 g) ΂ ΃ (70.2°C) ϭ 6.02 ϫ 104 J or 60.2 kJ
All units except joules cancel. To cool 205 g of water from 91.4°C to 21.2°C, it would be necessary
to remove exactly the same amount of heat, 60.2 kJ.
You should now work Exercises 54 and 55.
When two objects at different temperatures are brought into contact, heat flows from
the hotter to the colder body (Figure 1-18); this continues until the two are at the same
temperature. We say that the two objects are then in thermal equilibrium. The temperature
change that occurs for each object depends on the initial temperatures and the relative
masses and specific heats of the two materials.
EXAMPLE 1-19 Specific Heat
A 385-gram chunk of iron is heated to 97.5°C. Then it is immersed in 247 grams of water
originally at 20.7°C. When thermal equilibrium has been reached, the water and iron are both
at 31.6°C. Calculate the specific heat of iron.
Plan
The amount of heat gained by the water as it is warmed from 20.7°C to 31.6°C is the same as
the amount of heat lost by the iron as it cools from 97.5°C to 31.6°C. We can equate these
two amounts of heat and solve for the unknown specific heat.
Solution
Temperature change of water ϭ 31.6°C Ϫ 20.7°C ϭ 10.9°C
Temperature change of iron ϭ 97.5°C Ϫ 31.6°C ϭ 65.9°C
4.18 J
ᎏ
1 gи°C
In specific heat calculations, we use the
magnitude of the temperature change
(i.e., a positive number), so we subtract
the lower temperature from the higher
one in both cases.
38 CHAPTER 1: The Foundations of Chemistry
Figure 1-18 A hot object, such as a heated piece of metal (a), is placed into cooler water.
Heat is transferred from the hotter metal bar to the cooler water until the two reach the
same temperature (b). We say that they are then at thermal equilibrium.
(a) (b)
Number of joules gained by water ϭ (247 g)
΂4.18
΃(10.9°C)
Let x ϭ specific heat of iron
Number of joules lost by iron ϭ (385 g)
΂x
΃(65.9°C)
We set these two quantities equal to one another and solve for x.
(247g)
΂4.18
΃(10.9°C) ϭ (385 g)
΂x
΃(65.9°C)
x ϭ ϭ 0.444
You should now work Exercise 58.
The specific heat of iron is much smaller than the specific heat of water.
ϭ ϭ 0.106
The amount of heat required to raise the temperature of 205 g of iron by 70.2°C (as we
calculated for water in Example 1-18) is
Amount of heat ϭ (205 g) ΂ ΃(70.2°C) ϭ 6.39 ϫ 103 J, or 6.39 kJ
We see that the amount of heat required to accomplish a given change in temperature
for a given quantity of iron is less than that for the same quantity of water, by the same
ratio.
ϭ ϭ 0.106
It might not be necessary to carry out explicit calculations when we are looking only
for qualitative comparisons.
EXAMPLE 1-20 Comparing Specific Heats
We add the same amount of heat to 10.0 grams of each of the following substances starting at
20.0°C: liquid water, H2O(ᐉ); liquid mercury; Hg(ᐉ); liquid benzene, C6H6(ᐉ); and solid aluminum,
Al(s). Rank the samples from lowest to highest final temperature. Refer to Appendix
E for required data.
Plan
We can obtain the values of specific heats (Sp. Ht.) for these substances from Appendix E. The
higher the specific heat for a substance, the more heat is required to raise a given mass of sample
by a given temperature change, so the less its temperature changes by a given amount of
heat. The substance with the lowest specific heat undergoes the largest temperature change,
and the one with the highest specific heat undergoes the smallest temperature change. It is not
necessary to calculate the amount of heat required to answer this question.
6.39 kJ
ᎏ
60.2 kJ
Number of joules required to warm 205 g of iron by 70.2°C
ᎏᎏᎏᎏᎏᎏᎏ
Number of joules required to warm 205 g of water by 70.2°C
0.444 J
ᎏ
gи°C
0.444 J/gи°C
ᎏᎏ
4.18 J/gи°C
Specific heat of iron
ᎏᎏᎏ
Specific heat of water
J
ᎏ
gи°C
(247 g) ΂4.18 ᎏ
gи
J
°C
ᎏ΃(10.9°C)
ᎏᎏᎏᎏ
(385 g)(65.9°C)
J
ᎏ
gи°C
J
ᎏ
gи°C
J
ᎏ
gи°C
J
ᎏ
gи°C
1-13 Heat Transfer and the Measurement of Heat 39
Solution
The specific heats obtained from Appendix E are as follows:
Sp. Ht.
Substance
΂ᎏ
gи
J
°C
ᎏ΃
H2O(ᐉ) 4.18
Hg(ᐉ) 0.138
C6H6(ᐉ) 1.74
Al(s) 0.900
Ranked from highest to lowest specific heats: H2O(ᐉ) Ͼ C6H6(ᐉ) Ͼ Al(s) Ͼ Hg(ᐉ). Adding the
same amount of heat to the same size sample of these substances changes the temperature of
H2O(ᐉ) the least and that of Hg(ᐉ) the most. The ranking from lowest to highest final temperature
is
H2O(ᐉ) Ͻ C6H6(ᐉ) Ͻ Al(s) Ͻ Hg(ᐉ)
You should now work Exercise 70.
40 CHAPTER 1: The Foundations of Chemistry
Key Terms
Accuracy How closely a measured value agrees with the correct
value.
Calorie Defined as exactly 4.184 joules. Originally defined as the
amount of heat required to raise the temperature of one gram
of water from 14.5°C to 15.5°C.
Chemical change A change in which one or more new substances
are formed.
Chemical property See Properties.
Compound A substance composed of two or more elements in
fixed proportions. Compounds can be decomposed into their
constituent elements.
Density Mass per unit volume, D ϭ m/V.
Element A substance that cannot be decomposed into simpler
substances by chemical means.
Endothermic Describes processes that absorb heat energy.
Energy The capacity to do work or transfer heat.
Exothermic Describes processes that release heat energy.
Extensive property A property that depends on the amount of
material in a sample.
Heat A form of energy that flows between two samples of
matter because of their difference in temperature.
Heat capacity The amount of heat required to raise the
temperature of a body (of whatever mass) one degree Celsius.
Heterogeneous mixture A mixture that does not have uniform
composition and properties throughout.
Homogeneous mixture A mixture that has uniform composition
and properties throughout.
Intensive property A property that is independent of the
amount of material in a sample.
Joule A unit of energy in the SI system. One joule is 1 kgиm2/s2,
which is also 0.2390 cal.
Kinetic energy Energy that matter possesses by virtue of its mo-
tion.
Law of Conservation of Energy Energy cannot be created or
destroyed in a chemical reaction or in a physical change; it may
be changed from one form to another.
Law of Conservation of Matter No detectable change occurs
in the total quantity of matter during a chemical reaction or
during a physical change.
Law of Conservation of Matter and Energy The combined
amount of matter and energy available in the universe is fixed.
Law of Constant Composition See Law of Definite Proportions.
Law of Definite Proportions Different samples of any pure
compound contain the same elements in the same proportions
by mass; also known as the Law of Constant Composition.
Mass A measure of the amount of matter in an object. Mass is
usually measured in grams or kilograms.
Matter Anything that has mass and occupies space.
Mixture A sample of matter composed of variable amounts of
two or more substances, each of which retains its identity and
properties.
Physical change A change in which a substance changes from
one physical state to another, but no substances with different
compositions are formed.
Exercises 41
Physical property See Properties.
Potential energy Energy that matter possesses by virtue of its
position, condition, or composition.
Precision How closely repeated measurements of the same
quantity agree with one another.
Properties Characteristics that describe samples of matter.
Chemical properties are exhibited as matter undergoes chemical
changes. Physical properties are exhibited by matter with
no changes in chemical composition.
Scientific (natural) law A general statement based on the
observed behavior of matter, to which no exceptions are known.
Significant figures Digits that indicate the precision of measurements—digits
of a measured number that have uncertainty
only in the last digit.
Specific gravity The ratio of the density of a substance to the
density of water at the same temperature.
Specific heat The amount of heat required to raise the temperature
of one gram of a substance one degree Celsius.
Substance Any kind of matter all specimens of which have the
same chemical composition and physical properties.
Symbol (of an element) A letter or group of letters that represents
(identifies) an element.
Temperature A measure of the intensity of heat, that is, the
hotness or coldness of a sample or object.
Unit factor A factor in which the numerator and denominator
are expressed in different units but represent the same or equivalent
amounts. Multiplying by a unit factor is the same as
multiplying by one.
Weight A measure of the gravitational attraction of the earth for
a body.
Exercises
Asterisks are used to denote some of the more challenging ex-
ercises.
Matter and Energy
*61. Define the following terms, and illustrate each with a specific
example: (a) matter; (b) energy; (c) mass; (d) exothermic
process; (e) intensive property.
*62. Define the following terms, and illustrate each with a specific
example: (a) weight; (b) potential energy; (c) kinetic
energy; (d) endothermic process; (e) extensive property.
*63. State the following laws, and illustrate each.
(a) the Law of Conservation of Matter
(b) the Law of Conservation of Energy
(c) the Law of Conservation of Matter and Energy
*64. List the three states of matter and some characteristics of
each. How are they alike? different?
*65. An incandescent light bulb functions because of the flow
of electric current. Does the incandescent light bulb convert
all of the electrical energy to light? Observe a functioning
incandescent light bulb, and explain what occurs
with reference to the Law of Conservation of Energy.
*66. All electrical motors are less than 100% efficient in converting
electrical energy into useable work. How can their
efficiency be less than 100% and the Law of Conservation
of Energy still be valid?
States of Matter
*67. What is a homogeneous mixture? Which of the following
are pure substances? Which of the following are homogeneous
mixtures? Explain your answers. (a) sugar dissolved
in water; (b) tea and ice; (c) french onion soup; (d) mud;
(e) gasoline; (f ) carbon dioxide; (g) a chocolate-chip cookie.
*68. Define the following terms clearly and concisely. Give two
illustrations of each: (a) substance; (b) mixture; (c) element;
(d) compound.
*69. Classify each of the following as an element, a compound,
or a mixture. Justify your classification: (a) a soft drink;
(b) water; (c) air; (d) chicken noodle soup; (e) table salt;
(f ) popcorn; (g) aluminum foil.
*10. Classify each of the following as an element, a compound,
or a mixture. Justify your classification: (a) coffee; (b) silver;
(c) calcium carbonate; (d) ink from a ballpoint pen;
(e) toothpaste.
*11. Sand, candle wax, and table sugar are placed in a beaker
and stirred. (a) Is the resulting combination a mixture?
If so, what kind of mixture? (b) Design an experiment
in which the sand, candle wax, and table sugar can be
separated.
*12. A $10 gold piece minted in the early part of the 1900s appeared
to have a dirty area. The dirty appearance could not
be removed by careful cleaning. Close examination of the
coin revealed that the “dirty” area was really pure copper.
Is the mixture of gold and copper in this coin a heterogeneous
or homogeneous mixture?
Chemical and Physical Properties
*13. Distinguish between the following pairs of terms and give
two specific examples of each: (a) chemical properties and
physical properties; (b) intensive properties and extensive
properties; (c) chemical changes and physical changes;
(d) mass and weight.
*14. Which of the following are chemical properties, and which
are physical properties? (a) Baking powder gives off bubbles
of carbon dioxide when added to water. (b) A particular
42 CHAPTER 1: The Foundations of Chemistry
type of steel consists of 95% iron, 4% carbon, and 1% miscellaneous
other elements. (c) The density of gold is 19.3
g/mL. (d) Iron dissolves in hydrochloric acid with the
evolution of hydrogen gas. (e) Fine steel wool burns in air.
(f ) Refrigeration slows the rate at which fruit ripens.
*15. Which of the following are chemical properties, and which
are physical properties? (a) Metallic sodium is soft enough
to be cut with a knife. (b) When sodium metal is cut, the
surface is at first shiny; after a few seconds of exposure
to air, it turns a dull gray. (c) The density of sodium is
0.97 g/mL. (d) Cork floats on water. (e) When sodium
comes in contact with water, it melts, evolves a flammable
gas, and eventually disappears altogether. (f ) Household
bleach changes the color of your favorite T-shirt from purple
to pink.
*16. Describe each of the following as a chemical change, a
physical change, or both. (a) A wet towel dries in the sun.
(b) Lemon juice is added to tea, causing its color to change.
(c) Hot air rises over a radiator. (d) Coffee is brewed by
passing hot water through ground coffee. (e) Dynamite ex-
plodes.
*17. Describe each of the following as a chemical change, a
physical change, or both. (a) Powdered sulfur is heated,
first melting and then burning. (b) Alcohol is evaporated
by heating. (c) Transparent rock candy (pure sugar crystals)
is finely ground into an opaque white powder.
(d) Chlorine gas is bubbled through concentrated sea
water, releasing liquid bromine. (e) Electricity is passed
through water, resulting in the evolution of hydrogen
and oxygen gases. (f ) An ice cube in your glass of water
melts.
*18. Which of the following processes are exothermic? endothermic?
How can you tell? (a) combustion; (b) freezing
water; (c) melting ice; (d) boiling water; (e) condensing
steam; (f ) burning paper.
*19. Which of the following illustrate the concept of potential
energy and which illustrate kinetic energy? (a) a spinning
gyroscope; (b) a rubber band stretched around a newspaper;
(c) a frozen pint of ice cream; (d) a comet moving
through space; (e) a basketball dropping through a net;
(f ) the roof of a house.
*20. Which of the following illustrate the concept of potential
energy and which illustrate kinetic energy? (a) the gasoline
in a car’s gas tank; (b) a car’s battery; (c) the car as it moves
along the highway; (d) a space vehicle in orbit about the
earth; (e) a river flowing; (f ) fatty tissue in your body.
*21. A sample of yellow sulfur powder is placed in a sealed flask
with the air removed and replaced with an inert gas. Heat
is applied by means of a flame from a Bunsen burner until
the sulfur melts and begins to boil. After cooling, the
material in the flask is reddish and has the consistency of
used chewing gum. Careful chemical analysis tells us that
the substance is pure sulfur. Is this a chemical or physical
change? Propose an explanation for the change.
*22. A weighed sample of yellow sulfur is placed in a flask. The
flask is gently heated using a Bunsen burner. Observation
indicates that nothing appears to happen to the sulfur during
the heating, but the mass of sulfur is less than before
the heating and there is a sharp odor that was not present
before the heating. Propose an explanation of what caused
the change in the mass of the sulfur. Is your hypothesis of
the mass change a chemical or physical change?
Measurements and Calculations
*23. Express the following numbers in scientific notation:
(a) 6500.; (b) 0.00630; (c) 860 (assume that this number is
measured to Ϯ10); (d) 860 (assume that this number is measured
to Ϯ1); (e) 186,000; (f ) 0.10010.
*24. Express the following exponentials as ordinary numbers:
(a) 5.26 ϫ 104; (b) 4.10 ϫ 10Ϫ6; (c) 16.00 ϫ 102;
(d) 8.206 ϫ 10Ϫ4; (e) 9.346 ϫ 103; (f ) 9.346 ϫ 10Ϫ3.
*25. Which of the following are likely to be exact numbers?
Why? (a) 554 in.; (b) 7 computers; (c) $20,355.47; (d) 25
lb of sugar; (e) 12.5 gal of diesel fuel; (f ) 5446 ants.
*26. To which of the quantities appearing in the following statements
would the concept of significant figures apply?
Where it would apply, indicate the number of significant
figures. (a) The density of platinum at 20°C is 21.45 g/cm3.
(b) Wilbur Shaw won the Indianapolis 500-mile race in
1940 with an average speed of 114.277 mi/h. (c) A mile is
defined as 5280 ft. (d) The International Committee for
Weights and Measures “accepts that the curie be . . .
retained as a unit of radioactivity, with the value 3.7 ϫ
1010 sϪ1.” (This resolution was passed in 1964.)
*27. The circumference of a circle is given by ␲d, where d is
the diameter of the circle. Calculate the circumference of
a circle with a diameter of 6.91 cm. Use the value of
3.141593 for ␲.
*28. What is the total weight of 75 cars weighing an average of
1532.5 lb?
*29. Indicate the multiple or fraction of 10 by which a quantity
is multiplied when it is preceded by each of the following
prefixes. (a) M; (b) m; (c) c; (d) d; (e) k; (f ) n.
*30. Carry out each of the following conversions. (a) 18.5 m to
km; (b) 16.3 km to m; (c) 247 kg to g; (d) 4.32 L to mL;
(e) 85.9 dL to L; (f ) 8251 L to cm3.
*31. Express 5.31 centimeters in meters, millimeters, kilometers,
and micrometers.
*32. Express (a) 1.00 ft3 in liters; (b) 1.00 L in pints; (c) miles
per gallon in kilometers per liter.
*33. The screen of a laptop computer measures 8.25 in. wide
and 6.25 in. tall. If this computer were being sold in
Europe, what would be the metric size of the screen used
in the specifications for the computer?
*34. If the price of gasoline is $1.229/gal, what is its price in
cents per liter?
*35. Suppose your automobile gas tank holds 16 gal and the
price of gasoline is $0.325/L. How much would it cost to
fill your gas tank?
*36. What is the density of silicon, if 50.6 g occupies 21.72 mL?
*37. What is the mass of a rectangular piece of copper
24.4 cm ϫ 11.4 cm ϫ 7.9 cm? The density of copper is
8.92 g/cm3.
*38. A small crystal of sucrose (table sugar) had a mass of
5.536 mg. The dimensions of the box-like crystal were
2.20 mm ϫ 1.36 mm ϫ 1.12 mm. What is the density of
sucrose expressed in g/cm3?
*39. Vinegar has a density of 1.0056 g/cm3. What is the mass
of three L of vinegar?
*40. The density of silver is 10.5 g/cm3. (a) What is the
volume, in cm3, of an ingot of silver with mass 0.615 kg?
(b) If this sample of silver is a cube, how long is each edge
in cm? (c) How long is the edge of this cube in inches?
*41. A container has a mass of 73.91 g when empty and
91.44 g when filled with water. The density of water is
1.0000 g/cm3. (a) Calculate the volume of the container.
(b) When filled with an unknown liquid, the container had
a mass of 88.42 g. Calculate the density of the unknown
liquid.
*42. The mass of an empty container is 77.664 g. The mass of
the container filled with water is 99.646 g. (a) Calculate
the volume of the container, using a density of 1.0000
g/cm3 for water. (b) A piece of metal was added to the
empty container, and the combined mass was 85.308 g.
Calculate the mass of the metal. (c) The container with the
metal was filled with water, and the mass of the entire system
was 106.442 g. What mass of water was added?
(d) What volume of water was added? (e) What is the volume
of the piece of metal? (f ) Calculate the density of the
metal.
*43. A solution is 40.0% acetic acid (the characteristic component
in vinegar) by mass. The density of this solution is
1.049 g/mL at 20°C. Calculate the mass of pure acetic acid
in 100.0 mL of this solution at 20°C.
Heat Transfer and Temperature Measurement
*44. Which represents a larger temperature interval: (a) a
Celsius degree or a Fahrenheit degree? (b) a kelvin or a
Fahrenheit degree?
*45. Express (a) 283°C in K; (b) 15.25 K in °C; (c) Ϫ32.0°C in
°F; (d) 100.0°F in K.
*46. Express (a) 0°F in °C; (b) 98.6°F in K; (c) 298 K in °F;
(d) 11.3°C in °F.
*47. Make each of the following temperature conversions:
(a) 27°C to °F, (b) Ϫ27°C to °F, and (c) 100°F to °C.
*48. On the Réamur scale, which is no longer used, water
freezes at 0°R and boils at 80°R. (a) Derive an equation
that relates this to the Celsius scale. (b) Derive an equation
that relates this to the Fahrenheit scale. (c) Mercury
is a liquid metal at room temperature. It boils at 356.6°C
(673.9°F). What is the boiling point of mercury on the
Réamur scale?
*49. Liquefied gases have boiling points well below room temperature.
On the Kelvin scale the boiling points of the following
gases are: He, 4.2 K; N2, 77.4 K. Convert these
temperatures to the Celsius and the Fahrenheit scales.
*50. Convert the temperatures at which the following metals
melt to the Celsius and Fahrenheit scales: Al, 933.6 K; Ag,
1235.1 K.
*51. What is the melting point of lead in °F (mp ϭ 327.5°C)?
*52. The average temperature of a healthy German shepherd is
101.5°F. Express this temperature in degrees Celsius.
Express this temperature in kelvins.
*53. Calculate the amount of heat required to raise the temperature
of 78.2 g of water from 10.0°C to 35.0°C. The
specific heat of water is 4.18 J/gи°C.
*54. The specific heat of aluminum is 0.895 J/gи°C. Calculate
the amount of heat required to raise the temperature of
22.1 g of aluminum from 27.0°C to 65.5°C.
*55. How much heat must be removed from 15.5 g of water at
90.0°C to cool it to 43.2°C?
*56. In some solar-heated homes, heat from the sun is stored
in rocks during the day and then released during the cooler
night. (a) Calculate the amount of heat required to raise
the temperature of 78.7 kg of rocks from 25.0°C to 43.0°C.
Assume that the rocks are limestone, which is essentially
pure calcium carbonate. The specific heat of calcium carbonate
is 0.818 J/gи°C. (b) Suppose that when the rocks in
part (a) cool to 30.0°C, all the heat released goes to warm
the 10,000 ft3 (2.83 ϫ 105 L) of air in the house, originally
at 10.0°C. To what final temperature would the air be
heated? The specific heat of air is 1.004 J/gи°C, and its
density is 1.20 ϫ 10Ϫ3 g/mL.
*57. A small immersion heater is used to heat water for a cup
of coffee. We wish to use it to heat 235 mL of water (about
a teacupful) from 25°C to 90°C in 2.00 min. What must
be the heat rating of the heater, in kJ/min, to accomplish
this? Ignore the heat that goes to heat the cup itself. The
density of water is 0.997 g/mL.
*58. When 75.0 grams of metal at 75.0°C is added to 150 grams
of water at 15.0°C, the temperature of the water rises to
18.3°C. Assume that no heat is lost to the surroundings.
What is the specific heat of the metal?
Exercises 43
Mixed Exercises
*59. A sample is marked as containing 22.8% calcium carbonate
by mass. (a) How many grams of calcium carbonate are
contained in 64.33 g of the sample? (b) How many grams
of the sample would contain 11.4 g of calcium carbonate?
*60. An iron ore is found to contain 9.24% hematite (a compound
that contains iron). (a) How many tons of this ore
would contain 8.40 tons of hematite? (b) How many kilograms
of this ore would contain 9.40 kg of hematite?
*61. A foundry releases 6.0 tons of gas into the atmosphere each
day. The gas contains 2.2% sulfur dioxide by mass. What
mass of sulfur dioxide is released in a five-day week?
*62. A certain chemical process requires 215 gal of pure water
each day. The available water contains 11 parts per million
(ppm) by mass of salt (i.e., for every 1,000,000 parts of
available water, 11 parts of salt). What mass of salt must
be removed each day? A gallon of water weighs 3.67 kg.
*63. The radius of a hydrogen atom is about 0.37 Å, and the
average radius of the earth’s orbit around the sun is about
1.5 ϫ 108 km. Find the ratio of the average radius of the
earth’s orbit to the radius of the hydrogen atom.
*64. A notice on a bridge informs drivers that the height of
the bridge is 26.5 ft. How tall in meters is an 18-wheel
tractor-trailer combination if it just touches the bridge?
*65. Some American car manufacturers install speedometers
that indicate speed in the English system and in the metric
system (mi/h and km/h). What is the metric speed if
the car is traveling at 65 mi/h?
*66. The lethal dose of potassium cyanide (KCN) taken orally
is 1.6 mg/kg of body weight. Calculate the lethal dose of
potassium cyanide taken orally by a 175-lb person.
*67. Suppose you ran a mile in 4.00 min. (a) What would be
your average speed in km/h? (b) What would be your average
speed in cm/s? (c) What would be your time (in minutes:seconds)
for 1500 m?
*68. The distance light travels through space in one year is
called one light-year. Using the speed of light in a vacuum
(listed in Appendix D), and assuming that one year is 365
days, determine the distance of a light-year in kilometers
and in miles.
CONCEPTUAL EXERCISES
*69. If you were given the job of choosing the materials from
which pots and pans were to be made, what kinds of
materials would you choose on the basis of specific heat?
Why?
*70. Fill a kitchen pan half-full with water. Place the pan on an
active burner. Place one finger on the edge of the pan and
one just in the water. Which finger will feel the heat first?
Explain your answer.
*71. Which is more dense at 0°C, ice or water? How do you
know?
*72. Which has the higher temperature, a sample of water at
65°C or a sample of iron at 65°F?
*73. The drawing in the circle (below) is a greatly expanded
representation of the molecules in the liquid of the
thermometer on the left. The thermometer registers
20°C. Which of the figures (a–d) is the best representation
of the liquid in this same thermometer at 10°C? (Assume
that the same volume of liquid is shown in each expanded
representation.)
44 CHAPTER 1: The Foundations of Chemistry
0˚
10˚
20˚
30˚
0˚
10˚
20˚
30˚
(a) (b) (c) (d)
Exercise 73
*74. During the past several years, you have gained chemical
vocabulary and understanding from a variety of academic
and entertainment venues. List three events that occurred
early in the development of your current chemical knowl-
edge.
*75. At what temperature will a Fahrenheit thermometer give:
(a) the same reading as a Celsius thermometer? (b) a reading
that is twice that on the Celsius thermometer? (c) a
reading that is numerically the same but opposite in sign
from that on the Celsius thermometer?
*76. Cesium atoms are the largest naturally occurring atoms.
The radius of a cesium atom is 2.62 Å. How many cesium
atoms would have to be laid side by side to give a row of
cesium atoms 1.00 in. long? Assume that the atoms are
spherical.
*77. Based on what you have learned during the study of this
chapter, write a question that requires you to use chemical
information but no mathematical calculations.
*78. As you write out the answer to an end-of-chapter exercise,
what chemical changes occur? Did your answer involve
knowledge not covered in Chapter 1?
*79. Combustion is discussed later in this textbook; however, you
probably already know what the term means. (Look it up
to be sure.) List two other chemical terms that were in your
vocabulary before you read Chapter 1.
Exercises 45
2 Chemical Formulas
and Composition
Stoichiometry
OUTLINE
2-1 Atoms and Molecules
2-2 Chemical Formulas
2-3 Ions and Ionic Compounds
2-4 Names and Formulas of Some
Ionic Compounds
2-5 Atomic Weights
2-6 The Mole
2-7 Formula Weights, Molecular
Weights, and Moles
2-8 Percent Composition and
Formulas of Compounds
2-9 Derivation of Formulas from
Elemental Composition
2-10 Determination of Molecular
Formulas
2-11 Some Other Interpretations of
Chemical Formulas
2-12 Purity of Samples
T
OBJECTIVES
After you have studied this chapter, you should be able to
• Understand some early concepts of atoms
• Use chemical formulas to solve various kinds of chemical problems
• Relate names to formulas and charges of simple ions
• Combine simple ions to write formulas and names of some ionic compounds
• Recognize and use formula weights and mole relationships
• Interconvert masses, moles, and formulas
• Determine percent compositions in compounds
• Determine formulas from composition
• Perform calculations of purity of substances
T
he language we use to describe the forms of matter and the changes in its composition
is not limited to use in chemistry courses; it appears throughout the
scientific world. Chemical symbols, formulas, and equations are used in such
diverse areas as agriculture, home economics, engineering, geology, physics, biology,
medicine, and dentistry. In this chapter we describe the simplest atomic theory. We shall
use it as we represent the chemical formulas of elements and compounds. Later, after
additional facts have been introduced, this theory will be expanded.
The word “stoichiometry” is derived from the Greek stoicheion, which means “first
principle or element,” and metron, which means “measure.” Stoichiometry describes the
quantitative relationships among elements in compounds (composition stoichiometry) and
among substances as they undergo chemical changes (reaction stoichiometry). In this chapter
we are concerned with chemical formulas and composition stoichiometry. In Chapter
3 we shall discuss chemical equations and reaction stoichiometry.
It is important to learn this
fundamental material well so that you
can use it correctly and effectively.
Some minerals and gems, left to
right: Galena, PbS; aquamarine,
Be3AlSi6O18 (colored by trace
amounts of iron); ruby, Al2O3 (with
some Cr3؉ ions replacing Al3؉
ions); sulfur, S8; silver, Ag; elbaite,
Na(Li,Al)3 Al6 Si6 O18(BO3 )3(OH)4
(the notation [Li,Al] indicates a
variable ratio of lithium and
aluminum among elbaite samples);
sapphire, Al2O3 (with various
colors due to different metal ions
substituting for Al3؉ ions);
fluorite, CaF2; copper, Cu; azurite,
Cu3(CO3 )2(OH)2; and malachite,
Cu2CO3(OH)2.
ATOMS AND MOLECULES
The Greek philosopher Democritus (470–400 BC) suggested that all matter is composed
of tiny, discrete, indivisible particles that he called atoms. His ideas, based entirely on
philosophical speculation rather than experimental evidence, were rejected for 2000 years.
By the late 1700s, scientists began to realize that the concept of atoms provided an explanation
for many experimental observations about the nature of matter.
By the early 1800s, the Law of Conservation of Matter (Section 1-1) and the Law of
Definite Proportions (Section 1-5) were both accepted as general descriptions of how matter
behaves. John Dalton (1766–1844), an English schoolteacher, tried to explain why
matter behaves in such systematic ways as those expressed here. In 1808, he published the
first “modern” ideas about the existence and nature of atoms. Dalton’s explanation summarized
and expanded the nebulous concepts of early philosophers and scientists; more
importantly, his ideas were based on reproducible experimental results of measurements by
many scientists. These ideas form the core of Dalton’s Atomic Theory, one of the highlights
in the history of scientific thought. In condensed form, Dalton’s ideas may be stated
as follows:
1. An element is composed of extremely small, indivisible particles called atoms.
2. All atoms of a given element have identical properties that differ from those of
other elements.
3. Atoms cannot be created, destroyed, or transformed into atoms of another ele-
ment.
4. Compounds are formed when atoms of different elements combine with one another
in small whole-number ratios.
5. The relative numbers and kinds of atoms are constant in a given compound.
Dalton believed that atoms were solid, indivisible spheres, an idea we now reject. But
he showed remarkable insight into the nature of matter and its interactions. Some of his
ideas could not be verified (or refuted) experimentally at the time. They were based on
the limited experimental observations of his day. Even with their shortcomings, Dalton’s
ideas provided a framework that could be modified and expanded by later scientists. Thus
John Dalton is often considered to be the father of modern atomic theory.
The smallest particle of an element that maintains its chemical identity through all
chemical and physical changes is called an atom (Figure 2-1). In Chapter 5, we shall study
the structure of the atom in detail; let us simply summarize here the main features of
atomic composition. Atoms, and therefore all matter, consist principally of three fundamental
particles: electrons, protons, and neutrons. These are the basic building blocks of
2-1
The term “atom” comes from the
Greek language and means “not
divided” or “indivisible.”
48 CHAPTER 2: Chemical Formulas and Composition Stoichiometry
Figure 2-1 Relative sizes of monatomic molecules (single atoms) of the noble gases.
He Ne Ar Kr Xe Rn
The radius of a calcium atom is only
1.97 ϫ 10Ϫ8 cm, and its mass is
6.66 ϫ 10Ϫ23g.
Statement 3 is true for chemical
reactions. It is not true, however, for
nuclear reactions (Chapter 26).
For Group VIIIA elements, the noble
gases, a molecule contains only one
atom, and so an atom and a molecule
are the same (see Figure 2-1).
You should remember the common
elements that occur as diatomic
molecules: H2, N2, O2, F2, Cl2, Br2, I2.
2-1 Atoms and Molecules 49
atoms. The masses and charges of the three fundamental particles are shown in Table
2-1. The masses of protons and neutrons are nearly equal, but the mass of an electron is
much smaller. Neutrons carry no charge. The charge on a proton is equal in magnitude,
but opposite in sign, to the charge on an electron. Because atoms are electrically neutral,
an atom contains equal numbers of electrons and protons.
The atomic number (symbol is Z) of an element is defined as the number of protons
in the nucleus. In the periodic table, elements are arranged in order of increasing atomic
numbers. These are the red numbers above the symbols for the elements in the periodic
table on the inside front cover. For example, the atomic number of silver is 47.
A molecule is the smallest particle of an element or compound that can have a stable
independent existence. In nearly all molecules, two or more atoms are bonded together
in very small, discrete units (particles) that are electrically neutral.
Individual oxygen atoms are not stable at room temperature and atmospheric pressure.
Single atoms of oxygen mixed under these conditions quickly combine to form pairs. The
oxygen with which we are all familiar is made up of two atoms of oxygen; it is a diatomic
molecule, O2. Hydrogen, nitrogen, fluorine, chlorine, bromine, and iodine are other examples
of diatomic molecules (Figure 2-2).
Some other elements exist as more complex molecules. One form of phosphorus molecules
consists of four atoms, and sulfur exists as eight-atom molecules at ordinary temperatures
and pressures. Molecules that contain two or more atoms are called polyatomic
molecules (Figure 2-3).
Figure 2-2 Models of diatomic molecules of some elements, approximately to scale.
TABLE 2-1 Fundamental Particles of Matter
Particle Approximate Mass Charge
(symbol) (amu)* (relative scale)
electron (eϪ) 0.0 1Ϫ
proton (p or pϩ) 1.0 1ϩ
neutron (n or n0) 1.0 none
*1 amu ϭ 1.6605 ϫ 10Ϫ24 g
H
H2
(hydrogen)
O2
(oxygen)
F2
(fluorine)
I2
(iodine)
H O O F F II
atomic number
symbol
47
Ag
In modern terminology, O2 is named dioxygen, H2 is dihydrogen, P4 is tetraphosphorus,
and so on. Even though such terminology is officially preferred, it has not yet gained
wide acceptance. Most chemists still refer to O2 as oxygen, H2 as hydrogen, P4 as phosphorus,
and so on.
Molecules of compounds are composed of more than one kind of atom. A water molecule
consists of two atoms of hydrogen and one atom of oxygen. A molecule of methane
consists of one carbon atom and four hydrogen atoms. The shapes of a few molecules are
shown in Figure 2-4.
Atoms are the components of molecules, and molecules are the stable forms of many
elements and compounds. We are able to study samples of compounds and elements that
consist of large numbers of atoms and molecules. With the scanning tunnelling microscope
it is now possible to “see” atoms (Figure 2-5). It would take millions of atoms to
make a row as long as the diameter of the period at the end of this sentence.
Methane is the principal component of
natural gas.
50 CHAPTER 2: Chemical Formulas and Composition Stoichiometry
P
P
P P
(a)
S S
S
S
S
S S
S
(c)
S S
S
S S
S S
S
(b)
Figure 2-3 (a) A model of the P4 molecule of white phosphorus. (b) A model of the S8
ring found in rhombic sulfur. (c) Top view of the S8 ring in rhombic sulfur.
Figure 2-4 Formulas and models for molecules of some compounds.
H2O
(water)
CO2
(carbon dioxide)
CH4
(methane)
C2H5OH
(ethyl alcohol)
CHEMICAL FORMULAS
The chemical formula for a substance shows its chemical composition. This represents
the elements present as well as the ratio in which the atoms of the elements occur. The
formula for a single atom is the same as the symbol for the element. Thus, Na can represent
a single sodium atom. It is unusual to find such isolated atoms in nature, with the
exception of the noble gases (He, Ne, Ar, Kr, Xe, and Rn). A subscript following the symbol
of an element indicates the number of atoms in a molecule. For instance, F2 indicates
a molecule containing two fluorine atoms, and P4 a molecule containing four phosphorus
atoms.
Some elements exist in more than one form. Familiar examples include (1) oxygen,
found as O2 molecules, and ozone, found as O3 molecules, and (2) two crystalline forms
of carbon—diamond and graphite (Figure 13-33). Different forms of the same element
in the same physical state are called allotropic modifications, or allotropes.
Compounds contain two or more elements in chemical combination in fixed proportions.
Many compounds exist as molecules (Table 2-2). Hence, each molecule of hydrogen
chloride, HCl, contains one atom of hydrogen and one atom of chlorine; each molecule
of carbon tetrachloride, CCl4, contains one carbon atom and four chlorine atoms.
2-2
2-2 Chemical Formulas 51
Figure 2-5 A computer reconstruction of the
surface of a sample of silicon, as observed with a
scanning tunnelling electron microscope (STM),
reveals the regular pattern of individual silicon
atoms. Many important reactions occur on the
surfaces of solids. Observations of the atomic
arrangements on surfaces help chemists understand
such reactions. New information available using the
STM will give many details about chemical bonding
in solids.
TABLE 2-2 Names and Formulas of Some Common Molecular Compounds
Name Formula Name Formula Name Formula
water H2O sulfur dioxide SO2 butane C4H10
hydrogen peroxide H2O2 sulfur trioxide SO3 pentane C5H12
hydrogen chloride* HCl carbon monoxide CO benzene C6H6
sulfuric acid H2SO4 carbon dioxide CO2 methanol (methyl alcohol) CH3OH
nitric acid HNO3 methane CH4 ethanol (ethyl alcohol) CH3CH2OH
acetic acid CH3COOH ethane C2H6 acetone CH3COCH3
ammonia NH3 propane C3H8 diethyl ether (ether) CH3CH2XOXCH2CH3
*Called hydrochloric acid if dissolved in water.
An O2 molecule.
OO
An O3 molecule.
O
O
O
117°
An aspirin molecule, C9H8O4, contains nine carbon atoms, eight hydrogen atoms, and
four oxygen atoms.
Many of the molecules found in nature are organic compounds. Organic compounds
contain CXC or CXH bonds or both. Eleven of the compounds listed in Table 2-2 are
organic compounds (acetic acid and the last ten entries). All of the other compounds in
the table are inorganic compounds.
Some groups of atoms behave chemically as single entities. For instance, an oxygen
atom that is bonded to a hydrogen atom and also to a carbon atom that is bonded to three
other atoms forms the reactive combination of atoms known as the alcohol group or molecule.
In formulas of compounds containing two or more of the same group, the group
formula is enclosed in parentheses. Thus, ethylene glycol contains two alcohol groups and
its formula is C2H4(OH)2 (see structure in the margin). When you count the number of
atoms in this molecule from its formula, you must multiply the numbers of hydrogen and
oxygen atoms in the OH group by 2. There are two carbon atoms, six hydrogen atoms
and two oxygen atoms in a molecule of ethylene glycol.
Compounds were first recognized as distinct substances because of their different
physical properties and because they could be separated from one another by physical
methods. Once the concept of atoms and molecules was established, the reason for these
differences in properties could be understood: Two compounds differ from each other
because their molecules are different. Conversely, if two molecules contain the same
number of the same kinds of atoms, arranged the same way, then both are molecules of
the same compound. Thus, the atomic theory explains the Law of Definite Proportions
(see Section 1-5).
This law, also known as the Law of Constant Composition, can now be extended to
include its interpretation in terms of atoms. It is so important for performing the calculation
in this chapter that we restate it here:
Different pure samples of a compound always contain the same elements in the same
proportion by mass; this corresponds to atoms of these elements combined in fixed
numerical ratios.
So we see that for a substance composed of molecules, the chemical formula gives
the number of atoms of each type in the molecule. But this formula does not express the
order in which the atoms in the molecules are bonded together. The structural formula
shows the order in which atoms are connected. The lines connecting atomic symbols represent
chemical bonds between atoms. The bonds are actually forces that tend to hold
atoms at certain distances and angles from one another. For instance, the structural formula
of propane shows that the three C atoms are linked in a chain, with three H atoms
bonded to each of the end C atoms and two H atoms bonded to the center C. Ball-andstick
molecular models and space-filling molecular models help us to see the shapes and
relative sizes of molecules. These four representations are shown in Figure 2-6. The balland-stick
and space-filling models show (1) the bonding sequence, that is the order in which
the atoms are connected to each other, and (2) the geometrical arrangements of the atoms
in the molecule. As we shall see later, both are extremely important because they determine
the properties of compounds.
52 CHAPTER 2: Chemical Formulas and Composition Stoichiometry
A model of ethylene glycol.
IONS AND IONIC COMPOUNDS
So far we have discussed only compounds that exist as discrete molecules. Some compounds,
such as sodium chloride, NaCl, consist of collections of large numbers of ions.
An ion is an atom or group of atoms that carries an electric charge. Ions that possess a
positive charge, such as the sodium ion, Naϩ, are called cations. Those carrying a negative
charge, such as the chloride ion, ClϪ, are called anions. The charge on an ion must
2-3
Figure 2-6 Formulas and models
for some molecules. Structural
formulas show the order in which
atoms are connected but do not
represent true molecular shapes.
Ball-and-stick models use balls of
different colors to represent atoms
and sticks to represent bonds; they
show the three-dimensional shapes
of molecules. Space-filling models
show the (approximate) relative sizes
of atoms and the shapes of
molecules.
2-3 Ions and Ionic Compounds 53
H2O, water H O H
H HO OH2O2, hydrogen
peroxide
CCl4, carbon
tetrachloride
CCl
Cl
Cl
Cl
C3H8, propane
C2H5OH,
ethanol
C C C H
HHH
H H H
H
C C O H
HH
H H
H
Chemical Formula Structural Formula Ball-and-Stick Model Space-Filling Model
The words “cation” (katЈ-i-on)
and “anion” (anЈ-i-on) and their
relationship to cathode and anode
will be described in Chapter 21.
be included as a superscript on the right side of the chemical symbol(s) when we write
the formula for the individual ion.
As discussed in detail in Chapter 5, an atom consists of a very small, very dense, positively
charged nucleus surrounded by a diffuse distribution of negatively charged particles
called electrons. The number of positive charges in the nucleus defines the identity of the
element to which the atom corresponds. Electrically neutral atoms contain the same number
of electrons outside the nucleus as positive charges (protons) within the nucleus. Ions
are formed when neutral atoms lose or gain electrons. An Naϩ ion is formed when a
sodium atom loses one electron, and a ClϪ ion is formed when a chlorine atom gains one
electron.
The compound NaCl consists of an extended array of Naϩ and ClϪ ions (Figure 2-7).
Within the crystal (though not on the surface) each Naϩ ion is surrounded at equal distances
by six ClϪ ions, and each ClϪ ion is similarly surrounded by six Naϩ ions. Any
compound, whether ionic or molecular, is electrically neutral; that is, it has no net charge.
In NaCl this means that the Naϩ and ClϪ ions are present in a 1Ϻ1 ratio, and this is indicated
by the formula NaCl.
Because there are no “molecules” of ionic substances, we should not refer to “a
molecule of sodium chloride, NaCl,” for example. Instead, we refer to a formula unit
of NaCl, which consists of one Naϩ ion and one ClϪ ion. Likewise, one formula unit
of CaCl2 consists of one Ca2ϩ ion and two ClϪ ions. As you will see in the next section,
we speak of the formula unit of all ionic compounds as the smallest, whole-number
ratios of ions that yield neutral representations. It is also acceptable to refer to a formula
unit of a molecular compound. One formula unit of propane, C3H8, is the same as one
molecule of C3H8; it contains three C atoms and eight H atoms bonded together into
a group.
For the present, we shall tell you which substances are ionic and which are molecular
when it is important to know. Later you will learn to make the distinction yourself.
Polyatomic ions are groups of atoms that bear an electric charge. Examples include
the ammonium ion, NH4
ϩ; the sulfate ion, SO4
2Ϫ; and the nitrate ion, NO3
Ϫ. Table 2-3
shows the formulas, ionic charges, and names of some common ions. When writing the
formula of a polyatomic compound, we show groups in parentheses when they appear
more than once. For example, (NH4)2SO4 represents a compound that has two NH4
ϩ
ions for each SO4
2Ϫ ion.
Figure 2-7 The arrangement of
ions in NaCl. (a) A crystal of sodium
chloride consists of an extended
array that contains equal numbers
of sodium ions (small spheres) and
chloride ions (large spheres). Within
the crystal, (b) each chloride ion is
surrounded by six sodium ions, and
(c) each sodium ion is surrounded by
six chloride ions.
54 CHAPTER 2: Chemical Formulas and Composition Stoichiometry
Na+
Cl–
Cl–
Na+
Na+
(a)
(b)
(c)
The general term “formula unit”
applies to molecular or ionic
compounds, whereas the more specific
term “molecule” applies only to
elements and compounds that exist
as discrete molecules.
In this text, we use the standard
convention of representing multiple
charges with the number before the
sign, e.g., Ca2ϩ, not Caϩ2 and SO4
2Ϫ,
not SO4
Ϫ2.
NAMES AND FORMULAS OF SOME IONIC COMPOUNDS
During your study of chemistry you will have many occasions to refer to compounds by
name. In this section, we shall see how a few compounds should be named. More comprehensive
rules for naming compounds are presented at the appropriate places later in
the text.
Table 2-2 includes examples of names for a few common molecular compounds. You
should learn that short list before proceeding much farther in this textbook. We shall
name many more molecular compounds as we encounter them in later chapters.
2-4
As we shall see, some metals can form
more than one kind of ion with a
positive charge. For such metals, we
specify which ion we mean with a
Roman numeral—e.g., iron(II) or
iron(III). Because zinc forms no stable
ions other than Zn2ϩ, we do not need
to use Roman numerals in its name.
TABLE 2-3 Formulas, Ionic Charges, and Names of Some Common Ions
Common Cations (positive ions) Common Anions (negative ions)
Formula Charge Name Formula Charge Name
Naϩ 1ϩ sodium FϪ 1Ϫ fluoride
Kϩ 1ϩ potassium ClϪ 1Ϫ chloride
NH4
ϩ 1ϩ ammonium BrϪ 1Ϫ bromide
Agϩ 1ϩ silver OHϪ 1Ϫ hydroxide
CH3COOϪ 1Ϫ acetate
Mg2ϩ 2ϩ magnesium NO3
Ϫ 1Ϫ nitrate
Ca2ϩ 2ϩ calcium
Zn2ϩ 2ϩ zinc O2Ϫ 2Ϫ oxide
Cuϩ 1ϩ copper(I) S2Ϫ 2Ϫ sulfide
Cu2ϩ 2ϩ copper(II) SO4
2Ϫ 2Ϫ sulfate
Fe2ϩ 2ϩ iron(II) SO3
2Ϫ 2Ϫ sulfite
CO3
2Ϫ 2Ϫ carbonate
Fe3ϩ 3ϩ iron(III)
Al3ϩ 3ϩ aluminum PO4
3Ϫ 3Ϫ phosphate
Ionic compounds (clockwise, from top): salt
(sodium chloride, NaCl), calcite (calcium
carbonate, CaCO3), cobalt(II) chloride
hexahydrate, (CoCl2 и 6H2O), fluorite
(calcium fluoride, CaF2).
The names of some common ions appear in Table 2-3. You will need to know the
names and formulas of these frequently encountered ions. They can be used to write the
formulas and names of many ionic compounds. We write the formula of an ionic compound
by adjusting the relative numbers of positive and negative ions so their total charges
cancel (i.e., add to zero). The name of an ionic compound is formed by giving the names
of the ions, with the positive ion named first.
56 CHAPTER 2: Chemical Formulas and Composition Stoichiometry
EXAMPLE 2-1 Formulas for Ionic Compounds
Write the formulas for the following ionic compounds: (a) sodium fluoride, (b) calcium fluoride,
(c) iron(II) sulfate, (d) zinc phosphate.
Plan
In each case, we identify the chemical formulas of the ions from Table 2-3. These ions must
be present in the simplest whole-number ratio that gives the compound no net charge. Recall
that the formulas and names of ionic compounds are written by giving the positively charged
ion first.
Solution
(a) The formula for the sodium ion is Naϩ, and the formula for the fluoride ion is FϪ (Table
2-3). Because the charges on these two ions are equal in magnitude, the ions must be present
in equal numbers, or in a 1Ϻ1 ratio. Thus, the formula for sodium fluoride is NaF.
(b) The formula for the calcium ion is Ca2ϩ and the formula for the fluoride ion is FϪ. Now
each positive ion (Ca2ϩ) provides twice as much charge as each negative ion (FϪ). So there
must be twice as many FϪ ions as Ca2ϩ ions to equalize the charge. This means that the ratio
of calcium to fluoride ions is 1Ϻ2. So the formula for calcium fluoride is CaF2.
(c) The iron(II) ion is Fe2ϩ, and the sulfate ion is SO4
2Ϫ. As in part (a), the equal magnitudes
of positive and negative charges tell us that the ions must be present in equal numbers, or in
a 1:1 ratio. The formula for iron(II) sulfate is FeSO4.
(d) The zinc ion is Zn2ϩ, and the phosphate ion is PO4
3Ϫ. Now it will take three Zn2ϩ ions to
account for as much charge (6ϩ total) as would be present in two PO4
3Ϫ ions (6Ϫ total). So
the formula for zinc phosphate is Zn3(PO4)2.
You should now work Exercises 14 and 21.
EXAMPLE 2-2 Names for Ionic Compounds
Name the following ionic compounds: (a) (NH4)2S, (b) Cu(NO3)2, (c) ZnCl2, (d) Fe2(CO3)3.
Plan
In naming ionic compounds, it is helpful to inspect the formula for atoms or groups of atoms
that we recognize as representing familiar ions.
Problem-Solving Tip: Where to Start in Learning to Name
Compounds
You may not be sure of the best point to start learning the naming of compounds. It has
been found that before rules for naming can make much sense or before we can expand
our knowledge to more complex compounds, we need to know the names and formulas
in Tables 2-2 and 2-3. If you are unsure of your ability to recall a name or a formula in
Tables 2-2 and 2-3 when given the other, prepare flash cards, lists, and so on that you
can use to learn these tables.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 3.13, Naming Ionic
Compounds.
We use the information that the
carbonate ion has a 2Ϫ charge to find
the charge on the iron ions. The total
charges must add up to zero.
2-6 The Mole 57
Solution
(a) The presence of the polyatomic grouping NH4 in the formula suggests to us the presence
of the ammonium ion, NH4
ϩ. There are two of these, each accounting for 1ϩ in charge. To
balance this, the single S must account for 2Ϫ in charge, or S2Ϫ, which we recognize as the
sulfide ion. Thus, the name of the compound is ammonium sulfide.
(b) The NO3 grouping in the formula tells us that the nitrate ion, NO3
Ϫ, is present. Two of
these nitrate ions account for 2 ϫ 1Ϫ ϭ 2Ϫ in negative charge. To balance this, copper must
account for 2ϩ charge and be the copper(II) ion. The name of the compound is copper(II)
nitrate.
(c) The positive ion present is zinc ion, Zn2ϩ, and the negative ion is chloride, ClϪ. The name
of the compound is zinc chloride.
(d) Each CO3 grouping in the formula must represent the carbonate ion, CO3
2Ϫ. The presence
of three such ions accounts for a total of 6Ϫ in negative charge, so there must be a total
of 6ϩ present in positive charge to balance this. It takes two iron ions to provide this 6ϩ, so
each ion must have a charge of 3ϩ and be Fe3ϩ, the iron(III) ion, or ferric ion. The name of
the compound is iron(III) carbonate.
You should now work Exercises 13 and 20.
A more extensive discussion on naming compounds appears in Chapter 4.
ATOMIC WEIGHTS
As the chemists of the eighteenth and nineteenth centuries painstakingly sought information
about the compositions of compounds and tried to systematize their knowledge,
it became apparent that each element has a characteristic mass relative to every other
element. Although these early scientists did not have the experimental means to measure
the mass of each kind of atom, they succeeded in defining a relative scale of atomic masses.
An early observation was that carbon and hydrogen have relative atomic masses, also
traditionally called atomic weights (AW), of approximately 12 and 1, respectively.
Thousands of experiments on the compositions of compounds have resulted in the establishment
of a scale of relative atomic weights based on the atomic mass unit (amu),
which is defined as exactly ᎏ
1
1
2
ᎏ of the mass of an atom of a particular kind of carbon atom, called
carbon-12.
On this scale, the atomic weight of hydrogen (H) is 1.00794 amu, that of sodium (Na)
is 22.989768 amu, and that of magnesium (Mg) is 24.3050 amu. This tells us that Na
atoms have nearly 23 times the mass of H atoms, and Mg atoms are about 24 times heavier
than H atoms.
When you need values of atomic weights, consult the periodic table or the alphabetical
listing of elements, both found on facing pages inside the front cover.
THE MOLE
Even the smallest bit of matter that can be handled reliably contains an enormous number
of atoms. So we must deal with large numbers of atoms in any real situation, and some
unit for conveniently describing a large number of atoms is desirable. The idea of using
a unit to describe a particular number (amount) of objects has been around for a long
time. You are already familiar with the dozen (12 items) and the gross (144 items).
2-6
2-5
The term “atomic weight” is widely
accepted because of its traditional use,
although it is properly a mass rather
than a weight. “Atomic mass” is often
used.
The SI unit for amount is the mole, abbreviated mol. It is defined as the amount of
substance that contains as many entities (atoms, molecules, or other particles) as there are
atoms in exactly 0.012 kg of pure carbon-12 atoms. Many experiments have refined the
number, and the currently accepted value is
1 mole ϭ 6.0221367 ϫ 1023 particles
This number, often rounded off to 6.022 ϫ 1023, is called Avogadro’s number in honor
of Amedeo Avogadro (1776–1856), whose contributions to chemistry are discussed in
Section 12-8.
According to its definition, the mole unit refers to a fixed number of items, the identities
of which must be specified. Just as we speak of a dozen eggs or a pair of aces, we
refer to a mole of atoms or a mole of molecules (or a mole of ions, electrons, or other
particles). We could even think about a mole of eggs, although the size of the required
carton staggers the imagination! Helium exists as discrete He atoms, so one mole of helium
consists of 6.022 ϫ 1023 He atoms. Hydrogen commonly exists as diatomic (twoatom)
molecules, so one mole of hydrogen is 6.022 ϫ 1023 H2 molecules and 2(6.022 ϫ
1023) H atoms.
Every kind of atom, molecule, or ion has a definite characteristic mass. It follows that
one mole of a given pure substance also has a definite mass, regardless of the source of
the sample. This idea is of central importance in many calculations throughout the study
of chemistry and the related sciences.
Because the mole is defined as the number of atoms in 0.012 kg (or 12 g) of carbon-
12, and the atomic mass unit is defined as ᎏ
1
1
2
ᎏ of the mass of a carbon-12 atom, the following
convenient relationship is true:
The mass of one mole of atoms of a pure element in grams is numerically equal to
the atomic weight of that element in atomic mass units. This is also called the molar
mass of the element; its units are grams/mole, also written as g/mol or gиmolϪ1.
For instance, if you obtain a pure sample of the metallic element titanium (Ti), whose
atomic weight is 47.88 amu, and measure out 47.88 g of it, you will have one mole, or
6.022 ϫ 1023 titanium atoms.
The symbol for an element can be used to (1) identify the element, (2) represent one
atom of the element, or (3) represent one mole of atoms of the element. The last interpretation
will be extremely useful in calculations in the next chapter.
A quantity of a substance may be expressed in a variety of ways. For example, consider
a dozen eggs and 55.847 grams (or one mole) of iron (Figure 2-8). We can express the
amount of eggs or iron present in any of several units. We can then construct unit factors
to relate an amount of the substance expressed in one kind of unit to the same amount
expressed in another unit.
Unit Factors for Eggs Unit Factors for Iron
ᎏ
1
1
d
2
oz
eg
e
g
g
s
gs
ᎏ
ᎏ
24
12
oz
eg
e
g
g
s
gs
ᎏ
and so on and so on
6.022 ϫ 1023 Fe atoms
ᎏᎏᎏ
55.847 g Fe
6.022 ϫ 1023 Fe atoms
ᎏᎏᎏ
1 mol Fe atoms
“Mole” is derived from the Latin word
moles, which means “a mass.”
“Molecule” is the diminutive form of
this word and means “a small mass.”
58 CHAPTER 2: Chemical Formulas and Composition Stoichiometry
The atomic weight of iron (Fe) is
55.847 amu. Suppose that one dozen
large eggs weighs 24 oz.
As Table 2-4 suggests, the concept of a mole as applied to atoms is especially useful.
It provides a convenient basis for comparing the masses of equal numbers of atoms of
different elements.
Figure 2-9 shows what one mole of atoms of each of some common elements looks
like. Each of the examples in Figure 2-9 represents 6.022 ϫ 1023 atoms of the element.
The relationship between the mass of a sample of an element and the number of moles
of atoms in the sample is illustrated in Example 2-3.
In this textbook we usually work problems involving atomic weights (masses) or formula
weights (masses) rounded to only one decimal place. We round the answer further
if initial data do not support the number of significant figures obtained using the rounded
atomic weights. Similarly, if the initial data indicate that more significant figures are justified,
we will rework such problems using atomic weights and formula weights containing
values beyond the tenths place.
Figure 2-8 Three ways of representing amounts.
2-6 The Mole 59
12 large eggs
or
1 dozen eggs
or
24 ounces of eggs
6.022 ϫ 1023 Fe atoms
or
1 mole of Fe atoms
or
55.847 grams of iron
TABLE 2-4 Mass of One Mole of Atoms of Some Common
Elements
A Sample with a
Element Mass of Contains
carbon 12.0 g C 6.02 ϫ 1023 C atoms or 1 mol of C atoms
titanium 47.9 g Ti 6.02 ϫ 1023 Ti atoms or 1 mol of Ti atoms
gold 197.0 g Au 6.02 ϫ 1023 Au atoms or 1 mol of Au atoms
hydrogen 1.0 g H2 6.02 ϫ 1023 H atoms or 1 mol of H atoms
(3.01 ϫ 1023 H2 molecules or ᎏ1
2
ᎏ mol
of H2 molecules)
sulfur 32.1 g S8 6.02 ϫ 1023 S atoms or 1 mol of S atoms
(0.753 ϫ 1023 S8 molecules or ᎏ1
8
ᎏ mol
of S8 molecules)
EXAMPLE 2-3 Moles of Atoms
How many moles of atoms does 136.9 g of iron metal contain?
Plan
The atomic weight of iron is 55.85 amu. This tells us that the molar mass of iron is
55.85 g/mol, or that one mole of iron atoms is 55.85 g of iron. We can express this as either
of two unit factors:
or
Because one mole of iron has a mass of 55.85 g, we expect that 136.9 g will be a fairly small
number of moles (greater than 1, but less than 10).
Solution
_?_ mol Fe atoms ϭ 136.9 g Fe ϫ ϭ 2.451 mol Fe atoms
You should now work Exercises 34 and 40.
Once the number of moles of atoms of an element is known, the number of atoms in
the sample can be calculated, as Example 2-4 illustrates.
EXAMPLE 2-4 Numbers of Atoms
How many atoms are contained in 2.451 mol of iron?
Plan
One mole of atoms of an element contains Avogadro’s number of atoms, or 6.022 ϫ 1023 atoms.
This lets us generate the two unit factors
1 mol Fe atoms
ᎏᎏ
55.85 g Fe
55.85 g Fe
ᎏᎏ
1 mol Fe atoms
1 mol Fe atoms
ᎏᎏ
55.85 g Fe
Figure 2-9 One mole of atoms of
some common elements. Back row
(left to right): bromine, aluminum,
mercury, copper. Front row (left to
right): sulfur, zinc, iron.
60 CHAPTER 2: Chemical Formulas and Composition Stoichiometry
To the required four significant
figures, 1 mol Fe atoms ϭ 55.85 g Fe.
and
Solution
_?_ Fe atoms ϭ 2.451 mol Fe atoms ϫ ϭ 1.476 ϫ 1024 Fe atoms
We expected the number of atoms in more than two moles of atoms to be a very
large number. Written in nonscientific notation, the answer to this example is:
1,476,000,000,000,000,000,000,000.
You should now work Exercise 42.
If we know the atomic weight of an element on the carbon-12 scale, we can use the
mole concept and Avogadro’s number to calculate the average mass of one atom of that
element in grams (or any other mass unit we choose).
6.022 ϫ 1023 Fe atoms
ᎏᎏᎏ
1 mol Fe atoms
1 mol atoms
ᎏᎏᎏ
6.022 ϫ 1023 atoms
6.022 ϫ 1023 atoms
ᎏᎏᎏ
1 mol atoms
Try to name this number with its
many zeroes.
2-6 The Mole 61
The Development of ScienceCCHEMISTRY IN USE
Avogadro’s Number
If you think that the value of Avogadro’s number, 6 ϫ 1023,
is too large to be useful to anyone but chemists, look up into
the sky on a clear night. You may be able to see about 3000
stars with the naked eye, but the total number of stars swirling
around you in the known universe is approximately equal to
Avogadro’s number. Just think, the known universe contains
approximately one mole of stars! You don’t have to leave earth
to encounter such large numbers. The water in the Pacific
Ocean has a volume of about 6 ϫ 1023 mL and a mass of
about 6 ϫ 1023 g.
Avogadro’s number is almost incomprehensibly large. For
example, if one mole of dollars given away at the rate of a
million per second beginning when the earth first formed
some 4.5 billion years ago, would any remain today? Surprisingly,
about three fourths of the original mole of dollars
would be left today; it would take about 14,500,000,000 more
years to give away the remaining money at $1 million per
second.
Computers can be used to provide another illustration
of the magnitude of Avogadro’s number. If a computer can
count up to one billion in one second, it would take that
computer about 20 million years to count up to 6 ϫ 1023. In
contrast, recorded human history goes back only a few thousand
years.
The impressively large size of Avogadro’s number can give
us very important insights into the very small sizes of individual
molecules. Suppose one drop of water evaporates in
one hour. There are about 20 drops in one milliliter of water,
which weighs one gram. So one drop of water is about
0.05 g of water. How many H2O molecules evaporate per
second?
ϭ ϫ ϫ
ϫ ϫ
ϭ 5 ϫ 1017 H2O molecules/s
5 ϫ 1017 H2O molecules evaporating per second is five hundred
million billion H2O molecules evaporating per second
—a number that is beyond our comprehension! This calculation
helps us to recognize that water molecules are incredibly
small. There are approximately 1.7 ϫ 1021 water molecules
in a single drop of water.
By gaining some appreciation of the vastness of Avogadro’s
number, we gain a greater appreciation of the extremely tiny
volumes occupied by individual atoms, molecules, and ions.
Ronald DeLorenzo
Middle Georgia College
Original concept by Larry Nordell
1 min
ᎏ
60 s
1 h
ᎏ
60 min
6 ϫ 1023 H2O molecules
ᎏᎏᎏ
1 mol H2O
1 mol H2O
ᎏᎏ
18 g H2O
0.05 g H2O
ᎏᎏ
1 h
_?_ H2O molecules
ᎏᎏᎏ
1 s
EXAMPLE 2-5 Masses of Atoms
Calculate the average mass of one iron atom in grams.
Plan
We expect that the mass of a single atom in grams would be a very small number. We know
that one mole of Fe atoms has a mass of 55.85 g and contains 6.022 ϫ 1023 Fe atoms. We use
this information to generate unit factors to carry out the desired conversion.
Solution
ϭ ϫ ϭ 9.274 ϫ 10Ϫ23 g Fe/Fe atom
Thus, we see that the average mass of one Fe atom is only 9.274 ϫ 10Ϫ23 g, that is,
0.00000000000000000000009274 g.
Example 2-5 demonstrates how small atoms are and why it is necessary to use large
numbers of atoms in practical work. The next example will help you to realize how large
Avogadro’s number is.
EXAMPLE 2-6 Avogadro’s Number
A stack of 500 sheets of typing paper is 1.9 inches thick. Calculate the thickness, in inches and
in miles, of a stack of typing paper that contains one mole (Avogadro’s number) of sheets.
Plan
We construct unit factors from the data given, from conversion factors in Table 1-7, and from
Avogadro’s number.
Solution
_?_ in ϭ 1 mol sheets ϫ ϫ ϭ 2.3 ϫ 1021 in.
_?_ mi ϭ 2.3 ϫ 1021 in. ϫ ϫ ϭ 3.6 ϫ 1016 mi
By comparison, the sun is about 93 million miles from the earth. This stack of paper would
make 390 million stacks that reach from the earth to the sun.
1 mi
ᎏ
5280 ft
1 ft
ᎏ
12 in.
1.9 in.
ᎏᎏ
500 sheets
6.022 ϫ 1023 sheets
ᎏᎏᎏ
1 mol sheets
1 mol Fe atoms
ᎏᎏᎏ
6.022 ϫ 1023 Fe atoms
55.85 g Fe
ᎏᎏ
1 mol Fe atoms
_?_ g Fe
ᎏ
Fe atom
To gain some appreciation of how little
this is, write 9.274 ϫ 10Ϫ23 gram as a
decimal fraction, and try to name the
fraction.
62 CHAPTER 2: Chemical Formulas and Composition Stoichiometry
Problem-Solving Tip: When Do We Round?
Even though the number 1.9 has two significant figures, we carry the other numbers in
Example 2-6 to more significant figures. Then we round at the end to the appropriate
number of significant figures. The numbers in the distance conversions are exact num-
bers.
Imagine the number of trees required
to make this much paper!
FORMULA WEIGHTS, MOLECULAR WEIGHTS,
AND MOLES
The formula weight (FW) of a substance is the sum of the atomic weights (AW)
of the elements in the formula, each taken the number of times the element occurs.
Hence a formula weight gives the mass of one formula unit in atomic mass units.
Formula weights, like the atomic weights on which they are based, are relative masses.
The formula weight for sodium hydroxide, NaOH, (rounded off to the nearest 0.1 amu)
is found as follows.
Number of Atoms of Stated Kind ؋ Mass of One Atom ‫؍‬ Mass Due to Element
1 ϫ Na ϭ 1 ϫ 23.0 amu ϭ 23.0 amu of Na
1 ϫ H ϭ 1 ϫ 01.0 amu ϭ 01.0 amu of H
1 ϫ O ϭ 1 ϫ 16.0 amu ϭ 16.0 amu of O
Formula weight of NaOH ϭ 40.0 amu
The term “formula weight” is correctly used for either ionic or molecular substances.
When we refer specifically to molecular (nonionic) substances, that is, substances that exist
as discrete molecules, we often substitute the term molecular weight (MW).
EXAMPLE 2-7 Formula Weights
Calculate the formula weight (molecular weight) of acetic acid (vinegar), CH3COOH, using
rounded values for atomic weights given in the International Table of Atomic Weights inside
the front cover of the text.
Plan
We add the atomic weights of the elements in the formula, each multiplied by the number of
times the element occurs.
Solution
Number of Atoms of Stated Kind ؋ Mass of One Atom ‫؍‬ Mass Due to Element
2 ϫ C ϭ 2 ϫ 12.0 amu ϭ 24.0 amu of C
4 ϫ H ϭ 4 ϫ 01.0 amu ϭ 04.0 amu of H
2 ϫ O ϭ 2 ϫ 16.0 amu ϭ 32.0 amu of O
Formula weight (molecular weight) of acetic acid (vinegar) ϭ 60.0 amu
You should now work Exercise 26.
2-7
A space-filling model of an acetic
acid (vinegar) molecule,
CH3COOH.
2-7 Formula Weights, Molecular Weights, and Moles 63
The amount of substance that contains the mass in grams numerically equal to its
formula weight in amu contains 6.022 ϫ 1023 formula units, or one mole of the substance.
This is sometimes called the molar mass of the substance. Molar mass is
numerically equal to the formula weight of the substance (the atomic weight for atoms
of elements) and has the units grams/mole.
One mole of sodium hydroxide is 40.0 g of NaOH, and one mole of acetic acid is 60.0 g
of CH3COOH. One mole of any molecular substance contains 6.02 ϫ 1023 molecules of
the substance, as Table 2-5 illustrates.
Because no simple NaCl molecules exist at ordinary temperatures, it is inappropriate
to refer to the “molecular weight” of NaCl or any ionic compound. One mole of an ionic
compound contains 6.02 ϫ 1023 formula units of the substance. Recall that one formula
unit of sodium chloride consists of one sodium ion, Naϩ, and one chloride ion, ClϪ. One
mole, or 58.4 g, of NaCl contains 6.02 ϫ 1023 Naϩ ions and 6.02 ϫ 1023 ClϪ ions (Table
2-6).
The mole concept, together with Avogadro’s number, provides important connections
among the extensive properties mass of substance, number of moles of substance, and
number of molecules or ions. These are summarized as follows.
64 CHAPTER 2: Chemical Formulas and Composition Stoichiometry
ᎏ
(mol
F
ar
W
m
A
ass A)
ᎏ ϭ ᎏ
m
g
ol
A
A
ᎏ ᎏ
Av
n
o
u
g
m
a
b
d
e
ro
r
’s
ᎏ ϭ
Formula units of A
ᎏᎏᎏ
mol A
Moles A
Moles A
Mass A
Formula units of A
h
666666666
g
h
666666666
g
TABLE 2-5 One Mole of Some Common Molecular Substances
Molecular A Sample with a
Substance Weight Mass of Contains
6.02 ϫ 1023 H2 molecules or
hydrogen 2.0 2.0 g H2
1 mol of H2 molecules
(contains 2 ϫ 6.02 ϫ 1023 H
atoms or 2 mol of H atoms)
6.02 ϫ 1023 O2 molecules or
oxygen 32.0 32.0 g O2
1 mol of O2 molecules
(contains 2 ϫ 6.02 ϫ 1023 O
atoms or 2 mol of O atoms)
6.02 ϫ 1023 CH4 molecules or
methane 16.0 16.0 g CH4
1 mol of CH4 molecules
(contains 6.02 ϫ 1023 C atoms
and 4 ϫ 6.02 ϫ 1023 H atoms)
6.02 ϫ 1023 CH3COOH
acetic acid 60.0 60.0 g CH3COOH molecules or 1 mol of
(vinegar) CH3COOH molecules


















The following examples show the relations between numbers of molecules, atoms, or
formula units and their masses.
EXAMPLE 2-8 Masses of Molecules
What is the mass in grams of 10.0 million SO2 molecules?
Plan
One mole of SO2 contains 6.02 ϫ 1023 SO2 molecules and has a mass of 64.1 grams.
Solution
_?_ g SO2 ϭ 10.0 ϫ 106 SO2 molecules ϫ
ϭ 1.06 ϫ 10Ϫ15 g SO2
Ten million SO2 molecules have a mass of only 0.00000000000000106 g. Commonly used
analytical balances are capable of weighing to Ϯ0.0001 g.
You should now work Exercise 44.
EXAMPLE 2-9 Moles
How many (a) moles of O2, (b) O2 molecules, and (c) O atoms are contained in 40.0 g of
oxygen gas (dioxygen) at 25°C?
Plan
We construct the needed unit factors from the following equalities: (a) the mass of one mole
of O2 is 32.0 g (molar mass O2 ϭ 32.0 g/mol); (b) one mole of O2 contains 6.02 ϫ 1023 O2
molecules; (c) one O2 molecule contains two O atoms.
64.1 g SO2
ᎏᎏᎏ
6.02 ϫ 1023 SO2 molecules
When fewer than four significant
figures are needed in calculations,
Avogadro’s number may be rounded
off to 6.02 ϫ 1023.
2-7 Formula Weights, Molecular Weights, and Moles 65
TABLE 2-6 One Mole of Some Ionic Compounds
Formula A Sample with a
Compound Weight Mass of 1 Mol Contains
6.02 ϫ 1023 Naϩ ions or
sodium 58.4 58.4 g NaCl
1 mol of Naϩ ions
chloride 6.02 ϫ 1023 ClϪ ions or
1 mol of ClϪ ions
6.02 ϫ 1023 Ca2ϩ ions or
calcium 111.0 111.0 g CaCl2
1 mol of Ca2ϩ ions
chloride 2(6.02 ϫ 1023) ClϪ ions or
2 mol of ClϪ ions
2(6.02 ϫ 1023) Al3ϩ ions or
aluminum 342.1 342.1 g Al2(SO4)3
2 mol of Al3ϩ ions
sulfate 3(6.02 ϫ 1023) SO4
2Ϫ ions or
3 mol of SO4
2Ϫ ions















Solution
One mole of O2 contains 6.02 ϫ 1023 O2 molecules, and its mass is 32.0 g.
(a) _?_ mol O2 ϭ 40.0 g O2 ϫ ϭ 1.25 mol O2
(b) _?_ O2 molecules ϭ 40.0 g O2 ϫ
ϭ 7.52 ϫ 1023 molecules
Or, we can use the number of moles of O2 calculated in part (a) to find the number of O2 mol-
ecules.
_?_ O2 molecules ϭ 1.25 mol O2 ϫ ϭ 7.52 ϫ 1023 O2 molecules
(c) _?_ O atoms ϭ 40.0 g O2 ϫ ϫ
ϭ 1.50 ϫ 1024 O atoms
You should now work Exercise 30.
EXAMPLE 2-10 Numbers of Atoms
Calculate the number of hydrogen atoms in 39.6 g of ammonium sulfate, (NH4)2SO4.
Plan
One mole of (NH4)2SO4 is 6.02 ϫ 1023 formula units and has a mass of 132.1 g.
88n 88n 88n H atoms
Solution
formula units of
ᎏᎏ
(NH4)2SO4
mol of
ᎏᎏ
(NH4)2SO4
g of
ᎏᎏ
(NH4)2SO4
2 O atoms
ᎏᎏ
1 O2 molecule
6.02 ϫ 1023 O2 molecules
ᎏᎏᎏ
32.0 g O2
6.02 ϫ 1023 O2 molecules
ᎏᎏᎏ
1 mol O2
6.02 ϫ 1023 O2 molecules
ᎏᎏᎏ
32.0 g O2
1 mol O2
ᎏᎏ
32.0 g O2
In Example 2-10, we relate (a) grams
to moles, (b) moles to formula units,
and (c) formula units to H atoms.
66 CHAPTER 2: Chemical Formulas and Composition Stoichiometry
You should now work Exercise 38.
PERCENT COMPOSITION AND FORMULAS
OF COMPOUNDS
If the formula of a compound is known, its chemical composition can be expressed as the
mass percent of each element in the compound (percent composition). For example, one
2-8
_?_ H atoms ϭ 39.6 g (NH4)2SO4 ϫ ϫ
ϫ
ϭ 1.44 ϫ 1024 H atoms
8 H atoms
ᎏᎏᎏᎏ
1 formula units (NH4)2SO4
6.02 ϫ 1023 formula units (NH4)2SO4
ᎏᎏᎏᎏᎏ
1 mol (NH4)2SO4
1 mol (NH4)2SO4
ᎏᎏᎏ
132.1 g (NH4)2SO4
carbon dioxide molecule, CO2, contains one C atom and two O atoms. Percentage is the
part divided by the whole times 100 percent (or simply parts per 100), so we can represent
the percent composition of carbon dioxide as follows:
2-8 Percent Composition and Formulas of Compounds 67
% C ϭ ϫ 100% ϭ ϫ 100% ϭ ϫ 100% ϭ 27.3%
% O ϭ ϫ 100% ϭ ϫ 100% ϭ ϫ 100% ϭ 72.7% O
2(16.0 amu)
ᎏᎏ
44.0 amu
2 ϫ AW of O
ᎏᎏ
MW of CO2
mass of O
ᎏᎏ
mass of CO2
12.0 amu
ᎏᎏ
44.0 amu
AW of C
ᎏᎏ
MW of CO2
mass of C
ᎏᎏ
mass of CO2
One mole of CO2 (44.0 g) contains one mole of C atoms (12.0 g) and two moles of O
atoms (32.0 g). We could therefore have used these masses in the preceding calculation.
These numbers are the same as the ones used—only the units are different. In Example
2-11 we shall base our calculation on one mole rather than one molecule.
EXAMPLE 2-11 Percent Composition
Calculate the percent composition by mass of HNO3.
Plan
We first calculate the mass of one mole as in Example 2-7. Then we express the mass of each
element as a percent of the total.
Solution
The molar mass of HNO3 is calculated first.
Number of Mol of Atoms ϫ Mass of One Mol of Atoms ϭ Mass Due to Element
1 ϫ H ϭ 1 ϫ 1.0 g ϭ 1.0 g of H
1 ϫ N ϭ 1 ϫ 14.0 g ϭ 14.0 g of N
3 ϫ O ϭ 3 ϫ 16.0 g ϭ 48.0 g of O
Mass of 1 mol of HNO3 ϭ 63.0 g
Now, its percent composition is
% H ϭ ϫ 100% ϭ ϫ 100% ϭ 1.6% H
% N ϭ ϫ 100% ϭ ϫ 100% ϭ 22.2% N
% O ϭ ϫ 100% ϭ ϫ 100% ϭ 76.2% O
Total ϭ 100.0%
You should now work Exercise 46.
Nitric acid is 1.6% H, 22.2% N, and 76.2% O by mass. All samples of pure HNO3
have this composition, according to the Law of Definite Proportions.
48.0 g
ᎏ
63.0 g
mass of O
ᎏᎏ
mass of HNO3
14.0 g
ᎏ
63.0 g
mass of N
ᎏᎏ
mass of HNO3
1.0 g
ᎏ
63.0 g
mass of H
ᎏᎏ
mass of HNO3
As a check, we see that the percentages
add to 100%.
When chemists use the % notation,
they mean percent by mass unless they
specify otherwise.
68 CHAPTER 2: Chemical Formulas and Composition Stoichiometry
The Development of ScienceCCHEMISTRY IN USE
Names of the Elements
If you were to discover a new element, how would you name
it? Throughout history, scientists have answered this question
in different ways. Most have chosen to honor a person
or place or to describe the new substance.
Until the Middle Ages only nine elements were known:
gold, silver, tin, mercury, copper, lead, iron, sulfur, and carbon.
The metals’ chemical symbols are taken from descriptive
Latin names: aurum (“yellow”), argentum (“shining”),
stannum (“dripping” or “easily melted”), hydrargyrum (“silvery
water”), cuprum (“Cyprus,” where many copper mines
were located), plumbum (exact meaning unknown—possibly
“heavy”), and ferrum (also unknown). Mercury is named after
the planet, one reminder that the ancients associated metals
with gods and celestial bodies. In turn, both the planet, which
moves rapidly across the sky, and the element, which is the
only metal that is liquid at room temperature and thus flows
rapidly, are named for the fleet god of messengers in Roman
mythology. In English, mercury is nicknamed “quicksilver.”
Prior to the reforms of Antoine Lavoisier (1743–1794),
chemistry was a largely nonquantitative, unsystematic science
in which experimenters had little contact with each other. In
1787, Lavoisier published his Methode de Nomenclature
Chimique, which proposed, among other changes, that all new
elements be named descriptively. For the next 125 years, most
elements were given names that corresponded to their properties.
Greek roots were one popular source, as evidenced by
hydrogen (hydros-gen, “water-producing”), oxygen (oksys-gen,
“acid-producing”), nitrogen (nitron-gen, “soda-producing”),
bromine (bromos, “stink”), and argon (a-er-gon, “no reaction”).
The discoverers of argon, Sir William Ramsay (1852–1916)
and Baron Rayleigh (1842–1919), originally proposed the
name aeron (from aer or air), but critics thought it was too
close to the biblical name Aaron! Latin roots, such as radius
(“ray”), were also used (radium and radon are both naturally
radioactive elements that emit “rays”). Color was often the
determining property, especially after the invention of the
spectroscope in 1859, because different elements (or the light
that they emit) have prominent characteristic colors. Cesium,
indium, iodine, rubidium, and thallium were all named in this
manner. Their respective Greek and Latin roots denote bluegray,
indigo, violet, red, and green (thallus means “tree
sprout”). Because of the great variety of colors of its compounds,
iridium takes its name from the Latin iris, meaning
“rainbow.” Alternatively, an element name might suggest a
mineral or the ore that contained it. One example is Wolfram
or tungsten (W), which was isolated from wolframite. Two
other “inconsistent” elemental symbols, K and Na, arose
from occurrence as well. Kalium was first obtained from the
saltwort plant, Salsola kali, and natrium from niter. Their
English names, potassium and sodium, are derived from the
ores potash and soda.
Other elements, contrary to Lavoisier’s suggestion, were
named after planets, mythological figures, places, or superstitions.
“Celestial elements” include helium (“sun”), tellurium
(“earth”), selenium (“moon”—the element was discovered
in close proximity to tellurium), cerium (the asteroid
Ceres, which was discovered only two years before the element),
and uranium (the planet Uranus, discovered a few
years earlier). The first two transuranium elements (those beyond
uranium) to be produced were named neptunium and
plutonium for the next two planets, Neptune and Pluto. The
names promethium (Prometheus, who stole fire from
heaven), vanadium (Scandinavian goddess Vanadis), titanium
(Titans, the first sons of the earth), tantalum (Tantalos,
father of the Greek goddess Niobe), and thorium (Thor,
Scandinavian god of war) all arise from Greek or Norse
mythology.
“Geographical elements,” shown on the map, sometimes
honored the discoverer’s native country or workplace. The
Latin names for Russia (ruthenium), France (gallium), Paris
(lutetium), and Germany (germanium) were among those
used. Marie Sklodowska Curie named one of the elements
that she discovered polonium, after her native Poland. Often
the locale of discovery lends its name to the element; the
record holder is certainly the Swedish village Ytterby, the
site of ores from which the four elements terbium, erbium,
ytterbium, and yttrium were isolated. Elements honoring
important scientists include curium, einsteinium, nobelium,
fermium, and lawrencium.
Most of the elements now known were given titles peacefully,
but a few were not. Niobium, isolated in 1803 by
Ekeberg from an ore that also contained tantalum, and named
after Niobe (daughter of Tantalos), was later found to be
identical to an 1802 discovery of C. Hatchett, columbium.
(Interestingly, Hatchett first found the element in an ore
sample that had been sent to England more than a century
2-8 Percent Composition and Formulas of Compounds 69
earlier by John Winthrop, the first governor of Connecticut.)
Although “niobium” became the accepted designation in
Europe, the Americans, not surprisingly, chose “columbium.”
It was not until 1949—when the International Union of Pure
and Applied Chemistry (IUPAC) ended more than a century
of controversy by ruling in favor of mythology—that element
41 received a unique name.
In 1978, the IUPAC recommended that elements beyond
103 be known temporarily by systematic names based on numerical
roots; element 104 is unnilquadium (un for 1, nil for
0, quad for 4, plus the -ium ending), followed by unnilpentium,
unnilhexium, and so on. Arguments over the names of
elements 104 and 105 prompted the IUPAC to begin hearing
claims of priority to numbers 104 to 109. The IUPAC’s
final recommendations for these element names were announced
in 1997. The names and symbols recommended by
that report are: element 104, rutherfordium, Rf; element 105,
dubnium, Db; element 106, seaborgium, Sg; element 107,
bohrium, Bh; element 108, hassium, Hs; and element 109,
meitnerium, Mt. Some of these (Rf and Bh) are derived from
the names of scientists prominent in the development of
atomic theory; others (Sg, Hs, and Mt) are named for scientists
who were involved in the discovery of heavy elements.
Dubnium is named in honor of the Dubna laboratory in the
former Soviet Union, where important contributions to the
creation of heavy elements have originated.
Lisa Saunders Boffa
Senior Chemist
Exxon Corporation
LutetiumCalifornium
Berkelium
Hafnium
Rhenium
Holmium
Magnesium
Mangan
Kadmium
Kobber
Erbium
Terbium
Yttrium
Ytterbium
Strontium
THULIUM
FRANCIUM
GALLIUM
GERMANIUM
POLONIUMAMERICIUM
RUTHENIUM
SCANDIUM
EUROPIUM
Many chemical elements were named
after places.
A ball-and-stick model of a molecule
of water, H2O.
70 CHAPTER 2: Chemical Formulas and Composition Stoichiometry
Problem-Solving Tip: The Whole Is Equal to the Sum of Its Parts
Percentages must add to 100%. Roundoff errors may not cancel, however, and totals
such as 99.9% or 100.1% may be obtained in calculations. As an alternative method of
calculation, if we know all of the percentages except one, we can subtract their sum from
100% to obtain the other value.
DERIVATION OF FORMULAS FROM ELEMENTAL
COMPOSITION
The simplest, or empirical, formula for a compound is the smallest whole-number ratio
of atoms present. For molecular compounds the molecular formula indicates the actual
numbers of atoms present in a molecule of the compound. It may be the same as the
simplest formula or else some whole-number multiple of it. For example, the simplest and
molecular formulas for water are both H2O; however, for hydrogen peroxide, they are
HO and H2O2, respectively.
Each year thousands of new compounds are made in laboratories or discovered in nature.
One of the first steps in characterizing a new compound is the determination of its
percent composition. A qualitative analysis is performed to determine which elements are
present in the compound. Then a quantitative analysis is performed to determine the
amount of each element.
Once the percent composition of a compound (or its elemental composition by mass)
is known, the simplest formula can be determined.
EXAMPLE 2-12 Simplest Formulas
Compounds containing sulfur and oxygen are serious air pollutants; they represent the major
cause of acid rain. Analysis of a sample of a pure compound reveals that it contains 50.1% sulfur
and 49.9% oxygen by mass. What is the simplest formula of the compound?
Plan
One mole of atoms of any element is 6.02 ϫ 1023 atoms, so the ratio of moles of atoms in any
sample of a compound is the same as the ratio of atoms in that compound. This calculation is
carried out in two steps.
Step 1: Let’s consider 100.0 g of compound, which contains 50.1 g of S and 49.9 g of O. We
calculate the number of moles of atoms of each.
Step 2: We then obtain a whole-number ratio between these numbers that gives the ratio of
atoms in the sample and hence in the simplest formula for the compound.
Solution
Step 1: _?_ mol S atoms ϭ 50.1 g S ϫ ϭ 1.56 mol S atoms
_?_ mol O atoms ϭ 49.9 g O ϫ ϭ 3.12 mol O atoms
1 mol O atoms
ᎏᎏ
16.0 g O
1 mol S atoms
ᎏᎏ
32.1 g S
2-9
Remember that percent means parts per
hundred.
A ball-and-stick model of a molecule
of hydrogen peroxide, H2O2.
Step 2: Now we know that 100.0 g of the compound contains 1.56 mol of S atoms and 3.12
mol of O atoms. We obtain a whole-number ratio between these numbers that gives the ratio
of atoms in the simplest formula.
ϭ 1.00 S
SO2
ϭ 2.00 O
You should now work Exercise 52.
The solution for Example 2-12 can be set up in tabular form.
Relative Relative Number Smallest WholeMass
of of Atoms Divide by Number Ratio
Element Element (divide mass by AW) Smaller Number of Atoms
S 50.1 ᎏ
5
3
0
2
.
.
1
1
ᎏ ϭ 1.56 ᎏ
1
1
.
.
5
5
6
6
ᎏ ϭ 1.00 S
SO2
O 49.9 ᎏ
4
1
9
6
.
.
9
0
ᎏ ϭ 3.12 ᎏ
3
1
.
.
1
5
2
6
ᎏ ϭ 2.00 O
This tabular format provides a convenient way to solve simplest-formula problems, as the
next example illustrates.
EXAMPLE 2-13 Simplest Formula
A 20.882-gram sample of an ionic compound is found to contain 6.072 grams of Na, 8.474
grams of S, and 6.336 grams of O. What is its simplest formula?
Plan
We reason as in Example 2-12, calculating the number of moles of each element and the ratio
among them. Here we use the tabular format that was introduced earlier.
Solution
3.12
ᎏ
1.56
1.56
ᎏ
1.56
A simple and useful way to obtain
whole-number ratios among several
numbers follows: (a) Divide each
number by the smallest, and then,
(b) if necessary, multiply all of the
resulting numbers by the smallest
whole number that will eliminate
fractions.
2-9 Derivation of Formulas from Elemental Composition 71
Relative Relative Number Convert Fractions Smallest WholeMass
of of Atoms Divide by to Whole Numbers Number Ratio
Element Element (divide mass by AW) Smallest Number (multiply by integer) of Atoms
Na 6.072 ϭ 0.264 ϭ 1.00 1.00 ϫ 2 ϭ 2 Na
S 8.474 ϭ 0.264 ϭ 1.00 1.00 ϫ 2 ϭ 2 S Na2S2O3
O 6.336 ϭ 0.396 ϭ 1.50 1.50 ϫ 2 ϭ 3 O
0.396
ᎏ
0.264
6.336
ᎏ
16.0
0.264
ᎏ
0.264
8.474
ᎏ
32.1
0.264
ᎏ
0.264
6.072
ᎏ
23.0
The “Relative Mass” column is
proportional to the mass of each
element in grams. With this
interpretation, the next column could
be headed “Relative Number of Moles
of Atoms.” Then the last column
would represent the smallest wholenumber
ratios of moles of atoms. But
because a mole is always the same
number of items (atoms), that ratio is
the same as the smallest whole-number
ratio of atoms.
The ratio of atoms in the simplest formula must be a whole-number ratio (by definition). To convert
the ratio 1Ϻ1Ϻ1.5 to a whole-number ratio, each number in the ratio was multiplied by 2,
which gave the simplest formula Na2S2O3.
You should now work Exercise 53.
72 CHAPTER 2: Chemical Formulas and Composition Stoichiometry
Millions of compounds are composed of carbon, hydrogen, and oxygen. Analyses for
C and H can be performed in a C-H combustion system (Figure 2-10). An accurately
Problem-Solving Tip: Know Common Fractions in Decimal Form
As Example 2-13 illustrates, sometimes we must convert a fraction to a whole number
by multiplication by the correct integer. But we must first recognize which fraction is
represented by a nonzero part of a number. The decimal equivalents of the following
fractions may be useful.
Decimal Equivalent To Convert to Integer,
(to 2 places) Fraction Multiply by
0.50 ᎏ1
2
ᎏ 2
0.33 ᎏ1
3
ᎏ 3
0.67 ᎏ2
3
ᎏ 3
0.25 ᎏ1
4
ᎏ 4
0.75 ᎏ3
4
ᎏ 4
0.20 ᎏ1
5
ᎏ 5
The fractions ᎏ2
5
ᎏ, ᎏ3
5
ᎏ, ᎏ4
5
ᎏ are equal to 0.40, 0.60, and 0.80, respectively; these should be multiplied
by 5.
When we use the procedure given in this section, we often obtain numbers such as
0.99 and 1.52. Because the results obtained by analysis of samples usually contain some
error (as well as roundoff errors), we would interpret 0.99 as 1.0 and 1.52 as 1.5.
Figure 2-10 A combustion train used for carbon-hydrogen analysis. The absorbent for
water is magnesium perchlorate, Mg(ClO4)2. Carbon dioxide is absorbed by finely divided
sodium hydroxide supported on glass wool. Only a few milligrams of sample is needed for
analysis.
O2
Sample
Furnace
H2O absorber
CO2 absorber
known mass of a compound is burned in a furnace in a stream of oxygen. The carbon and
hydrogen in the sample are converted to carbon dioxide and water vapor, respectively.
The resulting increases in masses of the CO2 and H2O absorbers can then be related to
the masses and percentages of carbon and hydrogen in the original sample.
EXAMPLE 2-14 Percent Composition
Hydrocarbons are organic compounds composed entirely of hydrogen and carbon. A 0.1647gram
sample of a pure hydrocarbon was burned in a C-H combustion train to produce 0.4931
gram of CO2 and 0.2691 gram of H2O. Determine the masses of C and H in the sample and
the percentages of these elements in this hydrocarbon.
Plan
Step 1: We use the observed mass of CO2, 0.4931 grams, to determine the mass of carbon in
the original sample. There is one mole of carbon atoms, 12.01 grams, in each mole of CO2,
44.01 grams; we use this information to construct the unit factor
Step 2: Likewise, we can use the observed mass of H2O, 0.2691 grams, to calculate the amount
of hydrogen in the original sample. We use the fact that there are two moles of hydrogen
atoms, 2.016 grams, in each mole of H2O, 18.02 grams, to construct the unit factor
Step 3: Then we calculate the percentages by mass of each element in turn, using the rela-
tionship
% element ϭ ϫ 100%
Solution
Step 1: _?_ g C ϭ 0.4931 g CO2 ϫ ϭ 0.1346 g C
Step 2: _?_ g H ϭ 0.2691 g H2O ϫ ϭ 0.03010 g H
Step 3: % C ϭ ϫ 100% ϭ 81.72% C
%H ϭ ϫ 100% ϭ 18.28% H
Total ϭ 100.00%
You should now work Exercise 58.
When the compound to be analyzed contains oxygen, the calculation of the amount
or percentage of oxygen in the sample is somewhat different. Part of the oxygen that goes
to form CO2 and H2O comes from the sample and part comes from the oxygen stream
supplied. For that reason, we cannot directly determine the amount of oxygen already in
0.03010 g H
ᎏᎏ
0.1647 g sample
0.1346 g C
ᎏᎏ
0.1647 g sample
2.016 g H
ᎏᎏ
18.02 g H2O
12.01 g C
ᎏᎏ
44.01 g CO2
g element
ᎏᎏ
g sample
2.016 g H
ᎏᎏ
18.02 g H2O
12.01 g C
ᎏᎏ
44.01 g CO2
Hydrocarbons are obtained from coal
and coal tar and from oil and gas wells.
The main use of hydrocarbons is as
fuels. The simplest hydrocarbons are
methane CH4
ethane C2H6
propane C3H8
butane C4H10
2-9 Derivation of Formulas from Elemental Composition 73
We could calculate the mass of H
by subtracting mass of C from mass
of sample. It is good experimental
practice, however, when possible,
to base both on experimental
measurements, as we have done here.
This would help to check for errors in
the analysis or calculation.
Can you show that the hydrocarbon
in Example 2-14 is propane, C3H8?
the sample. The approach is to analyze as we did in Example 2-14 for all elements except
oxygen. Then we subtract the sum of their masses from the mass of the original sample
to obtain the mass of oxygen. The next example illustrates such a calculation.
EXAMPLE 2-15 Percent Composition
A 0.1014-g sample of purified glucose was burned in a C-H combustion train to produce
0.1486 g of CO2 and 0.0609 g of H2O. An elemental analysis showed that glucose contains
only carbon, hydrogen, and oxygen. Determine the masses of C, H, and O in the sample and
the percentages of these elements in glucose.
Plan
Steps 1 and 2: We first calculate the masses of carbon and hydrogen as we did in Example
2-14.
Step 3: The rest of the sample must be oxygen because glucose has been shown to contain only
C, H, and O. So we subtract the masses of C and H from the total mass of sample.
Step 4: Then we calculate the percentage by mass for each element.
Solution
Step 1: _?_ g C ϭ 0.1486 g CO2 ϫ ϭ 0.04055 g C
Step 2: _?_ g H ϭ 0.0609 g H2O ϫ ϭ 0.00681 g H
Step 3: _?_ g O ϭ 0.1014 g sample Ϫ [0.04055 g C ϩ 0.00681 g H] ϭ 0.0540 g O
Step 4: Now we can calculate the percentages by mass for each element:
% C ϭ ϫ 100% ϭ 39.99% C
% H ϭ ϫ 100% ϭ 06.72% H
% O ϭ ϫ 100% ϭ 53.2% O
Total ϭ 099.9%
You should now work Exercise 60.
DETERMINATION OF MOLECULAR FORMULAS
Percent composition data yield only simplest formulas. To determine the molecular formula
for a molecular compound, both its simplest formula and its molecular weight must
be known. Some methods for experimental determination of molecular weights are introduced
in Chapters 12 and 14.
2-10
0.0540 g O
ᎏᎏ
0.1014 g
0.00681 g H
ᎏᎏ
0.1014 g
0.04055 g C
ᎏᎏ
0.1014 g
2.016 g H
ᎏᎏ
18.02 g H2O
12.01 g C
ᎏᎏ
44.01 g CO2
We say that the mass of O in the
sample is calculated by difference.
74 CHAPTER 2: Chemical Formulas and Composition Stoichiometry
Glucose, a simple sugar, is the main
component of intravenous feeding
liquids. Its common name is dextrose.
It is also one of the products
of carbohydrate metabolism.
For many compounds the molecular formula is a multiple of the simplest formula.
Consider butane, C4H10. The simplest formula for butane is C2H5, but the molecular formula
contains twice as many atoms; that is, 2 ϫ (C2H5) ϭ C4H10. Benzene, C6H6, is another
example. The simplest formula for benzene is CH, but the molecular formula contains
six times as many atoms; that is, 6 ϫ (CH) ϭ C6H6.
The molecular formula for a compound is either the same as, or an integer multiple of,
the simplest formula.
Molecular formula ϭ n ϫ simplest formula
So we can write
Molecular weight ϭ n ϫ simplest formula weight
n ϭ
The molecular formula is then obtained by multiplying the simplest formula by the integer,
n.
EXAMPLE 2-16 Molecular Formula
In Example 2-15, we found the elemental composition of glucose. Other experiments show
that its molecular weight is approximately 180 amu. Determine the simplest formula and the
molecular formula of glucose.
Plan
Step 1: We first use the masses of C, H, and O found in Example 2-15 to determine the simplest
formula.
Step 2: We can use the simplest formula to calculate the simplest formula weight. Because the
molecular weight of glucose is known (approximately 180 amu), we can determine the molecular
formula by dividing the molecular weight by the simplest formula weight.
n ϭ
The molecular weight is n times the simplest formula weight, so the molecular formula of glucose
is n times the simplest formula.
Solution
Step 1:
molecular weight
ᎏᎏᎏ
simplest formula weight
molecular weight
ᎏᎏᎏ
simplest formula weight
As an alternative, we could have used
the percentages by mass from Example
2-15. Using the earliest available
numbers helps to minimize the effects
of rounding-off errors.
2-10 Determination of Molecular Formulas 75
Mass of Moles of Element Divide by Smallest Whole-Number
Element Element (divide mass by AW) Smallest Ratio of Atoms
C 0.04055 g ϭ 0.003376 mol ϭ 1.00 C
H 0.00681 g ϭ 0.00676 mol ϭ 2.00 H CH2O
O 0.0540 g ϭ 0.00338 mol ϭ 1.00 O
0.00338
ᎏᎏ
0.003376
0.0540
ᎏ
16.00
0.00676
ᎏᎏ
0.003376
0.00681
ᎏ
1.008
0.003376
ᎏᎏ
0.003376
0.04055
ᎏ
12.01
Step 2: The simplest formula is CH2O, which has a formula weight of 30.03 amu. Because the
molecular weight of glucose is approximately 180 amu, we can determine the molecular formula
by dividing the molecular weight by the simplest formula weight.
n ϭ ϭ 6.00
The molecular weight is six times the simplest formula weight, 6 ϫ (CH2O) ϭ C6H12O6, so
the molecular formula of glucose is C6H12O6.
You should now work Exercises 63 and 64.
As we shall see when we discuss the composition of compounds in some detail, two
(and sometimes more) elements may form more than one compound. The Law of
Multiple Proportions summarizes many experiments on such compounds. It is usually
stated: When two elements, A and B, form more than one compound, the ratio of the
masses of element B that combine with a given mass of element A in each of the compounds
can be expressed by small whole numbers. Water, H2O, and hydrogen peroxide,
H2O2, provide an example. The ratio of masses of oxygen that combine with a given mass
of hydrogen in H2O and H2O2 is 1Ϻ2. Many similar examples, such as CO and CO2 (1Ϻ2
ratio) and SO2 and SO3 (2Ϻ3 ratio), are known. The Law of Multiple Proportions had
been recognized from studies of elemental composition before the time of Dalton. It provided
additional support for his atomic theory.
EXAMPLE 2-17 Law of Multiple Proportions
What is the ratio of the masses of oxygen that are combined with one gram of nitrogen in the
compounds N2O3 and NO?
Plan
First we calculate the mass of O that combines with one gram of N in each compound. Then
we determine the ratio of the values of ᎏ
g
g
N
O
ᎏ for the two compounds.
Solution
In N2O3: ϭ ϭ 1.71 g O/g N
In NO: ϭ ϭ 1.14 g O/g N
(in N2O3)
The ratio is ϭ ϭ
(in NO)
We see that the ratio is three mass units of O (in N2O3) to two mass units of O (in NO).
You should now work Exercises 67 and 68.
g O
ᎏ
g N
3
ᎏ
2
1.5
ᎏ
1.0
1.71 g O/g N
ᎏᎏ
1.14 g O/g N
g O
ᎏ
g N
16.0 g O
ᎏᎏ
14.0 g N
_?_ g O
ᎏ
g N
48.0 g O
ᎏᎏ
28.0 g N
_?_ g O
ᎏ
g N
180 amu
ᎏᎏ
30.03 amu
Many sugars are rich sources in our
diet. The most familiar is ordinary
table sugar, which is sucrose,
C12H22O11. An enzyme in our saliva
readily splits sucrose into two simple
sugars, glucose and fructose. The
simplest formula for both glucose
and fructose is C6H12O6. They have
different structures and different
properties, however, so they are
different compounds.
76 CHAPTER 2: Chemical Formulas and Composition Stoichiometry
8888n
88888n











SOME OTHER INTERPRETATIONS OF CHEMICAL
FORMULAS
Once we master the mole concept and the meaning of chemical formulas, we can use them
in many other ways. The examples in this section illustrate a few additional kinds of information
we can get from a chemical formula and the mole concept.
EXAMPLE 2-18 Composition of Compounds
What mass of chromium is contained in 35.8 g of (NH4)2Cr2O7?
Plan
Let us first solve the problem in several steps.
Step 1: The formula tells us that each mole of (NH4)2Cr2O7 contains two moles of Cr atoms,
so we first find the number of moles of (NH4)2Cr2O7, using the unit factor
Step 2: Then we convert the number of moles of (NH4)2Cr2O7 into the number of moles of
Cr atoms it contains, using the unit factor
Step 3: We then use the atomic weight of Cr to convert the number of moles of chromium
atoms to mass of chromium.
Mass (NH4)2Cr2O7 88n mol (NH4)2Cr2O7 88n mol Cr 88n Mass Cr
Solution
Step 1: _?_ mol (NH4)2Cr2O7 ϭ 35.8 g (NH4)2Cr2O7 ϫ
ϭ 0.142 mol (NH4)2Cr2O7
Step 2: _?_ mol Cr atoms ϭ 0.142 mol (NH4)2Cr2O7 ϫ
ϭ 0.284 mol Cr atoms
Step 3: _?_ g Cr ϭ 0.284 mol Cr atoms ϫ ϭ 14.8 g Cr
If you understand the reasoning in these conversions, you should be able to solve this problem
in a single setup:
52.0 g Cr
ᎏᎏ
1 mol Cr atoms
2 mol Cr atoms
ᎏᎏᎏ
1 mol (NH4)2Cr2O7
1 mol (NH4)2Cr2O7
ᎏᎏᎏ
252.0 g (NH4)2Cr2O7
2 mol Cr atoms
ᎏᎏᎏ
1 mol (NH4)2Cr2O7
1 mol (NH4)2Cr2O7
ᎏᎏᎏ
252.0 g (NH4)2Cr2O7
2-11
2-11 Some Other Interpretations of Chemical Formulas 77
_?_ g Cr ϭ 35.8 g (NH4)2Cr2O7 ϫ ϫ ϫ ϭ 14.8 g Cr
52.0 g Cr
ᎏᎏ
1 mol Cr
2 mol Cr atoms
ᎏᎏᎏ
1 mol (NH4)2Cr2O7
1 mol (NH4)2Cr2O7
ᎏᎏᎏ
252.0 g (NH4)2Cr2O7
You should now work Exercise 72.
EXAMPLE 2-19 Composition of Compounds
What mass of potassium chlorate, KClO3, would contain 40.0 grams of oxygen?
Plan
The formula KClO3 tells us that each mole of KClO3 contains three moles of oxygen atoms.
Each mole of oxygen atoms weighs 16.0 grams. So we can set up the solution to convert:
Mass O 88n mol O 88n mol KClO3 88n Mass KClO3
Solution
_?_ g KClO3 ϭ 40.0 g O ϫ ϫ ϫ
ϭ 102 g KClO3
You should now work Exercise 74.
122.6 g KClO3
ᎏᎏ
1 mol KClO3
1 mol KClO3
ᎏᎏ
3 mol O atoms
1 mol O atoms
ᎏᎏ
16.0 g O atoms
78 CHAPTER 2: Chemical Formulas and Composition Stoichiometry
EXAMPLE 2-20 Composition of Compounds
(a) What mass of sulfur dioxide, SO2, would contain the same mass of oxygen as is contained
in 33.7 g of arsenic pentoxide, As2O5?
(b) What mass of calcium chloride, CaCl2, would contain the same number of chloride ions
as are contained in 48.6 g of sodium chloride, NaCl?
Plan
(a) We could find explicitly the number of grams of O in 33.7 g of As2O5, and then find the
mass of SO2 that contains that same number of grams of O. But this method includes some
unnecessary calculation. We need only convert to moles of O (because this is the same amount
of O regardless of its environment) and then to moles of SO2 to obtain mass of SO2.
Mass As2O5 88n mol As2O5 88n mol O atoms 88n mol SO2 88n Mass SO2
(b) Because one mole always consists of the same number (Avogadro’s number) of items, we
can reason in terms of moles of ClϪ ions and solve as in part (a).
Mass NaCl 88n mol NaCl 88n mol ClϪ ions 88n mol CaCl2 88n Mass CaCl2
Solution
(a) _?_ g SO2 ϭ 33.7 g As2O5 ϫ ϫ
ϫ ϫ ϭ 23.5 SO2
64.1 g SO2
ᎏᎏ
1 mol SO2
1 mol SO2
ᎏᎏ
2 mol O atoms
5 mol O atoms
ᎏᎏ
1 mol As2O5
1 mol As2O5
ᎏᎏ
229.8 g As2O5
Problem-Solving Tip: How Do We Know When . . . ?
How do we know when to represent oxygen as O and when as O2? A compound that
contains oxygen does not contain O2 molecules. So we solve problems such as Example
2-19 in terms of moles of O atoms. Thus, we must use the formula weight for O, which
is 16.0 g O atoms/1 mol O atoms. Similar reasoning applies to compounds containing
other elements that are polyatomic molecules in pure elemental form, such as H2, Cl2, or P4.
(b) _?_ g CaCl2 ϭ 48.6 NaCl ϫ ϫ
ϫ ϫ ϭ 46.2 g CaCl2
You should now work Exercise 76.
111.0 g CaCl2
ᎏᎏ
1 mol CaCl2
1 mol CaCl2
ᎏᎏ
2 mol ClϪ
1 mol ClϪ
ᎏᎏ
1 mol NaCl
1 mol NaCl
ᎏᎏ
58.4 g NaCl
Figure 2-11 One mole of some
compounds. The colorless liquid is water,
H2O (1 mol ϭ 18.0 g ϭ 18.0 mL). The
white solid ( front left) is anhydrous oxalic
acid, (COOH)2 (1 mol ϭ 90.0 g). The
second white solid ( front right) is
hydrated oxalic acid, (COOH)2и2H2O
(1 mol ϭ 126.0 g). The blue solid is
hydrated copper(II) sulfate, CuSO4и5H2O
(1 mol ϭ 249.7 g). The red solid is
mercury(II) oxide (1 mol ϭ 216.6 g).
2-11 Some Other Interpretations of Chemical Formulas 79
The physical appearance of one mole of each of some compounds is illustrated in Figure
2-11. Two different forms of oxalic acid are shown. The formula unit (molecule) of oxalic
acid is (COOH)2 (FW ϭ 90.0 amu; molar mass ϭ 90.0 g/mol). When oxalic acid is
obtained by crystallization from a water solution, however, two molecules of water are
present for each molecule of oxalic acid, even though it appears dry. The formula of this
hydrate is (COOH)2 и 2H2O (FW ϭ 126.1 amu; molar mass ϭ 126.1 g/mol). The dot
shows that the crystals contain two H2O molecules per (COOH)2 molecule. The water
can be driven out of the crystals by heating to leave anhydrous oxalic acid, (COOH)2.
Anhydrous means “without water.” Copper(II) sulfate, an ionic compound, shows similar
behavior. Anhydrous copper(II) sulfate (CuSO4; FW ϭ 159.6 amu; molar mass ϭ 159.6
g/mol) is almost white. Hydrated copper(II) sulfate (CuSO4 и 5H2O; FW ϭ 249.7 amu;
molar mass ϭ 249.7 g/mol) is deep blue. The following example illustrates how we might
find and use the formula of a hydrate.
EXAMPLE 2-21 Composition of Compounds
A reaction requires pure anhydrous calcium sulfate, CaSO4. Only an unidentified hydrate of
calcium sulfate, CaSO4 и xH2O, is available.
(a) We heat 67.5 g of unknown hydrate until all the water has been driven off. The resulting
mass of pure CaSO4 is 53.4 g. What is the formula of the hydrate, and what is its formula
weight?
(b) Suppose we wish to obtain enough of this hydrate to supply 95.5 grams of CaSO4. How
many grams should we weigh out?
Plan
(a) To determine the formula of the hydrate, we must find the value of x in the formula
CaSO4 и xH2O. The mass of water removed from the sample is equal to the difference in the
two masses given. The value of x is the number of moles of H2O per mole of CaSO4 in the
hydrate.
Heating blue CuSO4и5H2O forms
anhydrous CuSO4, which is gray.
Some blue CuSO4и5H2O is visible
in the cooler center portion of the
crucible.
(b) The formula weights of CaSO4, 136.2 g/mol, and of CaSO4 и xH2O, (136.2 ϩ x18.0) g/mol,
allow us to write the conversion factor required for the calculation.
Solution
(a) _?_ g water driven off ϭ 67.5 g CaSO4 и xH2O Ϫ 53.4 g CaSO4 ϭ 14.1 g H2O
x ϭ ϭ ϫ ϫ ϭ
Thus, the formula of the hydrate is CaSO4 и 2H2O. Its formula weight is
FW ϭ 1 ϫ (formula weight CaSO4) ϩ 2 ϫ (formula weight H2O)
ϭ 136.2 g/mol ϩ 2(18.0 g/mol) ϭ 172.2 g/mol
(b) The formula weights of CaSO4 (136.2 g/mol) and of CaSO4 и 2H2O (172.2 g/mol) allow
us to write the unit factor
We use this factor to perform the required conversion:
_?_ g CaSO4 и 2H2O ϭ 95.5 g CaSO4 desired ϫ
ϭ 121 g CaSO4 и 2H2O
You should now work Exercise 78.
PURITY OF SAMPLES
Most substances obtained from laboratory reagent shelves are not 100% pure. The percent
purity is the mass percentage of a specified substance in an impure sample. When
impure samples are used for precise work, account must be taken of impurities. The photo
in the margin shows the label from reagent-grade sodium hydroxide, NaOH, which is
98.2% pure by mass. From this information we know that total impurities represent 1.8%
of the mass of this material. We can write several unit factors:
, , and
The inverse of each of these gives us a total of six unit factors.
EXAMPLE 2-22 Percent Purity
Calculate the masses of NaOH and impurities in 45.2 g of 98.2% pure NaOH.
Plan
The percentage of NaOH in the sample gives the unit factor ᎏ
9
1
8
0
.
0
2
g
g
s
N
am
aO
pl
H
e
ᎏ. The remainder
1.8 g impurities
ᎏᎏ
98.2 g NaOH
1.8 g impurities
ᎏᎏ
100 g sample
98.2 g NaOH
ᎏᎏ
100 g sample
2-12
172.2 g CaSO4 и 2H2O
ᎏᎏᎏ
136.2 g CaSO4
172.2 g CaSO4 и 2H2O
ᎏᎏᎏ
136.2 g CaSO4
2.00 mol H2O
ᎏᎏ
mol CaSO4
136.2 g CaSO4
ᎏᎏ
1 mol CaSO4
1 mol H2O
ᎏᎏ
18.0 g H2O
14.1 g H2O
ᎏᎏ
53.4 g CaSO4
_?_ mol H2O
ᎏᎏ
mol CaSO4
Impurities are not necessarily bad.
For example, inclusion of 0.02% KI,
potassium iodide, in ordinary table
salt has nearly eliminated goiter in the
United States. Goiter is a disorder
of the thyroid gland caused by a
deficiency of iodine. Mineral water
tastes better than purer, distilled water.
80 CHAPTER 2: Chemical Formulas and Composition Stoichiometry
of the sample is 100% Ϫ 98.2% ϭ 1.8% impurities; this gives the unit factor ᎏ
1.
1
8
00
g
g
im
s
p
am
ur
p
it
l
i
e
es
ᎏ.
Solution
_?_ g NaOH ϭ 45.2 g sample ϫ ϭ 44.4 g NaOH
_?_ g impurities ϭ 45.2 g sample ϫ ϭ 0.81 g impurities
You should now work Exercises 82 and 83.
1.8 g impurities
ᎏᎏ
100 g sample
98.2 g NaOH
ᎏᎏ
100 g sample
2-12 Purity of Samples 81
Problem-Solving Tip: Utility of the Unit Factor Method
Observe the beauty of the unit factor approach to problem solving! Such questions as
“do we multiply by 0.982 or divide by 0.982?” never arise. The units always point toward
the correct answer because we use unit factors constructed so that units always cancel
out until we arrive at the desired unit.
Mass mixture
Vol mixture
MIXTURE
CONTAINING A
Vol A
PURE A
Mass A
Moles A
F units A
Density
mixture
Mass mixture
Vol mixture
ϭ
% A
(by mass)
Parts A (by mass)
100 parts mixture (by mass)
Density A
Mass A
Vol A
FW A
g A
mol A
Avogadro’s
number
F units A
mol A
ϭ
ϭ
ϭ
ϭ
Many important relationships have been introduced in this chapter. Some of the most
important transformations you have seen in Chapters 1 and 2 are summarized in Figure
2-12.
Figure 2-12 Some important
relationships from Chapters 1
and 2. The relationships that
provide unit factors are enclosed
in green boxes.
A label from a bottle of sodium hydroxide,
NaOH.
82 CHAPTER 2: Chemical Formulas and Composition Stoichiometry
Key Terms
Allotropic modifications (allotropes) Different forms of the
same element in the same physical state.
Anhydrous Without water.
Anion An ion with a negative charge.
Atom The smallest particle of an element that maintains its
chemical identity through all chemical and physical changes.
Atomic mass unit (amu) One twelfth of the mass of an atom of
the carbon-12 isotope; a unit used for stating atomic and formula
weights.
Atomic number The number of protons in the nucleus of an
atom.
Atomic weight Weighted average of the masses of the constituent
isotopes of an element; the relative mass of atoms of
different elements.
Avogadro’s number 6.022 ϫ 1023 units of a specified item. See
Mole.
Cation An ion with a positive charge.
Chemical formula Combination of symbols that indicates the
chemical composition of a substance.
Composition stoichiometry Describes the quantitative (mass)
relationships among elements in compounds.
Empirical formula See Simplest formula.
Formula Combination of symbols that indicates the chemical
composition of a substance.
Formula unit The smallest repeating unit of a substance—for
non-ionic substances, the molecule.
Formula weight The mass, in atomic mass units, of one formula
unit of a substance. Numerically equal to the mass, in grams,
of one mole of the substance (see Molar mass). This number is
obtained by adding the atomic weights of the atoms specified
in the formula.
Hydrate A crystalline sample that contains water, H2O, and another
compound in a fixed mole ratio. Examples include
CuSO4 и 5H2O and (COOH)2 и 2H2O.
Ion An atom or group of atoms that carries an electric charge.
A positive ion is a cation; a negative ion is an anion.
Ionic compound A compound that is composed of cations and
anions. An example is sodium chloride, NaCl.
Law of Constant Composition See Law of Definite Proportions.
Law of Definite Proportions Different samples of a pure compound
always contain the same elements in the same proportions
by mass; this corresponds to atoms of these elements in
fixed numerical ratios. Also known as the Law of Constant
Composition.
Law of Multiple Proportions When two elements, A and B,
form more than one compound, the ratio of the masses of element
B that combine with a given mass of element A in each
of the compounds can be expressed by small whole numbers.
Molar mass The mass of substance in one mole of the substance;
numerically equal to the formula weight of the substance. See
Formula weight; see Molecular weight.
Mole 6.022 ϫ 1023 (Avogadro’s number of) formula units (or
molecules, for a molecular substance) of the substance under
discussion. The mass of one mole, in grams, is numerically equal
to the formula (molecular) weight of the substance.
Molecular formula A formula that indicates the actual number
of atoms present in a molecule of a molecular substance.
Compare with Simplest formula.
Molecular weight The mass, in atomic mass units, of one
molecule of a nonionic (molecular) substance. Numerically
equal to the mass, in grams, of one mole of such a substance.
This number is obtained by adding the atomic weights of the
atoms specified in the formula.
Molecule The smallest particle of an element or compound that
can have a stable independent existence.
Percent composition The mass percentage of each element in
a compound.
Percent purity The mass percentage of a specified compound
or element in an impure sample.
Polyatomic Consisting of more than one atom. Elements such
as Cl2, P4, and S8 exist as polyatomic molecules. Examples of
polyatomic ions are the ammonium ion, NH4
ϩ, and the sulfate
ion, SO4
2Ϫ.
Simplest formula The smallest whole-number ratio of atoms
present in a compound; also called empirical formula. Compare
with Molecular formula.
Stoichiometry Description of the quantitative relationships
among elements in compounds (composition stoichiometry)
and among substances as they undergo chemical changes (reaction
stoichiometry).
Structural formula A representation that shows how atoms are
connected in a compound.
Exercises
Basic Ideas
001. (a) What is the origin of the word “stoichiometry”?
(b) Distinguish between composition stoichiometry and
reaction stoichiometry.
002. List the basic ideas of Dalton’s atomic theory.
003. Give examples of molecules that contain (a) two atoms; (b)
three atoms; (c) four atoms; (d) eight atoms.
004. Give the formulas of two diatomic molecules, a triatomic
molecule, and two more complex molecules. Label each
formula as being either the formula of an element or of a
compound.
Exercises 83
005. Which of the formulas you selected for Exercise 4 represent
allotropes? If none, give two examples that are allotropes.
Select a different element for each example.
006. Which of the compounds in Table 2-2 are inorganic com-
pounds?
007. When can we correctly use the terms “formula weight”
and “molecular weight” interchangeably?
008. What structural feature distinguishes organic compounds
from inorganic compounds?
Names and Formulas
009. Write formulas for the following compounds: (a) nitric
acid; (b) methyl alcohol; (c) sulfur dioxide; (d) acetic acid;
(e) butane.
010. Name the following compounds: (a) H2SO4; (b) C3H8;
(c) NH3; (d) CH3CH2OH.
011. Name each of the following ions. Classify each as a
monatomic or polyatomic ion. Classify each as a cation or
an anion. (a) Naϩ; (b) OHϪ; (c) SO4
2Ϫ; (d) S2Ϫ; (e) Zn2ϩ;
(f ) Fe2ϩ.
012. Write the chemical symbol for each of the following ions.
Classify each as a monatomic or polyatomic ion. Classify
each as a cation or an anion. (a) potassium ion; (b) sulfate
ion; (c) copper(II) ion; (d) ammonium ion; (e) carbonate
ion.
013. Name each of the following compounds: (a) MgCl2;
(b) Cu(NO3)2; (c) Li2SO4; (d) Ca(OH)2; (e) FeSO4.
014. Write the chemical formula for each of the following ionic
compounds: (a) potassium acetate; (b) ammonium sulfate;
(c) zinc phosphate; (d) calcium oxide; (e) aluminum sul-
fide.
015. Write the chemical formula for the ionic compound
formed between each of the following pairs of ions. Name
each compound. (a) Naϩ and S2Ϫ; (b) Al3ϩ and SO4
2Ϫ;
(c) Naϩ and PO4
3Ϫ; (d) Mg2ϩ and NO3
Ϫ; (e) Fe3ϩ and
CO3
2Ϫ.
016. Write the chemical formula for the ionic compound
formed between each of the following pairs of ions. Name
each compound. (a) Cu2ϩ and CO3
2Ϫ; (b) Mg2ϩ and ClϪ;
(c) NH4
ϩ and CO3
2Ϫ; (d) Zn2ϩ and OHϪ; (e) Fe2ϩ and
CH3COOϪ.
017. Define and illustrate the following: (a) ion; (b) cation;
(c) anion; (d) polyatomic ion; (e) molecule.
018. (a) There are no molecules in ionic compounds. Why not?
(b) What is the difference between a formula unit of an
ionic compound and a polyatomic molecule?
019. Convert each of the following into a correct formula represented
with correct notation. (a) AlOH3; (b) Mg2CO3;
(c) Zn(CO3)2; (d) (NH4)2SO4; (e) Mg2(SO4)2.
020. Write the formula of the compound produced by the combination
of each of the following pairs of elements. Name
each compound. (a) potassium and chlorine; (b) magnesium
and chlorine; (c) sulfur and oxygen; (d) calcium and
oxygen; (e) sodium and sulfur; (f ) aluminum and sulfur.
021. Write the chemical formula of each of the following:
(a) calcium carbonate—major component of coral,
seashells, and limestone—found in antacid preparations;
(b) magnesium sulfate—found in Epsom salts; (c) acetic
acid—the acid in vinegar; (d) sodium hydroxide—common
name is lye; (e) zinc oxide—used to protect from sunlight’s
UV rays when blended in an ointment.
Atomic and Formula Weights
022. (a) What is the atomic weight of an element? (b) Why can
atomic weights be referred to as relative numbers?
023. (a) What is the atomic mass unit (amu)? (b) The atomic
weight of vanadium is 50.942 amu, and the atomic weight
of ruthenium is 101.07 amu. What can we say about the
relative masses of V and Ru atoms?
024. What is the mass ratio (four significant figures) of one
atom of Rb to one atom of Cl?
025. A sample of 6.68 g of calcium combines exactly with 6.33
g of fluorine, forming calcium fluoride, CaF2. Find the relative
masses of the atoms of calcium and fluorine. Check
your answer using a table of atomic weights. If the formula
were not known, could you still do this calculation?
026. Determine the formula weight of each of the following
substances: (a) bromine, Br2; (b) water, H2O; (c) saccharin,
C7H5NSO3; (d) potassium dichromate, K2Cr2O7.
027. Determine the formula weight of each of the following
substances: (a) calcium sulfate, CaSO4; (b) butane, C4H10;
(c) the sulfa drug sulfanilamide, C6H4SO2(NH2)2;
(d) uranyl phosphate, (UO2)3(PO4)2.
028. Determine the formula weight of each of the following
common acids: (a) hydrochloric acid, HCl; (b) nitric acid,
HNO3; (c) phosphoric acid, H3PO4; (d) sulfuric acid,
H2SO4.
Zinc oxide used as a sunscreen.
The Mole Concept
029. A large neon sign is to be filled with a mixture of gases,
including 8.575 g neon. What number of moles is this?
030. How many molecules are in 18.0 g of each of the following
substances? (a) CO; (b) N2; (c) P4; (d) P2. (e) Do parts
(c) and (d) contain the same number of atoms of phos-
phorus?
031. Sulfur molecules exist under various conditions as S8, S6,
S4, S2, and S. (a) Is the mass of one mole of each of these
molecules the same? (b) Is the number of molecules in one
mole of each of these molecules the same? (c) Is the mass
of sulfur in one mole of each of these molecules the same?
(d) Is the number of atoms of sulfur in one mole of each
of these molecules the same?
032. How many moles of substance are contained in each of the
following samples? (a) 18.3 g of NH3; (b) 5.32 g of ammonium
bromide; (c) 6.6 g of PCl5; (d) 215 g of Sn.
033. How many moles of substance are contained in each of the
following samples? (a) 36.2 g of diethyl ether; (b) 15.6 g
of calcium carbonate; (c) 16.7 g of acetic acid; (d) 19.3 g
of ethanol.
034. Complete the following table. Refer to a table of atomic
weights.
Mass of One Mole
Element Atomic Weight of Atoms
(a) Mg ______________ _________________
(b) ________ 79.904 amu _________________
(c) Cl ______________ _________________
(d) ________ ______________ 51.9961 g
035. Complete the following table. Refer to a table of atomic
weights.
Mass of One Mole
Element Formula of Molecules
(a) Br Br2 _________________
(b) ________ H2 _________________
(c) ________ P4 _________________
(d) ________ ________ 20.1797 g
(e) S ________ 256.528 g
(f ) O ________ _________________
036. Complete the following table.
Moles of Moles of Moles of
Compound Cations Anions
1 mol KCl _______________ _______________
2 mol Na2SO4 _______________ _______________
0.2 mol calcium
nitrate _______________ _______________
_______________ 0.50 mol NH4
ϩ 0.25 mol SO4
2ϩ
037. What mass, in grams, should be weighed for an experiment
that requires 1.54 mol of (NH4)2HPO4?
038. How many hydrogen atoms are contained in 125 grams of
propane, C3H8?
039. How many atoms of C, H, and O are in each of the following?
(a) 1.24 mol of glucose, C6H12O6; (b) 3.31 ϫ 1019
glucose (C6H12O6) molecules; (c) 0.275 g of glucose.
040. Calculate the mass in grams and kilograms of 1.458 moles
of gold.
041. An atom of an element has a mass ever-so-slightly greater
than twice the mass of an Ni atom. Identify the element.
042. Calculate the number of Ni atoms in 1.0 millionth of a
gram of nickel.
043. Calculate the number of Ni atoms in 1.0 trillionth of a
gram of nickel.
044. What is the mass of 10.0 million methane, CH4, mole-
cules?
045. A sample of ethane, C2H6, has the same mass as 10.0
million molecules of methane, CH4. How many C2H6
molecules does the sample contain?
Percent Composition
046. Calculate the percent composition of each of the following
compounds: (a) nicotine, C10H14N2; (b) vitamin E,
C29H50O2; (c) vanillin, C8H8O3.
047. Calculate the percent composition of each of the following
compounds: (a) menthol, C10H19OH; (b) carborundum,
SiC; (c) aspirin, C9H8O4.
048. Calculate the percent by mass of silver found in a particular
mineral that is determined to be silver carbonate.
049. What percent by mass of iron(II) phosphate is iron?
*50. Copper is obtained from ores containing the following
minerals: azurite, Cu3(CO3)2(OH)2; chalcocite, Cu2S;
chalcopyrite, CuFeS2; covelite, CuS; cuprite, Cu2O; and
malachite, Cu2CO3(OH)2. Which mineral has the highest
copper content as a percent by mass?
Determination of Simplest and Molecular Formulas
051. Determine the simplest formula for each of the following
compounds: (a) copper(II) tartrate: 30.03% Cu; 22.70% C;
1.91% H; 45.37% O. (b) nitrosyl fluoroborate: 11.99% N;
13.70% O; 9.25% B; 65.06% F.
052. The hormone norepinephrine is released in the human
body during stress and increases the body’s metabolic rate.
Like many biochemical compounds, norepinephrine is
composed of carbon, hydrogen, oxygen, and nitrogen. The
percent composition of this hormone is 56.8% C, 6.56%
H, 28.4% O, and 8.28% N. What is the simplest formula
of norepinephrine?
053. (a) A sample of a compound is found to contain 5.60 g N,
14.2 g Cl, and 0.800 g H. What is the simplest formula of
this compound? (b) A sample of another compound containing
the same elements is found to be 26.2% N, 66.4%
84 CHAPTER 2: Chemical Formulas and Composition Stoichiometry
Exercises 85
Cl, and 7.5% H. What is the simplest formula of this
compound?
054. A common product found in nearly every kitchen contains
27.37% sodium, 1.20% hydrogen, 14.30% carbon, and
57.14% oxygen. The simplest formula is the same as the
formula of the compound. Find the formula of this
compound.
055. Bupropion is present in a medication that is an antidepressant
and is also used to aid in quitting smoking.
The composition of bupropion is 65.13% carbon, 7.57%
hydrogen, 14.79% chlorine, 5.84% nitrogen, and 6.67%
oxygen. The simplest formula is the same as the molecular
formula of this compound. Determine the formula.
056. Lysine is an essential amino acid. One experiment showed
that each molecule of lysine contains two nitrogen atoms.
Another experiment showed that lysine contains 19.2% N,
9.64% H, 49.3% C, and 21.9% O by mass. What is the
molecular formula for lysine?
057. A 2.00-g sample of a compound gave 4.86 g of CO2 and
2.03 g of H2O on combustion in oxygen. The compound
is known to contain only C, H, and O. What is its simplest
formula?
058. A 0.1647-gram sample of a pure hydrocarbon was burned
in a C-H combustion train to produce 0.5694 gram of CO2
and 0.0826 gram of H2O. Determine the masses of C and
H in the sample and the percentages of these elements in
this hydrocarbon.
059. Naphthalene is a hydrocarbon that is used for mothballs.
A 0.3204-gram sample of naphthalene was burned in a
C-H combustion train to produce 1.100 grams of carbon
dioxide and 0.1802 grams of water. What masses and
percentages of C and H are present in naphthalene?
060. Combustion of 0.5707 mg of a hydrocarbon produces
1.790 mg of CO2. What is the simplest formula of the
hydrocarbon?
*61. Complicated chemical reactions occur at hot springs on
the ocean floor. One compound obtained from such a hot
spring consists of Mg, Si, H, and O. From a 0.334-g
sample, the Mg is recovered as 0.115 g of MgO; H is recovered
as 25.7 mg of H2O; and Si is recovered as 0.172 g
of SiO2. What is the simplest formula of this compound?
062. A 1.000-gram sample of an alcohol was burned in oxygen
to produce 1.913 g of CO2 and 1.174 g of H2O. The alcohol
contained only C, H, and O. What is the simplest
formula of the alcohol?
063. An alcohol is 64.81% C, 13.60% H, and 21.59% O by
mass. Another experiment shows that its molecular weight
is approximately 74 amu. What is the molecular formula
of the alcohol?
064. Skatole is found in coal tar and in human feces. It contains
three elements: C, H, and N. It is 82.40% C and 6.92%
H by mass. Its simplest formula is its molecular formula.
What are (a) the formula and (b) the molecular weight of
skatole?
065. Testosterone, the male sex hormone, contains only C, H,
and O. It is 79.12% C and 9.79% H by mass. Each molecule
contains two O atoms. What are (a) the molecular
weight and (b) the molecular formula for testosterone?
*66. The beta-blocker drug, timolol, is expected to reduce the
need for heart bypass surgery. Its composition by mass is
49.4% C, 7.64% H, 17.7% N, 15.2% O, and 10.1% S.
The mass of 0.0100 mol of timolol is 3.16 g. (a) What is
the simplest formula of timolol? (b) What is the molecular
formula of timolol?
The Law of Multiple Proportions
067. Show that the compounds water, H2O, and hydrogen peroxide,
H2O2, obey the Law of Multiple Proportions.
068. Nitric oxide, NO, is produced in internal combustion engines.
When NO comes in contact with air, it is quickly
converted into nitrogen dioxide, NO2, a very poisonous,
corrosive gas. What mass of O is combined with 3.00 g of
N in (a) NO and (b) NO2? Show that NO and NO2 obey
the Law of Multiple Proportions.
069. Sulfur forms two chlorides. A 30.00-gram sample of one
chloride decomposes to give 5.53 g of S and 24.47 g of Cl.
A 30.00-gram sample of the other chloride decomposes to
give 3.93 g of S and 26.07 g of Cl. Show that these compounds
obey the Law of Multiple Proportions.
070. What mass of oxygen is combined with 3.65 g of sulfur in
(a) sulfur dioxide, SO2, and in (b) sulfur trioxide, SO3?
Interpretation of Chemical Formulas
071. One prominent ore of copper contains chalcopyrite,
CuFeS2. How many pounds of copper are contained in
2.63 pounds of pure CuFeS2?
072. Mercury occurs as a sulfide ore called cinnabar, HgS. How
many grams of mercury are contained in 887 g of pure
HgS?
073. (a) How many grams of copper are contained in 325 g of
CuSO4? (b) How many grams of copper are contained in
325 g of CuSO4 и 5H2O?
074. What mass of KMnO4 would contain 15.0 g of manganese?
075. What mass of azurite, Cu3(CO3)2(OH)2, would contain
610 g of copper?
076. Two minerals that contain copper are chalcopyrite,
CuFeS2, and chalcocite, Cu2S. What mass of chalcocite
would contain the same mass of copper as is contained in
125 pounds of chalcopyrite?
077. Tungsten is a very dense metal (19.3 g/cm3) with extremely
high melting and boiling points (3370°C and 5900°C).
When a small amount of it is included in steel, the resulting
alloy is far harder and stronger than ordinary steel.
Two important ores of tungsten are FeWO4 and CaWO4.
How many grams of CaWO4 would contain the same mass
of tungsten that is present in 569 g of FeWO4?
*78. When a mole of CuSO4 и 5H2O is heated to 110°C, it loses
four moles of H2O to form CuSO4 и H2O. When it is
heated to temperatures above 150°C, the other mole of
H2O is lost. (a) How many grams of CuSO4 и H2O could
be obtained by heating 695 g of CuSO4 и 5H2O to 110°C?
(b) How many grams of anhydrous CuSO4 could be obtained
by heating 695 g of CuSO4 и 5H2O to 180°C?
Percent Purity
079. A particular ore of lead, galena, is 10.0% lead sulfide, PbS,
and 90.0% impurities by weight. What mass of lead is contained
in 50.0 grams of this ore?
080. What mass of chromium is present in 150 grams of an ore
of chromium that is 65.0% chromite, FeCr2O4, and 35.0%
impurities by mass? If 90.0% of the chromium can be
recovered from 100.0 grams of the ore, what mass of pure
chromium is obtained?
081. What masses of (a) Sr and (b) N are contained in 106.7 g
of 88.2% pure Sr(NO3)2? Assume that the impurities do
not contain the elements mentioned.
082. (a) What weight of magnesium carbonate is contained in
315 pounds of an ore that is 27.7% magnesium carbonate
by weight? (b) What weight of impurities is contained in
the sample? (c) What weight of magnesium is contained
in the sample? (Assume that no magnesium is present in
the impurities.)
083. Vinegar is 5.0% acetic acid, C2H4O2, by mass. (a) How
many grams of acetic acid are contained in 24.0 g of vinegar?
(b) How many pounds of acetic acid are contained in
24.0 pounds of vinegar? (c) How many grams of sodium
chloride, NaCl, are contained in 24.0 g of saline solution
that is 5.0% NaCl by mass?
*84. What is the percent by mass of copper sulfate, CuSO4, in
a sample of copper sulfate pentahydrate, CuSO4 и 5H2O?
(b) What is the percent by mass of CuSO4 in a sample that
is 72.4% CuSO4 и 5H2O by mass?
Mixed Examples
085. How many moles of chlorine atoms are contained in each
of the following? (a) 35.45 ϫ 1023 Cl atoms; (b) 35.45 ϫ
1023 Cl2 molecules; (c) 35.45 g of chlorine; (d) 35.45 mol
of Cl2.
086. What is the maximum number of moles of CO2 that could
be obtained from the carbon in each of the following?
(a) 4.00 mol of Ru2(CO3)3; (b) 4.00 mol of CaCO3;
(c) 4.00 mol of Co(CO)6.
087. (a) How many formula units are contained in 154.3 g of
K2MoO4? (b) How many potassium ions? (c) How many
MoO4
2Ϫ ions? (d) How many atoms of all kinds?
088. (a) How many moles of ozone molecules are contained in
64.0 g of ozone, O3? (b) How many moles of oxygen atoms
are contained in 64.0 g of ozone? (c) What mass of O2
would contain the same number of oxygen atoms as 64.0
g of ozone? (d) What mass of oxygen gas, O2, would contain
the same number of molecules as 64.0 g of ozone?
089. Cocaine has the following percent composition by mass:
67.30% C, 6.930% H, 21.15% O, and 4.62% N. What is
the simplest formula of cocaine?
090. A compound with the molecular weight of 56.0 g was
found as a component of photochemical smog. The compound
is composed of carbon and oxygen, 42.9% and
57.1%, respectively. What is the formula of this com-
pound?
091. A carbon-hydrogen-oxygen compound, MW ϭ 90 g, is
analyzed and found to be 40.0% carbon, 6.7% hydrogen,
and 53.3% oxygen. What is the formula of this compound?
092. Find the number of moles of Ag needed to form each of
the following: (a) 0.235 mol Ag2S; (b) 0.235 mol Ag2O; (c)
0.235 g Ag2S; (d) 2.35 ϫ 1020 formula units of Ag2S.
093. A metal, M, forms an oxide having the simplest formula
M2O3. This oxide contains 52.9% of the metal by mass.
(a) Calculate the atomic weight of the metal. (b) Identify
the metal.
094. Three samples of magnesium oxide were analyzed to determine
the mass ratios O/Mg, giving the following re-
sults:
, ,
Which law of chemical combination is illustrated by these
data?
*95. The molecular weight of hemoglobin is about 65,000 g/mol.
Hemoglobin contains 0.35% Fe by mass. How many iron
atoms are in a hemoglobin molecule?
*96. More than 1 billion pounds of adipic acid (MW 146.1 g/mol)
is manufactured in the United States each year. Most of it
is used to make synthetic fabrics. Adipic acid contains only
C, H, and O. Combustion of a 1.6380-g sample of adipic
acid gives 2.960 g of CO2 and 1.010 g of H2O. (a) What
is the simplest formula for adipic acid? (b) What is its molecular
formula?
097. A filled 25-L container of an unknown, unlabeled liquid
was found in a storeroom and had to be identified to determine
a method of disposal. The compound was found
to contain only hydrogen and carbon. A 1.750-g sample
of the compound was burned in a pure oxygen atmosphere;
1.211 g H2O and 5.916 g CO2 were collected. Determine
the simplest formula.
098. An unknown sample weighing 1.50 g was found to contain
only manganese and sulfur. The sample was completely
reacted with oxygen and produced 1.22 g manganese(II)
oxide, MnO, and 1.38 g sulfur trioxide. What
is the simplest formula for this compound?
099. Copper(II) sulfate exists as a baby-blue powder when anhydrous
and as a deep blue crystal when hydrated with five
water molecules, CuSO4 и 5H2O. Which of these two
compounds contains more copper per mole of compound?
2.29 g O
ᎏᎏ
3.48 g Mg
0.658 g O
ᎏᎏ
1.00 g Mg
1.60 g O
ᎏᎏ
2.43 g Mg
86 CHAPTER 2: Chemical Formulas and Composition Stoichiometry
Exercises 87
What is the ratio of percent by mass of copper in the anhydrous
compound to the percent by mass of copper in
the hydrated compound?
CONCEPTUAL EXERCISES
100. What mass of NaCl would contain the same total number
of ions as 245 g of MgCl2?
101. How many atoms of oxygen are in 17.9325 g of sulfuric
acid?
102. In the “button” and “button hole” analogy of writing formulas
for ionic compounds, one may think of positive
charges as buttons and negative charges as button holes.
One prepares formulas by combining the buttons (positive
charges) with an equal number of button holes (negative
charges) so that every button will be associated with a single
buttonhole and vice versa. Using this analogy, how
many buttons (positive charges) are associated with a single
cation (positive ion) in each of the following ionic compounds?
(a) NaCl; (b) Na2SO4; (c) CaSO4; (d) Al2(SO4)3.
103. Two deposits of minerals containing silver are found. One
of the deposits contains silver oxide, and the other contains
silver sulfide. The deposits can be mined at the same
price per ton of the original silver-containing compound,
but only one deposit can be mined by your company.
Which of the deposits would you recommend and why?
104. A decision is to be made as to the least expensive source
of zinc. One source of zinc is zinc sulfate, ZnSO4, and another
is zinc acetate dihydrate, Zn(CH3COO)2 и 2H2O.
These two sources of zinc can be purchased at the same
price per kilogram of compound. Which is the most economical
source of zinc and by how much?
105. Assume that a penny is ᎏ
1
1
6
ᎏ in. thick and that the moon is
222,000 mi at its closest approach to the earth (perigee).
Show by calculation whether or not a picomole of pennies
stacked on their faces would reach from the earth to the
moon.
BUILDING YOUR KNOWLEDGE
NOTE Beginning with this chapter, exercises under the “Building
Your Knowledge” heading will often require that you use skills, concepts,
or information that you should have mastered in earlier chapters. This
provides you an excellent opportunity to “tie things together” as you study.
106. A 22-mL (19-g) sample of an unknown liquid is analyzed,
and the percent composition is found to be 53% C, 11%
H, and 36% O. Is this compound likely to be ethanol,
CH3CH2OH? Give two reasons for your answer. (Hint:
See Table 1-8.)
107. Three allotropes of phosphorus are known, with molecular
weights of 62.0, 31.0, and 124.0 amu, respectively.
Write the molecular formula for each allotrope.
108. Near room temperature, the density of water is 1.00 g/mL,
and the density of ethanol (grain alcohol) is 0.789 g/mL.
What volume of ethanol contains the same number of molecules
as are present in 175 mL of H2O?
109. Calculate the volume of 2.00 mol of mercury, a liquid
metal. (Hint: See Table 1-8.)
110. In Chapter 1 you learned that the specific heat of water is
4.18 J/g и °C. The molar heat capacity is defined as the
specific heat or heat capacity per mole of material.
Calculate the molar heat capacity for water. What value(s)
limited the number of significant figures in your answer?
3 Chemical Equations
and Reaction
Stoichiometry
OUTLINE
3-1 Chemical Equations
3-2 Calculations Based on Chemical
Equations
3-3 The Limiting Reactant Concept
3-4 Percent Yields from Chemical
Reactions
3-5 Sequential Reactions
3-6 Concentrations of Solutions
3-7 Dilution of Solutions
3-8 Using Solutions in Chemical
Reactions
I
OBJECTIVES
After you have studied this chapter, you should be able to
• Write balanced chemical equations to describe chemical reactions
• Interpret balanced chemical equations to calculate the moles of reactants and products
involved in each of the reactions
• Interpret balanced chemical equations to calculate the masses of reactants and products
involved in each of the reactions
• Determine which reactant is the limiting reactant in reactions
• Use the limiting reactant concept in calculations recording chemical equations
• Compare the amount of substance actually formed in a reaction (actual yield) with the
predicted amount (theoretical yield), and determine the percent yield
• Work with sequential reactions
• Use the terminology of solutions—solute, solvent, concentration
• Calculate concentrations of solutions when they are diluted
• Carry out calculations related to the use of solutions in chemical reactions
I
n Chapter 2 we studied composition stoichiometry, the quantitative relationships
among elements in compounds. In this chapter as we study reaction stoichiometry—
the quantitative relationships among substances as they participate in chemical reactions—we
ask several important questions. How can we describe the reaction of one substance
with another? How much of one substance reacts with a given amount of another
substance? Which reactant determines the amounts of products formed in a chemical reaction?
How can we describe reactions in aqueous solutions?
Whether we are concerned with describing a reaction used in a chemical analysis, one
used industrially in the production of a plastic, or one that occurs during metabolism in
the body, we must describe it accurately. Chemical equations represent a very precise, yet
a very versatile, language that describes chemical changes. We begin our study by examining
chemical equations.
Methane, CH4, is the main component of natural gas.
Steel wool burns in pure oxygen
gas.
4Fe ϩ 3O2 88n 2Fe2O3














CHEMICAL EQUATIONS
Chemical reactions always involve changing one or more substances into one or more different
substances. In other words, chemical reactions rearrange atoms or ions to form
other substances.
Chemical equations are used to describe chemical reactions, and they show (1) the
substances that react, called reactants; (2) the substances formed, called products; and (3) the
relative amounts of the substances involved. We write the reactants to the left of an arrow and
the products to the right of the arrow. As a typical example, let’s consider the combustion
(burning) of natural gas, a reaction used to heat buildings and cook food. Natural gas is
a mixture of several substances, but the principal component is methane, CH4. The equation
that describes the reaction of methane with excess oxygen is
CH4 ϩ 2O2 88n CO2 ϩ 2H2O
reactants products
What does this equation tell us? In the simplest terms, it tells us that methane reacts with
oxygen to produce carbon dioxide, CO2, and water. More specifically, it says that for every
CH4 molecule that reacts, two molecules of O2 also react, and that one CO2 molecule
and two H2O molecules are formed. That is,
heat
CH4 ϩ 2O2 8888n CO2 ϩ 2H2O
1 molecule 2 molecules 1 molecule 2 molecules
This description of the reaction of CH4 with O2 is based on experimental observations. By
this we mean experiments have shown that when one CH4 molecule reacts with two O2
molecules, one CO2 molecule and two H2O molecules are formed. Chemical equations
are based on experimental observations. Special conditions required for some reactions are
indicated by notation over the arrow. Figure 3-1 is a pictorial representation of the
rearrangement of atoms described by this equation.
As we pointed out in Section 1-1, there is no detectable change in the quantity of matter
during an ordinary chemical reaction. This guiding principle, the Law of Conservation of
Matter, provides the basis for “balancing” chemical equations and for calculations based
on those equations. Because matter is neither created nor destroyed during a chemical re-
action,
a balanced chemical equation must always include the same number of each kind of
atom on both sides of the equation.
3-1
Sometimes it is not possible to
represent a chemical change with a
single chemical equation. For example,
when too little O2 is present, both
CO2 and CO are found as products,
and a second chemical equation must
be used to describe the process. In the
present case (excess oxygen), only one
equation is required.
The arrow may be read “yields.”
The capital Greek letter delta (⌬) is
sometimes used in place of the word
“heat.”
90 CHAPTER 3: Chemical Equations and Reaction Stoichiometry
See the Saunders Interactive
General Chemistry CD-ROM,
Screens 4.1–4.4, Equations.
Figure 3-1 Two representations of
the reaction of methane with oxygen
to form carbon dioxide and water.
Chemical bonds are broken and new
ones are formed in each
representation. Part (a) illustrates the
reaction using models, and (b) uses
chemical formulas.
H
H
ϩ ϩ
H
HH
H
H
H
C
O O
O
O
C
OO
OO
CH4 2O2 CO2 2H2Oϩ ϩ
(a)
(b)
Chemists usually write equations with the smallest possible whole-number coefficients.
Before we attempt to balance an equation, all substances must be represented by formulas
that describe them as they exist. For instance, we must write H2 to represent diatomic
hydrogen molecules—not H, which represents hydrogen atoms. Once the correct
formulas are written, the subscripts in the formulas may not be changed. Different subscripts
in formulas specify different compounds, so changing the formulas would mean
that the equation would no longer describe the same reaction.
Dimethyl ether, C2H6O, burns in an excess of oxygen to give carbon dioxide and water.
Let’s balance the equation for this reaction. In unbalanced form, the equation is
C2H6O ϩ O2 88n CO2 ϩ H2O
Carbon appears in only one compound on each side, and the same is true for hydrogen.
We begin by balancing these elements:
C2H6O ϩ O2 88n 2CO2 ϩ 3H2O
Now we have an odd number of atoms of O on each side. The single O in C2H6O balances
one of the atoms of O on the right. We balance the other six by placing a coefficient
of 3 before O2 on the left.
C2H6O ϩ 3O2 88n 2CO2 ϩ 3H2O
When we think we have finished the balancing, we should always do a complete check for
each element, as shown in red in the margin.
Let’s generate the balanced equation for the reaction of aluminum metal with hydrochloric
acid to produce aluminum chloride and hydrogen. The unbalanced “equation” is
Al ϩ HCl 88n AlCl3 ϩ H2
As it now stands, the “equation” does not satisfy the Law of Conservation of Matter
because there are two H atoms in the H2 molecule and three Cl atoms in one formula
unit of AlCl3 (right side), but only one H atom and one Cl atom in the HCl molecule
(left side).
Let us first balance chlorine by putting a coefficient of 3 in front of HCl.
Al ϩ 3HCl 88n AlCl3 ϩ H2
Now there are 3H on the left and 2H on the right. The least common multiple of 3 and
2 is 6; to balance H, we multiply the 3HCl by 2 and the H2 by 3.
Al ϩ 6HCl 88n AlCl3 ϩ 3H2
Now Cl is again unbalanced (6Cl on the left, 3 on the right), but we can fix this by putting
a coefficient of 2 in front of AlCl3 on the right.
Al ϩ 6HCl 88n 2AlCl3 ϩ 3H2
Now all elements except Al are balanced (1 on the left, 2 on the right); we complete the
balancing by putting a coefficient of 2 in front of Al on the left.
2Al ϩ 6HCl 88n 2AlCl3 ϩ 3H2
aluminum hydrochloric acid aluminum chloride hydrogen
Balancing chemical equations “by inspection” is a trial-and-error approach. It requires
a great deal of practice, but it is very important! Remember that we use the smallest wholenumber
coefficients. Some chemical equations are difficult to balance by inspection or
“trial and error.” In Chapter 11 we will learn methods for balancing complex equations.
In Reactants In Products
2C, 6H, 3O 1C, 2H, 3O
C, H are not balanced
In Reactants In Products
2C, 6H, 3O 2C, 6H, 7O
Now O is not balanced
In Reactants In Products
2C, 6H, 7O 2C, 6H, 7O
Now the equation is balanced
In Reactants In Products
1Al, 1H, 1Cl 1Al, 2H, 3Cl
H, Cl are not balanced
In Reactants In Products
1Al, 3H, 3Cl 1Al, 2H, 3Cl
Now H is not balanced
In Reactants In Products
1Al, 6H, 6Cl 1Al, 6H, 3Cl
Now Cl is not balanced
In Reactants In Products
1Al, 6H, 6Cl 2Al, 6H, 6Cl
Now Al is not balanced
In Reactants In Products
2Al, 6H, 6Cl 2Al, 6H, 6Cl
Now the equation is balanced
3-1 Chemical Equations 91
A balanced chemical equation may be
interpreted on a molecular basis.
92 CHAPTER 3: Chemical Equations and Reaction Stoichiometry
CALCULATIONS BASED ON CHEMICAL EQUATIONS
We are now ready to use chemical equations to calculate the relative amounts of substances
involved in chemical reactions. Let us again consider the combustion of methane in excess
oxygen. The balanced chemical equation for that reaction is
CH4 ϩ 2O2 88n CO2 ϩ 2H2O
On a quantitative basis, at the molecular level, the equation says
CH4 ϩ 2O2 88n CO2 ϩ 2H2O
1 molecule 2 molecules 1 molecule 2 molecules
of methane of oxygen of carbon dioxide of water
EXAMPLE 3-1 Number of Molecules
How many O2 molecules react with 47 CH4 molecules according to the preceding equation?
Plan
The balanced equation tells us that one CH4 molecule reacts with two O2 molecules. We can
construct two unit factors from this fact:
and
These expressions are unit factors for this reaction because the numerator and denominator
are chemically equivalent. In other words, the numerator and the denominator
represent the same amount of reaction. To convert CH4 molecules to O2 molecules, we
multiply by the second of the two factors.
Solution
__? O2 molecules ϭ 47 CH4 molecules ϫ ϭ 94 O2 molecules
You should now work Exercise 8.
A chemical equation also indicates the relative amounts of each reactant and product
in a given chemical reaction. We showed earlier that formulas can represent moles of sub-
2 O2 molecules
ᎏᎏ
1 CH4 molecule
2 O2 molecules
ᎏᎏ
1 CH4 molecule
1 CH4 molecule
ᎏᎏ
2 O2 molecules
3-2
Problem-Solving Tip: Balancing Chemical Equations
There is no one best place to start when balancing a chemical equation, but the following
suggestions might be helpful:
(1) Look for elements that appear in only one place on each side of the equation (in
only one reactant and in only one product), and balance those elements first.
(2) If free, uncombined elements appear on either side, balance them last.
Notice how these suggestions worked in the procedures illustrated in this section. Above
all, remember that we should never change subscripts in formulas, because doing so would
describe different substances. We only adjust the coefficients to balance the equation.
See the Saunders Interactive
General Chemistry CD-ROM,
Screens 5.1–5.3, Stoichiometry.
We usually cannot work with
individual molecules; a mole of a
substance is an amount we might
use in a laboratory experiment.


















stances. Suppose Avogadro’s number of CH4 molecules, rather than just one CH4 molecule,
undergo this reaction. Then the equation can be written
CH4 ϩ 2O2 88n CO2 ϩ 2H2O
6.02 ϫ 1023 molecules 2(6.02 ϫ 1023 molecules) 6.02 ϫ 1023 molecules 2(6.02 ϫ 1023molecules)
ϭ 1 mol ϭ 2 mol ϭ 1 mol ϭ 2 mol
This interpretation tells us that one mole of methane reacts with two moles of oxygen to
produce one mole of carbon dioxide and two moles of water.
A balanced chemical equation may
be interpreted in terms of moles of
reactants and products.
3-2 Calculations Based On Chemical Equations 93
EXAMPLE 3-2 Number of Moles Formed
How many moles of water could be produced by the reaction of 3.5 mol of methane with excess
oxygen (i.e., more than a sufficient amount of oxygen is present)?
Plan
The equation for the combustion of methane
CH4 ϩ 2O2 88n CO2 ϩ 2H2O
1 mol 2 mol 1 mol 2 mol
shows that 1 mol of methane reacts with 2 mol of oxygen to produce 2 mol of water. From
this information we construct two unit factors:
and
We use the second factor in this calculation.
Solution
__? mol H2O ϭ 3.5 mol CH4 ϫ ϭ 7.0 mol H2O
You should now work Exercises 10 and 14.
We know the mass of 1 mol of each of these substances, so we can also write
CH4 ϩ 2O2 88n CO2 ϩ 2H2O
1 mol 2 mol 1 mol 2 mol
16.0 g 2(32.0 g) 44.0 g 2(18.0 g)
16.0 g 64.0 g 44.0 g 36.0 g
80.0 g reactants 80.0 g products
The equation now tells us that 16.0 grams of CH4 reacts with 64.0 grams of O2 to form
44.0 grams of CO2 and 36.0 grams of H2O. The Law of Conservation of Matter is satisfied.
Chemical equations describe reaction ratios, that is, the mole ratios of reactants
and products as well as the relative masses of reactants and products.
2 mol H2O
ᎏᎏ
1 mol CH4
2 mol H2O
ᎏᎏ
1 mol CH4
1 mol CH4
ᎏᎏ
2 mol H2O
Problem-Solving Tip: Are you Using a Balanced Equation?
Never start a calculation involving a chemical reaction without first checking that the
equation is balanced.
Please don’t try to memorize
unit factors for chemical reactions;
rather, learn the general method for
constructing them from balanced
chemical equations.
A balanced equation may be
interpreted on a mass basis.
94 CHAPTER 3: Chemical Equations and Reaction Stoichiometry
EXAMPLE 3-3 Mass of a Reactant Required
What mass of oxygen is required to react completely with 1.20 mol of CH4?
Plan
The balanced equation
CH4 ϩ 2O2 88n CO2 ϩ 2H2O
1 mol 2 mol 1 mol 2 mol
16.0 g 2(32.0 g) 44.0 g 2(18.0 g)
gives the relationships among moles and grams of reactants and products.
mol CH4 88n mol O2 88n g O2
Solution
__? g O2 ϭ 1.20 mol CH4 ϫ ϫ ϭ 76.8 g O2
EXAMPLE 3-4 Mass of a Reactant Required
What mass of oxygen is required to react completely with 24.0 g of CH4?
Plan
Recall the balanced equation in Example 3-3.
CH4 ϩ 2O2 88n CO2 ϩ 2H2O
1 mol 2 mol 1 mol 2 mol
16.0 g 2(32.0) g 44.0 g 2(18.0) g
32.0 g O2
ᎏᎏ
1 mol O2
2 mol O2
ᎏᎏ
1 mol CH4
Problem-Solving Tip: Use the Mole Ratio in Calculations with
Balanced Chemical Equations
The most important way of interpretating the balanced chemical equation is in terms of
moles. We use the coefficients to get the mole ratio of any two substances we want to
relate. Then we apply it as
moles of moles of mole ratio
desired ϭ substance ϫ from balanced
substance given chemical equation
It is important to include the substance formulas as part of the units; this can help
us decide how to set up the unit factors. Notice that in Example 3-2, we want to cancel
the term mol CH4, so we know that mol CH4 must be in the denominator of the mole
ratio by which we multiply; we want mol O2 as the answer, so mol O2 must appear
in the numerator of the mole ratio. In other words, do not just write ᎏ
m
m
o
o
l
l
ᎏ; write
, giving the formulas of the two substances involved.
mol of something
ᎏᎏᎏ
mol of something else
















XXXXXXX
XXXXXXXXXXXXXXX
XXXXXXXXXXXXXXXX
XXXXXX
66
66
g
g
This shows that 1 mol of CH4 reacts with 2 mol of O2. These two quantities are chemically
equivalent, so we can construct unit factors.
Solution
CH4 ϩ 2O2 88n CO2 ϩ 2H2O
1 mol 2 mol 1 mol 2 mol
__? mol CH4 ϭ 24.0 g CH4 ϫ ϭ 1.50 mol CH4
__? mol O2 ϭ 1.50 mol CH4 ϫ ϭ 3.00 mol O2
__? g O2 ϭ 3.00 mol O2 ϫ ϭ 96.0 g O2
All these steps could be combined into one setup in which we convert
g of CH4 88n mol of CH4 88n mol of O2 88n g of O2
__? g O2 ϭ 24.0 g CH4 ϫ ϫ ϫ ϭ 96.0 g O2
The same answer, 96.0 g of O2, is obtained by both methods.
You should now work Exercise 18.
The question posed in Example 3-4 may be reversed, as in Example 3-5.
EXAMPLE 3-5 Mass of a Reactant Required
What mass of CH4, in grams, is required to react with 96.0 grams of O2?
Plan
We recall that one mole of CH4 reacts with two moles of O2.
Solution
__? g CH4 ϭ 96.0 g O2 ϫ ϫ ϫ ϭ 24.0 g CH4
You should now work Exercise 22.
This is the amount of CH4 in Example 3-4 that reacted with 96.0 grams of O2.
EXAMPLE 3-6 Mass of a Product Formed
Most combustion reactions occur in excess O2, that is, more than enough O2 to burn the substance
completely. Calculate the mass of CO2, in grams, that can be produced by burning 6.00
mol of CH4 in excess O2.
16.0 g CH4
ᎏᎏ
1 mol CH4
1 mol CH4
ᎏᎏ
2 mol O2
1 mol O2
ᎏᎏ
32.0 g O2
32.0 g O2
ᎏᎏ
1 mol O2
2 mol O2
ᎏᎏ
1 mol CH4
1 mol CH4
ᎏᎏ
16.0 g CH4
32.0 g O2
ᎏᎏ
1 mol O2
2 mol O2
ᎏᎏ
1 mol CH4
1 mol CH4
ᎏᎏ
16.0 g CH4
Here we solve the problem in three
steps and so we convert
1. g CH4 n mol CH4
2. mol CH4 n mol O2
3. mol O2 n g O2
3-2 Calculations Based On Chemical Equations 95
These unit factors are the reciprocals
of those used in Example 3-4.
It is important to recognize that the
reaction must stop when the 6.0 mol
of CH4 has been used up. Some O2
will remain unreacted.
Plan
The balanced equation tells us that one mole of CH4 produces one mole of CO2.
CH4 ϩ 2O2 88n CO2 ϩ 2H2O
1 mol 2 mol 1 mol 2 mol
16.0 g 2(32.0 g) 44.0 g 2(18.0 g)
Solution
__? g CO2 ϭ 6.00 mol CH4 ϫ ϫ ϭ 2.64 ϫ 102 g CO2
You should now work Exercise 20.
Reaction stoichiometry usually involves interpreting a balanced chemical equation to
relate a given bit of information to the desired bit of information.
EXAMPLE 3-7 Mass of a Reactant Required
Phosphorus, P4, burns with excess oxygen to form tetraphosphorus decoxide, P4O10. In this
reaction, what mass of P4 reacts with 1.50 moles of O2?
Plan
The balanced equation tells us that one mole of P4 reacts with five moles of O2.
P4 ϩ 5O2 88n P4O10
1 mol 5 mol 1 mol
mol O2 88n mol P4 88n mass P4
Solution
__? g P4 ϭ 1.50 mol O2 ϫ ϫ ϭ 37.2 g P4
You should now work Exercise 24.
The possibilities for this kind of problem solving are limitless. Before you continue, you
should work Exercises 8–25 at the end of the chapter.
THE LIMITING REACTANT CONCEPT
In the problems we have worked thus far, the presence of an excess of one reactant was
stated or implied. The calculations were based on the substance that was used up first,
called the limiting reactant. Before we study the concept of the limiting reactant in stoichiometry,
let’s develop the basic idea by considering a simple but analogous nonchemical
example.
Suppose you have four slices of ham and six slices of bread and you wish to make as
many ham sandwiches as possible using only one slice of ham and two slices of bread per
sandwich. Obviously, you can make only three sandwiches, at which point you run out of
3-3
124.0 g P4
ᎏᎏ
mol P4
1 mol P4
ᎏᎏ
5 mol O2
44.0 g CO2
ᎏᎏ
1 mol CO2
1 mol CO2
ᎏᎏ
1 mol CH4
96 CHAPTER 3: Chemical Equations and Reaction Stoichiometry
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 5.5, Limiting Reactants.
bread. (In a chemical reaction this would correspond to one of the reactants being used
up—so the reaction would stop.) The bread is therefore the “limiting reactant,” and the
extra slice of ham is the “excess reactant.” The amount of product, ham sandwiches, is
determined by the amount of the limiting reactant, bread in this case. The limiting reactant
is not necessarily the reactant present in the smallest amount. We have four slices of
ham, the smallest amount, and six slices of bread, but the “reaction ratio” is two slices of
bread to one piece of ham, and so bread is the limiting reactant.
EXAMPLE 3-8 Limiting Reactant
What mass of CO2 could be formed by the reaction of 16.0 g of CH4 with 48.0 g of O2?
Plan
The balanced equation tells us that one mole of CH4 reacts with two moles of O2.
CH4 ϩ 2O2 88n CO2 ϩ 2H2O
1 mol 2 mol 1 mol 2 mol
16.0 g 2(32.0 g) 44.0 g 2(18.0 g)
We are given masses of both CH4 and O2, so we calculate the number of moles of each reactant,
and then determine the number of moles of each reactant required to react with the other.
From these calculations we identify the limiting reactant. Then we base the calculation on it.
Solution
__? mol CH4 ϭ 16.0 g CH4 ϫ ϭ 1.00 mol CH4
__? mol O2 ϭ 48.0 g O2 ϫ ϭ 1.50 mol O2
Now we return to the balanced equation. First we calculate the number of moles of O2 that
would be required to react with 1.00 mole of CH4.
__? mol O2 ϭ 1.00 mol CH4 ϫ ϭ 2.00 mol O2
We see that 2.00 moles of O2 is required, but we have 1.50 moles of O2, so O2 is the limiting
reactant. Alternatively, we can calculate the number of moles of CH4 that would react with
1.50 moles of O2.
__? mol CH4 ϭ 1.50 mol O2 ϫ ϭ 0.750 mol CH4
This tells us that only 0.750 mole of CH4 would be required to react with 1.50 moles of O2.
But we have 1.00 mole of CH4, so we see again that O2 is the limiting reactant. The reaction
must stop when the limiting reactant, O2, is used up, so we base the calculation on O2.
g of O2 88n mol of O2 88n mol of CO2 88n g of CO2
__? g CO2 ϭ 48.0 g O2 ϫ ϫ ϫ ϭ 33.0 g CO2
You should now work Exercise 26.
44.0 g CO2
ᎏᎏ
1 mol CO2
1 mol CO2
ᎏᎏ
2 mol O2
1 mol O2
ᎏᎏ
32.0 O2
1 mol CH4
ᎏᎏ
2 mol O2
2 mol O2
ᎏᎏ
1 mol CH4
1 mol O2
ᎏᎏ
32.0 g O2
1 mol CH4
ᎏᎏ
16.0 g CH4
3-3 The Limiting Reactant Concept 97
Thus, 33.0 grams of CO2 is the most CO2 that can be produced from 16.0 grams of
CH4 and 48.0 grams of O2. If we had based our calculation on CH4 rather than O2, our
answer would be too big (44.0 grams) and wrong because more O2 than we have would
be required.
Another approach to problems like Example 3-8 is to calculate the number of moles
of each reactant:
__? mol CH4 ϭ 16.0 g CH4 ϫ ϭ 1.00 mol CH4
__? mol O2 ϭ 48.0 g O2 ϫ ϭ 1.50 mol O2
Then we return to the balanced equation. We first calculate the required ratio of reactants
as indicated by the balanced chemical equation. We then calculate the available ratio of
reactants and compare the two:
Required Ratio Available Ratio
ϭ ϭ
We see that each mole of O2 requires exactly 0.500 mol of CH4 to be completely used
up. We have 0.667 mol of CH4 for each mole of O2, so there is more than enough CH4
to react with the O2 present. That means that there is insufficient O2 to react with all of
the available CH4. The reaction must stop when the O2 is gone; O2 is the limiting reactant,
and we must base the calculation on it. (If the available ratio of CH4 to O2 had been
smaller than the required ratio, we would have concluded that there is not enough CH4
to react with all of the O2, and CH4 would have been the limiting reactant.)
0.667 mol CH4
ᎏᎏ
1.00 mol O2
1.00 mol CH4
ᎏᎏ
1.50 mol O2
0.500 mol CH4
ᎏᎏ
1.00 mol O2
1 mol CH4
ᎏᎏ
2 mol O2
1 mol O2
ᎏᎏ
32.0 g O2
1 mol CH4
ᎏᎏ
16.0 g CH4
98 CHAPTER 3: Chemical Equations and Reaction Stoichiometry
Problem-Solving Tip: Choosing the Limiting Reactant
Students often wonder how to know which ratio to calculate to help find the limiting
reactant.
(1) The ratio must involve the two reactants with amounts given in the problem.
(2) It doesn’t matter which way you calculate the ratio, as long as you calculate both
required ratio and available ratio in the same order. For example, we could
calculate the required and available ratio of in the approach we just
illustrated. If you can’t decide how to solve a limiting reactant problem, as a last
resort, do the entire calculation twice—once based on each reactant amount
given. The smaller answer is the right one.
mol O2
ᎏᎏ
mol CH4
EXAMPLE 3-9 Limiting Reactant
What is the maximum mass of Ni(OH)2 that could be prepared by mixing two solutions that
contain 25.9 g of NiCl2 and 10.0 g of NaOH, respectively?
NiCl2 ϩ 2NaOH 88n Ni(OH)2 ϩ 2NaCl
Even though the reaction occurs in
aqueous solution, this calculation is
similar to earlier examples because
we are given the amounts of pure
reactants.
3-4 Percent Yields from Chemical Reactions 99
Plan
Interpreting the balanced equation as usual, we have
NiCl2 ϩ 2NaOH 88n Ni(OH)2 ϩ 2NaCl
1 mol 2 mol 1 mol 2 mol
129.6 g 2(40.0 g) 92.7 g 2(58.4 g)
We determine the number of moles of NiCl2 and NaOH present. Then we find the number
of moles of each reactant required to react with the other reactant. These calculations identify
the limiting reactant. We base the calculation on it.
Solution
__? mol NiCl2 ϭ 25.9 g NiCl2 ϫ ϭ 0.200 mol NiCl2
__? mol NaOH ϭ 10.0 g NaOH ϫ ϭ 0.250 mol NaOH
We return to the balanced equation and calculate the number of moles of NaOH required to
react with 0.200 mol of NiCl2.
__? mol NaOH ϭ 0.200 mol NiCl2 ϫ ϭ 0.400 mol NaOH
But we have only 0.250 mol of NaOH, so NaOH is the limiting reactant.
g of NaOH 88n mol of NaOH 88n mol Ni(OH)2 88n g of Ni(OH)2
__? g Ni(OH)2 ϭ 10.0 g NaOH ϫ ϫ ϫ
ϭ 11.6 g Ni(OH)2
You should now work Exercises 30 and 34.
92.7 g Ni(OH)2
ᎏᎏ
1 mol Ni(OH)2
1 mol Ni(OH)2
ᎏᎏ
2 mol NaOH
1 mol NaOH
ᎏᎏ
40.0 g NaOH
2 mol NaOH
ᎏᎏ
1 mol NiCl2
1 mol NaOH
ᎏᎏ
40.0 g NaOH
1 mol NiCl2
ᎏᎏ
129.6 g NiCl2
Problem-Solving Tip: How Can We Recognize a Limiting Reactant
Problem?
When we are given the amounts of two (or more) reactants, we should suspect that we
are dealing with a limiting reactant problem. It is very unlikely that exactly the stoichiometric
amounts of both reactants are present in a reaction mixture.
PERCENT YIELDS FROM CHEMICAL REACTIONS
The theoretical yield from a chemical reaction is the yield calculated by assuming that
the reaction goes to completion. In practice we often do not obtain as much product from
a reaction mixture as is theoretically possible. This is true for several reasons. (1) Many
reactions do not go to completion; that is, the reactants are not completely converted to
products. (2) In some cases, a particular set of reactants undergoes two or more reactions
3-4
In the examples we have worked to
this point, the amounts of products
that we calculated were theoretical
yields.
A precipitate of solid Ni(OH)2 forms
when colorless NaOH solution is
added to green NiCl2 solution.
simultaneously, forming undesired products as well as desired products. Reactions other
than the desired one are called “side reactions.” (3) In some cases, separation of the desired
product from the reaction mixture is so difficult that not all of the product formed
is successfully isolated. The actual yield is the amount of a specified pure product actually
obtained from a given reaction.
The term percent yield is used to indicate how much of a desired product is obtained
from a reaction.
percent yield ϭ ϫ 100%
Consider the preparation of nitrobenzene, C6H5NO2, by the reaction of a limited
amount of benzene, C6H6, with excess nitric acid, HNO3. The balanced equation for the
reaction may be written as
C6H6 ϩ HNO3 88n C6H5NO2 ϩ H2O
1 mol 1 mol 1 mol 1 mol
78.1 g 63.0 g 123.1 g 18.0 g
EXAMPLE 3-10 Percent Yield
A 15.6-gram sample of C6H6 is mixed with excess HNO3. We isolate 18.0 grams of C6H5NO2.
What is the percent yield of C6H5NO2 in this reaction?
Plan
First we interpret the balanced chemical equation to calculate the theoretical yield of C6H5NO2.
Then we use the actual (isolated) yield and the previous definition to calculate the percent
yield.
Solution
We calculate the theoretical yield of C6H5NO2.
__? g C6H5NO2 ϭ 15.6 g C6H6 ϫ ϫ ϫ
ϭ 24.6 g C6H5NO2 m theoretical yield
This tells us that if all the C6H6 were converted to C6H5NO2 and isolated, we should obtain
24.6 grams of C6H5NO2 (100% yield). We isolate only 18.0 grams of C6H5NO2, however.
percent yield ϭ ϫ 100% ϭ ϫ 100%
ϭ 73.2 percent yield
You should now work Exercise 38.
The amount of nitrobenzene obtained in this experiment is 73.2% of the amount that
would be expected if the reaction had gone to completion, if there were no side reactions,
and if we could have recovered all of the product as pure substance.
18.0 g
ᎏ
24.6 g
actual yield of product
ᎏᎏᎏ
theoretical yield of product
123.1 g C6H5NO2
ᎏᎏᎏ
1 mol C6H5NO2
1 mol C6H5NO2
ᎏᎏ
1 mol C6H6
1 mol C6H6
ᎏᎏ
78.1 g C6H6
actual yield of product
ᎏᎏᎏ
theoretical yield of product
It is not necessary to know the mass of
one mole of HNO3 to solve this
problem.
100 CHAPTER 3: Chemical Equations and Reaction Stoichiometry
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 5.6, Percent Yield.
3-5 Sequential Reactions 101
SEQUENTIAL REACTIONS
Often more than one reaction is required to change starting materials into the desired
product. This is true for many reactions that we carry out in the laboratory and for many
industrial processes. These are called sequential reactions. The amount of desired product
from each reaction is taken as the starting material for the next reaction.
EXAMPLE 3-11 Sequential Reactions
At high temperatures, carbon reacts with water to produce a mixture of carbon monoxide, CO,
and hydrogen, H2.
heat
C ϩ H2O 8888n CO ϩ H2
Carbon monoxide is separated from H2 and then used to separate nickel from cobalt by
forming a gaseous compound, nickel tetracarbonyl, Ni(CO)4.
Ni ϩ 4CO 88n Ni(CO)4
What mass of Ni(CO)4 could be obtained from the CO produced by the reaction of 75.0 g of
carbon? Assume 100% yield.
Plan
We interpret both chemical equations in the usual way, and solve the problem in two steps.
They tell us that one mole of C produces one mole of CO and that four moles of CO is
required to produce one mole of Ni(CO)4.
1. We determine the number of moles of CO formed in the first reaction.
2. From the number of moles of CO produced in the first reaction, we calculate the number
of grams of Ni(CO)4 formed in the second reaction.
Solution
1. C ϩ H2O 88n CO ϩ H2
1 mol 1 mol 1 mol 1 mol
12.0 g
__? mol CO ϭ 75.0 g C ϫ ϫ ϭ 6.25 mol CO
2. Ni ϩ 4CO 88n Ni(CO)4
1 mol 4 mol 1 mol
171 g
1 mol CO
ᎏᎏ
1 mol C
1 mol C
ᎏ
12.0 g C
3-5
__? g Ni(CO)4 ϭ 6.25 mol CO ϫ ϫ ϭ 267 g Ni(CO)4
Alternatively, we can set up a series of unit factors based on the conversions in the reaction
sequence and solve the problem in one setup.
g C 88n mol C 88n mol CO 88n mol Ni(CO)4 88n g Ni(CO)4
__? g Ni(CO)4 ϭ 75.0 g C ϫ ϫ ϫ ϫ ϭ 267 g Ni(CO)4
You should now work Exercise 46.
171 g Ni(CO)4
ᎏᎏ
1 mol Ni(CO)4
1 mol Ni(CO)4
ᎏᎏ
4 mol CO
1 mol CO
ᎏᎏ
1 mol C
1 mol C
ᎏ
12.0 g C
171 g Ni(CO)4
ᎏᎏ
1 mol Ni(CO)4
1 mol Ni(CO)4
ᎏᎏ
4 mol CO
EXAMPLE 3-12 Sequential Reactions
Phosphoric acid, H3PO4, is a very important compound used to make fertilizers. It is also present
in cola drinks.
H3PO4 can be prepared in a two-step process.
Rxn 1: P4 ϩ 5O2 88n P4O10
Rxn 2: P4O10 ϩ 6H2O 88n 4H3PO4
We allow 272 grams of phosphorus to react with excess oxygen, which forms tetraphosphorus
decoxide, P4O10, in 89.5% yield. In the second step reaction, a 96.8% yield of H3PO4 is obtained.
What mass of H3PO4 is obtained?
Plan
1. We interpret the first equation as usual and calculate the amount of P4O10 obtained.
P4 ϩ 5O2 88n P4O10
1 mol 5 mol 1 mol
124 g 5(32.0 g) 284 g
g P4 88n mol P4 88n mol P4O10 88n g P4O10
2. Then we interpret the second equation and calculate the amount of H3PO4 obtained from
the P4O10 from the first step.
P4O10 ϩ 6H2O 88n 4H3PO4
1 mol 6 mol 4 mol
284 g 6(18.0 g) 4(98.0 g)
g P4O10 88n mol P4O10 88n mol H3PO4 88n g H3PO4
Solution
1. __? g P4O10 ϭ 272 g P4 ϫ ϫ ϫ
ϫ ΂ ΃ϭ 558 g P4O10
2. __? g H3PO4 ϭ 558 g P4O10 ϫ ϫ
ϫ ϫ ΂ ΃ϭ 746 g H3PO4
You should now work Exercises 48 and 50.
Chemists have determined the structures of many naturally occurring compounds. One
way of proving the structure of such a compound is by synthesizing it from available starting
materials. Professor Grieco, now at Indiana University, was assisted by Majetich and
Ohfune in the synthesis of helenalin, a powerful anticancer drug, in a 40-step process.
This 40-step synthesis gave a remarkable average yield of about 90% for each step, which
still resulted in an overall yield of only about 1.5%.
96.8 g H3PO4 actual
ᎏᎏᎏ
100 g H3PO4 theoretical
98.0 g H3PO4 theoretical
ᎏᎏᎏ
1 mol H3PO4 theoretical
4 mol H3PO4
ᎏᎏ
1 mol P4O10
1 mol P4O10
ᎏᎏ
284 g P4O10
89.5 g P4O10 actual
ᎏᎏᎏ
100 g P4O10 theoretical
284 g P4O10 theoretical
ᎏᎏᎏ
1 mol P4O10 theoretical
1 mol P4O10
ᎏᎏ
1 mol P4
1 mol P4
ᎏᎏ
124 g P4
102 CHAPTER 3: Chemical Equations and Reaction Stoichiometry
Two large uses for H3PO4 are
in fertilizers and cola drinks.
Approximately 100 lb of
H3PO4– based fertilizer are used
per year per person in America.
The unit factors that account for less
than 100% reaction and less than
100% recovery are included in
parentheses.
In some solutions, such as a nearly
equal mixture of ethyl alcohol and
water, the distinction between solute
and solvent is arbitrary.
3-6 Concentrations of Solutions 103
CONCENTRATIONS OF SOLUTIONS
Many chemical reactions are more conveniently carried out with the reactants mixed in
solution rather than as pure substances. A solution is a homogeneous mixture, at the molecular
level, of two or more substances. Simple solutions usually consist of one substance,
the solute, dissolved in another substance, the solvent. The solutions used in the laboratory
are usually liquids, and the solvent is often water. These are called aqueous solutions.
For example, solutions of hydrochloric acid can be prepared by dissolving hydrogen
chloride (HCl, a gas at room temperature and atmospheric pressure) in water.
Solutions of sodium hydroxide are prepared by dissolving solid NaOH in water.
We often use solutions to supply the reactants for chemical reactions. Solutions allow
the most intimate mixing of the reacting substances at the molecular level, much more
than would be possible in solid form. (A practical example is drain cleaner, shown in the
photo.) We sometimes adjust the concentrations of solutions to speed up or slow down
the rate of a reaction. In this section we study methods for expressing the quantities of
the various components present in a given amount of solution.
Concentrations of solutions are expressed in terms of either the amount of solute present
in a given mass or volume of solution, or the amount of solute dissolved in a given
mass or volume of solvent.
Percent by Mass
Concentrations of solutions may be expressed in terms of percent by mass of solute,
which gives the mass of solute per 100 mass units of solution. The gram is the usual mass
unit.
percent solute ϭ ϫ 100%
percent ϭ ϫ 100%
Thus, a solution that is 10.0% calcium gluconate, Ca(C6H11O7)2, by mass contains 10.0
grams of calcium gluconate in 100.0 grams of solution. This could be described as 10.0
grams of calcium gluconate in 90.0 grams of water. The density of a 10.0% solution of
calcium gluconate is 1.07 g/mL, so 100 mL of a 10.0% solution of calcium gluconate has
a mass of 107 grams. Observe that 100 grams of a solution usually does not occupy 100 mL.
Unless otherwise specified, percent means percent by mass, and water is the solvent.
EXAMPLE 3-13 Percent of Solute
Calculate the mass of nickel(II) sulfate, NiSO4, contained in 200. g of a 6.00% solution of
NiSO4.
Plan
The percentage information tells us that the solution contains 6.00 grams of NiSO4 per
100. grams of solution. The desired information is the mass of NiSO4 in 200. grams of solution.
A unit factor is constructed by placing 6.00 grams NiSO4 over 100. grams of solution.
mass of solute
ᎏᎏᎏᎏ
mass of solute ϩ mass of solvent
mass of solute
ᎏᎏ
mass of solution
3-6
A 10.0% solution of Ca(C6H11O7)2
is sometimes administered intravenously
in emergency treatment
for black widow spider bites.
The sodium hydroxide and aluminum
in some drain cleaners do
not react while they are stored in
solid form. When water is added,
the NaOH dissolves and begins to
act on trapped grease. At the same
time, NaOH and Al react to produce
H2 gas; the resulting turbulence
helps to dislodge the blockage. Do
you see why the container should be
kept tightly closed?



















Multiplication of the mass of the solution, 200. grams, by this unit factor gives the mass of
NiSO4 in the solution.
Solution
__? g NiSO4 ϭ 200. g soln ϫ ϭ 12.0 g NiSO4
EXAMPLE 3-14 Mass of Solution
A 6.00% NiSO4 solution contains 40.0 g of NiSO4. Calculate the mass of the solution.
Plan
Placing 100. g of solution over 6.00 g of NiSO4 gives the desired unit factor.
Solution
__? g soln ϭ 40.0 g NiSO4 ϫ ϭ 667 g soln
EXAMPLE 3-15 Mass of Solute
Calculate the mass of NiSO4 present in 200. mL of a 6.00% solution of NiSO4. The density
of the solution is 1.06 g/mL at 25°C.
Plan
The volume of a solution multiplied by its density gives the mass of solution (see Section
1-11). The mass of solution is then multiplied by the mass fraction due to NiSO4 (6.00 g
NiSO4/100. g soln) to give the mass of NiSO4 in 200. mL of solution.
Solution
__? g NiSO4 ϭ 200. mL soln ϫ ϫ ϭ 12.7 g NiSO4
212 g soln
You should now work Exercise 55.
EXAMPLE 3-16 Percent Solute and Density
What volume of a solution that is 15.0% iron(III) nitrate contains 30.0 g of Fe(NO3)3? The
density of the solution is 1.16 g/mL at 25°C.
Plan
Two unit factors relate mass of Fe(NO3)3 and mass of solution, 15.0 g Fe(NO3)3/100 g and
100 g/15.0 g Fe(NO3)3. The second factor converts grams of Fe(NO3)3 to grams of solution.
Solution
__? mL soln ϭ 30.0 g Fe(NO3)3 ϫ ϫ ϭ 172 mL
200 g soln
1.00 mL soln
ᎏᎏ
1.16 g soln
100. g soln
ᎏᎏ
15.0 g Fe(NO3)3
6.00 g NiSO4
ᎏᎏ
100. g soln
1.06 g soln
ᎏᎏ
1.00 mL soln
100. g soln
ᎏᎏ
6.00 g NiSO4
6.00 g NiSO4
ᎏᎏ
100. g soln
104 CHAPTER 3: Chemical Equations and Reaction Stoichiometry















The definition of molarity specifies the
amount of solute per unit volume of
solution, whereas percent specifies
the amount of solute per unit mass of
solution. Molarity therefore depends
on temperature and pressure, whereas
percent by mass does not.
3-6 Concentrations of Solutions 105
Note that the answer is not 200. mL but considerably less because 1.00 mL of solution has a
mass of 1.16 grams; however, 172 mL of the solution has a mass of 200 g.
You should now work Exercise 58.
Molarity
Molarity (M), or molar concentration, is a common unit for expressing the concentrations
of solutions. Molarity is defined as the number of moles of solute per liter of solution:
molarity ϭ
To prepare one liter of a one molar solution, one mole of solute is placed in a oneliter
volumetric flask, enough solvent is added to dissolve the solute, and solvent is then
added until the volume of the solution is exactly one liter. Students sometimes make the
mistake of assuming that a one molar solution contains one mole of solute in a liter of
solvent. This is not the case; one liter of solvent plus one mole of solute usually has a total
volume of more than one liter. A 0.100 M solution contains 0.100 mol of solute per
liter of solution, and a 0.0100 M solution contains 0.0100 mol of solute per liter of solution
(Figure 3-2).
number of moles of solute
ᎏᎏᎏᎏ
number of liters of solution
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 5.11, Preparing Solutions, Direct
Addition.
Figure 3-2 Preparation of 0.0100 M solution of KMnO4, potassium permanganate. A
250.-mL sample of 0.0100 M KMnO4 solution contains 0.395 g of KMnO4 (1 mol ϭ 158 g).
(a) 0.395 g of KMnO4 (0.00250 mol) is weighed out carefully and transferred into a 250.-mL
volumetric flask. (b) The KMnO4 is dissolved in water. (c) Distilled H2O is added to the
volumetric flask until the volume of solution is 250. mL. The flask is then stoppered, and
its contents are mixed thoroughly to give a homogeneous solution.
(a) (b) (c)
Water is the solvent in most of the solutions that we encounter. Unless otherwise
indicated, we assume that water is the solvent. When the solvent is other than water, we
state this explicitly.
EXAMPLE 3-17 Molarity
Calculate the molarity (M) of a solution that contains 3.65 grams of HCl in 2.00 liters of
solution.
Plan
We are given the number of grams of HCl in 2.00 liters of solution. We apply the definition
of molarity, remembering to convert grams of HCl to moles of HCl.
Solution
ϭ ϫ ϭ 0.0500 mol HCl/L soln
The concentration of the HCl solution is 0.0500 molar, and the solution is called 0.0500 M
hydrochloric acid. One liter of the solution contains 0.0500 mol of HCl.
You should now work Exercise 60.
EXAMPLE 3-18 Mass of Solute
Calculate the mass of Ba(OH)2 required to prepare 2.50 L of a 0.0600 M solution of barium
hydroxide.
Plan
The volume of the solution, 2.50 L, is multiplied by the concentration, 0.0600 mol Ba(OH)2/L,
to give the number of moles of Ba(OH)2. The number of moles of Ba(OH)2 is then multiplied
by the mass of Ba(OH)2 in one mole, 171.3 g Ba(OH)2/mol Ba(OH)2, to give the mass of
Ba(OH)2 in the solution.
Solution
__? g Ba(OH)2 ϭ 2.50 L soln ϫ ϫ
ϭ 25.7 g Ba(OH)2
You should now work Exercise 62.
The solutions of acids and bases that are sold commercially are too concentrated for
most laboratory uses. We often dilute these solutions before we use them. We must know
the molar concentration of a stock solution before it is diluted. This can be calculated
from the specific gravity and the percentage data given on the label of the bottle.
EXAMPLE 3-19 Molarity
A sample of commercial sulfuric acid is 96.4% H2SO4 by mass, and its specific gravity is 1.84.
Calculate the molarity of this sulfuric acid solution.
171.3 g Ba(OH)2
ᎏᎏ
1 mol Ba(OH)2
0.0600 mol Ba(OH)2
ᎏᎏᎏ
1 L soln
1 mol HCl
ᎏᎏ
36.5 g HCl
3.65 g HCl
ᎏᎏ
2.00 L soln
__? mol HCl
ᎏᎏ
L soln
We place 3.65 g HCl over 2.00 L of
solution, and then convert g HCl to
mol HCl.
106 CHAPTER 3: Chemical Equations and Reaction Stoichiometry
A label that shows the analysis of
sulfuric acid.
3-7 Dilution of Solutions 107
Plan
The density of a solution, grams per milliliter, is numerically equal to its specific gravity, so
the density of the solution is 1.84 g/mL. The solution is 96.4% H2SO4 by mass; therefore 100. g
of solution contains 96.4 g of pure H2SO4. From this information, we can find the molarity of
the solution. First, we calculate the mass of one liter of solution.
Solution
ϭ ϫ ϭ 1.84 ϫ 103 g soln/L soln
The solution is 96.4% H2SO4 by mass, so the mass of H2SO4 in one liter is
ϭ ϫ ϭ 1.77 ϫ 103 g H2SO4/L soln
The molarity is the number of moles of H2SO4 per liter of solution.
ϭ ϫ ϭ 18.0 mol H2SO4/L soln
Thus, the solution is an 18.0 M H2SO4 solution. This problem can also be solved by using a
series of three unit factors.
ϭ ϫ ϫ ϫ
ϭ 18.1 mol H2SO4/L soln ϭ 18.1 M H2SO4
You should now work Exercise 68.
1 mol H2SO4
ᎏᎏ
98.1 g H2SO4
96.4 g H2SO4
ᎏᎏ
100 g soln
1000 mL soln
ᎏᎏ
L soln
1.84 g soln
ᎏᎏ
mL soln
__? mol H2SO4
ᎏᎏ
L soln
1 mol H2SO4
ᎏᎏ
98.1 g H2SO4
1.77 ϫ 103 g H2SO4
ᎏᎏᎏ
L soln
__? mol H2SO4
ᎏᎏ
L soln
96.4 g H2SO4
ᎏᎏ
100.0 g soln
1.84 ϫ 103 g soln
ᎏᎏ
L soln
__? g H2SO4
ᎏᎏ
L soln
1000 mL soln
ᎏᎏ
L soln
1.84 g soln
ᎏᎏ
mL soln
__? g soln
ᎏᎏ
L soln
Problem-Solving Tip: Write Complete Units
A common pitfall is to write units that are not complete enough to be helpful. For instance,
writing the density in Example 3-19 as just doesn’t help us figure
out the required conversions. It is much safer to write , ,
ᎏ
96
1
.
0
4
0
g
g
H
s
2
o
S
ln
O4
ᎏ, and so on. In Example 3-19, we have written complete units to help
guide us through the problem.
1000 mL soln
ᎏᎏ
L soln
1.84 g soln
ᎏᎏ
mL soln
1.84 g
ᎏ
mL
DILUTION OF SOLUTIONS
Recall that the definition of molarity is the number of moles of solute divided by the
volume of the solution in liters:
molarity ϭ
Multiplying both sides of the equation by the volume, we obtain
volume (in L) ϫ molarity ϭ number of moles of solute
number of moles of solute
ᎏᎏᎏᎏ
number of liters of solution
3-7
ANALYSIS
MAXIMUM LIMITS OF IMPURITIES
Appearance........Passes A.C.S.Test
Color (APHA)....................10 Max.
Residue after Ignition............4 ppm
Chloride (Cl)....................0.2 ppm
Nitrate (NO3)...................0.5 ppm
Ammonium (NH4).................1 ppm
Substances Reducing KMnO4 (limit about
2ppm as SO2).........Passes A.C.S.Test
Arsenic (As)..................0.004 ppm
Heavy Metals (as Pb)..........0.8 ppm
Iron (Fe).........................0.2 ppm
Mercury (Hg).......................5 ppb
Specific Gravity...................~1.84
Normality.............................~36
Assay(H2SO4)W/W...Min.95.0%--Max.98.0%
Suitable for Mercury Determinations
The small difference is due to
rounding.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 5.12, Preparing Solutions,
Dilution.
Multiplication of the volume of a solution, in liters, by its molar concentration gives
the amount of solute in the solution.
When we dilute a solution by mixing it with more solvent, the amount of solute present
does not change. But the volume and the concentration of the solution do change.
Because the same number of moles of solute is divided by a larger number of liters of
solution, the molarity decreases. Using a subscript 1 to represent the original concentrated
solution and a subscript 2 to represent the dilute solution, we obtain
volume1 ϫ molarity1 ϭ number of moles of solute ϭ volume2 ϫ molarity2
or
V1M1 ϭ V2M2 (for dilution only)
This expression can be used to calculate any one of four quantities when the other
three are known (Figure 3-3). We frequently need a certain volume of dilute solution of
a given molarity for use in the laboratory, and we know the concentration of the initial
solution available. Then we can calculate the amount of initial solution that must be used
to make the dilute solution.
A can of frozen orange juice contains a
certain mass (or moles) of vitamin C.
After the frozen contents of the can
are diluted by addition of water, the
amount of vitamin C in the resulting
total amount of solution will be
unchanged. The concentration, or
amount per a selected volume, will be
less in the final solution, however.
We could use any volume unit as long
as we use the same unit on both sides
of the equation. This relationship also
applies when the concentration is
changed by evaporating some solvent.
108 CHAPTER 3: Chemical Equations and Reaction Stoichiometry
EXAMPLE 3-20 Dilution
How many milliliters of 18.0 M H2SO4 are required to prepare 1.00 L of a 0.900 M solution
of H2SO4?
Plan
The volume (1.00 L) and molarity (0.900 M) of the final solution, as well as the molarity (18.0
M) of the original solution, are given. Therefore, the relationship V1M1 ϭ V2M2 can be used,
with subscript 1 for the initial acid solution and subscript 2 for the dilute solution. We solve
V1M1 ϭ V2M2 for V1
Solution
V1 ϭ ϭ ϭ 0.0500 L ϭ 50.0 mL
1.00 L ϫ 0.900 M
ᎏᎏᎏ
18.0 M
V2M2
ᎏ
M1
Dilution of a concentrated solution, especially of a strong acid or
base, frequently liberates a great deal of heat. This can vaporize
drops of water as they hit the concentrated solution and can cause dangerous spattering.
As a safety precaution, concentrated solutions of acids or bases are always poured slowly into
water, allowing the heat to be absorbed by the larger quantity of water. Calculations are
usually simpler to visualize, however, by assuming that the water is added to the concentrated
solution.
C A U T I O N !
Figure 3-3 Dilution of a solution. (a) A 100.-mL volumetric flask is filled to the
calibration line with a 0.100 M potassium chromate, K2CrO4 solution. (b) The 0.100 M
K2CrO4 solution is transferred into a 1.00-L volumetric flask. The smaller flask is rinsed
with a small amount of distilled H2O. The rinse solution is added to the solution in the
larger flask. To make sure that all the original K2CrO4 solution is transferred to the larger
flask, the smaller flask is rinsed twice more and each rinse is added to the solution in the
larger flask. (c) Distilled water is added to the 1.00-L flask until the liquid level coincides
with its calibration line. The flask is stoppered and its contents are mixed thoroughly. The
new solution is 0.0100 M K2CrO4. (100. mL of 0.100 M K2CrO4 solution has been diluted
to 1000. mL.) The 100. mL of original solution and the 1000. mL of final solution both
contain the amount of K2CrO4 dissolved in the original 100. mL of 0.100 M K2CrO4.
3-7 Dilution of Solutions 109
The dilute solution contains 1.00 L ϫ 0.900 M ϭ 0.900 mol of H2SO4, so 0.900 mol of H2SO4
must also be present in the original concentrated solution. Indeed, 0.0500 L ϫ 18.0 M ϭ 0.900
mol of H2SO4.
You should now work Exercises 70 and 72.
(a)
(c)
K+
=
CrO2–
4
=
(b)
USING SOLUTIONS IN CHEMICAL REACTIONS
If we plan to carry out a reaction in a solution, we must calculate the amounts of solutions
that we need. If we know the molarity of a solution, we can calculate the amount of
solute contained in a specified volume of that solution. This procedure is illustrated in
Example 3-21.
EXAMPLE 3-21 Amount of Solute
Calculate (a) the number of moles of H2SO4 and (b) the number of grams of H2SO4 in 500. mL
of 0.324 M H2SO4 solution.
Plan
Because we have two parallel calculations in this example, we state the plan for each step just
before the calculation is done.
Solution
(a) The volume of a solution in liters multiplied by its molarity gives the number of moles of
solute, H2SO4 in this case.
__? mol H2SO4 ϭ 0.500 L soln ϫ ϭ 0.162 mol H2SO4
(b) We may use the results of part (a) to calculate the mass of H2SO4 in the solution.
__? g H2SO4 ϭ 0.162 mol H2SO4 ϫ ϭ 15.9 g H2SO4
The mass of H2SO4 in the solution can be calculated without solving explicitly for the number
of moles of H2SO4.
__? g H2SO4 ϭ 0.500 L soln ϫ ᎏ
0.324
L
m
s
o
o
l
l
H
n
2SO4
ᎏ ϫ ᎏ
9
1
8
m
.1
o
g
l H
H
2
2
S
S
O
O
4
4
ᎏ ϭ 15.9 g H2SO4
One of the most important uses of molarity relates the volume of a solution of known
concentration of one reactant to the mass of the other reactant.
EXAMPLE 3-22 Solution Stoichiometry
Calculate the volume in liters and in milliliters of a 0.324 M solution of sulfuric acid required
to react completely with 2.792 grams of Na2CO3 according to the equation
H2SO4 ϩ Na2CO3 88n Na2SO4 ϩ CO2 ϩ H2O
Plan
The balanced equation tells us that one mole of H2SO4 reacts with one mole of Na2CO3, and
we can write
H2SO4 ϩ Na2CO3 88n Na2SO4 ϩ CO2 ϩ H2O
1 mol 1 mol 1 mol 1 mol 1 mol
106.0 g
98.1 g H2SO4
ᎏᎏ
1 mol H2SO4
0.324 mol H2SO4
ᎏᎏ
L soln
3-8
500. mL is more conveniently
expressed as 0.500 L in this problem.
By now, you should be able to convert
mL to L (and the reverse) without
writing out the conversion.
A mole of H2SO4 is 98.1 g.
110 CHAPTER 3: Chemical Equations and Reaction Stoichiometry
See the Saunders Interactive
General Chemistry CD-ROM,
Screens 5.13 and 5.15, Stoichiometry
of Reactions in Solution.
We convert (1) grams of Na2CO3 to moles of Na2CO3, (2) moles of Na2CO3 to moles of
H2SO4, and (3) moles of H2SO4 to liters of H2SO4 solution.
g Na2CO3 88n mol Na2CO3 88n mol H2SO4 88n L H2SO4 soln
Solution
__? L H2SO4 ϭ 2.792 g Na2CO3 ϫ ϫ ϫ
ϭ 0.0813 L H2SO4 soln or 81.3 mL H2SO4 soln
You should now work Exercise 76.
Often we must calculate the volume of solution of known molarity that is required to
react with a specified volume of another solution. We always examine the balanced chemical
equation for the reaction to determine the reaction ratio, that is, the relative numbers
of moles of reactants.
EXAMPLE 3-23 Volume of Solution Required
Find the volume in liters and in milliliters of a 0.505 M NaOH solution required to react with
40.0 mL of 0.505 M H2SO4 solution according to the reaction
H2SO4 ϩ 2NaOH 88n Na2SO4 ϩ 2H2O
Plan
We shall work this example in several steps, stating the “plan,” or reasoning, just before each
step in the calculation. Then we shall use a single setup to solve the problem.
1 L H2SO4 soln
ᎏᎏ
0.324 mol H2SO4
1 mol H2SO4
ᎏᎏ
1 mol Na2CO3
1 mol Na2CO3
ᎏᎏ
106.0 g Na2CO3
The indicator methyl orange
changes from yellow, its color in basic
solutions, to orange, its color in
acidic solutions, when the reaction
in Example 3-22 reaches completion.
3-8 Using Solutions in Chemical Reactions 111
Solution
The balanced equation tells us that the reaction ratio is 1 mol of H2SO4 to 2 mol of NaOH.
H2SO4 ϩ 2NaOH 88n Na2SO4 ϩ 2H2O
1 mol 2 mol 1 mol 2 mol
From the volume and the molarity of the H2SO4 solution, we can calculate the number of
moles of H2SO4.
__? mol H2SO4 ϭ 0.0400 L H2SO4 soln ϫ ϭ 0.0202 mol H2SO4
The number of moles of H2SO4 is related to the number of moles of NaOH by the reaction
ratio, 1 mol H2SO4/2 mol NaOH:
__? mol NaOH ϭ 0.0202 mol H2SO4 ϫ ϭ 0.0404 mol NaOH
Now we can calculate the volume of 0.505 M NaOH solution that contains 0.0404 mol of
NaOH:
__? L NaOH soln ϭ 0.0404 mol NaOH ϫ ϭ 0.0800 L NaOH soln
which we usually call 80.0 mL of NaOH solution.
We have worked through the problem stepwise; let us solve it in a single setup.
ᎏ
L H
av
2
a
S
i
O
lab
4
l
s
e
oln
ᎏ 88n ᎏ
m
a
o
v
l
a
H
ila
2
b
S
l
O
e
4ᎏ 88n ᎏ
s
m
o
o
ln
l N
ne
a
e
O
de
H
d
ᎏ 88n ᎏ
so
L
ln
N
n
a
e
O
ed
H
ed
ᎏ
__? L NaOH soln ϭ 0.0400 L H2SO4 soln ϫ ϫ
ϫ
ϭ 0.0800 L NaOH soln or 80.0 mL NaOH soln
You should now work Exercise 78.
1.00 L NaOH soln
ᎏᎏᎏ
0.505 mol NaOH
2 mol NaOH
ᎏᎏ
1 mol H2SO4
0.505 mol H2SO4
ᎏᎏ
L H2SO4 soln
1.00 L NaOH soln
ᎏᎏᎏ
0.505 mol NaOH
2 mol NaOH
ᎏᎏ
1 mol H2SO4
0.505 mol H2SO4
ᎏᎏ
L soln
The volume of H2SO4 solution is
expressed as 0.0400 L rather than
40.0 mL.
112 CHAPTER 3: Chemical Equations and Reaction Stoichiometry
Key Terms
Actual yield The amount of a specified pure product actually
obtained from a given reaction. Compare with Theoretical
yield.
Chemical equation Description of a chemical reaction by placing
the formulas of reactants on the left and the formulas of
products on the right of an arrow. A chemical equation must
be balanced; that is, it must have the same number of each kind
of atom on both sides.
Concentration The amount of solute per unit volume or mass
of solvent or of solution.
Dilution The process of reducing the concentration of a solute
in solution, usually simply by adding more solvent.
Limiting reactant A substance that stoichiometrically limits the
amount of product(s) that can be formed.
Molarity (M) The number of moles of solute per liter of solu-
tion.
Percent by mass 100% multiplied by the mass of a solute divided
by the mass of the solution in which it is contained.
Percent yield 100% times actual yield divided by theoretical
yield.
Again we see that molarity is a unit
factor. In this case,
1.00 L NaOH soln
ᎏᎏᎏ
0.505 mol NaOH
Exercises 113
Exercises
Chemical Equations
001. What is a chemical equation? What information does it
contain?
002. When balancing chemical equations, you make certain that
the same number of atoms of each element are on both
sides of the equation. What scientific (natural) law requires
that there be equal numbers of atoms of each element in
both the products and the reactants?
003. Use words to state explicitly the relationships among numbers
of molecules of reactants and products in the equation
for the combustion of hexane, C6H14.
2C6H14 ϩ 19O2 88n 12CO2 ϩ 14H2O
Balance each “equation” in Exercises 4–7 by inspection.
004. (a) Al ϩ Cl2 88n Al2Cl6
(b) N2 ϩ H2 88n NH3
(c) K ϩ KNO3 88n K2O ϩ N2
(d) H2O ϩ KO2 88n KOH ϩ O2
(e) H2SO4 ϩ NH3 88n (NH4)2SO4
005. (a) P4 ϩ O2 88n P4O6
(b) P4 ϩ O2 88n P4O10
(c) K2CO3 ϩ Al2Cl6 88n Al2(CO3)3 ϩ KCl
(d) KClO3 ϩ C12H22O11 88n KCl ϩ CO2 ϩ H2O
(e) KOH ϩ H3PO4 88n KH2PO4 ϩ H2O
006. (a) Fe2O3 ϩ CO 88n Fe ϩ CO2
(b) Mg3N2 ϩ H2O 88n NH3 ϩ Mg(OH)2
(c) Ca3(PO4)2 ϩ H2SO4 88n
Ca(H2PO4)2 ϩ Ca(HSO4)2
(d) (NH4)2Cr2O7 88n N2 ϩ H2O ϩ Cr2O3
(e) Al ϩ Cr2O3 88n Al2O3 ϩ Cr
007. (a) UO2 ϩ HF 88n UF4 ϩ H2O
(b) NaCl ϩ H2O ϩ SiO2 88n HCl ϩ Na2SiO3
(c) Ca(HCO3)2 ϩ Na2CO3 88n CaCO3 ϩ NaHCO3
(d) NH3 ϩ O2 88n NO ϩ H2O
(e) PCl3 ϩ O2 88n POCl3
Calculations Based on Chemical Equations
In Exercises 8–11, (a) write the balanced chemical equation that
represents the reaction described by words, and then perform calculations
to answer parts (b) and (c).
008. (a) Nitrogen, N2, combines with hydrogen, H2, to form
ammonia, NH3.
(b) How many hydrogen molecules are required to react
with 600 nitrogen molecules?
(c) How many ammonia molecules are formed in part (b)?
009. (a) Sulfur, S8, combines with oxygen at elevated temperatures
to form sulfur dioxide.
(b) If 250 oxygen molecules are used up in this reaction,
how many sulfur molecules react?
(c) How many sulfur dioxide molecules are formed in
part (b)?
010. (a) Lime, CaO, dissolves in muriatic acid, HCl, to form
calcium chloride, CaCl2, and water.
(b) How many moles of HCl are required to dissolve 6.7
mol of CaO?
(c) How many moles of water are formed in part (b)?
011. (a) Aluminum building materials have a hard, transparent,
protective coating of aluminum oxide, Al2O3,
formed by reaction with oxygen in the air. The sulfuric
acid, H2SO4, in acid rain dissolves this protective
coating and forms aluminum sulfate, Al2(SO4)3, and
water.
(b) How many moles of H2SO4 are required to react with
6.8 mol of Al2O3?
(c) How many moles of Al2(SO4)3 are formed in part (b)?
012. Calculate the number of grams of baking soda, NaHCO3,
that contain 14.0 moles of carbon.
013. Limestone, coral, and seashells are composed primarily of
calcium carbonate. The test for the identification of a carbonate
is to use a few drops of hydrochloric acid. The unbalanced
equation is
CaCO3 ϩ HCl 88n CaCl2 ϩ CO2 ϩ H2O
Products Substances produced in a chemical reaction.
Reactants Substances consumed in a chemical reaction.
Reaction ratio The relative amounts of reactants and products
involved in a reaction; may be the ratio of moles, or masses.
Reaction stoichiometry Description of the quantitative relationships
among substances as they participate in chemical re-
actions.
Sequential reaction A chemical process in which several reaction
steps are required to convert starting materials into products.
Solute The dispersed (dissolved) phase of a solution.
Solution A homogeneous mixture of two or more substances.
Solvent The dispersing medium of a solution.
Stoichiometry Description of the quantitative relationships
among elements and compounds as they undergo chemical
changes.
Theoretical yield The maximum amount of a specified product
that could be obtained from specified amounts of reactants, assuming
complete consumption of the limiting reactant according
to only one reaction and complete recovery of the product.
Compare with Actual yield.
(a) Balance the equation.
(b) How many atoms are in 0.250 moles of calcium car-
bonate?
(c) What number of carbon dioxide molecules is released
on the reaction of 0.250 moles of calcium carbonate?
014. How many moles of oxygen can be obtained by the decomposition
of 8.0 mol of reactant in each of the following
reactions?
(a) 2KClO3 88n 2KCl ϩ 3O2
(b) 2H2O2 88n 2H2O ϩ O2
(c) 2HgO 88n 2Hg ϩ O2
(d) 2NaNO3 88n 2NaNO2 ϩ O2
(e) KClO4 88n KCl ϩ 2O2
015. For the formation of 8.0 mol of water, which reaction uses
the most nitric acid?
(a) 3Cu ϩ 8HNO3 88n 3Cu(NO3)2 ϩ 2NO ϩ 4H2O
(b) Al2O3 ϩ 6HNO3 88n 2Al(NO3)3 ϩ 3H2O
(c) 4Zn ϩ 10HNO3 88n
4Zn(NO3)2 ϩ NH4NO3 ϩ 3H2O
016. Consider the reaction
not balanced
NH3 ϩ O2 8888888888n NO ϩ H2O
For every 25.00 mol of NH3, (a) how many moles of O2
are required, (b) how many moles of NO are produced,
and (c) how many moles of H2O are produced?
017. What masses of cobalt(II) chloride and of hydrogen fluoride
are needed to prepare 15.0 mol of cobalt(II) fluoride
by the following reaction?
CoCl2 ϩ 2HF 88n CoF2 ϩ 2HCl
018. We allow 24.0 g of methane, CH4, to react as completely
as possible with excess oxygen, O2, to form CO2 and water.
Write the balanced equation for this reaction. What
mass of oxygen reacts?
*19. Calculate the mass of calcium required to react with
3.770 g of carbon during the production of calcium carbide,
CaC2.
020. Sodium iodide, NaI, is a source of iodine used to produce
iodized salt.
(a) Write the balanced chemical equation for the reaction
of sodium and iodine.
(b) How many grams of sodium iodide are produced by
the reaction of 93.25 grams of iodine?
021. Consider the reaction
2NO ϩ Br2 88n 2NOBr
For every 7.50 mol of bromine that reacts, how many
moles of (a) NO react and (b) NOBr are produced?
022. A sample of magnetic iron oxide, Fe3O4, reacts completely
with hydrogen at red heat. The water vapor formed by the
reaction
heat
Fe3O4 ϩ 4H2 8888n 3Fe ϩ 4H2O
is condensed and found to weigh 18.75 g. Calculate the
mass of Fe3O4 that reacted.
023. Iron(III) oxide, Fe2O3, is a result of the reaction of iron
with the oxygen in air.
(a) What is the balanced equation for this reaction?
(b) What number of moles of iron react with 15.25 mol
of oxygen from the air?
(c) What mass of iron is required to react with 15.25 mol
of oxygen?
024. Calculate the number of molecules of propane, C3H8, that
will produce 4.80 grams of water when burned in excess
oxygen, O2.
025. What mass of pentane, C5H12, produces 4.52 ϫ 1022 CO2
molecules when burned in excess oxygen, O2?
Limiting Reactant
026. How many grams of NH3 can be prepared from 59.85 g
of N2 and 12.11 g of H2?
N2 ϩ 3H2 88n 2NH3
*27. Silver nitrate solution reacts with calcium chloride solution
according to the equation
2AgNO3 ϩ CaCl2 88n Ca(NO3)2 ϩ 2AgCl
All of the substances involved in this reaction are soluble
in water except silver chloride, AgCl, which forms a solid
(precipitate) at the bottom of the flask. Suppose we mix
together a solution containing 12.6 g of AgNO3 and one
containing 8.40 g of CaCl2. What mass of AgCl is formed?
*28. “Superphosphate,” a water-soluble fertilizer, is sometimes
marketed as “triple phosphate.” It is a mixture of
Ca(H2PO4)2 and CaSO4 on a 1:2 mole basis. It is formed
by the reaction
Ca3(PO4)2 ϩ 2H2SO4 88n Ca(H2PO4)2 ϩ 2CaSO4
We treat 300 g of Ca3(PO4)2 with 200 g of H2SO4. How
many grams of superphosphate could be formed?
114 CHAPTER 3: Chemical Equations and Reaction Stoichiometry
029. Gasoline is produced from crude oil, a nonrenewable resource.
Ethanol is mixed with gasoline to produce a fuel
called gasohol. Calculate the mass of water produced when
Exercises 115
66.89 g of ethanol, C2H5OH, is burned in 55.21 g of oxy-
gen.
030. What mass of potassium can be produced by the reaction
of 125.0 g of Na with 125.0 g of KCl?
heat
Na ϩ KCl 8888n NaCl ϩ K
031. Silicon carbide, an abrasive, is made by the reaction of silicon
dioxide with graphite.
heat
SiO2 ϩ C 8888n SiC ϩ CO (balanced?)
We mix 452 g of SiO2 and 306 g of C. If the reaction proceeds
as far as possible, which reactant is left over? How
much of this reactant remains?
032. Octane, C8H18, is a component of gasoline. A spark is used
to ignite 1.563 grams of octane and 6.778 grams of oxygen
in a sealed container. What mass of carbon dioxide is
produced?
033. What mass of Ca(NO3)2 can be prepared by the reaction
of 18.9 g of HNO3 with 7.4 g of Ca(OH)2?
2HNO3 ϩ Ca(OH)2 88n Ca(NO3)2 ϩ 2H2O
034. What is the maximum amount of Ca3(PO4)2 that can be
prepared from 7.4 g of Ca(OH)2 and 9.8 g of H3PO4?
3Ca(OH)2 ϩ 2H3PO4 88n Ca3(PO4)2 ϩ 6 H2O
035. A reaction mixture contains 11.0 g of PCl3 and 7.00 g of
PbF2. What mass of PbCl2 can be obtained from the following
reaction?
3PbF2 ϩ 2PCl3 88n 2PF3 ϩ 3PbCl2
How much of which reactant is left unchanged?
Percent Yield from Chemical Reactions
036. The percent yield for the reaction
PCl3 ϩ Cl2 88n PCl5
is 83.2%. What mass of PCl5 is expected from the reaction
of 73.7 g of PCl3 with excess chlorine?
037. The percent yield for the following reaction carried out in
carbon tetrachloride solution is 59.0%.
Br2 ϩ Cl2 88n 2BrCl
(a) What amount of BrCl is formed from the reaction of
0.0250 mol Br2 with 0.0250 mol Cl2?
(b) What amount of Br2 is left unchanged?
038. Solid silver nitrate undergoes thermal decomposition to
form silver metal, nitrogen dioxide, and oxygen. Write the
chemical equation for this reaction. A 0.443-g sample
of silver metal is obtained from the decomposition of a
0.784-g sample of AgNO3. What is the percent yield of
the reaction?
039. Tin(IV) chloride is produced in 85.0% yield by the reaction
of tin with chlorine. How much tin is required to produce
a kilogram of tin(IV) chloride?
040. Ethylene oxide, C2H4O, a fumigant sometimes used by exterminators,
is synthesized in 88.1% yield by reaction of
ethylene bromohydrin, C2H5OBr, with sodium hydroxide:
C2H5OBr ϩ NaOH 88n C2H4O ϩ NaBr ϩ H2O
How many grams of ethylene bromohydrin are consumed
in the production of 353 g of ethylene oxide, at 88.1%
yield?
041. If 15.0 g sodium carbonate is obtained from the thermal
decomposition of 75.0 g of sodium hydrogen carbonate,
2NaHCO3 88n Na2CO3 ϩ H2O ϩ CO2
what is the percent yield?
042. What is the percent yield if 112 mg SO2 is obtained from
the combustion of 78.1 mg of carbon disulfide according
to the reaction
CS2 ϩ 3O2 88n CO2 ϩ 2SO2?
043. From a 60.0-g sample of an iron ore containing Fe3O4,
2.09 g of Fe is obtained by the reaction
Fe3O4 ϩ 2C 88n 3Fe ϩ 2CO2
What is the percent of Fe3O4 in the ore?
044. The reaction of finely divided aluminum and iron(III) oxide,
Fe2O3, is called the thermite reaction. It produces a
tremendous amount of heat, making the welding of railroad
track possible. The reaction of 750. grams of aluminum
and 750. grams of iron(III) oxide produces 247.5
grams of iron.
(a) Calculate the mass of iron that should be released by
this reaction.
(b) What is the percent yield of iron?
Fe2O3 ϩ 2Al 88n 2Fe ϩ Al2O3 ϩ heat
045. Lime, Ca(OH)2, can be used to neutralize an acid spill. A
5.06-g sample of Ca(OH)2 reacts with an excess of hydrochloric
acid; 6.74 g of calcium chloride is collected.
What is the percent yield of this experiment?
Ca(OH)2 ϩ 2HCl 88n CaCl2 ϩ 2H2O
Sequential Reactions
046. Consider the two-step process for the formation of tellurous
acid described by the following equations:
TeO2 ϩ 2OHϪ 88n TeO3
2Ϫ ϩ H2O
TeO3
2Ϫ ϩ 2Hϩ 88n H2TeO3
What mass of H2TeO3 is formed from 95.2 g of TeO2, assuming
100% yield?
116 CHAPTER 3: Chemical Equations and Reaction Stoichiometry
047. Consider the formation of cyanogen, C2N2, and its subsequent
decomposition in water given by the equations
2Cu2ϩ ϩ 6CNϪ 88n 2[Cu(CN)2]Ϫ ϩ C2N2
C2N2 ϩ H2O 88n HCN ϩ HOCN
How much hydrocyanic acid, HCN, can be produced from
35.00 g of KCN, assuming 100% yield?
048. What mass of potassium chlorate is required to supply the
proper amount of oxygen needed to burn 44.2 g of
methane, CH4?
2KClO3 88n 2KCl ϩ 3O2
CH4 ϩ 2O2 88n CO2 ϩ 2H2O
049. Hydrogen, obtained by the electrical decomposition of water,
is combined with chlorine to produce 84.2 g of hydrogen
chloride. Calculate the mass of water decomposed.
2H2O 88n 2H2 ϩ O2
H2 ϩ Cl2 88n 2HCl
050. Ammonium nitrate, known for its use in agriculture, can
be produced from ammonia by the following sequence of
reactions:
NH3(g) ϩ O2(g) 88n NO(g) ϩ H2O(g)
NO(g) ϩ O2(g) 88n NO2(g)
NO2(g) ϩ H2O(ᐉ) 88n HNO3(aq) ϩ NO(g)
HNO3(aq) ϩ NH3(g) 88n NH4NO3(aq)
(a) Balance each equation.
(b) How many moles of nitrogen atoms are required for
every mole of ammonium nitrate (NH4NO3)?
(c) How much ammonia is needed to prepare 100.0 grams
of ammonium nitrate (NH4NO3)?
051. Calcium sulfate is the essential component of plaster and
sheet rock. Waste calcium sulfate can be converted into
quicklime, CaO, by reaction with carbon at high temperatures.
The following two reactions represent a sequence
of reactions that might take place:
CaSO4(s) ϩ 4C(s) 88n CaS(ᐉ) ϩ 4CO(g)
CaS(ᐉ) ϩ 3CaSO4(s) 88n 4CaO(s) ϩ 4SO2(g)
What weight of sulfur dioxide (in grams) could be obtained
from 1.000 kg of calcium sulfate?
*52. The Grignard reaction is a two-step reaction used to prepare
pure hydrocarbons. Consider the preparation of pure
ethane, CH3CH3, from ethyl chloride, CH3CH2Cl.
Step 1: CH3CH2Cl ϩ Mg 88n CH3CH2MgCl
Step 2: CH3CH2MgCl ϩ H2O 88n
CH3CH3 ϩ Mg(OH)Cl
We allow 27.2 g of CH3CH2Cl (64.4 g/mol) to react
with excess magnesium. From the first step reaction,
CH3CH2MgCl (88.7 g/mol) is obtained in 79.5% yield.
In the second step reaction, a 85.7% yield of CH3CH3
(30.0 g/mol) is obtained. What mass of CH3CH3 is ob-
tained?
*53. When sulfuric acid dissolves in water, the following reactions
take place:
H2SO4 88n Hϩ ϩ HSO4
Ϫ
HSO4
Ϫ 88n Hϩ ϩ SO4
2Ϫ
The first reaction is 100.0% complete, and the second
reaction is 10.0% complete. Calculate the concentrations
of the various ions in a 0.150 M aqueous solution of
H2SO4.
*54. The chief ore of zinc is the sulfide, ZnS. The ore is concentrated
by flotation and then heated in air, which converts
the ZnS to ZnO.
2ZnS ϩ 3O2 88n 2ZnO ϩ 2SO2
The ZnO is then treated with dilute H2SO4
ZnO ϩ H2SO4 88n ZnSO4 ϩ H2O
to produce an aqueous solution containing the zinc as
ZnSO4. An electric current is passed through the solution
to produce the metal.
2ZnSO4 ϩ 2H2O 88n 2Zn ϩ 2H2SO4 ϩ O2
What mass of Zn is obtained from an ore containing
225 kg of ZnS? Assume the flotation process to be 90.6%
efficient, the electrolysis step to be 92.2% efficient, and
the other steps to be 100% efficient.
Concentrations of Solutions—Percent by Mass
055. (a) How many moles of solute are contained in 500 g of
a 2.00% aqueous solution of K2Cr2O7?
(b) How many grams of solute are contained in the solution
of part (a)?
(c) How many grams of water (the solvent) are contained
in the solution of part (a)?
056. The density of an 18.0% solution of ammonium sulfate,
(NH4)2SO4, is 1.10 g/mL. What mass of (NH4)2SO4 is required
to prepare 275 mL of this solution?
057. The density of an 18.0% solution of ammonium chloride,
NH4Cl, solution is 1.05 g/mL. What mass of NH4Cl does
275 mL of this solution contain?
058. What volume of the solution of (NH4)2SO4 described in
Exercise 56 contains 90.0 g of (NH4)2SO4?
*59. A reaction requires 33.6 g of NH4Cl. What volume of the
solution described in Exercise 57 do you need if you wish
to use a 25.0% excess of NH4Cl?
Exercises 117
Concentrations of Solutions—Molarity
060. What is the molarity of a solution that contains 555 g of
phosphoric acid, H3PO4, in 3.00 L of solution?
061. What is the molarity of a solution that contains 4.50 g of
sodium chloride in 40.0 mL of solution?
062. How many grams of the cleansing agent Na3PO4 (a) are
needed to prepare 250 mL of 0.50 M solution, and (b) are
in 250 mL of 0.50 M solution?
063. How many kilograms of ethylene glycol, C2H6O2, are
needed to prepare a 9.50 M solution to protect a 14.0-L
car radiator against freezing? What is the mass of C2H6O2
in 14.0 L of 9.50 M solution?
064. A solution made by dissolving 16.0 g of CaCl2 in 64.0 g
of water has a density of 1.180 g/mL at 20°C.
(a) What is the percent by mass of CaCl2 in the solution?
(b) What is the molarity of CaCl2 in the solution?
065. A solution contains 0.100 mol/L of each of the following
acids: HCl, H2SO4, H3PO4.
(a) Is the molarity the same for each acid?
(b) Is the number of molecules per liter the same for each
acid?
(c) Is the mass per liter the same for each acid?
066. What is the molarity of a barium chloride solution prepared
by dissolving 1.50 g of BaCl2 и 2H2O in enough
water to make 600 mL of solution?
067. How many grams of potassium benzoate trihydrate,
KC7H5O2 и 3H2O, are needed to prepare 1 L of a
0.250 M solution of potassium benzoate?
068. Stock hydrofluoric acid solution is 49.0% HF and has a
specific gravity of 1.17. What is the molarity of the solu-
tion?
069. Stock phosphoric acid solution is 85.0% H3PO4 and has
a specific gravity of 1.70. What is the molarity of the
solution?
Dilution of Solutions
070. Commercial concentrated hydrochloric acid is 12.0 M
HCl. What volume of concentrated hydrochloric acid is
required to prepare 4.50 L of 2.25 M HCl solution?
071. Commercially available concentrated sulfuric acid is 18.0 M
H2SO4. Calculate the volume of concentrated sulfuric acid
required to prepare 4.50 L of 2.25 M H2SO4 solution.
072. Calculate the volume of 0.0600 M Ba(OH)2 solution that
contains the same number of moles of Ba(OH)2 as 225 mL
of 0.0900 M Ba(OH)2 solution.
073. Calculate the volume of 4.00 M NaOH solution required
to prepare 200 mL of a 0.800 M solution of NaOH.
074. Calculate the volume of pure water that dilutes 100. mL
of 12 M NaOH to 4.75 M.
075. In the laboratory preparation room is a reagent bottle that
contains 5.0 L of 12 M NaOH. Write a set of instructions
for the production of 250 mL of 3.0 M NaOH from the
12 M solution.
Using Solutions in Chemical Reactions
076. Calculate the volume of a 0.225 M solution of potassium
hydroxide, KOH, required to react with 0.215 g of acetic
acid, CH3COOH, according to the following reaction.
KOH ϩ CH3COOH 88n KCH3COO ϩ H2O
077. Calculate the number of grams of carbon dioxide, CO2,
that can react with 135 mL of a 0.357 M solution of potassium
hydroxide, KOH, according to the following reac-
tion.
2KOH ϩ CO2 88n K2CO3 ϩ H2O
078. What volume of 0.246 M HNO3 solution is required to
react completely with 38.6 mL of 0.0515 M Ba(OH)2?
Ba(OH)2 ϩ 2HNO3 88n Ba(NO3)2 ϩ 2H2O
079. What volume of 0.55 M HBr is required to react completely
with 0.80 mol of Ca(OH)2?
2HBr ϩ Ca(OH)2 88n CaBr2 ϩ 2H2O
080. An excess of AgNO3 reacts with 185.5 mL of an AlCl3 solution
to give 0.325 g of AgCl. What is the concentration,
in moles per liter, of the AlCl3 solution?
AlCl3 ϩ 3AgNO3 88n 3AgCl ϩ Al(NO3)3
081. An impure sample of solid Na2CO3 is allowed to react with
0.1755 M HCl.
Na2CO3 ϩ 2HCl 88n 2NaCl ϩ CO2 ϩ H2O
A 0.2337-g sample of sodium carbonate requires 15.55 mL
of HCl solution. What is the purity of the sodium car-
bonate?
082. Calculate the volume of 12 M HCl that just reacts with
15 g of aluminum.
083. What volume of 18 M H2SO4 do you need to measure in
order to neutralize 2 L of 5.35 M NaOH? The products
are sodium sulfate and water.
084. What volume of 0.0496 M HClO4 reacts with 35.9 mL of
a 0.505 M KOH solution according to the following re-
action?
KOH ϩ HClO4 88n KClO4 ϩ H2O
*85. What is the molarity of a solution of sodium hydroxide,
NaOH, if 36.9 mL of this solution is required to react with
29.2 mL of 0.101 M hydrochloric acid solution according
to the following reaction?
HCl ϩ NaOH 88n NaCl ϩ H2O
118 CHAPTER 3: Chemical Equations and Reaction Stoichiometry
*86. What is the molarity of a solution of sodium hydroxide,
NaOH, if 29.8 mL of this solution is required to react with
25.0 mL of 0.0513 M nitric acid solution according to the
following reaction?
HNO3 ϩ NaOH 88n NaNO3 ϩ H2O
Mixed Exercises
087. What mass of sulfuric acid can be obtained from 1.00 kg
of sulfur by the following series of reactions?
98% yield
S8 ϩ 8O2 888888888n 8SO2
96% yield
2SO2 ϩ O2 888888888n 2SO3
100% yield
SO3 ϩ H2SO4 888888888n H2S2O7
97% yield
H2S2O7 ϩ H2O 888888888n 2H2SO4
*88. What is the total mass of products formed when 38.8 g of
carbon disulfide is burned in air? What mass of carbon
disulfide must be burned to produce a mixture of carbon
dioxide and sulfur dioxide that has a mass of 54.2 g?
heat
CS2 ϩ 3O2 8888n CO2 ϩ 2SO2
*89. Iron(II) chloride, FeCl2, reacts with ammonia, NH3, and
water, H2O, to produce iron(II) hydroxide, Fe(OH)2, and
ammonium chloride, NH4Cl.
(a) Write the balanced equation for this reaction.
(b) We mix 78.5 g FeCl2, 25.0 g NH3, and 25.0 g H2O,
which then react as completely as possible. Which is
the limiting reactant?
(c) How many grams of ammonium chloride, NH4Cl, are
formed?
(d) How many grams of each of the two leftover reactants
remain at the completion of the reaction?
*90. An iron ore that contains Fe3O4 reacts according to the
reaction
Fe3O4 ϩ 2C 88n 3Fe ϩ 2CO2
We obtain 2.09 g of Fe from the reaction of 55.0 g of the
ore. What is the percent Fe3O4 in the ore?
*91. If 86.3% of the iron can be recovered from an ore that is
43.2% magnetic iron oxide, Fe3O4, what mass of iron
could be recovered from 2.00 kg of this ore? The reduction
of magnetic iron oxide is a complex process that can
be represented in simplified form as
Fe3O4 ϩ 4CO 88n 3Fe ϩ 4CO2
092. Gaseous chlorine will displace bromide ion from an aqueous
solution of potassium bromide to form aqueous potassium
chloride and aqueous bromine. Write the chemical
equation for this reaction. What mass of bromine is produced
if 0.381 g of chlorine undergoes reaction?
093. Calculate the volume of 2.50 M phosphoric acid solution
necessary to react with 45.0 mL of 0.150 M Mg(OH)2.
2H3PO4 ϩ 3Mg(OH)2 88n Mg3(PO4)2 ϩ 6H2O
CONCEPTUAL EXERCISES
094. Using your own words, give a definition of a chemical re-
action.
095. Magnesium burns with a bright, white flame in air.
(a) Write and balance the equation for the combination
of magnesium and oxygen.
(b) Explain the meaning of the numbers used to balance
the equation.
(c) How is a balanced equation an expression of the Law
of Conservation of Matter?
096. A properly written and balanced chemical reaction is critical
for the purposes of accurate communication. The oxidation
of beryllium is proposed to follow the equation
Be ϩ O 88n BeO
(a) Identify the basic error in the writing of this equation.
(b) Provide the balanced equation.
097. How would you prepare 1 L of 1.25 ϫ 10Ϫ6 M NaCl (molecular
weight ϭ 58.44 g/mol) solution by using a balance
that can measure mass only to 0.01 g?
098. The drawings shown below represent beakers of aqueous
solutions. Each sphere represents a dissolved solute parti-
cle.
(a) Which solution is most concentrated?
(b) Which solution is least concentrated?
(c) Which two solutions have the same concentration?
(d) When solutions E and F are combined, the resulting
solution has the same concentration as solution
______.
500. mL
Solution A
500. mL
Solution B
500. mL
Solution C
500. mL
Solution D
250. mL
Solution E
250. mL
Solution F
Exercises 119
099. You prepared a NaCl solution by adding 58.44 g of NaCl
to a 1-L volumetric flask and then adding water to dissolve
it. When finished, the final volume in your flask looked
like the illustration.
The solution you prepared is
(a) greater than 1 M because you added more solvent than
necessary.
(b) less than 1 M because you added less solvent than nec-
essary.
(c) greater than 1 M because you added less solvent than
necessary.
(d) less than 1 M because you added more solvent than
necessary.
(e) is 1 M because the amount of solute, not solvent, determines
the concentration.
For the reaction of the given sample, which of the following
is true?
(a) N2 is the limiting reactant.
(b) H2 is the limiting reactant.
(c) NH3 is the limiting reactant.
(d) No reactant is limiting; they are present in the correct
stoichiometric ratio.
BUILDING YOUR KNOWLEDGE
102. Acetic acid, CH3COOH, reacts with ethanol, CH3CH2OH,
to form ethyl acetate, CH3COOCH2CH3, (density ϭ
0.902 g/mL) by the following reaction.
CH3COOH ϩ CH3CH2OH 88n
CH3COOCH2CH3 ϩ H2O
We combine 20.2 mL of acetic acid with 20.1 mL of
ethanol.
(a) Which compound is the limiting reactant?
(b) If 27.5 mL of pure ethyl acetate is produced, what is
the percent yield? [Hint: See Tables 1-1 and 1-8.]
103. Concentrated hydrochloric acid solution is 37.0% HCl
and has a density of 1.19 g/mL. A dilute solution of HCl
100. Zinc is more active chemically than is silver; it may be used
to remove ionic silver from solution.
Zn(s) ϩ 2AgNO3(aq) 88n Zn(NO3)2(aq) ϩ 2Ag(s)
The concentration of a silver nitrate solution is determined
to be 1.330 mol/L. Pieces of zinc totaling 100.0 g
are added to 1.000 L of the solution; 90.0 g of silver is
collected.
(a) Calculate the percent yield of silver.
(b) Suggest a reason why the yield is less than 100.0%.
101. Ammonia is formed in a direct reaction of nitrogen and
hydrogen.
N2(g) ϩ 3H2(g) 88n 2NH3(g)
A tiny portion of the starting mixture is represented by the
diagram, where the blue spheres represent N and the white
spheres represent H. Which of the following represents
the product mixture?
43
21
5 6
120 CHAPTER 3: Chemical Equations and Reaction Stoichiometry
is prepared by diluting 4.50 mL of this concentrated
HCl solution to 100.00 mL with water. Then 10.0 mL of
this dilute HCl solution reacts with an AgNO3 solution
according to the following reaction.
HCl(aq) ϩ AgNO3(aq) 88n HNO3(aq) ϩ AgCl(s)
How many milliliters of 0.108 M AgNO3 solution is
required to precipitate all of the chloride as AgCl(s)?
104. In a particular experiment, 272 g of phosphorus, P4,
reacted with excess oxygen to form tetraphosphorus
decoxide, P4O10, in 89.5% yield. In the second step
reaction, a 97.8% yield of H3PO4 was obtained.
(a) Write the balanced equations for these two reaction
steps.
(b) What mass of H3PO4 was obtained?
105. Magnesium displaces copper from a dilute solution of copper(II)
sulfate; the pure copper will settle out of the solu-
tion.
Mg(s) ϩ CuSO4(aq) 88n MgSO4(aq) ϩ Cu(s)
A copper(II) sulfate solution is mixed by dissolving 25.000 g
of copper(II) sulfate, and then it is treated with an excess
of magnesium metal. The mass of copper collected is 8.786
g after drying. Calculate the percent yield of copper.
106. Suppose you are designing an experiment for the preparation
of hydrogen. For the production of equal amounts
of hydrogen, which metal, Zn or Al, is less expensive if Zn
costs about half as much as Al on a mass basis?
Zn ϩ 2HCl 88n ZnCl2 ϩ H2
2Al ϩ 6HCl 88n 2AlCl3 ϩ 3H2
107. Gaseous chlorine and gaseous fluorine undergo a combination
reaction to form the interhalogen compound
ClF.
(a) Write the chemical equation for this reaction.
(b) Calculate the mass of fluorine needed to react with
3.47 g of Cl2.
(c) How many grams of ClF are formed?
108. It has been estimated that 93% of all atoms in the entire
universe are hydrogen and that the vast majority of those
remaining are helium. Based on only these two elements
and the above-mentioned values, estimate the mass percentage
composition of the universe.
4Some Types of
Chemical Reactions
OUTLINE
4-1 The Periodic Table: Metals,
Nonmetals, and Metalloids
4-2 Aqueous Solutions: An
Introduction
1 Electrolytes and Extent of
Ionization
2 Strong and Weak Acids
3 Reversible Reactions
4 Strong Bases, Insoluble Bases,
and Weak Bases
5 Solubility Guidelines for
Compounds in Aqueous Solution
4-3 Reactions in Aqueous Solutions
4-4 Oxidation Numbers
Classifying Chemical Reactions
4-5 Oxidation-Reduction Reactions:
An Introduction
4-6 Combination Reactions
1 Element ϩ Element n Compound
2 Compound ϩ Element n Compound
3 Compound ϩ Compound n Compound
4-7 Decomposition Reactions
1 Compound n Element ϩ Element
2 Compound n Element ϩ Compound
3 Compound n Compound ϩ Compound
4-8 Displacement Reactions
1 More Active Metal ϩ
Salt of Less Active 88n
Metal
Less Active Metal ϩ
Salt of More Active Metal
2 Active Metal ϩ
Nonoxidizing Acid
88n
Hydrogen ϩ
Salt of Acid
3 Active Nonmetal ϩ
Salt of Less Active 88n
Nonmetal
Less Active Nonmetal ϩ
Salt of More Active
Nonmetal
4-9 Metathesis Reactions
1 Acid–Base (Neutralization)
Reactions: Formation of a
Nonelectrolyte
2 Precipitation Reactions
3 Gas-Formation Reactions
4-10 Summary of Reaction Types
Naming Inorganic Compounds
4-11 Naming Binary Compounds
4-12 Naming Ternary Acids and
Their Salts
OBJECTIVES
After you have studied this chapter, you should be able to
• Describe the periodic table and some of the relationships that it summarizes
• Recognize and describe nonelectrolytes, strong electrolytes, and weak electrolytes
• Recognize and classify acids (strong, weak), bases (strong, weak, insoluble), and salts
(soluble, insoluble); use the solubility guidelines
• Describe reactions in aqueous solutions by writing formula unit equations, total ionic
equations, and net ionic equations
• Assign oxidation numbers to elements when they are free, in compounds, or in ions
• Recognize oxidation–reduction reactions and identify which species are oxidized,
reduced, oxidizing agents, and reducing agents
• Recognize and describe classes of reactions: decomposition reactions, displacement
reactions, and various types of metathesis reactions
• Name and write formulas for common binary and ternary inorganic compounds










































The active metal sodium, Na, reacts
vigorously with water.
2Na(s) ϩ 2H2O(l) 88n
H2(g) ϩ 2NaOH(aq)
Heat released by the reaction causes
unreacted sodium to give off a
characteristic yellow glow and
ignites the hydrogen gas that is
formed.
2H2(g) ϩ O2(g) 88n 2H2O(g)
II
n this chapter we examine some types of chemical reactions. Millions of reactions are
known, so it is useful to group them into classes, or types, so that we can deal systematically
with these massive amounts of information. We will describe how some
compounds behave in aqueous solution, including how well their solutions conduct electricity
and whether or not the compounds dissolve in water. We introduce several ways
to represent chemical reactions in aqueous solution—formula unit equations, total ionic
equations, and net ionic equations—and the advantages and disadvantages of these methods.
Let us first take a brief look at the periodic table, which helps us to organize many
properties of the elements, including their chemical reactions.
THE PERIODIC TABLE: METALS, NONMETALS, AND
METALLOIDS
In 1869, the Russian chemist Dmitri Mendeleev (1834–1907) and the German chemist
Lothar Meyer (1830–1895) independently published arrangements of known elements
that are much like the periodic table in use today. Mendeleev’s classification was based
primarily on chemical properties of the elements, whereas Meyer’s classification was based
largely on physical properties. The tabulations were surprisingly similar. Both emphasized
the periodicity, or regular periodic repetition, of properties with increasing atomic weight.
Mendeleev arranged the known elements in order of increasing atomic weight in successive
sequences so that elements with similar chemical properties fell in the same column.
He noted that both physical and chemical properties of the elements vary in a periodic
fashion with atomic weight. His periodic table of 1872 contained the 62 known elements
(Figure 4-1). Mendeleev placed H, Li, Na, and K in his table as “Gruppe I.” These were
4-1
Pronounced “men-del-lay-ev.”
᭣ Three of the halogens, elements
from Group VIIA (left to right):
chlorine, bromine, iodine.
Silicon (top), germanium (middle),
and tin (bottom), three elements from
Group IVA.
known to combine with F, Cl, Br, and I of “Gruppe VII” to produce compounds that have
similar formulas such as HF, LiCl, NaCl, and KI. All these compounds dissolve in water
to produce solutions that conduct electricity. The “Gruppe II” elements were known to
form compounds such as BeCl2, MgBr2, and CaCl2, as well as compounds with O and S
from “Gruppe VI” such as MgO, CaO, MgS, and CaS. These and other chemical properties
led him to devise a table in which the elements were arranged by increasing atomic
weights and grouped into vertical families.
In most areas of human endeavor progress is slow and faltering. Occasionally, however
an individual develops concepts and techniques that clarify confused situations. Mendeleev
was such an individual. One of the brilliant successes of his periodic table was that it
provided for elements that were unknown at the time. When he encountered “missing”
elements, Mendeleev left blank spaces. Some appreciation of his genius in constructing
the table as he did can be gained by comparing the predicted (1871) and observed properties
of germanium, which was not discovered until 1886. Mendeleev called the
undiscovered element eka-silicon because it fell below silicon in his table. He was familiar
with the properties of germanium’s neighboring elements. They served as the basis for
his predictions of properties of germanium (Table 4-1). Some modern values for properties
of germanium differ significantly from those reported in 1886. But many of the values
on which Mendeleev based his predictions were also inaccurate.
Because Mendeleev’s arrangement of the elements was based on increasing atomic
weights, several elements would have been out of place in his table. Mendeleev put the
controversial elements (Te and I, Co and Ni) in locations consistent with their properties,
however. He thought the apparent reversal of atomic weights was due to inaccurate
values for those weights. Careful redetermination showed that the values were correct.
Explanation of the locations of these “out-of-place” elements had to await the development
of the concept of atomic number, approximately 50 years after Mendeleev’s work. The
atomic number (Section 5-5) of an element is the number of protons in the nucleus of
its atoms. (It is also the number of electrons in a neutral atom of an element). This quantity
is fundamental to the identity of each element because it is related to the electrical
make-up of atoms. Elements are arranged in the periodic table in order of increasing
atomic number. With the development of this concept, the periodic law attained essentially
its present form:
Copper is drawn into wire, which is
then collected into cables for use as
an electric conductor.
REIHEN
1 H ϭ 1
2
3
4
5
6
7
8
9
10
11
12
Li ϭ 7 Be ϭ 9,4 B ϭ 11 C ϭ 12
Na ϭ 23 Mg ϭ 24 Al ϭ 27,3 Si ϭ 28 P ϭ 31
K ϭ 39 Ca ϭ 40 – ϭ 44 Ti ϭ 48 V ϭ 51
GRUPPE I
R2
O
GRUPPE II
RO
GRUPPE III
R2
O3
GRUPPE V
R2
O5
GRUPPE IV
RO2
RH4
GRUPPE VI
RO3
RH2
RH3
GRUPPE VII
R2
O7
RH
GRUPPE VIII
RO4
N ϭ 14 O ϭ 16 F ϭ 19
S ϭ 32 Cl ϭ 35,5
Cr ϭ 52 Mn ϭ 55 Fe ϭ 56, Co ϭ 59,
Ni ϭ 59, Cu ϭ 63.
(Cu ϭ 63) Zn ϭ 65 – ϭ 68 – ϭ 72 As ϭ 75
Sb ϭ 122
Bi ϭ 208
Rb ϭ 85
(Ag ϭ 108)
(–)
Cs ϭ 133
(Au ϭ 199)
–
–
– – – –
– –
–
– – –
– – – – – – –
– – – –
–
––
–
–
–
– –
Sr ϭ 87
Cd ϭ 112
Ba ϭ 137
Hg ϭ 200
?Yt ϭ 88
In ϭ 113
Tl ϭ 204
?Di ϭ 138
?Er ϭ 178
Zr ϭ 90
Sn ϭ 118
?Ce ϭ 140
?La ϭ 180 Ta ϭ 182 W ϭ 184
Pb ϭ 207
Th ϭ 231
Nb ϭ 94
Se ϭ 78
Te ϭ 125
Br ϭ 80
J ϭ 127
Mo ϭ 96
U ϭ 240
– ϭ 100 Ru ϭ 104, Rh ϭ 104,
Pd ϭ 106, Ag ϭ 108.
Os ϭ 195, Ir ϭ 197,
Pt ϭ 198, Au ϭ 199.
Figure 4-1 Mendeleev’s early periodic table (1872). “J” is the German symbol for iodine.
The properties of the elements are periodic functions of their atomic numbers.
The periodic law tells us that if we arrange the elements in order of increasing atomic
number, we periodically encounter elements that have similar chemical and physical properties.
The presently used “long form” of the periodic table (Table 4-2 and inside the
front cover) is such an arrangement. The vertical columns are referred to as groups or
families, and the horizontal rows are called periods. Elements in a group have similar
chemical and physical properties, and those within a period have properties that change
progressively across the table. Several groups of elements have common names that are
used so frequently they should be learned. The Group IA elements, except H, are referred
to as alkali metals, and the Group IIA elements are called the alkaline earth metals.
The Group VIIA elements are called halogens, which means “salt formers,” and the
Group VIIIA elements are called noble (or rare) gases.
The general properties of metals and nonmetals are distinct. Physical and chemical
properties that distinguish metals from nonmetals are summarized in Tables 4-3 and 4-4.
Not all metals and nonmetals possess all these properties, but they share most of them to
varying degrees. The physical properties of metals can be explained on the basis of metallic
bonding in solids (Section 13-17).
Table 4-2, The Periodic Table, shows how we classify the known elements as metals
(shown in blue), nonmetals (tan), and metalloids (green). The elements to the left of those
Alkaline means basic. The character
of basic compounds is described in
Section 10-4.
About 80% of the elements are metals.
4-1 The Periodic Table: Metals, Nonmetals, and Metalloids 125
TABLE 4-1 Predicted and Observed Properties of Germanium
Eka-Silicon Germanium
Property Predicted, 1871 Reported, 1886 Modern Values
Atomic weight 72 72.32 72.61
Atomic volume 13 cm3 13.22 cm3 13.5 cm3
Specific gravity 5.5 5.47 5.35
Specific heat 0.073 cal/g°C 0.076 cal/g°C 0.074 cal/g°C
Maximum valence* 4 4 4
Color Dark gray Grayish white Grayish white
Reaction with water Will decompose steam Does not decompose water Does not decompose
with difficulty water
Reactions with acids and Slight with acids; Not attacked by HCl or Not dissolved by HCl or
alkalis more pronounced dilute aqueous NaOH; H2SO4 or dilute
with alkalis reacts vigorously NaOH; dissolved by
with molten NaOH concentrated NaOH
Formula of oxide EsO2 GeO2 GeO2
Specific gravity of oxide 4.7 4.703 4.228
Specific gravity of 1.9 at 0°C 1.887 at 18°C 1.8443 at 30°C
tetrachloride
Boiling point of 100°C 86°C 84°C
tetrachloride
Boiling point of 160°C 160°C 186°C
tetraethyl derivative
*“Valence” refers to the combining power of a specific element.
126 CHAPTER 4: Some Types of Chemical Reactions
3
Li
11
Na
19
K
37
Rb
55
Cs
87
Fr
12
Mg
20
Ca
38
Sr
56
Ba
88
Ra
21
Sc
39
Y
57
La
89
Ac
22
Ti
40
Zr
72
Hf
104
Rf
23
V
41
Nb
73
Ta
105
Db
24
Cr
42
Mo
74
W
106
Sg
25
Mn
43
Tc
75
Re
107
Bh
26
Fe
44
Ru
76
Os
108
Hs
27
Co
45
Rh
77
Ir
109
Mt
28
Ni
46
Pd
78
Pt
29
Cu
47
Ag
79
Au
30
Zn
48
Cd
80
Hg
5
B
13
Al
31
Ga
49
In
81
Tl
6
C
14
Si
32
Ge
50
Sn
82
Pb
7
N
15
P
33
As
51
Sb
83
Bi
8
O
16
S
34
Se
52
Te
84
Po
9
F
17
Cl
35
Br
53
I
85
At
10
Ne
18
Ar
36
Kr
54
Xe
86
Rn
1
H
4
Be
2
He
VIIA
VIIIA
(17)
VIA
(16)
VA
(15)
IVA
(14)
IIIA
(13)
IIB
(12)
IB
(11)(10)(9)(8)
VIIB
(7)
VIB
(6)
VB
(5)
IVB
(4)
IIIB
(3)
IIA
(2)
IA
(1)
VIIIB
Transition metals
58
Ce
59
Pr
60
Nd
61
Pm
62
Sm
63
Eu
64
Gd
65
Tb
66
Dy
67
Ho
68
Er
69
Tm
70
Yb
71
Lu
90
Th
91
Pa
92
U
93
Np
94
Pu
95
Am
96
Cm
97
Bk
98
Cf
99
Es
100
Fm
101
Md
102
No
103
Lr
Alkali
metals
Alkaline
earth
metals
(18)
Noble
gases
Halogens
Metals
Nonmetals
Metalloids
1
2
3
4
5
6
7
*
†
*
†
TABLE 4-2 The Periodic Table
There are other systems for numbering the groups in the periodic table. We number the groups by the standard American system of A and B
groups. An alternative system in which the groups are numbered 1 through 18 is shown in parentheses.
TABLE 4-3 Some Physical Properties of Metals and Nonmetals
Metals Nonmetals
1. High electrical conductivity that 1. Poor electrical conductivity (except
1. decreases with increasing temperature 1. carbon in the form of graphite)
2. High thermal conductivity 2. Good heat insulators (except carbon in
1. the form of diamond)
3. Metallic gray or silver luster* 3. No metallic luster
4. Almost all are solids† 4. Solids, liquids, or gases
5. Malleable (can be hammered into sheets) 5. Brittle in solid state
6. Ductile (can be drawn into wires) 6. Nonductile
*Except copper and gold.
†Except mercury; cesium and gallium melt in a protected hand.
General trends in metallic character
of A group elements with position in
the periodic table.
4-1 The Periodic Table: Metals, Nonmetals, and Metalloids 127
touching the heavy stairstep line are metals (except hydrogen), and those to the right are
nonmetals. Such a classification is somewhat arbitrary, and several elements do not fit
neatly into either class. Most elements adjacent to the heavy line are often called metalloids
(or semimetals), because they are metallic (or nonmetallic) only to a limited degree.
Metallic character increases from top to bottom and decreases from left to right with
respect to position in the periodic table.
Cesium, atomic number 55, is the most active naturally occurring metal. Francium and
radium are radioactive and do not occur in nature in appreciable amounts. Noble gases
seldom bond with other elements. They are unreactive, monatomic gases. The most active
nonmetal is fluorine, atomic number 9.
Nonmetallic character decreases from top to bottom and increases from left to right
in the periodic table.
Metalloids show some properties that are characteristic of both metals and nonmetals.
Many of the metalloids, such as silicon, germanium, and antimony, act as semiconductors,
which are important in solid-state electronic circuits. Semiconductors are insulators at
lower temperatures, but become conductors at higher temperatures (Section 13-17). The
conductivities of metals, by contrast, decrease with increasing temperature.
Aluminum is the least metallic of the metals and is sometimes classified as a metalloid.
It is metallic in appearance, and an excellent conductor of electricity.
In this and later chapters we will study some chemical reactions of elements and their
compounds and relate the reactions to the locations of the elements in the periodic table.
First, we will describe some important properties of solutions and what they tell us about
the nature and behavior of the dissolved substances, the solutes.
TABLE 4-4 Some Chemical Properties of Metals and Nonmetals
Metals Nonmetals
1. Outer shells contain few electrons— 1. Outer shells contain four or more
2. usually three or fewer 2. electrons*
2. Form cations (positive ions) by losing 2. Form anions (negative ions) by gaining
2. electrons 2. electrons†
3. Form ionic compounds with 3. Form ionic compounds with metals† and
2. nonmetals 2. molecular (covalent) compounds with other
2. nonmetals
4. Solid state characterized by metallic 4. Covalently bonded molecules; noble
2. bonding 2. gases are monatomic
*Except hydrogen and helium.
†Except the noble gases.
Aluminum is the most abundant metal
in the earth’s crust (7.5% by mass).
Silicon, a metalloid, is widely used in
the manufacture of electronic chips.
Decrease
Increase
Increase
Metallic character
The Development of ScienceCCHEMISTRY IN USE
The Discovery of Phosphorus
Technology and its impact on society have always been
intriguing subjects for artists. This was particularly true during
the industrial revolution, when chemistry was on the
verge of transforming itself from alchemical “magic” into a
scientific discipline. It is easy to see how the scientist, toiling
away in a laboratory full of strange equipment and trying
to make sense of the natural world, held a certain heroic
appeal to artists.
One of the more romantic accounts of chemical activity
during that period is The Alchymist in Search of the Philosophers’
Stone Discovers Phosphorus (1771) by the English painter
Joseph Wright of Derby (1734–1797). In Wright’s depiction,
a stately, bearded alchemist has just isolated a new element
by distillation. As the substance collects in the flask it begins
to glow in the dark, illuminating the laboratory with an eerie
white light and bringing the imaginary scientist to his knees
in wonder. The element phosphorus was in fact named for
this property—phosphorescence—with both words deriving
from the Greek phosphoros, or “giving light.”
The actual discovery of elemental phosphorus was probably
not quite as dramatic as Joseph Wright envisioned. It
was first isolated from urine by the German chemist Henning
Brand in 1669, by a much more laborious process than
the one represented by the tidy distillation apparatus in
Wright’s painting. The first step of the preparation, as
described in a 1726 treatise entitled “Phosphoros Elementalis,”
in fact involved steeping 50 or 60 pails of urine in tubs
for two weeks “. . . till it putrify and breed Worms”—hardly
a fitting subject for eighteenth century artwork!
The glowing material was of such novelty that two of
Brand’s scientific contemporaries offered to find a royal buyer
for his process. Expecting a bigger reward at a later date,
Brand gave the two the recipe for phosphorus in exchange
for some small gifts. However, one man instead claimed the
discovery for himself after repeating Brand’s work in his own
laboratory. Through the other, Brand did receive a contract
with the Duke of Hanover for the preparation of phosphorus;
however, he was dissatisfied with his pay, and it was only
after writing a number of complaint letters (and enlisting his
wife to do the same) that he finally received what he felt was
fair compensation for his discovery.
A number of other eighteenth-century scientific tableaux
were immortalized by Wright. He was particularly fascinated
by light and shadow effects. This, combined with his interest
in technological subjects (the town of Derby played an
important part in the beginnings of the industrial revolution),
led him to use other unusual objects, such as glowing iron
ingots (Iron Forge, 1772) and laboratory candles (Experiment
on a Bird in an Air Pump, 1786), as focal points in paintings
of industrial or scientific scenes.
Lisa S. Boffa
Senior Chemist
Exxon Corporation
The Alchymist in Search of the Philosophers’ Stone Discovers
Phosphorus, by Joseph Wright (1771).
128 CHAPTER 4: Some Types of Chemical Reactions
AQUEOUS SOLUTIONS: AN INTRODUCTION
Approximately three fourths of the earth’s surface is covered with water. The body fluids
of all plants and animals are mainly water. Thus we can see that many important chemical
reactions occur in aqueous (water) solutions, or in contact with water. In Chapter 3,
we introduced solutions and methods of expressing concentrations of solutions. It is useful
to know the kinds of substances that are soluble in water, and the forms in which they
exist, before we begin our systematic study of chemical reactions.
1 Electrolytes and Extent of Ionization
Solutes that are water-soluble can be classified as either electrolytes or nonelectrolytes.
Electrolytes are substances whose aqueous solutions conduct electric current. Strong
electrolytes are substances that conduct electricity well in dilute aqueous solution. Weak
electrolytes conduct electricity poorly in dilute aqueous solution. Aqueous solutions of
nonelectrolytes do not conduct electricity. Electric current is carried through aqueous
solution by the movement of ions. The strength of an electrolyte depends on the number
of ions in solution and also on the charges on these ions (Figure 4-2).
4-2
4-2 Aqueous Solutions: An Introduction 129
Figure 4-2 An experiment to demonstrate the presence of ions in solution. Two copper
electrodes are dipped into a liquid in a beaker. When the liquid contains significant
concentrations of ions, the ions move between the electrodes to complete the circuit (which
includes a light bulb). (a) Pure water and sugar are nonelectrolytes. (b) A solution of a weak
electrolyte, acetic acid (CH3COOH); it contains low concentrations of ions, and so the bulb
glows dimly. (c) A solution of a strong electrolyte, potassium chromate (K2CrO4); it contains
a high concentration of ions, and so the bulb glows brightly.
(a) (b) (c)
H2O molecule Sugar molecule Hydrogen ion,
H+
Acetic acid molecule,
CH3COOH
Acetate ion,
CH3COO–
Potassium ion,
K+
Chromate ion,
CrO2–
4
130 CHAPTER 4: Some Types of Chemical Reactions
Dissociation refers to the process in which a solid ionic compound, such as NaCl, separates
into its ions in solution:
H2O
NaCl(s) 8888n Naϩ(aq) ϩ ClϪ(aq)
Molecular compounds, for example pure HCl, exist as discrete molecules and do not
contain ions; however, many such compounds form ions in solution. Ionization refers to
the process in which a molecular compound separates or reacts with water to form ions in
solution:
H2O
HCl(g) 8888n Hϩ(aq) ϩ ClϪ(aq)
Three major classes of solutes are strong electrolytes: (1) strong acids, (2) strong
bases, and (3) most soluble salts. These compounds are completely or nearly completely
ionized (or dissociated) in dilute aqueous solutions, and therefore are strong electrolytes.
An acid can be defined as a substance that produces hydrogen ions, Hϩ, in aqueous
solutions. We usually write the formulas of inorganic acids with hydrogen written first.
Organic acids can often be recognized by the presence of the COOH group in the formula.
A base is a substance that produces hydroxide ions, OHϪ, in aqueous solutions. A salt is
a compound that contains a cation other than Hϩ and an anion other than hydroxide ion,
OHϪ, or oxide ion, O2Ϫ (see Table 2-3 on page 55). As we will see later in this chapter,
salts are formed when acids react with bases.
2 Strong and Weak Acids
As a matter of convenience we place acids into two classes: strong acids and weak acids.
Strong acids ionize (separate into hydrogen ions and stable anions) completely, or very
nearly completely, in dilute aqueous solution. The seven common strong acids and their
anions are listed in Table 4-5. Please learn this short list; you can then assume that other
acids you encounter are weak.
TABLE 4-5 Common Strong Acids and Their Anions
Common Strong Acids Anions of These Strong Acids
Formula Name Formula Name
HCl hydrochloric acid ClϪ chloride ion
HBr hydrobromic acid BrϪ bromide ion
HI hydroiodic acid IϪ iodide ion
HNO3 nitric acid NO3
Ϫ nitrate ion
HClO4 perchloric acid ClO4
Ϫ perchlorate ion
HClO3 chloric acid ClO3
Ϫ chlorate ion
HSO4
Ϫ hydrogen sulfate ion
H2SO4 sulfuric acid
SO4
2Ϫ sulfate ion



Recall that ions are charged particles.
The movement of charged particles
conducts electricity.
In Chapter 10, we will see that it is
appropriate to represent Hϩ(aq) as
H3Oϩ to emphasize its interaction
with water.
Acids and bases are further identified
in Subsections 2, 3, and 4. They are
discussed in more detail in Chapter 10.
Positively charged ions are called
cations, and negatively charged ions are
called anions (Section 2-3). The
formula for a salt may include H or
OH, but it must contain another cation
and another anion. For example,
NaHSO4 and Al(OH)2Cl are salts.
Many properties of aqueous solutions
of acids are due to Hϩ(aq) ions. These
are described in Section 10-4.
To give a more complete description of
reactions, we indicate the physical
states of reactants and products: (g) for
gases, (ᐉ) for liquids, and (s) for solids.
The notation (aq) following ions
indicates that they are hydrated in
aqueous solution; that is, they interact
with water molecules in solution. The
complete ionization of a strong
electrolyte is indicated by a single
arrow (n).
Acetic acid is the most familiar organic
acid.
Our stomachs have linings that are
much more resistant to attack by acids
than are our other tissues.
The carboxylate group XCOOH is
4-2 Aqueous Solutions: An Introduction 131
Because strong acids ionize completely or very nearly completely in dilute solutions,
their solutions contain predominantly ions rather than acid molecules. Consider the ionization
of hydrochloric acid. Pure hydrogen chloride, HCl, is a molecular compound that is
a gas at room temperature and atmospheric pressure. When it dissolves in water, it reacts
nearly 100% to produce a solution that contains hydrogen ions and chloride ions:
H2O
HCl(g) 8888n Hϩ(aq) ϩ ClϪ(aq) (to completion)
Similar equations can be written for all strong acids.
Weak acids ionize only slightly (usually less than 5%) in dilute aqueous solution. Some
common weak acids are listed in Appendix F. Several of them and their anions are given
in Table 4-6.
The equation for the ionization of acetic acid, CH3COOH, in water is typical of weak
acids:
CH3COOH(aq) 34 Hϩ(aq) ϩ CH3COOϪ(aq) (reversible)
The double arrow 34 generally signifies that the reaction occurs in both directions and
that the forward reaction does not go to completion. All of us are familiar with solutions
of acetic acid. Vinegar is 5% acetic acid by mass. Our use of oil and vinegar as a salad
dressing suggests that acetic acid is a weak acid; we could not safely drink a 5% solution
of any strong acid. To be specific, acetic acid is 0.5% ionized (and 99.5% nonionized) in
5% solution.
A multitude of organic acids occur in living systems. Organic acids contain the carboxylate
grouping of atoms, XCOOH. Most common organic acids are weak. They can ionize
slightly by breaking the OXH bond, as shown on the following page for acetic acid:
TABLE 4-6 Some Common Weak Acids and Their Anions
Common Weak Acids Anions of These Weak Acids
Formula Name Formula Name
HF* hydrofluoric acid FϪ fluoride ion
CH3COOH acetic acid CH3COOϪ acetate ion
HCN hydrocyanic acid CNϪ cyanide ion
HNO2
† nitrous acid NO2
Ϫ nitrite ion
HCO3
Ϫ hydrogen carbonate ion
H2CO3
† carbonic acid
CO3
2Ϫ carbonate ion
HSO3
Ϫ hydrogen sulfite ion
H2SO3
† sulfurous acid
SO3
2Ϫ sulfite ion
H2PO4
Ϫ dihydrogen phosphate ion
H3PO4 phosphoric acid HPO4
2Ϫ hydrogen phosphate ion
PO4
3Ϫ phosphate ion
H(COO)2
Ϫ hydrogen oxalate ion
(COOH)2 oxalic acid
(COO)2
2Ϫ oxalate ion
*HF is a weak acid, whereas HCl, HBr, and HI are strong acids.
†Free acid molecules exist only in dilute aqueous solution or not at all. Many salts of these acids are common, stable
compounds, however.














Citrus fruits contain citric acid, so
their juices are acidic. This is shown
here by the color changes on the
indicator paper. Acids taste sour.
The names given here correspond to
the aqueous solutions.
O H
C
O
Many common food and household
products are acidic (orange juice,
vinegar, soft drink, citrus fruits) or
basic (cleaning preparations, baking
soda).
132 CHAPTER 4: Some Types of Chemical Reactions
Organic acids are discussed in Chapter 27. Some naturally occurring organic weak acids
are tartaric acid (grapes), lactic acid (sour milk), and formic acid (ants). Carbonic acid,
H2CO3, and hydrocyanic acid, HCN(aq), are two common acids that contain carbon but
that are considered to be inorganic acids. Inorganic acids are often called mineral acids
because they are obtained primarily from nonliving sources.
EXAMPLE 4-1 Strong and Weak Acids
In the following lists of common acids, which are strong and which are weak? (a) H3PO4,
HCl, H2CO3, HNO3; (b) HClO4, H2SO4, HClO, HF.
Plan
We recall that Table 4-5 lists the common strong acids. Other common acids are assumed
to be weak.
Solution
(a) HCl and HNO3 are strong acids; H3PO4 and H2CO3 are weak acids.
(b) HClO4 and H2SO4 are strong acids; HClO and HF are weak acids.
You should now work Exercises 17 and 19.
3 Reversible Reactions
Reactions that can occur in both directions are reversible reactions. We use a double
arrow (34) to indicate that a reaction is reversible. What is the fundamental difference
between reactions that go to completion and those that are reversible? We have seen that
the ionization of HCl in water is nearly complete. Suppose we dissolve some table salt,
NaCl, in water and then add some dilute nitric acid to it. The resulting solution contains
Naϩ and ClϪ ions (from the dissociation of NaCl) as well as Hϩ and NO3
Ϫ (from the
ionization of HNO3). The Hϩ and ClϪ ions do not react significantly to form nonionized
HCl molecules; this would be the reverse of the ionization of HCl.
Hϩ(aq) ϩ ClϪ(aq) 88n no reaction
In contrast, when a sample of sodium acetate, NaCH3COO, is dissolved in H2O and
mixed with nitric acid, the resulting solution initially contains Naϩ, CH3COOϪ, Hϩ, and
NO3
Ϫ ions. But most of the Hϩ and CH3COOϪ ions combine to produce nonionized
molecules of acetic acid, the reverse of the ionization of the acid. Thus, the ionization of
acetic acid, like that of any other weak electrolyte, is reversible.
Hϩ(aq) ϩ CH3COOϪ(aq) 34 CH3COOH(aq) (reversible)
4 Strong Bases, Insoluble Bases, and Weak Bases
Most common bases are ionic metal hydroxides. Strong bases are soluble in water and
are dissociated completely in dilute aqueous solution. The common strong bases are listed
H3C C (aq) (aq) ϩ
O
O H
H3C C
O
OϪ
Hϩ(aq)
Inorganic acids may be strong or weak.
Naϩ and NO3
Ϫ ions do not combine
because NaNO3 is a soluble ionic
compound.
Solutions of bases have a set of
common properties due to the OHϪ
ion. These are described in Section
10-4.
Other organic acids have other groups
in the position of the H3CX group in
acetic acid.
Strong bases are ionic compounds in
the solid state.
The weak bases are molecular
substances that dissolve in water to
give slightly basic solutions; they are
sometimes called molecular bases.
4-2 Aqueous Solutions: An Introduction 133
in Table 4-7. They are the hydroxides of the Group IA metals and the heavier members
of Group IIA. The equation for the dissociation of sodium hydroxide in water is typical.
Similar equations can be written for other strong bases.
H2O
NaOH(s) 8888n Naϩ(aq) ϩ OHϪ(aq) (to completion)
Other metals form ionic hydroxides, but these are so sparingly soluble in water that
they cannot produce strongly basic solutions. They are called insoluble bases or sometimes
sparingly soluble bases. Typical examples include Cu(OH)2, Zn(OH)2, Fe(OH)2,
and Fe(OH)3.
Common weak bases are molecular substances that are soluble in water but form only
low concentrations of ions in solution. The most common weak base is ammonia, NH3.
NH3(aq) ϩ H2O(ᐉ) 34 NH4
ϩ(aq) ϩ OHϪ(aq) (reversible)
Closely related N-containing compounds, the amines, such as methylamine, CH3NH2,
and aniline, C6H5NH2, are also weak bases. Nicotine (found in tobacco) and caffeine
(found in coffee, tea, and cola drinks) are naturally occurring amines.
EXAMPLE 4-2 Classifying Bases
From the following lists, choose (i) the strong bases, (ii) the insoluble bases, and (iii) the weak
bases. (a) NaOH, Cu(OH)2, Pb(OH)2, Ba(OH)2; (b) Fe(OH)3, KOH, Mg(OH)2, Sr(OH)2,
NH3.
Plan
(i) We recall that Table 4-7 lists the common strong bases. (ii) Other common metal hydroxides
are assumed to be insoluble bases. (iii) Ammonia and closely related nitrogen-containing
compounds, the amines, are the common weak bases.
Solution
(a) (i) The strong bases are NaOH and Ba(OH)2, so
(ii) the insoluble bases are Cu(OH)2 and Pb(OH)2.
(b) (i) The strong bases are KOH and Sr(OH)2, so
(ii) the insoluble bases are Fe(OH)3 and Mg(OH)2, and
(iii) the weak base is NH3.
You should now work Exercises 20 and 22.
TABLE 4-7 Common Strong Bases
Group IA Group IIA
LiOH lithium hydroxide
NaOH sodium hydroxide
KOH potassium hydroxide Ca(OH)2 calcium hydroxide
RbOH rubidium hydroxide Sr(OH)2 strontium hydroxide
CsOH cesium hydroxide Ba(OH)2 barium hydroxide
5 Solubility Guidelines for Compounds in Aqueous Solution
Solubility is a complex phenomenon, and it is not possible to give a complete summary
of all of our observations. The following brief summary for solutes in aqueous solutions
will be very useful. These generalizations are often called the solubility guidelines.
Compounds whose solubility in water is less than about 0.02 mol/L are usually classified
as insoluble compounds, whereas those that are more soluble are classified as soluble
compounds. No gaseous or solid substances are infinitely soluble in water. You may wish
to review Tables 2-3 (page 55), 4-5, and 4-6. They list some common ions. Table 4-15 on
page 164 contains a more comprehensive list.
1. The common inorganic acids are soluble in water. Low-molecular-weight organic
acids are also soluble.
2. All common compounds of the Group IA metal ions (Liϩ, Naϩ, Kϩ, Rbϩ, Csϩ)
and the ammonium ion, NH4
ϩ, are soluble in water.
3. The common nitrates, NO3
Ϫ; acetates, CH3COOϪ; chlorates, ClO3
Ϫ; and perchlorates,
ClO4
Ϫ, are soluble in water.
4. (a) The common chlorides, ClϪ, are soluble in water except AgCl, Hg2Cl2, and
PbCl2.
(b) The common bromides, BrϪ, and iodides, IϪ, show approximately the same
solubility behavior as chlorides, but there are some exceptions. As these halide
ions (ClϪ, BrϪ, IϪ) increase in size, the solubilities of their slightly soluble
compounds decrease.
(c) The common fluorides, FϪ, are soluble in water except MgF2, CaF2, SrF2, BaF2,
and PbF2.
5. The common sulfates, SO4
2Ϫ, are soluble in water except PbSO4, BaSO4, and
HgSO4; CaSO4, SrSO4, and Ag2SO4 are moderately soluble.
6. The common metal hydroxides, OHϪ, are insoluble in water except those of the
Group IA metals and the heavier members of the Group IIA metals, beginning with
Ca(OH)2.
7. The common carbonates, CO3
2Ϫ, phosphates, PO4
3Ϫ, and arsenates, AsO4
3Ϫ, are
insoluble in water except those of the Group IA metals and NH4
ϩ. MgCO3 is moderately
soluble.
8. The common sulfides, S2Ϫ, are insoluble in water except those of the Group IA and
Group IIA metals and the ammonium ion.
Table 4-8 summarizes much of the information about the solubility guidelines.
We have distinguished between strong and weak electrolytes and between soluble and
insoluble compounds. Let us now see how we can describe chemical reactions in aqueous
solutions.
REACTIONS IN AQUEOUS SOLUTIONS
Many important chemical reactions occur in aqueous solutions. In this chapter you should
learn to describe such aqueous reactions and to predict the products of many reactions.
Let us first look at how we write chemical equations that describe reactions in aqueous
solutions. We use three kinds of chemical equations. Table 4-9 shows the kinds of information
about each substance that we use in writing equations for reactions in aqueous
4-3
There is no sharp dividing line
between “soluble” and “insoluble”
compounds. Compounds whose
solubilities fall near the arbitrary
dividing line are called “moderately
soluble” compounds.
134 CHAPTER 4: Some Types of Chemical Reactions
The White Cliffs of Dover,
England, are composed mainly
of calcium carbonate (CaCO3).
TABLE 4-8 Solubility Guidelines for Common Ionic Compounds in Water
Generally Soluble Exceptions
Naϩ, Kϩ, NH4
ϩ compounds No common exceptions
fluorides (FϪ) Insoluble: MgF2, CaF2, SrF2, BaF2, PbF2
chlorides (ClϪ) Insoluble: AgCl, Hg2Cl2
Soluble in hot water: PbCl2
bromides (BrϪ) Insoluble: AgBr, Hg2Br2, PbBr2
Moderately soluble: HgBr2
iodides (IϪ) Insoluble: many heavy-metal iodides
sulfates (SO4
2Ϫ) Insoluble: BaSO4, PbSO4, HgSO4
Moderately soluble: CaSO4, SrSO4, Ag2SO4
nitrates (NO3
Ϫ), nitrites (NO2
Ϫ) Moderately soluble: AgNO2
chlorates (ClO3
Ϫ), Moderately soluble: KClO4
perchlorates (ClO4
Ϫ)
acetates (CH3COOϪ) Moderately soluble: AgCH3COO
Generally Insoluble Exceptions
sulfides (S2Ϫ) Soluble: those of NH4
ϩ, Naϩ, Kϩ, Mg2ϩ, Ca2ϩ
oxides (O2Ϫ), hydroxides (OHϪ) Soluble: Li2O*, LiOH, Na2O*, NaOH, K2O*, KOH,
BaO*, Ba(OH)2
Moderately soluble: CaO*, Ca(OH)2, SrO*,
Sr(OH)2
carbonates (CO3
2Ϫ), phosphates Soluble: those of NH4
ϩ, Naϩ, Kϩ
(PO4
3Ϫ), arsenates (AsO4
3Ϫ)
*Dissolves with evolution of heat and formation of hydroxides.
TABLE 4-9 Bonding, Solubility, Electrolyte Characteristics, and Predominant Forms of Solutes in Contact with Water
Acids Bases Salts
Strong Weak Strong Insoluble Weak Soluble Insoluble
acids acids bases bases bases salts salts
Examples HCl CH3COOH NaOH Mg(OH)2 NH3 KCl, NaNO3, BaSO4, AgCl,
HNO3 HF Ca(OH)2 Al(OH)3 CH3NH2 NH4Br Ca3(PO4)2
Pure compound ionic Molecular Molecular Ionic Ionic Molecular Ionic Ionic
or molecular?
Water-soluble or Soluble* Soluble* Soluble Insoluble Soluble† Soluble Insoluble
insoluble?
Ϸ 100% ionized or Yes No Yes (footnote‡) No Yes§ (footnote‡)
dissociated in dilute
aqueous solution?
Written in ionic Separate Molecules Separate Complete Molecules Separate ions Complete
equations as ions ions formulas formulas
*Most common inorganic acids and the low-molecular-weight organic acids (XCOOH) are water-soluble.
†The low-molecular-weight amines are water-soluble.
‡The very small concentrations of “insoluble” metal hydroxides and insoluble salts in saturated aqueous solutions are nearly completely dissociated.
§There are a few exceptions. A few soluble salts are molecular (and not ionic) compounds.
Because we have not studied periodic
trends in properties of transition
metals, it would be difficult for you
to predict that Cu is more active than
Ag. The fact that this reaction occurs
(see Figure 4-3) shows that it is.
136 CHAPTER 4: Some Types of Chemical Reactions
solutions. Some typical examples are included. Refer to Table 4-9 often as you study the
following sections.
1. In formula unit equations, we show complete formulas for all compounds. When
metallic copper is added to a solution of (colorless) silver nitrate, the more active
metal—copper—displaces silver ions from the solution. The resulting solution
contains blue copper(II) nitrate, and metallic silver forms as a finely divided solid
(Figure 4-3):
2AgNO3(aq) ϩ Cu(s) 88n 2Ag(s) ϩ Cu(NO3)2(aq)
Both silver nitrate and copper(II) nitrate are soluble ionic compounds (for solubility
guidelines see page 134 and Table 4-8).
2. In total ionic equations, formulas are written to show the (predominant) form in
which each substance exists when it is in contact with aqueous solution. We often
use brackets in total ionic equations to show ions that have a common source or
that remain in solution after the reaction is complete. The total ionic equation for
this reaction is
2[Agϩ(aq) ϩ NO3
Ϫ(aq)] ϩ Cu(s) 88n 2Ag(s) ϩ [Cu2ϩ(aq) ϩ 2NO3
Ϫ(aq)]
Examination of the total ionic equation shows that NO3
Ϫ ions do not participate
in the reaction. Because they do not change, they are often called “spectator” ions.
3. In net ionic equations, we show only the species that react. The net ionic equation
is obtained by eliminating the spectator ions and the brackets from the total
ionic equation.
2Agϩ(aq) ϩ Cu(s) 88n 2Ag(s) ϩ Cu2ϩ(aq)
Net ionic equations allow us to focus on the essence of a chemical reaction in aqueous
solutions. On the other hand, if we are dealing with stoichiometric calculations we
frequently must deal with formula weights and therefore with the complete formulas of all
species. In such cases, formula unit equations are more useful. Total ionic equations provide
the bridge between the two.
(a) (b)
Figure 4-3 (a) Copper wire and
a silver nitrate solution. (b) The
copper wire has been placed in the
solution and some finely divided
silver has deposited on the wire. The
solution is blue because it contains
copper(II) nitrate.
Brackets are not used in net ionic
equations.
This is why it is important to know
how and when to construct net ionic
equations from formula unit equations.
Recall the lists of strong acids (Table 4-5) and strong bases (Table 4-7). These acids
and bases are completely or almost completely ionized or dissociated in dilute aqueous
solutions. Other common acids and bases are either insoluble or only slightly ionized or
dissociated. In addition, the solubility guidelines (page 134 and Table 4-8) allow you to
determine which salts are soluble in water. Most salts that are soluble in water are also
strong electrolytes. Exceptions such as lead acetate, Pb(CH3COO)2, which is soluble but
does not ionize appreciably, will be noted as they are encountered.
The only common substances that should be written as ions in ionic equations are
(1) strong acids, (2) strong bases, and (3) soluble ionic salts.
OXIDATION NUMBERS
The many reactions that involve the transfer of electrons from one species to another are
called oxidation–reduction reactions, or simply, redox reactions. We use oxidation
numbers to keep track of electron transfers. The systematic naming of compounds
(Sections 4-11 and 4-12) also makes use of oxidation numbers.
The oxidation number, or oxidation state, of an element in a simple binary ionic
compound is the number of electrons gained or lost by an atom of that element when it
forms the compound. In the case of a single-atom ion, it corresponds to the actual charge
4-4
Oxidation–reduction and displacement
reactions are discussed in Sections 4-5
and 4-8.
Binary means two. Binary compounds
contain two elements.
4-4 Oxidation Numbers 137
Problem-Solving Tip: Writing Ionic Equations
The following chart will help in deciding which formula units are to be written as separate
ions in the total ionic equation and which ones are to be written as unchanged
formula units. You must answer two questions about a substance to determine whether
it should be written in ionic form or as a formula unit in the total and net ionic equa-
tions.
1. Does it dissolve in water? If not, write the full formula.
2. (a) If it dissolves, does it ionize (a strong acid)?
(b) If it dissolves, does it dissociate (a strong base or a soluble salt)?
If the answer to either part of the second question is yes, the substance is a soluble strong
electrolyte, and its formula is written in ionic form.
Write in ionic form;
e.g., [Hϩ(aq) ϩ ClϪ(aq)] (because
HCl is completely ionized)
e.g., [2Naϩ(aq) ϩ SO4
2Ϫ] (because
Na2SO4 is a soluble ionic salt)
Write as full formula;
e.g., PbSO4(s)
Is it mostly
either ionized
or dissociated
in H2O?
Is it soluble
in water?
Write as full formula;
e.g., CH3COOH(aq)
Yes Yes
No No
6666666g
6666g
88888888n 8888n
on the ion. In molecular compounds, oxidation numbers do not have the same significance
they have in binary ionic compounds. Oxidation numbers, however, are very useful
aids in writing formulas and in balancing equations. In molecular species, the oxidation
numbers are assigned according to an arbitrary set of rules. The element farther to the
right and higher up in the periodic table is assigned a negative oxidation number, and the
element farther to the left and lower down in the periodic table is assigned a positive
oxidation number.
Some rules for assigning oxidation numbers follow. These rules are not comprehensive,
but they cover most cases. In applying these rules, keep in mind two important points.
First, oxidation numbers are always assigned on a per atom basis; second, treat the rules
in order of decreasing importance—the first rule that applies takes precedence over any
subsequent rules that seem to apply.
1. The oxidation number of the atoms in any free, uncombined element is zero. This
includes polyatomic elements such as H2, O2, O3, P4, and S8.
2. The oxidation number of an element in a simple (monatomic) ion is equal to the
charge on the ion.
3. The sum of the oxidation numbers of all atoms in a compound is zero.
4. In a polyatomic ion, the sum of the oxidation numbers of the constituent atoms is
equal to the charge on the ion.
5. Fluorine has an oxidation number of Ϫ1 in its compounds.
6. Hydrogen has an oxidation number of ϩ1 in compounds unless it is combined with
metals, in which case it has an oxidation number of Ϫ1. Examples of these exceptions
are NaH and CaH2.
7. Oxygen usually has an oxidation number of Ϫ2 in its compounds. There are some
exceptions:
a. Oxygen has an oxidation number of Ϫ1 in hydrogen peroxide, H2O2, and in
peroxides, which contain the O2
2Ϫ ion; examples are CaO2 and Na2O2.
b. Oxygen has an oxidation number of Ϫᎏ1
2
ᎏ in superoxides, which contain the O2
Ϫ
ion; examples are KO2 and RbO2.
c. When combined with fluorine in OF2, oxygen has an oxidation number of ϩ2.
8. The position of the element in the periodic table helps to assign its oxidation
number:
a. Group IA elements have oxidation numbers of ϩ1 in all of their compounds.
b. Group IIA elements have oxidation numbers of ϩ2 in all of their compounds.
c. Group IIIA elements have oxidation numbers of ϩ3 in all of their compounds,
with a few rare exceptions.
d. Group VA elements have oxidation numbers of Ϫ3 in binary compounds with
metals, with H, or with NH4
ϩ. Exceptions are compounds with a Group VA
element combined with an element to its right in the periodic table; in this case,
their oxidation numbers can be found by using rules 3 and 4.
e. Group VIA elements below oxygen have oxidation numbers of Ϫ2 in binary
compounds with metals, with H, or with NH4
ϩ. When these elements are
combined with oxygen or with a lighter halogen, their oxidation numbers can
be found by using rules 3 and 4.
Polyatomic elements have two or more
atoms per molecule.
138 CHAPTER 4: Some Types of Chemical Reactions
f. Group VIIA elements have oxidation numbers of Ϫ1 in binary compounds with
metals, with H, with NH4
ϩ, or with a heavier halogen. When these elements
except fluorine (i.e., Cl, Br, I) are combined with oxygen or with a lighter halogen,
their oxidation numbers can be found by using rules 3 and 4.
Table 4-10 summarizes rules 5 through 8, with many examples.
EXAMPLE 4-3 Oxidation Numbers
Determine the oxidation numbers of nitrogen in the following species: (a) N2O4, (b) NH3,
(c) HNO3, (d) NO3
Ϫ, (e) N2.
Plan
We first assign oxidation numbers to elements that exhibit a single common oxidation number
(see Table 4-10). We recall that oxidation numbers are represented per atom and that the sum
of the oxidation numbers in a compound is zero, and the sum of the oxidation numbers in an
ion equals the charge on the ion.
Aqueous solutions of some
compounds that contain chromium.
Left to right: chromium(II) chloride,
CrCl2, is blue; chromium(III)
chloride, CrCl3, is green; potassium
chromate, K2CrO4, is yellowish;
potassium dichromate, K2Cr2O7,
is orangish.
4-4 Oxidation Numbers 139
TABLE 4-10 Common Oxidation Numbers (States) for Group A Elements in Compounds and Ions
Element(s) Common Ox. Nos. Examples Other Ox. Nos.
H ϩ1 H2O, CH4, NH4Cl Ϫ1 in metal hydrides, e.g., NaH, CaH2
Group IA ϩ1 KCl, NaH, RbNO3, K2SO4 None
Group IIA ϩ2 CaCl2, MgH2, Ba(NO3)2, SrSO4 None
Group IIIA ϩ3 AlCl3, BF3, Al(NO3)3, GaI3 None in common compounds
Group IVA ϩ2 CO, PbO, SnCl2, Pb(NO3)2 Many others are also seen for C and Si
ϩ4 CCl4, SiO2, SiO3
2Ϫ, SnCl4
Group VA Ϫ3 in binary compounds Mg3N2, Na3P, Cs3As ϩ3, e.g., NO2
Ϫ, PCl3
with metals
Ϫ3 in NH4
ϩ, binary NH3, PH3, AsH3, NH4
ϩ ϩ5, e.g., NO3
Ϫ, PO4
3Ϫ, AsF5, P4O10
compounds with H
O Ϫ2 H2O, P4O10, Fe2O3, CaO, ClO3
Ϫ ϩ2 in OF2
Ϫ1 in peroxides, e.g., H2O2, Na2O2
Ϫᎏ1
2
ᎏ in superoxides, e.g., KO2, RbO2
Group VIA Ϫ2 in binary compounds H2S, CaS, Fe2S3, Na2Se ϩ4 with O and the lighter halogens,
(other than O) with metals and H e.g., SO2, SeO2, Na2SO3, SO3
2Ϫ,
Ϫ2 in binary compounds (NH4)2S, (NH4)2Se SF4
with NH4
ϩ ϩ6 with O and the lighter halogens,
e.g., SO3, TeO3, H2SO4, SO4
2Ϫ,
SF6
Group VIIA Ϫ1 in binary compounds MgF2, KI, ZnCl2, FeBr3 Cl, Br, or I with O or with a lighter
with metals and H halogen
Ϫ1 in binary compounds NH4Cl, NH4Br ϩ1, e.g., BrF, ClOϪ, BrOϪ
with NH4
ϩ ϩ3, e.g., ICl3, ClO2
Ϫ, BrO2
Ϫ
ϩ5, e.g., BrF5, ClO3
Ϫ, BrO3
Ϫ
ϩ7, e.g., IF7, ClO4
Ϫ, BrO4
Ϫ
By convention, oxidation numbers are
represented as ϩn and Ϫn, but ionic
charges are represented as nϩ and nϪ.
We shall circle oxidation numbers
associated with formulas and show
them in red. Both oxidation numbers
and ionic charges can be combined
algebraically.
Usually the element with the positive
oxidation number is written first.
For historic reasons, however, in
compounds containing nitrogen and
hydrogen, such as NH3, and many
compounds containing carbon and
hydrogen, such as CH4, hydrogen is
written last, although it has a positive
oxidation number.
140 CHAPTER 4: Some Types of Chemical Reactions
x Ϫ2
Solution
(a) The oxidation number of O is Ϫ2. The sum of the oxidation numbers for all atoms in a
compound must be zero:
ox. no./atom:
N2O4
total ox. no.: 2x ϩ 4(Ϫ2) ϭ 0 or x ϭ ϩ4
(b) The oxidation number of H is ϩ1:
ox. no./atom:
NH3
total ox. no.: x ϩ 3(1) ϭ 0 or x ϭ Ϫ3
(c) The oxidation number of H is ϩ1 and the oxidation number of O is Ϫ2.
ox. no./atom:
HNO3
total ox. no.: 1 ϩ x ϩ 3(Ϫ2) ϭ 0 or x ϭ ϩ5
(d) The sum of the oxidation numbers for all atoms in an ion equals the charge on the ion:
ox. no./atom:
NO3
Ϫ
total ox. no.: x ϩ 3(Ϫ2) ϭ Ϫ1 or x ϭ ϩ5
(e) The oxidation number of any free element is zero.
You should now work Exercise 38.
CLASSIFYING CHEMICAL REACTIONS
We now discuss chemical reactions in further detail. We classify them as oxidation–
reduction reactions, combination reactions, decomposition reactions, displacement reactions,
and metathesis reactions. The last type can be further described as precipitation
reactions, acid–base (neutralization) reactions, and gas-formation reactions. We will see
that many reactions, especially oxidation–reduction reactions, fit into more than one category,
and that some reactions do not fit neatly into any of them. As we study different
kinds of chemical reactions, we will learn to predict the products of other similar reactions.
In Chapter 6 we will describe typical reactions of hydrogen, oxygen, and their
compounds. These reactions will illustrate periodic relationships with respect to chemical
properties. It should be emphasized that our system is not an attempt to transform nature
so that it fits into small categories but rather an effort to give some order to our many
observations of nature.
OXIDATION–REDUCTION REACTIONS: AN
INTRODUCTION
The term “oxidation” originally referred to the combination of a substance with oxygen.
This results in an increase in the oxidation number of an element in that substance.
According to the original definition, the following reactions involve oxidation of the
4-5
x Ϫ2
x ϩ1
x Ϫ2ϩ1
Oxidation number is a formal concept
adopted for our convenience. The
numbers are determined by relying
on rules. These rules can result in a
fractional oxidation number, as shown
here. This does not mean that
electronic charges are split.
The terms “oxidation number” and
“oxidation state” are used
interchangeably.
In biological systems, reduction often
corresponds to the addition of
hydrogen to molecules or polyatomic
ions and oxidation often corresponds to
the removal of hydrogen.
4-5 Oxidation–Reduction Reactions: An Introduction 141
substance shown on the far left of each equation. Oxidation numbers are shown for one
atom of the indicated kind.
1. The formation of rust, Fe2O3, iron(III) oxide: oxidation state of Fe
4Fe(s) ϩ 3O2(g) 88n 2Fe2O3(s) 0 88n ϩ3
2. Combustion reactions: oxidation state of C
C(s) ϩ O2(g) 88n CO2(g) 0 88n ϩ4
2CO(g) ϩ O2(g) 88n 2CO2(g) ϩ2 88n ϩ4
C3H8(g) ϩ 5O2(g) 88n 3CO2(g) ϩ 4H2O(g) Ϫ8/3 88n ϩ4
Originally, reduction described the removal of oxygen from a compound. Oxide ores
are reduced to metals (a very real reduction in mass). For example, tungsten for use in
light bulb filaments can be prepared by reduction of tungsten(VI) oxide with hydrogen at
1200°C:
oxidation number of W
WO3(s) ϩ 3H2(g) 88n W(s) ϩ 3H2O(g) ϩ6 88n 0
Tungsten is reduced, and its oxidation state decreases from ϩ6 to zero. Hydrogen is
oxidized from zero to the ϩ1 oxidation state. The terms “oxidation” and “reduction” are
now applied much more broadly.
Oxidation is an increase in oxidation number and corresponds to the loss, or
apparent loss, of electrons. Reduction is a decrease in oxidation number and corresponds
to a gain, or apparent gain, of electrons.
Electrons are neither created nor destroyed in chemical reactions. So oxidation and
reduction always occur simultaneously, and to the same extent, in ordinary chemical reactions.
In the four equations cited previously as examples of oxidation, the oxidation numbers
of iron and carbon atoms increase as they are oxidized. In each case oxygen is reduced as
its oxidation number decreases from zero to Ϫ2.
Because oxidation and reduction occur simultaneously in all of these reactions, they
are referred to as oxidation–reduction reactions. For brevity, we usually call them redox
reactions. Redox reactions occur in nearly every area of chemistry and biochemistry. We
need to be able to identify oxidizing agents and reducing agents and to balance oxidation–reduction
equations. These skills are necessary for the study of electrochemistry in
Chapter 21. Electrochemistry involves electron transfer between physically separated
oxidizing and reducing agents and interconversions between chemical energy and electric
energy. These skills are also fundamental to the study of biology, biochemistry, environmental
science, and materials science.
Oxidizing agents are species that (1) oxidize other substances, (2) contain atoms that
are reduced, and (3) gain (or appear to gain) electrons. Reducing agents are species
that (1) reduce other substances, (2) contain atoms that are oxidized, and (3) lose
(or appear to lose) electrons.
The following abbreviations are widely
used:
ox. no. ϭ oxidation number
ox. agt. ϭ oxidizing agent
red. agt. ϭ reducing agent
142 CHAPTER 4: Some Types of Chemical Reactions
0 Ϫ1 ϩ1
ϩ1 0 0ϩ2ϩ5 Ϫ2 ϩ5 Ϫ2
8n
888n
888888
888
88888888888888888888888888888888888888888888888888888888
888888888888
The following equations represent examples of redox reactions. Oxidation numbers are
shown above the formulas, and oxidizing and reducing agents are indicated:
2Fe(s) ϩ 3Cl2(g) 88n 2FeCl3(s)
red. agt. ox. agt.
2FeBr3(aq) ϩ 3Cl2(g) 88n 2FeCl3(aq) ϩ 3Br2(ᐉ)
red. agt. ox. agt.
Equations for redox reactions can also be written as total ionic and net ionic equations.
For example, the second equation may also be written as:
2[Fe3ϩ(aq) ϩ 3BrϪ(aq)] ϩ 3Cl2(g) 88n 2[Fe3ϩ(aq) ϩ 3ClϪ(aq)] ϩ 3Br2(ᐉ)
We distinguish between oxidation numbers and actual charges on ions by denoting oxidation
numbers as ϩn or Ϫn in red circles just above the symbols of the elements, and actual
charges as nϩ or nϪ above and to the right of formulas of ions. The spectator ions, Fe3ϩ,
do not participate in electron transfer. Their cancellation allows us to focus on the oxidizing
agent, Cl2(g), and the reducing agent, BrϪ(aq).
2BrϪ(aq) ϩ Cl2(g) 88n 2ClϪ(aq) ϩ Br2(ᐉ)
A disproportionation reaction is a redox reaction in which the same element is
oxidized and reduced. An example is:
Cl2 ϩ H2O 88n HCl ϩ HClO
EXAMPLE 4-4 Redox Reactions
Write each of the following formula unit equations as a net ionic equation if the two differ.
Which ones are redox reactions? For the redox reactions, identify the oxidizing agent, the
reducing agent, the species oxidized, and the species reduced.
(a) 2AgNO3(aq) ϩ Cu(s) 88n Cu(NO3)2(aq) ϩ 2Ag(s)
heat
(b) 4KClO3(s) 888n KCl(s) ϩ 3KClO4(s)
(c) 3 AgNO3(aq) ϩ K3PO4(aq) 88n Ag3PO4(s) ϩ 3KNO3(aq)
Plan
To write ionic equations, we must recognize compounds that are (1) soluble in water and (2)
ionized or dissociated in aqueous solutions. To determine which are oxidation–reduction reactions,
we should assign an oxidation number to each element.
Solution
(a) According to the solubility guidelines (page 134), both silver nitrate, AgNO3, and copper(II)
nitrate, Cu(NO3)2, are water-soluble ionic compounds. The total ionic equation and oxidation
numbers are
2[Agϩ(aq) ϩ NO3
Ϫ(aq)] ϩ Cu(s) 88n [Cu2ϩ(aq) ϩ 2NO3
Ϫ(aq)] ϩ 2Ag(s)
ϩ2
Ϫ1
0 0 ϩ3 Ϫ1
00 ϩ3ϩ3 Ϫ1Ϫ1
Iron reacting with chlorine to form
iron(III) chloride.
Metallic silver formed by immersing
a spiral of copper wire in a silver
nitrate solution (see Example 4-4a).
The reaction of AgNO3(aq) and K3PO4(aq) is a precipitation reaction (see Example 4-4c). ᭤
4-5 Oxidation–Reduction Reactions: An Introduction 143
The nitrate ions, NO3
Ϫ, are spectator ions. Canceling them from both sides gives the net ionic
equation:
2Agϩ(aq) ϩ Cu(s) 88n Cu2ϩ(aq) ϩ 2Ag(s)
This is a redox equation. The oxidation number of silver decreases from ϩ1 to zero; silver ion
is reduced and is the oxidizing agent. The oxidation number of copper increases from zero to
ϩ2; copper is oxidized and is the reducing agent.
(b) This reaction involves only solids, so there are no ions in solution and the formula unit
and net ionic equations are identical. It is a redox reaction:
4KClO3(s) 88n KCl(s) ϩ 3KClO4(s)
Ϫ6
ϩ2
Chlorine is reduced from ϩ5 in KClO3 to the Ϫ1 oxidation state in KCl; the oxidizing agent
is KClO3. Chlorine is oxidized from ϩ5 in KClO3 to the ϩ7 oxidation state in KClO4. KClO3
is also the reducing agent. This is a disproportionation reaction. We see that KClO3 is both
the oxidizing agent and the reducing agent.
(c) The solubility guidelines indicate that all these salts are soluble except for silver phosphate,
Ag3PO4. The total ionic equation is
3[Agϩ(aq) ϩ NO3
Ϫ(aq)] ϩ [3Kϩ(aq) ϩ PO4
3Ϫ(aq)] 88n Ag3PO4(s) ϩ 3[Kϩ(aq) ϩ NO3
Ϫ(aq)]
Eliminating the spectator ions gives the net ionic equation:
3Agϩ(aq) ϩ PO4
3Ϫ(aq) 88n Ag3PO4(s)
There are no changes in oxidation numbers; this is not a redox reaction.
You should now work Exercises 44 and 47.
ϩ1 0 0ϩ2
ϩ1 ϩ5 Ϫ2 ϩ1 ϩ7 Ϫ2ϩ1 Ϫ1
8n
888n
8888888
8888888888888888888888888
888888888888888
In Chapter 11 we will learn to balance redox equations and to carry out stoichiometric
calculations using the balanced equations.
Problem-Solving Tip: A Foolproof Way to Recognize a Redox Reaction
You can always recognize a redox reaction by analyzing oxidation numbers. First determine
the oxidation number of each element wherever it appears in the reaction. If no
elements change in oxidation numbers, the reaction is not an oxidation–reduction reaction.
If changes do occur, the reaction is an oxidation–reduction reaction. Remember
that oxidation and reduction must always occur together; if some atoms increase in oxidation
numbers, then others must decrease.
ϩ1 ϩ5 Ϫ2ϩ1ϩ5 Ϫ2
Potassium, a metal, reacts with
chlorine, a nonmetal, to form
potassium chloride, KCl. The
reaction releases energy in the form
of heat and light.
COMBINATION REACTIONS
Reactions in which two or more substances combine to form a compound are called
combination reactions.
They may involve (1) the combination of two elements to form a compound, (2) the combination
of an element and a compound to form a single new compound, or (3) the
combination of two compounds to form a single new compound. Let’s examine some of
these reactions.
1 Element ؉ Element n Compound
For this type of combination reaction, each element goes from an uncombined state, where
its oxidation state is zero, to a combined state in a compound, where its oxidation state is
not zero. Thus reactions of this type are oxidation–reduction reactions (see Section 4-5).
Metal ϩ Nonmetal n Binary Ionic Compound
Most metals react with most nonmetals to form binary ionic compounds. The Group IA
metals combine with the Group VIIA nonmetals to form binary ionic compounds with the
general formula MX (Section 7-2):
2M(s) ϩ X2 88n 2(MϩXϪ)(s) M ϭ Li, Na, K, Rb, Cs
X ϭ F, Cl, Br, I
This general equation thus represents the 20 combination reactions that form the ionic
compounds listed in Table 4-11. Sodium, a silvery-white metal, combines with chlorine,
a pale green gas, to form sodium chloride, or ordinary table salt. All members of both
families undergo similar reactions.
2Na(s) ϩ Cl2(g) 88n 2NaCl(s) sodium chloride (mp 801°C)
As we might expect, the Group IIA metals also combine with the Group VIIA nonmetals
to form binary compounds. Except for BeCl2, BeBr2, and BeI2, these are ionic compounds.
In general terms these combination reactions may be represented as:
M(s) ϩ X2 88n MX2(s) M ϭ Be, Mg, Ca, Sr, Ba
X ϭ F, Cl, Br, I
Consider the reaction of magnesium with fluorine to form magnesium fluoride:
Mg(s) ϩ F2(g) 88n MgF2(s) magnesium fluoride (mp 1266°C)
Because all the IIA and VIIA elements undergo similar reactions, the general equation,
written above, represents 20 reactions. We omit radium and astatine, the rare and highly
radioactive members of the families.
Nonmetal ϩ Nonmetal n Binary Covalent Compound
When two nonmetals combine with each other they form binary covalent compounds. In
such reactions, the oxidation number of the element with the more positive oxidation
4-6
Another important reaction of this
kind is the formation of metal oxides
(Section 6-8).
TABLE 4-11 Alkali Metal
Halides:
Compounds
Formed by
Group IA and
VIIA
Elements
LiF LiCl LiBr LiI
NaF NaCl NaBr NaI
KF KCl KBr KI
RbF RbCl RbBr RbI
CsF CsCl CsBr CsI
Nonmetals in odd-numbered periodic
groups favor odd oxidation numbers,
whereas those in even-numbered
groups favor even oxidation numbers
in their compounds. The maximum
oxidation number for a representative
element is equal to its periodic group
number. For example, sulfur (Group
VIA) can form both SF4 and SF6.
4-6 Combination Reactions 145
number is often variable, depending on reaction conditions. For example, phosphorus
(Group VA) combines with a limited amount of chlorine to form phosphorus trichloride,
in which phosphorus exhibits the ϩ3 oxidation state.
P4(s) ϩ 6Cl2(g) 88n 4PCl3(ᐉ) (with limited Cl2) (mp Ϫ112°C)
With an excess of chlorine, the product is phosphorus pentachloride, which contains phosphorus
in the ϩ5 oxidation state:
P4(s) ϩ 10Cl2(g) 88n 4PCl5(s) (with excess Cl2) (decomposes at 167°C)
In general, a higher oxidation state of a nonmetal is formed when it reacts with an excess of
another nonmetal. There are many more reactions in which two elements combine to form
a compound (see Sections 6-7 and 6-8).
2 Compound ؉ Element n Compound
Phosphorus in the ϩ3 oxidation state in PCl3 molecules can be converted to the ϩ5 state
in PCl5 by combination with chlorine:
PCl3(ᐉ) ϩ Cl2(g) 88n PCl5(s)
Likewise, sulfur in the ϩ4 state is converted to the ϩ6 state when SF4 reacts with fluorine
to form SF6:
SF4(g) ϩ F2(g) 88n SF6(g) sulfur hexafluoride (mp Ϫ50.5°C)
Combination reactions of this type are also oxidation–reduction reactions.
3 Compound ؉ Compound n Compound
An example of reactions in this category is the combination of calcium oxide with carbon
dioxide to produce calcium carbonate:
ϩ3
ϩ5
ϩ3 ϩ5
ϩ6ϩ4
Phosphorus and chlorine, two nonmetals, react to
form phosphorus pentachloride, PCl5.
CaO(s) ϩ CO2(g) 88n CaCO3(s)
Pyrosulfuric acid is produced by dissolving sulfur trioxide in concentrated sulfuric acid:
SO3(g) ϩ H2SO4(ᐉ) 88n H2S2O7(ᐉ)
Pyrosulfuric acid, H2S2O7, is then diluted with water to make H2SO4:
H2S2O7(ᐉ) ϩ H2O(ᐉ) 88n 2H2SO4(ᐉ)
Oxides of the Group IA and IIA metals react with water to form metal hydroxides, e.g.:
CaO(s) ϩ H2O(ᐉ) 88n Ca(OH)2(aq)
DECOMPOSITION REACTIONS
Decomposition reactions are those in which a compound decomposes to produce
(1) two elements, (2) one or more elements and one or more compounds, or (3) two
or more compounds.
Examples of each type follow.
1 Compound n Element ؉ Element
The electrolysis of water produces two elements by the decomposition of a compound. A
compound that ionizes, such as H2SO4, is added to increase the conductivity of water and
the rate of the reaction (Figure 1-8), but it does not participate in the reaction:
electrolysis
2H2O(ᐉ) 8888888888n 2H2(g) ϩ O2(g)
Small amounts of oxygen can be prepared by the thermal decomposition of certain
oxygen-containing compounds. Some metal oxides, such as mercury(II) oxide, HgO,
decompose on heating to produce oxygen:
heat
2HgO(s) 8888n 2Hg(ᐉ) ϩ O2(g)
mercury(II) oxide
2 Compound n Compound ؉ Element
The alkali metal chlorates, such as KClO3, decompose when heated to produce the corresponding
chlorides and liberate oxygen. Potassium chlorate is a common laboratory source
of small amounts of oxygen:
heat
2KClO3(s) 8888n 2KCl(s) ϩ 3O2(g)
potassium chlorate MnO2 potassium chloride
Nitrate salts of alkali metals or alkaline earth metals decompose to form metal nitrites
and oxygen gas.
2NaNO3(s) 88n 2NaNO2(s) ϩ O2(g)
Hydrogen peroxide decomposes to form water and oxygen.
H2O2(ᐉ) 88n 2H2O(ᐉ) ϩ O2(g)
4-7
Decomposition reactions can be
considered as the opposite of
combination reactions.
146 CHAPTER 4: Some Types of Chemical Reactions
Mercury(II) oxide, a red compound,
decomposes when heated into the
two elements: mercury (a metal) and
oxygen (a nonmetal). Mercury vapor
condenses on the cooler upper
portion of the test tube.
Manganese dioxide, MnO2, is used as
a catalyst, a substance that speeds up a
chemical reaction but is not consumed.
Here it allows the decomposition to
occur at a lower temperature.
3 Compound n Compound ؉ Compound
The thermal decomposition of calcium carbonate (limestone) and other carbonates
produces two compounds, a metal oxide and carbon dioxide:
heat
CaCO3(s) 8888n CaO(s) ϩ CO2(g)
This is an important reaction in the production of cement. Calcium oxide is also used as
a base in industrial processes.
When some solid hydroxides are heated, they decompose to form a metal oxide and
water vapor.
heat
Mg(OH)2(s) 8888n MgO(s) ϩ H2O(g)
Magnesium oxide, MgO, is pressed into sheets for use as a thermal insulating material in
oven walls.
Ammonium salts lose ammonia.
heat
(NH4)2SO4(s) 8888n 2NH3(g) ϩ H2SO4(ᐉ)
If the ammonium salt contains an anion that is a strong oxidizing agent (e.g., nitrate,
nitrite, or dichromate), its decomposition reaction produces an oxide, water (as vapor at
high temperatures), and nitrogen gas. Such a reaction is a redox reaction.
heat
(NH4)2Cr2O7(s) 8888n Cr2O3(s) ϩ 4H2O(g) ϩ N2(g)
DISPLACEMENT REACTIONS
Reactions in which one element displaces another from a compound are called
displacement reactions.
These reactions are always redox reactions. The more readily a metal undergoes oxidation,
the more active we say it is.
Active metals displace less active metals or hydrogen from their compounds in
aqueous solution to form the oxidized form of the more active metal and the reduced
(free metal) form of the other metal or hydrogen.
In Table 4-12, the most active metals are listed at the top of the first column. These metals
tend to react to form their oxidized forms (cations). Elements at the bottom of the activity
series (the first column of Table 4-12) tend to remain in their reduced form. They are
easily converted from their oxidized forms to their reduced forms.
1 More Active Metal ؉ Less Active Metal ؉
[Salt of Less Active Metal]88n [Salt of More Active Metal]
The reaction of copper with silver nitrate that was described in detail in Section 4-3 is
typical. Please refer to it.
4-8
Solid ammonium dichromate,
[(NH4)2Cr2O7, orange] decomposes
when heated into chromium(II)
oxide, (Cr2O3, green), nitrogen, and
steam (water vapor). This reaction
is sometimes demonstrated as the
“classroom volcano,” but it must be
done with extreme caution due to
the carcinogenic (cancer-causing)
nature of Cr2O3.
4-8 Displacement Reactions 147
Alkali metal carbonates do not
decompose when heated.
A decomposition reaction may or may
not also be an oxidation-reduction
reaction. You can always identify a
redox reaction by determining the
oxidation state of each element in each
occurrence in the reaction (see the
Problem-Solving Tip in Section 4-5).
EXAMPLE 4-5 Displacement Reaction
A large piece of zinc metal is placed in a copper(II) sulfate, CuSO4, solution. The blue solution
becomes colorless as copper metal falls to the bottom of the container. The resulting
solution contains zinc sulfate, ZnSO4. Write balanced formula unit, total ionic, and net ionic
equations for the reaction.
Plan
The metals zinc and copper are not ionized or dissociated in contact with H2O. Both CuSO4
and ZnSO4 are soluble salts (solubility guideline 5), and so they are written in ionic form.
Solution
Zn(s) ϩ CuSO4(aq) 88n Cu(s) ϩ ZnSO4(aq)
Zn(s) ϩ [Cu2ϩ(aq) ϩ SO4
2Ϫ(aq)] 88n Cu(s) ϩ [Zn2ϩ(aq) ϩ SO4
2Ϫ(aq)]
Zn(s) ϩ Cu2ϩ(aq) 88n Cu(s) ϩ Zn2ϩ(aq)
In this displacement reaction, the more active metal, zinc, displaces the ions of the less active
metal, copper, from aqueous solution.
You should now work Exercise 58.
A strip of zinc metal was placed in a
blue solution of copper(II) sulfate,
CuSO4. The copper has been
displaced from solution and has
fallen to the bottom of the beaker.
The resulting zinc sulfate, ZnSO4,
solution is colorless.
148 CHAPTER 4: Some Types of Chemical Reactions
TABLE 4-12 Activity Series of Some Elements
Common Common
Element Reduced Form Oxidized Forms
Li Li Liϩ
K K Kϩ
Ca Ca Ca2ϩ
Na Na Naϩ
Mg Mg Mg2ϩ
Al Al Al3ϩ
Mn Mn Mn2ϩ
Zn Zn Zn2ϩ
Cr Cr Cr3ϩ, Cr6ϩ
Fe Fe Fe2ϩ, Fe3ϩ
Cd Cd Cd2ϩ
Co Co Co2ϩ
Ni Ni Ni2ϩ
Sn Sn Sn2ϩ, Sn4ϩ
Pb Pb Pb2ϩ, Pb4ϩ
H (a nonmetal) H2 Hϩ
Sb (a metalloid) Sb Sb3ϩ
Cu Cu Cuϩ, Cu2ϩ
Hg Hg Hg2
2ϩ, Hg2ϩ
Ag Ag Agϩ
Pt Pt Pt2ϩ, Pt4ϩ
Au Au Auϩ, Au3ϩ
Displacehydrogen
fromnonoxidizingacids
Displacehydrogen
fromsteam
Displace
hydrogen
from
cold
water
nm
666666666
g
h
666666666
6666
g
h
6666
Our Daily LivesCCHEMISTRY IN USE
Troublesome Displacement Reactions
The deterioration of the Statue of Liberty and the damage
done at the Three Mile Island and Chernobyl nuclear facilities
are just a few of the major problems that have resulted
from ignorance about chemical reactivity.
When originally constructed over one hundred years ago
the Statue of Liberty had a 200,000-pound outer copper skin
supported by a framework of 2000 iron bars. First, oxygen in
the air oxidized the copper skin to form copper oxide. In a
series of reactions, iron (the more active metal) then reduced
the Cu2ϩ ions in copper oxide.
2Fe ϩ 3Cu2ϩ 88n 2Fe3ϩ ϩ 3Cu
Over the years, the supporting iron frame was reduced to less
than half its original thickness; this made necessary the repairs
done to the statue before the celebration of its 100th birthday
on July 4, 1986.
Two major nuclear power plant accidents, one at Three
Mile Island near Harrisburg, Pennsylvania, in 1979, and the
other at Chernobyl in Ukraine in 1986, were also unexpected
consequences of chemical reactivity. In each case, cooling
pump failures sent temperatures soaring above 340°C. Like
aluminum, zirconium (used in building the reactors) forms a
protective oxide coating that protects it from further reactions.
However, that protective coating breaks down at high
temperatures. Without its protective coating, zirconium
reacts with steam.
Zr(s) ϩ 2H2O(g) 88n ZrO2(s) ϩ 2H2(g)
At Three Mile Island, this displacement reaction produced a
1000-cubic foot bubble of hydrogen gas. Because hydrogen
is easily ignited by a spark, the nuclear power plant was in
real danger of a complete meltdown until the hydrogen could
be removed.
During the Middle Ages (ϳ400–1400 AD), another displacement
reaction completely misled alchemists into
foolishly pursuing a philosopher’s stone that was believed to
have the power to turn base metals such as iron and lead into
more precious metals such as silver and gold. The alchemists’
ignorance of relative activities of metals led them to believe
that they had turned iron into a more precious metal when
they inserted an iron rod into a blue copper(II) sulfate solution.
In fact, the following displacement reaction had
occurred, plating shiny copper metal onto the iron rod.
Fe(s) ϩ 3Cu2ϩ(aq) 88n 2Fe3ϩ(aq) ϩ Cu(s)
In the 1960s and 1970s, some automobile manufacturers
showed their ignorance of chemical reactivity by building cars
with aluminum water pumps and aluminum engine heads
attached to cast-iron engine blocks. These water pumps often
leaked and the engine heads quickly deteriorated. These
problems occurred as the more active aluminum reacted with
iron(II) oxide (formed when the iron engine reacted with
atmospheric oxygen).
Al ϩ Fe3ϩ 88n Al3ϩ ϩ Fe
Some dentists have made similar mistakes by placing gold
caps over teeth that are adjacent to existing fillings. The
slightly oxidized gold can react with a dental amalgam filling
(an alloy of silver, tin, copper, and mercury). As the dental
amalgam is oxidized, it dissolves in saliva to produce a persistent
metallic taste in the patient’s mouth.
When plumbers connect galvanized pipes (iron pipes
coated with zinc) to copper pipes, copper ions oxidize the
zinc coating and expose the underlying iron, allowing it to
rust. The displacement reaction that occurs is
Zn ϩ Cu2ϩ 88n Zn2ϩ ϩ Cu
Once the zinc coating has been punctured on an iron pipe,
oxidation of the iron pipes occurs rapidly because iron is a
more active metal than copper.
It is important to keep in mind that a variety of other reactions
probably take place in addition to the above
displacement reactions. For example, less active metals (such
as copper) can conduct electrons from the metals being oxidized
to oxidizing agents (such as oxygen or the oxide of
nitrogen and sulfur) that are present in the atmosphere. Oxygen
plays an important role in all of these displacement
examples.
Ronald DeLorenzo
Middle Georgia College
2 [Active Metal ؉ Nonoxidizing Acid] 88n [Hydrogen ؉ Salt of Acid]
A common method for the preparation of small amounts of hydrogen involves the reaction
of active metals with nonoxidizing acids, such as HCl and H2SO4. For example, when
zinc is dissolved in H2SO4, the reaction produces zinc sulfate; hydrogen is displaced from
the acid, and it bubbles off as gaseous H2. The formula unit equation for this reaction is
Zn(s) ϩ H2SO4(aq) 88n ZnSO4(aq) ϩ H2(g)
strong acid soluble salt
Both sulfuric acid (in very dilute solution) and zinc sulfate exist primarily as ions; so the
total ionic equation is
Zn(s) ϩ [2Hϩ(aq) ϩ SO4
2Ϫ(aq)] 88n [Zn2ϩ(aq) ϩ SO4
2Ϫ(aq)] ϩ H2(g)
Elimination of spectator ions from the total ionic equation gives the net ionic equation:
Zn(s) ϩ 2Hϩ(aq) 88n Zn2ϩ(aq) ϩ H2(g)
Table 4-12 lists the activity series. When any metal listed above hydrogen in this series
is added to a solution of a nonoxidizing acid such as hydrochloric acid, HCl, and sulfuric
acid, H2SO4, the metal dissolves to produce hydrogen, and a salt is formed. HNO3 is the
common oxidizing acid. It reacts with active metals to produce oxides of nitrogen, but not
hydrogen, H2.
EXAMPLE 4-6 Displacement Reaction
Which of the following metals can displace hydrogen from hydrochloric acid solution? Write
balanced formula unit, total ionic, and net ionic equations for reactions that can occur.
Al, Cu, Ag
Plan
The activity series of the metals, Table 4-12, tells us that copper and silver do not displace
hydrogen from solutions of nonoxidizing acids. Aluminum is an active metal that can displace
H2 from HCl and form aluminum chloride.
Solution
2Al(s) ϩ 6HCl(aq) 88n 3H2(g) ϩ 2AlCl3(aq)
2Al(s) ϩ 6[Hϩ(aq) ϩ ClϪ(aq)] 88n 3H2(g) ϩ 2[Al3ϩ(aq) ϩ 3ClϪ(aq)]
2Al(s) ϩ 6Hϩ(aq) 88n 3H2(g) ϩ 2Al3ϩ(aq)
You should now work Exercises 57 and 59.
Very active metals can even displace hydrogen from water. The reaction of potassium,
or another metal of Group IA, with water is also a displacement reaction:
2K(s) ϩ 2H2O(ᐉ) 88n 2[Kϩ(aq) ϩ OHϪ(aq)] ϩ H2(g)
Such reactions of very active metals of Group IA are dangerous, however, because they
generate enough heat to cause explosive ignition of the hydrogen (Figure 4-4).
H2SO4 can function as an oxidizing
agent with other substances, but it is
not an oxidizing agent in its reaction
with active metals.
150 CHAPTER 4: Some Types of Chemical Reactions
Zinc reacts with dilute H2SO4 to
produce H2 and a solution that
contains ZnSO4. This is a
displacement reaction.
Aluminum displaces H2 from a
hydrochloric acid solution.
EXAMPLE 4-7 Displacement Reaction
Which of the following metals can displace hydrogen from water at room temperature? Write
balanced formula unit, total ionic, and net ionic equations for reactions that can occur.
Sn, Ca, Hg
Plan
The activity series, Table 4-12, tells us that tin and mercury cannot displace hydrogen from
water. Calcium is a very active metal (see Table 4-12) that displaces hydrogen from cold water
and forms calcium hydroxide, a strong base.
Solution
Ca(s) ϩ 2H2O(ᐉ) 88n H2(g) ϩ Ca(OH)2(aq)
Ca(s) ϩ 2H2O(ᐉ) 88n H2(g) ϩ [Ca2ϩ(aq) ϩ 2OHϪ(aq)]
Ca(s) ϩ 2H2O(ᐉ) 88n H2(g) ϩ Ca2ϩ(aq) ϩ 2OHϪ(aq)
You should now work Exercise 65.
3 Active Nonmetal ؉ Less Active Nonmetal ؉
[Salt of Less Active Nonmetal]88n [Salt of More Active Nonmetal]
Many nonmetals displace less active nonmetals from combination with a metal or other
cation. For example, when chlorine is bubbled through a solution containing bromide ions
(derived from a soluble ionic salt such as sodium bromide, NaBr), chlorine displaces
bromide ions to form elemental bromine and chloride ions (as aqueous sodium chloride):
Cl2(g) ϩ 2[Naϩ(aq) ϩ BrϪ(aq)] 88n 2[Naϩ(aq) ϩ ClϪ(aq)] ϩ Br2(ᐉ)
chlorine sodium bromide sodium chloride bromine
The displacement reaction of
calcium with water at room
temperature produces bubbles of
hydrogen.
4-8 Displacement Reactions 151
Figure 4-4 Potassium, like other Group IA metals,
reacts vigorously with water. For this photograph,
the room was completely dark, and all of the light
you see here was produced by dropping a small
piece of potassium into a beaker of water.
Similarly, when bromine is added to a solution containing iodide ions, the iodide ions are
displaced by bromine to form iodine and bromide ions:
Br2(ᐉ) ϩ 2[Naϩ(aq) ϩ IϪ(aq)] 88n 2[Naϩ(aq) ϩ BrϪ(aq)] ϩ I2(s)
bromine sodium iodide sodium bromide iodine
Each halogen will displace less active (heavier) halogens from their binary salts; that is, the
order of decreasing activities is
F2 Ͼ Cl2 Ͼ Br2 Ͼ I2
Conversely, a halogen will not displace more active (lighter) members from their salts:
I2(s) ϩ 2FϪ 88n no reaction
EXAMPLE 4-8 Displacement Reactions
Which of the following combinations would result in a displacement reaction? Write balanced
formula unit, total ionic, and net ionic equations for reactions that occur.
(a) I2(s) ϩ NaBr(aq) 88n
(b) Cl2(g) ϩ NaI(aq) 88n
(c) Br2(ᐉ) ϩ NaCl(aq) 88n
Plan
The activity of the halogens decreases from top to bottom in the periodic table. We see (a)
that Br is above I and (c) that Cl is above Br in the periodic table; therefore, neither combination
(a) nor combination (c) could result in reaction. Cl is above I in the periodic table, and
so combination (b) results in a displacement reaction.
(a) Bromine, Br2, in water ( pale
orange) is poured into an aqueous
solution of NaI, the top layer in the
cylinder. (b) Br2 displaces IϪ from
solution and forms solid iodine, I2.
The I2 dissolves in water to give a
brown solution but is more soluble
in many organic liquids (purple
bottom layer).
152 CHAPTER 4: Some Types of Chemical Reactions
(a) (b)
Activity of the halogens decreases
going down the group in the
periodic table.
Solution
The more active halogen, Cl2, displaces the less active halogen, I2, from its compounds.
Cl2(g) ϩ 2NaI(aq) 88n I2(s) ϩ 2NaCl(aq)
Cl2(g) ϩ 2[Naϩ(aq) ϩ IϪ(aq)] 88n I2(s) ϩ 2[Naϩ(aq) ϩ ClϪ(aq)]
Cl2(g) ϩ 2IϪ(aq) 88n I2(s) ϩ 2ClϪ(aq)
You should now work Exercise 64.
METATHESIS REACTIONS
In many reactions between two compounds in aqueous solution, the positive and
negative ions appear to “change partners” to form two new compounds, with no
change in oxidation numbers. Such reactions are called metathesis reactions.
We can represent such reactions by the following general equation, where A and B represent
positive ions (cations) and X and Y represent negative ions (anions):
AX ϩ BY 88n AY ϩ BX
For example, when we mix silver nitrate and sodium chloride solutions, solid silver chloride
is formed and sodium nitrate remains dissolved in water:
AgNO3(aq) ϩ NaCl(aq) 88n AgCl(s) ϩ NaNO3(aq)
Metathesis reactions result in the removal of ions from solution; this removal of ions
can be thought of as the driving force for the reaction—the reason it occurs. The removal
of ions can occur in three ways, which can be used to classify three types of metathesis
reactions:
1. Formation of predominantly nonionized molecules (weak or nonelectrolytes) in
solution; the most common such nonelectrolyte product is water
2. Formation of an insoluble solid, called a precipitate (which separates from the solu-
tion)
3. Formation of a gas (which is evolved from the solution)
1 Acid–Base (Neutralization) Reactions: Formation of a Nonelectrolyte
Acid–base reactions are among the most important kinds of chemical reactions. Many
acid–base reactions occur in nature in both plants and animals. Many acids and bases are
essential compounds in an industrialized society (see Table 4-13). For example, approximately
350 pounds of sulfuric acid, H2SO4, and approximately 135 pounds of ammonia,
NH3, is required to support the lifestyle of an average American for one year.
The reaction of an acid with a metal hydroxide base produces a salt and water. Such
reactions are called neutralization reactions because the typical properties of acids and
bases are neutralized.
4-9
Pronounced “meh-tath-uh-sis.”
4-9 Metathesis Reactions 153
The manufacture of fertilizers
consumes more H2SO4 and more NH3
than any other single use.
In nearly all neutralization reactions, the driving force is the combination of Hϩ(aq)
from an acid and OHϪ(aq) from a base (or a base plus water) to form water mole-
cules.
Consider the reaction of hydrochloric acid, HCl(aq), with aqueous sodium hydroxide,
NaOH. Table 4-5 tells us that HCl is a strong acid, and Table 4-7 tells us that NaOH is
a strong base. The salt sodium chloride, NaCl, is formed in this reaction. It contains the
cation of its parent base, Naϩ, and the anion of its parent acid, ClϪ. Solubility guidelines
2 and 4 tell us that NaCl is a soluble salt.
HCl(aq) ϩ NaOH(aq) 88n H2O(ᐉ) ϩ NaCl(aq)
[Hϩ(aq) ϩ ClϪ(aq)] ϩ [Naϩ(aq) ϩ OHϪ(aq)] 88n H2O(ᐉ) ϩ [Naϩ(aq) ϩ ClϪ(aq)]
Hϩ(aq) ϩ OHϪ(aq) 88n H2O(ᐉ)
The net ionic equation for all reactions of strong acids with strong bases that form
soluble salts and water is
Hϩ(aq) ϩ OHϪ(aq) 88n H2O(ᐉ)
When a base such as ammonia or an
amine reacts with an acid, a salt, but
no water, is formed. This is still called
an acid–base, or neutralization,
reaction.
154 CHAPTER 4: Some Types of Chemical Reactions
TABLE 4-13 1997 Production of Inorganic Acids, Bases, and Salts in the United States
Billions
of
Formula Name Pounds Major Uses
H2SO4 sulfuric acid 95.84 Manufacture of fertilizers and other chemicals
CaO, Ca(OH)2 lime (calcium oxide and 47.56 Manufacture of other chemicals, steelmaking, water treatment
calcium hydroxide)
NH3 ammonia 38.39 Fertilizer; manufacture of fertilizers and other chemicals
H3PO4 phosphoric acid 26.83 Manufacture of fertilizers
Na2CO3 sodium carbonate (soda ash) 22.93 Manufacture of glass, other chemicals, detergents, pulp, and
paper
NaOH sodium hydroxide 22.74 Manufacture of other chemicals, pulp and paper, soap and
detergents, aluminum, textiles
HNO3 nitric acid 18.15 Manufacture of fertilizers, explosives, plastics, and lacquers
NH4NO3 ammonium nitrate 16.52 Fertilizer and explosive
HCl hydrochloric acid 8.44 Manufacture of other chemicals and rubber; metal cleaning
(NH4)2SO4 ammonium sulfate 5.62 Fertilizer
KOH, K2CO3 potash 3.15 Manufacture of fertilizers
Al2(SO4)2 aluminum sulfate 2.36 Water treatment, dyeing textiles
Na2SiO3 sodium silicate 2.15 Manufacture of detergents, cleaning agents, and adhesives
CaCl2 calcium chloride 1.40 De-icing roads in winter, controlling dust in summer, concrete
additive
NaClO3 sodium chlorate 1.31 Manufacture of other chemicals, explosives, plastics
Na2SO4 sodium sulfate 1.28 Manufacture of paper, glass, and detergents
Recall that in balanced equations we
show the smallest whole-number
coefficients possible.
4-9 Metathesis Reactions 155
Problem-Solving Tip: Salt Formation
The salt that is formed in a neutralization reaction is composed of the cation of the base
and the anion of the acid. The salt may be soluble or insoluble. If our goal were to obtain
the salt from the reaction of aqueous HCl with aqueous NaOH, we could evaporate the
water and obtain solid NaCl.
EXAMPLE 4-9 Neutralization Reactions
Predict the products of the reaction between HI(aq) and Ca(OH)2(aq). Write balanced formula
unit, total ionic, and net ionic equations.
Plan
This is an acid–base neutralization reaction; the products are H2O and the salt that contains
the cation of the base, Ca2ϩ, and the anion of the acid, IϪ; CaI2 is a soluble salt (solubility
guideline 4). HI is a strong acid (see Table 4-5), Ca(OH)2 is a strong base (see Table 4-7), and
CaI2 is a soluble ionic salt, so all are written in ionic form.
Solution
2HI(aq) ϩ Ca(OH)2(aq) 88n CaI2(aq) ϩ 2H2O(ᐉ)
2[Hϩ(aq) ϩ IϪ(aq)] ϩ [Ca2ϩ(aq) ϩ 2OHϪ(aq)] 88n [Ca2ϩ(aq) ϩ 2IϪ(aq)] ϩ 2H2O(ᐉ)
We cancel the spectator ions.
2Hϩ(aq) ϩ 2OHϪ(aq) 88n 2H2O(ᐉ)
Dividing by 2 gives the net ionic equation
Hϩ(aq) ϩ OHϪ(aq) 88n H2O(ᐉ)
You should now work Exercise 71.
Reactions of weak acids with strong bases also produce salts and water, but there is a significant
difference in the balanced ionic equations because weak acids are only slightly ionized.
EXAMPLE 4-10 Neutralization Reactions
Write balanced formula unit, total ionic, and net ionic equations for the reaction of acetic acid
with potassium hydroxide.
Plan
Neutralization reactions involving metal hydroxide bases produce a salt and water. CH3COOH
is a weak acid (see Table 4-6), and so it is written as formula units. KOH is a strong base (see
Table 4-7) and KCH3COO is a soluble salt (solubility guidelines 2 and 3), and so both are
written in ionic form.
Solution
CH3COOH(aq) ϩ KOH(aq) 88n KCH3COO(aq) ϩ H2O(ᐉ)
CH3COOH(aq) ϩ [Kϩ(aq) ϩ OHϪ(aq)] 88n [Kϩ(aq) ϩ CH3COOϪ(aq)] ϩ H2O(ᐉ)
A monoprotic acid contains one acidic H
per formula unit.
156 CHAPTER 4: Some Types of Chemical Reactions
The spectator ion is Kϩ, the cation of the strong base, KOH.
CH3COOH(aq) ϩ OHϪ(aq) 88n CH3COOϪ(aq) ϩ H2O(ᐉ)
Thus, we see that this net ionic equation includes molecules of the weak acid and anions of the
weak acid.
You should now work Exercise 72.
The reactions of weak monoprotic acids with strong bases that form soluble salts can be
represented in general terms as
HA(aq) ϩ OHϪ(aq) 88n AϪ(aq) ϩ H2O(ᐉ)
where HA represents the weak acid and AϪ represents its anion.
EXAMPLE 4-11 Salt Formation
Write balanced formula unit, total ionic, and net ionic equations for an acid–base reaction that
will produce the salt, barium chloride.
Plan
Neutralization reactions produce a salt. The salt contains the cation from the base and the
anion from the acid. The base must therefore contain Ba2ϩ, that is, Ba(OH)2, and the acid
must contain ClϪ, that is, HCl. We write equations that represent the reaction between the
strong base, Ba(OH)2, and the strong acid, HCl.
Solution
2HCl(aq) ϩ Ba(OH)2(aq) 88n BaCl2(aq) ϩ 2H2O(ᐉ)
2[Hϩ(aq) ϩ ClϪ(aq)] ϩ [Ba2ϩ(aq) ϩ 2OHϪ(aq)] 88n [Ba2ϩ(aq) ϩ 2ClϪ(aq)] ϩ 2H2O(ᐉ)
We cancel the spectator ions.
2Hϩ(aq) ϩ 2OHϪ(aq) 88n 2H2O(ᐉ)
Dividing by 2 gives the net ionic equation:
Hϩ(aq) ϩ OHϪ(aq) 88n H2O(ᐉ)
You should now work Exercise 80.
2 Precipitation Reactions
In precipitation reactions an insoluble solid, a precipitate, forms and then settles out
of solution. The driving force for these reactions is the strong attraction between cations
and anions. This results in the removal of ions from solution by the formation of a precipitate.
Our teeth and bones were formed by very slow precipitation reactions in which
mostly calcium phosphate Ca3(PO4)2 was deposited in the correct geometric arrange-
ments.
An example of a precipitation reaction is the formation of bright yellow insoluble
lead(II) chromate when we mix solutions of the soluble ionic compounds lead(II) nitrate
To understand the discussion of
precipitation reactions, you must know
the solubility guidelines (page 134) and
Table 4-8.
The net ionic equation shows the
driving force for this reaction. The
formula unit equation shows the salt
formed or that could be isolated if the
water were evaporated.
Figure 4-5 A precipitation reaction.
When K2CrO4 solution is added to
aqueous Pb(NO3)2 solution, the yellow
compound PbCrO4 precipitates. The
resulting solution contains Kϩ and
NO3
Ϫ ions, the ions of KNO3.
and potassium chromate (Figure 4-5). The other product of the reaction is KNO3, a
soluble ionic salt.
The balanced formula unit, total ionic, and net ionic equations for this reaction follow.
Pb(NO3)2(aq) ϩ K2CrO4(aq) 88n PbCrO4(s) ϩ 2KNO3(aq)
[Pb2ϩ(aq) ϩ 2 NO3
Ϫ(aq)] ϩ [2Kϩ(aq) ϩ CrO4
2Ϫ(aq)] 88n
PbCrO4(s) ϩ 2[Kϩ(aq) ϩ NO3
Ϫ(aq)]
Pb2ϩ(aq) ϩ CrO4
2Ϫ(aq) 88n PbCrO4(s)
Another important precipitation reaction involves the formation of insoluble carbonates
(solubility guideline 7). Limestone deposits are mostly calcium carbonate, CaCO3,
although many also contain significant amounts of magnesium carbonate, MgCO3.
Suppose we mix together aqueous solutions of sodium carbonate, Na2CO3, and calcium
chloride, CaCl2. We recognize that both Na2CO3 and CaCl2 (solubility guidelines 2, 4a,
and 7) are soluble ionic compounds. At the instant of mixing, the resulting solution contains
four ions:
Naϩ(aq), CO3
2Ϫ(aq), Ca2ϩ(aq), ClϪ(aq)
One pair of ions, Naϩ and ClϪ, cannot form an insoluble compound (solubility guidelines
2 and 4). We look for a pair of ions that could form an insoluble compound. Ca2ϩ ions
and CO3
2Ϫ ions are such a combination; they form insoluble CaCO3 (solubility guideline
7). The equations for the reaction follow.
CaCl2(aq) ϩ Na2CO3(aq) 88n CaCO3(s) ϩ 2 NaCl(aq)
[Ca2ϩ(aq) ϩ 2 ClϪ(aq)] ϩ [2Naϩ(aq) ϩ CO3
2Ϫ(aq)] 88n
CaCO3(s) ϩ 2[Naϩ(aq) ϩ ClϪ(aq)]
Ca2ϩ(aq) ϩ CO3
2Ϫ(aq) 88n CaCO3(s)
Seashells, which are formed in very
slow precipitation reactions, are
mostly calcium carbonate (CaCO3),
a white compound. Traces of
transition metal ions give them
color.
158 CHAPTER 4: Some Types of Chemical Reactions
EXAMPLE 4-12 Solubility Guidelines and Precipitation Reactions
Will a precipitate form when aqueous solutions of Ca(NO3)2 and NaCl are mixed in reasonable
concentrations? Write balanced formula unit, total ionic, and net ionic equations for any
reaction.
Plan
We recognize that both Ca(NO3)2 (solubility guideline 3) and NaCl (solubility guidelines 2
and 4) are soluble compounds. We use the solubility guidelines to determine whether any of
the possible products are insoluble.
Solution
At the instant of mixing, the resulting solution contains four ions:
Ca2ϩ(aq), NO3
Ϫ(aq), Naϩ(aq), ClϪ(aq)
New combinations of ions could be CaCl2 and NaNO3. But solubility guideline 4 tells us that
CaCl2 is a soluble compound, and solubility guidelines 2 and 3 tell us that NaNO3 is a soluble
compound. Therefore no precipitate forms in this solution.
You should now work Exercise 84.
EXAMPLE 4-13 Solubility Guidelines and Precipitation Reactions
Will a precipitate form when aqueous solutions of CaCl2 and K3PO4 are mixed in reasonable
concentrations? Write balanced formula unit, total ionic, and net ionic equations for any reac-
tion.
Plan
We recognize that both CaCl2 (solubility guideline 4) and K3PO4 (solubility guideline 2) are
soluble compounds. We use the solubility guidelines to determine whether any of the possible
products are insoluble.
Solution
At the instant of mixing, the resulting solution contains four ions:
Ca2ϩ(aq), ClϪ(aq), Kϩ(aq), PO4
3Ϫ(aq)
New combinations of ions could be KCl and Ca3(PO4)2. Solubility guidelines 2 and 4 tell us
that KCl is a soluble compound; solubility guideline 7 tells us that Ca3(PO4)2 is an insoluble
compound, so a precipitate of Ca3(PO4)2 forms in this solution.
The equations for the formation of calcium phosphate follow.
3CaCl2(aq) ϩ 2K3PO4(aq) 88n Ca3(PO4)2(s) ϩ 6KCl(aq)
3[Ca2ϩ(aq) ϩ 2ClϪ(aq)] ϩ 2[3Kϩ(aq) ϩ PO4
3Ϫ(aq)] 88n
Ca3(PO4)2(s) ϩ 6[Kϩ(aq) ϩ ClϪ(aq)]
3Ca2ϩ(aq) ϩ 2PO4
3Ϫ(aq) 88n Ca3(PO4)2(s)
You should now work Exercise 90.
Hydrogen sulfide, H2S, is responsible
for the smell of rotten eggs.
4-9 Metathesis Reactions 159
3 Gas-Formation Reactions
The formation of an insoluble or slightly soluble gas provides a driving force for a third
type of metathesis reaction that we call a gas-formation reaction. The only common
gases that are very soluble in water are HCl(g) and NH3(g). All other gases are sufficiently
insoluble to force a reaction to proceed if they are formed as a reaction product.
When an acid—for example, hydrochloric acid—is added to solid calcium carbonate,
a reaction occurs in which carbonic acid, a weak acid, is produced.
2HCl(aq) ϩ CaCO3(s) 88n H2CO3(aq) ϩ CaCl2(aq)
2[Hϩ(aq) ϩ ClϪ(aq)] ϩ CaCO3(s) 88n H2CO3(aq) ϩ [Ca2ϩ(aq) ϩ 2ClϪ(aq)]
2Hϩ(aq) ϩ CaCO3(s) 88n H2CO3(aq) ϩ Ca2ϩ(aq)
The heat generated in the reaction causes thermal decomposition of carbonic acid to
gaseous carbon dioxide and water:
H2CO3(aq) 88n CO2(g) ϩ H2O(ᐉ)
Most of the CO2 bubbles off, and the reaction goes to completion (with respect to the
limiting reactant). The net effect is the conversion of ionic species into nonionized molecules
of a gas (CO2) and water.
2HCl(aq) ϩ CaCO3(s) 88n CO2(g) ϩ H2O(ᐉ) ϩ CaCl2(aq)
Salts containing the sulfite ion, SO3
2Ϫ, react with acids in a similar manner to form
sulfur dioxide gas, SO2(g).
SO3
2Ϫ(aq) ϩ 2Hϩ(aq) 88n SO2(g) ϩ H2O(ᐉ)
Many sulfide salts react with acids to form gaseous hydrogen sulfide, H2S. The low
solubility of H2S in water helps the reaction to proceed.
MnS(s) ϩ 2HCl(aq) 88n MnCl2(aq) ϩ H2S(g)
Blackboard chalk is mostly calcium carbonate,
CaCO3. Bubbles of carbon dioxide, CO2, are clearly
visible in this photograph of CaCO3 reacting with
HCl in a gas-forming metathesis reaction.
Millions of compounds are known, so
it is important to be able to associate
names and formulas in a systematic
way.
160 CHAPTER 4: Some Types of Chemical Reactions
SUMMARY OF REACTION TYPES
Table 4-14 summarizes the reaction types we have presented. Remember that a reaction
might be classified in more than one category.
EXAMPLE 4-14 Classifying Reactions
Classify each of the following reactions.
(a) Zn(s) ϩ AgNO3(aq) 88n Zn(NO3)2(aq) ϩ 2Ag(s)
heat
(b) Ca(OH)2(s) 888n CaO(s) ϩ H2O(g)
heat
(c) 2HI(g) 888n H2(g) ϩ I2(g)
(d) Cu(NO3)2(aq) ϩ Na2S(aq) 88n CuS(s) ϩ 2NaNO3(aq)
(e) SO2(g) ϩ H2O(ᐉ) 88n H2SO3(aq)
(f ) H2SO3(aq) ϩ 2KOH(aq) 88n K2SO3(aq) ϩ 2H2O(ᐉ)
Plan
We identify each reaction type by its characteristics, using Table 4-13 and the appropriate
sections as a guide.
Solution
(a) One element, Zn, displaces another, Ag, from a compound; this is a displacement reac-
tion.
(b) A single compound breaks apart into two compounds; this is a decomposition reaction.
(c) A single compound breaks apart into two elements; this is another decomposition reac-
tion.
However, now there are changes in oxidation numbers; H changes from ϩ1 in HI
to 0 in H2, and I changes from Ϫ1 in HI to 0 in I2. So this is also an oxidationreduction
(redox) reaction.
(d) The positive and negative ions in the two reactant compounds change partners; this is a
metathesis reaction. An insoluble product, CuS(s), is formed, so the reaction is a precipitation
reaction.
(e) Two compounds combine to form a single product; this is a combination reaction.
(f ) The positive and negative ions change partners; this is a metathesis reaction. An acid and
a base react to form a salt and water; this is an acid-base (neutralization) reaction.
You should now work Exercises 93 through 101.
NAMING INORGANIC COMPOUNDS
The rules for naming inorganic compounds are set down by the Committee on Inorganic
Nomenclature of the International Union of Pure and Applied Chemistry (IUPAC). The
names and formulas of a few organic compounds were given in Table 2-2, and more
systematic rules for naming them will appear in Chapter 27.
4-10
4-10 Summary of Reaction Types 161
TABLE 4-14 Summary and Examples of Reaction Types
Section Reaction Type, Examples Characteristics
4-5 Oxidation-Reduction (Redox) Oxidation numbers (Sec. 4-4) of some elements change;
at least one element must increase and at least one
must decrease in oxidation number.
4-6 Combination More than one reactant, single product
1. Element ϩ element n compound
2Al(s) ϩ 3Cl2(g) n 2AlCl3(s)*
P4(s) ϩ 10Cl2(g) n 4PCl5(s)*
2. Compound ϩ element n compound
SF4(g) ϩ F2(g) n SF6(g)*
2SO2(g) ϩ O2(g) n 2SO3(ᐉ)*
3. Compound ϩ compound n compound
CaO(s) ϩ CO2(g) n CaCO3(s)
Na2O(s) ϩ H2O(ᐉ) n 2NaOH(aq)
4-7 Decomposition Single reactant, more than one product
1. Compound n element ϩ element
2HgO(s) n 2Hg(g) ϩ O2(g)*
2H2O(ᐉ) n 2H2(g) ϩ O2(g)*
2. Compound n compound ϩ element
2NaNO3(s) n 2NaNO2(s) ϩ O2(g)*
2H2O2(ᐉ) n 2H2O(ᐉ) ϩ O2(g)*
3. Compound n compound ϩ compound
CaCO3(s) n CaO(s) ϩ CO2(g)
Mg(OH)2(s) n MgO(s) ϩ H2O(ᐉ)
4-8 Displacement One element displaces another from a compound:
Zn(s) ϩ CuSO4(aq) n Cu(s) ϩ ZnSO4(aq)* Element ϩ compound n element ϩ compound
Zn(s) ϩ H2SO4(aq) n H2(g) ϩ ZnSO4(aq)* Activity series (Table 4-12) summarizes metals and
Cl2(g) ϩ 2NaI(aq) n I2(s) ϩ 2NaCl(aq)* hydrogen; halogen activities (Group VIIA) decrease
going down the group
4-9 Metathesis Positive and negative ions in two compounds appear to
“change partners” to form two new compounds; no
change in oxidation numbers
1. Acid-base neutralization Product is a salt; water is often formed
HCl(aq) ϩ NaOH(aq) n NaCl(aq) ϩ H2O(ᐉ)
CH3COOH(aq) ϩ KOH(aq) n KCH3COO(aq) ϩ H2O(ᐉ)
HCl(aq) ϩ NH3(aq) n NH4Cl(aq)
2H3PO4(aq) ϩ 3Ca(OH)2(aq) n Ca3(PO4)2(s) ϩ 6H2O(ᐉ)†
2. Precipitation Products include an insoluble substance, which
CaCl2(aq) ϩ Na2CO3(aq) n CaCO3(s) ϩ 2NaCl(aq) precipitates from solution as a solid; solubility
Pb(NO3)2(aq) ϩ K2CrO4(aq) n PbCrO4(s) ϩ 2KNO3(aq) guidelines assist in predicting, recognizing
2H3PO4(aq) ϩ 3Ca(OH)2(aq) n Ca3(PO4)2(s) ϩ 6H2O(ᐉ)†
3. Gas-formation Products include an insoluble or slightly soluble gas,
2HCl(aq) ϩ CaCO3(s) n CO2(g) ϩ H2O(ᐉ) ϩ CaCl2(aq) which escapes from solution
MnS(s) ϩ 2HCl(aq) n MnCl2(aq) ϩ H2S(g)
*These examples are also oxidation-reduction (redox) reactions.
†This reaction is both an acid-base neutralization reaction and a precipitation reaction.
NAMING BINARY COMPOUNDS
Binary compounds consist of two elements; they may be either ionic or molecular. The
rule is to name the more metallic element first and the less metallic element second. The
less metallic element is named by adding an “-ide” suffix to the element’s unambiguous
stem. Stems for the nonmetals follow.
4-11
The stem for each element is derived
from the name of the element.
162 CHAPTER 4: Some Types of Chemical Reactions
IIIA IVA VA VIA VIIA
H hydr
B bor C carb N nitr O ox F fluor
Si silic P phosph S sulf Cl chlor
As arsen Se selen Br brom
Sb antimon Te tellur I iod
Formula Name Formula Name
KBr potassium bromide Rb2S rubidium sulfide
CaCl2 calcium chloride Ba3N2 barium nitride
NaH sodium hydride SrO strontium oxide
Binary ionic compounds contain metal cations and nonmetal anions. The cation is
named first and the anion second.
The preceding method is sufficient for naming binary ionic compounds containing
metals that exhibit only one oxidation number other than zero (Section 4-4). Most transition
metals and the metals of Groups IIIA (except Al), IVA, and VA, exhibit more than
one oxidation number. These metals may form two or more binary compounds with the
same nonmetal. To distinguish among all the possibilities, the oxidation number of the
metal is indicated by a Roman numeral in parentheses following its name. This method
can be applied to any binary compound of a metal and a nonmetal.
The advantage of the IUPAC system is that if you know the formula you can write the
exact and unambiguous name; if you are given the name you can write the formula at
once. An older method, still in use but not recommended by the IUPAC, uses “-ous” and
“-ic” suffixes to indicate lower and higher oxidation numbers, respectively. This system
can distinguish between only two different oxidation numbers for a metal. It is therefore
not as useful as the Roman numeral system.
Ox. No. Ox. No.
Formula of Metal Name Formula of Metal Name
Cu2O ϩ1 copper(I) oxide SnCl2 ϩ2 tin(II) chloride
CuF2 ϩ2 copper(II) fluoride SnCl4 ϩ4 tin(IV) chloride
FeS ϩ2 iron(II) sulfide PbO ϩ2 lead(II) oxide
Fe2O3 ϩ3 iron(III) oxide PbO2 ϩ4 lead(lV) oxide
Familiarity with the older system is
still necessary. It is still widely used
in many scientific, engineering, and
medical fields.
Roman numerals are not necessary for
metals that commonly exhibit only one
oxidation number in their compounds.
If you don’t already know them, you
should learn these common prefixes.
Number Prefix
2 di
3 tri
4 tetra
5 penta
6 hexa
7 hepta
8 octa
9 nona
10 deca
4-11 Naming Binary Compounds 163
Some compounds contain polyatomic ions that behave much like monatomic anions.
Compounds that contain these ions are called pseudobinary ionic compounds. The
prefix “pseudo-” means “false”; these compounds are named as though they were binary
compounds. The common examples of such polyatomic anions are the hydroxide ion,
OHϪ, and the cyanide ion, CNϪ. The ammonium ion, NH4
ϩ, is the common cation that
behaves like a simple metal cation.
Ox. No. Ox. No.
Formula of Metal Name Formula of Metal Name
CuCl ϩ1 cuprous chloride SnF2 ϩ2 stannous fluoride
CuCl2 ϩ2 cupric chloride SnF4 ϩ4 stannic fluoride
FeO ϩ2 ferrous oxide Hg2Cl2 ϩ1 mercurous chloride
FeBr3 ϩ3 ferric bromide HgCl2 ϩ2 mercuric chloride
Formula Name Formula Name
NH4I ammonium iodide NH4CN ammonium cyanide
Ca(CN)2 calcium cyanide Cu(OH)2 copper(II) hydroxide or cupric hydroxide
NaOH sodium hydroxide Fe(OH)3 iron(III) hydroxide or ferric hydroxide
A list of common cations and anions appears in Table 4-15. It will enable you to name
many of the ionic compounds you encounter.
Nearly all binary molecular compounds involve two nonmetals bonded together.
Although many nonmetals can exhibit different oxidation numbers, their oxidation
numbers are not properly indicated by Roman numerals or suffixes. Instead, elemental
proportions in binary covalent compounds are indicated by using a prefix system for
both elements. The Greek and Latin prefixes for one through ten are mono, di, tri, tetra,
penta, hexa, hepta, octa, nona, and deca. The prefix “mono-” is omitted for both elements
except in the common name for CO, carbon monoxide. We use the minimum
number of prefixes needed to name a compound unambiguously. The final “a” in a prefix
is omitted when the nonmetal stem begins with the letter “o”; we write “heptoxide,” not
“heptaoxide.”
Formula Name Formula Name
SO2 sulfur dioxide Cl2O7 dichlorine heptoxide
SO3 sulfur trioxide CS2 carbon disulfide
N2O4 dinitrogen tetroxide SF4 sulfur tetrafluoride
As4O6 tetraarsenic hexoxide SF6 sulfur hexafluoride
Binary acids are compounds in which H is bonded to a Group VIA element other
than O or to a Group VIIA element; they act as acids when dissolved in water. The pure
compounds are named as typical binary compounds. Their aqueous solutions are named by
modifying the characteristic stem of the nonmetal with the prefix “hydro-” and the suffix
“-ic” followed by the word “acid.” The stem for sulfur in this instance is “sulfur” rather
than “sulf.”
164 CHAPTER 4: Some Types of Chemical Reactions
Formula Name of Compound Name of Aqueous Solution
HCl hydrogen chloride hydrochloric acid, HCl(aq)
HF hydrogen fluoride hydrofluoric acid, HF(aq)
H2S hydrogen sulfide hydrosulfuric acid, H2S(aq)
HCN hydrogen cyanide hydrocyanic acid, HCN(aq)
In later chapters we will learn additional systematic rules for naming more complex
compounds.
TABLE 4-15 Formulas, Ionic Charges, and Names for Some Common Ions
Common Cations Common Anions
Formula Charge Name Formula Charge Name
Liϩ 1ϩ lithium ion FϪ 1Ϫ fluoride ion
Naϩ 1ϩ sodium ion ClϪ 1Ϫ chloride ion
Kϩ 1ϩ potassium ion BrϪ 1Ϫ bromide ion
NH4
ϩ 1ϩ ammonium ion IϪ 1Ϫ iodide ion
Agϩ 1ϩ silver ion OHϪ 1Ϫ hydroxide ion
CNϪ 1Ϫ cyanide ion
Mg2ϩ 2ϩ magnesium ion ClOϪ 1Ϫ hypochlorite ion
Ca2ϩ 2ϩ calcium ion ClO2
Ϫ 1Ϫ chlorite ion
Ba2ϩ 2ϩ barium ion ClO3
Ϫ 1Ϫ chlorate ion
Cd2ϩ 2ϩ cadmium ion ClO4
Ϫ 1Ϫ perchlorate ion
Zn2ϩ 2ϩ zinc ion CH3COOϪ 1Ϫ acetate ion
Cu2ϩ 2ϩ copper(II) ion or cupric ion MnO4
Ϫ 1Ϫ permanganate ion
Hg2
2ϩ 2ϩ mercury(I) ion or mercurous ion NO2
Ϫ 1Ϫ nitrite ion
Hg2ϩ 2ϩ mercury(II) ion or mercuric ion NO3
Ϫ 1Ϫ nitrate ion
Mn2ϩ 2ϩ manganese(II) ion or manganous ion SCNϪ 1Ϫ thiocyanate ion
Co2ϩ 2ϩ cobalt(II) ion or cobaltous ion
Ni2ϩ 2ϩ nickel(II) ion or nickelous ion O2Ϫ 2Ϫ oxide ion
Pb2ϩ 2ϩ lead(II) ion or plumbous ion S2Ϫ 2Ϫ sulfide ion
Sn2ϩ 2ϩ tin(II) ion or stannous ion HSO3
Ϫ 1Ϫ hydrogen sulfite ion or bisulfite ion
Fe2ϩ 2ϩ iron(II) ion or ferrous ion SO3
2Ϫ 2Ϫ sulfite ion
HSO4
Ϫ 1Ϫ hydrogen sulfate ion or bisulfate ion
Fe3ϩ 3ϩ iron(III) ion or ferric ion SO4
2Ϫ 2Ϫ sulfate ion
Al3ϩ 3ϩ aluminum ion HCO3
Ϫ 1Ϫ hydrogen carbonate ion or bicarbonate ion
Cr3ϩ 3ϩ chromium(III) ion or chromic ion CO3
2Ϫ 2Ϫ carbonate ion
CrO4
2Ϫ 2Ϫ chromate ion
Cr2O7
2Ϫ 2Ϫ dichromate ion
PO4
3Ϫ 3Ϫ phosphate ion
AsO4
3Ϫ 3Ϫ arsenate ion
The oxoacid with the central element
in the highest oxidation state usually
contains more O atoms. Oxoacids
with their central elements in lower
oxidation states usually have fewer O
atoms.
4-12 Naming Ternary Acids and Their Salts 165
NAMING TERNARY ACIDS AND THEIR SALTS
A ternary compound consists of three elements. Ternary acids (oxoacids) are compounds
of hydrogen, oxygen, and (usually) a nonmetal. Nonmetals that exhibit more than one
oxidation state form more than one ternary acid. These ternary acids differ in the number
of oxygen atoms they contain. The suffixes “-ous” and “-ic” following the stem name of
the central element indicate lower and higher oxidation states, respectively. One common
ternary acid of each nonmetal is (somewhat arbitrarily) designated as the “-ic” acid. That
is, it is named by putting the element stem before the “-ic” suffix. The common ternary
“-ic acids” are shown in Table 4-16. It is important to learn the names and formulas of
these acids, because the names of all other ternary acids and salts are derived from them.
There are no common “-ic” ternary acids for the omitted nonmetals.
Acids containing one fewer oxygen atom per central atom are named in the same way
except that the “-ic” suffix is changed to “-ous.” The oxidation number of the central
element is lower by 2 in the “-ous” acid than in the “-ic” acid.
4-12
Formula Ox. No. Name Formula Ox. No. Name
H2SO3 ϩ4 sulfurous acid H2SO4 ϩ6 sulfuric acid
HNO2 ϩ3 nitrous acid HNO3 ϩ5 nitric acid
H2SeO3 ϩ4 selenous acid H2SeO4 ϩ6 selenic acid
HBrO2 ϩ3 bromous acid HBrO3 ϩ5 bromic acid
TABLE 4-16 Formulas of Some “-ic” Acids
Periodic Group of Central Elements
IIA IVA VA VIA VIIA
H3BO3 H2CO3 HNO3
boric acid carbonic acid nitric acid
H4SiO4 H3PO4 H2SO4 HClO3
silicic acid phosphoric acid sulfuric acid chloric acid
H3AsO4 H2SeO4 HBrO3
arsenic acid selenic acid bromic acid
H6TeO6 HIO3
telluric acid iodic acid
Note that the oxidation state of the central atom is equal to its periodic group number, except for the halogens.
ϩ3 ϩ4 ϩ5
ϩ5 ϩ6 ϩ5
ϩ5
ϩ5
ϩ6
ϩ6
ϩ5
ϩ4
Notice that H2N2O2 has a 1Ϻ1 ratio of
nitrogen to oxygen, as would the
hypothetical HNO.
166 CHAPTER 4: Some Types of Chemical Reactions
Ternary acids that have one fewer O atom than the “-ous” acids (two fewer O atoms
than the “-ic” acids) are named using the prefix “hypo-” and the suffix “-ous.” These are
acids in which the oxidation state of the central nonmetal is lower by 2 than that of the
central nonmetal in the “-ous acids.”
Formula Ox. No. Name
HClO ϩ1 hypochlorous acid
H3PO2 ϩ1 hypophosphorous acid
HIO ϩ1 hypoiodous acid
H2N2O2 ϩ1 hyponitrous acid
Acids containing one more oxygen atom per central nonmetal atom than the normal
“-ic acid” are named “perstemic” acids.
Formula Ox. No. Name
HClO4 ϩ7 perchloric acid
HBrO4 ϩ7 perbromic acid
HIO4 ϩ7 periodic acid
The oxoacids of chlorine can be summarized as follows:
Formula Ox. No. Name
HClO ϩ1 hypochlorous acid
HClO2 ϩ3 chlorous acid
HClO3 ϩ5 chloric acid
HClO4 ϩ7 perchloric acid
Ternary salts are compounds that result from replacing the hydrogen in a ternary acid
with another ion. They usually contain metal cations or the ammonium ion. As with binary
compounds, the cation is named first. The name of the anion is based on the name of the
ternary acid from which it is derived.
An anion derived from a ternary acid with an “-ic” ending is named by dropping the
“-ic acid” and replacing it with “-ate.” An anion derived from an “-ous acid” is named by
replacing the suffix “-ous acid” with “-ite.” The “per-” and “hypo-” prefixes are retained.
Formula Name
(NH4)2SO4 ammonium sulfate (SO4
2Ϫ, from sulfuric acid, H2SO4)
KNO3 potassium nitrate (NO3
Ϫ, from nitric acid, HNO3)
Ca(NO2)2 calcium nitrite (NO2
Ϫ, from nitrous acid, HNO2)
LiClO4 lithium perchlorate (ClO4
Ϫ, from perchloric acid, HClO4)
FePO4 iron(III) phosphate (PO4
3Ϫ, from phosphoric acid, H3PO4)
NaClO sodium hypochlorite (ClOϪ, from hypochlorous acid, HClO)
Acidic salts contain anions derived from ternary polyprotic acids in which one or more
acidic hydrogen atoms remain. These salts are named as if they were the usual type of
ternary salt, with the word “hydrogen” or “dihydrogen” inserted after the name of the
cation to show the number of acidic hydrogen atoms.
Key Terms 167
An older, commonly used method (which is not recommended by the IUPAC, but
which is widely used in commerce) involves the use of the prefix “bi-” attached to the
name of the anion to indicate the presence of an acidic hydrogen. According to this system,
NaHSO4 is called sodium bisulfate and NaHCO3 is called sodium bicarbonate.
Formula Name Formula Name
NaHSO4 sodium hydrogen sulfate KH2PO4 potassium dihydrogen phosphate
NaHSO3 sodium hydrogen sulfite K2HPO4 potassium hydrogen phosphate
NaHCO3 sodium hydrogen carbonate
Problem-Solving Tip: Naming Ternary Acids and Their Anions
The following table might help you to remember the names of the ternary acids and
their ions. First learn the formulas of the acids mentioned earlier that end with “-ic acid.”
Then relate possible other acids to the following table. The stem (XXX) represents the
stem of the name, for example, “nitr,” “sulfur,” or “chlor.”
Ternary Acid Anion
perXXXic acid perXXXate
XXXic acid XXXate
XXXous acid XXXite
hypoXXXous acid hypoXXXite
Decreasing
oxidationnumber
ofcentralatom
Decreasingnumber
ofoxygenatoms
oncentralatom
6666666666666666
g
6666666666666666
g
Key Terms
Acid A substance that produces Hϩ(aq) ions in aqueous solution.
Strong acids ionize completely or almost completely in dilute
aqueous solution. Weak acids ionize only slightly.
Acid-base reaction See Neutralization reaction.
Active metal A metal that readily loses electrons to form cations.
Activity series A listing of metals (and hydrogen) in order of
decreasing activity.
Alkali metals Elements of Group IA in the periodic table, except
hydrogen.
Alkaline earth metals Group IIA elements in the periodic table.
Atomic number The number of protons in the nucleus of an
atom of an element.
Base A substance that produces OHϪ(aq) ions in aqueous solution.
Strong bases are soluble in water and are completely
dissociated. Weak bases ionize only slightly.
Binary acid A binary compound in which H is bonded to a nonmetal
in Group VIIA or a nonmetal other than oxygen in Group
VIA.
Binary compound A compound consisting of two elements; may
be ionic or molecular.
Chemical periodicity The variation in properties of elements
with their positions in the periodic table.
Combination reaction Reaction in which two substances (elements
or compounds) combine to form one compound.
Decomposition reaction Reaction in which a compound
decomposes to form two or more products (elements, compounds,
or some combination of these).
Displacement reaction A reaction in which one element displaces
another from a compound.
Disproportionation reaction A redox reaction in which the oxidizing
agent and the reducing agent are the same element.
Dissociation In aqueous solution, the process in which a solid
ionic compound separates into its ions.
Electrolyte A substance whose aqueous solutions conduct elec-
tricity.
Formula unit equation An equation for a chemical reaction in
which all formulas are written as complete formulas.
Gas-formation reaction A metathesis reaction in which an
insoluble or slightly soluble gas is formed as a product.
Group (family) The elements in a vertical column of the periodic
table.
Halogens Group VIIA elements in the periodic table.
Ionization In aqueous solution, the process in which a molecular
compound separates to form ions.
168 CHAPTER 4: Some Types of Chemical Reactions
Exercises
The Periodic Table
1. State the periodic law. What does it mean?
2. What was Mendeleev’s contribution to the construction of
the modern periodic table?
3. Consult a handbook of chemistry, and look up melting
points of the elements of Periods 2 and 3. Show that melting
point is a property that varies periodically for these
elements.
*4. Mendeleev’s periodic table was based on increasing atomic
weight. Argon has a higher atomic weight than potassium,
yet in the modern table argon appears before potassium.
Explain how this can be.
5. Estimate the density of antimony from the following densities
(g/cm3): As, 5.72; Bi, 9.8; Sn, 7.30; Te, 6.24. Show
how you arrived at your answer. Using a reference other
than your textbook, look up the density of antimony. How
does your predicted value compare with the reported
value?
6. Given the following melting points in °C, estimate the
value for CBr4: CF4, Ϫ184; CCl4, Ϫ23; CI4, 171 (decomposes).
Using a reference other than your textbook, look
Metal An element below and to the left of the stepwise division
(metalloids) of the periodic table; about 80% of the known elements
are metals.
Metalloids Elements with properties intermediate between metals
and nonmetals: B, Si, Ge, As, Sb, Te, Po, and At.
Metathesis reaction A reaction in which the positive and negative
ions in two compounds “change partners,” with no change
in oxidation numbers, to form two new compounds.
Net ionic equation An equation that results from canceling
spectator ions from a total ionic equation.
Neutralization reaction The reaction of an acid with a base to
form a salt. Often, the reaction of hydrogen ions with hydroxide
ions to form water molecules.
Noble (rare) gases Elements of Group VIIIA in the periodic
table.
Nonelectrolyte A substance whose aqueous solutions do not
conduct electricity.
Nonmetals Elements above and to the right of the metalloids in
the periodic table.
Oxidation An increase in oxidation number; corresponds to a
loss of electrons.
Oxidation numbers Arbitrary numbers that can be used as
mechanical aids in writing formulas and balancing equations;
for single-atom ions they correspond to the charge on the ion;
less metallic atoms are assigned negative oxidation numbers in
compounds and polyatomic ions.
Oxidation–reduction reaction A reaction in which oxidation
and reduction occur; also called a redox reaction.
Oxidation states See Oxidation numbers.
Oxidizing agent The substance that oxidizes another substance
and is reduced.
Period The elements in a horizontal row of the periodic table.
Periodic law The properties of the elements are periodic functions
of their atomic numbers.
Periodic table An arrangement of elements in order of increasing
atomic number that also emphasizes periodicity.
Periodicity Regular periodic variations of properties of elements
with atomic number (and position in the periodic table).
Precipitate An insoluble solid that forms and separates from a
solution.
Precipitation reaction A reaction in which a solid (precipitate)
forms.
Pseudobinary ionic compound A compound that contains
more than two elements but is named like a binary compound.
Redox reaction See Oxidation–reduction reaction.
Reducing agent The substance that reduces another substance
and is oxidized.
Reduction A decrease in oxidation number; corresponds to a
gain of electrons.
Reversible reaction A reaction that occurs in both directions;
described with double arrows (34).
Salt A compound that contains a cation other than Hϩ and an
anion other than OHϪ or O2Ϫ.
Semiconductor A substance that does not conduct electricity at
low temperatures but does so at higher temperatures.
Spectator ions Ions in solution that do not participate in a chemical
reaction. They do not appear in net ionic equations.
Strong acid An acid that ionizes (separates into ions) completely,
or very nearly completely, in dilute aqueous solution.
Strong electrolyte A substance that conducts electricity well in
dilute aqueous solution.
Strong base Metal hydroxide that is soluble in water and dissociates
completely in dilute aqueous solution.
Ternary acid A ternary compound containing H, O, and another
element, usually a nonmetal.
Ternary compound A compound consisting of three elements;
may be ionic or molecular.
Total ionic equation An equation for a chemical reaction written
to show the predominant form of all species in aqueous
solution or in contact with water.
Weak acid An acid that ionizes only slightly in dilute aqueous
solution.
Weak base A molecular substance that ionizes only slightly in
water to produce an alkaline (base) solution.
Weak electrolyte A substance that conducts electricity poorly
in dilute aqueous solution.
Exercises 169
up the melting point of CBr4. How does your predicted
value compare with the reported value?
7. Calcium and magnesium form the following compounds:
CaCl2, MgCl2, CaO, MgO, Ca3N2, and Mg3N2. Predict
the formula for a compound of (a) magnesium and sulfur,
(b) barium and bromine.
8. The formulas of some hydrides of second-period representative
elements are as follows: BeH2, BH3, CH4, NH3,
H2O, HF. A famous test in criminology laboratories for
the presence of arsenic (As) involves the formation of
arsine, the hydride of arsenic. Predict the formula of arsine.
Which ones ionize only slightly? (a) HF, (b) H2SO3,
(c) CH3COOH, (d) HNO3.
22. The most common weak base is present in a common
household chemical. Write the equation for the ionization
of this weak base.
23. Summarize the electrical properties of strong electrolytes,
weak electrolytes, and nonelectrolytes.
24. What is the difference between ionization and dissociation
in aqueous solution?
25. List names and formulas of five insoluble bases.
26. Many household “chemicals” are acidic or basic. List a few
of each kind.
27. What are reversible reactions? Give some examples.
28. Which of the following are strong electrolytes? Weak
electrolytes? Nonelectrolytes? (a) NaNO3, (b) Ba(OH)2,
(c) CH3OH, (d) HCN, (e) Al(NO3)3.
29. Classify the following as strong electrolytes, weak electrolytes,
or nonelectrolytes: (a) HClO4, (b) HClO2,
(c) CH3CH2OH, (d) CH3COOH, (e) HNO3.
30. Write the formulas of two soluble and two insoluble chlorides,
sulfates, and hydroxides.
31. Describe an experiment for classifying each of these compounds
as a strong electrolyte, a weak electrolyte, or a
nonelectrolyte: Na2CO3, HCN, CH3OH, H2S, H2SO4,
NH3. Predict and explain the expected results.
32. (a) Which of these are acids? HI, NH3, H2SeO4, BF3,
Fe(OH)3, H2S, C6H6, CsOH, H3PO3, HCN. (b) Which
of these are bases? NaOH, H2Se, BCl3, NH3.
*33. Classify each substance as either an electrolyte or a
nonelectrolyte: NH4Cl, HBr, C6H6, Zn(CH3COO)2,
Cu(NO3)2, CH3COOH, C12H22O11 (sugar), LiOH,
KHCO3, CCl4, La2(SO4)3, I2.
*34. Classify each substance as either a strong or weak electrolyte,
and then list (a) the strong acids, (b) the strong
bases, (c) the weak acids, and (d) the weak bases. NaCl,
MgSO4, HCl, H2C2O4, Ba(NO3)2, H3PO4, CsOH,
HNO3, HI, Ba(OH)2, LiOH, C2H5COOH, NH3, KOH,
Mg(CH3COO)2, HCN, HClO4.
35. Some chemical reactions reach an equilibrium, rather than
going to completion. What is “equal” in such an equilib-
rium?
36. Vinegar is 5% acetic acid, an organic acid, by mass. Many
organic acids occur in living systems. What conclusion can
be drawn from this information as to the strengths of
organic acids?
Oxidation Numbers
37. Assign oxidation numbers to the element specified in each
group of compounds.
(a) N in NO, N2O3, N2O4, NH3, N2H4, NH2OH,
HNO2, HNO3
(b) C in CO, CO2, CH2O, CH4O, C2H6O, (COOH)2,
Na2CO3
(c) S in S8, H2S, SO2, SO3, Na2SO3, H2SO4, K2SO4
9. Distinguish between the following terms clearly and concisely,
and provide specific examples of each: groups
(families) of elements, and periods of elements.
10. Write names and symbols for (a) the alkaline earth metals,
(b) the Group IVA elements, (c) the Group VIB
elements.
11. Write names and symbols for (a) the alkali metals, (b) the
noble gases, (c) the Group IIIA elements.
12. Define and illustrate the following terms clearly and concisely:
(a) metals, (b) nonmetals, (c) halogens.
Aqueous Solutions
13. Define and distinguish among (a) strong electrolytes,
(b) weak electrolytes, and (c) nonelectrolytes.
14. Three common classes of compounds are electrolytes.
Name them, and give an example of each.
15. Define (a) acids, (b) bases, (c) salts, and (d) molecular com-
pounds.
16. How can a salt be related to a particular acid and a particular
base?
17. List the names and formulas of the common strong acids.
18. Write equations for the ionization of the following acids:
(a) hydrochloric acid, (b) nitric acid, (c) chloric acid.
19. List names and formulas of five weak acids.
20. List names and formulas of the common strong bases.
21. Write equations for the ionization of the following acids.
Arsine burns to form a dark spot.
170 CHAPTER 4: Some Types of Chemical Reactions
38. Assign oxidation numbers to the element specified in each
group of compounds.
(a) P in PCl3, P4O6, P4O10, HPO3, H3PO3, POCl3,
H4P2O7, Mg3(PO4)2
(b) Cl in Cl2, HCl, HClO, HClO2, KClO3, Cl2O7,
Ca(ClO4)2
(c) Mn in MnO, MnO2, Mn(OH)2, K2MnO4, KMnO4,
Mn2O7
(d) O in OF2, Na2O, Na2O2, KO2
39. Assign oxidation numbers to the element specified in each
group of ions.
(a) S in S2Ϫ, SO3
2Ϫ, SO4
2Ϫ, S2O3
2Ϫ, S4O6
2Ϫ, HSϪ
(b) Cr in CrO2
Ϫ, Cr(OH)4
Ϫ, CrO4
2Ϫ, Cr2O7
2Ϫ
(c) B in BO2
Ϫ, BO3
3Ϫ, B4O7
2Ϫ
40. Assign oxidation numbers to the element specified in each
group of ions.
(a) N in N3Ϫ, NO2
Ϫ, NO3
Ϫ, N3
Ϫ, NH4
ϩ
(b) Br in BrϪ, BrOϪ, BrO2
Ϫ, BrO3
Ϫ, BrO4
Ϫ
Oxidation – Reduction Reactions
41. Define and illustrate the following terms: (a) oxidation,
(b) reduction, (c) oxidizing agent, (d) reducing agent.
42. Why must oxidation and reduction always occur simultaneously
in chemical reactions?
43. Determine which of the following are oxidation–reduction
reactions. For those that are, identify the oxidizing and
reducing agents.
(a) 3Zn(s) ϩ 2CoCl3(aq) n 3ZnCl2(aq) ϩ 2Co(s)
(b) ICl(s) ϩ H2O(ᐉ) n HCl(aq) ϩ HIO(aq)
(c) 3HCl(aq) ϩ HNO3(aq) n
Cl2(g) ϩ NOCl(g) ϩ 2H2O(ᐉ)
heat
(d) Fe2O3(s) ϩ 3CO(g) 888n 2Fe(s) ϩ 3CO2(g)
44. Determine which of the following are oxidation–reduction
reactions. For those that are, identify the oxidizing and
reducing agents.
(a) HgCl2(aq) ϩ 2KI(aq) n HgI2(s) ϩ 2KCl(aq)
(b) 4NH3(g) ϩ 3O2(g) n 2N2(g) ϩ 6H2O(g)
(c) CaCO3(s) ϩ 2HNO3(aq) n
Ca(NO3)2(aq) ϩ CO2(g) ϩ H2O(ᐉ)
(d) PCl3(ᐉ) ϩ 3H2O(ᐉ) n 3HCl(aq) ϩ H3PO3(aq)
45. Write balanced formula unit equations for the following
redox reactions:
(a) aluminum reacts with sulfuric acid, H2SO4, to produce
aluminum sulfate, Al2(SO4)3, and hydrogen
(b) nitrogen, N2, reacts with hydrogen, H2, to form
ammonia, NH3
(c) zinc sulfide, ZnS, reacts with oxygen, O2, to form zinc
oxide, ZnO, and sulfur dioxide, SO2
(d) carbon reacts with nitric acid, HNO3, to produce
nitrogen dioxide, NO2, carbon dioxide, CO2, and
water
(e) sulfuric acid reacts with hydrogen iodide, HI, to produce
sulfur dioxide, SO2, iodine, I2, and water
46. Identify the oxidizing agents and reducing agents in the
oxidation–reduction reactions given in Exercise 45.
47. Write total ionic and net ionic equations for the following
redox reactions occurring in aqueous solution or in
contact with water:
(a) Fe ϩ 2HCl 88n FeCl2 ϩ H2
(b) 2KMnO4 ϩ 16HCl 88n
2MnCl2 ϩ 2KCl ϩ 5Cl2 ϩ 8H2O
(Note: MnCl2 is water-soluble.)
(c) 4Zn ϩ 10HNO3 88n
4Zn(NO3)2 ϩ NH4NO3 ϩ 3H2O.
48. The air we inhale contains O2. We exhale CO2 and H2O.
Does this suggest that our bodily processes involve oxidation?
Why?
Combination Reactions
49. Write balanced equations that show the combination reactions
of the following Group IA metals combining with
the Group VIIA nonmetals. (a) Li and Cl2, (b) K and F2,
(c) Na and I2
50. Write balanced equations that show the combination reactions
of the following Group IIA metals and Group VIIA
nonmetals. (a) Be and F2, (b) Ca and Br2, (c) Ba and Cl2
51. Phosphorus and fluorine can react to form two compounds.
Write balanced equations for these reactions.
Which reaction requires excess phosphorus and which
requires excess fluorine?
52. Two binary compounds contain arsenic and chlorine.
What are their formulas? Under what conditions could
each be formed?
In Exercises 53 and 54, some combination reactions are described
by words. Write the balanced chemical equation for each,
and assign oxidation numbers to elements other than H and O.
53. (a) Antimony reacts with a limited amount of chlorine to
form antimony(III) chloride.
(b) Antimony(III) chloride reacts with excess chlorine to
form antimony(V) chloride.
(c) Carbon burns in a limited amount of oxygen to form
carbon monoxide.
54. (a) Sulfur trioxide reacts with aluminum oxide to form
aluminum sulfate.
(b) Dichlorine heptoxide reacts with water to form perchloric
acid.
(c) When cement “sets,” the main reaction is the combination
of calcium oxide with silicon dioxide to form
calcium silicate, CaSiO3.
Decomposition Reactions
In Exercises 55 and 56, write balanced formula unit equations
for the reactions described by words. Assign oxidation numbers
to all elements.
55. (a) Hydrogen peroxide, H2O2, is used as an antiseptic.
Blood causes it to decompose into water and oxygen.
Exercises 171
(b) When heated, ammonium nitrate can decompose
explosively to form nitrogen oxide and steam.
56. (a) A “classroom volcano” is made by heating solid ammonium
dichromate, (NH4)2Cr2O7, which decomposes
into nitrogen, chromium(III) oxide, and steam.
(b) At high temperatures, sodium nitrate (a fertilizer)
forms sodium nitrite and oxygen.
Displacement Reactions
57. Which of the following would displace hydrogen when a
piece of the metal is dropped into dilute H2SO4 solution?
Write balanced total ionic and net ionic equations for the
reactions: Zn, Cu, Sn, Al.
58. Which of the following metals would displace copper from
an aqueous solution of copper(II) sulfate? Write balanced
total ionic and net ionic equations for the reactions: Hg,
Zn, Fe, Pt.
59. Arrange the metals listed in Exercise 57 in order of increasing
activity.
60. Arrange the metals listed in Exercise 58 in order of increasing
activity.
61. Which of the following metals would displace hydrogen
from cold water? Write balanced net ionic equations for
the reactions: Zn, Na, Ca, Cr.
62. Arrange the metals listed in Exercise 61 in order of increasing
activity.
63. What is the order of decreasing activity of the halogens?
64. Of the possible displacement reactions shown, which
one(s) could occur?
(a) 2ClϪ(aq) ϩ Br2(ᐉ) n 2BrϪ(aq) ϩ Cl2(g)
(b) 2BrϪ(aq) ϩ F2(g) n 2FϪ(aq) ϩ Br2(ᐉ)
(c) 2IϪ(aq) ϩ Cl2(g) n 2ClϪ(aq) ϩ I2(s)
(d) 2BrϪ(aq) ϩ I2(s) n 2IϪ(aq) ϩ Br2(ᐉ)
65. (a) Name two common metals: one that does not displace
hydrogen from water, and one that does not displace
hydrogen from water or acid solutions.
(b) Name two common metals: one that does displace
hydrogen from water, and one that displaces hydrogen
from acid solutions but not from water. Write net
ionic equations for the reactions that occur.
66. Predict the products of each mixture. If a reaction occurs,
write the net ionic equation. If no reaction occurs, write
“no reaction.”
(a) Cd2ϩ(aq) ϩ Al n
(b) Li ϩ H2O n
(c) Ni ϩ H2O n
(d) Hg ϩ HCl(aq) n
(e) Ni ϩ H2SO4(aq) n
(f ) Fe ϩ H2SO4(aq) n
67. Use the activity series to predict whether or not the following
reactions will occur:
(a) Cu(s) ϩ Mg2ϩ n Mg(s) ϩ Cu2ϩ
(b) Ni(s) ϩ Cu2ϩ n Ni2ϩ ϩ Cu(s)
(c) Cu(s) ϩ 2Hϩ n Cu2ϩ ϩ H2(g)
(d) Mg(s) ϩ H2O(g) n MgO(s) ϩ H2(g)
68. Repeat Exercise 67 for
(a) Sn(s) ϩ Ca2ϩ n Sn2ϩ ϩ Ca(s)
heat
(b) Al2O3(s) ϩ 3H2(g) 888n 2Al(s) ϩ 3H2O(g)
(c) Cu(s) ϩ 2Hϩ n Cu2ϩ ϩ H2(g)
(d) Cu(s) ϩ Pb2ϩ n Cu2ϩ ϩ Pb(s)
Metathesis Reactions
Exercises 69 and 70 describe precipitation reactions in aqueous
solutions. For each, write balanced (i) formula unit, (ii) total ionic,
and (iii) net ionic equations. Refer to the solubility guidelines as
necessary.
69. (a) Black-and-white photographic film contains some silver
bromide, which can be formed by the reaction of
sodium bromide with silver nitrate.
(b) Barium sulfate is used when x-rays of the gastrointestinal
tract are made. Barium sulfate can be prepared
by reacting barium chloride with dilute sulfuric acid.
(c) In water purification systems small solid particles are
often “trapped” as aluminum hydroxide precipitates
and falls to the bottom of the sedimentation pool. Aluminum
sulfate reacts with calcium hydroxide (from
lime) to form aluminum hydroxide and calcium sul-
fate.
70. (a) Our bones are mostly calcium phosphate. Calcium
chloride reacts with potassium phosphate to form calcium
phosphate and potassium chloride.
(b) Mercury compounds are very poisonous. Mercury(II)
nitrate reacts with sodium sulfide to form mercury(II)
sulfide, which is very insoluble, and sodium nitrate.
(c) Chromium(III) ions are very poisonous. They can be
removed from solution by precipitating very insoluble
chromium(III) hydroxide. Chromium(III) chloride
reacts with calcium hydroxide to form chromium(III)
hydroxide and calcium chloride.
In Exercises 71 through 74, write balanced (i) formula unit, (ii)
total ionic, and (iii) net ionic equations for the reactions that
occur between the acid and the base. Assume that all reactions
occur in water or in contact with water.
71. (a) hydrochloric acid ϩ calcium hydroxide
(b) dilute sulfuric acid ϩ potassium hydroxide
(c) perchloric acid ϩ aqueous ammonia
72. (a) acetic acid ϩ potassium hydroxide
(b) sulfurous acid ϩ sodium hydroxide
(c) hydrofluoric acid ϩ lithium hydroxide
*73. (a) potassium hydroxide ϩ hydrosulfuric acid
(b) barium hydroxide ϩ hydrosulfuric acid
(c) lead(II) hydroxide ϩ hydrosulfuric acid
74. (a) sodium hydroxide ϩ sulfuric acid
(b) calcium hydroxide ϩ phosphoric acid
(c) copper(II) hydroxide ϩ nitric acid
172 CHAPTER 4: Some Types of Chemical Reactions
In Exercises 75 through 78, write balanced (i) formula unit,
(ii) total ionic, and (iii) net ionic equations for the reaction of
an acid and a base that will produce the indicated salts.
75. (a) sodium chloride, (b) sodium phosphate, (c) barium
acetate
76. (a) calcium perchlorate, (b) ammonium sulfate, (c) copper(II)
acetate
*77. (a) sodium carbonate, (b) barium carbonate, (c) nickel(II)
nitrate
*78. (a) sodium sulfide, (b) aluminum phosphate, (c) lead(II)
arsenate
79. (a) Propose a definition for salts, as a class of compounds,
on the basis of how they are formed.
(b) Provide an example, in the form of a chemical reaction,
to illustrate your definition of salts.
80. We can tell from the formula of a salt how it can be
produced. Write a balanced chemical equation for the
production of each of the following salts: (a) magnesium
sulfate, (b) aluminum sulfite, (c) potassium carbonate,
(d) zinc chlorate, (e) lithium acetate.
81. Magnesium hydroxide is a gelatinous material that forms
during the water purification process in some water treatment
plants because of magnesium ions in the water.
(a) Write the chemical equation for the reaction of
hydrochloric acid with magnesium hydroxide. (b) Explain
what drives this reaction to completion.
Precipitation Reactions
82. A common test for the presence of chloride ions is the formation
of a heavy, white precipitate when a solution of
silver nitrate is added. (a) Write the balanced chemical
equation for the production of silver chloride from silver
nitrate solution and calcium chloride solution. (b) Explain
why this reaction goes to completion.
83. Based on the solubility guidelines given in Table 4-8, how
would you write the formulas for the following substances
in a total ionic equation? (a) PbSO4, (b) Na(CH3COO),
(c) Na2CO3, (d) MnS, (e) BaCl2.
84. Repeat Exercise 83 for the following: (a) (NH4)2SO4,
(b) NaBr, (c) Sr(OH)2, (d) Mg(OH)2, (e) Na2CO3.
Refer to the solubility guidelines given in Table 4-8. Classify the
compounds in Exercises 85 through 88 as soluble, moderately
soluble, or insoluble in water.
85. (a) NaClO4, (b) AgCl, (c) Pb(NO3)2, (d) KOH,
(e) CaSO4
86. (a) BaSO4, (b) Al(NO3)3, (c) CuS, (d) Na2S,
(e) Ca(CH3COO)2
87. (a) Fe(NO3)3, (b) Hg(CH3COO)2, (c) BeCl2, (d) CuSO4,
(e) CaCO3
88. (a) KClO3, (b) NH4Cl, (c) NH3, (d) HNO2, (e) PbS
In Exercises 89 and 90, write balanced (i) formula unit, (ii) total
ionic, and (iii) net ionic equations for the reactions that occur
when aqueous solutions of the compounds are mixed.
89. (a) Ba(NO3)2 ϩ K2CO3 n
(b) NaOH ϩ NiCl2 n
(c) Al2(SO4)3 ϩ NaOH n
90. (a) Cu(NO3)2 ϩ Na2S n
(b) CdSO4 ϩ H2S n
(c) Bi2(SO4)3 ϩ (NH4)2S n
91. In each of the following, both compounds are watersoluble.
Predict whether a precipitate will form when solutions
of the two are mixed, and, if so, identify the
compound that precipitates. (a) Pb(NO3)2, NaI; (b)
Ba(NO3)2, KCl; (c) (NH4)2S, AgNO3
92. In each of the following, both compounds are watersoluble.
Predict whether a precipitate will form when solutions
of the two are mixed, and, if so, identify the
compound that precipitates. (a) NH4Br, Hg2(NO3)2; (b)
KOH, Na2S; (c) Cs2SO4, MgCl2
Identifying Reaction Types
The following reactions apply to Exercises 93 through 101.
(a) H2SO4(aq) ϩ 2KOH(aq) n K2SO4(aq) ϩ 2H2O(ᐉ)
heat
(b) 2Rb(s) ϩ Br2(ᐉ) 888n 2RbBr(s)
(c) 2KI(aq) ϩ F2(g) n 2KF(aq) ϩ I2(s)
heat
(d) CaO(s) ϩ SiO2(s) 888n CaSiO3(s)
heat
(e) S(s) ϩ O2(g) 888n SO2(g)
heat
(f ) BaCO3(s) 888n BaO(s) ϩ CO2(g)
heat
(g) HgS(s) ϩ O2(g) 888n Hg(ᐉ) ϩ SO2(g)
(h) AgNO3(aq) ϩ HCl(aq) n AgCl(s) ϩ HNO3(aq)
(i) Pb(s) ϩ 2HBr(aq) n PbBr2(s) ϩ H2(g)
(j) 2HI(aq) ϩ H2O2(aq) n I2(s) ϩ 2H2O(ᐉ)
(k) RbOH(aq) ϩ HNO3(aq) n RbNO3(aq) ϩ H2O(ᐉ)
(l) N2O5(s) ϩ H2O(ᐉ) n 2HNO3(aq)
heat
(m) H2O(g) ϩ CO(g) 888n H2(g) ϩ CO2(g)
(n) MgO(s) ϩ H2O(ᐉ) n Mg(OH)2(s)
heat
(o) PbSO4(s) ϩ PbS(s) 888n 2Pb(s) ϩ 2SO2(g)
93. Identify the precipitation reactions.
94. Identify the acid–base reactions.
95. Identify the oxidation–reduction reactions.
96. Identify the oxidizing agent and reducing agent for each
oxidation–reduction reaction.
97. Identify the displacement reactions.
98. Identify the metathesis reactions.
99. Identify the combination reactions.
100. Identify the decomposition reactions.
101. (a) Do any of these reactions fit into more than one class?
Which ones? (b) Do any of these reactions not fit into any
of our classes of reactions? Which ones?
102. Predict whether or not a solid is formed when we mix the
following; identify any solid product by name and identify
the reaction type: (a) copper(II) nitrate solution and
Exercises 173
magnesium metal, (b) barium nitrate and sodium phosphate
solutions, (c) calcium acetate solution with
aluminum metal, (d) silver nitrate and sodium iodide
solutions.
103. Predict whether or not a solid is formed when we mix the
following; identify any solid product by formula and by
name: (a) potassium permanganate and sodium phosphate
solutions, (b) lithium carbonate and cadmium nitrate,
(c) stannous fluoride and bismuth chloride, (d) strontium
sulfate with barium chloride solutions.
Naming Compounds
104. Name the following monatomic cations, using the IUPAC
system of nomenclature: (a) Liϩ, (b) Au3ϩ, (c) Ca2ϩ,
(d) Zn2ϩ, (e) Agϩ.
105. Write the chemical symbol for each of the following:
(a) sodium ion, (b) lead(II) ion, (c) silver ion, (d) mercury(II)
ion, (e) bismuth(III) ion.
106. Write the chemical formula for each of the following:
(a) chloride ion, (b) sulfide ion, (c) telluride ion, (d) oxide
ion, (e) nitrite ion.
107. Name the following ionic compounds: (a) K2S, (b) SnO2,
(c) RbBr, (d) Li2O, (e) Ba3N2.
108. Name the following ionic compounds: (a) CuI2, (b) Hg2S,
(c) Li3N, (d) MnCl2, (e) CuCO3, (f ) FeO.
109. Write the chemical formula for each of the following compounds:
(a) sodium fluoride, (b) zinc oxide, (c) barium
oxide, (d) magnesium bromide, (e) hydrogen cyanide,
(f ) copper(I) chloride.
110. Write the chemical formula for each of the following
compounds: (a) copper(II) chlorate, (b) potassium nitrite,
(c) barium phosphate, (d) copper(I) sulfate, (e) sodium sul-
fite.
111. What is the name of the acid with the formula H2CO3?
Write the formulas of the two anions derived from it and
name these ions.
112. What is the name of the acid with the formula H3PO3?
What is the name of the HPO3
2Ϫ ion?
113. Name the following binary molecular compounds: (a) NO,
(b) CO2, (c) SF6, (d) SiCl4, (e) IF.
114. Name the following binary molecular compounds: (a)
AsF3, (b) Br2O, (c) BrF5, (d) CSe2, (e) Cl2O7.
115. Write the chemical formula for each of the following
compounds: (a) iodine bromide, (b) silicon dioxide,
(c) phosphorus trichloride, (d) tetrasulfur dinitride,
(e) bromine trifluoride, (f ) hydrogen telluride, (g) xenon
tetrafluoride.
116. Write the chemical formula for each of the following compounds:
(a) diboron trioxide, (b) dinitrogen pentasulfide,
(c) phosphorus triiodide, (d) sulfur tetrachloride, (e) silicon
sulfide, (f) hydrogen sulfide, (g) tetraphosphorus
hexoxide.
117. Write formulas for the compounds that are expected to be
formed by the following pairs of ions:
118. Write the names for the compounds of Exercise 117.
119. Write balanced chemical equations for each of the following
processes: (a) Calcium phosphate reacts with
sulfuric acid to produce calcium sulfate and phosphoric
acid. (b) Calcium phosphate reacts with water containing
dissolved carbon dioxide to produce calcium hydrogen carbonate
and calcium hydrogen phosphate.
120. Write balanced chemical equations for each of the following
processes: (a) When heated, nitrogen and oxygen
combine to form nitrogen oxide. (b) Heating a mixture of
lead(II) sulfide and lead(II) sulfate produces metallic lead
and sulfur dioxide.
CONCEPTUAL EXERCISES
As we have seen, two substances may react to form different
products when they are mixed in different proportions under
different conditions. In Exercises 121 and 122, write balanced
equations for the reactions described by words. Assign oxidation
numbers.
121. (a) Ethane burns in excess air to form carbon dioxide and
water.
(b) Ethane burns in a limited amount of air to form carbon
monoxide and water.
(c) Ethane burns (poorly) in a very limited amount of air
to form elemental carbon and water.
122. (a) Butane (C4H10) burns in excess air to form carbon
dioxide and water.
(b) Butane burns in a limited amount of air to form carbon
monoxide and water.
(c) When heated in the presence of very little air, butane
“cracks” to form acetylene, C2H2, carbon monoxide,
and hydrogen.
123. Use the general metallic trend of the periodic table to predict
which element in each of the following pairs is more
1. NH4
A. ClϪ
B. OHϪ
Omit –
see note
C. SO4
2Ϫ D. PO4
3Ϫ E. NO3
Ϫ
ϩ
2. Naϩ
3. Mg2ϩ
4. Ni2ϩ
5. Fe3ϩ
6. Agϩ
NOTE: The compound NH4OH does not exist. The solution
commonly labeled “NH4OH” is aqueous ammonia, NH3(aq).
174 CHAPTER 4: Some Types of Chemical Reactions
metallic: (a) aluminum or sodium, (b) aluminum or zinc,
(c) sodium or potassium, (d) silver or sulfur.
124. Use the general metallic trend of the periodic table to predict
which element in each of the following pairs is more
metallic: (a) phosphorus or sulfur, (b) francium or uranium,
(c) oxygen or sulfur, (d) selenium or chlorine.
125. Calcium phosphate is the component of human bone that
provides rigidity. Fallout from a nuclear bomb can contain
radioactive strontium-90. These two facts are closely tied
together when one considers human health. Explain.
126. Limestone consists mainly of the mineral calcite, which is
calcium carbonate. A very similar deposit called dolostone
is composed primarily of the mineral dolomite, an ionic
substance that contains carbonate ions and a mixture of
magnesium and calcium ions. (a) Is this a surprising mixture
of ions? Explain, based on the periodic table. (b) A
test for limestone is to apply cold dilute hydrochloric acid,
which causes the rapid formation of bubbles. What causes
these bubbles?
129. When the following pairs of reactants are combined in a
beaker: (a) describe in words what the contents of the
beaker would look like before and after any reaction that
might occur, (b) use different circles for atoms, molecules,
and ions to draw a nanoscale (particulate-level) diagram of
what the contents would look like, and (c) write a chemical
equation for any reactions that might occur.
LiCl(aq) and AgNO3(aq)
NaOH(aq) and HCl(aq)
CaCO3(s) and HCl(aq)
Na2CO3(aq) and Ca(OH)2(aq)
130. Explain how you could prepare barium sulfate by (a) an
acid–base reaction, (b) a precipitation reaction, and (c) a
gas-forming reaction. The materials you have to start with
are BaCO3, Ba(OH)2, Na2SO4, and H2SO4.
BUILDING YOUR KNOWLEDGE
131. All of the following useful ammonium salts dissolve easily
in water. Suggest the appropriate aqueous acid for preparing
each salt, and describe each preparation by a balanced
“complete formula” equation. (a) Ammonium chloride:
Used in medicine as an expectorant and in certain soldering
operations as a “flux.” (b) Ammonium nitrate: A
valuable fertilizer, but potentially explosive; as in the 1995
bombing of the Federal Building in Oklahoma City.
(c) Ammonium sulfate: A common fertilizer often recommended
for alkaline soils. (d) Ammonium phosphate: A
particularly valuable fertilizer providing both nitrogen and
phosphorus, essential elements for plant growth.
132. How many moles of oxygen can be obtained by the decomposition
of 10.0 grams of reactant in each of the following
reactions?
cat.
(a) 2KClO3(s) 888n 2KCl(s) ϩ 3O2(g)
(b) 2H2O2(aq) n 2H2O(ᐉ) ϩ O2(g)
heat
(c) 2HgO(s) 888n 2Hg(ᐉ) ϩ O2(g)
133. Magnesium oxide, marketed as tablets or as an aqueous
slurry called “milk of magnesia,” is a common commercial
antacid. What volume, in milliliters, of fresh gastric juice,
corresponding in acidity to 0.17 M HCl, could be neutralized
by 125 mg of magnesium oxide?
MgO(s) ϩ 2HCl(aq) 88n MgCl2(aq) ϩ H2O(ᐉ)
134. What mass of Zn is needed to displace 20.6 grams of Cu
from CuSO4 и 5H2O?
The Dolomite Alps of Italy.
127. Chemical equations can be interpreted on either a particulate
level (atoms, molecules, ions) or a mole level (moles
of reactants and products). Write word statements to
describe the combustion of butane on a particulate level
and a mole level.
2C4H10(g) ϩ 13O2(g) 88n 8CO2(g) ϩ 10H2O(ᐉ)
128. Write word statements to describe the following reaction
on a particulate level and a mole level.
P4(s) ϩ 6Cl2(g) 88n 4PCl3(ᐉ)
5The Structure
of Atoms
OUTLINE
Subatomic Particles
5-1 Fundamental Particles
5-2 The Discovery of Electrons
5-3 Canal Rays and Protons
5-4 Rutherford and the Nuclear
Atom
5-5 Atomic Number
5-6 Neutrons
5-7 Mass Number and Isotopes
5-8 Mass Spectrometry and Isotopic
Abundance
5-9 The Atomic Weight Scale and
Atomic Weights
The Electronic Structures of Atoms
5-10 Electromagnetic Radiation
5-11 The Photoelectric Effect
5-12 Atomic Spectra and the Bohr
Atom
5-13 The Wave Nature of the
Electron
5-14 The Quantum Mechanical
Picture of the Atom
5-15 Quantum Numbers
5-16 Atomic Orbitals
5-17 Electron Configurations
5-18 The Periodic Table and
Electron Configurations
OBJECTIVES
After you have studied this chapter, you should be able to
• Describe the evidence for the existence and properties of electrons, protons, and
neutrons
• Predict the arrangements of the particles in atoms
• Describe isotopes and their composition
• Calculate atomic weights from isotopic abundance
• Describe the wave properties of light and how wavelength, frequency, and speed are
related
• Use the particle description of light, and explain how it is related to the wave
description
• Relate atomic emission and absorption spectra to important advances in atomic
theory
• Describe the main features of the quantum mechanical picture of the atom
• Describe the four quantum numbers, and give possible combinations of their values for
specific atomic orbitals
• Describe the shapes of orbitals and recall the usual order of their relative energies
• Write the electron configurations of atoms
• Relate the electron configuration of an atom to its position in the periodic table
Descriptions of waves play an
important role in our theories of
light and of atomic structure.
TT
he Dalton theory of the atom and related ideas were the basis for our study of
composition stoichiometry (Chapter 2) and reaction stoichiometry (Chapter 3), but that
level of atomic theory leaves many questions unanswered. Why do atoms combine
to form compounds? Why do they combine only in simple numerical ratios? Why are
particular numerical ratios of atoms observed in compounds? Why do different elements
have different properties? Why are they gases, liquids, solids, metals, nonmetals, and so
on? Why do some groups of elements have similar properties and form compounds with
similar formulas? The answers to these and many other fascinating questions in chemistry
are supplied by our modern understanding of the nature of atoms. But how can we
study something as small as an atom?
Much of the development of modern atomic theory was based on two broad types of
research carried out by dozens of scientists just before and after 1900. The first type dealt
with the electrical nature of matter. These studies led scientists to recognize that atoms
are composed of more fundamental particles and helped them to describe the approximate
arrangements of these particles in atoms. The second broad area of research dealt
with the interaction of matter with energy in the form of light. Such research included
studies of the colors of the light that substances give off or absorb. These studies led to
a much more detailed understanding of the arrangements of particles in atoms. It became
clear that the arrangement of the particles determines the chemical and physical properties
of each element. As we learn more about the structures of atoms, we are able to
organize chemical facts in ways that help us to understand the behavior of matter.
We will first study the particles that make up atoms and the basic structure of atoms.
Then we will take a look at the quantum mechanical theory of atoms and see how this
theory describes the arrangement of the electrons in atoms. Current atomic theory is
considerably less than complete. Even so, it is a powerful tool that helps us describe the
forces holding atoms in chemical combination with one another.
SUBATOMIC PARTICLES
FUNDAMENTAL PARTICLES
In our study of atomic structure, we look first at the fundamental particles. These are
the basic building blocks of all atoms. Atoms, and hence all matter, consist principally of
three fundamental particles: electrons, protons, and neutrons. Knowledge of the nature and
functions of these particles is essential to understanding chemical interactions. The relative
masses and charges of the three fundamental particles are shown in Table 5-1. The
5-1
5-1 Fundamental Particles 177
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 7.2, Structure of Atoms.
Many other subatomic particles, such
as quarks, positrons, neutrinos, pions,
and muons, have also been discovered.
It is not necessary to study their
characteristics to learn the
fundamentals of atomic structure that
are important in chemical reactions.
TABLE 5-1 Fundamental Particles of Matter
Charge
Particle Mass (relative scale)
electron (eϪ) 0.00054858 amu 1Ϫ
proton (p or pϩ) 1.0073 amu 1ϩ
neutron (n or n0) 1.0087 amu none
Accounts of some important
developments in atomic theory appear
on the World Wide Web; for example,
The National Museum of Science and
Industry can be found at
http://www.nmsi.ac.uk/on-line/
Electron
mass of an electron is very small compared with the mass of either a proton or a neutron.
The charge on a proton is equal in magnitude, but opposite in sign, to the charge on an
electron. Let’s examine these particles in more detail.
THE DISCOVERY OF ELECTRONS
Some of the earliest evidence about atomic structure was supplied in the early 1800s by
the English chemist Humphry Davy (1778–1829). He found that when he passed electric
current through some substances, the substances decomposed. He therefore suggested
that the elements of a chemical compound are held together by electrical forces. In 1832–
1833, Michael Faraday (1791–1867), Davy’s student, determined the quantitative relationship
between the amount of electricity used in electrolysis and the amount of chemical
reaction that occurs. Studies of Faraday’s work by George Stoney (1826–1911) led him
to suggest in 1874 that units of electric charge are associated with atoms. In 1891, he
suggested that they be named electrons.
The most convincing evidence for the existence of electrons came from experiments
using cathode-ray tubes (Figure 5-1). Two electrodes are sealed in a glass tube containing
5-2
The process is called chemical
electrolysis. Lysis means “splitting
apart.”
Study Figures 5-1 and 5-2 carefully as
you read this section.
178 CHAPTER 5: The Structure of Atoms
Figure 5-1 Some experiments with cathode-ray tubes that show the nature
of cathode rays. (a) A cathode-ray (discharge) tube, showing the production
of a beam of electrons (cathode rays). The beam is detected by observing
the glow of a fluorescent screen. (b) A small object placed in a beam of
cathode rays casts a shadow. This shows that cathode rays travel in straight
lines. (c) Cathode rays have negative electric charge, as demonstrated by
their deflection in an electric field. (The electrically charged plates produce
an electric field.) (d) Interaction of cathode rays with a magnetic field is also
consistent with negative charge. The magnetic field goes from one pole to
the other. (e) Cathode rays have mass, as shown by their ability to turn a
small paddle wheel in their path.
–
+
–
+
–
+
–
+
–
+
Anode
Cathode
Collimator
Electron beam
(cathode rays)
Cathode
Anode
Fluorescent
screen
Shadow
Charged metal
plates –
+
Magnet
S
N
Paddle wheel
(a)
(b)
(c)
(d)
(e)
gas at a very low pressure. When a high voltage is applied, current flows and rays are
given off by the cathode (negative electrode). These rays travel in straight lines toward
the anode (positive electrode) and cause the walls opposite the cathode to glow. An object
placed in the path of the cathode rays casts a shadow on a zinc sulfide screen placed near
the anode. The shadow shows that the rays travel from the cathode toward the anode.
The rays must therefore be negatively charged. Furthermore, they are deflected by electric
and magnetic fields in the directions expected for negatively charged particles.
In 1897 J. J. Thomson (1856–1940) studied these negatively charged particles more
carefully. He called them electrons, the name Stoney had suggested in 1891. By studying
the degree of deflections of cathode rays in different electric and magnetic fields, Thomson
determined the ratio of the charge (e) of the electron to its mass (m). The modern value
for this ratio is
e/m ϭ 1.75882 ϫ 108 coulomb (C)/gram
This ratio is the same regardless of the type of gas in the tube, the composition of the
electrodes, or the nature of the electric power source. The clear implication of Thomson’s
work was that electrons are fundamental particles present in all atoms. We now know that
this is true and that all atoms contain integral numbers of electrons.
Once the charge-to-mass ratio for the electron had been determined, additional experiments
were necessary to determine the value of either its mass or its charge, so that the
other could be calculated. In 1909, Robert Millikan (1868–1953) solved this dilemma with
the famous “oil-drop experiment,” in which he determined the charge of the electron.
This experiment is described in Figure 5-2. All of the charges measured by Millikan turned
The coulomb (C) is the standard unit
of quantity of electric charge. It is
defined as the quantity of electricity
transported in one second by a current
of one ampere. It corresponds to the
amount of electricity that will deposit
0.00111798 g of silver in an apparatus
set up for plating silver.
5-2 The Discovery of Electrons 179
Figure 5-2 The Millikan oil-drop experiment. Tiny spherical oil droplets are produced by
an atomizer. The mass of the spherical drop can be calculated from its volume (obtained
from a measurement of the radius of the drop with a microscope) and the known density
of the oil. A few droplets fall through the hole in the upper plate. Irradiation with X-rays
gives some of these oil droplets a negative charge. When the voltage between the plates is
increased, a negatively charged drop falls more slowly because it is attracted by the positively
charged upper plate and repelled by the negatively charged lower plate. At one particular
voltage, the electrical force (up) and the gravitational force (down) on the drop are exactly
balanced, and the drop remains stationary. Knowing this voltage and the mass of the drop,
we can calculate the charge on the drop.
Oil droplets
Atomizer
Microscope
Oil droplet
under observation
Charged plate (+)
Small hole
X-ray beam
Charged plate (–)
Robert A. Millikan (left) was an
American physicist who was a
professor at the University of
Chicago and later director of the
physics laboratory at the California
Institute of Technology. He won the
1923 Nobel Prize in physics.
X-rays are a form of radiation of much
shorter wavelength than visible light
(see Section 5-10). They are
sufficiently energetic to knock
electrons out of the atoms in the air.
In Millikan’s experiment these free
electrons became attached to some of
the oil droplets.
out to be integral multiples of the same number. He assumed that this smallest charge
was the charge on one electron. This value is 1.60218 ϫ 10Ϫ19 coulomb (modern value).
The charge-to-mass ratio, e/m ϭ 1.75882 ϫ 108 C/g, can be used in inverse form to
calculate the mass of the electron:
m ϭ ϫ 1.60218 ϫ 10Ϫ19 C
ϭ 9.10940 ϫ 10Ϫ28 g
This is only about 1/1836 the mass of a hydrogen atom, the lightest of all atoms. Millikan’s
simple oil-drop experiment stands as one of the cleverest, yet most fundamental, of all
classic scientific experiments. It was the first experiment to suggest that atoms contain
integral numbers of electrons; we now know this to be true.
CANAL RAYS AND PROTONS
In 1886, Eugen Goldstein (1850–1930) first observed that a cathode-ray tube also generates
a stream of positively charged particles that moves toward the cathode. These were
called canal rays because they were observed occasionally to pass through a channel, or
“canal,” drilled in the negative electrode (Figure 5-3). These positive rays, or positive ions,
are created when the gaseous atoms in the tube lose electrons. Positive ions are formed
by the process
Atom 88n cationϩ ϩ eϪ or X 88n Xϩ ϩ eϪ (energy absorbed)
Different elements give positive ions with different e/m ratios. The regularity of the e/m
values for different ions led to the idea that there is a unit of positive charge and that it
resides in the proton. The proton is a fundamental particle with a charge equal in magnitude
but opposite in sign to the charge on the electron. Its mass is almost 1836 times that
of the electron.
5-3
1 g
ᎏᎏ
1.75882 ϫ 108 C
The charge on one mole (Avogadro’s
number) of electrons is 96,485
coulombs.
180 CHAPTER 5: The Structure of Atoms
Figure 5-3 A cathode-ray tube with a different design and with a perforated cathode. Such
a tube was used to produce canal rays and to demonstrate that they travel toward the
cathode. Like cathode rays, these positive rays are deflected by electric or magnetic fields, but
in the opposite direction from cathode rays. Canal ray particles have e/m ratios many times
smaller than those of electrons, due to their much greater masses. When different elements
are in the tube, positive ions with different e/m ratios are observed.
+
+
+
+
+
+
+
+
Cathode, with holes
(pierced disk) –
Positive ion
Electron
Anode
+
The proton was observed by
Rutherford and James Chadwick in
1919 as a particle that is emitted by
bombardment of certain atoms with
␣-particles.
The value of e/m obtained by
Thomson and the values of e and m
obtained by Millikan differ slightly
from the modern values given in this
text because early measurements were
not as accurate as modern ones.
RUTHERFORD AND THE NUCLEAR ATOM
By the early 1900s, it was clear that each atom contains regions of both positive and negative
charge. The question was, how are these charges distributed? The dominant view of
that time was summarized in J. J. Thomson’s model of the atom; the positive charge was
assumed to be distributed evenly throughout the atom. The negative charges were pictured
as being imbedded in the atom like plums in a pudding (the “plum pudding model”).
Soon after Thomson developed his model, tremendous insight into atomic structure
was provided by one of Thomson’s former students, Ernest Rutherford (1871–1937), who
was the outstanding experimental physicist of his time.
By 1909, Ernest Rutherford had established that alpha (␣) particles are positively
charged particles. They are emitted at high kinetic energies by some radioactive atoms,
that is, atoms that disintegrate spontaneously. In 1910, Rutherford’s research group carried
out a series of experiments that had an enormous impact on the scientific world. They
bombarded a very thin piece of gold foil with ␣-particles from a radioactive source. A
fluorescent zinc sulfide screen was placed behind the foil to indicate the scattering of the
␣-particles by the gold foil (Figure 5-4). Scintillations (flashes) on the screen, caused by
the individual ␣-particles, were counted to determine the relative numbers of ␣-particles
deflected at various angles. Alpha particles were known to be extremely dense, much denser
than gold.
If the Thomson model of the atom were correct, any ␣-particles passing through the
foil would have been deflected by very small angles. Quite unexpectedly, nearly all of the
␣-particles passed through the foil with little or no deflection. A few, however, were
5-4
␣-Particles are now known to be He2ϩ
ions, that is, helium atoms without
their two electrons. (See Chapter 26.)
Radioactivity is contrary to the
Daltonian idea of the indivisibility of
atoms.
5-4 Rutherford and the Nuclear Atom 181
Figure 5-4 The Rutherford scattering experiment. A narrow beam of ␣-particles from a
radioactive source was directed at a very thin sheet of gold foil. Most of the particles passed
right through the gold foil (brown). Many were deflected through moderate angles (shown in
red). These deflections were surprising, but the 0.001% of the total that were reflected at
acute angles (shown in blue) were totally unexpected. Similar results were observed using foils
of other metals.
Lead block
(for shielding)
Slits
Alpha source
Gold foil
Scintillation screen
Alpha beam
deflected through large angles, and a very few ␣-particles even returned from the gold
foil in the direction from which they had come! Rutherford was astounded. In his own
words,
It was quite the most incredible event that has ever happened to me in my life. It was almost
as if you fired a 15-inch shell into a piece of tissue paper and it came back and hit you.
Rutherford’s mathematical analysis of his results showed that the scattering of positively
charged ␣-particles was caused by repulsion from very dense regions of positive charge in
the gold foil. He concluded that the mass of one of these regions is nearly equal to that
of a gold atom, but that the diameter is no more than 1/10,000 that of an atom. Many
experiments with foils of different metals yielded similar results. Realizing that these observations
were inconsistent with previous theories about atomic structure, Rutherford
discarded the old theory and proposed a better one. He suggested that each atom contains
a tiny, positively charged, massive center that he called an atomic nucleus. Most ␣-particles
pass through metal foils undeflected because atoms are primarily empty space populated
only by the very light electrons. The few particles that are deflected are the ones that
come close to the heavy, highly charged metal nuclei (Figure 5-5).
Rutherford was able to determine the magnitudes of the positive charges on the atomic
nuclei. The picture of atomic structure that he developed is called the Rutherford model
of the atom.
Atoms consist of very small, very dense positively charged nuclei surrounded by
clouds of electrons at relatively great distances from the nuclei.
Ernest Rutherford was one of the
giants in the development of our
understanding of atomic structure.
While working with J. J. Thomson
at Cambridge University, he
discovered ␣ and ␤ radiation. He
spent the years 1899–1907 at McGill
University in Canada where he
proved the nature of these two
radiations, for which he received the
Nobel Prize in chemistry in 1908.
He returned to England in 1908,
and it was there, at Manchester
University, that he and his coworkers
Hans Geiger and Ernst Marsden
performed the famous gold foil
experiments that led to the nuclear
model of the atom. Not only did he
perform much important research in
physics and chemistry, but he also
guided the work of ten future
recipients of the Nobel Prize.
182 CHAPTER 5: The Structure of Atoms
Figure 5-5 An interpretation of the Rutherford scattering experiment. The atom is
pictured as consisting mostly of “open” space. At the center is a tiny and extremely dense
nucleus that contains all of the atom’s positive charge and nearly all of its mass. The
electrons are thinly distributed throughout the “open” space. Most of the positively charged
␣-particles (black arrows) pass through the open space undeflected, not coming near any gold
nuclei. The few that pass fairly close to a nucleus (red arrows) are repelled by electrostatic
forces and thereby deflected. The very few particles that are on a “collision course” with
gold nuclei are repelled backward at acute angles (blue arrows). Calculations based on the
results of the experiment indicated that the diameter of the open-space portion of the atom
is from 10,000 to 100,000 times greater than the diameter of the nucleus.
This representation is not to scale. If
nuclei were as large as the black dots
that represent them, each white region,
which represents the size of an atom,
would have a diameter of more than
30 feet!
+Metal
anode
(target)
X-rays
e –
–
ATOMIC NUMBER
Only a few years after Rutherford’s scattering experiments, H. G. J. Moseley (1887–1915)
studied X-rays given off by various elements. Max von Laue (1879–1960) had shown that
X-rays could be diffracted by crystals into a spectrum in much the same way that visible
light can be separated into its component colors. Moseley generated X-rays by aiming a
beam of high-energy electrons at a solid target made of a single pure element (Figure 5-6).
The spectra of X-rays produced by targets of different elements were recorded photographically.
Each photograph consisted of a series of lines representing X-rays at various
wavelengths; each element produced its own distinct set of wavelengths. Comparison of
results from different elements revealed that corresponding lines were displaced toward
shorter wavelengths as atomic weights of the target materials increased, with a few exceptions.
Moseley showed that the X-ray wavelengths could be better correlated with the
atomic number. On the basis of his mathematical analysis of these X-ray data, he concluded
that
each element differs from the preceding element by having one more positive charge
in its nucleus.
For the first time it was possible to arrange all known elements in order of increasing
nuclear charge. A plot illustrating this interpretation of Moseley’s data appears in
Figure 5-7.
We now know that every nucleus contains an integral number of protons exactly equal
to the number of electrons in a neutral atom of the element. Every hydrogen atom contains
5-5
In a modern technique known as
“X-ray fluorescence spectroscopy,” the
wavelengths of X-rays given off by a
sample target indicate which elements
are present in the sample.
5-5 Atomic Number 183
Figure 5-6 A simplified representation of the production of X-rays by bombardment of a
solid target with a high-energy beam of electrons.
H. G. J. Moseley was one of the
many remarkable scientists who
worked with Ernest Rutherford. In
1913, Moseley found that the
wavelengths of X-rays emitted by an
element are related in a precise way
to the atomic number of the
element. This discovery led to the
realization that atomic number,
related to the electrical properties of
the atom, was more fundamental to
determining the properties of the
elements than atomic weight. This
put the ideas of the periodic table on
a more fundamental footing.
Moseley’s scientific career was very
short. He enlisted in the British
army during World War I and died
in battle in the Gallipoli campaign in
1915.
one proton, every helium atom contains two protons, and every lithium atom contains
three protons. The number of protons in the nucleus of an atom determines its identity;
this number is known as the atomic number of that element.
NEUTRONS
The third fundamental particle, the neutron, eluded discovery until 1932. James Chadwick
(1891–1974) correctly interpreted experiments on the bombardment of beryllium
with high-energy ␣-particles. Later experiments showed that nearly all elements up to
potassium, element 19, produce neutrons when they are bombarded with high-energy
␣-particles. The neutron is an uncharged particle with a mass slightly greater than that
of the proton.
Atoms consist of very small, very dense nuclei surrounded by clouds of electrons at
relatively great distances from the nuclei. All nuclei contain protons; nuclei of all
atoms except the common form of hydrogen also contain neutrons.
Nuclear diameters are about 10Ϫ5 nanometers (nm); atomic diameters are about 10Ϫ1
nm. To put this difference in perspective, suppose that you wish to build a model of an
atom using a basketball (diameter about 9.5 inches) as the nucleus; on this scale, the atomic
model would be nearly 6 miles across!
MASS NUMBER AND ISOTOPES
Most elements consist of atoms of different masses, called isotopes. The isotopes of a
given element contain the same number of protons (and also the same number of electrons)
because they are atoms of the same element. They differ in mass because they
contain different numbers of neutrons in their nuclei.
5-7
5-6
Figure 5-7 A plot of some of
Moseley’s X-ray data. The atomic
number of an element is found to be
directly proportional to the square
root of the reciprocal of the
wavelength of a particular X-ray
spectral line. Wavelength (Section
5-10) is represented by ␭.
184 CHAPTER 5: The Structure of Atoms
Atomicnumber,Z
10
20
30
40
50
Al
Si
Cl
K
Ca
Ti
Cr
V
Mn
Fe
Ni
Co
Cu
Zn
Y
Nb
Ru
Ag
Pd
Mo
Zr
ͱසස1
␭
This does not mean that elements
above number 19 do not have
neutrons, only that neutrons are not
generally knocked out of atoms of
higher atomic number by ␣-particle
bombardment.
Isotopes are atoms of the same element with different masses; they are atoms
containing the same number of protons but different numbers of neutrons.
For example, there are three distinct kinds of hydrogen atoms, commonly called
hydrogen, deuterium, tritium. (This is the only element for which we give each isotope
a different name.) Each contains one proton in the atomic nucleus. The predominant form
of hydrogen contains no neutrons, but each deuterium atom contains one neutron and
each tritium atom contains two neutrons in its nucleus (Table 5-2). All three forms of
hydrogen display very similar chemical properties.
The mass number of an atom is the sum of the number of protons and the number
of neutrons in its nucleus; that is
Mass number ϭ number of protons ϩ number of neutrons
ϭ atomic number ϩ neutron number
The mass number for normal hydrogen atoms is 1; for deuterium, 2; and for tritium, 3.
The composition of a nucleus is indicated by its nuclide symbol. This consists of the
symbol for the element (E), with the atomic number (Z) written as a subscript at the lower
left and the mass number (A) as a superscript at the upper left, A
ZE. By this system, the
three isotopes of hydrogen are designated as 1
1H, 2
1H, and 3
1H.
EXAMPLE 5-1 Determination of Atomic Makeup
Determine the number of protons, neutrons, and electrons in each of the following species.
Are the members within each pair isotopes?
(a) 35
17Cl and 37
17Cl (b) 63
29Cu and 65
29Cu
Plan
Knowing that the number at the bottom left of the nuclide symbol is the atomic number or
number of protons, we can verify the identity of the element in addition to knowing the number
or protons per nuclide. From the mass number at the top left, we know the number of protons
plus neutrons. The number of protons (atomic number) minus the number of electrons must
equal the charge, if any, shown at the top right. From these data one can determine if two
nuclides have the same number of protons and are therefore the same element. If they are the
same element, they will be isotopes only if their mass numbers differ.
A mass number is a count of the
number of protons plus neutrons
present, so it must be a whole number.
Because the masses of the proton and
the neutron are both about 1 amu, the
mass number is approximately equal to
the actual mass of the isotope (which is
not a whole number).
5-7 Mass Number and Isotopes 185
TABLE 5-2 The Three Isotopes of Hydrogen
Atomic No. of Electrons
Nuclide Mass Abundance No. of No. of (in neutral)
Name Symbol Symbol (amu) in Nature Protons Neutrons atoms)
hydrogen H 1
1H 1.007825 99.985% 1 0 1
deuterium D 2
1H 2.01400 0.015% 1 1 1
tritium* T 3
1H 3.01605 0.000% 1 2 1
*No known natural sources; produced by decomposition of artificial isotopes.
Solution
(a) For 35
17Cl: Atomic number ϭ 17. There are therefore 17 protons per nucleus.
Mass number ϭ 35. There are therefore 35 protons plus neutrons or, because
we know that there are 17 protons, there are 18 neutrons.
Because no charge is indicated, there must be equal numbers of protons and
electrons, or 17 electrons.
For 37
17Cl: There are 17 protons, 20 neutrons, and 17 electrons per atom.
These are isotopes of the same element. Both have 17 protons, but they differ in their numbers
of neutrons: one has 18 neutrons and the other has 20.
(b) For 63
29Cu: Atomic number ϭ 29. There are 29 protons per nucleus.
Mass number ϭ 63. There are 29 protons plus 34 neutrons.
Because no charge is indicated, there must be equal numbers of protons and
electrons, or 29 electrons.
For 65
29Cu: There are 29 protons, 36 neutrons, and 29 electrons per atom.
These are isotopes. Both have 29 protons, but they differ in their numbers of neutrons: one
isotope has 34 neutrons and the other has 36.
You should now work Exercises 16 and 18.
MASS SPECTROMETRY AND ISOTOPIC ABUNDANCE
Mass spectrometers are instruments that measure the charge-to-mass ratio of charged
particles (Figures 5-8). A gas sample at very low pressure is bombarded with high-energy
electrons. This causes electrons to be ejected from some of the gas molecules, creating
positive ions. The positive ions are then focused into a very narrow beam and accelerated
by an electric field toward a magnetic field. The magnetic field deflects the ions from
their straight-line path. The extent to which the beam of ions is deflected depends on
four factors:
1. Magnitude of the accelerating voltage (electric field strength). Higher voltages result in
beams of more rapidly moving particles that are deflected less than the beams of
the more slowly moving particles produced by lower voltages.
2. Magnetic field strength. Stronger fields deflect a given beam more than weaker fields.
3. Masses of the particles. Because of their inertia, heavier particles are deflected less
than lighter particles that carry the same charge.
4. Charges on the particles. Particles with higher charges interact more strongly with
magnetic fields and are thus deflected more than particles of equal mass with smaller
charges.
The mass spectrometer is used to measure masses of isotopes as well as isotopic abundances,
that is, the relative amounts of the isotopes. Helium occurs in nature almost
exclusively as 4
2He. Its atomic mass can be determined in an experiment such as that illustrated
in Figure 5-8.
A beam of Neϩ ions in the mass spectrometer is split into three segments. The mass
spectrum of these ions (a graph of the relative numbers of ions of each mass) is shown in
Figure 5-9. This indicates that neon occurs in nature as three isotopes: 20
10Ne, 21
10Ne, and
5-8
186 CHAPTER 5: The Structure of Atoms
22
10Ne. In Figure 5-9 we see that the isotope 20
10Ne, mass 19.99244 amu, is the most abundant
isotope (has the tallest peak). It accounts for 90.48% of the atoms. 22
10Ne, mass
21.99138, accounts for 9.25%, and 21
10Ne, mass 20.99384, for only 0.27% of the atoms.
Figure 5-10 shows a modern mass spectrometer. In nature, some elements, such as
fluorine and phosphorus, exist in only one form, but most elements occur as isotopic
mixtures. Some examples of natural isotopic abundances are given in Table 5-3. The
percentages are based on the numbers of naturally occurring atoms of each isotope, not
on their masses.
Isotopes are two or more forms of
atoms of the same element with
different masses; the atoms contain the
same number of protons but different
numbers of neutrons.
5-8 Mass Spectrometry and Isotopic Abundance 187
Figure 5-8 The mass spectrometer. In the mass spectrometer, gas molecules at low
pressure are ionized and accelerated by an electric field. The ion beam is then passed
through a magnetic field. In that field the beam is resolved into components, each
containing particles of equal charge-to-mass ratio. Lighter particles are deflected more
strongly than heavy ones with the same charge. In a beam containing 12
06Cϩ and 4
2Heϩ ions,
the lighter 4
2Heϩ ions would be deflected more than the heavier 12
06Cϩ ions. The
spectrometer shown is adjusted to detect the 12
06Cϩ ions. By changing the magnitude of the
magnetic or electric field, we can move the beam of 4
2Heϩ ions striking the collector from B
to A, where it would be detected. The relative masses of the ions are calculated from the
changes required to refocus the beam.
Figure 5-9 Mass spectrum of neon (1ϩ ions only). Neon consists of three isotopes, of
which neon-20 is by far the most abundant (90.48%). The mass of that isotope, to five
decimal places, is 19.99244 amu on the carbon-12 scale. The number by each peak
corresponds to the fraction of all Neϩ ions represented by the isotope with that mass.
A
B
Slit
Collector
Detector
Magnet
Electron
gun
Gas
inlet
Accelerating
plates
+
–
Beam of 12
6
C+
ions
Beam of 4
2
He+
ions
20
0.9048
0.0027 0.0925
22 2321
Mass
Relativeabundance
188 CHAPTER 5: The Structure of Atoms
The Development of ScienceCCHEMISTRY IN USE
Stable Isotope Ratio Analysis
Many elements exist as two or more stable isotopes, although
one isotope is usually present in far greater abundance. For example,
there are two stable isotopes of carbon, 13C and 12C, of
which 12C is the more abundant, constituting 98.89% of all
carbon. Similarly, there are two stable isotopes of nitrogen,
14N and 15N, of which 14N makes up 99.63% of all nitrogen.
Differences in chemical and physical properties that arise
from differences in atomic mass of an element are known as
isotope effects. We know that the extranuclear structure of
an element (the number of electrons and their arrangement)
essentially determines its chemical behavior, whereas the
nucleus has more influence on many of the physical properties
of the element. Because all isotopes of a given element
contain the same number and arrangement of electrons, it
was assumed for a long time that isotopes would behave identically
in chemical reactions. In reality, although isotopes
behave very similarly in chemical reactions, the correspondence
is not perfect. The mass differences between different
isotopes of the same element cause them to have slightly different
physical and chemical properties. For example, the
presence of only one additional neutron in the nucleus of the
heavier isotope can cause it to react a little more slowly than
its lighter counterpart. Such an effect often results in a ratio
of heavy isotope to light isotope in the product of a reaction
that is different from the ratio found in the reactant.
Stable isotope ratio analysis (SIRA) is an analytical technique
that takes advantage of the chemical and physical
properties of isotopes. In SIRA the isotopic composition of
a sample is measured using a mass spectrometer. This composition
is then expressed as the relative ratios of two or more
of the stable isotopes of a specific element. For instance, the
ratio of 13C to 12C in a sample can be determined. This ratio
is then compared with the isotope ratio of a defined standard.
Because mass differences are most pronounced among the
lightest elements, those elements experience the greatest isotope
effects. Thus, the isotopes of the elements H, C, N, O,
and S are used most frequently for SIRA. These elements
have further significance because they are among the most
abundant elements in biological systems.
The isotopic composition of a sample is usually expressed
as a “del” value (∂), defined as
∂Xsample (0/00) ϭ ϫ 1000
(Rsample Ϫ Rstandard)
ᎏᎏᎏ
Rstandard
where ∂Xsample is the isotope ratio relative to a standard, and
Rsample and Rstandard are the absolute isotope ratios of the sample
and standard, respectively. Multiplying by 1000 allows the
values to be expressed in parts per thousand (0/00). If the del
value is a positive number, the sample has a greater amount
of the heavier isotope than does the standard. In such cases
the sample is said to be “heavier” than the standard, or to
have been “enriched” in the heavy isotope. Similarly, if the
del value is negative, the sample has a higher proportion of
the lighter isotope and thus is described as “lighter” than the
standard.
The most frequently used element for SIRA is carbon.
The first limited data on 13C/12C isotope ratios in natural
materials were published in 1939. At that time, it was established
that limestones, atmospheric CO2, marine plants, and
terrestrial plants each possessed characteristic carbon isotope
ratios. In the succeeding years, 13C/12C ratios were determined
for a wide variety of things, including petroleum, coal,
diamonds, marine organisms, and terrestrial organisms. Such
data led to the important conclusion that a biological organism
has an isotope ratio that depends on the main source of
carbon to that organism—that is, its food source. For example,
if an herbivore (an animal that feeds on plants) feeds
exclusively on one type of plant, that animal’s carbon isotope
ratio will be almost identical to that of the plant. If another
animal were to feed exclusively on that herbivore, it would
5-8 Mass Spectrometry and Isotopic Abundance 189
also have a similar carbon isotope ratio. Suppose now that an
animal, say a rabbit, has a diet comprising two plants, A and
B. Plant A has a ∂13C value of Ϫ240/00, and plant B has a del
value of Ϫ100/00. If the rabbit eats equal amounts of the two
plants, the ∂13C value of the rabbit will be the average of the
two values, or Ϫ170/00. Values more positive than Ϫ170/00
would indicate a higher consumption of plant B than of plant
A, whereas more negative values would reflect a preference
for plant A.
Similar studies have been conducted with the stable isotopes
of nitrogen. A major way in which nitrogen differs from
carbon in isotopic studies relates to how ∂13C and ∂15N values
change as organic matter moves along the food chain—
from inorganic nutrient to plant, then to herbivore, to carnivore,
and on to higher carnivores. It has been pointed out
that ∂13C remains nearly constant throughout successive levels
of the food chain. In contrast, on average there is a ϩ3
to ϩ50/00 shift in the value of ∂15N at each successive level
of the food chain. For instance, suppose a plant has a ∂15N
value of 10/00. If an herbivore, such as a rabbit, feeds exclusively
on that one type of plant, it will have a ∂15N value of
40/00. If another animal, such as a fox, feeds exclusively on
that particular type of rabbit, it in turn will have a ∂15N value
of 70/00. An important implication of this phenomenon is that
an organism’s nitrogen isotope ratio can be used as an indicator
of the level in the food chain at which that species of
animal feeds.
An interesting application of SIRA is the determination
of the adulteration of food. As already mentioned, the isotope
ratios of different plants and animals have been determined.
For instance, corn has a ∂13C value of about Ϫ120/00
and most flowering plants have ∂13C values of about Ϫ260/00.
The difference in these ∂13C values arises because these plants
carry out photosynthesis by slightly different chemical reactions.
In the first reaction of photosynthesis, corn produces
a molecule that contains four carbons, whereas flowering
plants produce a molecule that has only three carbons. Highfructose
corn syrup (HFCS) is thus derived from a “C4” plant,
whereas the nectar that bees gather comes from “C3” plants.
The slight differences in the photosynthetic pathways of C3
and C4 plants create the major differences in their ∂13C values.
Brokers who buy and sell huge quantities of “sweet”
products are able to monitor HFCS adulteration of honey,
maple syrup, apple juice, and so on by taking advantage of
the SIRA technique. If the ∂13C value of one of these products
is not appropriate, then the product obviously has had
other substances added to it, i.e., has been adulterated. The
U.S. Department of Agriculture conducts routine isotope
analyses to ensure the purity of those products submitted for
subsidy programs. Similarly, the honey industry monitors
itself with the SIRA technique.
Another interesting use of SIRA is in the determination
of the diets of prehistoric human populations. It is known
that marine plants have higher ∂15N values than terrestrial
plants. This difference in ∂15N is carried up food chains, causing
marine animals to have higher ∂15N values than terrestrial
animals. The ∂15N values of humans feeding on marine food
sources are therefore higher than those of people feeding on
terrestrial food. This phenomenon has been used to estimate
the marine and terrestrial components of the diets of historic
and prehistoric human groups through the simple determination
of the ∂15N value of bone collagen collected from
excavated skeletons.
Stable isotope ratio analysis is a powerful tool; many of
its potential uses are only slowly being recognized by
researchers. In the meantime, the use of stable isotope methods
in research is becoming increasingly common, and
through these methods scientists are attaining new levels
of understanding of chemical, biological, and geological
processes.
Beth A. Trust
The distribution of isotopic masses, although nearly constant, does vary somewhat
depending on the source of the element. For example, the abundance of 13
06C in atmospheric
CO2 is slightly different from that in seashells. The chemical history of a compound
can be inferred from small differences in isotope ratios.
Figure 5-10 (a) A modern mass spectrometer. (b) The mass spectrum of Xeϩ ions. The
isotope 126
054Xe is at too low an abundance (0.090%) to appear in this experiment.
190 CHAPTER 5: The Structure of Atoms
(a) (b)
TABLE 5-3 Some Naturally Occurring Isotopic Abundances
Element Atomic Weight (amu) Isotope % Natural Abundance Mass (amu)
boron 10.811 010
05B 19.91 10.01294
011
05B 80.09 11.00931
oxygen 15.9994 016
08O 99.762 15.99492
017
08O 0.038 16.99913
018
08O 0.200 17.99916
chlorine 35.4527 035
17Cl 75.770 34.96885
037
17Cl 24.230 36.96590
uranium 238.0289 234
092U 0.0055 234.0409
235
092U 0.720 235.0439
238
092U 99.2745 238.0508
The 20 elements that have only one naturally occurring isotope are 9
4Be, 19
09F, 23
11Na, 27
13Al, 31
15P, 45
21Sc, 55
25Mn, 59
27Co,
75
33As, 89
39Y, 93
41Nb, 103
045Rh, 127
053I, 133
055Cs, 141
059Pr, 159
065Tb, 165
067Ho, 169
069Tm, 197
079Au, and 209
083Bi. There are however other,
artificially produced isotopes of these elements.
THE ATOMIC WEIGHT SCALE AND ATOMIC WEIGHTS
We said in Section 2-5 that the atomic weight scale is based on the mass of the carbon-
12 isotope. As a result of action taken by the International Union of Pure and Applied
Chemistry in 1962,
one amu is exactly 1/12 of the mass of a carbon-12 atom.
This is approximately the mass of one atom of 1H, the lightest isotope of the element
with lowest mass.
In Section 2-6 we said that one mole of atoms, contains 6.022 ϫ 1023 atoms. The mass
of one mole of atoms of any element, in grams, is numerically equal to the atomic weight
of the element. Because the mass of one carbon-12 atom is exactly 12 amu, the mass of
one mole of carbon-12 atoms is exactly 12 grams.
To show the relationship between atomic mass units and grams, let us calculate the
mass, in amu, of 1.000 gram of 12
06C atoms.
5-9
Described another way, the mass of
one atom of 12
06C is exactly 12 amu.
5-9 The Atomic Weight Scale and Atomic Weights 191
_?_ amu ϭ 1.000 g 12
06C atoms ϫ ϫ ϫ
ϭ 6.022 ϫ 1023 amu (in 1 gram)
12 amu
ᎏᎏ12
06C atom
6.022 ϫ 1023 12
06C atoms
ᎏᎏᎏ
1 mol 12
06C atoms
1 mol 12
06C
ᎏᎏ
12 g 12
06C atoms
Thus,
1 g ϭ 6.022 ϫ 1023 amu or 1 amu ϭ 1.660 ϫ 10Ϫ24 g
At this point, we emphasize the following:
1. The atomic number, Z, is an integer equal to the number of protons in the nucleus
of an atom of the element. It is also equal to the number of electrons in a neutral
atom. It is the same for all atoms of an element.
2. The mass number, A, is an integer equal to the sum of the number of protons and
the number of neutrons in the nucleus of an atom of a particular isotope of an element.
It is different for different isotopes of the same element.
3. Many elements occur in nature as mixtures of isotopes. The atomic weight of such
an element is the weighted average of the masses of its isotopes. Atomic weights are
fractional numbers, not integers.
The atomic weight that we determine experimentally (for an element that consists of more
than one isotope) is such a weighted average. The following example shows how an atomic
weight can be calculated from measured isotopic abundances.
EXAMPLE 5-2 Calculation of Atomic Weight
Three isotopes of magnesium occur in nature. Their abundances and masses, determined by
mass spectrometry, are listed in the following table. Use this information to calculate the atomic
weight of magnesium.
We saw in Chapter 2 that Avogadro’s
number is the number of particles of a
substance in one mole of that
substance. We now see that Avogadro’s
number also represents the number of
amu in one gram. You may wish to
verify that the same result is obtained
regardless of the element or isotope
chosen.
Isotope % Abundance Mass (amu)
24
12Mg 78.99 23.98504
25
12Mg 10.00 24.98584
26
12Mg 11.01 25.98259
Plan
We multiply the fraction of each isotope by its mass and add these numbers to obtain the
atomic weight of magnesium.
Solution
Atomic weight ϭ 0.7899(23.98504 amu) ϩ 0.1000(24.98584 amu) ϩ 0.1101(25.98259 amu)
ϭ 18.946 amu ϩ 2.4986 amu ϩ 2.8607 amu
ϭ 24.30 amu (to four significant figures)
The two heavier isotopes make small contributions to the atomic weight of magnesium because
most magnesium atoms are the lightest isotope.
You should now work Exercises 26 and 28.
192 CHAPTER 5: The Structure of Atoms
Problem-Solving Tip: “Weighted” Averages
Consider the following analogy to the calculation of atomic weights. Suppose you want
to calculate the average weight of your classmates. Imagine that one half of them weigh
100 pounds each, and the other half weigh 200 pounds each. The average weight would
be
Average weight ϭ ᎏ
1
2
ᎏ (100 lb) ϩ ᎏ
1
2
ᎏ (200 lb) ϭ 150 lb
Imagine, however, that three quarters of the class members weigh 100 pounds each, and
the other quarter weigh 200 pounds each. Now, the average weight would be
Average weight ϭ ᎏ
3
4
ᎏ (100 lb) ϩ ᎏ
1
4
ᎏ (200 lb) ϭ 125 lb
We can express the fractions in this calculation in decimal form:
Average weight ϭ 0.750(100 lb) ϩ 0.250(200 lb) ϭ 125 lb
In such a calculation, the value (in this case, the weight) of each thing (people, atoms)
is multiplied by the fraction of things that have that value. In Example 5-2 we expressed
each percentage as a decimal fraction, such as
78.99% ϭ ᎏ
10
7
0
8.
p
9
a
9
rt
p
s
a
t
r
o
ts
tal
ᎏ ϭ 0.7899
Example 5-3 shows how the process can be reversed. Isotopic abundances can be calculated
from isotopic masses and from the atomic weight of an element that occurs in nature
as a mixture of only two isotopes.
EXAMPLE 5-3 Calculation of Isotopic Abundance
The atomic weight of gallium is 69.72 amu. The masses of the naturally occurring isotopes
are 68.9257 amu for 69
31Ga and 70.9249 amu for 71
31Ga. Calculate the percent abundance of each
isotope.
Plan
We represent the fraction of each isotope algebraically. Atomic weight is the weighted average
of the masses of the constituent isotopes. So the fraction of each isotope is multiplied by its
mass, and the sum of the results is equal to the atomic weight.
Solution
Let x ϭ fraction of 69
31Ga. Then (1 Ϫ x) ϭ fraction of 71
31Ga.
x(68.9257 amu) ϩ (1 Ϫ x)(70.9249 amu) ϭ 69.72 amu
68.9257x ϩ 70.9249 Ϫ 70.9249x ϭ 69.72
Ϫ1.9992x ϭ Ϫ1.20
x ϭ 0.600
x ϭ 0.600 ϭ fraction of 69
31Ga І 60.0% 69
31Ga
(1 Ϫ x) ϭ 0.400 ϭ fraction of 71
31Ga І 40.0% 71
31Ga
You should now work Exercise 30.
THE ELECTRONIC STRUCTURES OF ATOMS
The Rutherford model of the atom is consistent with the evidence presented so far, but
it has some serious limitations. It does not answer such important questions as: Why do
different elements have such different chemical and physical properties? Why does chemical
bonding occur at all? Why does each element form compounds with characteristic
formulas? How can atoms of different elements give off or absorb light only of characteristic
colors (as was known long before 1900)?
To improve our understanding, we must first learn more about the arrangements of
electrons in atoms. The theory of these arrangements is based largely on the study of the
light given off and absorbed by atoms. Then we will develop a detailed picture of the electron
configurations of different elements. A knowledge of these arrangements will help us
to understand the periodic table and chemical bonding.
ELECTROMAGNETIC RADIATION
Our ideas about the arrangements of electrons in atoms have evolved slowly. Much of the
information has been derived from atomic emission spectra. These are the lines, or
bands, produced on photographic film by radiation that has passed through a refracting
glass prism after being emitted from electrically or thermally excited atoms. To help us
understand the nature of atomic spectra, we first describe electromagnetic radiation.
All types of electromagnetic radiation, or radiant energy, can be described in the terminology
of waves. To help characterize any wave, we specify its wavelength (or its frequency).
5-10
When a quantity is represented by
fractions, the sum of the fractions
must always be unity. In this case,
x ϩ (1 Ϫ x) ϭ 1.
5-10 Electromagnetic Radiation 193
White light is dispersed by a prism
into a continuous spectrum.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 7.3, Electromagnetic Radiation.
Let us use a familiar kind of wave, that on the surface of water (Figure 5-11), to illustrate
these terms. The significant feature of wave motion is its repetitive nature. The wavelength,
␭, is the distance between any two adjacent identical points of the wave, for
instance, two adjacent crests. The frequency is the number of wave crests passing a given
point per unit time; it is represented by the symbol ␯ (Greek letter “nu”) and is usually
expressed in cycles per second or, more commonly, simply as 1/s or sϪ1 with “cycles”
understood. For a wave that is “traveling” at some speed, the wavelength and the frequency
are related to each other by
␭␯ ϭ speed of propagation of the wave or ␭␯ ϭ c
Thus, wavelength and frequency are inversely proportional to each other; for the same
wave speed, shorter wavelengths correspond to higher frequencies.
For water waves, it is the surface of the water that changes repetitively; for a vibrating
violin string, it is the displacement of any point on the string. Electromagnetic radiation
is a form of energy that consists of electric and magnetic fields that vary repetitively. The
electromagnetic radiation most obvious to us is visible light. It has wavelengths ranging
from about 4.0 ϫ 10Ϫ7 m (violet) to about 7.5 ϫ 10Ϫ7 m (red). Expressed in frequencies,
this range is about 7.5 ϫ 1014 Hz (violet) to about 4.0 ϫ 1014 Hz (red).
Isaac Newton (1642–1727) first recorded the separation of sunlight into its component
colors by allowing it to pass through a prism. Because sunlight (white light) contains all
wavelengths of visible light, it gives the continuous spectrum observed in a rainbow (Figure
5-12a). Visible light represents only a tiny segment of the electromagnetic radiation spectrum
(Figure 5-12b). In addition to all wavelengths of visible light, sunlight also contains
shorter wavelength (ultraviolet) radiation as well as longer wavelength (infrared) radiation.
Neither of these can be detected by the human eye. Both may be detected and
recorded photographically or by detectors designed for that purpose. Many other familiar
kinds of radiation are simply electromagnetic radiation of longer or shorter wavelengths.
In a vacuum, the speed of electromagnetic radiation, c, is the same for all wavelengths,
2.99792458 ϫ 108 m/s. The relationship between the wavelength and frequency of electromagnetic
radiation, with c rounded to three significant figures, is
␭␯ ϭ c ϭ 3.00 ϫ 108 m/s
Figure 5-11 Illustrations of the
wavelength and frequency of water
waves. The distance between any
two identical points, such as crests, is
the wavelength, ␭. We could
measure the frequency, ␯, of the
wave by observing how often the
level rises and falls at a fixed point in
its path—for instance, at the post—
or how often crests hit the post. (a)
and (b) represent two waves that are
traveling at the same speed. In (a)
the wave has long wavelength and
low frequency; in (b) the wave has
shorter wavelength and higher
frequency.
One cycle per second is also called one
hertz (Hz), after Heinrich Hertz
(1857–1894). In 1887, Hertz
discovered electromagnetic radiation
outside the visible range and measured
its speed and wavelengths.
The diffraction of white light by the
closely spaced grooves of a compact
disk spreads the light into its
component colors. Diffraction is
described as the constructive and
destructive interference of light
waves.
(a)
(b)
Sir Isaac Newton, one of the giants of science. You probably know of him from his theory of
gravitation. In addition, he made enormous contributions to the understanding of many
other aspects of physics, including the nature and behavior of light, optics, and the laws of
motion. He is credited with the discoveries of differential calculus and of expansions into
infinite series.
5-10 Electromagnetic Radiation 195
Figure 5-12 (a) Dispersion of visible light by a prism. Light from a source of white light
is passed through a slit and then through a prism. It is spread into a continuous spectrum
of all wavelengths of visible light. (b) Visible light is only a very small portion of the
electromagnetic spectrum. Some radiant energy has longer or shorter wavelengths than our
eyes can detect. The upper part shows the approximate ranges of the electromagnetic
spectrum on a logarithmic scale. The lower part shows the visible region on an expanded
scale. Note that wavelength increases as frequency decreases.
4000 5000 6000 7000
4.28 × 1014
5.00 × 1014
6.00 × 1014
7.50 × 1014
λ(Å)
v(Hz)
White
light
Prism
Increasing
wavelength
Gamma rays
X-rays
Ultraviolet
rays
Infrared rays
(heat)
Microwaves
Hertzian waves
(radio, TV)
λ(Å)
v(Hz)
10–3
10–1
10 103
105
107
109
1011
3 × 1019
3 × 1015
3 × 1011
3 × 107
Low energy
Low frequency
Long wavelength
High energy
High frequency
Short wavelength
Visible light
(a)
(b)
EXAMPLE 5-4 Wavelength of Light
Light near the middle of the ultraviolet region of the electromagnetic radiation spectrum has
a frequency of 2.73 ϫ 1016 sϪ1. Yellow light near the middle of the visible region of the spectrum
has a frequency of 5.26 ϫ 1014 sϪ1. Calculate the wavelength that corresponds to each of
these two frequencies of light.
Plan
Wavelength and frequency are inversely proportional to each other, ␭␯ ϭ c. We solve this relationship
for ␭ and calculate the wavelengths.
Solution
(ultraviolet light) ␭ ϭ ϭ ϭ 1.10 ϫ 10Ϫ8 m (1.10 ϫ 102 Å)
(yellow light) ␭ ϭ ϭ ϭ 5.70 ϫ 10Ϫ7 m (5.70 ϫ 103 Å)
You should now work Exercise 38.
We have described light in terms of wave behavior. Under certain conditions, it is also
possible to describe light as composed of particles, or photons. According to the ideas
presented by Max Planck (1858–1947) in 1900, each photon of light has a particular
amount (a quantum) of energy. The amount of energy possessed by a photon depends on
the frequency of the light. The energy of a photon of light is given by Planck’s equation
E ϭ h␯ or E ϭ ᎏ
h
␭
c
ᎏ
where h is Planck’s constant, 6.6260755 ϫ 10Ϫ34 Jиs, and ␯ is the frequency of the light.
Thus, energy is directly proportional to frequency. Planck’s equation is used in Example
5-5 to show that a photon of ultraviolet light has more energy than a photon of yellow light.
EXAMPLE 5-5 Energy of Light
In Example 5-4 we calculated the wavelengths of ultraviolet light of frequency 2.73 ϫ 1016 sϪ1
and of yellow light of frequency 5.26 ϫ 1014 sϪ1. Calculate the energy, in joules, of an individual
photon of each. Compare these photons by calculating the ratio of their energies.
Plan
We use each frequency to calculate the photon energy from the relationship E ϭ h␯. Then we
calculate the required ratio.
Solution
(ultraviolet light) E ϭ h␯ ϭ (6.626 ϫ 10Ϫ34 Jиs)(2.73 ϫ 1016 sϪ1) ϭ 1.81 ϫ 10Ϫ17 J
(yellow light) E ϭ h␯ ϭ (6.626 ϫ 10Ϫ34 Jиs)(5.26 ϫ 1014 sϪ1) ϭ 3.49 ϫ 10Ϫ19 J
(You can check these answers by calculating the energies directly from the wavelengths, using
the equation E ϭ hc/␭.)
Now, we compare the energies of these two photons.
3.00 ϫ 108 mиsϪ1
ᎏᎏ
5.26 ϫ 1014 sϪ1
c
ᎏ
␯
3.00 ϫ 108 mиsϪ1
ᎏᎏ
2.73 ϫ 1016 sϪ1
c
ᎏ
␯
196 CHAPTER 5: The Structure of Atoms
ϭ ϭ 51.9
A photon of light near the middle of the ultraviolet region is more than 51 times more energetic
than a photon of light near the middle of the visible region.
You should now work Exercise 39.
THE PHOTOELECTRIC EFFECT
One experiment that had not been satisfactorily explained with the wave theory of light
was the photoelectric effect. The apparatus for the photoelectric effect is shown in Figure
5-13. The negative electrode in the evacuated tube is made of a pure metal such as cesium.
When light of a sufficiently high energy strikes the metal, electrons are knocked off its
surface. They then travel to the positive electrode and form a current flowing through
the circuit. The important observations follow.
1. Electrons can be ejected only if the light is of sufficiently short wavelength (has
sufficiently high energy), no matter how long or how brightly the light shines. This
wavelength limit is different for different metals.
2. The current (the number of electrons emitted per second) increases with increasing
brightness (intensity) of the light. The current, however, does not depend on the color
of the light, as long as the wavelength is short enough (has high enough energy).
Classical theory said that even “low-” energy light should cause current to flow if the
metal is irradiated long enough. Electrons should accumulate energy and be released when
they have enough energy to escape from the metal atoms. According to the old theory, if
the light is made more energetic, then the current should increase even though the light
intensity remains the same. Such is not the case.
The answer to the puzzle was provided by Albert Einstein (1879–1955). In 1905, he
extended Planck’s idea that light behaves as though it were composed of photons, each with
a particular amount (a quantum) of energy. According to Einstein, each photon can transfer
its energy to a single electron during a collision. When we say that the intensity of light
is increased, we mean that the number of photons striking a given area per second is
increased. The picture is now one of a particle of light striking an electron near the surface
of the metal and giving up its energy to the electron. If that energy is equal to or greater
than the amount needed to liberate the electron, it can escape to join the photoelectric
current. For this explanation, Einstein received the 1921 Nobel Prize in physics.
5-11
1.81 ϫ 10Ϫ17 J
ᎏᎏ
3.49 ϫ 10Ϫ19 J
Euv
ᎏ
Eyellow
This is one reason why ultraviolet
(UV) light damages your skin much
more rapidly than visible light.
Another reason is that many of the
organic compounds in the skin absorb
UV light more readily than visible
light. The absorbed ultraviolet light
breaks bonds in many biologically
important molecules.
5-11 The Photoelectric Effect 197
The intensity of light is the brightness
of the light. In wave terms, it is related
to the amplitude of the light waves.
Figure 5-13 The photoelectric
effect. When electromagnetic
radiation of sufficient minimum
energy strikes the surface of a metal
(negative electrode) inside an
evacuated tube, electrons are
stripped off the metal to create an
electric current. The current
increases with increasing radiation
intensity.
The photoelectric effect is used in the
photocells of automatic cameras. The
photoelectric sensors that open some
supermarket and elevator doors also utilize
this effect.
Light
Positive plate
Stream of
electrons
Active metal
(cathode)
Battery Ammeter
ϩϪ
ATOMIC SPECTRA AND THE BOHR ATOM
Incandescent (“red hot” or “white hot”) solids, liquids, and high-pressure gases give continuous
spectra. When an electric current is passed through a gas in a vacuum tube at very
low pressures, however, the light that the gas emits can be dispersed by a prism into distinct
lines (Figure 5-14a). Such an emission spectrum is described as a bright line spectrum.
The lines can be recorded photographically, and the wavelength of light that produced
each line can be calculated from the position of that line on the photograph.
Similarly, we can shine a beam of white light (containing a continuous distribution of
wavelengths) through a gas and analyze the beam that emerges. We find that only certain
wavelengths have been absorbed (Figure 5-14b). The wavelengths that are absorbed in
this absorption spectrum are also given off in the emission experiment. Each spectral
line corresponds to a specific wavelength of light and thus to a specific amount of energy
that is either absorbed or emitted. An atom of each element displays its own characteristic
set of lines in its emission or absorption spectrum (Figure 5-15). These spectra can
serve as “fingerprints” that allow us to identify different elements present in a sample,
even in trace amounts.
5-12
Figure 5-14 (a) Atomic emission. The light emitted by a sample of excited hydrogen atoms
(or any other element) can be passed through a prism and separated into certain discrete
wavelengths. Thus, an emission spectrum, which is a photographic recording of the
separated wavelengths, is called a line spectrum. Any sample of reasonable size contains an
enormous number of atoms. Although a single atom can be in only one excited state at a
time, the collection of atoms contains all possible excited states. The light emitted as these
atoms fall to lower energy states is responsible for the spectrum. (b) Atomic absorption. When
white light is passed through unexcited hydrogen and then through a slit and a prism, the
transmitted light is lacking in intensity at the same wavelengths as are emitted in part (a).
The recorded absorption spectrum is also a line spectrum and the photographic negative of
the emission spectrum.
198 CHAPTER 5: The Structure of Atoms
Film or detector
Film or detector
Prism
Prism
White
light
source
Absorbing sample
Excited
sample
(a)
(b)
Increasing wavelength
Absorption spectrum
Increasing wavelength
Emission spectrum
EXAMPLE 5-6 Energy of Light
A green line of wavelength 4.86 ϫ 10Ϫ7 m is observed in the emission spectrum of hydrogen.
Calculate the energy of one photon of this green light.
Plan
We know the wavelength of the light, and we calculate its frequency so that we can then calculate
the energy of each photon.
Solution
E ϭ ϭ ϭ 4.09 ϫ 10Ϫ19 J/photon
To gain a better appreciation of the amount of energy involved, let’s calculate the total energy,
in kilojoules, emitted by one mole of atoms. (Each atom emits one photon.)
ϭ 4.09 ϫ 10Ϫ19 ϫ ϫ ϭ 2.46 ϫ 102 kJ/mol
This calculation shows that when each atom in one mole of hydrogen atoms emits light of
wavelength 4.86 ϫ 10Ϫ7 m, the mole of atoms loses 246 kJ of energy as green light. (This
would be enough energy to operate a 100-watt light bulb for more than 40 minutes.)
You should now work Exercises 40 and 42.
6.02 ϫ 1023 atoms
ᎏᎏᎏ
mol
1 kJ
ᎏᎏ
1 ϫ 103 J
J
ᎏ
atom
_?_ kJ
ᎏ
mol
(6.626 ϫ 10Ϫ34 Jиs)(3.00 ϫ 108 m/s)
ᎏᎏᎏᎏ
(4.86 ϫ 10Ϫ7 m)
hc
ᎏ
␭
Figure 5-15 Atomic spectra in the visible region for some elements. Figure 5-14a shows
how such spectra are produced. (a) Emission spectra for some elements. (b) Absorption
spectrum for hydrogen. Compare the positions of these lines with those in the emission
spectrum for H in (a).
4000 5000 6000 7000λ(Å)
(b)
H
4000 5000 6000 7000
Ne
Hg
H
(a)
λ(Å)
When an electric current is passed through hydrogen gas at very low pressures, several
series of lines in the spectrum of hydrogen are produced. These lines were studied intensely
by many scientists. In the late nineteenth century, Johann Balmer (1825–1898) and
Johannes Rydberg (1854–1919) showed that the wavelengths of the various lines in the
hydrogen spectrum can be related by a mathematical equation:
ϭ R ΂ Ϫ ΃
Here R is 1.097 ϫ 107 mϪ1 and is known as the Rydberg constant. The n’s are positive
integers, and n1 is smaller than n2. The Balmer-Rydberg equation was derived from
numerous observations, not theory. It is thus an empirical equation.
In 1913, Niels Bohr (1885–1962), a Danish physicist, provided an explanation for
Balmer and Rydberg’s observations. He wrote equations that described the electron of a
hydrogen atom as revolving around its nucleus in circular orbits. He included the assumption
that the electronic energy is quantized; that is, only certain values of electronic energy
are possible. This led him to suggest that electrons can only be in certain discrete orbits,
and that they absorb or emit energy in discrete amounts as they move from one orbit to
another. Each orbit thus corresponds to a definite energy level for the electron. When an
electron is promoted from a lower energy level to a higher one, it absorbs a definite (or
quantized) amount of energy. When the electron falls back to the original energy level,
1
ᎏ
n2
2
1
ᎏ
n1
2
1
ᎏ
␭
The lightning flashes produced in
electrical storms and the light
produced by neon gas in neon signs
are two familiar examples of visible
light produced by electronic
transitions.
200 CHAPTER 5: The Structure of Atoms
Figure 5-16 (a) The radii of the first four Bohr orbits for a hydrogen atom. The dot at
the center represents the nuclear position. The radius of each orbit is proportional to n2,
so these four are in the ratio 1Ϻ4Ϻ9Ϻ16. (b) Relative values for the energies associated with
the various energy levels in a hydrogen atom. The energies become closer together as n
increases. They are so close together for large values of n that they form a continuum. By
convention, potential energy is defined as zero when the electron is at an infinite distance
from the atom. Any more stable arrangement would have a lower energy. Potential energies
of electrons in atoms are therefore always negative. Some possible electronic transitions
corresponding to lines in the hydrogen emission spectrum are indicated by arrows.
Transitions in the opposite directions account for lines in the absorption spectrum.
n = 1 n = 2
n = 3
n = 4
n = 3
n = 2
n = 1
n = 4
n = 5
n = ∞0
Potentialenergy
(b)
(a)
(b)
(a)
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 7.7, Bohr’s Model of the
Hydrogen Atom.
it emits exactly the same amount of energy it absorbed in moving from the lower to the
higher energy level. Figure 5-16 illustrates these transactions schematically. The values of
n1 and n2 in the Balmer-Rydberg equation identify the lower and higher levels, respectively,
of these electronic transitions.
The Danish physicist Niels Bohr was one of the most influential scientists of the twentieth
century. Like many other now-famous physicists of his time, he worked for a time in
England with J. J. Thomson and later with Ernest Rutherford. During this period, he began
to develop the ideas that led to the publication of his explanation of atomic spectra and his
theory of atomic structure, for which he received the Nobel Prize in 1922. After escaping
from German-occupied Denmark to Sweden in 1943, he helped to arrange the escape of
hundreds of Danish Jews from the Hitler regime. He later went to the United States, where,
until 1945, he worked with other scientists at Los Alamos, New Mexico, on the
development of the atomic bomb. From then until his death in 1962, he worked for the
development and use of atomic energy for peaceful purposes.
5-12 Atomic Spectra and the Bohr Atom 201
The Bohr Theory and the Balmer-Rydberg Equation
From mathematical equations describing the orbits for the hydrogen atom, together with
the assumption of quantization of energy, Bohr was able to determine two significant aspects
of each allowed orbit:
1. Where the electron can be with respect to the nucleus—that is, the radius, r, of the
circular orbit. This is given by
r ϭ n2a0
where n is a positive integer (1, 2, 3, . . .) that tells which orbit is being described and
a0 is the Bohr radius. Bohr was able to calculate the value of a0 from a combination
of Planck’s constant, the charge of the electron, and the mass of the electron as
a0 ϭ 5.292 ϫ 10Ϫ11 m ϭ 0.5292 Å
2. How stable the electron would be in that orbit—that is, its potential energy, E. This
is given by
E ϭ Ϫ ΂ᎏ
8␲2
h
m
2
a0
2
ᎏ΃ϭ Ϫ
where h ϭ Planck’s constant, m ϭ the mass of the electron, and the other symbols
have the same meaning as before. E is always negative when the electron is in the
atom; E ϭ 0 when the electron is completely removed from the atom (n ϭ infinity).
Results of evaluating these equations for some of the possible values of n (1, 2, 3, . . .)
are shown in Figure 5-17. The larger the value of n, the farther from the nucleus is the
orbit being described, and the radius of this orbit increases as the square of n increases. As
n increases, n2 increases, 1/n2 decreases, and thus the electronic energy increases (becomes
less negative and smaller in magnitude). For orbits farther from the nucleus, the electronic
potential energy is higher (less negative—the electron is in a higher energy level or in a less
2.180 ϫ 10Ϫ18 J
ᎏᎏ
n2
1
ᎏ
n2
E n r i c h m e n t
Note: r is proportional to n2.
Note: E is proportional to Ϫᎏ
n
1
2
ᎏ.
We define the potential energy of a set
of charged particles to be zero when
the particles are infinitely far apart.
Figure 5-17 (a) The energy levels that the electron can occupy in a hydrogen atom and
a few of the transitions that cause the emission spectrum of hydrogen. The numbers on
the vertical lines show the wavelengths of light emitted when the electron falls to a lower
energy level. (Light of the same wavelength is absorbed when the electron is promoted to
the higher energy level.) The difference in energy between two given levels is exactly the
same for all hydrogen atoms, so it corresponds to a specific wavelength and to a specific
line in the emission spectrum of hydrogen. In a given sample, some hydrogen atoms
could have their electrons excited to the n ϭ 2 level. Some of these electrons could then
fall to the n ϭ 1 energy level, giving off the difference in energy in the form of light (the
1216-Å transition). Other hydrogen atoms might have their electrons excited to the n ϭ 3
level; subsequently some could fall to the n ϭ 1 level (the 1026-Å transition). Because
higher energy levels become closer and closer in energy, differences in energy between
successive transitions become smaller and smaller. The corresponding lines in the
emission spectrum become closer together and eventually result in a continuum, a series
of lines so close together that they are indistinguishable. (b) The emission spectrum of
hydrogen. The series of lines produced by the electron falling to the n ϭ 1 level is known
as the Lyman series; it is in the ultraviolet region. A transition in which the electron falls
to the n ϭ 2 level gives rise to a similar set of lines in the visible region of the spectrum,
known as the Balmer series. Not shown are series involving transitions to energy levels
with higher values of n. (c) The Balmer series shown on an expanded scale. The line at
6563 Å (the n ϭ 3 n n ϭ 2 transition) is much more intense than the line at 4861 Å (the
n ϭ 4 n n ϭ 2 transition) because the first transition occurs much more frequently than
the second. Successive lines in the spectrum become less intense as the series limit is
approached because the transitions that correspond to these lines are less probable.
202 CHAPTER 5: The Structure of Atoms
n
∞
5
4
3
2
1
Ground
state
6563A°
4861A°
4340A°
1216A°
1026A°
973A°
950A°
°
Balmer
series
Balmer series
Excitedstates
Lyman series
Lyman series
red Visible region violet Ultraviolet
Series limit
6563 4861 4340 4102 3646 ␭(A)
␭(A)
°
(a)
(b)
(c)
5000 2500 2000 1500 1250 1000
Increasingenergy
(Enrichment, continued)
(c)
(b)
(a)
5-12 Atomic Spectra and the Bohr Atom 203
stable state). Going away from the nucleus, the allowable orbits are farther apart in distance,
but closer together in energy. Consider the two possible limits of these equations. One limit
is when n ϭ 1; this describes the electron at the smallest possible distance from the nucleus
and at its lowest (most negative) energy. The other limit is for very large values of n, that
is, as n approaches infinity. As this limit is approached, the electron is very far from the
nucleus, or effectively removed from the atom; the potential energy is as high as possible,
approaching zero.
Each line in the emission spectrum represents the difference in energies between two
allowed energy levels for the electron. When the electron goes from energy level n2 to
energy level n1, the difference in energy is given off as a single photon. The energy of this
photon can be calculated from Bohr’s equation for the energy, as follows.
E of photon ϭ E2 Ϫ E1 ϭ ΂Ϫ ΃Ϫ ΂Ϫ ΃
Factoring out the constant 2.180 ϫ 10Ϫ18 J and rearranging, we get
E of photon ϭ 2.180 ϫ 10Ϫ18 J΂ Ϫ ΃
The Planck equation, E ϭ hc/␭, relates the energy of the photon to the wavelength of the
light, so
ᎏ
h
␭
c
ᎏ ϭ 2.180 ϫ 10Ϫ18 J΂ Ϫ ΃
Rearranging for 1/␭, we obtain
ϭ ᎏ
2.180 ϫ
hc
10Ϫ18 J
ᎏ΂ Ϫ ΃
Comparing this to the Balmer-Rydberg equation, Bohr showed that the Rydberg constant
is equivalent to 2.180 ϫ 10Ϫ18 J/hc. We can use the values for h and c to obtain the same
value, 1.097 ϫ 107 mϪ1, that was obtained by Rydberg on a solely empirical basis. Furthermore,
Bohr showed the physical meaning of the two whole numbers n1 and n2; they represent
the two energy states between which the transition takes place. Using this approach, Bohr
was able to use fundamental constants to calculate the wavelengths of the observed lines in
the hydrogen emission spectrum. Thus, Bohr’s application of the idea of quantization of
energy to the electron in an atom provided the answer to a half-century-old puzzle
concerning the discrete colors given off in the spectrum.
1
ᎏ
n2
2
1
ᎏ
n1
2
1
ᎏ
␭
1
ᎏ
n2
2
1
ᎏ
n1
2
1
ᎏ
n2
2
1
ᎏ
n1
2
2.180 ϫ 10Ϫ18 J
ᎏᎏ
n1
2
2.180 ϫ 10Ϫ18 J
ᎏᎏ
n2
2
We now accept the fact that electrons occupy only certain energy levels in atoms. In
most atoms, some of the energy differences between levels correspond to the energy of
visible light. Thus, colors associated with electronic transitions in such elements can be
observed by the human eye.
Although the Bohr theory satisfactorily explained the spectra of hydrogen and of other
species containing one electron (Heϩ, Li2ϩ, etc.) the wavelengths in the observed spectra
of more complex species could not be calculated. Bohr’s assumption of circular orbits was
modified in 1916 by Arnold Sommerfeld (1868–1951), who assumed elliptical orbits. Even
so, the Bohr approach was doomed to failure, because it modified classical mechanics to
solve a problem that could not be solved by classical mechanics. It was a contrived solution.
This failure of classical mechanics set the stage for the development of a new physics,
quantum mechanics, to deal with small particles. The Bohr theory, however, did introduce
the ideas that only certain energy levels are possible, that these energy levels are
described by quantum numbers that can have only certain allowed values, and that the
quantum numbers indicate something about where and how stable the electrons are in
these energy levels. The ideas of modern atomic theory have replaced Bohr’s original
theory. But his achievement in showing a link between electronic arrangements and Balmer
and Rydberg’s empirical description of light absorption, and in establishing the quantization
of electronic energy, was a very important step toward an understanding of atomic
structure.
Two big questions remained about electrons in atoms: (1) How are electrons arranged
in atoms? (2) How do these electrons behave? We now have the background to consider
how modern atomic theory answers these questions.
THE WAVE NATURE OF THE ELECTRON
Einstein’s idea that light can exhibit both wave properties and particle properties suggested
to Louis de Broglie (1892–1987) that very small particles, such as electrons, might also
display wave properties under the proper circumstances. In his doctoral thesis in 1925, de
Broglie predicted that a particle with a mass m and velocity v should have the wavelength
associated with it. The numerical value of this de Broglie wavelength is given by
␭ ϭ h/mv (where h ϭ Planck’s constant)
Two years after de Broglie’s prediction, C. Davisson (1882–1958) and L. H. Germer
(1896–1971) at the Bell Telephone Laboratories demonstrated diffraction of electrons by
a crystal of nickel. This behavior is an important characteristic of waves. It shows conclusively
that electrons do have wave properties. Davisson and Germer found that the
wavelength associated with electrons of known energy is exactly that predicted by de
Broglie. Similar diffraction experiments have been successfully performed with other particles,
such as neutrons.
EXAMPLE 5-7 de Broglie Equation
(a) Calculate the wavelength in meters of an electron traveling at 1.24 ϫ 107 m/s. The mass of
an electron is 9.11 ϫ 10Ϫ28 g. (b) Calculate the wavelength of a baseball of mass 5.25 oz traveling
at 92.5 mph. Recall that 1 J ϭ 1 kgиm2/s2.
Plan
For each calculation, we use the de Broglie equation
␭ ϭ ᎏ
m
h
v
ᎏ
where
h (Planck’s constant) ϭ 6.626 ϫ 10Ϫ34 Jиs ϫ
ϭ 6.626 ϫ 10Ϫ34 ᎏ
kgи
s
m2
ᎏ
For consistency of units, mass must be expressed in kilograms. In part (b), we must also convert
the speed to meters per second.
1ᎏ
kg
s
и
2
m2
ᎏ
ᎏ
1 J
5-13
Materials scientists study electron
diffraction patterns to learn about
the surfaces of solids.
204 CHAPTER 5: The Structure of Atoms
Be careful to distinguish between the
letter v, which represents velocity, and
the Greek letter nu, ␯, which
represents frequency. (See Section
5-10.)
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 7.8, Wave Properties of the
Electron.
Solution
(a) m ϭ 9.11 ϫ 10Ϫ28 g ϫ ϭ 9.11 ϫ 10Ϫ31 kg
Substituting into the de Broglie equation,
␭ ϭ ᎏ
m
h
v
ᎏ ϭ ϭ 5.87 ϫ 10Ϫ11 m
Though this seems like a very short wavelength, it is similar to the spacing between atoms in
many crystals. A stream of such electrons hitting a crystal gives measurable diffraction patterns.
(b) m ϭ 5.25 oz ϫ ϫ ϭ 0.149 kg
v ϭ ϫ ϫ ϫ ϭ 41.3
Now, we substitute into the de Broglie equation.
␭ ϭ ᎏ
m
h
v
ᎏ ϭ ϭ 1.08 ϫ 10Ϫ34 m
This wavelength is far too short to give any measurable effects. Recall that atomic diameters
are in the order of 10Ϫ10 m, which is 24 powers of 10 greater than the baseball “wavelength.”
You should now work Exercise 64.
As you can see from the results of Example 5-7, the particles of the subatomic world
behave very differently from the macroscopic objects with which we are familiar. To talk
about the behavior of atoms and their particles, we must give up many of our long-held
views about the behavior of matter. We must be willing to visualize a world of new and
unfamiliar properties, such as the ability to act in some ways like a particle and in other
ways like a wave.
The wave behavior of electrons is exploited in the electron microscope. This instrument
allows magnification of objects far too small to be seen with an ordinary light
microscope.
THE QUANTUM MECHANICAL PICTURE OF THE ATOM
Through the work of de Broglie, Davisson and Germer, and others, we now know that
electrons in atoms can be treated as waves more effectively than as small compact particles
traveling in circular or elliptical orbits. Large objects such as golf balls and moving
automobiles obey the laws of classical mechanics (Isaac Newton’s laws), but very small
5-14
6.626 ϫ 10Ϫ34 ᎏ
kgи
s
m2
ᎏ
ᎏᎏᎏ
(0.149 kg)΂41.3 ᎏ
m
s
ᎏ΃
m
ᎏ
s
1000 m
ᎏ
1 km
1.609 km
ᎏᎏ
1 mile
1 h
ᎏ
3600 s
92.5 miles
ᎏᎏ
h
1 kg
ᎏ
2.205 lb
1 lb
ᎏ
16 oz
6.626 ϫ 10Ϫ34 ᎏ
kgи
s
m2
ᎏ
ᎏᎏᎏᎏ
(9.11 ϫ 10Ϫ31 kg)΂1.24 ϫ 107 ᎏ
m
s
ᎏ΃
1 kg
ᎏ
1000 g
A color-enhanced scanning electron
micrograph of human red blood
cells, magnified 1200ϫ.
5-14 The Quantum Mechanical Picture of the Atom 205
A modern electron microscope.
particles such as electrons, atoms, and molecules do not. A different kind of mechanics,
called quantum mechanics, which is based on the wave properties of matter, describes
the behavior of very small particles much better. Quantization of energy is a consequence
of these properties.
One of the underlying principles of quantum mechanics is that we cannot determine
precisely the paths that electrons follow as they move about atomic nuclei. The Heisenberg
Uncertainty Principle, stated in 1927 by Werner Heisenberg (1901–1976), is a
theoretical assertion that is consistent with all experimental observations.
It is impossible to determine accurately both the momentum and the position of an
electron (or any other very small particle) simultaneously.
Momentum is mass times velocity, mv. Because electrons are so small and move so rapidly,
their motion is usually detected by electromagnetic radiation. Photons that interact with
electrons have about the same energies as the electrons. Consequently, the interaction of
a photon with an electron severely disturbs the motion of the electron. It is not possible
to determine simultaneously both the position and the velocity of an electron, so we resort
to a statistical approach and speak of the probability of finding an electron within specified
regions in space.
With these ideas in mind, we list some basic ideas of quantum mechanics.
1. Atoms and molecules can exist only in certain energy states. In each energy state,
the atom or molecule has a definite energy. When an atom or molecule changes its
energy state, it must emit or absorb just enough energy to bring it to the new energy
state (the quantum condition).
Atoms and molecules possess various forms of energy. Let us focus our attention on
their electronic energies.
2. When atoms or molecules emit or absorb radiation (light), they change their energies.
The energy change in the atom or molecule is related to the frequency or
wavelength of the light emitted or absorbed by the equations:
⌬E ϭ h␯ or ⌬E ϭ hc/␭
This gives a relationship between the energy change, ⌬E, and the wavelength, ␭, of
the radiation emitted or absorbed. The energy lost (or gained) by an atom as it goes from higher
to lower (or lower to higher) energy states is equal to the energy of the photon emitted (or absorbed)
during the transition.
3. The allowed energy states of atoms and molecules can be described by sets of
numbers called quantum numbers.
The mathematical approach of quantum mechanics involves treating the electron in an
atom as a standing wave. A standing wave is a wave that does not travel and therefore has
at least one point at which it has zero amplitude, called a node. As an example, consider
the various ways that a guitar string can vibrate when it is plucked (Figure 5-18). Because
both ends are fixed (nodes), the string can vibrate only in ways in which there is a whole
number of half-wavelengths in the length of the string (Figure 5-18a). Any possible motion
of the string can be described as some combination of these allowed vibrations. In a similar
way, we can imagine that the electron in the hydrogen atom behaves as a wave (recall the
de Broglie relationship in the last section). The electron can be described by the same
This is like trying to locate the
position of a moving automobile by
driving another automobile into it.
206 CHAPTER 5: The Structure of Atoms
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 7.9, Heisenberg’s Uncertainty
Principle.
Recall that ␭␯ ϭ c, so ␯ ϭ c/␭.
kind of standing-wave mathematics that is applied to the vibrating guitar string. In this
approach, the electron is characterized by a three-dimensional wave function, ␺. In a given
space around the nucleus, only certain “waves” can exist. Each “allowed wave” corresponds
to a stable energy state for the electron and is described by a particular set of quantum
numbers.
The quantum mechanical treatment of atoms and molecules is highly mathematical.
The important point is that each solution of the Schrödinger wave equation (see the
following Enrichment section) describes a possible energy state for the electrons in the
atom. Each solution is described by a set of quantum numbers. These numbers are in
accord with those deduced from experiment and from empirical equations such as the
Balmer-Rydberg equation. Solutions of the Schrödinger equation also tell us about the
shapes and orientations of the probability distributions of the electrons. (The Heisenberg
Principle implies that this is how we must describe the positions of the electrons.) These
atomic orbitals (which are described in Section 5-16) are deduced from the solutions of the
Schrödinger equation. The orbitals are directly related to the quantum numbers.
Figure 5-18 When a string that is fixed at both ends—such as (a) a guitar string—is
plucked, it has a number of natural patterns of vibration, called normal modes. Because the
string is fixed at both ends, the ends must be stationary. Each different possible vibration is a
standing wave, and can be described by a wave function. The only waves that are possible
are those in which a whole number of half-wavelengths fits into the string length. These
allowed waves constitute a harmonic series. Any total motion of the string is some
combination of these allowed harmonics. (b) Some of the ways in which a plucked guitar
string can vibrate. The position of the string at one extreme of each vibration is shown as a
solid line, and at the other extreme as a dashed line. (c) An example of vibration that is not
possible for a plucked string. In such a vibration, an end of the string would move; this is
not possible because the ends are fixed.
5-14 The Quantum Mechanical Picture of the Atom 207
The Schrödinger Equation
In 1926, Erwin Schrödinger (1887–1961) modified an existing equation that described a
three-dimensional standing wave by imposing wavelength restrictions suggested by
de Broglie’s ideas. The modified equation allowed him to calculate the energy levels in the
hydrogen atom. It is a differential equation that need not be memorized or even understood
to read this book. A knowledge of differential calculus would be necessary.
Ϫ ΂ ϩ ϩ ΃ϩ V␺ ϭ E␺
This equation has been solved exactly only for one-electron species such as the hydrogen
atom and the ions Heϩ and Li2ϩ. Simplifying assumptions are necessary to solve the equation
for more complex atoms and molecules. Chemists and physicists have used their
intuition and ingenuity (and modern computers), however, to apply this equation to more
complex systems.
∂2␺
ᎏ
∂z2
∂2␺
ᎏ
∂y2
∂2␺
ᎏ
∂x2
h2
ᎏ
8␲2m
E n r i c h m e n t
In 1928, Paul A. M. Dirac (1902–1984) reformulated electron quantum mechanics to take
into account the effects of relativity. This gave rise to a fourth quantum number.
A stationary string
1 half-wavelength
2 half-wavelengths (or one full wave)
3 half-wavelengths
2– half-wavelengths (not possible)
(a)
(b)
(c) 1
2
Schrödinger’s equation and wave
functions.
QUANTUM NUMBERS
The solutions of the Schrödinger and Dirac equations for hydrogen atoms give wave functions,
␺, that describe the various states available to hydrogen’s single electron. Each of
these possible states is described by four quantum numbers. We can use these quantum
numbers to designate the electronic arrangements in all atoms, their so-called electron
configurations. These quantum numbers play important roles in describing the energy
levels of electrons and the shapes of the orbitals that describe distributions of electrons
in space. The interpretation will become clearer when we discuss atomic orbitals in the
following section. For now, let’s say that
an atomic orbital is a region of space in which the probability of finding an electron
is high.
We define each quantum number and describe the range of values it may take.
1. The principal quantum number, n, describes the main energy level, or shell, an
electron occupies. It may be any positive integer:
n ϭ 1, 2, 3, 4, . . .
2. The angular momentum quantum number, ᐉ, designates the shape of the region
in space that an electron occupies. Within a shell (defined by the value of n, the
principal quantum number) different sublevels or subshells are possible, each with
a characteristic shape. The angular momentum quantum number designates a
sublevel, or specific shape of atomic orbital that an electron may occupy. This number,
ᐉ, may take integral values from 0 up to and including (n Ϫ 1):
ᐉ ϭ 0, 1, 2, . . . , (n Ϫ 1)
Thus, the maximum value of ᐉ is (n Ϫ 1). We give a letter notation to each value
of ᐉ. Each letter corresponds to a different sublevel (subshell).
ᐉ ϭ 0, 1, 2, 3, . . . , (n Ϫ 1)
s p d f
In the first shell, the maximum value of ᐉ is zero, which tells us that there is only
an s subshell and no p subshell. In the second shell, the permissible values of ᐉ are
0 and 1, which tells us that there are only s and p subshells.
3. The magnetic quantum number, mᐉ, designates the specific orbital within a
subshell. Orbitals within a given subshell differ in their orientations in space, but
not in their energies. Within each subshell, mᐉ may take any integral values from
Ϫᐉ through zero up to and including ϩᐉ:
mᐉ ϭ (Ϫᐉ), . . . , 0, . . . , (ϩᐉ)
The maximum value of mᐉ depends on the value of ᐉ. For example, when ᐉ ϭ 1,
which designates the p subshell, there are three permissible values of mᐉ: Ϫ1, 0, and
ϩ1. Thus, three distinct regions of space, called atomic orbitals, are associated
with a p subshell. We refer to these orbitals as the px, py, and pz orbitals (see Section
5-16).
5-15
The s, p, d, f designations arise from
the characteristics of spectral emission
lines produced by electrons occupying
the orbitals: s (sharp), p (principal), d
(diffuse), and f (fundamental).
208 CHAPTER 5: The Structure of Atoms
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 7.11, Shells, Subshells, and
Orbitals.
4. The spin quantum number, ms, refers to the spin of an electron and the orientation
of the magnetic field produced by this spin. For every set of n, ᐉ, and mᐉ values,
ms can take the value ϩᎏ1
2
ᎏ or Ϫᎏ1
2
ᎏ:
ms ϭ Ϯᎏ1
2
ᎏ
The values of n, ᐉ, and mᐉ describe a particular atomic orbital. Each atomic orbital can
accommodate no more than two electrons, one with ms ϭ ϩᎏ1
2
ᎏ and another with ms ϭ Ϫᎏ1
2
ᎏ.
Table 5-4 summarizes some permissible values for the four quantum numbers. Spectroscopic
evidence confirms the quantum mechanical predictions about the number of
atomic orbitals in each shell.
ATOMIC ORBITALS
Let us now describe the distributions of electrons in atoms. For each neutral atom, we
must account for a number of electrons equal to the number of protons in the nucleus,
that is, the atomic number of the atom. Each electron is said to occupy an atomic orbital
defined by a set of quantum numbers n, ᐉ, and mᐉ. In any atom, each orbital can hold a
maximum of two electrons. Within each atom, these atomic orbitals, taken together, can
be represented as a diffuse cloud of electrons (Figure 5-19).
The main shell of each atomic orbital in an atom is indicated by the principal quantum
number n (from the Schrödinger equation). As we have seen, the principal quantum
number takes integral values: n ϭ 1, 2, 3, 4, . . . . The value n ϭ 1 describes the first, or
innermost, shell. These shells have been referred to as electron energy levels. Successive
shells are at increasingly greater distances from the nucleus. For example, the n ϭ 2 shell
is farther from the nucleus than the n ϭ 1 shell. The electron capacity of each shell is
indicated in the right-hand column of Table 5-4. For a given n, the capacity is 2n2.
By the rules of Section 5-15, each shell has an s subshell (defined by ᐉ ϭ 0) consisting
of one s atomic orbital (defined by mᐉ ϭ 0). We distinguish among orbitals in different
5-16
As you study the next two sections,
keep in mind that the wave function,
␺, for an orbital characterizes two
features of an electron in that orbital:
(1) where (the region in space) the
probability of finding the electron is
high and (2) how stable that electron is
(its energy).
5-16 Atomic Orbitals 209
TABLE 5-4 Permissible Values of the Quantum Numbers Through n ‫؍‬ 4
Electron Capacity of Electron Capacity of
n ᐉ mᐉ ms Subshell ‫؍‬ 4ᐉ ؉ 2 Shell ‫؍‬ 2n2
1 0 (1s) 0 ϩᎏ1
2
ᎏ, Ϫᎏ1
2
ᎏ 2 2
2 0 (2s) 0 ϩᎏ1
2
ᎏ, Ϫᎏ1
2
ᎏ 2 8
1 (2p) Ϫ1, 0, ϩ1 Ϯᎏ1
2
ᎏ for each value of mᐉ 6
3 0 (3s) 0 ϩᎏ1
2
ᎏ, Ϫᎏ1
2
ᎏ 2 18
1 (3p) Ϫ1, 0, ϩ1 Ϯᎏ1
2
ᎏ for each value of mᐉ 6
2 (3d) Ϫ2, Ϫ1, 0, ϩ1, ϩ2 Ϯᎏ1
2
ᎏ for each value of mᐉ 10
4 0 (4s) 0 ϩᎏ1
2
ᎏ, Ϫᎏ1
2
ᎏ 2 32
1 (4p) Ϫ1, 0, ϩ1 Ϯᎏ1
2
ᎏ for each value of mᐉ 6
2 (4d) Ϫ2, Ϫ1, 0, ϩ1, ϩ2 Ϯᎏ1
2
ᎏ for each value of mᐉ 10
3 (4f) Ϫ3, Ϫ2, Ϫ1, 0, ϩ1, ϩ2, ϩ3 Ϯᎏ1
2
ᎏ for each value of mᐉ 14
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 7.12, Quantum Numbers and
Orbitals.
principal shells (main energy levels) by using the principal quantum number as a coefficient;
1s indicates the s orbital in the first shell, 2s is the s orbital in the second shell, 2p
is a p orbital in the second shell, and so on (Table 5-4).
For each solution to the quantum mechanical equation, we can calculate the electron
probability density (sometimes just called the electron density) at each point in the atom.
This is the probability of finding an electron at that point. It can be shown that this electron
density is proportional to r2␺2, where r is the distance from the nucleus.
In the graphs in Figure 5-20, the electron probability density at a given distance from
the nucleus is plotted against distance from the nucleus, for s orbitals. It is found that the
electron probability density curve is the same regardless of the direction in the atom. We
Figure 5-19 An electron cloud surrounding an atomic nucleus. The electron density drops
off rapidly but smoothly as distance from the nucleus increases.
z
y
x
Figure 5-20 Plots of the electron density distributions associated with s orbitals. For any
s orbital, this plot is the same in any direction (spherically symmetrical). The sketch below
each plot shows a cross-section, in the plane of the atomic nucleus, of the electron cloud
associated with that orbital. Electron density is proportional to r2␺2.
r 2
␺2
r 2 ␺2
radius radius radius
r 2
␺2
ᐉ = 0
ls
ᐉ = 0
2s
ᐉ = 0
3s
describe an s orbital as spherically symmetrical; that is, it is round like a basketball (Figure
5-21). The electron clouds (electron densities) associated with the 1s, 2s, and 3s atomic
orbitals are shown just below the plots. The electron clouds are three-dimensional, and
only cross-sections are shown here. The regions shown in some figures (Figures 5-21
through 5-25) appear to have surfaces or skins only because they are arbitrarily “cut off”
so that there is a 90% probability of finding an electron occupying the orbital somewhere
within the surfaces.
Beginning with the second shell, each shell also contains a p subshell, defined by ᐉ ϭ
1. Each of these subshells consists of a set of three p atomic orbitals, corresponding to the
three allowed values of mᐉ (Ϫ1, 0, and ϩ1) when ᐉ ϭ 1. The sets are referred to as 2p,
3p, 4p, 5p, . . . orbitals to indicate the main shells in which they are found. Each set of
atomic p orbitals resembles three mutually perpendicular equal-arm dumbbells (see Figure
5-22). The nucleus defines the origin of a set of Cartesian coordinates with the usual x,
y, and z axes (see Figure 5-23a). The subscript x, y, or z indicates the axis along which
each of the three two-lobed orbitals is directed. A set of three p atomic orbitals may be
represented as in Figure 5-23b.
Beginning at the third shell, each shell also contains a third subshell (ᐉ ϭ 2) composed
of a set of five d atomic orbitals (mᐉ ϭ Ϫ2, Ϫ1, 0, ϩ1, ϩ2). They are designated 3d, 4d,
5d, . . . to indicate the shell in which they are found. The shapes of the members of a set
are indicated in Figure 5-24.
In each of the fourth and larger shells, there is also a fourth subshell, containing a set
of seven f atomic orbitals (ᐉ ϭ 3, mᐉ ϭ Ϫ3, Ϫ2, Ϫ1, 0, ϩ1, ϩ2, ϩ3). These are shown in
Figure 5-25.
Thus, we see the first shell contains only the 1s orbital; the second shell contains the
2s and three 2p orbitals; the third shell contains the 3s, three 3p, and five 3d orbitals; and
the fourth shell consists of a 4s, three 4p, five 4d, and seven 4f orbitals. All subsequent
shells contain s, p, d, and f subshells as well as others that are not occupied in any presently
known elements in their lowest energy states.
Figure 5-21 The shape of an s
orbital.
5-16 Atomic Orbitals 211
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 7.13, Shapes of Atomic Orbitals.
x
y
z
Figure 5-22 Three representations
of the shape of a p orbital. The plot
at the bottom is along the axis of
maximum electron density for this
orbital. A plot along any other
direction would be different, because
a p orbital is not spherically
symmetrical.
x
x
r2
␺ 2
px
The sizes of orbitals increase with increasing n and the true shapes of p orbitals are
“diffuse,” as shown in Figure 5-26. The directions of p, d, and f orbitals, however, are
easier to visualize in drawings such as those in Figures 5-23, 5-24, and 5-25; therefore,
these “slender” representations are usually used.
In this section, we haven’t yet discussed the fourth quantum number, the spin quantum
number, ms. Because ms has two possible values, ϩᎏ1
2
ᎏ and Ϫᎏ1
2
ᎏ, each atomic orbital, defined
Figure 5-23 (a) The relative directional character of a set of p orbitals. (b) A model of
three p orbitals (px, py, and pz) of a single set of orbitals. The nucleus is at the center. (The
lobes are actually more diffuse (“fatter”) than depicted. See Figure 5-26.)
212 CHAPTER 5: The Structure of Atoms
Figure 5-24 Spatial orientation of d orbitals. Note that the lobes of the dx2Ϫy2 and dz2
orbitals lie along the axes, whereas the lobes of the others lie along diagonals between the
axes.
x
y
z
py
(a)
y
x
z
px
x
y
z
pz
(b)
x
y
z
dx 2
– y 2
z
y
x
dz 2
dxzdxy dyz
by the values of n, ᐉ, and mᐉ, has a capacity of two electrons. Electrons are negatively
charged, and they behave as though they were spinning about axes through their centers,
so they act like tiny magnets. The motions of electrons produce magnetic fields, and these
can interact with one another. Two electrons in the same orbital having opposite ms values
are said to be spin-paired, or simply paired (Figure 5-27).
Let us summarize, in tabular form, some of the information we have developed to this
point. The principal quantum number n indicates the main shell. The number of subshells
per shell is equal to n, the number of atomic orbitals per shell is n2, and the maximum
number of electrons per shell is 2n2, because each atomic orbital can hold two electrons.
Figure 5-26 Shapes and
approximate relative sizes of several
orbitals in an atom.
5-16 Atomic Orbitals 213
Figure 5-25 Relative directional character of f orbitals. The seven orbitals are shown
within cubes as an aid to visualization.
z
y
x fz3 – – zr2
fxyz
3
5
fy3 – – yr23
5
fy (x2 – z2)
fz (x2 – y2)fx (z2 – y2)
fx3 – – xr23
5
2s1s 2p
3s 3p
Number of Subshells Number of Maximum Number
Shell per Shell Atomic Orbitals of Electrons
n n n2 2n2
1 1 1 (1s) 2
2 2 4 (2s, 2px, 2py, 2pz) 8
3 3 9 (3s, three 3p’s, five 3d’s) 18
4 4 16 32
5 5 25 50
ELECTRON CONFIGURATIONS
The wave function for an atom simultaneously depends on (describes) all of the electrons
in the atom. The Schrödinger equation is much more complicated for atoms with more
than one electron than for a one-electron species such as a hydrogen atom, and an explicit
solution to this equation is not possible even for helium, let alone for more complicated
atoms. We must therefore rely on approximations to solutions of the many-electron
Schrödinger equation. We shall use one of the most common and useful, called the orbital
approximation. In this approximation, the electron cloud of an atom is assumed to be
the superposition of charge clouds, or orbitals, arising from the individual electrons; these
orbitals resemble the atomic orbitals of hydrogen (for which exact solutions are known),
which we described in some detail in the previous section. Each electron is described by
the same allowed combinations of quantum numbers (n, ᐉ, mᐉ, and ms) that we used for
the hydrogen atom; however, the order of energies of the orbitals is often different from
that in hydrogen.
Let us now examine the electronic structures of atoms of different elements. The electronic
arrangement that we will describe for each atom is called the ground state electron
configuration. This corresponds to the isolated atom in its lowest energy, or unexcited,
state. Electron configurations for the elements, as determined by experiment, are given
in Appendix B. We will consider the elements in order of increasing atomic number, using
as our guide the periodic table on the inside front cover of this text.
In describing ground state electron configuration, the guiding idea is that the total
energy of the atom is as low as possible. To determine these configurations, we use the
Aufbau Principle as a guide:
5-17
Figure 5-27 Electron spin. Electrons act as though they
spin about an axis through their centers. Because electrons
may spin in two directions, the spin quantum number has
two possible values, ϩᎏ1
2
ᎏ and Ϫᎏ1
2
ᎏ. Each electron spin
produces a magnetic field. When two electrons have
opposite spins, the attraction due to their opposite
magnetic fields (gray arrows) helps to overcome the
repulsion of their like charges. This permits two electrons
to occupy the same region (orbital).
214 CHAPTER 5: The Structure of Atoms
One electron has ms = + –;
the other has ms = – –.1
2
1
2
The great power of modern computers
has allowed scientists to make
numerical approximations to this
solution to very high accuracy for
simple atoms such as helium. As the
number of electrons increases,
however, even such numerical
approaches become quite difficult to
apply and interpret. For multielectron
atoms, more quantitative
approximations are used.
The German verb aufbauen means “to
build up.”
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 8.2, Spinning Electrons.
Each atom is “built up” by (1) adding the appropriate numbers of protons and
neutrons as specified by the atomic number and the mass number, and (2) adding
the necessary number of electrons into orbitals in the way that gives the lowest total
energy for the atom.
As we apply this principle, we will focus on the difference in electronic arrangement
between a given element and the element with an atomic number that is one lower. In
doing this, we emphasize the particular electron that distinguishes each element from the
previous one; however, we should remember that this distinction is artificial, because electrons
are not really distinguishable. Though we do not always point it out, we must keep
in mind that the atomic number (the charge on the nucleus) also differs.
The orbitals increase in energy with increasing value of the quantum number n. For a
given value of n, energy increases with increasing value of ᐉ. In other words, within a
particular main shell, the s subshell is lowest in energy, the p subshell is the next lowest,
then the d, then the f, and so on. As a result of changes in the nuclear charge and interactions
among the electrons in the atom, the order of energies of the orbitals can vary
somewhat from atom to atom.
Two general rules help us to predict electron configurations.
1. Electrons are assigned to orbitals in order of increasing value of (n ϩ ᐉ).
2. For subshells with the same value of (n ϩ ᐉ), electrons are assigned first to the
subshell with lower n.
For example, the 2s subshell has (n ϩ ᐉ ϭ 2 ϩ 0 ϭ 2), and the 2p subshell has (n ϩ ᐉ ϭ
2 ϩ 1 ϭ 3), so we would expect to fill the 2s subshell before the 2p subshell (rule 1). This
rule also predicts that the 4s subshell (n ϩ ᐉ ϭ 4 ϩ 0 ϭ 4) will fill before the 3d subshell
(n ϩ ᐉ ϭ 3 ϩ 2 ϭ 5). Rule 2 reminds us to fill 2p (n ϩ ᐉ ϭ 2 ϩ 1 ϭ 3) before 3s (n ϩ ᐉ ϭ
3 ϩ 0 ϭ 3), because 2p has a lower value of n. The usual order of energies of orbitals of
an atom and a helpful device for remembering this order are shown in Figures 5-28 and
5-29.
But we should consider these only as a guide to predicting electron arrangements. The
observed electron configurations of lowest total energy do not always match those
predicted by the Aufbau guide, and we will see a number of exceptions, especially for
elements in the B groups of the periodic table.
The electronic structures of atoms are governed by the Pauli Exclusion Principle:
No two electrons in an atom may have identical sets of four quantum numbers.
An orbital is described by a particular allowed set of values for n, ᐉ, and mᐉ. Two electrons
can occupy the same orbital only if they have opposite spins, ms. Two such electrons
in the same orbital are paired. For simplicity, we shall indicate atomic orbitals as __ and
show an unpaired electron as __h and spin-paired electrons as __hg. By “unpaired electron”
we mean an electron that occupies an orbital singly.
Row 1. The first shell consists of only one atomic orbital, 1s. This can hold a maximum
of two electrons. Hydrogen, as we have already noted, contains just one electron. Helium,
5-17 Electron Configurations 215
216 CHAPTER 5: The Structure of Atoms
Figure 5-28 The usual order of filling (Aufbau order) of the orbitals of an atom. The
relative energies are different for different elements, but the following main features should
be noted: (1) The largest energy gap is between the 1s and 2s orbitals. (2) The energies of
orbitals are generally closer together at higher energies. (3) The gap between np and (n ϩ 1)s
(e.g., between 2p and 3s or between 3p and 4s) is fairly large. (4) The gap between (n Ϫ 1)d
and ns (e.g., between 3d and 4s) is quite small. (5) The gap between (n Ϫ 2)f and ns (e.g.,
between 4f and 6s) is even smaller.
Figure 5-29 An aid to
remembering the usual order of
filling of atomic orbitals. Write each
shell (value of n) on one horizontal
line, starting with n ϭ 1 at the
bottom. Write all like subshells
(same ᐉ values) in the same vertical
column. Subshells are filled in order
of increasing (n ϩ ᐉ). When
subshells have the same (n ϩ ᐉ), the
subshell with the lower n fills first.
To use the diagram, we follow the
diagonal arrows in order, reading
bottom to top.
0 1 2 3ᐉ =
7s
6s
5s
4s
3s
2s
1s
6p
5p
4p
3p
2p
5d
4d
3d
4f
Increasingenergy
n value
8s
7p
6d
5f
6s
5p
4d
5s
4p
3d
4s
3p3s
2p2s
1s
0 1 2 3
ഞ value
7s
6p
5d
4f4
5
6
7
8
3
2
1
n + ഞ = 8
n + ഞ = 7n + ഞ = 6
n + ഞ = 5n + ഞ = 4
n + ഞ = 3n + ഞ = 2
n + ഞ = 1
a noble gas, has a filled first main shell (two electrons). The atom is so stable that no
chemical reactions of helium are known.
Orbital Notation
1s Simplified Notation
1H __h 1s1
2He __hg 1s2
Row 2. Elements of atomic numbers 3 through 10 occupy the second period, or horizontal
row, in the periodic table. In neon atoms the second main shell is filled completely.
Neon, a noble gas, is extremely stable. No reactions of it are known.
Orbital Notation
1s 2s 2p Simplified Notation
3Li __hg __h 1s22s1 or [He] 2s1
4Be __hg __hg 1s22s2 [He] 2s2
5B __hg __hg __h __ __ 1s22s22p1 [He] 2s22p1
6C __hg __hg __h __h __ 1s22s22p2 [He] 2s22p2
7N __hg __hg __h __h __h 1s22s22p3 [He] 2s22p3
8O __hg __hg __hg __h __h 1s22s22p4 [He] 2s22p4
9F __hg __hg __hg __hg __h 1s22s22p5 [He] 2s22p5
10Ne __hg __hg __hg __hg __hg 1s22s22p6 [He] 2s22p6
We see that some atoms have unpaired electrons in the same set of energetically equivalent,
or degenerate, orbitals. We have already seen that two electrons can occupy a given
atomic orbital (with the same values of n, ᐉ, and mᐉ) only if their spins are paired (have
opposite values of ms). Even with pairing of spins, however, two electrons that are in the
same orbital repel each other more strongly than do two electrons in different (but equalenergy)
orbitals. Thus, both theory and experimental observations (see the following
Enrichment section) lead to Hund’s Rule:
Electrons occupy all the orbitals of a given subshell singly before pairing begins.
These unpaired electrons have parallel spins.
Thus, carbon has two unpaired electrons in its 2p orbitals, and nitrogen has three.
Helium’s electrons can be displaced
only by very strong forces, as in
excitation by high-voltage discharge.
In the simplified notation, we indicate
with superscripts the number of
electrons in each subshell.
5-17 Electron Configurations 217
As with helium, neon’s electrons can be displaced by
high-voltage electric discharge, as is observed in
neon signs.
In writing electron configurations of
atoms, we frequently simplify the
notations. The abbreviation [He]
indicates that the 1s orbital is
completely filled, 1s2, as in helium.
218 CHAPTER 5: The Structure of Atoms
E n r i c h m e n t
Paramagnetism and Diamagnetism
Substances that contain unpaired electrons are weakly attracted into magnetic fields and are
said to be paramagnetic. By contrast, those in which all electrons are paired are very weakly
repelled by magnetic fields and are called diamagnetic. The magnetic effect can be measured
by hanging a test tube full of a substance on a balance by a long thread and suspending it
above the gap of an electromagnet (Figure 5-30). When the current is switched on, a paramagnetic
substance such as copper(II) sulfate is pulled into the strong field. The
Iron displays ferromagnetism.
Figure 5-30 Diagram of an apparatus for measuring the paramagnetism of a substance.
The tube contains a measured amount of the substance, often in solution. (a) Before the
magnetic field is turned on, the position and mass of the sample are determined. (b)
When the field is on, a paramagnetic substance is attracted into the field. (c) A
diamagnetic substance would be repelled very weakly by the field.
(a) (b) (c)
Sample
Magnet Magnet
No magnetic field
Both paramagnetism and diamagnetism
are hundreds to thousands of times
weaker than ferromagnetism, the effect
seen in iron bar magnets.
(a) (b) (c)
5-17 Electron Configurations 219
Row 3. The next element beyond neon is sodium. Here we begin to add electrons to the
third shell. Elements 11 through 18 occupy the third period in the periodic table.
Orbital Notation
3s 3p Simplified Notation
11Na [Ne] __h [Ne] 3s1
12Mg [Ne] __hg [Ne] 3s2
13Al [Ne] __hg __h __ __ [Ne] 3s23p1
14Si [Ne] __hg __h __h __ [Ne] 3s23p2
15P [Ne] __hg __h __h __h [Ne] 3s23p3
16S [Ne] __hg __hg __h __h [Ne] 3s23p4
17Cl [Ne] __hg __hg __hg __h [Ne] 3s23p5
18Ar [Ne] __hg __hg __hg __hg [Ne] 3s23p6
Although the third shell is not yet filled (the d orbitals are still empty), argon is a noble
gas. All noble gases except helium have ns2np6 electron configurations (where n indicates
the largest occupied shell). The noble gases are quite unreactive.
Rows 4 and 5. It is an experimentally observed fact that an electron occupies the available
orbital that gives the atom the lowest total energy. It is observed that filling the 4s orbitals
before electrons enter the 3d orbitals usually leads to a lower total energy for the atom
than some other arrangements. We therefore fill the orbitals in this order (see Figure
5-28). According to the Aufbau order (recall Figures 5-28 and 5-29), 4s fills before 3d. In
general, the (n ϩ 1)s orbital fills before the nd orbital. This is sometimes referred to as the
(n ϩ 1) rule.
After the 3d sublevel is filled to its capacity of 10 electrons, the 4p orbitals fill next,
taking us to the noble gas krypton. Then the 5s orbital, the five 4d orbitals, and the three
5p orbitals fill to take us to xenon, a noble gas.
Let us now examine the electronic structure of the 18 elements in the fourth period
in some detail. Some of these have electrons in d orbitals.
paramagnetic attraction per mole of substance can be measured by weighing the sample
before and after energizing the magnet. The paramagnetism per mole increases with
increasing number of unpaired electrons per formula unit. Many transition metals and ions
have one or more unpaired electrons and are paramagnetic.
The metals of the iron triad (Fe, Co, and Ni) are the only free elements that exhibit
ferromagnetism. This property is much stronger than paramagnetism; it allows a substance
to become permanently magnetized when placed in a magnetic field. This happens as
randomly oriented electron spins align themselves with an applied field. To exhibit ferromagnetism,
the atoms must be within the proper range of sizes so that unpaired electrons
on adjacent atoms can interact cooperatively with one another, but not to the extent that
they pair. Experimental evidence suggests that in ferromagnets, atoms cluster together into
domains that contain large numbers of atoms in fairly small volumes. The atoms within each
domain interact cooperatively with one another.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 8.3, Spinning Electrons and
Magnetism.
Orbital Notation
3d 4s 4p Simplified Notation
19K [Ar] __h [Ar] 4s1
20Ca [Ar] __hg [Ar] 4s2
21Sc [Ar] __h __ __ __ __ __hg [Ar] 3d14s2
22Ti [Ar] __h __h __ __ __ __hg [Ar] 3d24s2
23V [Ar] __h __h __h __ __ __hg [Ar] 3d34s2
24Cr [Ar] __h __h __h __h __h __h [Ar] 3d54s1
25Mn [Ar] __h __h __h __h __h __hg [Ar] 3d54s2
26Fe [Ar] __hg __h __h __h __h __hg [Ar] 3d64s2
27Co [Ar] __hg __hg __h __h __h __hg [Ar] 3d74s2
28Ni [Ar] __hg __hg __hg __h __h __hg [Ar] 3d84s2
29Cu [Ar] __hg __hg __hg __hg __hg __h [Ar] 3d104s1
30Zn [Ar] __hg __hg __hg __hg __hg __hg [Ar] 3d104s2
31Ga [Ar] __hg __hg __hg __hg __hg __hg __h __ __ [Ar] 3d104s24p1
32Ge [Ar] __hg __hg __hg __hg __hg __hg __h __h __ [Ar] 3d104s24p2
33As [Ar] __hg __hg __hg __hg __hg __hg __h __h __h [Ar] 3d104s24p3
34Se [Ar] __hg __hg __hg __hg __hg __hg __hg __h __h [Ar] 3d104s24p4
35Br [Ar] __hg __hg __hg __hg __hg __hg __hg __hg __h [Ar] 3d104s24p5
36Kr [Ar] __hg __hg __hg __hg __hg __hg __hg __hg __hg [Ar] 3d104s24p6
As you study these electron configurations, you should be able to see how most of them
are predicted from the Aufbau order. However, as we fill the 3d set of orbitals, from 21Sc
to 30Zn, we see that these orbitals are not filled quite regularly. As the 3d orbitals are
filled, their energies get closer to that of the 4s orbital and eventually become lower. If
the order of filling of electrons on chromium gave the expected configuration, it would
be: [Ar] 4s23d4. Chemical and spectroscopic evidence indicates, however, that the configuration
of Cr has only one electron in the 4s orbital, [Ar] 4s13d5. For this element, the 4s
and 3d orbitals are nearly equal in energy. Six electrons in these six orbitals of nearly the
same energy are more stable with the electrons all unpaired, [Ar] 3d __h __h __h __h __h 4s __h
rather than the predicted order [Ar] 3d __h ____h __h __h __ 4s __hg.
The next elements, Mn to Ni, have configurations as predicted by the Aufbau order,
presumably because forming a pair of electrons in the larger 4s orbital is easier than in a
smaller, less diffuse, 3d orbital. By the time Cu is reached, the energy of 3d is sufficiently
lower than that of 4s, so that the total energy of the configuration [Ar] 4s13d10 is lower
than that of [Ar] 4s23d9.
We notice that the exceptions for Cr and Cu give half-filled or filled sets of equivalent
orbitals (d5 and d10, respectively), and this is also true for several other exceptions to
the Aufbau order. You may wonder why such an exception does not occur in, for example,
14Si or 32Ge, where we could have an s1p3 configuration that would have half-filled sets
of s and p orbitals. It does not occur because of the very large energy gap between ns and
np orbitals. There is some evidence that does, however, suggest an enhanced stability of
half-filled sets of p orbitals.
End-of-chapter Exercises 81–111
provide much valuable practice in
writing electron configurations.
Problem-Solving Tip: Exceptions to the Aufbau Order
In Appendix B, you will find a number of exceptions to the electron configurations
predicted from the Aufbau Principle. You should realize that statements such as the
Aufbau Principle and the (n ϩ 1) rule merely represent general guidelines and should
The electron configurations of
elements 1 through 109 are given in
Appendix B.
Electrons are indistinguishable. We
have numbered them 1, 2, 3, and so on
as an aid to counting them.
5-17 Electron Configurations 221
Let us now write the quantum numbers to describe each electron in an atom of nitrogen.
Keep in mind the fact that Hund’s Rule must be obeyed. Thus, there is only one (unpaired)
electron in each 2p orbital in a nitrogen atom.
EXAMPLE 5-8 Electron Configurations and Quantum Numbers
Write an acceptable set of four quantum numbers for each electron in a nitrogen atom.
Plan
Nitrogen has seven electrons, which occupy the lowest energy orbitals available. Two electrons
can occupy the first shell, n ϭ 1, in which there is only one s orbital; when n ϭ 1, then ᐉ must
be zero, and therefore mᐉ ϭ 0. The two electrons differ only in spin quantum number, ms. The
next five electrons can all fit into the second shell, for which n ϭ 2 and ᐉ may be either 0 or
1. The ᐉ ϭ 0 (s) subshell fills first, and the ᐉ ϭ 1 (p) subshell is occupied next.
Solution
not be viewed as hard-and-fast rules; the total energy of the atom is as low as possible.
Some of the reasons for exceptions are
1. The Aufbau order of orbital energies is based on calculations for the hydrogen atom,
which contains only one electron. The orbital energies also depend on additional
factors such as the nuclear charge and interactions of electrons in different occupied
orbitals.
2. The energy scale varies with the atomic number.
3. Some orbitals are very close together, so their order can change, depending on the
occupancies of other orbitals.
Some types of exceptions to the Aufbau order are general enough to remember easily,
for example, those based on the special stability of filled or half-filled sets of orbitals.
Other exceptions are quite unpredictable. Your instructor may expect you to remember
some of the exceptions.
EXAMPLE 5-9 Electron Configurations and Quantum Numbers
Write an acceptable set of four quantum numbers for each electron in a chlorine atom.
Plan
Chlorine is element number 17. Its first seven electrons have the same quantum numbers as
those of nitrogen in Example 5-8. Electrons 8, 9, and 10 complete the filling of the 2p subshell
Electron n ᐉ mᐉ ms eϪ Configuration
1,2
1 0 0 ϩᎏ1
2
ᎏ
1s2
1 0 0 Ϫᎏ1
2
ᎏ
3,4
2 0 0 ϩᎏ1
2
ᎏ
2s2
2 0 0 Ϫᎏ1
2
ᎏ
2 1 Ϫ1 ϩᎏ1
2
ᎏ or Ϫᎏ1
2
ᎏ 2px
1
5, 6, 7 2 1 0 ϩᎏ1
2
ᎏ or Ϫᎏ1
2
ᎏ 2py
1 or 2p3
2 1 ϩ1 ϩᎏ1
2
ᎏ or Ϫᎏ1
2
ᎏ 2pz
1



























In the lowest energy configurations,
the three 2p electrons either have ms ϭ
ϩᎏ1
2
ᎏ or all have ms ϭ Ϫᎏ1
2
ᎏ.
222 CHAPTER 5: The Structure of Atoms
(n ϭ 2, ᐉ ϭ 1) and therefore also the second energy level. Electrons 11 through 17 fill the 3s
subshell (n ϭ 3, ᐉ ϭ 0) and partially fill the 3p subshell (n ϭ 3, ᐉ ϭ 1).
Solution
Electron n ᐉ mᐉ ms eϪ Configuration
1, 2 1 0 0 Ϯᎏ1
2
ᎏ 1s2
3, 4 2 0 0 Ϯᎏ1
2
ᎏ 2s2
2 1 Ϫ1 Ϯᎏ1
2
ᎏ
5–10 2 1 0 Ϯᎏ1
2
ᎏ 2p6
2 1 ϩ1 Ϯᎏ1
2
ᎏ
11, 12 3 0 0 Ϯᎏ1
2
ᎏ 3s2
3 1 Ϫ1 Ϯᎏ1
2
ᎏ
13–17 3 1 0 Ϯᎏ1
2
ᎏ 3p5
3 1 ϩ1 ϩᎏ1
2
ᎏ or Ϫᎏ1
2
ᎏ*
*The 3p orbital with only a single electron can be any one of the set, not necessarily
the one with mᐉ ϭ ϩ1.




















You should now work Exercises 104 and 108.
THE PERIODIC TABLE AND ELECTRON
CONFIGURATIONS
In this section, we view the periodic table (see inside front cover and Section 4-1) from a
modern, much more useful perspective—as a systematic representation of the electron
configurations of the elements. In the periodic table, elements are arranged in blocks based
on the kinds of atomic orbitals that are being filled (Figure 5-31). The periodic tables in
this text are divided into “A” and “B” groups. The A groups contain elements in which s
and p orbitals are being filled. Elements within any particular A group have similar electron
configurations and chemical properties, as we shall see in the next chapter. The B
groups are those in which there are one or two electrons in the s orbital of the outermost
occupied shell, and the d orbitals, one shell smaller, are being filled.
Lithium, sodium, and potassium, elements of the leftmost column of the periodic table
(Group IA), have a single electron in their outermost s orbital (ns1). Beryllium and magnesium,
of Group IIA, have two electrons in their outermost shell, ns2, while boron and
aluminum (Group IIIA) have three electrons in their outermost shell, ns2np1. Similar observations
can be made for other A group elements.
The electron configurations of the A group elements and the noble gases can be
predicted reliably from Figures 5-28 and 5-29. However, there are some more pronounced
irregularities in the B groups below the fourth period. In the heavier B group elements,
the higher energy subshells in different principal shells have energies that are very nearly
equal (Figure 5-29). It is easy for an electron to jump from one orbital to another of nearly
the same energy, even in a different set. This is because the orbital energies are perturbed
(change slightly) as the nuclear charge changes, and an extra electron is added in going
from one element to the next. This phenomenon gives rise to other irregularities that are
analogous to those of Cr and Cu, described earlier.
5-18
We can extend the information in Figure 5-31 to indicate the electron configurations
that are represented by each group (column) of the periodic table. Table 5-5 shows this
interpretation of the periodic table, along with the most important exceptions (A more
complete listing of electron configurations is given in Appendix B.) We can use this interpretation
of the periodic table to write, quickly and reliably, the electron configurations
for elements.
Figure 5-31 A periodic table colored to show the kinds of atomic orbitals (subshells) being
filled and the symbols of blocks of elements. The electronic structures of the A group
elements are quite regular and can be predicted from their positions in the periodic table,
but many exceptions occur in the d and f blocks. The colors in this figure are the same as
those in Figure 5-28.
Hydrogen and helium are shown here in their usual positions in the periodic table. These
may seem somewhat unusual based just on their electron configurations. We should
remember, however, that the first shell (n ϭ 1) can hold a maximum of only two electrons.
This shell is entirely filled in helium, so He behaves as a noble gas, and we put it in the
column with the other noble gases (Group VIIIA). Hydrogen has one electron that is easily
lost, like the metals in Group IA, so we put it in Group IA even though it is not a metal.
Furthermore, hydrogen is one electron short of a noble gas configuration (He), so we could
also place it with the other such elements in Group VIIA.
5-18 The Periodic Table and Electron Configurations 223
11
Na
12
Mg
90
Th
91
Pa
92
U
93
Np
94
Pu
95
Am
96
Cm
58
Ce
59
Pr
60
Nd
61
Pm
62
Sm
63
Eu
64
Gd
19
K
20
Ca
21
Sc
22
Ti
23
V
24
Cr
25
Mn
26
Fe
27
Co
28
Ni
29
Cu
30
Zn
37
Rb
38
Sr
39
Y
40
Zr
41
Nb
42
Mo
43
Tc
44
Ru
45
Rh
46
Pd
47
Ag
48
Cd
49
In
50
Sn
51
Sb
52
Te
53
I
54
Xe
55
Cs
56
Ba
57
La
72
Hf
73
Ta
74
W
75
Re
76
Os
77
Ir
78
Pt
79
Au
80
Hg
81
Tl
82
Pb
83
Bi
84
Po
85
At
86
Rn
87
Fr
88
Ra
89
Ac
104 105
Rf
106
Sg
107
Bh
108 109 110 111 112
1
H
2
He
5
B
6
C
7
N
8
O
9
F
10
Ne
13
Al
14
Si
15
P
16
S
17
Cl
18
Ar
1
H
3
Li
4
Be
66
Dy
67
Ho
68
Er
69
Tm
70
Yb
71
Lu
65
Tb
98
Cf
99
Es
100
Fm
101
Md
102
No
103
Lr
97
Bk
Period
1
2
3
4
5
6
7
6
7
(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15) (16) (17) (18)
IA IIA IIIB IVB VB VIB VIIB VIIIB IB IIB IIIA IVA VA VIA VIIA VIIIAGroup
31
Ga
32
Ge
33
As
34
Se
35
Br
36
Kr
1s
2s
3s
4s
6s
5s
7s
5f
4f
6d
5d
4d
3d
6p
5p
4p
3p
2p
1s
Db Hs Mt
EXAMPLE 5-10 Electron Configurations
Use Table 5-5 to determine the electron configurations of (a) magnesium, Mg; (b) germanium,
Ge; and (c) molybdenum, Mo.
Plan
We will use the electron configurations indicated in Table 5-5 for each group. Each period (row)
begins filling a new shell (new value of n). Elements to the right of the d orbital block have
the d orbitals in the (n Ϫ 1) shell already filled. We often find it convenient to collect all sets
of orbitals with the same value of n together, to emphasize the number of electrons in the outermost
shell, that is, the shell with the highest value of n.
Solution
(a) Magnesium, Mg, is in Group IIA, which has the general configuration s2; it is in Period
3 (third row). The last filled noble gas configuration is that of neon, or [Ne]. The electron
configuration of Mg is [Ne] 3s2.
(b) Germanium, Ge, is in Group IVA, for which Table 5-5 shows the general configuration
s2p2. It is in Period 4 (the fourth row), so we interpret this as 4s24p2. The last filled noble gas
configuration is that of argon, Ar, accounting for 18 electrons. In addition, Ge lies beyond the
224 CHAPTER 5: The Structure of Atoms
11
Na
12
Mg
90
Th
91
Pa
92
U
93
Np
94
Pu
95
Am
96
Cm
58
Ce
59
Pr
60
Nd
61
Pm
62
Sm
63
Eu
64
Gd
19
K
20
Ca
21
Sc
22
Ti
23
V
24
Cr
25
Mn
26
Fe
27
Co
28
Ni
29
Cu
30
Zn
37
Rb
38
Sr
39
Y
40
Zr
41
Nb
42
Mo
43
Tc
44
Ru
45
Rh
46
Pd
47
Ag
48
Cd
49
In
50
Sn
51
Sb
52
Te
53
I
54
Xe
55
Cs
56
Ba
57
La
72
Hf
73
Ta
74
W
75
Re
76
Os
77
Ir
78
Pt
79
Au
80
Hg
81
Tl
82
Pb
83
Bi
84
Po
85
At
86
Rn
87
Fr
88
Ra
89
Ac
104
Rf
105
Db
106
Sg
107
Bh Hs Mt
108 109 110 111 112
2
He
5
B
6
C
7
N
8
O
9
F
10
Ne
13
Al
14
Si
15
P
16
S
17
Cl
18
Ar
1
H
3
Li
4
Be
66
Dy
67
Ho
68
Er
69
Tm
70
Yb
71
Lu
65
Tb
98
Cf
99
Es
100
Fm
101
Md
102
No
103
Lr
97
Bk
LANTHANIDE
SERIES
(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15) (16) (17) (18)
IA IIA IIIB IVB VB VIB VIIB VIIIB IB IIB IIIA IVA VA VIA VIIA VIIIA
31
Ga
32
Ge
33
As
34
Se
35
Br
36
Kr
58
Ce
71
Lu
90
Th
103
Lr
111 112
f
orbital
block
d orbital block
d2s2 d3s2 d5s1 d5s2 d6s2 d7s2 d8s2 d10s1 d10s2
s2
p1
s2
p2
s2
p3
s2
p4
s2
p5
s2
p6
p orbital block
d4
s1
d4
s2
d7
s1
d8
s1
d9
s1
d10s0
ACTINIDE
SERIES
4f subshell
being filled
5f subshell
being filled
n = 7
n = 6
n = 7
n = 6
n = 5
n = 4
n = 3
n = 2
n = 1
d1s2
GROUPS
s orbital block
s1 s2
s2
TABLE 5-5 The s, p, d, and f Blocks of the Periodic Table*
*n is the principal quantum number. The d1s2, d2s2, . . . designations represent known configurations. They refer to (n Ϫ 1)d
and ns orbitals. Several exceptions to the configurations indicated above each group are shown in gray.
d orbital block, so we know that the 3d orbitals are completely filled. The electron configuration
of Ge is [Ar] 4s23d104p2 or [Ar] 3d104s24p2.
(c) Molybdenum, Mo, is in Group VIB, with the general configuration d5s1; it is in Period 5,
which begins with 5s and is beyond the noble gas krypton. The electron configuration of Mo is
[Kr] 5s14d5 or [Kr] 4d55s1. The electron configuration of molybdenum is analogous to that
of chromium, Cr, the element just above it. The configuration of Cr was discussed in Section
5-17 as one of the exceptions to the Aufbau order of filling.
You should now work Exercise 106.
EXAMPLE 5-11 Unpaired Electrons
Determine the number of unpaired electrons in an atom of tellurium, Te.
Plan
Te is in Group VIA in the periodic table, which tells us that its configuration is s2p4. All other
shells are completely filled, so they contain only paired electrons. We need only to find out
how many unpaired electrons are represented by s2p4.
Solution
The notation s2p4 is a short representation for s __hg p __hg __h __h . This shows that an atom of Te
contains two unpaired electrons.
You should now work Exercises 100 and 102.
The periodic table has been described as “the chemist’s best friend.” Chemical reactions
involve loss, gain, or sharing of electrons. In this chapter, we have seen that the
fundamental basis of the periodic table is that it reflects similarities and trends in electron
configurations. It is easy to use the periodic table to determine many important aspects
of electron configurations of atoms. Practice until you can use the periodic table with
confidence to answer many questions about electron configurations. As we continue our
study, we will learn many other useful ways to interpret the periodic table. We should
always keep in mind that the many trends in chemical and physical properties that we correlate
with the periodic table are ultimately based on the trends in electron configurations.
Key Terms 225
Key Terms
Absorption spectrum The spectrum associated with absorption
of electromagnetic radiation by atoms (or other species) resulting
from transitions from lower to higher energy states.
Alpha (␣) particle A helium ion with a 2ϩ charge; an assembly
of two protons and two neutrons.
amu See Atomic mass unit.
Angular momentum quantum number (ᐉ) The quantum
mechanical solution to a wave equation that designates the subshell,
or set of orbitals (s, p, d, f), within a given main shell in
which an electron resides.
Anode In a cathode-ray tube, the positive electrode.
Atomic mass unit An arbitrary mass unit defined to be exactly
one-twelfth the mass of the carbon-12 isotope.
Atomic number The integral number of protons in the nucleus;
defines the identity of an element.
Atomic orbital The region or volume in space in which the
probability of finding electrons is highest.
Aufbau (“building up”) Principle A guide for predicting the
order in which electrons fill subshells and shells in atoms.
Balmer-Rydberg equation An empirical equation that relates
wavelengths in the hydrogen emission spectrum to
integers.
Canal ray A stream of positively charged particles (cations) that
moves toward the negative electrode in a cathode-ray tube;
observed to pass through canals in the negative electrode.
Cathode In a cathode-ray tube, the negative electrode.
226 CHAPTER 5: The Structure of Atoms
Cathode ray The beam of electrons going from the negative
electrode toward the positive electrode in a cathode-ray tube.
Cathode-ray tube A closed glass tube containing a gas under
low pressure, with electrodes near the ends and a luminescent
screen at the end near the positive electrode; produces cathode
rays when high voltage is applied.
Continuous spectrum The spectrum that contains all wavelengths
in a specified region of the electromagnetic spectrum.
d orbitals Beginning in the third shell, a set of five degenerate
orbitals per shell, higher in energy than s and p orbitals in the
same shell.
Degenerate orbitals Two or more orbitals that have the same
energy.
Diamagnetism Weak repulsion by a magnetic field; associated
with all electrons being paired.
Electromagnetic radiation Energy that is propagated by means
of electric and magnetic fields that oscillate in directions perpendicular
to the direction of travel of the energy.
Electron A subatomic particle having a mass of 0.00054858 amu
and a charge of 1Ϫ.
Electron configuration The specific distribution of electrons in
the atomic orbitals of atoms and ions.
Electron transition The transfer of an electron from one energy
level to another.
Emission spectrum The spectrum associated with emission of
electromagnetic radiation by atoms (or other species) resulting
from electron transitions from higher to lower energy states.
Excited state Any energy state other than the ground state of
an atom, ion, or molecule.
f orbitals Beginning in the fourth shell, a set of seven degenerate
orbitals per shell, higher in energy than s, p, and d orbitals
in the same shell.
Ferromagnetism The property that allows a substance to
become permanently magnetized when placed in a magnetic
field; exhibited by iron, cobalt, and nickel.
Frequency (␯) The number of crests of a wave that pass a given
point per unit time.
Fundamental particles Subatomic particles of which all matter
is composed; protons, electrons, and neutrons are fundamental
particles.
Ground state The lowest energy state or most stable state of an
atom, molecule, or ion.
Group A vertical column in the periodic table; also called a fam-
ily.
Heisenberg Uncertainty Principle It is impossible to determine
accurately both the momentum and position of an electron
simultaneously.
Hund’s Rule Each orbital of a given subshell is occupied by a
single electron before pairing begins. See Aufbau Principle.
Isotopes Two or more forms of atoms of the same element with
different masses; that is, atoms containing the same number of
protons but different numbers of neutrons.
Line spectrum An atomic emission or absorption spectrum.
Magnetic quantum number (mᐉ) Quantum mechanical solution
to a wave equation that designates the particular orbital
within a given subshell (s, p, d, f ) in which an electron resides.
Mass number The integral sum of the numbers of protons and
neutrons in an atom.
Mass spectrometer An instrument that measures the charge-tomass
ratios of charged particles.
Natural radioactivity Spontaneous decomposition of an atom.
Neutron A subatomic nuclear particle having a mass of 1.0087
amu and no charge.
Nucleus The very small, very dense, positively charged center
of an atom containing protons and neutrons, except for 1
1H.
Nuclide symbol The symbol for an atom, A
ZE, in which E is the
symbol for an element, Z is its atomic number, and A is its mass
number.
p orbitals Beginning with the second shell, a set of three degenerate
mutually perpendicular, equal-arm, dumbbell-shaped
atomic orbitals per shell.
Pairing of electrons Interaction of two electrons with opposite
ms values in the same orbital (__hg ).
Paramagnetism Attraction toward a magnetic field, stronger
than diamagnetism, but still very weak compared with ferromagnetism;
due to presence of unpaired electrons.
Pauli Exclusion Principle No two electrons in the same atom
may have identical sets of four quantum numbers.
Period A horizontal row in the periodic table.
Photoelectric effect Emission of an electron from the surface
of a metal, caused by impinging electromagnetic radiation of
certain minimum energy; the resulting current increases with
increasing intensity of radiation.
Photon A “packet” of light or electromagnetic radiation; also
called a quantum of light.
Principal quantum number (n) The quantum mechanical solution
to a wave equation that designates the main shell, or energy
level, in which an electron resides.
Proton A subatomic particle having a mass of 1.0073 amu and a
charge of 1ϩ, found in the nuclei of atoms.
Quantum A “packet” of energy. See Photon.
Quantum mechanics A mathematical method of treating particles
on the basis of quantum theory, which assumes that energy
(of small particles) is not infinitely divisible.
Quantum numbers Numbers that describe the energies of electrons
in atoms; they are derived from quantum mechanical
treatment.
Radiant energy See Electromagnetic radiation.
s orbital A spherically symmetrical atomic orbital; one per
shell.
Spectral line Any of a number of lines corresponding to definite
wavelengths in an atomic emission or absorption spectrum;
these lines represent the energy difference between two energy
levels.
Spectrum Display of component wavelengths of electromagnetic
radiation.
Spin quantum number (ms) The quantum mechanical solution
to a wave equation that indicates the relative spins of
electrons.
Wavelength (␭) The distance between two identical points of a
wave.
Exercises 227
Exercises
Particles and the Nuclear Atom
*001. List the three fundamental particles of matter, and indicate
the mass and charge associated with each.
*002. (a) How do we know that canal rays have charges opposite
in sign to cathode rays? What are canal rays? (b) Why
are cathode rays from all samples of gases identical,
whereas canal rays are not?
00*3. The following data are measurements of the charges on
oil droplets using an apparatus similar to that used by Mil-
likan:
13.458 ϫ 10Ϫ19 C 17.308 ϫ 10Ϫ19 C
15.373 ϫ 10Ϫ19 C 28.844 ϫ 10Ϫ19 C
17.303 ϫ 10Ϫ19 C 11.545 ϫ 10Ϫ19 C
15.378 ϫ 10Ϫ19 C 19.214 ϫ 10Ϫ19 C
Each should be a whole-number ratio of some fundamental
charge. Using these data, determine the value of
the fundamental charge.
00*4. Suppose we discover a new positively charged particle,
which we call the “whizatron.” We want to determine its
charge.
(a) What modifications would we have to make to the
Millikan oil-drop apparatus to carry out the corresponding
experiment on whizatrons?
(b) In such an experiment, we observe the following
charges on five different droplets:
5.76 ϫ 10Ϫ19 C 7.20 ϫ 10Ϫ19 C
2.88 ϫ 10Ϫ19 C 10.08 ϫ 10Ϫ19 C
8.64 ϫ 10Ϫ19 C
What is the charge on the whizatron?
*005. Outline Rutherford’s contribution to understanding the
nature of atoms.
*006. Why was Rutherford so surprised that some of the
␣-particles were scattered backward in the gold foil
experiment?
*007. Summarize Moseley’s contribution to our knowledge of
the structure of atoms.
*008. The approximate radius of a hydrogen atom is 0.0529 nm,
and that of a proton is 1.5 ϫ 10Ϫ15 m. Assuming both the
hydrogen atom and the proton to be spherical, calculate
the fraction of the space in an atom of hydrogen that is
occupied by the nucleus. V ϭ (4/3)␲r3 for a sphere.
*009. The approximate radius of a neutron is 1.5 ϫ 10Ϫ15 m,
and the mass is 1.675 ϫ 10Ϫ27 kg. Calculate the density
of a neutron. V ϭ (4/3)␲r3 for a sphere.
Atom Composition, Isotopes, and Atomic Weights
*010. Arrange the following in order of increasing ratio of
charge to mass: 12Cϩ, 12C2ϩ, 14Nϩ, 14N2ϩ.
*011. Refer to Exercise 10. Suppose all of these high-energy
ions are present in a mass spectrometer. For which one
will its path be changed (a) the most and (b) the least by
increasing the external magnetic field?
*012. Estimate the percentage of the total mass of a 65
29Cu atom
that is due to (a) electrons, (b) protons, and (c) neutrons
by assuming that the mass of the atom is simply the sum
of the masses of the appropriate numbers of subatomic
particles.
*013. (a) How are isotopic abundances determined experimentally?
(b) How do the isotopes of a given element differ?
*014. Define and illustrate the following terms clearly and concisely:
(a) atomic number, (b) isotope, (c) mass number,
(d) nuclear charge.
*015. Write the composition of one atom of each of the three
isotopes of silicon: 28Si, 29Si, 30Si.
*016. Write the composition of one atom of each of the four
isotopes of sulfur: 32S, 33S, 34S, 36S.
*017. Complete Chart A for neutral atoms.
*018. Complete Chart B for neutral atoms.
Chart A
Kind of Atomic Mass Number of Number of Number of
Atom Number Number Isotope Protons Electrons Neutrons
________ ________ ________ 21
10Ne ___________ ___________ ___________
chlorine ________ 35 _______ ___________ ___________ ___________
________ 28 58 _______ ___________ ___________ ___________
________ ________ 40 _______ 18 ___________ ___________
Chart B
Kind of Atomic Mass Number of Number of Number of
Atom Number Number Isotope Protons Electrons Neutrons
selenium ________ ________ _______ ___________ ___________ 40
________ ________ ________ 11
05B ___________ ___________ ___________
________ ________ ________ _______ ___________ 35 46
________ ________ 104 _______ ___________ 45 ___________
228 CHAPTER 5: The Structure of Atoms
0*19. Prior to 1962, the atomic weight scale was based on the
assignment of an atomic weight of exactly 16 amu to the
naturally occurring mixture of oxygen. The atomic weight
of cobalt is 58.9332 amu on the carbon-12 scale. What
was it on the older scale?
*020. Determine the number of protons, neutrons, and electrons
in each of the following species: (a) 40
20Ca; (b) 45
21Sc;
(c) 91
40Zr; (d) 39
19Kϩ; (e) 65
30Zn2ϩ; (f ) 108
047Agϩ.
*021. Determine the number of protons, neutrons, and electrons
in each of the following species: (a) 34
16Sϩ; (b) 93
41Nb;
(c) 27
13Al; (d) 63
29Cuϩ; (e) 56
26Fe2ϩ; (f ) 55
26Fe3ϩ.
*022. What is the symbol of the species composed of each of
the following sets of subatomic particles? (a) 24p, 27n,
24e; (b) 20p, 20n, 18e; (c) 34p, 44n, 34e; (d) 53p, 74n, 54e.
*023. What is the symbol of the species composed of each of
the following sets of subatomic particles? (a) 94p, 150n,
94e; (b) 79p, 118n, 76e; (c) 34p, 45n, 36e; (d) 52p, 76n, 54e.
*024. The atomic weight of lithium is 6.941 amu. The two naturally
occurring isotopes of lithium have the following
masses: 6Li, 6.01512 amu; 7Li, 7.01600 amu. Calculate
the percent of 6Li in naturally occurring lithium.
*025. The atomic weight of rubidium is 85.4678 amu. The two
naturally occurring isotopes of rubidium have the following
masses: 85Rb, 84.9118 amu; 87Rb, 86.9092 amu.
Calculate the percent of 85Rb in naturally occurring
rubidium.
*026. Bromine is composed of 79
35Br, 78.9183 amu, and 81
35Br,
80.9163 amu. The percent composition of a sample is
50.69% Br-79 and 49.31% Br-81. Based on this sample,
calculate the atomic weight of bromine.
*027. What is the atomic weight of a hypothetical element that
consists of the following isotopes in the indicated relative
abundances?
Isotopic % Natural
Isotope Mass (amu) Abundance
1 94.9 12.4
2 95.9 73.6
3 97.9 14.0
*028. Naturally occurring iron consists of four isotopes with the
abundances indicated here. From the masses and relative
abundances of these isotopes, calculate the atomic weight
of naturally occurring iron.
Isotopic % Natural
Isotope Mass (amu) Abundance
54Fe 53.9396 5.82
56Fe 55.9349 91.66
57Fe 56.9354 2.19
58Fe 57.9333 0.33
*029. Calculate the atomic weight of nickel from the following
information.
Isotopic % Natural
Isotope Mass (amu) Abundance
58Ni 57.9353 67.88
60Ni 59.9332 26.23
61Ni 60.9310 1.19
62Ni 61.9283 3.66
64Ni 63.9280 1.08
*030. The atomic weight of copper is 63.546 amu. The two naturally
occurring isotopes of copper have the following
masses: 63Cu, 62.9298 amu; 65Cu, 64.9278 amu. Calculate
the percent of 63Cu in naturally occurring copper.
*031. Silver consists of two naturally occurring isotopes: 107Ag,
which has a mass of 106.90509 amu, and 109Ag, which has
a mass of 108.9047 amu. The atomic weight of silver is
107.8682 amu. Determine the percent abundance of each
isotope in naturally occurring silver.
*032. Refer to Table 5-3 only and calculate the atomic weights
of oxygen and chlorine. Do your answers agree with the
atomic weights given in that table?
*033. The following is a mass spectrum of the 1ϩ charged ions
of an element. Calculate the atomic weight of the element.
What is the element?
*034. Suppose you measure the mass spectrum of the 1ϩ
charged ions of germanium, atomic weight 72.61 amu.
Unfortunately, the recorder on the mass spectrometer
jams at the beginning and again at the end of your experiment.
You obtain only the following partial spectrum,
which may or may not be complete. From the information
given here, can you tell whether one of the germanium
isotopes is missing? If one is missing, at which end of the
plot should it appear?
0.7215
0.2785
84.9117 86.9085
Mass (amu)
Relativeabundance
Exercises 229
*035. Calculate the atomic weight of silicon using the following
data for the percent natural abundance and mass of
each isotope: (a) 92.23% 28Si (27.9769 amu); (b) 4.67%
29Si (28.9765 amu); (c) 3.10% 30Si (29.9738 amu).
*036. Calculate the atomic weight of chromium using the following
data for the percent natural abundance and mass
of each isotope: (a) 4.35% 50Cr (49.9461 amu); (b) 83.79%
52Cr (51.9405 amu); (c) 9.50% 53Cr (52.9406 amu);
(d) 2.36% 54Cr (53.9389 amu).
Electromagnetic Radiation
*037. Calculate the wavelengths, in meters, of radiation of the
following frequencies: (a) 5.60 ϫ 1015 sϪ1; (b) 2.11 ϫ
1014 sϪ1; (c) 3.89 ϫ 1012 sϪ1.
*038. Calculate the frequency of radiation of each of the following
wavelengths: (a) 8973 Å; (b) 492 nm; (c) 4.92 cm;
(d) 4.55 ϫ 10Ϫ9 cm.
*039. What is the energy of a photon of each of the radiations
in Exercise 37? Express your answer in joules per photon.
In which regions of the electromagnetic spectrum do
these radiations fall?
*040. Excited lithium ions emit radiation at a wavelength of
670.8 nm in the visible range of the spectrum. (This characteristic
color is often used as a qualitative analysis test
for the presence of Liϩ.) Calculate (a) the frequency and
(b) the energy of a photon of this radiation. (c) What color
is this light?
*041. Calculate the energy, in joules per photon, of X-rays having
a wavelength of 3.0 Å. How does the energy of these
X-rays compare with the energy calculated for yellow
light in Example 5-5?
*042. Ozone in the upper atmosphere absorbs ultraviolet radiation,
which induces the following chemical reaction:
O3(g) 88n O2(g) ϩ O(g)
What is the energy of a 3400-Å photon that is absorbed?
What is the energy of a mole of these photons?
0*43. During photosynthesis, chlorophyll-␣ absorbs light of
wavelength 440 nm and emits light of wavelength
670 nm. What is the energy available for photosynthesis
from the absorption–emission of a mole of photons?
5.49
1.55
7.31
1.55
71.9217 72.9234 73.9219 75.9219
Mass (amu)
Relativeabundance
*044. Classical music radio station KMFA in Austin broadcasts
at a frequency of 89.5 MHz. What is the wavelength of
its signal in meters?
*045. Describe the following terms clearly and concisely in
relation to electromagnetic radiation: (a) wavelength,
(b) frequency, (c) amplitude, (d) color.
0*46. Assume that 10Ϫ17 J of light energy is needed by the interior
of the human eye to “see” an object. How many
photons of green light (wavelength ϭ 495 nm) are needed
to generate this minimum energy?
0*47. The human eye receives a 2.500 ϫ 10Ϫ14 J signal consisting
of photons of orange light, ␭ ϭ 6150 Å. How many
photons reach the eye?
*048. Alpha Centauri is the star closest to our solar system. It
is 4.3 light-years away. How many miles in this? A lightyear
is the distance that light travels (in a vacuum) in one
year. Assume that space is essentially a vacuum.
The Photoelectric Effect
*049. What evidence supports the idea that electromagnetic
radiation is (a) wave-like; (b) particle-like?
*050. Describe the influence of frequency and intensity of electromagnetic
radiation on the current in the photoelectric
effect.
0*51. Cesium is often used in “electric eyes” for self-opening
doors in an application of the photoelectric effect. The
amount of energy required to ionize (remove an electron
from) a cesium atom is 3.89 electron volts (1 eV ϭ 1.60 ϫ
10Ϫ19 J). Show by calculation whether a beam of yellow
light with wavelength 5830 Å would ionize a cesium atom.
0*52. Refer to Exercise 51. What would be the wavelength, in
nanometers, of light with just sufficient energy to ionize
a cesium atom? What color would this light be?
Photosynthesis.
230 CHAPTER 5: The Structure of Atoms
Atomic Spectra and the Bohr Theory
*053. (a) Distinguish between an atomic emission spectrum and
an atomic absorption spectrum. (b) Distinguish between
a continuous spectrum and a line spectrum.
*054. Prepare a sketch similar to Figure 5-16b that shows a
ground energy state and three excited energy states. Using
vertical arrows, indicate the transitions that would correspond
to the absorption spectrum for this system.
*055. Why is the Bohr model of the hydrogen atom referred
to as the solar system model?
0*56. If each atom in one mole of atoms emits a photon of wavelength
5.50 ϫ 103 Å, how much energy is lost? Express
the answer in kJ/mol. As a reference point, burning one
mole (16 g) of CH4 produces 819 kJ of heat.
*057. What is the Balmer-Rydberg equation? Why is it called
an empirical equation?
*058. Hydrogen atoms absorb energy so that the electrons are
excited to the energy level n ϭ 7. Electrons then undergo
these transitions: (1) n ϭ 7 n n ϭ 1; (2) n ϭ 7 n n ϭ 2;
(3) n ϭ 2 n n ϭ 1. Which of these transitions will produce
the photon with (a) the smallest energy; (b) the
highest frequency; (c) the shortest wavelength? (d) What
is the frequency of a photon resulting from the transition
n ϭ 6 n n ϭ 1?
0*59. Five energy levels of the He atom are given in joules per
atom above an arbitrary reference energy: (1) 6.000 ϫ
10Ϫ19; (2) 8.812 ϫ 10Ϫ19; (3) 9.381 ϫ 10Ϫ19; (4) 10.443 ϫ
10Ϫ19; (5) 10.934 ϫ 10Ϫ19. Construct an energy level
diagram for He and find the energy of the photon
(a) absorbed for the electron transition from level 1 to
level 5 and (b) emitted for the electron transition from
level 4 to level 1.
*060. The following are prominent lines in the visible region
of the emission spectra of the elements listed. The lines
can be used to identify the elements. What color is
the light responsible for each line? (a) lithium, 4603 Å;
(b) neon, 540.0 nm; (c) calcium, 6573 Å; (d) potassium,
␯ ϭ 3.90 ϫ 1014 Hz.
*061. Hydrogen atoms have an absorption line at 1026 Å. What
is the frequency of the photons absorbed, and what is the
energy difference, in joules, between the ground state and
this excited state of the atom?
0*62. An argon laser emits blue light with a wavelength of 488.0
nm. How many photons are emitted by this laser in 2.00
seconds, operating at a power of 515 milliwatts? One watt
(a unit of power) is equal to 1 joule/second.
The Wave–Particle View of Matter
*063. (a) What evidence supports the idea that electrons are
particle-like? (b) What evidence supports the idea that
electrons are wave-like?
*064. (a) What is the de Broglie wavelength of a proton moving
at a speed of 2.50 ϫ 107 m/s? The proton mass is
1.67 ϫ 10Ϫ24 g. (b) What is the de Broglie wavelength of
a stone with a mass of 30.0 g moving at 2.00 ϫ 103 m/h
(Ϸ100 mph)? (c) How do the wavelengths in parts (a) and
(b) compare with the typical radii of atoms? (See the
atomic radii in Figure 6-1).
*065. What is the wavelength corresponding to a neutron of
mass 1.67 ϫ 10Ϫ27 kg moving at 2360 m/s?
*066. What is the velocity of an ␣-particle (a helium nucleus)
that has a de Broglie wavelength of 0.529 Å?
Quantum Numbers and Atomic Orbitals
*067. (a) What is a quantum number? What is an atomic
orbital? (b) How many quantum numbers are required to
specify a single atomic orbital? What are they?
*068. How are the possible values for the angular momentum
quantum number for a given electron restricted by the
value of n?
*069. Without giving the ranges of possible values of the four
quantum numbers, n, ᐉ, mᐉ, and ms, describe briefly what
information each one gives.
Lasers.
Exercises 231
*070. An electron is in one of the 3p orbitals. What are the possible
values of the quantum numbers n, ᐉ, mᐉ, and ms for
the electron?
*071. What is the maximum number of electrons in an atom
that can have the following quantum numbers? (a) n ϭ 3;
(b) n ϭ 3 and ᐉ ϭ 1; (c) n ϭ 3, ᐉ ϭ 1, and mᐉ ϭ Ϫ1;
(d) n ϭ 3, ᐉ ϭ 1, mᐉ ϭ Ϫ1, and ms ϭ Ϫᎏ1
2
ᎏ.
*072. What is the maximum number of electrons in an atom
that can have the following quantum numbers? (a) n ϭ 3
and ᐉ ϭ 1; (b) n ϭ 3 and ᐉ ϭ 2; (c) n ϭ 3, ᐉ ϭ 2 and mᐉ ϭ
Ϫ1; (d) n ϭ 3, ᐉ ϭ 1, and mᐉ ϭ Ϫ1; (e) n ϭ 3, ᐉ ϭ 2, mᐉ ϭ
0, and ms ϭ Ϫᎏ1
2
ᎏ.
*073. What are the values of n and ᐉ for the following subshells?
(a) 1s; (b) 4s; (c) 3p; (d) 3d; (e) 4f.
*074. How many individual orbitals are there in the third shell?
Write out n, ᐉ, and mᐉ quantum numbers for each one,
and label each set by the s, p, d, f designations.
*075. (a) Write the possible values of ᐉ when n ϭ 5. (b) Write
the allowed number of orbitals (1) with the quantum numbers
n ϭ 4, ᐉ ϭ 3; (2) with the quantum number n ϭ 4;
(3) with the quantum numbers n ϭ 4, ᐉ ϭ 2, mᐉ ϭ 0;
(4) with the quantum numbers n ϭ 6, ᐉ ϭ 5.
*076. Write the subshell notations that correspond to (a) n ϭ
3, ᐉ ϭ 0; (b) n ϭ 3, ᐉ ϭ 1; (c) n ϭ 7, ᐉ ϭ 0; (d) n ϭ 3,
ᐉ ϭ 2.
*077. What values can mᐉ take for (a) a 3d orbital, (b) a 1s orbital,
and (c) a 3p orbital?
*078. How many orbitals in any atom can have the given
quantum number or designation? (a) 4p; (b) 3p; (c) 3px;
(d) n ϭ 5; (e) 6d; (f ) 5d; (g) 5f; (h) 7s.
*079. The following incorrect sets of quantum numbers in the
order n, ᐉ, mᐉ, ms are written for paired electrons or for
one electron in an orbital. Correct them, assuming n values
are correct. (a) 1, 0, 0, ϩᎏ1
2
ᎏ, ϩᎏ1
2
ᎏ; (b) 2, 2, 1, Ϯ ᎏ1
2
ᎏ; (c) 3,
2, 3, Ϯ ᎏ1
2
ᎏ; (d) 3, 1, 2, ϩᎏ1
2
ᎏ; (e) 2, 1, Ϫ1, 0; (f) 3, 0, Ϫ1, Ϫᎏ1
2
ᎏ.
*080. (a) How are a 1s orbital and a 2s orbital in an atom similar?
How do they differ? (b) How are a 2px orbital and
a 2py orbital in an atom similar? How do they differ?
Electron Configurations and the Periodic Table
You should be able to use the positions of elements in the periodic
table to answer the exercises in this section.
*081. Draw representations of ground state electron configurations
using the orbital notation (__hg) for the following
elements. (a) N; (b) Fe; (c) Cl; (d) Rh.
*082. Draw representations of ground state electron configurations
using the orbital notation (__hg) for the following
elements. (a) P; (b) Ni; (c) Ga; (d) Zr.
*083. Determine the number of electrons in the outer occupied
shell of each of the following elements, and indicate the
principal quantum number of that shell. (a) Na; (b) Al;
(c) Ca; (d) Sr; (e) Ba; (f ) Br.
*084. With the help of Appendix B, list the symbols for the first
eight elements, by atomic number, that have an unpaired
electron in an s orbital. Identify the group in which most
of these are found in the periodic table.
*085. List the elements having an atomic number of 20 or less
that have one or more unpaired p orbital electrons. Indicate
the group to which each of these elements belongs
in the periodic table.
*086. Identify the element, or elements possible, given only the
number of electrons in the outermost shell and the principal
quantum number of that shell. (a) 1 electron, first
shell; (b) 2 electrons, second shell; (c) 3 electrons, third
shell; (d) 2 electrons, seventh shell; (e) 4 electrons, fourth
shell; (f ) 8 electrons, sixth shell.
*087. Give the ground state electron configurations for the elements
of Exercise 81 using shorthand notation—that is,
1s22s22p6, and so on.
*088. Give the ground state electron configurations for the elements
of Exercise 82 using shorthand notation—that is,
1s22s22p6, and so on.
*089. State the Pauli Exclusion Principle. Would any of the following
electron configurations violate this rule: (a) 1s2;
(b) 1s22p7; (c) 1s3? Explain.
*090. State Hund’s Rule. Would any of the following electron
configurations violate this rule: (a) 1s2; (b) 1s22s22px
2;
(c) 1s22s22px
12py
1; (d) 1s22s22px
12pz
1; (e) 1s22px
22py
12pz
1?
Explain.
0*91. Classify each of the following atomic electron configurations
as (i) a ground state, (ii) an excited state, or (iii)
a forbidden state: (a) 1s22s22p53s1; (b) [Kr] 4d105s3;
(c) 1s22s22p63s23p63d84s2; (d) 1s22s22p63s23p63d1;
(e) 1s22s22p103s23p5.
*092. Which of the elements with atomic numbers of 10 or less
are paramagnetic when in the atomic state?
*093. Semiconductor industries depend on such elements as Si,
Ga, As, Ge, Al, Cd, and Se. Write the predicted electron
configuration of each element.
*094. The manufacture of high-temperature ceramic superconductors
depends on such elements as Cu, O, La, Y,
Ba, Tl, and Bi. Write the predicted electron configuration
of each element. (Consult Appendix B if necessary.)
*095. In nature, potassium and sodium are often found together.
(a) Write the electron configurations for potassium and
for sodium. (b) How are they similar? (c) How do they
differ?
*096. Which elements are represented by the following electron
configurations?
(a) 1s22s22p63s23p63d104s24p5
(b) [Kr] 4d104f145s25p65d105f146s26p66d17s2
(c) [Kr] 4d104f145s25p65d106s2 6p6
(d) [Kr] 4d55s2
(e) 1s22s22p63s23p63d34s2
*097. Repeat Exercise 96 for
(a) 1s22s22p63s23p63d24s2
(b) [Kr] 4d104f145s25p65d106s26p3
(c) 1s22s22p63s23p2
(d) [Kr] 4d104f145s25p65d106s26p67s2
232 CHAPTER 5: The Structure of Atoms
*098. Find the total number of s, p, and d electrons in each of
the following: (a) Si; (b) Ar; (c) Ni; (d) Zn; (e) V.
*099. (a) Distinguish between the terms “diamagnetic” and
“paramagnetic,” and provide an example that illustrates
the meaning of each. (b) How is paramagnetism measured
experimentally?
*100. How many unpaired electrons are in atoms of Na, Ne,
B, Be, As, and Ti?
*101. Which of the following ions or atoms possess paramagnetic
properties? (a) ClϪ; (b) Naϩ; (c) Co; (d) ArϪ; (e) Si.
*102. Which of the following ions or atoms possess paramagnetic
properties? (a) F; (b) Ne; (c) Neϩ; (d) Fe; (e) ClϪ.
*103. Write the electron configurations of the Group IIA elements
Be, Mg, and Ca (see inside front cover). What
similarity do you observe?
*104. Construct a table in which you list a possible set of values
for the four quantum numbers for each electron in
the following atoms in their ground states. (a) N; (b) S;
(c) Ca.
*105. Construct a table in which you list a possible set of values
for the four quantum numbers for each electron in
the following atoms in their ground states. (a) Mg; (b) Cl;
(c) Cu.
*106. Draw general electron structures for the A group elements
using the __hg notation, where n is the principal
quantum number for the highest occupied energy level.
ns np
IA — — — —
IIA — — — —
and so on
*107. Repeat Exercise 106 using nsxnpy notation.
*108. List n, ᐉ, and mᐉ quantum numbers for the highest energy
electron (or one of the highest energy electrons if there
are more than one) in the following atoms in their ground
states. (a) P; (b) Zn; (c) Cl; (d) Pr.
*109. List n, ᐉ, and mᐉ quantum numbers for the highest energy
electron (or one of the highest energy electrons if there
are more than one) in the following atoms in their ground
states. (a) Se; (b) Zn; (c) Mg; (d) Pu.
*110. Write the ground state electron configurations for elements
A–E.
*111. Repeat Exercise 110 for elements F–J.
CONCEPTUAL EXERCISES
*112. Draw a three-dimensional representation of each of the
following orbitals. (a) 3pz; (b) 2s; (c) 3dxy; (d) 3dz2.
*113. We often show the shapes of orbitals as drawings. What
are some of the limitations of these drawings?
*114. For a lithium atom, give: (a) its ground state electron configuration;
(b) the electron configuration for one of its
lowest energy excited states; and (c) an electron configuration
for a forbidden or impossible state.
*115. In the first 100 elements, do more elements (as isolated
atoms) have ground state configurations that are paramagnetic
or diamagnetic? Justify your answer.
*116. Element 123 has not been discovered. Will its atoms be
paramagnetic in the ground state?
*117. Suppose we could excite all of the electrons in a sample
of hydrogen atoms to the n ϭ 6 level. They would then
emit light as they relaxed to lower energy states. Some
atoms might undergo the transition n ϭ 6 to n ϭ 1, and
others might go from n ϭ 6 to n ϭ 5, then from n ϭ 5 to
n ϭ 4, and so on. How many lines would we expect to
observe in the resulting emission spectrum?
BUILDING YOUR KNOWLEDGE
*118. Two isotopes of hydrogen occur naturally (1H, Ͼ99%,
and 2H, Ͻ1%) and two of chlorine occur naturally (35Cl,
76%, and 37Cl, 24%). (a) How many different masses of
HCl molecules can be formed from these isotopes? (b)
What is the approximate mass of each of the molecules,
expressed in atomic mass units? (Use atomic weights
rounded to the nearest whole number.) (c) List these HCl
molecules in order of decreasing relative abundance.
*119. CH4 is methane. If 1H, 2H, 12C, and 13C were the only
isotopes present in a given sample of methane, show the
different formulas and formula weights that might exist
in that sample. (Use atomic weights rounded to the nearest
whole number.)
*120. The density of platinum is 21.09 g/cm3. Recall that naturally
occurring platinum consists of a single isotope.
Suppose we have a teaspoon-sized sample, 5.00 cm3, of
platinum. (a) What is the mass of the sample? (b) What
is the mass of the electrons in this sample?
*121. A helium atom (4He) contains 2 protons, 2 neutrons, and
2 electrons. Using the masses listed in Table 5-1, calculate
the mass of a mole of helium atoms. Compare the
calculated value to the listed atomic weight of helium.
From your calculated value, which isotopes of helium
might you find in a natural sample of helium? (This question
ignores binding energy, a topic discussed in Chapter
26, Nuclear Chemistry.)
E
J
B
G A
I
H
F
D
C
Exercises 233
*122. When compounds of barium are heated in a flame, green
light of wavelength 554 nm is emitted. How much energy
is lost when one mole of barium atoms each emit one
photon of this wavelength?
*123. A 60-watt light bulb consumes energy at the rate of 60
JиsϪ1. Much of the light is emitted in the infrared region,
and less than 5% of the energy appears as visible light.
Calculate the number of visible photons emitted per second.
Make the simplifying assumptions that 5.0% of the
light is visible and that all visible light has a wavelength
of 550 nm (yellow-green).
6 Chemical
Periodicity
OUTLINE
6-1 More About the Periodic Table
Periodic Properties of the Elements
6-2 Atomic Radii
6-3 Ionization Energy
6-4 Electron Affinity
6-5 Ionic Radii
6-6 Electronegativity
Chemical Reactions and Periodicity
6-7 Hydrogen and the Hydrides
Hydrogen
Reactions of Hydrogen and the
Hydrides
6-8 Oxygen and the Oxides
Oxygen and Ozone
Reactions of Oxygen and the
Oxides
Combustion Reactions
Combustion of Fossil Fuels and
Air Pollution
OBJECTIVES
After you have studied this chapter, you should be able to
• More effectively use the periodic table
• Discuss chemical periodicity of the following physical properties:
Atomic radii
Ionization energy
Electron affinity
Ionic radii
Electronegativity
• Describe chemical periodicity in the reactions of
Hydrogen
Oxygen
• Describe chemical periodicity in the compounds of
Hydrogen
Oxygen
MORE ABOUT THE PERIODIC TABLE
In Chapter 4 we described the development of the periodic table, some terminology for
it, and its guiding principle, the periodic law.
The properties of the elements are periodic functions of their atomic numbers.
In Chapter 5 we described electron configurations of the elements. In the long form
of the periodic table, elements are arranged in blocks based on the kinds of atomic orbitals
being filled. (Please review Table 5-5 and Figure 5-31 carefully.) We saw that electron
6-1
The bright colors in fireworks
displays are produced when various
metals react with oxygen.
configurations of elements in the A groups are entirely predictable from their positions
in the periodic table. We also noted, however, that some irregularities occur within the B
groups.
Now we classify the elements according to their electron configurations, which is a
very useful system.
Noble Gases. For many years the Group VIIIA elements—the noble gases—were called
inert gases because no chemical reactions were known for them. We now know that the
heavier members do form compounds, mostly with fluorine and oxygen. Except for helium,
each of these elements has eight electrons in its outermost occupied shell. Their outer
shell may be represented as having the electron configuration . . . ns2np6.
Representative Elements. The A group elements in the periodic table are called representative
elements. Their “last” electron is assigned to an outer shell s or p orbital. These
elements show distinct and fairly regular variations in their properties with changes in
atomic number.
d-Transition Elements. Elements in the B groups in the periodic table are known as the
d-transition elements or, more simply, as transition elements or transition metals. The
elements of four transition series are all metals and are characterized by electrons being
assigned to d orbitals. Stated differently, the d-transition elements contain electrons in
both the ns and (n Ϫ 1)d orbitals, but not in the np orbitals. The first transition series, Sc
through Zn, has electrons in the 4s and 3d orbitals, but not in the 4p orbitals. They are
referred to as
First transition series: 21Sc through 30Zn
Second transition series: 39Y through 48Cd
Third transition series: 57La and 72Hf through 80Hg
Fourth transition series: 89Ac and 104Rf through element 112
The properties of elements are
correlated with their positions in
the periodic table. Chemists use the
periodic table as an invaluable guide
in their search for new, useful
materials. A barium sodium niobate
crystal can convert infrared laser
light into visible green light. This
harmonic generation or “frequency
doubling” is very important in
chemical research using lasers and in
the telecommunications industry.
236 CHAPTER 6: Chemical Periodicity
[He] ϭ 1s2
Some transition metals (left to right): Ti, V, Cr, Mn, Fe, Co, Ni, Cu.
6-1 More About the Periodic Table 237
The Development of ScienceCCHEMISTRY IN USE
The Periodic Table
The periodic table is one of the first things a student
of chemistry encounters. It appears invariably
in textbooks, in lecture halls, and in laboratories.
Scientists consider it an indispensable reference.
And yet, less than 150 years ago, the idea of arranging
the elements by atomic weight or number was
considered absurd. At an 1866 meeting of the
Chemical Society at Burlington House, England,
J. A. R. Newlands (1837–1898) presented a theory
he called the law of octaves. It stated that when the
known elements were listed by increasing atomic
weights, those that were eight places apart would
be similar, much like notes on a piano keyboard.
His colleagues’ reactions are probably summed up
best by the remark of a Professor Foster: “Have
you thought of arranging the elements according
to their initial letters? Maybe some better connections
would come to light that way.”
It is not surprising that poor Newlands was not taken seriously.
In the 1860s, little information was available to
illustrate relationships among the elements. Only 62 of them
had been distinguished from more complex substances when
Mendeleev first announced his discovery of the periodic law
in 1869. As advances in atomic theory were made, however,
and as new experiments contributed to the understanding of
chemical behavior, some scientists had begun to see similarities
and patterns among the elements. In 1869, Lothar Meyer
and Dmitri Mendeleev independently published similar versions
of the now-famous periodic table.
Mendeleev’s discovery was the result of many years of hard
work. He gathered information on the elements from all corners
of the earth—by corresponding with colleagues,
studying books and papers, and redoing experiments to confirm
data. He put the statistics of each element on a small
card and pinned the cards to his laboratory wall, where he
arranged and rearranged them many times until he was sure
that they were in the right order. One especially farsighted
feature of Mendeleev’s accomplishment was his realization
that some elements were missing from the table. He predicted
the properties of these substances (gallium, scandium,
and germanium). (It is important to remember that
Mendeleev’s periodic table organization was devised more
than 50 years before the discovery and characterization of
subatomic particles.)
Since its birth in 1869, the periodic table has been discussed
and revised many times. Spectroscopic and other
discoveries have filled in the blanks left by Mendeleev and
added a new column consisting of the noble gases. As scientists
learned more about atomic structure, the basis for
ordering was changed from atomic weight to atomic number.
The perplexing rare earths were sorted out and given a
special place, along with many of the elements created by
atomic bombardment. Even the form of the table has been
experimented with, resulting in everything from spiral and
circular tables to exotic shapes such as the one suggested by
Charles Janet. A three-dimensional periodic table that takes
into account valence-shell energies has been proposed by
Professor Leland C. Allen of Princeton University.
During the past century, chemistry has become a fastmoving
science in which methods and instruments are often
outdated within a few years. But it is doubtful that our old
friend, the periodic table, will ever become obsolete. It may
be modified, but it will always stand as a statement of basic
relationships in chemistry and as a monument to the wisdom
and insight of its creator, Dmitri Mendeleev.
Lisa Saunders Boffa
Senior Chemist
Exxon Corporation
Vla
ll6
Po
Bi
Li
He
HFCl B Al
Sc
Ga
In
Tl
Y
La Ce
Pr
Eu Gd
Cm
Bk
Cf
Tb
Dy
Ho
Er
Tm
Yb
Lu
Ti
Zr
V
Nb
Cr
Mo
Mn
Tc
Re
FeCo
Rh
Ir
Ni
Pd
Pt
Cu
Ag
Au
Zn
Cd
Hg RuOs
W
Ta
Mv
Fm
E
Hf
Am
Pu
Np
U
Pa
Th
Sm
Pm
Nd
Ac
Be
Mg
Ca
Sr
Bg
Ra
BrIAt
Ne
A
Kr
Xe
Rn
Na
K
Rb
Cs
Fr
Sb
As
GeP
N
C
Si
Sn
Pb
Te
Se
S
O
ll7Vlla
ll8
O
ll9
la
Va
120
lla
lllc
lllb
llla
lVc
Vc
Vlc
Vllc
Vb
Vlb
Vllb
lVa
lllb
lVb
llc lc
ll5
ll3
llla
llb
lb
-
-
-
-
-
-
-
-
-
ll2
lll
ll0
l09
l08
l07
l06
l05
l04l03
l02
ll4
An alternative representation of the periodic table, as proposed by Charles Janet,
1928.
The Development of ScienceCCHEMISTRY IN USE
Glenn Seaborg: A Human Side to the
Modern Periodic Table
If it were possible to associate a human face with the modern
periodic table, that face would most likely belong to
Glenn Seaborg (1912–1999), the codiscoverer of ten
transuranium elements and the name behind Element 106.
Seaborg’s contributions to heavy-element chemistry began in
1940, when he and coworkers at the University of California
at Berkeley produced the first sample of plutonium by bombarding
uranium with deuterons (2
1H nuclei) in a particle
accelerator. They found that the isotope plutonium-239
undergoes nuclear fission (see Chapter 26), making it a potential
energy source for nuclear power or nuclear weapons.
As American involvement in World War II grew, President
Franklin Roosevelt called Seaborg and other eminent
scientists to the Wartime Metallurgical Laboratory at the
University of Chicago, where they figured out how to prepare
and purify plutonium-239 in useful quantities for the
Manhatten Project, the making of the atom bomb. In 1945,
Seaborg was one of the signers of the Franck Report, a
document recommending that a safe demonstration test of
the atomic bomb might persuade Japan to surrender without
the bomb actually being used. Professor Seaborg served
as a scientific advisor for nine other presidents following
Roosevelt and was chairman of the U.S. Atomic Energy
Commission under
Kennedy, Johnson,
and Nixon.
Seaborg’s contributions
illustrate
how certain areas
of science can be
highly influenced
by a particular institution
or even
a national tradition
over time. His
discovery of plutonium
at the University
of California
at Berkeley
followed the 1940
synthesis of neptunium
at the
same site by Edwin
238 CHAPTER 6: Chemical Periodicity
McMillan, who shared the 1951 Nobel Prize in chemistry
with Seaborg for these accomplishments. Since that time,
Seaborg and other teams involving Berkeley researchers at
the University’s Lawrence Berkeley Laboratory have prepared
nine more heavy elements. He and coworkers hold the
world’s only patents on chemical elements, for americium and
curium. The original location of the first transuranium laboratory
on the Berkeley campus (a few yards from the later
site of Professor Seaborg’s reserved “Nobel Laureate” parking
space) is now a national historic landmark.
Laboratories in the United States, Russia, and Germany
have been the most active in the synthesis of new elements.
In 1994, nationalistic feelings invaded what should have been
impartial decisions by the International Union of Pure and
Applied Chemistry (IUPAC) regarding the official names for
elements 101 and 109. In some cases, researchers from different
countries had proposed different names for these
elements based on where credit for their discovery was felt
to be deserved. For example, the name “hahnium” was proposed
for element 105 by American researchers, while the
Russians preferred the more snappy “nielsbohrium.” The
American Chemical Society proposed to name element 106
seaborgium (Sg), but the IUPAC’s nomenclature committee
rejected the choice, objecting to the fact that Seaborg was
still alive (“and they can prove it,” he quipped). Outrage at
this and some of the other naming decisions prompted many
scientists to ignore the IUPAC’s recommended name for 106,
rutherfordium, and to continue to use seaborgium. In 1997,
the IUPAC reversed its decision and endorsed Sg, saving the
chemical literature from future confusion caused by different
naming practices in the scientific journals and conferences of
different countries.
If Professor Seaborg had been nominated for a different
honor—appearance on a U.S. postage stamp—the story
would have had an unhappier ending. Although surely not as
rare a commodity as the names of new chemical elements,
United States stamps are not permitted to honor living individuals.
Seaborg would have been the only person in the
world who could have received mail addressed entirely in elements:
Seaborgium, Lawrencium (for the Lawrence Berkeley
Laboratory), Berkelium, Californium, Americium—and
don’t forget the ZIP code, 94720.
Additional information about Seaborg and his research is
available through the Web site for this textbook.
Lisa Saunders Boffa
Senior Chemist
Exxon Corporation
f-Transition Elements. Sometimes known as inner transition elements, these are elements
in which electrons are being added to f orbitals. In these elements, the second from the
outermost occupied shell is building from 18 to 32 electrons. All are metals. The ftransition
elements are located between Groups IIIB and IVB in the periodic table.
They are
First f-transition series (lanthanides): 58Ce through 71Lu
Second f-transition series (actinides): 90Th through 103Lr
The A and B designations for groups of elements in the periodic table are somewhat
arbitrary, and they are reversed in some periodic tables. In another designation, the groups
are numbered 1 through 18. The system used in this text is the one commonly used in
the United States. Elements with the same group numbers, but with different letters, have
relatively few similar properties. The origin of the A and B designations is the fact that
some compounds of elements with the same group numbers have similar formulas but
quite different properties, for example, NaCl (IA) and AgCl (IB), MgCl2 (IIA) and ZnCl2
(IIB). As we shall see, variations in the properties of the B groups across a row are not
nearly as regular and dramatic as the variations observed across a row of A group elements.
The outermost electrons have the greatest influence on the properties of elements.
Adding an electron to an inner d orbital results in less striking changes in properties
than adding an electron to an outer s or p orbital.
PERIODIC PROPERTIES OF THE ELEMENTS
Now we investigate the nature of periodicity. Knowledge of periodicity is valuable in
understanding bonding in simple compounds. Many physical properties, such as melting
points, boiling points, and atomic volumes, show periodic variations. For now, we describe
the variations that are most useful in predicting chemical behavior. The variations in these
properties depend on electron configurations, especially the configurations in the outermost
occupied shell, and on how far away that shell is from the nucleus.
6-1 More About the Periodic Table 239
The elements of Period 3.
Properties progress (left to right)
from solids (Na, Mg, Al, Si, P, S) to
gases (Cl, Ar) and from the most
metallic (Na) to the most
nonmetallic (Ar).
In any atom the outermost electrons are
those that have the highest value of the
principal quantum number, n.
http://webbook.nist.gov
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 8.9, Atomic Properties and
Periodic Trends.
240 CHAPTER 6: Chemical Periodicity
ATOMIC RADII
In Section 5-16 we described individual atomic orbitals in terms of probabilities of distributions
of electrons over certain regions in space. Similarly, we can visualize the total
electron cloud that surrounds an atomic nucleus as somewhat indefinite. We cannot isolate
a single atom and measure its diameter the way we can measure the diameter of a golf
ball. For all practical purposes, the size of an individual atom cannot be uniquely defined.
An indirect approach is required. The size of an atom is determined by its immediate
environment, especially its interaction with surrounding atoms. By analogy, suppose we
arrange some golf balls in an orderly array in a box. If we know how the balls are positioned,
the number of balls, and the dimensions of the box, we can calculate the diameter
of an individual ball. Application of this reasoning to solids and their densities leads us to
values for the atomic sizes of many elements. In other cases, we derive atomic radii from
the observed distances between atoms that are combined with one another. For example,
the distance between atomic centers (nuclei) in the Cl2 molecule is measured to be 2.00 Å.
This suggests that the radius of each Cl atom is half the interatomic distance, or 1.00 Å.
We collect the data obtained from many such measurements to indicate the relative sizes
of individual atoms.
The top of Figure 6-1 displays the relative sizes of atoms of the representative elements
and the noble gases. It shows the periodicity in atomic radii. (The ionic radii at the bottom
of Figure 6-1 are discussed in Section 6-5.)
The effective nuclear charge, Zeff, experienced by an electron in an outer shell is less
than the actual nuclear charge, Z. This is because the attraction of outer-shell electrons
by the nucleus is partly counterbalanced by the repulsion of these outer-shell electrons by
electrons in inner shells. We say that the electrons in inner shells screen, or shield, electrons
in outer shells from the full effect of the nuclear charge. This concept of a screening,
or shielding, effect helps us understand many periodic trends in atomic properties.
Consider an atom of lithium; it has two electrons in a filled shell, 1s2, and one electron
in the 2s orbital, 2s1. The electron in the 2s orbital is fairly effectively screened from
the nucleus by the two electrons in the filled 1s orbital, so the 2s electron does not “feel”
the full 3ϩ charge of the nucleus. The effective nuclear charge, Zeff, experienced by the
electron in the 2s orbital, however, is not 1 (3 minus 2) either. The electron in the outer
shell of lithium has some probability of being found close to the nucleus (see Figure 5-20).
We say that, to some extent, it penetrates the region of the 1s electrons; that is, the 1s electrons
do not completely shield the outer-shell electrons from the nucleus. The electron
in the 2s shell “feels” an effective nuclear charge a little larger than 1ϩ. Sodium, element
number 11, has ten electrons in inner shells, 1s22s22p6, and one electron in an outer shell,
3s1. The ten inner-shell electrons of the sodium atom screen (shield) the outer-shell electron
somewhat from the nucleus, counteracting some of the 11ϩ nuclear charge. But the
3s electron of sodium penetrates the inner shells to a significant extent, so the effective
nuclear charge felt by the outermost (3s) electron is actually greater than it is for lithium
(2s). The somewhat increased attraction for the outermost electron in sodium is
outweighed, however, by the fact that the “outer” electron in a sodium atom is in the third
shell, whereas in lithium it is in the second shell. Recall from Chapter 5 that the third
shell (n ϭ 3) is farther from the nucleus than the second shell (n ϭ 2). Thus, we see why
sodium atoms are larger than lithium atoms. Similar reasoning explains why potassium
atoms are larger than sodium atoms and why the sizes of the elements in each column of
the periodic table are related in a similar way.
6-2
The radius of an atom, r, is taken as
half of the distance between nuclei
in homonuclear molecules such as Cl2.
Cl Cl
2r
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 8.10, Atomic Properties and
periodic Trends: Size.
Atomic radii are often stated in
angstroms (1 Å ϭ 10Ϫ10 m) or in the
SI units nanometers (1 nm ϭ 10Ϫ9 m)
or picometers (1 pm ϭ 10Ϫ12 m).
To convert from Å to nm, move the
decimal point to the left one place
(1 Å ϭ 0.1 nm). For example, the
atomic radius of Li is 1.52 Å, or
0.152 nm.
IA IIA IIIA IVA VA VIA VIIA VIIIA
Atomic radii
Li
1.52
Be
1.12
B
0.85
C
0.77
N
0.75
O
0.73
F
0.72
Ne
0.71
H
0.37
He
0.31
Al
1.43
Si
1.18
P
1.10
S
1.03
Cl
1.00
Ar
0.98
Na
1.86 1.60
Mg
Kr
1.12
Br
1.14
Se
1.19
As
1.20
Ge
1.22
Ga
1.35
K
2.27 1.97
Ca
In
1.67
Sn
1.40
Sb
1.40
Te
1.42
I
1.33
Xe
1.31
Rb
2.48 2.15
Sr
Rn
1.41
At
1.40
Po
1.68
Bi
1.50
Pb
1.46
Tl
1.70
Cs
2.65
2.22
Ba
Ionic radii
Rbϩ
1.66 1.32
Sr2ϩ
Csϩ
1.81 1.49
Ba2ϩ
Naϩ
1.16
Mg2ϩ
0.85
Kϩ
1.52
Ca2ϩ
1.14
Liϩ
0.90
Be2ϩ
0.59
N3Ϫ
1.71
O2Ϫ
1.26
S2Ϫ
1.70
Se2Ϫ
1.84
BrϪ
1.82
IϪ
2.06
2 Å
Te2Ϫ
2.07
ClϪ
1.67
FϪ
1.19
Al3ϩ
0.68
Ga3ϩ
0.76
In3ϩ
0.94
Tl3ϩ
1.03
Figure 6-1 (Top) Atomic radii of the A group (representative) elements and the noble
gases, in angstroms, Å (Section 6-2). Atomic radii increase going down a group because
electrons are being added to shells farther from the nucleus. Atomic radii decrease from left
to right within a given period owing to increasing effective nuclear charge. Hydrogen atoms
are the smallest and cesium atoms are the largest naturally occurring atoms.
(Bottom) Sizes of ions of the A group elements, in angstroms (Section 6-5). Positive ions
(cations) are always smaller than the neutral atoms from which they are formed. Negative
ions (anions) are always larger than the neutral atoms from which they are formed.
Within a family (vertical group on the periodic table) of representative elements,
atomic radii increase from top to bottom as electrons are added to shells farther
from the nucleus.
As we move across the periodic table, atoms become smaller due to increasing effective
nuclear charges. Consider the elements B (Z ϭ 5, 1s22s22p1) to F (Z ϭ 9, 1s22s22p5). In B
there are two electrons in a noble gas configuration, 1s2, and three electrons in the second
shell, 2s22p1. The two electrons in the noble gas configuration fairly effectively screen out
the effect of two protons in the nucleus. So the electrons in the second shell of B “feel”
a greater effective nuclear charge than do those of Be. By similar arguments, we see that
in carbon (Z ϭ 6, 1s22s22p2) the electrons in the second shell “feel” an effective nuclear
charge greater than those of B. So we expect C atoms to be smaller than B atoms, and
they are. In nitrogen (Z ϭ 7, 1s22s22p3), the electrons in the second shell “feel” an even
greater effective nuclear charge, and so N atoms are smaller than C atoms.
As we move from left to right across a period in the periodic table, atomic radii of
representative elements decrease as a proton is added to the nucleus and an electron
is added to a particular shell.
For the transition elements, the variations are not so regular because electrons are being
added to an inner shell. All transition elements have smaller radii than the preceding
Group IA and IIA elements in the same period.
EXAMPLE 6-1 Trends in Atomic Radii
Arrange the following elements in order of increasing atomic radii. Justify your order.
Cs, F, K, Cl
Plan
Both K and Cs are Group IA metals, whereas F and Cl are halogens (VIIA nonmetals). Figure
6-1 shows that atomic radii increase as we descend a group, so K Ͻ Cs and F Ͻ Cl. Atomic
radii decrease from left to right.
Solution
The order of increasing atomic radii is F Ͻ Cl Ͻ K Ͻ Cs.
You should now work Exercise 18.
IONIZATION ENERGY
The first ionization energy (IE1), also called first ionization potential, is
the minimum amount of energy required to remove the most loosely bound electron
from an isolated gaseous atom to form an ion with a 1ϩ charge.
6-3
General trends in atomic radii of A
group elements with position in the
periodic table.
242 CHAPTER 6: Chemical Periodicity
Atomic radii
Decrease
Increase
Figure 6-2 A plot of first
ionization energies for the first 38
elements versus atomic number.
The noble gases have very high first
ionization energies, and the IA
metals have low first ionization
energies. Note the similarities in the
variations for the Period 2 elements,
3 through 10, to those for the
Period 3 elements, 11 through 18,
as well as for the later A group
elements. Variations for B group
elements are not nearly so
pronounced as those for A group
elements.
6-3 Ionization Energy 243
For calcium, for example, the first ionization energy, IE1, is 590 kJ/mol:
Ca(g) ϩ 590 kJ 88n Caϩ(g) ϩ eϪ
The second ionization energy (IE2) is the amount of energy required to remove the
second electron. For calcium, it may be represented as
Caϩ(g) ϩ 1145 kJ 88n Ca2ϩ(g) ϩ eϪ
For a given element, IE2 is always greater than IE1 because it is always more difficult to
remove an electron from a positively charged ion than from the corresponding neutral
atom. Table 6-1 gives first ionization energies.
Ionization energies measure how tightly electrons are bound to atoms. Ionization always
requires energy to remove an electron from the attractive force of the nucleus. Low ionization
energies indicate ease of removal of electrons, and hence ease of positive ion (cation)
formation. Figure 6-2 shows a plot of first ionization energy versus atomic number for
several elements.
H
1312
Li
520
Na
496
K
419
Rb
403
Cs
377
Be
899
Mg
738
Ca
599
Sr
550
Ba
503
Sc
631
Y
617
La
538
Ti
658
Zr
661
Hf
681
V
650
Nb
664
Ta
761
Cr
652
Mo
685
W
770
Mn
717
Tc
702
Re
760
Fe
759
Ru
711
Os
840
Co
758
Rh
720
Ir
880
Ni
757
Pd
804
Pt
870
Cu
745
Ag
731
Au
890
Zn
906
Cd
868
Hg
1007
B
801
Al
578
Ga
579
In
558
Tl
589
C
1086
Si
786
Ge
762
Sn
709
Pb
715
N
1402
P
1012
As
947
Sb
834
Bi
703
O
1314
S
1000
Se
941
Te
869
Po
812
F
1681
Cl
1251
Br
1140
I
1008
At
890
He
2372
Ne
2081
Ar
1521
Kr
1351
Xe
1170
Rn
1037
TABLE 6-1 First Ionization Energies (kJ/mol of atoms) of Some Elements
2500
2000
1500
1000
500
0
5 10 15 20 25 30 35
Atomic number
Firstionizationenergy(kJ/mol)
Period 2 Period 3 Period 4
H
He
Li
Be
B
C
N
O
F
Ne
Na
Mg
Al
Si
P
S
Cl
Ar
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co Cu
Zn
Ga
Ge
As
Br
Kr
Sr
Rb
SeNi
As one goes across a period on the
periodic table, the slight breaks in the
increasing ionization energies occur
between Groups IIA and IIIA
(electrons first enter the np subshell)
and again between Groups VA and
VIA (electrons are first paired in the
np subshell).
244 CHAPTER 6: Chemical Periodicity
Elements with low ionization energies (IE) lose electrons easily to form cations.
We see that in each period of Figure 6-2, the noble gases have the highest first ionization
energies. This should not be surprising, because the noble gases are known to be very
unreactive elements. It requires more energy to remove an electron from a helium atom
(slightly less than 4.0 ϫ 10Ϫ18 J/atom, or 2372 kJ/mol) than to remove one from a neutral
atom of any other element.
He(g) ϩ 2372 kJ 88n Heϩ(g) ϩ eϪ
The Group IA metals (Li, Na, K, Rb, Cs) have very low first ionization energies. Each
of these elements has only one electron in its outermost shell (. . . ns1), and they are the
largest atoms in their periods. The first electron added to a shell is easily removed to form
a noble gas configuration. As we move down the group, the first ionization energies become
smaller. The force of attraction of the positively charged nucleus for electrons decreases
as the square of the distance between them increases. So as atomic radii increase in a given
group, first ionization energies decrease because the outermost electrons are farther from
the nucleus.
Effective nuclear charge, Zeff, increases going from left to right across a period. The
increase in effective nuclear charge causes the outermost electrons to be held more tightly,
making them harder to remove. The first ionization energies therefore generally increase
from left to right across the periodic table. The reason for the trend in first ionization
energies is the same as that used in Section 6-2 to explain trends in atomic radii. The first
ionization energies of the Group IIA elements (Be, Mg, Ca, Sr, Ba) are significantly higher
than those of the Group IA elements in the same periods. This is because the Group IIA
elements have higher Zeff values and smaller atomic radii. Thus, their outermost electrons
are held more tightly than those of the neighboring IA metals. It is harder to remove an
electron from a pair in the filled outermost s orbitals of the Group IIA elements than to
remove the single electron from the half-filled outermost s orbitals of the Group IA
elements.
The first ionization energies for the Group IIIA elements (B, Al, Ga, In, Tl) are exceptions
to the general horizontal trends. They are lower than those of the IIA elements in
the same periods because the IIIA elements have only a single electron in their outermost
p orbitals. Less energy is required to remove the first p electron than the second s electron
from the outermost shell, because the p orbital is at a higher energy (less stable) than
an s orbital within the same shell (n value).
Going from Groups IIIA to VA, electrons are going singly into separate np orbitals,
where they do not shield one another significantly. The general left-to-right increase in
IE1 for each period is interrupted by a dip between Groups VA (N, P, As, Sb, Bi) and VIA
elements (O, S, Se, Te, Po). Presumably, this behavior is because the fourth np electron
in the Group VIA elements is paired with another electron in the same orbital, so it experiences
greater repulsion than it would in an orbital by itself. This increased repulsion
apparently outweighs the increase in Zeff, so the fourth np electron in an outer shell (Group
VIA elements) is somewhat easier to remove (lower ionization energy) than is the third
np electron in an outer shell (Group VA elements). After the dip between Groups VA and
VIA, the importance of the increasing Zeff outweighs the repulsion of electrons needing
to be paired, and the general left-to-right increases in first ionization energies resume.
Knowledge of the relative values of ionization energies assists us in predicting whether
an element is likely to form ionic or molecular (covalent) compounds. Elements with low
ionization energies form ionic compounds by losing electrons to form cations (positively
By Coulomb’s Law, F ϰ ᎏ
(qϩ
d
)(
2
qϪ)
ᎏ,
the attraction for the outer shell
electrons is directly proportional to
the effective charges and inversely
proportional to the square of the
distance between the charges. Even
though the effective nuclear charge
increases going down a group, the
greatly increased size results in a
weaker net attraction for the outer
electrons and thus in a lower first
ionization energy.
General trends in first ionization
energies of A group elements with
position in the periodic table.
Exceptions occur at Groups IIIA
and VIA.
First IE
Increase
Decrease
Here is one reason why trends in
ionization energies are important.
Noble gas electron configurations are
stable only for ions in compounds. In
fact, Liϩ(g) is less stable than Li(g) by
520 kJ/mol.
6-4 Electron Affinity 245
charged ions). Elements with intermediate ionization energies generally form molecular
compounds by sharing electrons with other elements. Elements with very high ionization
energies, such as Groups VIA and VIIA, often gain electrons to form anions (negatively
charged ions).
One factor that favors an atom of a representative element forming a monatomic ion in
a compound is the formation of a stable noble gas electron configuration. Energy considerations
are consistent with this observation. For example, as one mole of Li from Group
IA forms one mole of Liϩ ions, it absorbs 520 kJ per mole of Li atoms. The IE2 value is
14 times greater, 7298 kJ/mol, and is prohibitively large for the formation of Li2ϩ ions
under ordinary conditions. For Li2ϩ ions to form, an electron would have to be removed
from the filled first shell. We recognize that this is unlikely. The other alkali metals behave
in the same way, for similar reasons.
The first two ionization energies of Be (Group IIA) are 899 and 1757 kJ/mol, but IE3
is more than eight times larger, 14,849 kJ/mol. So Be forms Be2ϩ ions, but not Be3ϩ ions.
The other alkaline earth metals—Mg, Ca, Sr, Ba, and Ra—behave in a similar way. Only
the lower members of Group IIIA, beginning with Al, form 3ϩ ions. Bi and some d- and
f-transition metals do so, too. We see that the magnitudes of successive ionization energies
support the ideas of electron configurations discussed in Chapter 5.
Due to the high energy required, simple monatomic cations with charges greater than
3ϩ do not form under ordinary circumstances.
EXAMPLE 6-2 Trends in First IEs
Arrange the following elements in order of increasing first ionization energy. Justify your order.
Na, Mg, Al, Si
Plan
Table 6-1 shows that first ionization energies generally increase from left to right in the same
period, but there are exceptions at Groups IIIA and VIA. Al is a IIIA element with only one
electron in its outer p orbitals, 1s22s22p63s23p1.
Solution
There is a slight dip at Group IIIA in the plot of first IE versus atomic number (see Figure
6-2). The order of increasing first ionization energy is Na Ͻ Al Ͻ Mg Ͻ Si.
You should now work Exercise 26.
ELECTRON AFFINITY
The electron affinity (EA) of an element may be defined as
the amount of energy absorbed when an electron is added to an isolated gaseous atom
to form an ion with a 1Ϫ charge.
The convention is to assign a positive value when energy is absorbed and a negative value
when energy is released. Most elements have no affinity for an additional electron and
6-4
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 8.13, Electron Affinity.
This is consistent with thermodynamic
convention.
The value of EA for Cl can also be
represented as Ϫ5.79 ϫ 10Ϫ19 J/atom
or Ϫ3.61 eV/atom. The electron
volt (eV) is a unit of energy
(1 eV ϭ 1.6022 ϫ 10Ϫ19 J).
246 CHAPTER 6: Chemical Periodicity
thus have an electron affinity (EA) equal to zero. We can represent the electron affinities
of helium and chlorine as
He(g) ϩ eϪ 88nϫ HeϪ(g) EA ϭ Ϫ000 kJ/mol
Cl(g) ϩ eϪ 88n ClϪ(g) ϩ 349 kJ EA ϭ Ϫ349 kJ/mol
The first equation tells us that helium will not add an electron. The second equation tells
us that when one mole of gaseous chlorine atoms gain one electron each to form gaseous
chloride ions, 349 kJ of energy is released (exothermic). Figure 6-3 shows a plot of electron
affinity versus atomic number for several elements.
Electron affinity involves the addition of an electron to a neutral gaseous atom. The
process by which a neutral atom X gains an electron (EA),
X(g) ϩ eϪ 88n XϪ(g) (EA)
is not the reverse of the ionization process,
Xϩ(g) ϩ eϪ 88n X(g) (reverse of IE1)
The first process begins with a neutral atom, whereas the second begins with a positive
ion. Thus, IE1 and EA are not simply equal in value with the signs reversed. We see from
Figure 6-3 that electron affinities generally become more negative from left to right across
a row in the periodic table (excluding the noble gases). This means that most representative
elements in Groups IA to VIIA show a greater attraction for an extra electron from
left to right. Halogen atoms, which have the outer electron configuration ns2np5, have the
most negative electron affinities. They form stable anions with noble gas configurations,
. . . ns2np6, by gaining one electron.
General trends in electron affinities
of A group elements with position in
the periodic table. There are many
exceptions.
Electron affinity
Increase
(Lessnegative)
Decrease
(More negative)
Figure 6-3 A plot of electron affinity versus atomic number for the first 20 elements. The
general horizontal trend is that electron affinities become more negative (more energy is
released as an extra electron is added) from Group IA through Group VIIA for a given
period. Exceptions occur at the IIA and VA elements.
Period
Atomic number
F
OC
NBe
B
LiH
He
K
Ar
Cl
S
P
Si
Al
Mg
Na
Ne
4321
Ca
Electronaffinity(kJ/mol)
0 5 10 15 20 25
Ϫ400
300
200
100
0
Ϫ100
Ϫ200
Ϫ300
This reasoning is similar to that used
to explain the low IE1 values for
Group VIA elements.
6-4 Electron Affinity 247
Elements with very negative electron affinities gain electrons easily to form negative
ions (anions).
“Electron affinity” is a precise and quantitative term, like “ionization energy,” but it is
difficult to measure. Table 6-2 shows electron affinities for several elements.
For many reasons, the variations in electron affinities are not regular across a period.
The general trend is: the electron affinities of the elements become more negative from
left to right in each period. Noteworthy exceptions are the elements of Groups IIA and
VA, which have less negative values than the trends suggest (see Figure 6-3). It is very
difficult to add an electron to a IIA metal atom because its outer s subshell is filled. The
values for the VA elements are slightly less negative than expected because they apply to
the addition of an electron to a half-filled set of np orbitals (ns2np3 n ns2np4), which
requires pairing. The resulting repulsion overcomes the increased attractive force of the
nucleus.
Energy is always required to bring a negative charge (electron) closer to another negative
charge (anion). So the addition of a second electron to a 1Ϫ anion to form an ion
with a 2Ϫ charge is always endothermic. Thus, electron affinities of anions are always posi-
tive.
EXAMPLE 6-3 Trends in EAs
Arrange the following elements in order of increasing values of electron affinity, that is, from
most negative to least negative.
K, Br, Cs, Cl
Plan
Table 6-2 shows that electron affinity values generally become more negative from left to right
across a period, with major exceptions at Groups IIA (Be) and VA (N). They generally become
more negative from bottom to top.
Cu
–118
Ag
–125
Au
–282
H
–73
Li
–60
Na
–53
K
–48
Rb
–47
Cs
–45
Be
(~0)
Mg
(~0)
Ca
Sr
Ba
B
–29
Al
–43
Ga
–29
In
–29
Tl
–19
C
–122
Si
–134
Ge
–119
Sn
–107
Pb
–35
N
0
P
–72
As
–78
Sb
–101
Bi
–91
O
–141
S
–200
Se
–195
Te
–190
F
–328
Cl
–349
Br
–324
I
–295
He
0
Ne
0
Ar
0
Kr
0
Xe
0
(~0)
(~0)
(~0)
1
2
3
4
5
6
TABLE 6-2 Electron Affinity Values (kJ/mol ) of Some Elements*
*Estimated values are in parentheses.
Solution
The order of increasing values of electron affinity is
(most negative EA) Cl Ͻ Br Ͻ K Ͻ Cs (least negative EA)
You should now work Exercises 32 and 33.
IONIC RADII
Many elements on the left side of the periodic table react with other elements by losing
electrons to form positively charged ions. Each of the Group IA elements (Li, Na, K, Rb,
Cs) has only one electron in its outermost shell (electron configuration . . . ns1). These
elements react with other elements by losing one electron to attain noble gas configurations.
They form the ions Liϩ, Naϩ, Kϩ, Rbϩ, and Csϩ. A neutral lithium atom, Li,
contains three protons in its nucleus and three electrons, with its outermost electron in
the 2s orbital. A lithium ion, Liϩ, however, contains three protons in its nucleus but only
two electrons, both in the 1s orbital. So a Liϩ ion is much smaller than a neutral Li atom
(see figure in the margin). Likewise, a sodium ion, Naϩ, is considerably smaller than a
sodium atom, Na. The relative sizes of atoms and common ions of some representative
elements are shown in Figure 6-1.
Isoelectronic species have the same number of electrons. We see that the ions formed
by the Group IIA elements (Be2ϩ, Mg2ϩ, Ca2ϩ, Sr2ϩ, Ba2ϩ) are significantly smaller than
the isoelectronic ions formed by the Group IA elements in the same period. The radius of
the Liϩ ion is 0.90 Å, whereas the radius of the Be2ϩ ion is only 0.59 Å. This is what we
should expect. A beryllium ion, Be2ϩ, is formed when a beryllium atom, Be, loses both of
its 2s electrons while the 4ϩ nuclear charge remains constant. We expect the 4ϩ nuclear
charge in Be2ϩ to attract the remaining two electrons quite strongly. Comparison of the
ionic radii of the IIA elements with their atomic radii indicates the validity of our reasoning.
Similar reasoning indicates that the ions of the Group IIIA metals (Al3ϩ, Ga3ϩ, In3ϩ,
Tl3ϩ) should be even smaller than the ions of Group IA and Group IIA elements in the
same periods.
Now consider the Group VIIA elements (F, Cl, Br, I). These have the outermost electron
configuration . . . ns2np5. These elements can completely fill their outermost p orbitals
by gaining one electron to attain noble gas configurations. Thus, when a fluorine atom
(with seven electrons in its outer shell) gains one electron, it becomes a fluoride ion, FϪ,
with eight electrons in its outer shell. These eight electrons repel one another more
strongly than the original seven, so the electron cloud expands. The FϪ ion is much larger
than the neutral F atom (see figure in the margin). Similar reasoning indicates that a chloride
ion, ClϪ, should be larger than a neutral chlorine atom, Cl. Observed ionic radii (see
Figure 6-1) verify this prediction.
Comparing the sizes of an oxygen atom (Group VIA) and an oxide ion, O2Ϫ, again we
find that the negatively charged ion is larger than the neutral atom. The oxide ion is also
larger than the isoelectronic fluoride ion because the oxide ion contains ten electrons held
6-5
The nuclear charge remains constant
when the ion is formed.
248 CHAPTER 6: Chemical Periodicity
᭣ General trends in ionic radii of A group elements with position in the periodic table.
Ionic radii
Decrease
Increase
r = 1.52 Å r = 0.90 Å
r = 1.86 Å r = 1.16 Å
Liϩ
Li
Na Naϩ
r ϭ 1.12 Å
r ϭ 1.60 Å r ϭ 0.85 Å
r ϭ 0.72 Å r ϭ 1.19 Å
r ϭ 1.00 Å r ϭ 1.67 Å
r ϭ 0.59 Å
Be2ϩ
Mg2ϩ
Be
F
Cl
Mg
ClϪ
FϪ
by a nuclear charge of only 8ϩ, whereas the fluoride ion has ten electrons held by a nuclear
charge of 9ϩ. Comparison of radii is not a simple issue when we try to compare atoms,
positive and negative ions, and ions with varying charge. Sometimes we compare atoms
to their ions, atoms or ions that are vertically or horizontally positioned on the periodic
table, or isoelectronic species. The following guidelines are often considered in the order
given.
1. Simple positively charged ions (cations) are always smaller than the neutral atoms
from which they are formed.
2. Simple negatively charged ions (anions) are always larger than the neutral atoms
from which they are formed.
3. The sizes of cations decrease from left to right across a period.
4. The sizes of anions decrease from left to right across a period.
5. Within an isoelectronic series, radii decrease with increasing atomic number because
of increasing nuclear charge.
6. Both cation and anion sizes increase going down a group.
Large ring compounds can selectively
trap ions based on the ability of
various ions to fit within a cavity in
the large compound. This results in
selectivity such as occurs in transport
across membranes in biological
systems.
6-5 Ionic Radii 249
Naϩ
Mg2ϩ
1.71
10
ϩ7
1.26
10
ϩ8
1.19
10
ϩ9
1.16
10
ϩ11
0.85
10
ϩ12
0.68
10
ϩ13
Ionic radius (Å)
No. of electrons
Nuclear charge
An isoelectronic series of ions
N3Ϫ
O2Ϫ FϪ
Al3ϩ
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 8.11, Atomic Properties and
Periodic Trends: Ion Sizes.
EXAMPLE 6-4 Trends in Ionic Radii
Arrange the following ions in order of increasing ionic radii: (a) Ca2ϩ, Kϩ, Al3ϩ; (b) Se2Ϫ, BrϪ,
Te2Ϫ.
Plan
Some of the pairs of ions are isoelectronic, so we can compare their sizes on the basis of nuclear
charges. Other comparisons can be made based on the outermost occupied shell (highest value
of n).
Solution
(a) Ca2ϩ and Kϩ are isoelectronic (18 electrons each) with an outer-shell electron configuration
of 3s23p6. Because Ca2ϩ has a higher nuclear charge (20ϩ) than Kϩ (19ϩ), Ca2ϩ holds its
18 electrons more tightly, and Ca2ϩ is smaller than Kϩ. Al3ϩ has electrons only in the second
main shell (outer-shell electron configuration of 2s22p6), so it is smaller than either of the other
two ions.
Al3ϩ Ͻ Ca2ϩ Ͻ Kϩ
(b) BrϪ and Se2Ϫ are isoelectronic (36 electrons each) with an outer-shell electron configuration
of 4s24p6. Because BrϪ has a higher nuclear charge (35ϩ) than Se2Ϫ (34ϩ), BrϪ holds its
Because the noble gases form few
compounds, they are not included in
this discussion.
250 CHAPTER 6: Chemical Periodicity
36 electrons more tightly, and BrϪ is smaller than Se2Ϫ. Te2Ϫ has electrons in the fifth main
shell (outer configuration of 5s25p6), so it is larger than either of the other two ions.
BrϪ Ͻ Se2Ϫ Ͻ Te2Ϫ
You should now work Exercises 38 and 40.
ELECTRONEGATIVITY
The electronegativity (EN) of an element is a measure of the relative tendency of an
atom to attract electrons to itself when it is chemically combined with another atom.
Elements with high electronegativities (nonmetals) often gain electrons to form
anions. Elements with low electronegativities (metals) often lose electrons to form
cations.
Electronegativities of the elements are expressed on a somewhat arbitrary scale, called the
Pauling scale (Table 6-3). The electronegativity of fluorine (4.0) is higher than that of any
other element. This tells us that when fluorine is chemically bonded to other elements,
6-6
H
2.1
1
Li
1.0
3
Na
1.0
11
K
0.9
19
Rb
0.9
37
Cs
0.8
55
Fr
0.8
87
Be
1.5
4
Mg
1.2
12
Ca
1.0
20
Sr
1.0
38
Ba
1.0
56
Ra
1.0
88
Sc
1.3
21
Y
1.2
39
La
1.1
57
Ac
1.1
89
Ti
1.4
22
Zr
1.3
40
Hf
1.3
72
V
1.5
23
Nb
1.5
41
Ta
1.4
73
Cr
1.6
24
Mo
1.6
42
W
1.5
74
Mn
1.6
25
Tc
1.7
43
Re
1.7
75
Fe
1.7
26
Ru
1.8
44
Os
1.9
76
Co
1.7
27
Rh
1.8
45
Ir
1.9
77
Ni
1.8
28
Pd
1.8
46
Pt
1.8
78
Cu
1.8
29
Ag
1.6
47
Au
1.9
79
Zn
1.6
30
Cd
1.6
48
Hg
1.7
80
B
2.0
5
Al
1.5
13
Ga
1.7
31
In
1.6
49
Tl
1.6
81
C
2.5
6
Si
1.8
14
Ge
1.9
32
Sn
1.8
50
Pb
1.7
82
N
3.0
7
P
2.1
15
As
2.1
33
Sb
1.9
51
Bi
1.8
83
O
3.5
8
S
2.5
16
Se
2.4
34
Te
2.1
52
Po
1.9
84
F
4.0
9
Cl
3.0
17
Br
2.8
35
I
2.5
53
At
2.1
85
He
2
Ne
10
Ar
18
Kr
36
Xe
54
Rn
86
1
2
3
4
5
6
7
IIIB IVB VB VIB IIBVIIB IB
IIIA IVA VA VIA VIIA
VIIIA
VIIIB
Ce
1.1
58
Pr
1.1
59
Nd
1.1
60
Pm
1.1
61
Sm
1.1
62
Eu
1.1
63
Gd
1.1
64
Tb
1.1
65
Dy
1.1
66
Ho
1.1
67
Er
1.1
68
Tm
1.1
69
Yb
1.0
70
Lu
1.2
71
Th
1.2
90
Pa
1.3
91
U
1.5
92
Np
1.3
93
Pu
1.3
94
Am
1.3
95
Cm
1.3
96
Bk
1.3
97
Cf
1.3
98
Es
1.3
99
Fm
1.3
100
Md
1.3
101
No
1.3
102
Lr
1.5
103
*
†
*
†
IIA
IA
Metals
Nonmetals
Metalloids
TABLE 6-3 Electronegativity Values of the Elementsa
a
Electronegativity values are given at the bottoms of the boxes.
General trends in electronegativities
of A group elements with position in
the periodic table.
6-7 Hydrogen and the Hydrides 251
it has a greater tendency to attract electron density to itself than does any other element.
Oxygen is the second most electronegative element.
For the representative elements, electronegativities usually increase from left to right
across periods and decrease from top to bottom within groups.
Variations among the transition metals are not as regular. In general, both ionization energies
and electronegativities are low for elements at the lower left of the periodic table and
high for those at the upper right.
EXAMPLE 6-5 Trends in ENs
Arrange the following elements in order of increasing electronegativity.
B, Na, F, O
Plan
Table 6-3 shows that electronegativities increase from left to right across a period and decrease
from top to bottom within a group.
Solution
The order of increasing electronegativity is Na Ͻ B Ͻ O Ͻ F.
You should now work Exercise 44.
Although the electronegativity scale is somewhat arbitrary, we can use it with reasonable
confidence to make predictions about bonding. Two elements with quite different
electronegativities (a metal and a nonmetal) tend to react with each other to form ionic
compounds. The less electronegative element gives up its electron(s) to the more electronegative
element. Two nonmetals with similar electronegativities tend to form covalent
bonds with each other. That is, they share their electrons. In this sharing, the more electronegative
element attains a greater share. This is discussed in detail in Chapters 7 and 8.
CHEMICAL REACTIONS AND PERIODICITY
Now we will illustrate the periodicity of chemical properties by considering some reactions
of hydrogen, oxygen, and their compounds. We choose to discuss hydrogen and
oxygen because, of all the elements, they form the most kinds of compounds with other
elements. Additionally, compounds of hydrogen and oxygen are very important in such
diverse phenomena as all life processes and most corrosion processes.
HYDROGEN AND THE HYDRIDES
Hydrogen
Elemental hydrogen is a colorless, odorless, tasteless diatomic gas with the lowest molecular
weight and density of any known substance. Discovery of the element is attributed
to the Englishman Henry Cavendish (1731–1810), who prepared it in 1766 by passing
steam through a red-hot gun barrel (mostly iron) and by the reaction of acids with active
6-7
Electronegativity
Increase
Decrease
Ionization energy (Section 6-3) and
electron affinity (Section 6-4) are
precise quantitative concepts. We find,
however, that the more qualitative
concept of electronegativity is more
useful in describing chemical bonding.
The name “hydrogen” means “water
former.”
Can you write the net ionic equation
for the reaction of Zn with HCl(aq)?
252 CHAPTER 6: Chemical Periodicity
metals. The latter is still the method commonly used for the preparation of small amounts
of H2 in the laboratory. In each case, H2 is liberated by a displacement (and redox) reaction,
of the kind described in Section 4-8. (See the activity series, Table 4-12.)
heat
3Fe(s) ϩ 4H2O(g) 888n Fe3O4(s) ϩ 4H2(g)
Zn(s) ϩ 2HCl(aq) 888n ZnCl2(aq) ϩ H2(g)
Hydrogen also can be prepared by electrolysis of water.
electricity
2H2O(ᐉ) 8888888n 2H2(g) ϩ O2(g)
In the future, if it becomes economical to convert solar energy into electrical energy
that can be used to electrolyze water, H2 could become an important fuel (although the
dangers of storage and transportation would have to be overcome). The combustion of H2
liberates a great deal of heat. Combustion is the highly exothermic combination of a
substance with oxygen, usually with a flame. (See Section 6-8, Combustion Reactions.)
spark
2H2(g) ϩ O2(g) 88888n 2H2O(ᐉ) ϩ energy
or heat
Hydrogen is very flammable; it was responsible for the Hindenburg airship disaster in
1937. A spark is all it takes to initiate the combustion reaction, which is exothermic
enough to provide the heat necessary to sustain the reaction.
Hydrogen is prepared by the “water gas reaction,” which results from the passage of
steam over white-hot coke (impure carbon, a nonmetal) at 1500°C. The mixture of products
commonly called “water gas” is used industrially as a fuel. Both components, CO and
H2, undergo combustion.
C(s) ϩ H2O(g) 88n CO(g) ϩ H2(g)
in coke steam “water gas”
Vast quantities of hydrogen are produced commercially each year by a process called
steam cracking. Methane reacts with steam at 830°C in the presence of a nickel catalyst.
heat
CH4(g) ϩ H2O(g) 888n CO(g) ϩ 3H2(g)
Ni
Reactions of Hydrogen and Hydrides
Atomic hydrogen has the 1s1 electron configuration. It reacts with metals and with other
nonmetals to form binary compounds called hydrides. These can be (a) ionic hydrides,
which contain hydride ions, HϪ, formed when hydrogen gains one electron per atom from
an active metal; or (2) molecular hydrides, in which hydrogen shares electrons with an
atom of another nonmetal.
The ionic or molecular character of the binary compounds of hydrogen depends on
the position of the other element in the periodic table (Figure 6-4). The reactions of H2
with the alkali (IA) and the heavier (more active) alkaline earth (IIA) metals result in solid
ionic hydrides. The reaction with the molten (liquid) IA metals may be represented in
general terms as
high temperatures
2M(ᐉ) ϩ H2(g) 88888888888888n 2(Mϩ, HϪ)(s) M ϭ Li, Na, K, Rb, Cs
high pressures
This is the reverse of the
decomposition of H2O.
Hydrogen is no longer used in blimps
and dirigibles. It has been replaced by
helium, which is slightly denser,
nonflammable, and much safer.
The use of the term “hydride” does
not necessarily imply the presence of
the hydride ion, HϪ.









Figure 6-4 Common hydrides of the representative elements. The ionic hydrides are
shaded blue, molecular hydrides are shaded red, and those of intermediate character are
shaded purple.
6-7 Hydrogen and the Hydrides 253
Thus, hydrogen combines with lithium to form lithium hydride and with sodium to form
sodium hydride.
2Li(ᐉ) ϩ H2(g) 88n 2LiH(s) lithium hydride (mp 680°C)
2Na(ᐉ) ϩ H2(g) 88n 2NaH(s) sodium hydride (mp 800°C)
In general terms, the reactions of the heavier (more active) IIA metals may be represented
as
M(ᐉ) ϩ H2(g) 88n (M2ϩ, 2HϪ)(s) M ϭ Ca, Sr, Ba
Thus, calcium combines with hydrogen to form calcium hydride:
Ca(ᐉ) ϩ H2(g) 88n CaH2(s) calcium hydride (mp 816°C)
These ionic hydrides are all basic because they react with water to form hydroxide ions.
When water is added by drops to lithium hydride, for example, lithium hydroxide and
hydrogen are produced. The reaction of calcium hydride is similar.
LiH(s) ϩ H2O(ᐉ) 88n LiOH(s) ϩ H2(g)
CaH2(s) ϩ 2H2O(ᐉ) 88n Ca(OH)2(s) ϩ 2H2(g)
Hydrogen reacts with nonmetals to form binary molecular hydrides. For example, H2
combines with the halogens to form colorless, gaseous hydrogen halides (Figure 6-5):
H2(g) ϩ X2 88n 2HX(g) X ϭ F, Cl, Br, I
hydrogen halides
Specifically, hydrogen reacts with fluorine to form hydrogen fluoride and with chlorine
to form hydrogen chloride:
H2(g) ϩ F2(g) 88n 2HF(g) hydrogen fluoride
H2(g) ϩ Cl2(g) 88n 2HCl(g) hydrogen chloride
Hydrogen combines with Group VIA elements to form molecular compounds:
heat
2H2(g) ϩ O2(g) 888n 2H2O(g)
The heavier members of this family also combine with hydrogen to form binary
compounds that are gases at room temperature. Their formulas resemble that of water.
LiH
NaH
KH
RbH
CsH
BeH2
MgH2
CaH2
SrH2
BaH2
B2H6
(AlH3)x
Ga2H6
InH3
TlH
CH4
SiH4
GeH4
SnH4
PbH4
NH3
PH3
AsH3
SbH3
BiH3
H2O
H2S
H2Se
H2Te
H2Po
HF
IA IIA IIIA IVA VA VIA VIIA
HCl
HBr
HI
HAt
The ionic hydrides are named by
naming the metal first, followed by
“hydride.”
Ionic hydrides can serve as sources of
hydrogen. They must be stored in
environments free of moisture and O2.
We show LiOH and Ca(OH)2 as solids
here because not enough water is
available to act as a solvent.
The hydrogen halides are named by
the word “hydrogen” followed by the
stem for the halogen with an “-ide”
ending.
These compounds are named:
H2O, hydrogen oxide (water)
H2S, hydrogen sulfide
H2Se, hydrogen selenide
H2Te, hydrogen telluride
All except H2O are very toxic.
The primary industrial use of H2 is in the synthesis of ammonia, a molecular hydride,
by the Haber process (Section 17-7). Most of the NH3 is used in liquid form as a fertilizer
(Figure 6-6) or to make other fertilizers, such as ammonium nitrate, NH4NO3, and
ammonium sulfate, (NH4)2SO4:
catalysts
N2(g) ϩ 3H2(g) 88888888888888n 2NH3(g)
heat, high pressure
Many of the molecular (nonmetal) hydrides are acidic; their aqueous solutions contain
hydrogen ions. These include HF, HCl, HBr, HI, H2S, H2Se, and H2Te.
EXAMPLE 6-6 Predicting Products of Reactions
Predict the products of the reactions involving the reactants shown. Write a balanced formula
unit equation for each.
heat
(a) H2(g) ϩ I2(g) 888n
heat
(b) K(ᐉ) ϩ H2(g) 888n
(c) NaH(s) ϩ H2O(ᐉ) (excess) 88n
Plan
(a) Hydrogen reacts with the halogens (Group VIIA) to form hydrogen halides—in this
example, HI.
(b) Hydrogen reacts with active metals to produce hydrides—in this case, KH.
(c) Active metal hydrides react with water to produce a metal hydroxide and H2.
Solution
heat
(a) H2(g) ϩ I2(g) 888n 2HI(g)
heat
(b) 2K(ᐉ) ϩ H2(g) 888n 2KH(s)
(c) NaH(s) ϩ H2O(ᐉ) 88n NaOH(aq) ϩ H2(g)
Figure 6-5 Hydrogen, H2, burns
in an atmosphere of pure chlorine,
Cl2, to produce hydrogen chloride.
H2 ϩ Cl2 88n 2HCl
254 CHAPTER 6: Chemical Periodicity
Figure 6-6 Ammonia may be applied directly to the soil as a fertilizer.
Remember that hydride ions, HϪ,
react with (reduce) water to produce
OHϪ ions and H2(g).
The name “oxygen” means “acid
former.”
Liquid O2 is used as an oxidizer for
rocket fuels. O2 also is used in the
health fields for oxygen-enriched air.
Allotropes are different forms of the
same element in the same physical
state (Section 2-2).
A radical is a species containing one or
more unpaired electrons; many radicals
are very reactive.
6-8 Oxygen and the Oxides 255
EXAMPLE 6-7 Ionic and Molecular Properties
Predict the ionic or molecular character of the products in Example 6-6.
Plan
We refer to Figure 6-4, which displays the nature of hydrides.
Solution
Reaction (a) is a reaction between hydrogen and another nonmetal. The product, HI, must
be molecular. Reaction (b) is the reaction of hydrogen with an active Group IA metal. Thus,
KH must be ionic. The products of reaction (c) are molecular H2(g) and the strong base,
NaOH, which is ionic.
You should now work Exercises 58 and 59.
OXYGEN AND THE OXIDES
Oxygen and Ozone
Oxygen was discovered in 1774 by an English minister and scientist, Joseph Priestley
(1733–1804). He observed the thermal decomposition of mercury(II) oxide, a red powder:
heat
2HgO(s) 888n 2Hg(ᐉ) ϩ O2(g)
That part of the earth we see—land, water, and air—is approximately 50% oxygen by
mass. About two thirds of the mass of the human body is due to oxygen in H2O. Elemental
oxygen, O2, is an odorless and colorless gas that makes up about 21% by volume of dry
air. In the liquid and solid states it is pale blue. Oxygen is only slightly soluble in water;
only about 0.04 gram dissolves in 1 liter of water at 25°C. This is sufficient to sustain fish
and other marine organisms. Oxygen is obtained commercially by the fractional distillation
of liquid air. The greatest single industrial use of O2 is for oxygen-enrichment in
blast furnaces for the conversion of pig iron (reduced, high-carbon iron) to steel.
Oxygen also exists in a second allotropic form, ozone, O3. Ozone is an unstable, pale
blue gas at room temperature. It is formed by passing an electrical discharge through
gaseous oxygen. Its unique, pungent odor is often noticed during electrical storms and in
the vicinity of electrical equipment. Not surprisingly, its density is about 1ᎏ1
2
ᎏ times that of
O2. At Ϫ112°C it condenses to a deep blue liquid. It is a very strong oxidizing agent. As
a concentrated gas or a liquid, ozone can easily decompose explosively:
2O3(g) 88n 3O2(g)
Oxygen atoms, or radicals, are intermediates in this exothermic decomposition of O3 to
O2. They act as strong oxidizing agents in such applications as destroying bacteria in water
purification.
The ozone molecule is angular (page 51). The two oxygen–oxygen bond lengths (1.28 Å)
are identical and are intermediate between typical single and double bond lengths.
Reactions of Oxygen and the Oxides
Oxygen forms oxides by direct combination with all other elements except the noble gases
and noble (unreactive) metals (Au, Pd, Pt). Oxides are binary compounds that contain
oxygen. Although such reactions are generally very exothermic, many proceed quite slowly
6-8
256 CHAPTER 6: Chemical Periodicity
and require heating to supply the energy necessary to break the strong bonds in O2 molecules.
Once these reactions are initiated, most release more than enough energy to be
self-sustaining and sometimes become “red hot.”
Reactions of O2 with Metals
In general, metallic oxides (including peroxides and superoxides) are ionic solids. The Group IA
metals combine with oxygen to form three kinds of solid ionic products called oxides,
peroxides, and superoxides. Lithium combines with oxygen to form lithium oxide.
4Li(s) ϩ O2(g) 88n 2Li2O(s) lithium oxide (mp Ͼ 1700°C)
By contrast, sodium reacts with an excess of oxygen to form sodium peroxide, Na2O2,
rather than sodium oxide, Na2O, as the major product.
2Na(s) ϩ O2(g) 88n Na2O2(g) sodium peroxide (decomposes at 460°C)
Peroxides contain the OXO2Ϫ, O2
2Ϫ group, in which the oxidation number of each
oxygen is Ϫ1, whereas normal oxides such as lithium oxide, Li2O, contain oxide ions, O2Ϫ.
The heavier members of the family (K, Rb, Cs) react with excess oxygen to form superoxides.
These contain the superoxide ion, O2
Ϫ, in which the oxidation number of each
oxygen is Ϫᎏ1
2
ᎏ. The reaction with K is
K(s) ϩ O2(g) 88n KO2(s) potassium superoxide (mp 430°C)
The tendency of the Group IA metals to form oxygen-rich compounds increases going
down the group. This is because cation radii increase going down the group. You can
recognize these classes of compounds as
Class Contains Ions Oxidation No. of Oxygen
normal oxides O2Ϫ Ϫ2
peroxides O2
2Ϫ Ϫ1
superoxides O2
Ϫ Ϫᎏ1
2
ᎏ
The Group IIA metals react with oxygen to form normal ionic oxides, MO, but at high
pressures of oxygen the heavier ones form ionic peroxides, MO2 (Table 6-4).
2M(s) ϩ O2(g) 88n 2(M2ϩ, O2Ϫ)(s) M ϭ Be, Mg, Ca, Sr, Ba
M(s) ϩ O2(g) 88n (M2ϩ, O2
2Ϫ)(s) M ϭ Ca, Sr, Ba
IA IIA
Li Na K Rb Cs Be Mg Ca Sr Ba
normal oxides Li2O Na2O K2O Rb2O Cs2O BeO MgO CaO SrO BaO
peroxides Li2O2 Na2O2 K2O2 Rb2O2 Cs2O2 CaO2 SrO2 BaO2
superoxides NaO2 KO2 RbO2 CsO2
TABLE 6-4 Oxygen Compounds of the IA and IIA Metals*
*The shaded compounds represent the principal products of the direct reaction of the metal with oxygen.
Beryllium reacts with oxygen only at
elevated temperatures and forms only
the normal oxide, BeO. The other
Group IIA metals form normal oxides
at moderate temperatures.
Figure 6-7 Iron powder burns
brilliantly to form iron(III) oxide,
Fe2O3.
6-8 Oxygen and the Oxides 257
For example, the equations for the reactions of calcium and oxygen are
2Ca(s) ϩ O2(g) 88n 2CaO(s) calcium oxide (mp 2580°C)
Ca(s) ϩ O2(g) 88n CaO2(s) calcium peroxide (decomposes at 275°C)
The other metals, with the exceptions noted previously (Au, Pd, and Pt), react with
oxygen to form solid metal oxides. Many metals to the right of Group IIA show variable
oxidation states, so they may form several oxides. For example, iron combines with oxygen
in the following series of reactions to form three different oxides (Figure 6-7).
heat
2Fe(s) ϩ O2(g) 888n 2FeO(s) iron(II) oxide
heat
6FeO(s) ϩ O2(g) 888n 2Fe3O4(s) magnetic iron oxide (a mixed oxide)
heat
4Fe3O4(s) ϩ O2(g) 888n 6Fe2O3(s) iron(III) oxide
Copper reacts with a limited amount of oxygen to form red Cu2O, whereas with excess
oxygen it forms black CuO.
heat
4Cu(s) ϩ O2(g) 888n 2Cu2O(s) copper(I) oxide
heat
2Cu(s) ϩ O2(g) 888n 2CuO(s) copper(II) oxide
Metals that exhibit variable oxidation states react with a limited amount of oxygen
to give oxides with lower oxidation states (such as FeO and Cu2O). They react with
an excess of oxygen to give oxides with higher oxidation states (such as Fe2O3 and
CuO).
Reactions of Metal Oxides with Water
Oxides of metals are called basic anhydrides (or basic oxides) because many of them
combine with water to form bases with no change in oxidation state of the metal (Figure
6-8). “Anhydride” means “without water”; in a sense, the metal oxide is a hydroxide base
with the water “removed.” Metal oxides that are soluble in water react to produce the
corresponding hydroxides.
Figure 6-8 The normal oxides of the representative elements in their maximum oxidation
states. Acidic oxides (acid anhydrides) are shaded red, amphoteric oxides are shaded purple,
and basic oxides (basic anhydrides) are shaded blue. An amphoteric oxide is one that shows
some acidic and some basic properties.
Increasingbasecharacter
Increasing acidic character
IA IIA IIIA IVA VA VIA VIIA
Li2O
Na2O
K2O Ga2O3 GeO2CaO
Al2O3 SiO2 P4O10 SO3
SeO3
Cl2O7MgO
Rb2O In2O3 SnO2 Sb2O5
As2O5
TeO3 I2O7
Br2O7
SrO
Cs2O Tl2O3 PbO2 Bi2O5 PoO3 At2O7BaO
B2O3 N2O5 OF2CO2BeO
Metal Oxide ؉ Water 88n Metal Hydroxide (base)
sodium oxide Na2O(s) ϩ H2O(ᐉ) 88n 2 NaOH(aq) sodium hydroxide
calcium oxide CaO(s) ϩ H2O(ᐉ) 88n Ca(OH)2(aq) calcium hydroxide
barium oxide BaO(s) ϩ H2O(ᐉ) 88n Ba(OH)2(aq) barium hydroxide
Recall that oxidation states are
indicated by red numbers in red
circles.
258 CHAPTER 6: Chemical Periodicity
ϩ2
ϩ4
ϩ3
Carbon burns brilliantly in pure O2
to form CO2.
The oxides of the Group IA metals and the heavier Group IIA metals dissolve in water
to give solutions of strong bases. Most other metal oxides are relatively insoluble in water.
Reactions of O2 with Nonmetals
Oxygen combines with many nonmetals to form molecular oxides. For example, carbon burns
in oxygen to form carbon monoxide or carbon dioxide, depending on the relative amounts
of carbon and oxygen.
2C(s) ϩ O2(g) 88n 2CO(s) (excess C and limited O2)
C(s) ϩ O2(g) 88n CO2(g) (limited C and excess O2)
Carbon monoxide is a very poisonous gas because it forms a stronger bond with the
iron atom in hemoglobin than does an oxygen molecule. Attachment of the CO molecule
to the iron atom destroys the ability of hemoglobin to pick up oxygen in the lungs and
carry it to the brain and muscle tissue. Carbon monoxide poisoning is particularly insidious
because the gas has no odor and because the victim first becomes drowsy.
Unlike carbon monoxide, carbon dioxide is not toxic. It is one of the products of the
respiratory process. It is used to make carbonated beverages, which are mostly saturated
solutions of carbon dioxide in water; a small amount of the carbon dioxide combines with
the water to form carbonic acid (H2CO3), a very weak acid.
Phosphorus reacts with a limited amount of oxygen to form tetraphosphorus hexoxide,
P4O6:
P4(s) ϩ 3O2(g) 88n P4O6(s) tetraphosphorus hexoxide
whereas reaction with an excess of oxygen gives tetraphosphorus decoxide, P4O10:
P4(s) ϩ 5O2(g) 88n P4O10(s) tetraphosphorus decoxide
Sulfur burns in oxygen to form primarily sulfur dioxide (Figure 6-9) and only very small
amounts of sulfur trioxide.
S8(s) ϩ 8O2(g) 88n 8SO2(g) sulfur dioxide (bp Ϫ10.0°C)
S8(s) ϩ 12O2(g) 88n 8SO3(ᐉ) sulfur trioxide (bp 43.4°C) The production of SO3 at a reasonable
rate requires the presence of a catalyst.
6-8 Oxygen and the Oxides 259
ϩ5
Nearly all nonmetals exhibit more
than one oxidation state in their
compounds. In general, the most common oxidation states of a nonmetal are (1) its periodic
group number, (2) its periodic group number minus two, and (3) its periodic group
number minus eight. The reactions of nonmetals with a limited amount of oxygen usually
give products that contain the nonmetals (other than oxygen) in lower oxidation states,
usually case (2). Reactions with excess oxygen give products in which the nonmetals
exhibit higher oxidation states, case (1). The examples we have cited are CO and CO2,
P4O6 and P4O10, and SO2 and SO3. The molecular formulas of the oxides are sometimes
not easily predictable, but the simplest formulas are. For example, the two most
common oxidation states of phosphorus in molecular compounds are ϩ3 and ϩ5. The
simplest formulas for the corresponding phosphorus oxides therefore are P2O3 and P2O5,
respectively. The molecular (true) formulas are twice these, P4O6 and P4O10.
OXIDATION STATES OF NONMETALS
ϩ4
ϩ6
Reactions of Nonmetal Oxides with Water
Nonmetal oxides are called acid anhydrides (or acidic oxides) because many of them
dissolve in water to form acids with no change in oxidation state of the nonmetal (see Figure
6-8). Several ternary acids can be prepared by reaction of the appropriate nonmetal oxides
with water. Ternary acids contain three elements, usually H, O, and another nonmetal.
Nonmetal
Oxide
؉ Water 88n Ternary Acid
carbon dioxide CO2(g) ϩ H2O(ᐉ) 88n H2CO3(aq) carbonic acid
sulfur dioxide SO2(g) ϩ H2O(ᐉ) 88n H2SO3(aq) sulfurous acid
sulfur trioxide SO3(ᐉ) ϩ H2O(ᐉ) 88n H2SO4(aq) sulfuric acid
dinitrogen pentoxide N2O5(s) ϩ H2O(ᐉ) 88n 2HNO3(aq) nitric acid
tetraphosphorus decoxide P4O10(s) ϩ 6H2O(ᐉ) 88n 4H3PO4(aq) phosphoric acid
ϩ4
ϩ4
ϩ6
ϩ5
ϩ5
ϩ4
ϩ4
ϩ6
ϩ5
ϩ5
Nearly all oxides of nonmetals react with water to give solutions of ternary acids. The
oxides of boron and silicon, which are insoluble, are two exceptions.
Figure 6-9 Sulfur burns in oxygen
to form sulfur dioxide.
260 CHAPTER 6: Chemical Periodicity
Metal Nonmetal
Oxide
؉
Oxide
88n Salt
calcium oxide ϩ
sulfur trioxide CaO(s) ϩ SO3(ᐉ) 88n CaSO4(s) calcium sulfate
magnesium oxide ϩ
carbon dioxide MgO(s) ϩ CO2(g) 88n MgCO3(s) magnesium carbonate
sodium oxide ϩ
tetraphosphorus decoxide 6Na2O(s) ϩ P4O10(s) 88n 4Na3PO4(s) sodium phosphate
ϩ2
ϩ2
ϩ1
ϩ6
ϩ4
ϩ5
ϩ2
ϩ2
ϩ1
ϩ6
ϩ4
ϩ5
EXAMPLE 6-8 Acidic Character of Oxides
Arrange the following oxides in order of increasing molecular (acidic) character: SO3, Cl2O7,
CaO, and PbO2.
Plan
Molecular (acidic) character of oxides increases as nonmetallic character of the element that is
combined with oxygen increases (see Figure 6-8).
increasing nonmetallic character
888888888888888888888888n
Ca Ͻ Pb Ͻ S Ͻ Cl
Periodic group: IIA IVA VIA VIIA
Solution
increasing molecular character
888888888888888888888888n
Thus, the order is CaO Ͻ PbO2 Ͻ SO3 Ͻ Cl2O7
EXAMPLE 6-9 Basic Character of Oxides
Arrange the oxides in Example 6-8 in order of increasing basicity.
Plan
The greater the molecular character of an oxide, the more acidic it is. Thus, the most basic
oxides have the least molecular (most ionic) character (see Figure 6-8).
Solution
increasing basic character
8888888888888888888n
molecular Cl2O7 Ͻ SO3 Ͻ PbO2 Ͻ CaO ionic
EXAMPLE 6-10 Predicting Reaction Products
Predict the products of the following pairs of reactants. Write a balanced equation for each
reaction.
(a) Cl2O7(ᐉ) ϩ H2O(ᐉ) 88n
Reactions of Metal Oxides with Nonmetal Oxides
Another common kind of reaction of oxides is the combination of metal oxides (basic anhydrides)
with nonmetal oxides (acid anhydrides), with no change in oxidation states, to form salts.
CaO is called quicklime. Ca(OH)2 is
called slaked lime.
6-8 Oxygen and the Oxides 261
heat
(b) As4(s) ϩ O2(g) (excess) 888n
heat
(c) Mg(s) ϩ O2(g) 888n
Plan
(a) The reaction of a nonmetal oxide (acid anhydride) with water forms a ternary acid in which
the nonmetal (Cl) has the same oxidation state (ϩ7) as in the oxide. Thus, the acid is perchloric
acid, HClO4.
(b) Arsenic, a Group VA nonmetal, exhibits common oxidation states of ϩ5 and ϩ5 Ϫ 2 ϭ ϩ3.
Reaction of arsenic with excess oxygen produces the higher-oxidation-state oxide, As2O5. By
analogy with the oxide of phosphorus in the ϩ5 oxidation state, P4O10, we might write the
formula as As4O10, but this oxide exists as As2O5.
(c) The reaction of a Group IIA metal with oxygen produces the normal metal oxide—MgO
in this case.
Solution
(a) Cl2O7(ᐉ) ϩ H2O(ᐉ) 88n 2HClO4(aq)
heat
(b) As4(s) ϩ 5O2(g) 888n 2As2O5(s)
heat
(c) 2Mg(s) ϩ O2(g) 888n 2MgO(s)
EXAMPLE 6-11 Predicting Reaction Products
Predict the products of the following pairs of reactants. Write a balanced equation for each
reaction.
(a) CaO(s) ϩ H2O(ᐉ) 88n
(b) Li2O(s) ϩ SO3(ᐉ) 88n
Plan
(a) The reaction of a metal oxide with water produces the metal hydroxide.
(b) The reaction of a metal oxide with a nonmetal oxide produces a salt containing the cation
of the metal oxide and the anion of the acid for which the nonmetal oxide is the anhydride.
SO3 is the acid anhydride of sulfuric acid, H2SO4.
Solution
(a) Calcium oxide reacts with water to form calcium hydroxide.
CaO(s) ϩ H2O(ᐉ) 88n Ca(OH)2(aq)
(b) Lithium oxide reacts with sulfur trioxide to form lithium sulfate.
Li2O(s) ϩ SO3(ᐉ) 88n Li2SO4(s)
You should now work Exercises 70–73.
Combustion Reactions
Combustion, or burning, is an oxidation–reduction reaction in which oxygen combines
rapidly with oxidizable materials in highly exothermic reactions, usually with a visible
flame. The complete combustion of hydrocarbons, in fossil fuels for example, produces
carbon dioxide and water (steam) as the major products:
Hydrocarbons are compounds that
contain only hydrogen and carbon.
Figure 6-10 Georgia Power
Company’s Plant Bowen at
Taylorsville, Georgia. Plants such as
this one burn more than 107 tons of
coal and produce over 2 ϫ 108
megawatt-hours of electricity each
year.
262 CHAPTER 6: Chemical Periodicity
Ϫ4 ϩ1 0 ϩ4 Ϫ2 ϩ1 Ϫ2
Ϫ2 ϩ1 0 ϩ4 Ϫ2 ϩ1 Ϫ2
heat
CH4(g) ϩ 2O2(g) 888n CO2(g) ϩ 2H2O(g) ϩ heat
excess
heat
C6H12(g) ϩ 9O2(g) 888n 6CO2(g) ϩ 6H2O(g) ϩ heat
cyclohexane excess
As we have seen, the origin of the term “oxidation” lies in just such reactions, in which
oxygen “oxidizes” another species.
Combustion of Fossil Fuels and Air Pollution
Fossil fuels are mixtures of variable composition that consist primarily of hydrocarbons.
We burn them to use the energy that is released rather than to obtain chemical products
(Figure 6-10). For example, burning octane, C8H18, in an excess of oxygen (plenty of air)
produces carbon dioxide and water. There are many similar compounds in gasoline and
diesel fuels.
2C8H18(ᐉ) ϩ 25O2(g) 88n 16CO2(g) ϩ 18H2O(ᐉ)
excess
Carbon monoxide is produced by the incomplete burning of carbon-containing
compounds in a limited amount of oxygen.
2C8H18(ᐉ) ϩ 17O2(g) 88n 16CO(g) ϩ 18H2O(ᐉ)
limited amount
In very limited oxygen, carbon (soot) is produced by partially burning hydrocarbons.
For octane, the reaction is
2C8H18(ᐉ) ϩ 9O2(g) 88n 16C(s) ϩ 18H2O(ᐉ)
very limited amount
When you see blue or black smoke (carbon) coming from an internal combustion engine,
(or smell unburned fuel in the air) you may be quite sure that lots of carbon monoxide is
also being produced and released into the air.
We see that the incomplete combustion of hydrocarbons yields undesirable products
—carbon monoxide and elemental carbon (soot), which pollute the air. Unfortunately, all
fossil fuels—natural gas, coal, gasoline, kerosene, oil, and so on—also have undesirable
nonhydrocarbon impurities that burn to produce oxides that act as additional air pollutants.
At this time it is not economically feasible to remove all of these impurities from
the fuels before burning them.
Fossil fuels result from the decay of animal and vegetable matter (Figure 6-11). All
living matter contains some sulfur and nitrogen, so fossil fuels also contain sulfur and
nitrogen impurities to varying degrees. Table 6-5 gives composition data for some common
kinds of coal.
Combustion of sulfur produces sulfur dioxide, SO2, probably the single most harmful
pollutant.
heat
S8(s) ϩ 8O2(g) 888n 8SO2(g)
Large amounts of SO2 are produced by the burning of sulfur-containing coal.
0 ϩ4
“White smoke” from auto exhaust
systems is tiny droplets of water
condensing in the cooler air.
Carbon in the form of soot is one of
the many kinds of particulate matter in
polluted air.
Figure 6-11 The luxuriant growth
of vegetation that occurred during
the carboniferous age is the source
of our coal deposits.
6-8 Oxygen and the Oxides 263
Many metals occur in nature as sulfides. The process of extracting the free (elemental)
metals involves roasting—heating an ore in the presence of air. For many metal sulfides
this produces a metal oxide and SO2. The metal oxides are then reduced to the free metals.
Consider lead sulfide, PbS, as an example.
2PbS(s) ϩ 3O2(g) 88n 2PbO(s) ϩ 2SO2(g)
Sulfur dioxide is corrosive; it damages plants, structural materials, and humans. It is a
nasal, throat, and lung irritant. Sulfur dioxide is slowly oxidized to sulfur trioxide, SO3,
by oxygen in air:
2SO2(g) ϩ O2(g) 88n 2SO3(ᐉ)
Sulfur trioxide combines with moisture in the air to form the strong, corrosive acid, sulfuric
acid:
SO3(ᐉ) ϩ H2O(ᐉ) 88n H2SO4(ᐉ)
Oxides of sulfur are the main cause of acid rain.
Compounds of nitrogen are also impurities in fossil fuels; they burn to form nitric
oxide, NO. Most of the nitrogen in the NO in exhaust gases from furnaces, automobiles,
airplanes, and so on, however, comes from the air that is mixed with the fuel.
N2(g) ϩ O2(g) 88n 2NO(g)
NO can be further oxidized by oxygen to nitrogen dioxide, NO2; this reaction is enhanced
in the presence of ultraviolet light from the sun.
UV
2NO(g) ϩ O2(g) 88n 2NO2(g) (a reddish-brown gas)
light
NO2 is responsible for the reddish-brown haze that hangs over many cities on sunny afternoons
(Figure 6-12) and probably for most of the respiratory problems associated with
this kind of air pollution. It can react to produce other oxides of nitrogen and other
secondary pollutants.
In addition to being a pollutant itself, nitrogen dioxide reacts with water in the air to
form nitric acid, another major contributor to acid rain:
3NO2(g) ϩ H2O(ᐉ) 88n 2HNO3(aq) ϩ NO(g)
0 ϩ2
ϩ2 ϩ4
TABLE 6-5 Some Typical Coal Compositions in
Percent (dry, ash-free)
C H O N S
lignite 70.59 4.47 23.13 1.04 0.74
subbituminous 77.2 5.01 15.92 1.30 0.51
bituminous 80.2 5.80 7.53 1.39 5.11
anthracite 92.7 2.80 2.70 1.00 0.90
Remember that “clean air” is about
80% N2 and 20% O2 by mass. This
reaction does not occur at room
temperature but does occur at the high
temperatures of furnaces, internal
combustion engines, and jet engines.
Figure 6-12 Photochemical
pollution (a brown haze) enveloping
a city.
264 CHAPTER 6: Chemical Periodicity
Key Terms
Acid anhydride A nonmetal oxide that reacts with water to form
an acid.
Acidic oxide See Acid anhydride.
Actinides Elements 90 through 103 (after actinium).
Amphoteric oxide An oxide that shows some acidic and some
basic properties.
Amphoterism The ability of a substance to react with both acids
and bases.
Angstrom (Å) 10Ϫ10 meter, 10Ϫ1 nm, or 102 pm.
Atomic radius The radius of an atom.
Basic anhydride A metal oxide that reacts with water to form a
base.
Basic oxide See Basic anhydride.
Catalyst A substance that speeds up a chemical reaction without
itself being consumed in the reaction.
Combustion reaction The reaction of a substance with oxygen
in a highly exothermic reaction, usually with a visible flame.
d-Transition elements (metals) The B group elements in the
periodic table; sometimes called simply transition elements.
Effective nuclear charge (Zeff) The nuclear charge experienced
by the outermost electrons of an atom; the actual nuclear charge
minus the effects of shielding due to inner shell electrons.
Electron affinity The amount of energy absorbed in the process
in which an electron is added to a neutral isolated gaseous atom
to form a gaseous ion with a 1Ϫ charge; has a negative value if
energy is released.
Electronegativity A measure of the relative tendency of an atom
to attract electrons to itself when chemically combined with
another atom.
f-Transition elements (metals) Elements 58 through 71 and 90
through 103; also called inner transition elements (metals).
Hydride A binary compound of hydrogen.
Inner transition elements See f-Transition elements.
Ionic radius The radius of an ion.
Ionization energy The amount of energy required to remove
the most loosely held electron of an isolated gaseous atom or
ion.
Isoelectronic Having the same number of electrons.
Lanthanides Elements 58 through 71 (after lanthanum).
Noble gas configuration The stable electron configuration of
a noble gas.
Noble gases Elements of periodic Group VIIIA; also called rare
gases; formerly called inert gases.
Normal oxide A metal oxide containing the oxide ion, O2Ϫ (oxygen
in the Ϫ2 oxidation state).
Nuclear shielding See Shielding effect.
Oxide A binary compound of oxygen.
Periodicity Regular periodic variations of properties of elements
with atomic number and position in the periodic table.
Periodic law The properties of the elements are periodic functions
of their atomic numbers.
Peroxide A compound containing oxygen in the Ϫ1 oxidation
state. Metal peroxides contain the peroxide ion, O2
2Ϫ.
Radical A species containing one or more unpaired electrons;
many radicals are very reactive.
Rare earths Inner transition elements.
Representative elements The A group elements in the periodic
table.
Exercises 265
Roasting Heating an ore of an element in the presence of air.
Shielding effect Electrons in filled sets of s and p orbitals
between the nucleus and outer shell electrons shield the outer
shell electrons somewhat from the effect of protons in the
nucleus; also called screening effect.
Superoxide A compound containing the superoxide ion, O2
Ϫ
(oxygen in the Ϫᎏ1
2
ᎏ oxidation state).
Ternary acid An acid containing three elements: H, O, and (usually)
another nonmetal.
Exercises
Classification of the Elements
*01. Define and illustrate the following terms clearly and
concisely: (a) representative elements; (b) d-transition elements;
(c) inner transition elements.
*02. Explain why Period 1 contains two elements and Period 2
contains eight elements.
*03. Explain why Period 4 contains 18 elements.
*04. The third shell (n ϭ 3) has s, p, and d subshells. Why does
Period 3 contain only eight elements?
*05. Account for the number of elements in Period 6.
0*6. What would be the atomic number of the as-yetundiscovered
alkaline earth element of Period 8?
*07. Identify the group, family, or other periodic table location
of each element with the outer electron configuration
(a) ns2np4
(b) ns2
(c) ns2(n Ϫ 1)d0Ϫ2(n Ϫ 2)f1Ϫ14
*08. Repeat Exercise 7 for
(a) ns2np6
(b) ns1
(c) ns2(n Ϫ 1)d1Ϫ10
(d) ns2np1
*09. Write the outer electron configurations for the (a) alkaline
earth metals; (b) first column of d-transition metals; and
(c) halogens.
*10. Which of the elements in the following periodic table is
(are) (a) alkali metals; (b) an element with the outer configuration
of d7s2; (c) lanthanides; (d) p-block representative
elements; (e) elements with partially filled f-subshells;
(f ) halogens; (g) s-block representative elements;
(h) actinides; (i) d-transition elements; (j) noble gases?
*11. Identify the elements and the part of the periodic table in
which the elements with the following configurations are
found.
(a) 1s22s22p63s23p64s2
(b) [Kr] 4d85s2
(c) [Xe] 4f 145d56s1
(d) [Xe] 4f 126s2
(e) [Kr] 4d105s25p3
(f ) [Kr] 4d104f 145s25p65d106s26p2
*12. Which of the following species are iso-electronic?
O2Ϫ, FϪ, Ne, Na, Mg2ϩ, Al3ϩ
*13. Which of the following species are iso-electronic?
P3ϩ, S2Ϫ, ClϪ, Ar, Kϩ, Caϩ
Atomic Radii
*14. What is meant by nuclear shielding? What effect does it
have on trends in atomic radii?
*15. Why do atomic radii decrease from left to right within a
period in the periodic table?
*16. Consider the elements in Group VIA. Even though it has
not been isolated or synthesized, what can be predicted
about the ionic radius of element number 116?
*17. Variations in the atomic radii of the transition elements are
not so pronounced as those of the representative elements.
Why?
*18. Arrange each of the following sets of atoms in order of
increasing atomic radii: (a) the alkaline earth elements;
(b) the noble gases; (c) the representative elements in the
third period; (d) N, Ba, B, Sr, and Sb.
*19. Arrange each of the following sets of atoms in order of
increasing atomic volume: (a) O, Mg, Al, Si; (b) O, S, Se,
Te; (c) Ca, Sr, Ga, As
Ionization Energy
*20. Define (a) first ionization energy and (b) second ionization
energy.
*21. Why is the second ionization energy for a given element
always greater than the first ionization energy?
*22. What is the usual relationship between atomic radius and
first ionization energy, other factors being equal?
*23. What is the usual relationship between nuclear charge and
first ionization energy, other factors being equal?
*24. Going across a period on the periodic table, what is the
relationship between shielding and first ionization energy?
B
J
K
A
D
H G
C
E
IF
266 CHAPTER 6: Chemical Periodicity
*25. Within a group on the periodic table, what is the relationship
between shielding and first ionization energy?
*26. Arrange the members of each of the following sets of
elements in order of increasing first ionization energies:
(a) the alkali metals; (b) the halogens; (c) the elements in
the second period; (d) Br, F, B, Ga, Cs, and H.
*27. The second ionization energy value is generally much
larger than the first ionization energy. Explain why this is so.
*28. What is the general relationship between the sizes of the
atoms of Period 2 and their first ionization energies? Rationalize
the relationship.
*29. In a plot of first ionization energy versus atomic number
for Periods 2 and 3, “dips” occur at the IIIA and VIA elements.
Account for these dips.
*30. On the basis of electron configurations, would you expect
a Na2ϩ ion to exist in compounds? Why or why not? How
about Mg2ϩ?
*31. How much energy, in kilojoules, must be absorbed by
1.25 mol of gaseous lithium atoms to convert all of them
to gaseous Liϩ ions?
Electron Affinity
*32. Arrange the following elements in order of increasing negative
values of electron affinity: P, S, Cl, and Br.
*33. Arrange the members of each of the following sets of elements
in order of increasingly negative electron affinities:
(a) the Group IA metals; (b) the Group VIIA elements;
(c) the elements in the second period; (d) Li, K, C, F,
and Cl.
*34. The electron affinities of the halogens are much more negative
than those of the Group VIA elements. Why is this so?
*35. The addition of a second electron to form an ion with a
2Ϫ charge is always endothermic. Why is this so?
*36. Write the equation for the change described by each of the
following, and write the electron configuration for each
atom or ion shown: (a) the electron affinity of oxygen;
(b) the electron affinity of chlorine; (c) the electron affinity
of magnesium.
Ionic Radii
*37. Compare the sizes of cations and the neutral atoms from
which they are formed by citing three specific examples.
*38. Arrange the members of each of the following sets of
cations in order of increasing ionic radii: (a) Kϩ, Ca2ϩ,
Ga3ϩ; (b) Ca2ϩ, Be2ϩ, Ba2ϩ, Mg2ϩ; (c) Al3ϩ, Sr2ϩ, Rbϩ,
Kϩ; (d) Kϩ, Ca2ϩ, Rbϩ.
*39. Compare the sizes of anions and the neutral atoms from
which they are formed by citing three specific examples.
*40. Arrange the following sets of anions in order of increasing
ionic radii: (a) ClϪ, S2Ϫ, P3Ϫ; (b) O2Ϫ, S2Ϫ, Se2Ϫ; (c) N3Ϫ,
S2Ϫ, BrϪ, P3Ϫ; (d) ClϪ, BrϪ, IϪ.
*41. Explain the trend in size of either the atom or ion as one
moves down a group.
*42. Most transition metals can form more than one simple positive
ion. For example, iron forms both Fe2ϩ and Fe3ϩ ions,
and tin forms both Sn2ϩ and Sn4ϩ ions. Which is the
smaller ion of each pair, and why?
Electronegativity
*43. What is electronegativity?
*44. Arrange the members of each of the following sets of elements
in order of increasing electronegativities: (a) Pb, C,
Sn, Ge; (b) S, Na, Mg, Cl; (c) P, N, Sb, Bi; (d) Se, Ba, F,
Si, Sc.
*45. Which of the following statements is better? Why?
(a) Magnesium has a weak attraction for electrons in a
chemical bond because it has a low electronegativity.
(b) The electronegativity of magnesium is low because
magnesium has a weak attraction for electrons in a chemical
bond.
*46. Some of the second-period elements show a similarity to
the element one column to the right and one row down.
For instance, Li is similar in many respects to Mg, and Be
is similar to Al. This has been attributed to the charge density
on the stable ions (Liϩ vs. Mg2ϩ; Be2ϩ vs. Al3ϩ). From
the values of electronic charge (Chapter 5) and ionic radii
(Chapter 6), calculate the charge density for these four ions,
in coulombs per cubic angstrom.
*47. One element takes on only a negative oxidation number
when combined with other elements. From the table of
electronegativity values, determine which element this is.
*48. Compare the electronegativity values of the metalloids.
Additional Exercises on the Periodic Table
*49. The chemical reactivities of carbon and lead are similar,
but there are also major differences. Using their electron
configurations, explain why these similarities and differences
may exist.
*50. The bond lengths in F2 and Cl2 molecules are 1.42 Å and
1.98 Å, respectively. Calculate the atomic radii for these
elements. Predict the ClXF bond length. (The actual
ClXF bond length is 1.64 Å.)
*51. The atoms in crystalline nickel are arranged so that they
touch one another in a plane as shown in the sketch:
r
r
r
a
r
Exercises 267
From plane geometry, we can see that 4r ϭ a͙2ෆ.
Calculate the radius of a nickel atom given that a ϭ
3.5238 Å.
*52. Compare the respective values of the first ionization energy
(see Table 6-1) and electron affinity (see Table 6-2) for several
elements. Which energy is greater? Why?
*53. Compare the respective values of the first ionization energy
(see Table 6-1) and electron affinity (see Table 6-2) for
nitrogen to those for carbon and oxygen. Explain why the
nitrogen values are considerably different.
*54. Based on general trends, the electron affinity of fluorine
would be expected to be greater than that of chlorine; however,
the value is less and is similar to the value for bromine.
Explain.
*55. Atomic number 91 is the element protactinium, an
extremely rare element. There is very little known about
protactinium; even the boiling point is unknown. Thorium,
atomic number 90, has the boiling point of 4788°C and
uranium’s boiling point is 4131°C. Predict the boiling point
of protactinium, and discuss how close the prediction is
likely to be.
Hydrogen and the Hydrides
*56. Summarize the physical properties of hydrogen.
*57. Write balanced formula unit equations for (a) the reaction
of iron with steam, (b) the reaction of calcium with
hydrochloric acid, (c) the electrolysis of water, and (d) the
“water gas” reaction.
*58. Write a balanced formula unit equation for the preparation
of (a) an ionic hydride and (b) a molecular hydride.
*59. Classify the following hydrides as molecular or ionic:
(a) NaH, (b) H2S, (c) AlH3, (d) RbH, (e) NH3.
*60. Explain why NaH and H2S are different kinds of hydrides.
*61. Write formula unit equations for the reactions of (a) NaH
and (b) AlH3 with water.
*62. Name the following (pure) compounds: (a) H2S, (b) HCl,
(c) KH, (d) NH3, (e) H2Se, (f) MgH2, (g) CaH2.
Oxygen and the Oxides
*63. How are O2 and O3 similar? How are they different?
*64. Briefly compare and contrast the properties of oxygen with
those of hydrogen.
*65. Write molecular equations to show how oxygen can be prepared
from (a) mercury(II) oxide, HgO, (b) hydrogen
peroxide, H2O2, and (c) potassium chlorate, KClO3.
*66. Which of the following elements form normal oxides as
the major products of reactions with oxygen? (a) Li, (b) Na,
(c) Rb, (d) Mg, (e) Zn (exhibits only one common oxidation
state), (f) Al.
*67. Oxygen has a positive oxidation number when combined
with which element? Compare the electronegativity values
of oxygen and that element.
*68. Write formula unit equations for the reactions of the following
elements with a limited amount of oxygen: (a) C,
(b) As4, (c) Ge.
*69. Write formula unit equations for the reactions of the
following elements with an excess of oxygen: (a) C, (b) As4,
(c) Ge.
*70. Distinguish among normal oxides, peroxides, and superoxides.
What is the oxidation state of oxygen in each
case?
*71. Which of the following can be classified as basic anhydrides?
(a) CO2, (b) Li2O, (c) SeO3, (d) CaO, (e) N2O5.
*72. Write balanced formula unit equations for the following
reactions and name the products:
(a) carbon dioxide, CO2, with water
(b) sulfur trioxide, SO3, with water
(c) selenium trioxide, SeO3, with water
(d) dinitrogen pentoxide, N2O5, with water
(e) dichlorine heptoxide, Cl2O7, with water
*73. Write balanced formula unit equations for the following
reactions and name the products:
(a) sodium oxide, Na2O, with water
(b) calcium oxide, CaO, with water
(c) lithium oxide, Li2O, with water
(d) magnesium oxide, MgO, with sulfur dioxide, SO2
(e) calcium oxide, CaO, with carbon dioxide, CO2
*74. Identify the acid anhydrides of the following ternary acids:
(a) H2SO4, (b) H2CO3, (c) H2SO3, (d) H3PO4, (e) HNO2.
*75. Identify the basic anhydrides of the following metal
hydroxides: (a) NaOH, (b) Mg(OH)2, (c) Fe(OH)2,
(d) Al(OH)3.
Combustion Reactions
*76. Define combustion. Why are all combustion reactions also
redox reactions?
*77. Write equations for the complete combustion of the following
compounds: (a) methane, CH4(g); (b) propane,
C3H8(g); (c) ethanol, C2H5OH(ᐉ).
*78. Write equations for the incomplete combustion of the
following compounds to produce carbon monoxide:
(a) methane, CH4(g); (b) propane, C3H8(g).
As we have seen, two substances may react to form different products
when they are mixed in different proportions under different
conditions. In Exercises 79 and 80, write balanced equations for
the reactions described. Assign oxidation numbers.
*79. (a) Ethane burns in excess air to form carbon dioxide and
water.
(b) Ethane burns in a limited amount of air to form carbon
monoxide and water.
(c) Ethane burns (poorly) in a very limited amount of air
to form elemental carbon and water.
*80. (a) Butane (C4H10) burns in excess air to form carbon dioxide
and water.
partial replacement of calcium by magnesium in an original
limestone sediment. Is this explanation reasonable,
given what you know of the ionic radii of magnesium and
calcium ions?
BUILDING YOUR KNOWLEDGE
*88. Hydrogen can be obtained from water by electrolysis.
Hydrogen may someday be an important replacement
for current fuels. Describe some of the problems that
you would predict if hydrogen were used in today’s motor
vehicles.
*89. The only chemically stable ion of rubidium is Rbϩ. The
most stable monatomic ion of bromine is BrϪ. Krypton
(Kr) is among the least reactive of all elements. Compare
the electron configurations of Rbϩ, BrϪ, and Kr. Then predict
the most stable monatomic ions of strontium (Sr) and
selenium (Se).
*90. The first ionization energy of potassium, K, is 419 kJ/mol.
What is the minimum frequency of light required to ionize
gaseous potassium atoms?
*91. Potassium and argon would be anomalies in a periodic table
in which elements are arranged in order of increasing
atomic weights. Identify two other elements among the
transition elements whose positions in the periodic table
would have been reversed in a “weight-sequence” arrangement.
Which pair of elements would most obviously be
out of place on the basis of their chemical behavior?
Explain your answer in terms of the current atomic model,
showing electron configurations for these elements.
*92. The second ionization energy for magnesium is 1451 kJ/
mol. How much energy, in kilojoules, must be absorbed by
1.50 g of gaseous magnesium atoms to convert them to
gaseous Mg2ϩ ions?
(b) Butane burns in a limited amount of air to form carbon
monoxide and water.
(c) When heated in the presence of very little air, butane
“cracks” to form acetylene, C2H2; carbon monoxide; and
hydrogen.
*81. (a) How much SO2 would be formed by burning 1.00 kg
of bituminous coal that is 5.15% sulfur by mass? Assume
that all of the sulfur is converted to SO2.
(b) If 27.0% of the SO2 escaped into the atmosphere and
75.0% of the escaped SO2 were converted to H2SO4, how
many grams of H2SO4 would be produced in the atmo-
sphere?
*82. Write equations for the complete combustion of the
following compounds. Assume that sulfur is converted to
SO2 and nitrogen is converted to NO. (a) C6H5NH2(ᐉ),
(b) C2H5SH(ᐉ), (c) C7H10NO2S(ᐉ).
CONCEPTUAL EXERCISES
*83. Write the electron configuration for the product of the second
ionization of the third largest alkaline earth metal.
*84. You are given the atomic radii of 110 pm, 118 pm, 120 pm,
122 pm, and 135 pm, but do not know to which element
(As, Ga, Ge, P, and Si) these values correspond. Which
must be the value of Ge?
*85. What compound will most likely form between potassium
and element X, if element X has the electronic configuration
1s22s22p63s23p4?
*86. Write the electron configurations of beryllium and
magnesium. What similarities in their chemical properties
can you predict on the basis of their electron configu-
rations?
*87. Dolostone is often more porous than limestone. One explanation
of the origin of dolostone is that it results from
268 CHAPTER 6: Chemical Periodicity
7Chemical
Bonding
OUTLINE
7-1 Lewis Dot Formulas of Atoms
Ionic Bonding
7-2 Formation of Ionic Compounds
Covalent Bonding
7-3 Formation of Covalent Bonds
7-4 Lewis Formulas for Molecules
and Polyatomic Ions
7-5 The Octet Rule
7-6 Resonance
7-7 Limitations of the Octet Rule
for Lewis Formulas
7-8 Polar and Nonpolar Covalent
Bonds
7-9 Dipole Moments
7-10 The Continuous Range of
Bonding Types
OBJECTIVES
After you have studied this chapter, you should be able to
• Write Lewis dot representations of atoms
• Predict whether bonding between specified elements will be primarily ionic, covalent, or
polar covalent
• Compare and contrast characteristics of ionic and covalent compounds
• Describe how the properties of compounds depend on their bonding
• Describe how the elements bond by electron transfer (ionic bonding)
• Describe energy relationships in ionic compounds
• Predict the formulas of ionic compounds
• Describe how elements bond by sharing electrons (covalent bonding)
• Write Lewis dot and dash formulas for molecules and polyatomic ions
• Recognize exceptions to the octet rule
• Write formal charges for atoms in covalent structures
• Describe resonance, and know when to write resonance structures and how
to do so
• Relate the nature of bonding to electronegativity differences
Carbon atoms are covalently bonded
together in a three-dimensional
array to make diamond, the hardest
substance known.
CC
hemical bonding refers to the attractive forces that hold atoms together in
compounds. There are two major classes of bonding. (1) Ionic bonding results
from electrostatic interactions among ions, which often results from the net
transfer of one or more electrons from one atom or group of atoms to another. (2) Covalent
bonding results from sharing one or more electron pairs between two atoms. These
two classes represent two extremes; all bonds between atoms of different elements have
at least some degree of both ionic and covalent character. Compounds containing predominantly
ionic bonding are called ionic compounds. Those that are held together mainly
by covalent bonds are called covalent compounds. Some nonmetallic elements, such as
H2, Cl2, N2, and P4, also involve covalent bonding. Some properties usually associated
with many simple ionic and covalent compounds are summarized in the following list.
The differences in these properties can be accounted for by the differences in bonding
between the atoms or ions.
Ionic Compounds Covalent Compounds
1. They are solids with high melting 1. They are gases, liquids, or solids with
2. points (typically Ͼ400°C). 2. low melting points (typically Ͻ300°C).
2. Many are soluble in polar solvents 2. Many are insoluble in polar solvents.
2. such as water.
3. Most are insoluble in nonpolar 3. Most are soluble in nonpolar solvents,
2. solvents, such as hexane, C6H14, and 2. such as hexane, C6H14, and carbon
2. carbon tetrachloride, CCl4. 2. tetrachloride, CCl4.
4. Molten compounds conduct 4. Liquid and molten compounds do not
2. electricity well because they contain 2. conduct electricity.
2. mobile charged particles (ions).
5. Aqueous solutions conduct electricity 5. Aqueous solutions are usually poor
2. well because they contain mobile 2. conductors of electricity because most
2. charged particles (ions). 2. do not contain charged particles.
6. They are often formed between two 6. They are often formed between two
2. elements with quite different 2. elements with similar electronega-
2. electronegativities, usually a metal 2. tivities, usually nonmetals.
2. and a nonmetal. 2.
LEWIS DOT FORMULAS OF ATOMS
The number and arrangements of electrons in the outermost shells of atoms determine
the chemical and physical properties of the elements as well as the kinds of chemical bonds
they form. We write Lewis dot formulas (or Lewis dot representations, or just Lewis
formulas) as a convenient bookkeeping method for keeping track of these “chemically
important electrons.” We now introduce this method for atoms of elements; in our discussion
of chemical bonding in subsequent sections, we will frequently use such formulas for
atoms, molecules, and ions.
7-1
As you study Chapters 7 and 8, keep in
mind the periodic similarities that you
learned in Chapters 4 and 6. What you
learn about the bonding of an element
usually applies to the other elements in
the same column of the periodic table,
with minor variations.
7-1 Lewis Dot Formulas of Atoms 271
As we saw in Section 4-2, aqueous
solutions of some covalent compounds
do conduct electricity, because they
react with water to some extent to
form ions.
The distinction between polar and
nonpolar molecules is made in Section
7-8.
Chemical bonding usually involves only the outermost electrons of atoms, also called
valence electrons. In Lewis dot representations, only the electrons in the outermost occupied
s and p orbitals are shown as dots. Paired and unpaired electrons are also indicated.
Table 7-1 shows Lewis dot formulas for the representative elements. All elements in a
given group have the same outer-shell electron configuration. It is somewhat arbitrary on
which side of the atom symbol we write the electron dots. We do, however, represent an
electron pair as a pair of dots and an unpaired electron as a single dot.
IONIC BONDING
FORMATION OF IONIC COMPOUNDS
The first kind of chemical bonding we shall describe is ionic bonding. We recall (Section
2-3) that an ion is an atom or a group of atoms that carries an electrical charge. An ion
in which the atom or group of atoms has fewer electrons than protons is positively charged,
and is called a cation; one that has more electrons than protons is negatively charged,
and is called an anion. An ion that consists of only one atom is described as a monatomic
ion. Examples include the chloride ion, ClϪ, and the magnesium ion, Mg2ϩ. An ion that
contains more than one atom is called a polyatomic ion. Examples include the ammonium
ion, NH4
ϩ; the hydroxide ion, OHϪ; and the sulfate ion, SO4
2Ϫ. The atoms of a
polyatomic ion are held together by covalent bonds. In this section we shall discuss how
ions can be formed from individual atoms; polyatomic ions will be discussed along with
other covalently bonded species.
Ionic bonding is the attraction of oppositely charged ions (cations and anions) in
large numbers to form a solid. Such a solid compound is called an ionic solid.
7-2
The chemical and physical properties
of an ion are quite different from those
of the atom from which the ion is
derived. For example, an atom of Na
and an Naϩ ion are quite different.
Lewis Dot Formulas for Representative Elements
Group
Number of
electrons in
valence shell
IA
H
Li
Na
K
Rb
Cs
Fr
1
IIA
Be
Mg
Ca
Sr
Ba
Ra
2
IIIA
B
Al
Ga
In
Tl
3
Period 7
Period 6
Period 5
Period 4
Period 3
Period 2
Period 1
IVA
4
C
Si
Ge
Sn
Pb
VA
5
N
P
As
Sb
Bi
VIA
6
O
S
Se
Te
Po
F
Cl
Br
I
At
VIIA
7
TABLE 7-1TABLE 7-1
VIIIA
8
(except
He)
He
Ne
Ar
Kr
Xe
Rn
For the Group A elements, the
number of valence electrons (dots in
the Lewis formula) for the neutral
atom is equal to the group number.
Exceptions: H (one valence electron)
and He (two valence electrons).
For example, Al has the electron
configuration [Ar] 3s23p1. The three
dots in the Lewis dot formula for
Al represent the two s electrons (the
pair of dots) and the p electron (the
single dot) beyond the noble gas
configuration. Because of the large
numbers of dots, such formulas are not
as useful for the transition and inner
transition elements.
As our previous discussions of ionization energy, electronegativity, and electron affinity
would suggest, ionic bonding can occur easily when elements that have low ionization
energies (metals) react with elements having high electronegativities and very negative
electron affinities (nonmetals). Many metals are easily oxidized—that is, they lose electrons
to form cations; and many nonmetals are readily reduced—that is, they gain electrons
to form anions.
When the electronegativity difference, ⌬(EN), between two elements is large, as
between a metal and a nonmetal, the elements are likely to form a compound by
ionic bonding (transfer of electrons).
Let us describe some combinations of metals with nonmetals to form ionic compounds.
Group IA Metals and Group VIIA Nonmetals
Consider the reaction of sodium (a Group IA metal) with chlorine (a Group VIIA
nonmetal). Sodium is a soft silvery metal (mp 98°C), and chlorine is a yellowish-green
corrosive gas at room temperature. Both sodium and chlorine react with water, sodium
vigorously. By contrast, sodium chloride is a white solid (mp 801°C) that dissolves in water
with no reaction and with the absorption of just a little heat. We can represent the reaction
for its formation as
2Na(s) ϩ Cl2(g) 88n 2NaCl(s)
sodium chlorine sodium chloride
We can understand this reaction better by showing electron configurations for all species.
We represent chlorine as individual atoms rather than molecules, for simplicity.
11Na [Ne] __h Naϩ [Ne] 1eϪ lost
3s
88n
17Cl [Ne] __hg __hg __hg __h ClϪ [Ne] __hg __hg __hg __hg 1eϪ gained
3s 3p 3s 3p
Freshly cut sodium has a metallic
luster. A little while after being cut,
the sodium metal surface turns white
as it reacts with the air.
7-2 Formation of Ionic Compounds 273


















Some ionic compounds. Clockwise
from front right: sodium chloride
(NaCl, white); copper(II) sulfate
pentahydrate (CuSO4и5H2O, blue);
nickel(II) chloride hexahydrate
(NiCl2и6H2O, green); potassium
dichromate (K2Cr2O7, orange); and
cobalt(II) chloride hexahydrate
(CoCl2и6H2O, red). One mole of
each substance is shown.
In this reaction, Na atoms lose one electron each to form Naϩ ions, which contain only
ten electrons, the same number as the preceding noble gas, neon. We say that sodium ions
have the neon electronic structure: Naϩ is isoelectronic with Ne (Section 6-5). In contrast,
Cl atoms gain one electron each to form ClϪ ions, which contain 18 electrons. This is
the same number as the following noble gas, argon; ClϪ is isoelectronic with Ar. These
processes can be represented compactly as
Na 88n Naϩ ϩ eϪ and Cl ϩ eϪ 88n ClϪ
Similar observations apply to most ionic compounds formed by reactions between representative
metals and representative nonmetals.
We can use Lewis dot formulas (Section 7-1) to represent the reaction.
The formula for sodium chloride is NaCl because the compound contains Naϩ and
ClϪ in a 1:1 ratio. This is the formula we predict based on the fact that each Na atom
contains only one electron in its outermost occupied shell and each Cl atom needs only
one electron to fill completely its outermost p orbitals.
The chemical formula NaCl does not explicitly indicate the ionic nature of the
compound, only the ratio of ions. Furthermore, values of electronegativities are not always
available. So we must learn to recognize, from positions of elements in the periodic table
and known trends in electronegativity, when the difference in electronegativity is large
enough to favor ionic bonding.
The farther apart across the periodic table two Group A elements are, the more
ionic their bonding will be.
The greatest difference in electronegativity occurs from the lower left to upper right, so
CsF (⌬[EN] ϭ 3.2) is more ionic than LiI (⌬[EN] ϭ 1.5).
All the Group IA metals (Li, Na, K, Rb, Cs) will react with the Group VIIA elements
(F, Cl, Br, I) to form ionic compounds of the same general formula, MX. All the resulting
ions, Mϩ and XϪ, have noble gas configurations. Once we understand the bonding of one
member of a group (column) in the periodic table, we know a great deal about the others
in the family. Combining each of the five common alkali metals with each of the four
common halogens gives 5 ϫ 4 ϭ 20 possible compounds. The discussion of NaCl
presented here applies also to the other 19 such compounds.
The collection of isolated positive and negative ions occurs at a higher energy than the
elements from which they are formed. The ion formation alone is not sufficient to account
for the formation of ionic compounds. Some other favorable factor must account for the
observed stability of these compounds. Because of the opposite charges on Naϩ and ClϪ,
an attractive force is developed. According to Coulomb’s Law, the force of attraction, F,
between two oppositely charged particles of charge magnitudes qϩ and qϪ is directly
proportional to the product of the charges and inversely proportional to the square of the
distance separating their centers, d. Thus, the greater the charges on the ions and the
smaller the ions are, the stronger the resulting ionic bonding. Of course, like-charged ions
repel each other, so the distances separating the ions in ionic solids are those at which the
attractions exceed the repulsions by the greatest amount.
The energy associated with the attraction of separated gaseous positive and negative
ions to form an ionic solid is the crystal lattice energy of the solid. For NaCl, this energy
ClNa Na
ϩ
[ Clϩ ]
Ϫ
The loss of electrons is oxidation
(Section 4-5). Na atoms are oxidized to
form Naϩ ions.
The gain of electrons is reduction
(Section 4-5). Cl atoms are reduced to
form ClϪ ions.
274 CHAPTER 7: Chemical Bonding
Formulas for some of these are
LiF NaF KF
LiCl NaCl KCl
LiBr NaBr KBr
LiI NaI KI
Coulomb’s Law is F ϰ ᎏ
qϩ
d
q
2
Ϫ
ᎏ. The
symbol ϰ means “is proportional to.”
The noble gases are excluded from this
generalization.
is Ϫ789 kJ/mol; that is, one mole of NaCl solid is 789 kJ lower in energy (more stable)
than one mole of isolated Naϩ ions and one mole of isolated ClϪ ions. We could also say
that it would require 789 kJ of energy to separate one mole of NaCl solid into isolated
gaseous ions. The stability of ionic compounds is thus due to the interplay of the energy
cost of ion formation and the energy repaid by the crystal lattice energy. The best tradeoff
usually comes when the monatomic ions of representative elements have noble gas
configurations.
The structure of common table salt, sodium chloride (NaCl), is shown in Figure 7-1.
Like other simple ionic compounds, NaCl(s) exists in a regular, extended array of positive
and negative ions, Naϩ and ClϪ. Distinct molecules of solid ionic substances do not
exist, so we must refer to formula units (Section 2-3) instead of molecules. The forces that
hold all the particles together in an ionic solid are quite strong. This explains why such
substances have quite high melting and boiling points (a topic that we will discuss more
fully in Chapter 13). When an ionic compound is melted or dissolved in water, its charged
particles are free to move in an electric field, so such a liquid shows high electrical conductivity
(Section 4-2, part 1).
We can represent the general reaction of the IA metals with the VIIA elements as
follows:
2M(s) ϩ X2 88n 2MX(s) M ϭ Li, Na, K, Rb, Cs; X ϭ F, Cl, Br, I
The Lewis dot representation for the generalized reaction is
Group IA Metals and Group VIA Nonmetals
Next, consider the reaction of lithium (Group IA) with oxygen (Group VIA) to form
lithium oxide, a solid ionic compound (mp Ͼ 1700°C) (see next page). We may represent
the reaction as
4Li(s) ϩ O2(g) 88n 2Li2O(s)
lithium oxygen lithium oxide
The formula for lithium oxide, Li2O, indicates that two atoms of lithium combine with
one atom of oxygen. If we examine the structures of the atoms before reaction, we can
see the reason for this ratio.
3Li __hg __h Liϩ __hg __ 1 eϪ lost
1s 2s 1s 2s
3Li __hg __h
N 88n Liϩ __hg __ 1 eϪ lost
1s 2s 1s 2s
8O __hg __hg __hg __h __h O2Ϫ __hg __hg __hg __hg __hg 2 eϪ gained
1s 2s 2p 1s 2s 3p
In a compact representation,
2[Li 88n Liϩ ϩ eϪ] and O ϩ 2eϪ 88n O2Ϫ
The Lewis dot formulas for the atoms and ions are
Lithium ions, Liϩ, are isoelectronic with helium atoms (2 eϪ). Oxide ions, O2Ϫ, are isoelectronic
with neon atoms (10 eϪ).
O2Li 2Li
ϩ
[Oϩ ]
2Ϫ
XX2M 2 (M
ϩ
[Xϩ ]
Ϫ
)
Figure 7-1 A representation of the
crystal structure of NaCl. Each ClϪ
ion (green) is surrounded by six
sodium ions, and each Naϩ ion
(gray) is surrounded by six chloride
ions. Any NaCl crystal includes
billions of ions in the pattern shown.
Adjacent ions actually are
in contact with one another; in this
drawing, the structure has been
expanded to show the spatial
arrangement of ions. The lines do not
represent covalent bonds. Compare
with Figure 2-7, a space-filling
drawing of the NaCl structure.
7-2 Formation of Ionic Compounds 275
Naϩ
ClϪ
Although the oxides of the other
Group IA metals are prepared by
different methods, similar descriptions
apply to compounds between the
Group IA metals (Li, Na, K, Rb, Cs)
and the Group VIA nonmetals (O, S,
Se, Te, Po).
Each Li atom has 1 eϪ in its valence
shell, one more eϪ than a noble gas
configuration, [He]. Each O atom has
6 eϪ in its valence shell and needs
2 eϪ more to attain a noble gas
configuration [Ne]. The Liϩ ions are
formed by oxidation of Li atoms, and
the O2Ϫ ions are formed by reduction
of O atoms.
Isoelectronic species have the same
number of electrons (see Section 6-5).
Some isoelectronic species:
O2Ϫ 8 protons 10 electrons
FϪ 9 protons 10 electrons
Ne 10 protons 10 electrons
Naϩ 11 protons 10 electrons
Mg2ϩ 12 protons 10 electrons


















The very small size of the Liϩ ion gives it a much higher charge density (ratio of charge
to size) than that of the larger Naϩ ion (Figure 6-1). Similarly, the O2Ϫ ion is smaller
than the ClϪ ion, so its smaller size and double negative charge give it a much higher
charge density. These more concentrated charges and smaller sizes bring the Liϩ and O2Ϫ
ions closer together in Li2O than the Naϩ and ClϪ ions are in NaCl. Consequently, the
qϩqϪ product in the numerator of Coulomb’s Law is greater in Li2O, and the d2 term in
the denominator is smaller. The net result is that the ionic bonding is much stronger (the
lattice energy is much more negative) in Li2O than in NaCl. This is consistent with the
higher melting temperature of Li2O (Ͼ1700°C) compared to NaCl (801°C).
Group IIA Metals and Group VIA Nonmetals
As our final illustration of ionic bonding, consider the reaction of calcium (Group IIA)
with oxygen (Group VIA). This reaction forms calcium oxide, a white solid ionic
compound with a very high melting point, 2580°C.
2Ca(s) ϩ O2(g) 88n 2CaO(s)
calcium oxygen calcium oxide
Again, we show the electronic structure of the atoms and ions, representing the inner electrons
by the symbol of the preceding noble gas.
20Ca [Ar] __hg Ca2ϩ [Ar] __ 2 eϪ lost
4s 4s
88n
8O [He] __hg __hg __h __h O2Ϫ [He] __hg __hg __hg __hg 2 eϪ gained
2s 2p 2s 2p
The Lewis dot notation for the reacting atoms and the resulting ions is
Calcium ions, Ca2ϩ, are isoelectronic with argon (18 eϪ), the preceding noble gas. Oxide
ions, O2Ϫ, are isoelectronic with neon (10 eϪ), the following noble gas.
Ca2ϩ is about the same size as Naϩ (see Figure 6-1) but carries twice the charge, so
its charge density is higher. Because the attraction between the two small, highly charged
ions Ca2ϩ and O2Ϫ is quite high, the ionic bonding is very strong, accounting for the very
high melting point of CaO, 2580°C.
Binary Ionic Compounds: A Summary
Table 7-2 summarizes the general formulas of binary ionic compounds formed by the
representative elements. “M” represents metals, and “X” represents nonmetals from the
indicated groups. In these examples of ionic bonding, each of the metal atoms has lost
one, two, or three electrons, and each of the nonmetal atoms has gained one, two, or three
electrons. Simple (monatomic) ions rarely have charges greater than 3ϩ or 3Ϫ. Ions with
greater charges would interact so strongly with the electron clouds of other ions that the
electron clouds would be very distorted, and considerable covalent character in the bonds
would result. Distinct molecules of solid ionic substances do not exist. The sum of the
attractive forces of all the interactions in an ionic solid is substantial. Therefore, binary
ionic compounds have high melting and boiling points.
The d- and f-transition elements form many compounds that are essentially ionic in
character. Many simple ions of transition metals do not have noble gas configurations.
OCa Ca
2ϩ
[Oϩ ]
2Ϫ
Lithium is a metal, as the shiny
surface of freshly cut Li shows.
Where it has been exposed to air,
the surface is covered with lithium
oxide.
276 CHAPTER 7: Chemical Bonding


















This discussion also applies to other
ionic compounds between any Group
IIA metal (Be, Mg, Ca, Sr, Ba) and any
Group VIA nonmetal (O, S, Se, Te).
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 8.8, Electron Configurations
in Ions.
The distortion of the electron cloud of
an anion by a small, highly charged
cation is called polarization.
Binary compounds contain two
elements.
TABLE 7-2 Simple Binary Ionic Compounds
General Ions
Metal Nonmetal Formula Present Example mp (°C)
IA* ϩ VIIA 88n MX (Mϩ, XϪ) LiBr 547
IIA ϩ VIIA 88n MX2 (M2ϩ, 2XϪ) MgCl2 708
IIIA ϩ VIIA 88n MX3 (M3ϩ, 3XϪ) GaF3 800 (subl)
IA*† ϩ VIA 88n M2X (2Mϩ, X2Ϫ) Li2O Ͼ1700
IIA ϩ VIA 88n MX (M2ϩ, X2Ϫ) CaO 2580
IIA ϩ VIA 88n M2X3 (2M3ϩ, 3X2Ϫ) Al2O3 2045
IA* ϩ VA 88n M3X (3Mϩ, X3Ϫ) Li3N 840
IIA ϩ VA 88n M3X2 (3M2ϩ, 2X3Ϫ) Ca3P2 Ϸ1600
IIIA ϩ VA 88n MX (M3ϩ, X3Ϫ) AlP
*Hydrogen is a nonmetal. All binary compounds of hydrogen are covalent, except for certain metal hydrides such as NaH and CaH2,
which contain hydride, HϪ, ions.
†As we saw in Section 6-8, the metals in Groups IA and IIA also commonly form peroxides (containing the O2
2Ϫ ion) or superoxides
(containing the O2
Ϫ ion). See Table 6-4. The peroxide and superoxide ions contain atoms that are covalently bonded to one another.
Introduction to Energy Relationships in Ionic Bonding
The following discussion may help you to understand why ionic bonding occurs between
elements with low ionization energies and those with high electronegativities. There is a
general tendency in nature to achieve stability. One way to do this is by lowering potential
energy; lower energies generally represent more stable arrangements.
Let us use energy relationships to describe why the ionic solid NaCl is more stable than
a mixture of individual Na and Cl atoms. Consider a gaseous mixture of one mole of sodium
atoms and one mole of chlorine atoms, Na(g) ϩ Cl(g). The energy change associated with
the loss of one mole of electrons by one mole of Na atoms to form one mole of Naϩ ions
(step 1 in Figure 7-2) is given by the first ionization energy of Na (see Section 6-3).
Na(g) 88n Naϩ(g) ϩ eϪ first ionization energy ϭ 496 kJ/mol
This is a positive value, so the mixture Naϩ(g) ϩ eϪ ϩ Cl(g) is 496 kJ/mol higher in energy
than the original mixture of atoms (the mixture Naϩ ϩ eϪ ϩ Cl is less stable than the mixture
of atoms).
The energy change for the gain of one mole of electrons by one mole of Cl atoms to
form one mole of ClϪ ions (step 2) is given by the electron affinity of Cl (see Section 6-4).
Cl(g) ϩ eϪ 88n ClϪ(g) electron affinity ϭ Ϫ349 kJ/mol
This negative value, Ϫ349 kJ/mol, lowers the energy of the mixture, but the mixture of
separated ions, Naϩ ϩ ClϪ, is still higher in energy (less stable) by (496 Ϫ 349) kJ/mol ϭ 147
kJ/mol than the original mixture of atoms (the red arrow in Figure 7-2). Thus, just the
formation of ions does not explain why the process occurs. The strong attractive force
between ions of opposite charge draws the ions together into the regular array shown in
Figure 7-1. The energy associated with this attraction (step 3) is the crystal lattice energy of
NaCl, Ϫ789 kJ/mol.
Naϩ(g) ϩ ClϪ(g) 88n NaCl(s) crystal lattice energy ϭ Ϫ789 kJ/mol
The crystal (solid) formation thus further lowers the energy to (147 Ϫ 789) kJ/mol ϭ
Ϫ642 kJ/mol. The overall result is that one mole of NaCl(s) is 642 kJ/mol lower in energy
E n r i c h m e n t
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 8.17, Lattice Energy.
COVALENT BONDING
Ionic bonding cannot result from a reaction between two nonmetals, because their electronegativity
difference is not great enough for electron transfer to take place. Instead,
reactions between two nonmetals result in covalent bonding.
A covalent bond is formed when two atoms share one or more pairs of electrons.
Covalent bonding occurs when the electronegativity difference, ⌬(EN), between
elements (atoms) is zero or relatively small.
Figure 7-2 A schematic
representation of the energy changes
that accompany the process
Na(g) ϩ Cl(g) n NaCl(s).
The red arrow represents the positive
energy change (unfavorable) for the
process of ion formation,
Na(g) ϩ Cl(g) n Naϩ(g) ϩ ClϪ(g).
The blue arrow represents the
negative energy change (favorable)
for the overall process, including the
formation of the ionic solid.
278 CHAPTER 7: Chemical Bonding
(more stable) than the original mixture of atoms (the blue arrow in Figure 7-2). Thus, we
see that a major driving force for the formation of ionic compounds is the large electrostatic
stabilization due to the attraction of the ionic charges (step 3).
In this discussion we have not taken into account the fact that sodium is a solid metal
or that chlorine actually exists as diatomic molecules. The additional energy changes
involved when these are changed to gaseous Na and Cl atoms, respectively, are small enough
that the overall energy change starting from Na(s) and Cl2(g) is still negative.
Naϩ(g) ϩ eϪ ϩ Cl(g)
Na(g) ϩ Cl(g)
Increasingpotentialenergy(lessstable) Naϩ(g) ϩ ClϪ(g)
EA of Cl ϭ Ϫ349 kJ/mol
Net energy change for
Na(g) ϩ Cl(g) Naϩ(g) ϩ ClϪ(g)
ϭ ϩ147 kJ/mol
Net energy change for
ϭ Ϫ642 kJ/mol
ϩ Cl(g)Na(g) NaCl(s)
1st IE of Na
ϭ 496 kJ/mol
Crystal lattice
energy of NaCl
ϭϪ789 kJ/mol
NaCl(s)
1
2
3
(Enrichment, continued)
In predominantly covalent compounds the bonds between atoms within a molecule
(intramolecular bonds) are relatively strong, but the forces of attraction between molecules
(intermolecular forces) are relatively weak. As a result, covalent compounds have lower
melting and boiling points than ionic compounds. The relation of bonding types to physical
properties of liquids and solids will be developed more fully in Chapter 13.
FORMATION OF COVALENT BONDS
Let us look at the simplest case of covalent bonding, the reaction of two hydrogen atoms
to form the diatomic molecule H2. As you recall, an isolated hydrogen atom has the ground
state electron configuration 1s1, with the probability density for this one electron spherically
distributed about the hydrogen nucleus (Figure 7-3a). As two hydrogen atoms
approach each other, the electron of each hydrogen atom is attracted by the nucleus of
the other hydrogen atom as well as by its own nucleus (Figure 7-3b). If these two electrons
have opposite spins so that they can occupy the same region (orbital), both electrons
can now preferentially occupy the region between the two nuclei (Figure 7-3c), because
they are attracted by both nuclei. The electrons are shared between the two hydrogen
atoms, and a single covalent bond is formed. We say that the 1s orbitals overlap so that
both electrons are now in the orbitals of both hydrogen atoms. The closer together the
atoms come, the more nearly this is true. In that sense, each hydrogen atom then has the
helium configuration 1s2.
The bonded atoms are at lower energy (more stable) than the separated atoms. This
is shown in the plot of energy versus distance in Figure 7-4. As the two atoms get closer
together, however, the two nuclei, being positively charged, exert an increasing repulsion
on each other. At some distance, a minimum energy, Ϫ435 kJ/mol, is reached; it corresponds
to the most stable arrangement and occurs at 0.74 Å, the actual distance between
two hydrogen nuclei in an H2 molecule. At greater internuclear separation, the repulsive
forces diminish, but the attractive forces decrease even faster. At smaller separations, repulsive
forces increase more rapidly than attractive forces. The magnitude of this stabilization
7-3
Figure 7-3 A representation of the
formation of a covalent bond
between two hydrogen atoms. The
position of each positively charged
nucleus is represented by a black
dot. Electron density is indicated by
the depth of shading. (a) Two
hydrogen atoms separated by a large
distance. (b) As the atoms approach
each other, the electron of each
atom is attracted by the positively
charged nucleus of the other atom,
so the electron density begins to
shift. (c) The two electrons can both
occupy the region where the two 1s
orbitals overlap; the electron density
is highest in the region between the
nuclei of the two atoms.
7-3 Formation of Covalent Bonds 279
Figure 7-4 The potential energy of the H2 molecule as a function of the distance between
the two nuclei. The lowest point in the curve, Ϫ435 kJ/mol, corresponds to the internuclear
distance actually observed in the H2 molecule, 0.74 Å. (The minimum potential energy,
Ϫ435 kJ/mol, corresponds to the value of Ϫ7.23 ϫ 10Ϫ19 joule per H2 molecule.) Energy is
compared with that of two separated hydrogen atoms.
H H
H H
H2
(a)
(b)
(c)
Increasing internuclear distance
H H H H H H
Increasingenergy
0
H2 molecule
0.74 Å
–435
kJ
mol
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 9.3, Chemical Bond Formation:
Covalent Bonding, and Screen 10.3,
Valence Bond Theory.
is called the HXH bond energy. Bond energies for covalent bonds will be discussed more
extensively in Section 15-9.
Other pairs of nonmetal atoms share electron pairs to form covalent bonds. The result
of this sharing is that each atom attains a more stable electron configuration—frequently
the same as that of the nearest noble gas. This results in a more stable arrangement for
the bonded atoms. (This is discussed in Section 7-5.) Most covalent bonds involve sharing
of two, four, or six electrons—that is, one, two, or three pairs of electrons. Two atoms
form a single covalent bond when they share one pair of electrons, a double covalent
bond when they share two electron pairs, and a triple covalent bond when they share
three electron pairs. These are usually called simply single, double, and triple bonds. Covalent
bonds that involve sharing of one and three electrons are known, but are relatively rare.
In a Lewis formula, we represent a covalent bond by writing each shared electron pair
either as a pair of two dots between the two atom symbols or as a dash connecting them.
Thus, the formation of H2 from H atoms could be represented as
HT ϩ TH 88n HSH or HXH
where the dash represents a single bond. Similarly, the combination of two fluorine atoms
to form a molecule of fluorine, F2, can be shown as
The formation of a hydrogen fluoride, HF, molecule from a hydrogen atom and a fluorine
atom can be shown as
We will see many more examples of this representation.
In our discussion, we have postulated that bonds form by the overlap of two atomic
orbitals. This is the essence of the valence bond theory, which we will describe in more
detail in the next chapter. Another theory, molecular orbital theory, is discussed in
Chapter 9. For now, let us concentrate on the number of electron pairs shared and defer
the discussion of which orbitals are involved in the sharing until the next chapter.
LEWIS FORMULAS FOR MOLECULES AND
POLYATOMIC IONS
In Sections 7-1 and 7-2 we drew Lewis formulas for atoms and monatomic ions. In Section
7-3, we used Lewis formulas to show the valence electrons in three simple molecules. A
water molecule can be represented by either of the following diagrams.
In dash formulas, a shared pair of electrons is indicated by a dash. There are two double
bonds in carbon dioxide, and its Lewis formula is
The covalent bonds in a polyatomic ion can be represented in the same way. The Lewis
formula for the ammonium ion, NH4
ϩ, shows only eight electrons, even though the
N atom has five electrons in its valence shell and each H atom has one, for a total of
CC O OO O or
dot formula dash formula
O
H
HO
H
H or
dot formula dash formula
7-4
FHFH HFϩ or
FFFF FFϩ orThe bonding in the other halogens,
Cl2, Br2, and I2, is analogous to that
in F2.
The bonding in gaseous HCl, HBr,
and HI is analogous to that in HF.
280 CHAPTER 7: Chemical Bonding
An H2O molecule has two shared electron pairs, that is, two
single covalent bonds. The O atom also has two unshared pairs.
A CO2 molecule has four shared electron pairs in two
double bonds. The central atom (C) has no unshared
pairs.
In H2O, two of the six valence
electrons of an oxygen atom are used
in covalent bonding; the one valence
electron of each hydrogen atom is used
in covalent bonding.
In CO2, the four valence electrons of a
carbon atom are used in covalent
bonding; two of the six valence
electrons of each oxygen atom are used
in covalent bonding.
A polyatomic ion is an ion that
contains more than one atom.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 9.4, Lewis Electron Dot
Structures.
5 ϩ 4(1) ϭ 9 electrons. The NH4
ϩ ion, with a charge of 1ϩ, has one less electron than
the original atoms.
The writing of Lewis formulas is an electron bookkeeping method that is useful as a first
approximation to suggest bonding schemes. It is important to remember that Lewis dot
formulas only show the number of valence electrons, the number and kinds of bonds,
and the order in which the atoms are connected. They are not intended to show the threedimensional
shapes of molecules and polyatomic ions. We will see in Chapter 8, however, that
the three-dimensional geometry of a molecule can be predicted from its Lewis formula.
THE OCTET RULE
Representative elements usually attain stable noble gas electron configurations when they
share electrons. In the water molecule eight electrons are in the outer shell of the O atom,
and it has the neon electron configuration; two electrons are in the valence shell of each
H atom, and each has the helium electron configuration. Likewise, the C and O of CO2
and the N of NH3 and the NH4
ϩ ion each have a share in eight electrons in their outer
shells. The H atoms in NH3 and NH4
ϩ each share two electrons. Many Lewis formulas
are based on the idea that
in most of their compounds, the representative elements achieve noble gas configu-
rations.
This statement is usually called the octet rule, because the noble gas configurations have
8 eϪ in their outermost shells (except for He, which has 2 eϪ).
For now, we restrict our discussion to compounds of the representative elements. The
octet rule alone does not let us write Lewis formulas. We still must decide how to place
the electrons around the bonded atoms—that is, how many of the available valence electrons
are bonding electrons (shared) and how many are unshared electrons (associated
with only one atom). A pair of unshared electrons in the same orbital is called a lone pair.
A simple mathematical relationship is helpful here:
S ϭ N Ϫ A
S is the total number of electrons shared in the molecule or polyatomic ion.
N is the total number of valence shell electrons needed by all the atoms in the
molecule or ion to achieve noble gas configurations (N ϭ 8 ϫ number of atoms that
are not H, plus 2 ϫ number of H atoms).
A is the number of electrons available in the valence shells of all of the (representative)
atoms. This is equal to the sum of their periodic group numbers. We must
adjust A, if necessary, for ionic charges. We add electrons to account for negative
charges and subtract electrons to account for positive charges.
7-5
N
H
H
HHN
H
H
H H
dot formula dash formula
ϩ
ϩ
The NH3 molecule, like the NH4
ϩ
ion, has eight valence electrons about
the N atom.
H
H HN
H
H HN or
7-5 The Octet Rule 281
In some compounds, the central atom
does not achieve a noble gas
configuration. Such exceptions to the
octet rule are discussed in Section 7-7.
The representative elements are those
in the A groups of the periodic table.
Let us see how this relationship applies to some species whose Lewis formulas we have
already shown.
For F2,
N ϭ 2 ϫ 8 (for two F atoms) ϭ 16 eϪ needed
A ϭ 2 ϫ 7 (for two F atoms) ϭ 14 eϪ available
S ϭ N Ϫ A ϭ 16 Ϫ 14 ϭ 2 eϪ shared
The Lewis formula for F2 shows 14 valence electrons total, with 2 eϪ shared in a single
bond.
For HF,
N ϭ 1 ϫ 2 (for one H atom) ϩ 1 ϫ 8 (for one F atom) ϭ 10 eϪ needed
A ϭ 1 ϫ 1 (for one H atom) ϩ 1 ϫ 7 (for one F atom) ϭ 8 eϪ available
S ϭ N Ϫ A ϭ 10 Ϫ 8 ϭ 2 eϪ shared
The Lewis formula for HF shows 8 valence electrons total, with 2 eϪ shared in a single
bond.
For H2O,
N ϭ 2 ϫ 2 (for two H atoms) ϩ 1 ϫ 8 (for one O atom) ϭ 12 eϪ needed
A ϭ 2 ϫ 1 (for two H atoms) ϩ 1 ϫ 6 (for one O atom) ϭ 8 eϪ available
S ϭ N Ϫ A ϭ 12 Ϫ 8 ϭ 4 eϪ shared
The Lewis formula for H2O shows 8 valence electrons total, with a total of 4 eϪ shared,
2 eϪ in each single bond.
For CO2,
N ϭ 1 ϫ 8 (for one C atom) ϩ 2 ϫ 8 (for two O atoms) ϭ 24 eϪ needed
A ϭ 1 ϫ 4 (for one C atom) ϩ 2 ϫ 6 (for two O atoms) ϭ 16 eϪ available
S ϭ N Ϫ A ϭ 24 Ϫ 16 ϭ 8 eϪ shared
The Lewis formula for CO2 shows 16 valence electrons total, with a total of 8 eϪ shared,
4 eϪ in each double bond.
For NH4
ϩ,
N ϭ 1 ϫ 8 (for one N atom) ϩ 4 ϫ 2 (for four H atoms) ϭ 16 eϪ needed
A ϭ 1 ϫ 5 (for one N atom) ϩ 4 ϫ 1 (for four H atoms) Ϫ 1 (for 1ϩ charge) ϭ 8 eϪ available
S ϭ N Ϫ A ϭ 16 Ϫ 8 ϭ 8 eϪ shared
The Lewis formula for NH4
ϩ shows 8 valence electrons total, with all 8 eϪ shared, 2 eϪ
in each single bond.
The following general steps describe the use of the S ϭ N Ϫ A relationship in
constructing dot formulas for molecules and polyatomic ions.
Writing Lewis Formulas
1. Select a reasonable (symmetrical) “skeleton” for the molecule or polyatomic ion.
a. The least electronegative element is usually the central element, except that H is
never the central element, because it forms only one bond. The least elec-
282 CHAPTER 7: Chemical Bonding
FF
FH
OH
H
O C O
The 1ϩ ionic charge is due to a
deficiency of one eϪ relative to the
neutral atoms.
H
H
H HN
ϩ
tronegative element is usually the one that needs the most electrons to fill its
octet. Example: CS2 has the skeleton S C S.
b. Oxygen atoms do not bond to each other except in (1) O2 and O3 molecules; (2)
hydrogen peroxide, H2O2, and its derivatives, the peroxides, which contain the
O2
2Ϫ group; and (3) the rare superoxides, which contain the O2
Ϫ group. Example:
The sulfate ion, SO4
2Ϫ, has the skeleton
O
O S O
O
c. In ternary oxoacids, hydrogen usually bonds to an O atom, not to the central atom.
Example: nitrous acid, HNO2, has the skeleton H O N O. There are a few
exceptions to this guideline, such as H3PO3 and H3PO2.
d. For ions or molecules that have more than one central atom, the most symmetrical
skeletons possible are used. Examples: C2H4 and P2O7
4Ϫ have the following
skeletons:
H H O O
C C and O P O P O
H H O O
2. Calculate N, the number of valence (outer) shell electrons needed by all atoms in the
molecule or ion to achieve noble gas configurations. Examples:
A ternary oxoacid contains three
elements—H, O, and another
element, often a nonmetal.
7-5 The Octet Rule 283
For H2SO4,
N ϭ 2 ϫ 2 (H atoms) ϩ 1 ϫ 8 (S atom) ϩ 4 ϫ 8 (O atoms) ϭ 4 ϩ 8 ϩ 32 ϭ 44 eϪ needed
For ClO4
Ϫ,
N ϭ 8 (Cl atoms) ϩ 32 (O atoms) ϭ 40 eϪ needed
For NO3
Ϫ,
N ϭ 1 ϫ 8 (N atom) ϩ 3 ϫ 8 (O atoms) ϭ 32 eϪ needed
Calculate A, the number of electrons available in the valence (outer) shells of all the
atoms. For negatively charged ions, add to this total the number of electrons equal
to the charge on the anion; for positively charged ions, subtract the number of electrons
equal to the charge on the cation. Examples:
For H2SO4,
A ϭ 2 ϫ 1 (H atoms) ϩ 1 ϫ 6 (S atom) ϩ 4 ϫ 6 (O atoms) ϭ 2 ϩ 6 ϩ 24 ϭ 32 eϪ available
For ClO4
Ϫ,
A ϭ 1 ϫ 7 (Cl atom) ϩ 4 ϫ 6 (O atoms) ϩ 1 (for 1Ϫ charge) ϭ 7 ϩ 24 ϩ 1 ϭ 32 eϪ available
For NO3
Ϫ,
A ϭ 1 ϫ 5 (N atom) ϩ 3 ϫ 6 (O atoms) ϩ 1 (for 1Ϫ charge) ϭ 5 ϩ 18 ϩ 1 ϭ 24 eϪ available
Calculate S, total number of electrons shared in the molecule or ion, using the relationship
S ϭ N Ϫ A. Examples:
  2Ϫ
 
 
 
 
 
  4Ϫ
 
 
 
 
 
For the representative elements, the
number of valence shell electrons in an
atom is equal to its periodic group
number. Exceptions: 1 for an H atom
and 2 for He.
For compounds containing only
representative elements, N is equal to
8 ϫ number of atoms that are not H,
plus 2 ϫ number of H atoms.
For H2SO4,
S ϭ N Ϫ A ϭ 44 Ϫ 32
ϭ 12 electrons shared (6 pairs of eϪ shared)
For ClO4
Ϫ,
S ϭ N Ϫ A ϭ 40 Ϫ 32
ϭ 8 electrons shared (4 pairs of eϪ shared)
For NO3
Ϫ,
S ϭ N Ϫ A ϭ 32 Ϫ 24 ϭ 8 eϪ shared (4 pairs of eϪ shared)
3. Place the S electrons into the skeleton as shared pairs. Use double and triple bonds
only when necessary. Lewis formulas may be shown as either dot formulas or dash
formulas.
Dot Formula Dash Formula
(“bonds” in place, (“bonds” in place,
Formula Skeleton but incomplete) but incomplete)
H2SO4
ClO4
Ϫ
NO3
Ϫ
4. Place the additional electrons into the skeleton as unshared (lone) pairs to fill the
octet of every A group element (except H, which can share only 2 eϪ). Check that
the total number of electrons is equal to A, from step 2. Examples:
For H2SO4,
Check: 16 pairs of eϪ have been used and all octets are satisfied. 2 ϫ 16 ϭ 32 eϪ
available.
For ClO4
Ϫ,
O
O
O O
O
O
O OCl
Ϫ
Cl
Ϫ
or
O
O
O OH HS or S
O
O
HH OO
O
O O
N
Ϫ
O
O O
N
Ϫ
O
O O
N
Ϫ
O
O
O OCl
Ϫ
O
O
O OCl
Ϫ
O
O
O OCl
Ϫ
S
O
O
HH OO
O
O
O OH HS
O
O
O OH HS
C, N, and O often form double and
triple bonds. S can form double bonds
with C, N, and O.
284 CHAPTER 7: Chemical Bonding
Check: 16 pairs of eϪ have been used and all octets are satisfied. 2 ϫ 16 ϭ 32 eϪ
available.
For NO3
Ϫ,
Check: 12 pairs of eϪ have been used and all octets are satisfied. 2 ϫ 12 ϭ 24 eϪ
available.
O
OO O
N
Ϫ
N
O
O
or
Ϫ
C is the central atom, or the element
in the middle of the molecule. It needs
four more electrons to acquire an
octet, and each S atom needs only two
more electrons.
7-5 The Octet Rule 285
EXAMPLE 7-1 Writing Lewis Formulas
Write the Lewis formula for the nitrogen molecule, N2.
Plan
We follow the stepwise procedure that was just presented for writing Lewis formulas.
Solution
Step 1: The skeleton is N N.
Step 2: N ϭ 2 ϫ 8 ϭ 16 eϪ needed (total) by both atoms
Step 2: A ϭ 2 ϫ 5 ϭ 10 eϪ available (total) for both atoms
Step 2: S ϭ N Ϫ A ϭ 16 eϪ Ϫ 10 eϪ ϭ 6 eϪ shared
Step 3: NSSSN 6 eϪ (3 pairs) are shared; a triple bond.
Step 4: The additional 4 eϪ are accounted for by a lone pair on each N. The complete
Lewis formula is
Check: 10 eϪ (5 pairs) have been used.
EXAMPLE 7-2 Writing Lewis Formulas
Write the Lewis formula for carbon disulfide, CS2, a foul-smelling liquid.
Plan
Again, we follow the stepwise procedure to apply the relationship S ϭ N Ϫ A.
Solution
Step 1: The skeleton is S C S.
Step 2: N ϭ 1 ϫ 8 (for C) ϩ 2 ϫ 8 (for S) ϭ 24 eϪ needed by all atoms
Step 2: A ϭ 1 ϫ 4 (for C) ϩ 2 ϫ 6 (for S) ϭ 16 eϪ available
Step 2: S ϭ N Ϫ A ϭ 24 eϪ Ϫ 16 eϪ ϭ 8 eϪ shared
Step 3: SSSCSSS 8 eϪ (4 pairs) are shared; two double bonds.
NN or NN
Problem-Solving Tip: Be Careful of the Total Number of Electrons
A very common error in writing Lewis formulas is showing the wrong number of electrons.
Always make a final check to be sure that the Lewis formula you write shows the
same number of electrons you calculated as A.
For a slightly different approach
than the one used in this text,
see the Saunders Interactive General
Chemistry CD-ROM, Screen 9.5,
Drawing Lewis Structures.
Step 4: C already has an octet, so the remaining 8 eϪ are distributed as lone pairs on
the S atoms to give each S an octet. The complete Lewis formula is
Check: 16 eϪ (8 pairs) have been used. The bonding picture is similar to that of CO2; this is not surprising,
because S is below O in Group VIA.
You should now work Exercise 29.
EXAMPLE 7-3 Writing Lewis Formulas
Write the Lewis formula for the carbonate ion, CO3
2Ϫ.
Plan
The same stepwise procedure can be applied to ions. We must remember to adjust A, the total
number of electrons, to account for the charge shown on the ion.
Solution
O 2Ϫ
Step 1: The skeleton is O C O
Step 2: N ϭ 1 ϫ 8 (for C) ϩ 3 ϫ 8 (for O) ϭ 8 ϩ 24 ϭ 32 eϪ needed by all atoms
Step 2: A ϭ 1 ϫ 4 (for C) ϩ 3 ϫ 6 (for O) ϩ 2 (for the 2Ϫ charge)
Step 2: ϭ 4 ϩ 18 ϩ 2 ϭ 24 eϪ available
Step 2: S ϭ N Ϫ A ϭ 32 eϪ Ϫ 24 eϪ ϭ 8 eϪ (4 pairs) shared
O 2Ϫ
hStep 3: OSCSSO (Four pairs are shared. At this point it doesn’t matter which O is doubly
bonded.)
Step 4: The Lewis formula is
Check: 24 eϪ (12 pairs) have been used.
You should now work Exercises 30 and 38.
You should practice writing many Lewis formulas. A few common types of organic
compounds and their Lewis formulas are shown here. Each follows the octet rule. Methane,
CH4, is the simplest of a huge family of organic compounds called hydrocarbons (composed
solely of hydrogen and carbon). Ethane, C2H6, is another hydrocarbon that contains only
single bonds. Ethylene, C2H4, has a carbon–carbon double bond, and acetylene, C2H2,
contains a carbon–carbon triple bond.
C
H
H
H C
H
H
H CH C HC
H
H
HH
ethane, C2H6methane, CH4 acetylene, C2H2ethylene, C2H4
CC
H H
HH
OO
O O
C
2Ϫ2Ϫ
C
OO
or
CC S SS S or
Calcium carbonate, CaCO3, occurs
in several forms in nature. The
mineral calcite forms very large clear
crystals (right); in microcrystalline
form it is the main constituent of
limestone (top left). Seashells (bottom)
are largely calcium carbonate.
286 CHAPTER 7: Chemical Bonding
Acetylene burns in oxygen with a
flame so hot that it is used to weld
metals.
For practice, apply the methods of this
section to write these Lewis formulas.
Halogen atoms can appear in place of hydrogen atoms in many organic compounds,
because hydrogen and halogen atoms each need one more electron to attain noble gas
configurations. An example is chloroform, CHCl3. Alcohols contain the group CXOXH;
the simplest alcohol is methanol, CH3OH. An organic compound that contains a carbon–
oxygen double bond is formaldehyde, H2CO.
H
H
H O HCC
Cl
H
Cl ClC
O
H H
C
formaldehyde, H2COmethanol, CH3OHchloroform, CHCl3
7-6 Resonance 287
RESONANCE
In addition to the Lewis formula shown in Example 7-3, two other Lewis formulas with
the same skeleton for the CO3
2Ϫ ion are equally acceptable. In these formulas, the double
bond could be between the carbon atom and either of the other two oxygen atoms.
A molecule or polyatomic ion for which two or more Lewis formulas with the same
arrangements of atoms can be drawn to describe the bonding is said to exhibit resonance.
The three structures above are resonance structures of the carbonate ion. The relationship
among them is indicated by the double-headed arrows, mn. This symbol does not
mean that the ion flips back and forth among these three structures. The true structure
can be described as an average, or hybrid, of the three.
O
O O O
C
2Ϫ
O
O O O
C
2Ϫ 2Ϫ
O
C
7-6
Problem-Solving Tip: Some Guidelines for Drawing Lewis Formulas
The following guidelines might help you draw Lewis formulas.
1. In most of their covalent compounds, the representative elements follow the octet
rule, except that hydrogen always shares only two electrons.
2. Carbon always forms four bonds. This can be accomplished as:
a. four single bonds
b. two double bonds
c. two single bonds and one double bond or
d. one single bond and one triple bond
3. Hydrogen forms only one bond to another element; thus hydrogen can never be a
central atom.
4. In neutral (uncharged) species, nitrogen forms three bonds, and oxygen forms two
bonds.
5. Nonmetals can form single, double, or triple bonds, but never quadruple bonds.
6. Carbon forms double or triple bonds to C, N, O, or S atoms; oxygen can form
double bonds with many other elements.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 9.6, Resonance Structures.
C, N, and O often form double and
triple bonds. S can form double bonds
with C, N, and O.
Experiments show that the CXO bonds in CO3
2Ϫ are neither double nor single bonds,
but are intermediate in bond length and strength. Based on measurements in many
compounds, the typical CXO single bond length is 1.43 Å, and the typical CUO double
bond length is 1.22 Å. The CXO bond length for each bond in the CO3
2Ϫ ion is intermediate
at 1.29 Å. Another way to represent this situation is by delocalization of bonding
electrons:
The dashed lines include that some of the electrons shared between C and O atoms are
delocalized among all four atoms; that is, the four pairs of shared electrons are equally
distributed among three CXO bonds.
EXAMPLE 7-4 Lewis Formulas, Resonance
Draw two resonance structures for the sulfur dioxide molecule, SO2.
Plan
The stepwise procedure (Section 7-5) can be used to write each resonance structure.
Solution
S O
q q
N ϭ 1(8) ϩ 2(8) ϭ 24 eϪ
A ϭ 1(6) ϩ 2(6) ϭ 18 eϪ
S ϭ N Ϫ A ϭ 6 eϪ shared
The resonance structures are
We could show delocalization of electrons as follows:
Remember that Lewis formulas do not necessarily show shapes. SO2 molecules are angular, not
linear.
You should now work Exercises 50 and 52.
SO O (lone pairs on O not shown)
SS O OO O orS OO SO O
O
2Ϫ
O
C
O
(lone pairs on O atoms not shown)
When electrons are shared among
more than two atoms, the electrons are
said to be delocalized. The concept of
delocalization is important in
molecular orbital theory (Chapter 9).
288 CHAPTER 7: Chemical Bonding
Problem-Solving Tip: Some Guidelines for Resonance Structures
Resonance involves several different acceptable Lewis formulas with the same arrangement
of atoms. They differ only in the arrangements of electron pairs, never in atom
positions. The actual structure of such a molecule or ion is the average, or composite,
of its resonance structures, but this does not mean that the electrons are moving from
one place to another. This average structure is more stable than any of the individual
resonance structures.
The rose on the left is in an
atmosphere of sulfur dioxide, SO2.
Gaseous SO2 and its aqueous
solutions are used as bleaching
agents. A similar process is used to
bleach wood pulp before it is
converted to paper.
7-6 Resonance 289
Formal Charges
An experimental determination of the structure of a molecule or polyatomic ion is necessary
to establish unequivocally its correct structure. We often do not have these results
available, however. Formal charge is the hypothetical charge on an atom in a molecule or
polyatomic ion; to find the formal charge, we count bonding electrons as though they were
equally shared between the two bonded atoms. The concept of formal charges helps us to
write correct Lewis formulas in most cases. The most energetically favorable formula for a
molecule is usually one in which the formal charge on each atom is zero or as near zero as
possible.
Consider the reaction of NH3 with hydrogen ion, Hϩ, to form the ammonium ion,
NH4
ϩ.
The previously unshared pair of electrons on the N atom in the NH3 molecule is shared
with the Hϩ ion to form the NH4
ϩ ion, in which the N atom has four covalent bonds.
Because N is a Group VA element, we expect it to form three covalent bonds to complete
its octet. How can we describe the fact that N has four covalent bonds in species like NH4
ϩ?
The answer is obtained by calculating the formal charge on each atom in NH4
ϩ by the
following rules:
Rules for Assigning Formal Charges to Atoms of A Group Elements
1. The formal charge, abbreviated FC, on an atom in a Lewis formula is given by
the relationship
FC ϭ (group number) Ϫ [(number of bonds) ϩ (number of unshared eϪ)]
Formal charges are represented by ᮍ and ᮎ to distinguish them from real charges
on ions.
2. In a Lewis formula, an atom that has the same number of bonds as its periodic group
number has a formal charge of zero.
3. a. In a molecule, the sum of the formal charges is zero.
b. In a polyatomic ion, the sum of the formal charges is equal to the charge.
Let us apply these rules to the ammonia molecule, NH3, and to the ammonium ion, NH4
ϩ.
Because N is a Group VA element, its group number is 5.
In NH3 the N atom has 3 bonds and 2 unshared eϪ, and so for N,
FC ϭ (group number) Ϫ [(number of bonds) ϩ (number of unshared eϪ)]
ϭ 5 Ϫ (3 ϩ 2) ϭ 0 (for N)
H
H
H HN
ϩ
H
H HN
H
H
H HN
ϩ
H
H H H
ϩ
N ϩ
E n r i c h m e n t
The Hϩ ion has a vacant orbital,
which accepts a share in the lone pair
on nitrogen. The formation of a
covalent bond by the sharing of an
electron pair that is provided by one
atom is called coordinate covalent bond
formation. This type of bond formation
is discussed again in Chapters 10 and
25.
FCs are indicated by ᮍ and ᮎ. The
sum of the formal charges in a
polyatomic ion is equal to the charge
on the ion: ϩ1 in NH4
ϩ.
290 CHAPTER 7: Chemical Bonding
For H,
FC ϭ (group number) Ϫ [(number of bonds) ϩ (number of unshared eϪ)]
ϭ 1 Ϫ (1 ϩ 0) ϭ 0 (for H)
The formal charges of N and H are both zero in NH3, so the sum of the formal charges
is 0 ϩ 3(0) ϭ 0, consistent with rule 3a.
In NH4
ϩ the atom has 4 bonds and no unshared eϪ, and so for N,
FC ϭ (group number) Ϫ [(number of bonds) ϩ (number of unshared eϪ)]
ϭ 5 Ϫ (4 ϩ 0) ϭ ϩ1 (for N)
Calculation of the FC for H atoms gives zero, as shown previously. The sum of the formal
charges in NH4
ϩ is (1ϩ) ϭ 4(0) ϭ 1ϩ. This is consistent with rule 3b.
Thus, we see the octet rule is obeyed in both NH3 and NH4
ϩ. The sum of the formal
charges in each case is that predicted by rule 3, even though nitrogen has four covalent
bonds in the NH4
ϩ ion.
This bookkeeping system helps us to choose among various Lewis formulas for a molecule
or ion, according to the following guidelines.
a. The most likely formula for a molecule or ion is usually one in which the
formal charge on each atom is zero or as near zero as possible.
b. Negative formal charges are more likely to occur on the more electronegative
elements.
c. Lewis formulas in which adjacent atoms have formal charges of the same sign
are usually not accurate representations (the adjacent charge rule).
Let us now write some Lewis formulas for, and assign formal charges to, the atoms in
nitrosyl chloride, NOCl, a compound often used in organic synthesis. The Cl atom and the
O atom are both bonded to the N atom. Two Lewis formulas that satisfy the octet rule are
(i) (ii)
For Cl, FC ϭ 7 Ϫ (2 ϩ 4) ϭ ϩ1 For Cl, FC ϭ 7 Ϫ (1 ϩ 6) ϭ 0
For N, FC ϭ 5 Ϫ (3 ϩ 2) ϭ 0 For N, FC ϭ 5 Ϫ (3 ϩ 2) ϭ 0
For O, FC ϭ 6 Ϫ (1 ϩ 6) ϭ Ϫ1 For O, FC ϭ 6 Ϫ (2 ϩ 4) ϭ 0
We believe that (ii) is a preferable Lewis formula, because it has smaller formal charges than
(i). We see that a double-bonded terminal Cl atom would have its electrons arranged as
/ÅClUX, with the formal charge of Cl equal to 7 Ϫ (2 ϩ 4) ϭ ϩ1. A positive formal charge
on such an electronegative element is quite unlikely, and double bonding to chlorine does
not occur. The same reasoning would apply to the other halogens.
NCl ONCl O
ϩ Ϫ
H
H
H HN
ϩ
ϩ
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 9.13, Formal Charge.
7-7 Limitations of the Octet Rule for Lewis Formulas 291
LIMITATIONS OF THE OCTET RULE FOR
LEWIS FORMULAS
Recall that representative elements achieve noble gas electron configurations in most of
their compounds. But when the octet rule is not applicable, the relationship S ϭ N Ϫ A
is not valid without modification. The following are general cases for which the procedure
in Section 7-5 must be modified—that is, there are four types of limitations of the
octet rule.
A. Most covalent compounds of beryllium, Be. Because Be contains only two valence
shell electrons, it usually forms only two covalent bonds when it bonds to two other
atoms. We therefore use four electrons as the number needed by Be in step 2, Section
7-5. In steps 3 and 4 we use only two pairs of electrons for Be.
B. Most covalent compounds of the Group IIIA elements, especially boron, B. The
IIIA elements contain only three valence shell electrons, so they often form three
covalent bonds when they bond to three other atoms. We therefore use six electrons
as the number needed by the IIIA elements in step 2; and in steps 3 and 4 we use
only three pairs of electrons for the IIIA elements.
C. Compounds or ions containing an odd number of electrons. Examples are NO,
with 11 valence shell electrons, and NO2, with 17 valence shell electrons.
D. Compounds or ions in which the central element needs a share in more than eight
valence shell electrons to hold all the available electrons, A. Extra rules are added
to steps 2 and 4 of the procedure in Section 7-5 when this is encountered.
Step 2a: If S, the number of electrons shared, is less than the number needed to
bond all atoms to the central atom, then S is increased to the number of
electrons needed.
Step 4a: If S must be increased in step 2a, then the octets of all the atoms might
be satisfied before all of the electrons (A) have been added. Place the
extra electrons on the central element.
Many species that violate the octet rule are quite reactive. For instance, compounds
containing atoms with only four valence shell electrons (limitation type A above) or six
valence shell electrons (limitation type B above) frequently react with other species that
supply electron pairs. Compounds such as these that accept a share in a pair of electrons
are called Lewis acids; a Lewis base is a species that makes available a share in a pair of electrons.
(This kind of behavior will be discussed in detail in Section 10-10.) Molecules with
an odd number of electrons often dimerize (combine in pairs) to give products that
do satisfy the octet rule. Examples are the dimerization of NO to form N2O2 (Section
24-15) and NO2 to form N2O4 (Section 24-15). Examples 7-5 through 7-9 illustrate some
limitations and show how such Lewis formulas are constructed.
EXAMPLE 7-5 Limitations of the Octet Rule
Write the Lewis formula for gaseous beryllium chloride, BeCl2, a covalent compound.
Plan
This is an example of limitation type A. So, as we follow the steps in writing the Lewis formula,
we must remember to use four electrons as the number needed by Be in step 2. Steps 3 and 4
should show only two pairs of electrons for Be.
7-7
Lewis formulas are not normally
written for compounds containing
d- and f-transition metals. The d- and
f-transition metals utilize d or f orbitals
(or both) in bonding as well as s and p
orbitals. Thus, they can accommodate
more than eight valence electrons.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 9.8, Free Radicals—Exceptions
to the Octet Rule.
BF3 and BCl3 are gases at room
temperature. Liquid BBr3 and solid
BI3 are shown here.
292 CHAPTER 7: Chemical Bonding
Solution
Step 1: The skeleton is Cl Be Cl
see limitation type A
g
Step 2: N ϭ 2 ϫ 8 (for Cl) ϩ 1 ϫ 4 (for Be) ϭ 20 eϪ needed
Step 3: A ϭ 2 ϫ 7 (for Cl) ϩ 1 ϫ 2 (for Be) ϭ 16 eϪ available
Step 3: S ϭ N Ϫ A ϭ 20 eϪ Ϫ 16 eϪ ϭ 4 eϪ shared
Step 3: ClSBeSCl
Step 4:
Calculation of formal charges shows that
for Be, FC ϭ 2 Ϫ (2 ϩ 0) ϭ 0 and for Cl, FC ϭ 7 Ϫ (1 ϩ 6) ϭ 0
In BeCl2, the chlorine atoms achieve the argon configuration, [Ar], and the beryllium
atom has a share of only four electrons. Compounds such as BeCl2, in which the central
atom shares fewer than 8 eϪ, are sometimes referred to as electron deficient compounds.
This “deficiency” refers only to satisfying the octet rule for the central atom. The term
does not imply that there are fewer electrons than there are protons in the nuclei, as in
the case of a cation, because the molecule is neutral.
A Lewis formula can be written for BeCl2 that does satisfy the octet rule (see structure
shown in the margin). Let us evaluate the formal charges for that formula:
for Be, FC ϭ 2 Ϫ (4 ϩ 0) ϭ Ϫ2 and for Cl, FC ϭ 7 Ϫ (2 ϩ 4) ϭ ϩ1
We have said that the most favorable structure for a molecule is one in which the formal
charge on each atom is zero, if possible. In case some atoms did have nonzero formal
charges, we would expect that the more electronegative atoms (Cl) would be the ones with
lowest formal charge. Thus, we prefer the Lewis structure shown in Example 7-5 over
the one in the margin.
One might expect a similar situation for compounds of the other IIA metals, Mg, Ca,
Sr, Ba, and Ra. These elements, however, have lower ionization energies and larger radii
than Be, so they usually form ions by losing two electrons.
EXAMPLE 7-6 Limitations of the Octet Rule
Write the Lewis formula for boron trichloride, BCl3, a covalent compound.
Plan
This covalent compound of boron is an example of limitation type B. As we follow the steps
in writing the Lewis formula, we use six electrons as the number needed by boron in step 2. Steps
3 and 4 should show only three pairs of electrons for boron.
Solution
Cl
Step 1: The skeleton is Cl B Cl
Be ClCl or Be ClCl
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 9.7, Electron-Deficient
Compounds.
Cl ClBe
see limitation type B
g
Step 2: N ϭ 3 ϫ 8 (for Cl) ϩ 1 ϫ 6 (for B) ϭ 30 eϪ needed
Step 3: A ϭ 3 ϫ 7 (for Cl) ϩ 1 ϫ 3 (for B) ϭ 24 eϪ available
Step 3: S ϭ N Ϫ A ϭ 30 eϪ Ϫ 24 eϪ ϭ 6 eϪ shared
Cl
hStep 3: ClSBSCl
Step 4:
Each chlorine atom achieves the Ne configuration. The boron (central) atom acquires a share
of only six valence shell electrons. Calculation of formal charges shows that
for B, FC ϭ 3 Ϫ (3 ϩ 0) ϭ 0 and for Cl, FC ϭ 7 Ϫ (1 ϩ 6) ϭ 0
You should now work Exercise 54.
EXAMPLE 7-7 Limitations of the Octet Rule
Write the Lewis formula for phosphorus pentafluoride, PF5, a covalent compound.
Plan
We apply the usual stepwise procedure to write the Lewis formula. In PF5, all five F atoms are
bonded to P. This requires the sharing of a minimum of 10 eϪ, so this is an example of limitation
type D. We therefore add the extra step 2a, and increase S from the calculated value of
8 eϪ to 10 eϪ.
Solution
Step 1: The skeleton is
Step 2: N ϭ 5 ϫ 8 (for F) ϩ 1 ϫ 8 (for P) ϭ 48 eϪ needed
Step 3: A ϭ 5 ϫ 7 (for F) ϩ 1 ϫ 5 (for P) ϭ 40 eϪ available
Step 3: S ϭ N Ϫ A ϭ 8 eϪ shared
Step 2: Five F atoms are bonded to P. This requires the sharing of a minimum of
10 eϪ. But only 8 eϪ have been calculated. This is therefore an example of
limitation type D.
Step 2a: Increase S from 8 eϪ to 10 eϪ, the number required to bond five F atoms to
one P atom. The number of electrons available, 40, does not change.
Step 3:
F
F
FF
F
P
B ClCl or
Cl Cl
B ClCl
7-7 Limitations of the Octet Rule for Lewis Formulas 293
F
F
FF
F
P
Step 4:
When the octets of the five F atoms have been satisfied, all 40 of the available electrons have
been added. The phosphorus (central) atom has a share of ten electrons.
Calculation of formal charges shows that
for P, FC ϭ 5 Ϫ (5 ϩ 0) ϭ 0 and for F, FC ϭ 7 Ϫ (1 ϩ 6) ϭ 0
When an atom has a share of more than eight electrons, as does P in PF5, we say that
it exhibits an expanded valence shell. The electronic basis of the octet rule is that one s and
three p orbitals in the valence shell of an atom can accommodate a maximum of eight
electrons. The valence shell of phosphorus has n ϭ 3, so it also has 3d orbitals available
that can be involved in bonding. It is for this reason that phosphorus (and many other
representative elements of Period 3 and beyond) can exhibit an expanded valence shell.
By contrast, elements in the second row of the periodic table can never exceed eight electrons
in their valence shells, because each atom has only one s and three p orbitals in that
shell. Thus, we understand why PF5 can exist but NF5 cannot.
EXAMPLE 7-8 Limitations of the Octet Rule
Write the Lewis formula for sulfur tetrafluoride, SF4.
Plan
We apply the usual stepwise procedure. The calculation of S ϭ N Ϫ A in step 2 shows only 6 eϪ
shared, but a minimum of 8 eϪ are required to bond four F atoms to the central S atom.
Limitation type D applies, and we proceed accordingly.
Solution
F F
Step 1: The skeleton is S
F F
Step 2: N ϭ 1 ϫ 8 (S atom) ϩ 4 ϫ 8 (F atoms) ϭ 40 eϪ needed
Step 3: A ϭ 1 ϫ 6 (S atom) ϩ 4 ϫ 7 (F atoms) ϭ 34 eϪ available
Step 3: S ϭ N Ϫ A ϭ 40 Ϫ 34 ϭ 6 eϪ shared. Four F atoms are bonded to the central
S. This requires a minimum of 8 eϪ, but only 6 eϪ have been calculated in
step 2. This is therefore an example of limitation type D.
Step 2a: We increase S from 6 eϪ to 8 eϪ.
Step 3:
Step 4: F
FF
F
S
F
FF
F
S
or
F
F
FF
F
P
F
F
FF
F
P
This is sometimes referred to as
hypervalence.
294 CHAPTER 7: Chemical Bonding
We first complete octets on the atoms around the central atom.
Step 4a: Now we have satisfied the octet rule, but we have used only 32 of the 34 eϪ
available. We place the other two electrons on the central S atom.
Calculation of the formal charge shows that
for S, FC ϭ 6 Ϫ (4 ϩ 2) ϭ 0
for F, FC ϭ 7 Ϫ (1 ϩ 6) ϭ 0
EXAMPLE 7-9 Limitations of the Octet Rule
Write the Lewis formula for the triiodide ion, I3
Ϫ.
Plan
We apply the usual stepwise procedure. The calculation of S ϭ N Ϫ A in step 2 shows only
2 eϪ shared, but a minimum of 4 eϪ are required to bond two I atoms to the central I. Limitation
type D applies, and we proceed accordingly.
Solution
Step 1: The skeleton is [I I I]Ϫ
Step 2: N ϭ 3 ϫ 8 (for I) ϭ 24 eϪ needed
Step 3: A ϭ 3 ϫ 7 (for I) ϩ 1 (for the 1Ϫ charge) ϭ 22 eϪ available
Step 3: S ϭ N Ϫ A ϭ 2 eϪ shared. Two I atoms are bonded to the central I. This
requires a minimum of 4 eϪ, but only 2 eϪ have been calculated in step 4.
This is therefore an example of limitation type D.
Step 2a: Increase S from 2 eϪ to 4 eϪ.
Step 3:
Step 4:
Step 4a: Now we have satisfied the octets of all atoms using only 20 of the 22 eϪ available.
We place the other two electrons on the central I atom.
Calculation of the formal charge shows that
for I on ends, FC ϭ 7 Ϫ (1 ϩ 6) ϭ 0
for I in middle, FC ϭ 7 Ϫ (2 ϩ 6) ϭ Ϫ1
You should now work Exercise 58.
We have seen that atoms attached to the central atom nearly always attain noble gas configurations,
even when the central atom does not.
Ϫ
orI I I
Ϫ
I I I
I ]
Ϫ
[ I I
I I]
Ϫ
[I
F
FF
F
S
F
FF
F
S In SF4, sulfur has an expanded valence
shell.
7-7 Limitations of the Octet Rule for Lewis Formula 295
The central iodine atom in I3
Ϫ has an
expanded valence shell.
POLAR AND NONPOLAR COVALENT BONDS
Covalent bonds may be either polar or nonpolar. In a nonpolar bond such as that in the
hydrogen molecule, H2, (HSH or HXH) the electron pair is shared equally between the
two hydrogen nuclei. We defined electronegativity as the tendency of an atom to attract
electrons to itself in a chemical bond (see Section 6-6). Both H atoms have the same electronegativity.
This means that the shared electrons are equally attracted to both hydrogen
nuclei and therefore spend equal amounts of time near each nucleus. In this nonpolar
covalent bond, the electron density is symmetrical about a plane that is perpendicular to
a line between the two nuclei. This is true for all homonuclear diatomic molecules, such as
H2, O2, N2, F2, and Cl2, because the two identical atoms have identical electronegativities.
We can generalize:
The covalent bonds in all homonuclear diatomic molecules must be nonpolar.
Let us now consider heteronuclear diatomic molecules. Start with the fact that hydrogen
fluoride, HF, is a gas at room temperature. This tells us that it is a covalent compound.
We also know that the HXF bond has some degree of polarity because H and F are not
identical atoms and therefore do not attract the electrons equally. But how polar will this
bond be?
The electronegativity of hydrogen is 2.1, and that of fluorine is 4.0 (see Table 6-3).
Clearly, the F atom, with its higher electronegativity, attracts the shared electron pair
much more strongly than does the H atom. We can represent the structure of HF as
shown in the margin. The electron density is distorted in the direction of the more electronegative
F atom. This small shift of electron density leaves H somewhat positive.
Covalent bonds, such as the one in HF, in which the electron pairs are shared unequally
are called polar covalent bonds. Two kinds of notation used to indicate polar bonds are
shown in the margin.
The ␦Ϫ over the F atom indicates a “partial negative charge.” This means that the F
end of the molecule is somewhat more negative than the H end. The ␦ϩ over the H atom
indicates a “partial positive charge,” or that the H end of the molecule is positive with
respect to the F end. We are not saying that H has a charge of ϩ1 or that F has a charge
of Ϫ1! A second way to indicate the polarity is to draw an arrow so that the head points
toward the negative end (F) of the bond and the crossed tail indicates the positive end (H).
The separation of charge in a polar covalent bond creates an electric dipole. We expect
the dipoles in the covalent molecules HF, HCl, HBr, and HI to be different because F,
Cl, Br, and I have different electronegativities. This tells us that atoms of these elements
have different tendencies to attract an electron pair that they share with hydrogen. We
indicate this difference as shown here, where ⌬(EN) is the difference in electronegativity
between two atoms that are bonded together.
|888n |88n |8n |n
HXF HXCl HXBr HXI
EN: 2.1 4.0 2.1 3.0 2.1 2.8 2.1 2.5
⌬(EN) 1.9 0.9 0.7 0.4
7-8
A homonuclear molecule contains
only one kind of atom. A molecule
that contains two or more kinds of
atoms is described as heteronuclear.
296 CHAPTER 7: Chemical Bonding
Remember that ionic compounds are
solids at room temperature.
FH
␦ϩ ␦Ϫ |88n
HXF or HXF
The word “dipole” means “two poles.”
Here it refers to the positive and
negative poles that result from the
separation of charge within a molecule.
The values of electronegativity are
obtained from Table 6-3.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 9.11, Bond Polarity and
Electronegativity.
H F




















Most polar Least polar
The longest arrow indicates the largest dipole, or greatest separation of electron density
in the molecule (see Table 7-3). For comparison, the ⌬(EN) values for some typical 1:1
ionic compounds are RbCl, 2.1; NaF, 3.0; and KCl, 2.1
EXAMPLE 7-10 Polar Bonds
Each halogen can form single covalent bonds with other halogens, to form compounds called
interhalogens; some examples are ClF and BrF. Use the electronegativity values in Table 6-3 to
rank the following single bonds from most polar to least polar: FXF, FXCl, FXBr, ClXBr,
ClXI, and ClXCl.
Plan
The bond polarity decreases as the electronegativity difference between the two atoms
decreases. We can calculate ⌬(EN) for each bond, and then arrange them according to
decreasing ⌬(EN) value.
Solution
We know that two F atoms have the same electronegativity, so ⌬(EN) for FXF must be zero,
and the FXF bond is nonpolar; the same reasoning applies to ClXCl. We use the values from
Table 6-3 to calculate ⌬(EN) for each of the other pairs, always subtracting the smaller from
the larger value:
Elements ⌬(EN)
F, Cl 4.0 Ϫ 3.0 ϭ 1.0
F, Br 4.0 Ϫ 2.8 ϭ 1.2
Cl, Br 3.0 Ϫ 2.8 ϭ 0.2
Cl, I 3.0 Ϫ 2.5 ϭ 0.5
The bonds, arranged from most polar to least polar, are:
FXBr FXCl ClXI ClXBr ClXCl FXF
EN: 4.0 2.8 4.0 3.0 3.0 2.5 3.0 2.8 3.0 3.0 4.0 4.0
⌬(EN): 1.2 1.0 0.5 0.2 0 0
You should now work Exercise 74.
DIPOLE MOMENTS
It is convenient to express bond polarities on a numerical scale. We indicate the polarity
of a molecule by its dipole moment, which measures the separation of charge within the
molecule. The dipole moment, ␮, is defined as the product of the distance, d, separating
charges of equal magnitude and opposite sign, and the magnitude of the charge, q. A
dipole moment is measured by placing a sample of the substance between two plates and
applying a voltage. This causes a small shift in electron density of any molecule, so the
applied voltage is diminished very slightly. Diatomic molecules that contain polar bonds,
however, such as HF, HCl, and CO, tend to orient themselves in the electric field (Figure
7-5). This causes the measured voltage between the plates to decrease more markedly for
these substances, and we say that these molecules are polar. Molecules such as F2 or N2
7-9
␮ ϭ d ϫ q
7-9 Dipole Moments 297






























TABLE 7-3 ⌬(EN) Values
and Dipole
Moments for
Some Pure
(Gaseous)
Substances
Dipole
Substance ⌬(EN) Moment (␮)*
HF 1.9 1.91 D
HCl 0.9 1.03 D
HBr 0.7 0.79 D
HI 0.4 0.38 D
HXH 0.0 0.00 D
*The magnitude of a dipole moment is given
by the product of charge ϫ distance of separation.
Molecular dipole moments are usually
expressed in debyes (D).
do not reorient, so the change in voltage between the plates remains slight; we say that
these molecules are nonpolar.
Generally, as electronegativity differences increase in diatomic molecules, the measured
dipole moments increase. This can be seen clearly from the data for the hydrogen halides
(see Table 7-3).
Unfortunately, the dipole moments associated with individual bonds can be measured
only in simple diatomic molecules. Entire molecules rather than selected pairs of atoms must
be subjected to measurement. Measured values of dipole moments reflect the overall polarities
of molecules. For polyatomic molecules they are the result of all the bond dipoles in
the molecules. In Chapter 8, we will see that structural features, such as molecular geometry
and the presence of lone (unshared) pairs of electrons, also affect the polarity of a
molecule.
THE CONTINUOUS RANGE OF BONDING TYPES
Let us now clarify our classification of bonding types. The degree of electron sharing or
transfer depends on the electronegativity difference between the bonding atoms. Nonpolar
covalent bonding (involving equal sharing of electron pairs) is one extreme, occurring when
the atoms are identical (⌬(EN) is zero). Ionic bonding (involving complete transfer of electrons)
represents the other extreme, and occurs when two elements with very different
electronegativities interact (⌬(EN) is large).
Polar covalent bonds may be thought of as intermediate between pure (nonpolar) covalent
bonds and pure ionic bonds. In fact, bond polarity is sometimes described in terms
of partial ionic character. This usually increases with increasing difference in electronegativity
between bonded atoms. Calculations based on the measured dipole moment
of gaseous HCl indicate about 17% “ionic character.”
When cations and anions interact strongly, some amount of electron sharing takes
place; in such cases we can consider the ionic compound as having some partial covalent
character. For instance, the high charge density of the very small Liϩ ion causes it to
distort large anions that it approaches. The distortion attracts electron density from the
anion to the region between it and the Liϩ ion, giving lithium compounds a higher degree
of covalent character than in other alkali metal compounds.
7-10
Figure 7-5 If polar molecules, such
as HF, are subjected to an electric
field, they tend to line up very
slightly in a direction opposite to
that of the field. This minimizes the
electrostatic energy of the molecules.
Nonpolar molecules are not oriented
by an electric field. The effect is
greatly exaggerated in this drawing.
298 CHAPTER 7: Chemical Bonding
Field off Field on
␦ϩ
␦ϩ
␦ϩ
␦ϩ
␦ϩ
␦ϩ
␦ϩ
␦ϩ
␦ϩ
␦ϩ ␦ϩ
␦ϩ
␦ϩ
␦ϩ ␦ϩ
␦ϩ
␦ϩ
␦ϩ
␦ϩ ␦ϩ
␦ϩ
␦Ϫ
␦ϩ ␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
ϩϪ
Almost all bonds have both ionic and covalent character. By experimental means, a
given type of bond can usually be identified as being “closer” to one or the other extreme
type. We find it useful and convenient to use the labels for the major classes of bonds to
describe simple substances, keeping in mind that they represent ranges of behavior.
Above all, we must recognize that any classification of a compound that we might
suggest based on electronic properties must be consistent with the physical properties of
ionic and covalent substances described at the beginning of the chapter. For instance, HCl
has a rather large electronegativity difference (0.9), and its aqueous solutions conduct electricity.
But we know that we cannot view it as an ionic compound because it is a gas, and
not a solid, at room temperature. Liquid HCl is a nonconductor.
Let us point out another aspect of the classification of compounds as ionic or covalent.
Not all ions consist of single charged atoms. Many are small groups of atoms that are
covalently bonded together, yet they still have excess positive or negative charge. Examples
of such polyatomic ions are ammonium ion, NH4
ϩ, sulfate ion, SO4
2Ϫ, and nitrate ion,
NO3
Ϫ. A compound such as potassium sulfate, K2SO4, contains potassium ions, Kϩ, and
sulfate ions, SO4
2Ϫ, in a 2:1 ratio. We should recognize that this compound contains both
covalent bonding (electron sharing within each sulfate ion) and ionic bonding (electrostatic
attractions between potassium and sulfate ions). We classify this compound as ionic,
however, because it is a high-melting solid (mp 1069°C), it conducts electricity both in
molten form and in aqueous solution, and it displays other properties that we generally
associate with ionic compounds. Put another way, covalent bonding holds each sulfate ion
together, but the forces that hold the entire substance together are ionic.
In summary, we can describe chemical bonding as a continuum that may be represented
as
⌬(EN) for the
bonding atoms zero 88n intermediate 88n large
Bonding types nonpolar covalent 88n polar covalent 88n ionic
HCl ionizes in aqueous solution. We
will study more about this behavior in
Chapter 10.
Key Terms 299
Ionic character increases
Covalent character increases
Key Terms
Anion A negatively charged ion; that is, an ion in which the atom
or group of atoms has more electrons than protons.
Binary compound A compound consisting of two elements; may
be ionic or covalent.
Bonding pair A pair of electrons involved in a covalent bond.
Also called shared pair.
Cation A positively charged ion; that is, an ion in which the atom
or group of atoms has fewer electrons than protons.
Chemical bonds Attractive forces that hold atoms together in
elements and compounds.
Covalent bond A chemical bond formed by the sharing of one
or more electron pairs between two atoms.
Covalent compound A compound containing predominantly
covalent bonds.
Debye The unit used to express dipole moments.
Exercises
Chemical Bonding: Basic Ideas
*01. What type of force is responsible for chemical bonding?
Explain the differences between ionic bonding and covalent
bonding.
*02. What kind of bonding (ionic or covalent) would you
predict for the products resulting from the following combinations
of elements?
(a) Na ϩ Cl2; (c) N2 ϩ O2,
(b) C ϩ O2; (d) S ϩ O2.
*03. Why are covalent bonds called directional bonds, whereas
ionic bonding is termed nondirectional?
*04. (a) What do Lewis dot formulas for atoms show? (b) Write
Lewis dot formulas for the following atoms: He; Si; P; Ne;
Mg; Br.
*05. Write Lewis dot formulas for the following atoms: Li; B;
As; K; Xe; Al.
*06. Describe the types of bonding in sodium chlorate, NaClO3.
O
O O
O
Na
ϩ
Cl
Ϫ
*07. Describe the types of bonding in ammonium chloride,
NH4Cl.
*08. Based on the positions in the periodic table of the following
pairs of elements, predict whether bonding between
the two would be primarily ionic or covalent. Justify your
answers. (a) Ca and Cl; (b) P and O; (c) Br and I; (d) Li
and I; (e) Si and Br; (f) Ba and F.
*09. Predict whether the bonding between the following pairs
of elements would be primarily ionic or covalent. Justify
your answers. (a) Rb and Cl; (b) N and O; (c) Ca and F;
(d) P and S; (e) C and F; (f) K and O.
*10. Classify the following compounds as ionic or covalent:
(a) Ca(NO3)2; (b) H2S; (c) KNO3; (d) CaCl2; (e) H2CO3;
(f ) PCl3; (g) Li2O; (h) N2H4; (i) SOCl2.
H
H
H HN Cl
Ϫ
ϩ
300 CHAPTER 7: Chemical Bonding
Delocalization of electrons Refers to bonding electrons distributed
among more than two atoms that are bonded together;
occurs in species that exhibit resonance.
Dipole Refers to the separation of charge between two covalently
bonded atoms.
Dipole moment (␮) The product of the distance separating
opposite charges of equal magnitude and the magnitude of the
charge; a measure of the polarity of a bond or molecule. A measured
dipole moment refers to the dipole moment of an entire
molecule.
Double bond A covalent bond resulting from the sharing of four
electrons (two pairs) between two atoms.
Electron-deficient compound A compound containing at
least one atom (other than H) that shares fewer than eight
electrons.
Formal charge The hypothetical charge on an atom in a covalently
bonded molecule or ion; bonding electrons are counted
as though they were shared equally between the two bonded
atoms.
Heteronuclear Consisting of different elements.
Homonuclear Consisting of only one element.
Ion An atom or a group of atoms that carries an electrical
charge.
Ionic bonding The attraction of oppositely charged ions (cations
and anions) in large numbers to form a solid. Ions result from
the transfer of one or more electrons from one atom or group
of atoms to another.
Ionic compound A compound containing predominantly ionic
bonding.
Isoelectronic Having the same number of electrons.
Lewis acid A substance that accepts a share in a pair of electrons
from another species.
Lewis base A substance that makes available a share in an electron
pair.
Lewis formula The representation of a molecule, ion, or formula
unit by showing atomic symbols and only outer-shell
electrons; does not represent the shape of the molecule or ion.
Lone pair A pair of electrons residing on one atom and not
shared by other atoms; unshared pair.
Monatomic ion An ion that consists of only one atom.
Nonpolar bond A covalent bond between two atoms with the
same electronegativity, so that the electron density is symmetrically
distributed.
Octet rule Many representative elements attain at least a share
of eight electrons in their valence shells when they form molecular
or ionic compounds; there are some limitations.
Polar bond A covalent bond between two atoms with different
electronegativities, so that the electron density is unsymmetrically
distributed.
Polyatomic ion An ion that consists of more than one atom.
Resonance A concept in which two or more Lewis formulas for
the same arrangement of atoms (resonance structures) are used
to describe the bonding in a molecule or ion.
Single bond A covalent bond resulting from the sharing of two
electrons (one pair) between two atoms.
Triple bond A covalent bond resulting from the sharing of six
electrons (three pairs) between two atoms.
Unshared pair See Lone pair.
Valence electrons The s and p electrons in the outermost shell
of an atom.
Exercises 301
Ionic Bonding
*11. Describe what happens to the valence electron(s) as a metal
atom and a nonmetal atom combine to form an ionic com-
pound.
*12. Describe an ionic crystal. What factors might determine
the geometrical arrangement of the ions?
*13. Why are solid ionic compounds rather poor conductors of
electricity? Why does conductivity increase when an ionic
compound is melted or dissolved in water?
*14. Write the formula for the ionic compound that forms
between each of the following pairs of elements: (a) Ca and
Br2; (b) Ba and Cl2; (c) Na and Cl2.
*15. Write the formula for the ionic compound that forms
between each of the following pairs of elements: (a) Cs and
F2; (b) Sr and S; (c) Na and Se.
*16. When a d-transition metal ionizes, it loses its outer s
electrons before it loses any d electrons. Using [nobel
gas] (n Ϫ 1) dx representations, write the outer-electron
configurations for the following ions: (a) Cr3ϩ;
(b) Mn2ϩ; (c) Agϩ; (d) Fe3ϩ; (e) Cu2ϩ; (f) Sc3ϩ;
(g) Fe2ϩ.
*17. Which of the following do not accurately represent stable
binary ionic compounds? Why? BaCl2; NaS; AlF4; SrS2;
Ca2O3; NaBr2; LiSe2.
*18. Which of the following do not accurately represent stable
binary ionic compounds? Why? MgI; Al(OH)2; InF2; CO2;
RbCl2; CsSe; Be2O.
*19. (a) Write Lewis formulas for the positive and negative
ions in these compounds: SrBr2; K2O; Ca3P2; PbCl2;
Bi2O3. (b) Which ions do not have a noble gas con-
figuration?
*20. Write formulas for two cations and two anions that are isoelectronic
with argon.
*21. Write formulas for two cations and two anions that are isoelectronic
with krypton.
*22. Write formulas for two cations that have each of the following
electron configurations in their highest occupied
energy level: (a) 3s23p6; (b) 4s24p6.
*23. Write formulas for two anions that have each of the electron
configurations listed in Exercise 22.
Covalent Bonding: General Concepts
*24. What does Figure 7-4 tell us about the attractive and repulsive
forces in a hydrogen molecule?
*25. Distinguish between heteronuclear and homonuclear
diatomic molecules.
*26. How many electrons are shared between two atoms in
(a) a single covalent bond, (b) a double covalent bond, and
(c) a triple covalent bond?
*27. What is the maximum number of covalent bonds that a
second-period element could form? How can the representative
elements beyond the second period form more
than this number of covalent bonds?
Lewis Formulas for Molecules and Polyatomic Ions
*28. What information about chemical bonding can a Lewis
formula give for a compound or ion? What information
about bonding is not directly represented by a Lewis
formula?
*29. Write Lewis formulas for the following: H2; N2; Cl2; HCl;
HBr.
*30. Write Lewis formulas for the following: H2O; NH3; OHϪ;
FϪ.
*31. Use Lewis formulas to represent the covalent molecules
formed by these pairs of elements. Write only structures
that satisfy the octet rule. (a) P and H; (b) Se and Br;
(c) N and Cl; (d) Si and Cl.
*32. Use Lewis formulas to represent the covalent molecules
formed by these pairs of elements. Write only structures
that satisfy the octet rule. (a) S and Cl; (b) As and F;
(c) I and Cl; (d) P and Cl.
*33. Find the total number of valence electrons in each of the
following molecules or ions.
(a) NH2
Ϫ (b) ClO3
Ϫ
(c) HCN (d) SnCl4
*34. How many valence electrons does each of these molecules
or ions have?
(a) H2S (b) PCl3
(c) NOCl (d) OHϪ
*35. Write Lewis structures for the molecules or ions in Exercise
33.
*36. Write Lewis structures for the molecules or ions in Exercise
34.
*37. Write Lewis structures for the following molecules or ions:
(a) ClO4
Ϫ; (b) C2H6O (two possibilities); (c) HOCl;
(d) SO3
2Ϫ.
*38. Write Lewis structures for the following molecules or ions:
(a) H2CO; (b) ClF; (c) BF4
Ϫ; (d) PO4
3Ϫ.
*39. (a) Write the Lewis formula for AlCl3, a molecular compound.
Note that in AlCl3, the aluminum atom is an
exception to the octet rule. (b) In the gaseous phase, two
molecules of AlCl3 join together (dimerize) to form
Al2Cl6. (The two molecules are joined by two “bridging”
AlXClXAl bonds.) Write the Lewis formula for this
molecule.
*40. Write the Lewis formula for molecular ClO2. There is
a single unshared electron on the chlorine atom in this
molecule.
*41. Write Lewis formulas for CH4 and SiH4; explain the sim-
ilarity.
*42. Write Lewis formulas for CCl4, SiF4, and PbI4; explain the
similarity.
*43. Write Lewis formulas for the fluorine molecule and for
sodium fluoride. Describe the nature of the chemical bonding
involved in each substance.
*44. Write Lewis formulas for butane, CH3CH2CH2CH3,
and propane, CH3CH2CH3. Describe the nature
of the bond indicated: CH3CH2XCH2CH3 and
CH3XCH2CH3.
302 CHAPTER 7: Chemical Bonding
*45. Write Lewis structures for the following molecules:
(a) Formic acid, HCOOH, in which the atomic arrangement
is
O
H C O H
(b) Acetonitrile, CH3CN;
(c) Vinyl chloride, CH2CHCl, the molecule from which
PVC plastics are made.
*46. Write Lewis structures for the following molecules:
(a) Tetrafluoroethylene, C2F4, the molecule from which
Teflon is made;
(b) Acrylonitrile, CH2CHCN, the molecule from which
Orlon is made.
*47. What do we mean by the term “resonance”? Do the resonance
structures that we draw actually represent the
bonding in the substance? Explain your answer.
*48. Careful examination of the ozone molecule indicates that
the two outer oxygens are the same distance from the central
oxygen. Write Lewis formulas or resonance structures
that are consistent with this finding.
*49. Draw resonance structures for the nitric acid molecule.
*50. We can write two resonance structures for toluene,
C6H5CH3:
How would you expect the carbon–carbon bond lengths
in the six-membered ring to compare with the
carbon–carbon bond length between the CH3 group and
the carbon atom on the ring?
*51. Write resonance structures for the formate ion,
HCOOϪ.
*52. Write resonance structures for the nitrate ion, NO3
Ϫ.
*53. Write Lewis formulas for (a) H2NOH (i.e., one H
bonded to O); (b) S8 (a ring of eight atoms); (c) PCl3;
(d) F2O2 (O atoms in center, F atoms on outside); (e) CO;
(f ) SeCl6.
*54. Write the Lewis formula for each of the following covalent
compounds. Which ones contain at least one atom with a
share in less than an octet of valence electrons? (a) BeBr2;
(b) BBr3; (c) NCl3; (d) AlCl3.
*55. Which of the following species contain at least one atom
that violates the octet rule?
(a) (b) (c) F FXeO OClF Cl
H
H
HH
H
H
H
H
C
C
C
CC
CC
H
H
HH
H
H
H
H
C
C
C
CC
CC
(d)
*56. Write the Lewis formula for each of the following molecules
or ions. Which ones contain at least one atom with
a share in less than an octet of valence electrons? (a)
CH2Cl2; (b) BF3; (c) BCl4
Ϫ; (d) AlF4
Ϫ.
*57. None of the following is known to exist. What is wrong
with each one?
(a) (d)
(b)
(c)
*58. Write the Lewis formula for each of the following molecules
or ions. Which ones contain at least one atom with
a share in more than an octet of valence electrons? (a) PF6
Ϫ;
(b) AsCl5; (c) ICl3; (d) C2H6.
*59. Suppose that “El” is the general symbol for a representative
element. In each case, in which periodic group is El
located? Justify your answers and cite a specific example
for each one.
(a) (c)
(b) (d)
*60. Suppose that “El” is the general symbol for a representative
element. In each case, in which periodic group is El
located? Justify your answers and cite a specific example
for each one.
(a) (c)
(b) (d)
*61. Many common stains, such as those of chocolate and other
fatty foods, can be removed by dry-cleaning solvents such
as tetrachloroethylene, C2Cl4. Is C2Cl4 ionic or covalent?
Write its Lewis formula.
H
H HEl
ϩ
O El O
H
H
H HEl
ϩ
Br
Br
Br El
ElOH H
HO
O OEl
2Ϫ
O
El OOH
O
O OEl
Ϫ
NO O
Ϫ
HH ClO P
Cl
Na O
Na
SCl O
O
O
O O
O
S
2Ϫ
(e) O
–
O Cl
Exercises 303
*62. Write acceptable Lewis formulas for the following common
air pollutants: (a) SO2; (b) NO2; (c) CO; (d) O3
(ozone); (e) SO3; (f ) (NH4)2SO4. Which one is a solid?
Which ones exhibit resonance? Which ones violate the
octet rule?
Formal Charges
*63. Assign a formal charge to each atom in the following:
(a) (d)
(b) S OO
O
O OC
2Ϫ
Cl
Cl O
*70. How does one predict that a covalent bond between two
atoms is polar?
*71. Ionic compounds generally have a higher melting point
than covalent compounds. What is the major difference in
the structures of ionic and covalent compounds that
explains the difference in melting points?
*72. Explain why the electrons in the carbon–fluorine covalent
bond tend to move more toward the halogen atom
than do the electrons in the carbon–bromine covalent
bond.
*73. Why do we show only partial charges, and not full charges,
on the atoms of a polar molecule?
*74. In each pair of bonds, indicate the more polar bond, and
use ␦ϩ and ␦Ϫ to show the direction of polarity in each
bond.
(a) CXO and CXN
(b) BXO and PXS
(c) PXH and PXN
(d) BXH and BXI
*75. The molecule below is urea, a compound used in plastics
and fertilizers.
(a) Which bonds in this molecule are polar and which are
nonpolar?
(b) Which is the most polar bond in the molecule? Which
atom is the partial negative end of this bond?
*76. (a) Which two of the following pairs of elements are most
likely to form ionic bonds? Te and H; C and F; Ba and F;
N and F; K and O. (b) Of the remaining three pairs, which
one forms the least polar, and which the most polar, covalent
bond?
*77. (a) List three reasonably nonpolar covalent bonds between
dissimilar atoms. (b) List three pairs of elements whose
compounds should exhibit extreme ionic character.
*78. Classify the bonding between the following pairs of
atoms as ionic, polar covalent, or nonpolar covalent. (a) Li
and O; (b) Br and I; (c) Na and H; (d) O and O; (e) H
and O.
*79. For each of the following, tell whether the bonding is primarily
ionic or covalent.
(a) potassium and iodine in potassium iodide;
(b) beryllium and the nitrate ion in beryllium nitrate;
(c) the carbon–carbon bond in CH3CH3;
(d) carbon and oxygen in carbon monoxide;
(e) phosphorus and oxygen in the phosphate ion.
*80. Identify the bond in each of the following bonded pairs
that is likely to have the greater proportion of “ionic char-
acter.”
(a) NaXCl or MgXCl; (b) CaXS or FeXS;
(c) AlXBr or OXBr; (d) RaXH or CXH.
N
H H
H H
C
O
N
(e)(c)
*64. Assign a formal charge to each atom in the following:
(a) (b) (c) O OC
FF
F
P
FF
F FAs
F
O OCl
O O
O O
Cl O O OCl
O
O
Ϫ
(d) (e)
*65. With the aid of formal charges, explain which Lewis formula
is more likely to be correct for each given molecule.
(a) For Cl2O,
(b) For HN3,
(c) For N2O,
*66. Write Lewis formulas for three different atomic arrangements
with the molecular formula HCNO. Indicate all
formal charges. Predict which arrangement is likely to be
the least stable and justify your selection.
Ionic Versus Covalent Character and Bond Polarities
*67. Distinguish between polar and nonpolar covalent bonds.
*68. Why is an HCl molecule polar but a Cl2 molecule is non-
polar?
*69. How does one predict that the chemical bonding between
two elements is likely to be ionic?
N O N or N N O
H N N N or H N N N
Cl O Cl Cl OClor
Cl ClAl
Cl
Cl
Ϫ
O ON
ϩ
304 CHAPTER 7: Chemical Bonding
*81. Identify the bond in each of the following bonded pairs
that is likely to have the greater proportion of “covalent
character.”
(a) KXBr or CrXBr; (b) LiXO or HXO;
(c) AlXCl or CXCl; (d) AsXS or OXS.
CONCEPTUAL EXERCISES
*82. When asked to give an example of resonance structures, a
student drew the following. Why is this example incorrect?
*83. Why is the following not an example of resonance struc-
tures?
*84. Describe the circumstances under which one would expect
a bond to exhibit 100% “covalent character” and 0% “ionic
character.” Give an example of two bonded atoms that
would be predicted to exhibit 0% “ionic character.”
*85. The following properties can be found in a handbook of
chemistry:
S C N S CN
H H
H
H
C
O
C H
H
H
C
OH
C
camphor, C10H16O—colorless crystals; specific gravity,
0.990 at 25°C; sublimes, 204°C; insoluble in water; very
soluble in alcohol and ether.
praseodymium chloride, PrCl3—blue-green needle crystals;
specific gravity, 4.02; melting point, 786°C; boiling
point, 1700°C; solubility in cold water, 103.9 g/100 mL
H2O; very soluble in hot water.
Would you describe each of these as ionic or covalent?
Why?
BUILDING YOUR KNOWLEDGE
*86. Look up the properties of NaCl and PCl3 in a handbook
of chemistry. Why do we describe NaCl as an ionic compound
and PCl3 as a covalent compound?
*87. (a) How many moles of electrons are transferred when
10.0 g of magnesium react as completely as possible with
10.0 g of fluorine to form MgF2? (b) How many electrons
is this? (c) Look up the charge on the electron in coulombs.
What is the total charge, in coulombs, that is transferred?
*88. Write the formula for the compound that forms between
(a) calcium and nitrogen, (b) aluminum and oxygen,
(c) potassium and selenium, and (d) strontium and chlorine.
Classify each compound as covalent or ionic.
*89. Write the Lewis formulas for the nitric acid molecule
(HNO3) that are consistent with the following bond length
data: 1.405 Å for the bond between the nitrogen atom and
the oxygen atom that is attached to the hydrogen atom;
1.206 Å for the bonds between the nitrogen atom and each
of the other oxygen atoms.
*90. Write the total ionic and net ionic equations for the reaction
between each of the following pairs of compounds in
aqueous solution. Then give the Lewis formula for each
species in these equations. (a) HCN and NaOH; (b) HCl
and NaOH; (c) CaCl2 and Na2CO3.
*91. Sketch a portion of an aqueous solution of NaCl. Show the
Lewis formulas of the solute and solvent species. Suggest
the relative location of each species with respect to the oth-
ers.
*92. Sketch a portion of an aqueous solution of CH3COOH.
Show the Lewis formulas of the solute and solvent species.
Suggest the relative location of each species with respect
to the others.
*93. Determine the oxidation number of each element in the
following compounds: (a) CO2; (b) CH4; (c) PF3; (d) PF5;
(e) Na2O; (f ) Na2O2.Camphor sublimes readily.
8Molecular Structure
and Covalent
Bonding Theories
WW
e know a great deal about the molecular structures of many thousands of
compounds, all based on reliable experiments. In our discussion of theories
of covalent bonding, we must keep in mind that the theories represent an
attempt to explain and organize experimental observations. For bonding theories to be valid,
they must be consistent with the large body of experimental observations about molec-
OUTLINE
8-1 A Preview of the Chapter
8-2 Valence Shell Electron Pair
Repulsion (VSEPR) Theory
8-3 Polar Molecules: The Influence
of Molecular Geometry
8-4 Valence Bond (VB) Theory
Molecular Shapes and Bonding
8-5 Linear Electronic Geometry:
AB2 Species (No Lone Pairs
of Electrons on A)
8-6 Trigonal Planar Electronic
Geometry: AB3 Species (No
Lone Pairs of Electrons on A)
8-7 Tetrahedral Electronic
Geometry: AB4 Species (No
Lone Pairs of Electrons on A)
8-8 Tetrahedral Electronic
Geometry: AB3U Species (One
Lone Pair of Electrons on A)
8-9 Tetrahedral Electronic
Geometry: AB2U2 Species (Two
Lone Pairs of Electrons on A)
8-10 Tetrahedral Electronic
Geometry: ABU3 Species
(Three Lone Pairs of Electrons
on A)
8-11 Trigonal Bipyramidal Electronic
Geometry: AB5, AB4U, AB3U2,
and AB2U3
8-12 Octahedral Electronic
Geometry: AB6, AB5U, and
AB4U2
8-13 Compounds Containing Double
Bonds
8-14 Compounds Containing Triple
Bonds
8-15 A Summary of Electronic and
Molecular Geometries
OBJECTIVES
After you have studied this chapter, you should be able to
• Describe the basic ideas of the valence shell electron pair repulsion (VSEPR) theory
• Use the VSEPR theory to predict the electronic geometry of polyatomic molecules and
ions
• Use the VSEPR theory to predict the molecular geometry of polyatomic molecules and
ions
• Describe the relationships between molecular shapes and molecular polarities
• Predict whether a molecule is polar or nonpolar
• Describe the basic ideas of the valence bond (VB) theory
• Analyze the hybrid orbitals used in bonding in polyatomic molecules and ions
• Use hybrid orbitals to describe the bonding in double and triple bonds
A computer model of a polar
molecule of wood alcohol, or
methanol, CH3OH (C, gray;
H, white; O, red). The ball-andstick
model is shown inside a
computer-generated molecular
surface. The surface is color-coded to
show how charge ranges from very
positive (red) to very negative
(blue).
In Chapter 7 we used valence bond
terminology to discuss the bonding in
H2 although we did not name the
theory there.
8-1 A Preview of the Chapter 307
ular structure. In this chapter we will study two theories of covalent bonding, which usually
allow us to predict correct structures and properties. Like any simplified theories, they
are not entirely satisfactory in describing every known structure; however, their successful
application to many thousands of structures justifies their continued use.
A PREVIEW OF THE CHAPTER
The electrons in the outer shell, or valence shell, of an atom are the electrons involved
in bonding. In most of our discussion of covalent bonding, we will focus attention on
these electrons. Valence shell electrons are those that were not present in the preceding
noble gas, ignoring filled sets of d and f orbitals. Lewis formulas show the number of
valence shell electrons in a polyatomic molecule or ion (Sections 7-4 through 7-7). We
will write Lewis formulas for each molecule or polyatomic ion we discuss. The theories
introduced in this chapter apply equally well to polyatomic molecules and to ions.
Two theories go hand in hand in a discussion of covalent bonding. The valence shell
electron pair repulsion (VSEPR) theory helps us to understand and predict the spatial arrangement
of atoms in a polyatomic molecule or ion. It does not, however, explain how bonding
occurs, just where it occurs and where unshared pairs of valence shell electrons are directed.
The valence bond (VB) theory describes how the bonding takes place, in terms of overlapping
atomic orbitals. In this theory, the atomic orbitals discussed in Chapter 5 are often
“mixed,” or hybridized, to form new orbitals with different spatial orientations. Used
together, these two simple ideas enable us to understand the bonding, molecular shapes,
and properties of a wide variety of polyatomic molecules and ions.
We will first discuss the basic ideas and application of these two theories. Then we will
learn how an important molecular property, polarity, depends on molecular shape. Most
of this chapter will then be devoted to studying how these ideas are applied to various
types of polyatomic molecules and ions.
8-1
Different instructors prefer to cover these two theories in
different ways. Your instructor will tell you the order in
which you should study the material in this chapter. Regardless of how you study this
chapter, Tables 8-1, 8-2, 8-3, and 8-4 are important summaries, and you should refer to
them often.
1. One approach is to discuss both the VSEPR theory and the VB theory together,
emphasizing how they complement each other. If your instructor prefers this
parallel approach, you should study the chapter in the order in which it is presented.
2. An alternative approach is to first master the VSEPR theory and the related topic
of molecular polarity for different structures, and then learn how the VB theory
describes the overlap of bonding orbitals in these structures. If your instructor
takes this approach, you should study this chapter in the following order:
a. Read the summary material under the main heading “Molecular Shapes and
Bonding” preceding Section 8-5.
b. VSEPR theory, molecular polarity. Study Sections 8-2 and 8-3; then in Sections
8-5 through 8-12, study only the subsections marked A and B.
c. VB theory. Study Section 8-4; then in Sections 8-5 through 8-12, study the
Valence Bond Theory subsections, marked C; then study Sections 8-13 and 8-14.
IMPORTANT NOTE
No matter which order your instructor prefers, the following procedure will help you
analyze the structure and bonding in any compound.
1. Write the Lewis formula for the molecule or polyatomic ion, and identify a central
atom—an atom that is bonded to more than one other atom (Section 8-2).
2. Count the number of regions of high electron density on the central atom (Section
8-2).
3. Apply the VSEPR theory to determine the arrangement of the regions of high
electron density (the electronic geometry) about the central atom (Section 8-2; Tables
8-1 and 8-4).
4. Using the Lewis formula as a guide, determine the arrangement of the bonded
atoms (the molecular geometry) about the central atom, as well as the location of
the unshared valence electron pairs on that atom (parts B of Sections 8-5 through
8-12; Tables 8-3 and 8-4). This description includes ideal bond angles.
5. If there are lone (unshared) pairs of valence shell electrons on the central atom,
consider how their presence might modify somewhat the ideal molecular geometry
and bond angles deduced in step 4 (Section 8-2; of Sections 8-8 through
8-12).
6. Use the VB theory to determine the hybrid orbitals utilized by the central atom;
describe the overlap of these orbitals to form bonds; describe the orbitals that
contain unshared pairs of valence shell electrons on the central atom (parts C of
Sections 8-5 through 8-12; Sections 8-13; 8-14; Tables 8-2 and 8-4).
7. If more than one atom can be identified as a central atom, repeat steps 2 through
6 for each central atom, to build up a picture of the geometry and bonding in
the entire molecule or ion.
8. When all central atoms in the molecule or ion have been accounted for, use the
entire molecular geometry, electronegativity differences, and the presence of lone
pairs of valence shell electrons on the central atom to predict molecular polarity
(Section 8-3; parts B of Sections 8-5 through 8-12).
The following diagram summarizes this procedure.
Never skip to step 4 until you have
done step 3. The electronic geometry
and the molecular geometry may or
may not be the same; knowing the
electronic geometry first will enable
you to find the correct molecular
geometry.
308 CHAPTER 8: Molecular Structure and Covalent Bonding Theories
7.8. 6. 5.
2.1. 3.Write Lewis
formula and
identify a
central atom
Count
regions of
high electron
density on
that central
atom
Determine
electronic
geometry
around that
central atom
Determine
molecular
geometry
around that
central atom
Determine
whether
molecule is
polar or
nonpolar
Can another
central atom
be identified?
Determine
hybrid orbitals,
describe bonding
Adjust
molecular
geometry for
any lone
pairs
4.
YES
NO
Learn this procedure, and use it as a
mental “checklist.” Trying to do this
reasoning in a different order often
leads to confusion or wrong answers.
Many chemists use the terms lone pair
and unshared pair interchangeably, as
we will do throughout this discussion.
In Section 7-4 we showed that Lewis formulas of polyatomic ions can be constructed
in the same way as those of neutral molecules. Once the Lewis formula of an ion is known,
we use the VSEPR and VB theories to deduce its electronic geometry, shape, and
hybridization, just as for neutral molecules.
VALENCE SHELL ELECTRON PAIR REPULSION
(VSEPR) THEORY
The basic ideas of the valence shell electron pair repulsion (VSEPR) theory are:
Each set of valence shell electrons on a central atom is significant. The sets of valence
shell electrons on the central atom repel one another. They are arranged about the
central atom so that repulsions among them are as small as possible.
This results in maximum separation of the regions of high electron density about the
central atom.
A central atom is any atom that is bonded to more than one other atom. In some
molecules, more than one central atom may be present. In such cases, we determine the
arrangement around each in turn, to build up a picture of the overall shape of the entire
molecule or ion. We first count the number of regions of high electron density around
the central atom, as follows:
1. Each bonded atom is counted as one region of high electron density, whether the
bonding is single, double, or triple.
2. Each unshared pair of valence electrons on the central atom is counted as one
region of high electron density.
Consider the following molecules and polyatomic ions as examples.
8-2
Recall that we must take into account
the “extra” electrons on anions and the
“missing” electrons of cations.
8-2 Valence Shell Electron Pair Repulsion (VSEPR) Theory 309
Lewis dot formula:
Formula:
Central atom:
Number of atoms
bonded to central atom:
Number of unshared
pairs on central atom:
Total number of regions
of high electron density
on central atom:
O
O
O OS
2Ϫ
CO O
H
H
H HC
H
H
HN
CO2 NH3 CH4 SO4
2Ϫ
C N C S
2
2
3 4
444
4
000 1
The VSEPR theory assumes that
regions of high electron density
(electron pairs) on the central atom
will be as far from one another as
possible.
According to VSEPR theory, the structure is most stable when the regions of high
electron density on the central atom are as far apart as possible. The arrangement
of these regions of high electron density around the central atom is referred to as the
electronic geometry of the central atom.
For instance, two regions of high electron density are most stable on opposite sides of the
central atom (the linear arrangement). Three regions are most stable when they are
arranged at the corners of an equilateral triangle. Each different number of regions of
high electron density corresponds to a most stable arrangement of those regions. Table
8-1 shows the relationship between the common numbers or regions of high electron
density and their corresponding electronic geometries. After we know the electronic geometry
(and only then), we consider how many of these regions of high electron density
connect (bond) the central atom to other atoms. This lets us deduce the arrangement of
atoms around the central atom, called the molecular geometry. If necessary, we repeat
this procedure for each central atom in the molecule or ion. These procedures are illustrated
in parts B of Section 8-5 through 8-12.
POLAR MOLECULES: THE INFLUENCE OF
MOLECULAR GEOMETRY
In Chapter 7 we saw that the unequal sharing of electrons between two atoms with different
electronegativities, ⌬(EN) Ͼ 0, results in a polar bond. For heteronuclear diatomic molecules
such as HF, this bond polarity results in a polar molecule. Then the entire molecule
acts as a dipole, and we would find that the molecule has a measurable dipole moment, that
is, greater than zero.
When a molecule consists of more than two atoms joined by polar bonds, we must also
take into account the arrangement of the resulting bond dipoles in deciding whether or not
a molecule is polar. For such a case, we first use the VSEPR theory to deduce the molecular
geometry (arrangement of atoms), as described in the preceding section and
exemplified in parts A and B of Sections 8-5 through 8-12. Then we determine whether
the bond dipoles are arranged in such a way that they cancel (so that the resulting molecule
is nonpolar) or do not cancel (so that the resulting molecule is polar).
In this section we will discuss the ideas of cancellation of dipoles in general terms, using
general atomic symbols A and B. Then we will apply these ideas to specific molecular
geometries and molecular polarities in parts B of Sections 8-5 through 8-12.
Let us consider a heteronuclear triatomic molecule with the formula AB2 (A is the
central atom). Such a molecule must have one of the following two molecular geometries:
Suppose that atom B has a higher electronegativity than atom A. Then each AXB bond
is polar, with the negative end of the bond dipole pointing toward B. We can view each
bond dipole as an electronic vector, with a magnitude and a direction. In the linear AB2 arrangement,
the two bond dipoles are equal in magnitude and opposite in direction. They therefore
cancel to give a nonpolar molecule (dipole moment equal to zero).
AB B B
B
Aor
linear angular
8-3
Although the terminology is not as
precise as we might wish, we use
“molecular geometry” to describe the
arrangement of atoms in polyatomic ions
as well as in molecules.
310 CHAPTER 8: Molecular Structure and Covalent Bonding Theories
The angular form could have different
angles, but either the molecule is
linear or it is not. The angular
arrangement is sometimes called
V-shaped or bent.
8-3 Polar Molecules: The Influence of Molecular Geometry 311
TABLE 8-1 Number of Regions of High Electron Density About a Central Atom
No. Regions
Electronic Geometry*
of High Electron Description;
Density Angles† Line Drawing‡ Stereo View§
2 linear;
180°
3 trigonal planar;
120°
4 tetrahedral;
109.5°
5 trigonal bipyramidal;
90°, 120°, 180°
6 octahedral;
90°, 180°
*Electronic geometries are illustrated here using only single pairs of electrons as regions of high electron density. Each orange sphere represents a region of high electron
density about the central atom (gray sphere). Each may represent either an atom bonded to the central atom or a lone pair on the central atom.
†Angles made by imaginary lines through the nucleus and the centers of regions of high electron density.
‡By convention, a line in the plane of the drawing is represented by a solid line XX, a line behind this plane is shown as a dashed line ,,, and a line in front of
this plane is shown as a wedge ! with the fat end of the wedge nearest the viewer.
§View the three-dimensional drawings with the aid of the viewer inside the cover of the text.
In the case of the angular arrangement, the two equal dipoles do not cancel, but add to give
a dipole moment greater than zero. The angular molecular arrangement represents a polar
molecule.
If the electronegativity differences were reversed in this BXAXB molecule—that is, if A
were more electronegative than B—the directions of all bond polarities would be reversed.
But the bond polarities would still cancel in the linear arrangement to give a nonpolar
molecule. In the angular arrangement, bond polarities would still add to give a polar molecule,
but with the net dipole pointing in the opposite direction from that described earlier.
We can make similar arguments based on addition of bond dipoles for other arrangements.
As we will see in Section 8-8, lone pairs on the central atom can also affect the
direction and the magnitude of the net molecular dipole, so the presence of lone pairs on
the central atom must always be taken into account.
For a molecule to be polar, two conditions must be met:
1. There must be at least one polar bond or one lone (unshared) pair on the central
atom.
and
2. a. The polar bonds, if there are more than one, must not be arranged so that
their polarities (bond dipoles) cancel.
or
b. If there are two or more lone (unshared) pairs on the central atom, they must
not be arranged so that their polarities cancel.
Put another way, if there are no polar bonds or unshared pairs of electrons on the central
atom, the molecule cannot be polar. Even if polar bonds or unshared pairs are present,
they may be arranged so that their polarities cancel one another, resulting in a nonpolar
molecule.
Carbon dioxide, CO2, is a three-atom molecule in which each carbon–oxygen bond is
polar because of the electronegativity difference between C and O. But the molecule as a
whole is shown by experiment (dipole moment measurement) to be nonpolar. This tells
us that the polar bonds are arranged in such a way that the bond polarities cancel. Water,
H2O, on the other hand, is a very polar molecule; this tells us that the HXO bond polarities
do not cancel one another. Molecular shapes clearly play a crucial role in determining
molecular dipole moments. We will develop a better understanding of molecular shapes
in order to understand molecular polarities.
The logic used in deducing whether a molecule is polar or nonpolar is outlined in
Figure 8-1. The approach described in this section will be applied to various electronic
and molecular geometries in parts B of Sections 8-5 through 8-12.
AB
B
Net dipole Ͼ 0
(polar molecule)
AB B
Net dipole ϭ 0
(nonpolar molecule)
OH
H
angular molecule;
bond dipoles do not cancel;
molecule is polar
CO O
linear molecule;
bond dipoles cancel;
molecule is nonpolar
312 CHAPTER 8: Molecular Structure and Covalent Bonding Theories
VALENCE BOND (VB) THEORY
In Chapter 7 we described covalent bonding as electron pair sharing that results from the
overlap of orbitals from two atoms. This is the basic idea of the valence bond (VB) theory
—it describes how bonding occurs. In many examples throughout this chapter, we first
use the VSEPR theory to describe the orientations of the regions of high electron density.
Then we use the VB theory to describe the atomic orbitals that overlap to produce the
bonding with that geometry. We also assume that each lone pair occupies a separate orbital.
Thus, the two theories work together to give a fuller description of the bonding.
We learned in Chapter 5 that an isolated atom has its electrons arranged in orbitals in
the way that leads to the lowest total energy for the atom. Usually, however, these “pure
atomic” orbitals do not have the correct energies or orientations to describe where the
electrons are when an atom is bonded to other atoms. When other atoms are nearby as
in a molecule or ion, an atom can combine its valence shell orbitals to form a new set of
orbitals that is at a lower total energy in the presence of the other atoms than the pure
atomic orbitals would be. This process is called hybridization, and the new orbitals that
are formed are called hybrid orbitals. These hybrid orbitals can overlap with orbitals on
other atoms to share electrons and form bonds. Such hybrid orbitals usually give an
improved description of the experimentally observed geometry of the molecule or ion.
The designation (label) given to a set of hybridized orbitals reflects the number and
kind of atomic orbitals that hybridize to produce the set (Table 8-2). Further details about
hybridization and hybrid orbitals appear in the following sections. Throughout the text,
hybrid orbitals are shaded in green.
8-4
Figure 8-1 A guide to determining
whether a polyatomic molecule is
polar or nonpolar. Study the more
detailed presentation in the text.
TABLE 8-2 Relation Between Electronic Geometries and Hybridization
Regions of Atomic Orbitals Mixed
High Electron Electronic from Valence Shell
Density Geometry of Central Atom Hybridization
2 linear one s, one p sp
3 trigonal planar one s, two p’s sp2
4 tetrahedral one s, three p’s sp3
5 trigonal bipyramidal one s, three p’s, one d sp3d
6 octahedral one s, three p’s, two d’s sp3d2
NO
Are polar
bonds
arranged so
that they
cancel?
Are there
lone pairs on
the central
atom?
Are lone pairs
on central
atom arranged
so that they
cancel?
Molecule is
POLAR
Are there
polar bonds
present?
Molecule is
POLAR
Molecule is
NONPOLAR
Molecule is
NONPOLAR
NONO
YES
YES
YES
NO
YES
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 10.2, Models of Chemical
Bonding, and Screen 10.3, Valence
Bond Theory.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 10.6, Determining Hybrid
Orbitals.
We can describe hybridization as the
mathematical combination of the
waves that represent the orbitals of
the atom. This is analogous to the
formation of new waves on the
surface of water when different
waves interact.
MOLECULAR SHAPES AND BONDING
We are now ready to study the structures of some simple molecules. We often refer to
generalized chemical formulas in which “A” represents the central atom and “B” represents
an atom bonded to A. We follow the eight steps of analysis outlined in Section 8-1.
We first give the known (experimentally determined) facts about polarity and shape and
write the Lewis formula (part A of each section). Then we explain these facts in terms of
the VSEPR and VB theories. The simpler VSEPR theory will be used to explain (or
predict) first the electronic geometry and then the molecular geometry in the molecule (part
B). We then show how the molecular polarity of a molecule is a result of bond polarities,
lone pairs, and molecular geometry. Finally, we use the VB theory to describe the bonding
in molecules in more detail, usually using hybrid orbitals (part C). As you study each
section, refer frequently to the summaries that appear in Table 8-4.
LINEAR ELECTRONIC GEOMETRY: AB2 SPECIES
(NO LONE PAIRS OF ELECTRONS ON A)
A. Experimental Facts and Lewis Formulas
Several linear molecules consist of a central atom plus two atoms of another element,
abbreviated as AB2. These compounds include gaseous BeCl2, BeBr2, and BeI2, as well
as CdX2 and HgX2, where X ϭ Cl, Br, or I. All of these are known to be linear (bond
angle ϭ 180°), nonpolar, covalent compounds.
Let’s focus on gaseous BeCl2 molecules (mp 405°C). We wrote the Lewis formula for
BeCl2 in Example 7-5. It shows two single covalent bonds, with Be and Cl each
contributing one electron to each bond.
In many of its compounds, Be does not satisfy the octet rule (see Section 7-7).
B. VSEPR Theory
Valence shell electron pair repulsion theory places the two electron pairs on Be 180° apart,
that is, with linear electronic geometry. Both electron pairs are bonding pairs, so VSEPR
also predicts a linear atomic arrangement, or linear molecular geometry, for BeCl2.
If we examine the bond dipoles, we see that the electronegativity difference (see Table
6-3) is large (1.5 units) and each bond is quite polar:
The two bond dipoles are equal in magnitude and opposite in direction. They therefore
cancel to give a nonpolar molecule.
Cl ClBe
Cl ClBe
Net dipole ϭ 0
3.03.0 1.5
1.5 1.5
ϭ
⌬(EN) ϭ
EN
Cl ClBe
180Њ
Cl ClBe Cl ClBeor
8-5
See the “Important Note” in Section
8-1; consult your instructor for
guidance on the order in which you
should study Sections 8-5 through
8-12.
314 CHAPTER 8: Molecular Structure and Covalent Bonding Theories
A model of BeCl2, a linear AB2
molecule.
It is important to distinguish between
nonpolar bonds and nonpolar molecules.
The high melting point of BeCl2 is
due to its polymeric nature in the solid
state.
The VSEPR theory describes the
locations of bonded atoms around the
central atom, as well as where its lone
pairs of valence shell electrons are
directed.
The difference in electronegativity between Be and Cl is so large that we might expect
ionic bonding. The radius of Be2ϩ is so small (0.59 Å) and its charge density (ratio of
charge to size) is so high, however, that most simple beryllium compounds are covalent
rather than ionic. The high charge density of Be2ϩ causes it to attract and distort the electron
cloud of monatomic anions of all but the most electronegative elements. As a result,
the bonds in BeCl2 are polar covalent rather than ionic. Two exceptions are BeF2 and
BeO. They are ionic compounds because they contain the two most electronegative
elements bonded to Be.
C. Valence Bond Theory
Consider the ground state electron configuration of Be. There are two electrons in the
1s orbital, but these nonvalence (inner) electrons are not involved in bonding. Two more
electrons are paired in the 2s orbital. How, then, will a Be atom bond to two Cl atoms?
The Be atom must somehow make available one orbital for each bonding Cl electron (the
unpaired p electrons). The following ground state electron configuration for Be is the
configuration for an isolated Be atom. Another configuration may be more stable when
the Be atom is covalently bonded. Suppose that the Be atom “promoted” one of the paired
2s electrons to one of the 2p orbitals, the next higher energy orbitals.
Then there would be two Be orbitals available for bonding. This description, however, is
still not fully consistent with experimental fact. The Be 2s and 2p orbitals could not overlap
a Cl 3p orbital with equal effectiveness; that is, this “promoted pure atomic” arrangement
would predict two nonequivalent BeXCl bonds. Yet we observe experimentally that the
BeXCl bonds are identical in bond length and bond strength.
For these two orbitals on Be to become equivalent, they must hybridize to give two
orbitals intermediate between the s and p orbitals. These are called sp hybrid orbitals.
Consistent with Hund’s Rule, each of these equivalent hybrid orbitals on Be would contain
one electron.
The sp hybrid orbitals are described as linear orbitals, and we say that Be has linear electronic
geometry.
2p
sp
2p
2s
[He]Be [He]Be
hybridize
2p2p
2s2s
Be [He][He]Be
promote
We say that the Be2ϩ ion polarizes the
anions, ClϪ.
8-5 Linear Electronic Geometry: AB2 Species (No Lone Pairs of Electrons on A) 315
Be +
2s 2p two sp hybrid orbitals simplified representation
of two sp hybrids
Be BeBe
As we did for pure atomic orbitals, we
often draw hybrid orbitals more
slender than they actually are. Such
drawings are intended to remind us of
the orientations and general shapes of
orbitals.
Hund’s Rule is discussed in Section
5-17.
Cl ground state configuration:
3s 3p[He]
We can imagine that there is one electron in each of these hybrid orbitals on the Be
atom. Recall that each Cl atom has a half-filled 3p orbital that can overlap with a halffilled
sp hybrid of Be. We picture the bonding in BeCl2 in the following diagram, in which
only the bonding electrons are represented.
Unshared pairs of electrons on Cl
atoms are not shown. The hybrid
orbitals on the central atom are shown
in green in this and subsequent
drawings.
316 CHAPTER 8: Molecular Structure and Covalent Bonding Theories
Problem-Solving Tip: Number and Kind of Hybrid Orbitals
One additional idea about hybridization is worth special emphasis:
The number of hybrid orbitals is always equal to the number of atomic
orbitals that hybridize.
Hybrid orbitals are named by indicating the number and kind of atomic orbitals
hybridized. Hybridization of one s orbital and one p orbital gives two sp hybrid orbitals. We
shall see presently that hybridization of one s and two p orbitals gives three sp2 hybrid
orbitals; hybridization of one s orbital and three p orbitals gives four sp3 hybrids, and so
on (see Table 8-2).
Thus, the Be and two Cl nuclei would lie on a straight line. This is consistent with the experimental
observation that the molecule is linear.
The structures of beryllium bromide, BeBr2, and beryllium iodide, BeI2, are similar to
that of BeCl2. The chlorides, bromides, and iodides of cadmium, CdX2, and mercury,
HgX2, are also linear, covalent molecules (where X ϭ Cl, Br, or I).
sp Hybridization occurs at the central atom whenever there are two regions of high
electron density around the central atom. AB2 molecules and ions with no lone pairs
on the central atom have linear electronic geometry, linear molecular geometry, and
sp hybridization on the central atom.
3p
two sp hybrids on Be
Cl Be Cl
3p
TRIGONAL PLANAR ELECTRONIC GEOMETRY: AB3
SPECIES (NO LONE PAIRS OF ELECTRONS ON A)
A. Experimental Facts and Lewis Formulas
Boron is a Group IIIA element that forms many covalent compounds by bonding to three
other atoms. Typical examples include boron trifluoride, BF3 (mp Ϫ127°C); boron trichloride,
BCl3 (mp Ϫ107°C); boron tribromide, BBr3 (mp Ϫ46°C); and boron triiodide, BI3
(mp 50°C). All are trigonal planar nonpolar molecules.
F F
F
B B
F
F
F
B
F
F
F120Њ
120Њ120Њ
8-6
Hybridization usually involves orbitals
from the same main shell (same n).
The two X’s within one structure are
identical.
A trigonal planar molecule is a flat
molecule in which all three bond
angles are 120°.
The solid lines represent bonds
between B and F atoms.
The Lewis formula for BF3 is derived from the following: (a) each B atom has three
electrons in its valence shell and (b) each B atom is bonded to three F (or Cl, Br, I) atoms.
In Example 7-6 we wrote the Lewis formula for BCl3. Both F and Cl are members of
Group VIIA, and so the Lewis formulas for BF3 and BCl3 should be similar.
We see that BF3 and other similar molecules have a central element that does not satisfy
the octet rule. Boron shares only six electrons.
B. VSEPR Theory
Boron, the central atom, has three regions of high electron density (three bonded atoms,
no lone pairs on B). The VSEPR theory predicts trigonal planar electronic geometry for
molecules such as BF3 because this structure gives maximum separation among the three
regions of high electron density. There are no lone pairs of electrons associated with the
boron atom, so a fluorine atom is at each corner of the equilateral triangle, and the molecular
geometry is also trigonal planar. The maximum separation of any three items (electron
pairs) around a fourth item (B atom) is at 120° angles in a single plane. All four atoms are
in the same plane. The three F atoms are at the corners of an equilateral triangle, with
the B atom in the center. The structures of BCl3, BBr3, and BI3 are similar.
Examination of the bond dipoles of BF3 shows that the electronegativity difference (see
Table 6-3) is very large (2.0 units) and that the bonds are very polar.
However, the three bond dipoles are symmetrical, so they cancel to give a nonpolar mole-
cule.
C. Valence Bond Theory
To be consistent with experimental findings and the predictions of the VSEPR theory, the
VB theory must explain three equivalent BXF bonds. Again we use the idea of hybridization.
Now the 2s orbital and two of the 2p orbitals of B hybridize to form a set of three
equivalent sp2 hybrid orbitals.
Three sp2 hybrid orbitals point toward the corners of an equilateral triangle:
2p
sp2
2p
2s
[He]B [He]B
hybridize
B F
2.0 4.0
2.0
ϭ
⌬(EN) ϭ
EN
F F
F
B
Net molecular dipole ϭ 0
A model of BF3, a trigonal planar
AB3 molecule.
8-6 Trigonal Planar Electronic Geometry: AB3 Species (No Lone Pairs of Electrons on A) 317
B
2s
+ +
2p
2p
or
three sp2
hybrid
orbitals
B B
120°
B
B
Trigonal planar geometry is sometimes
called plane triangular or, simply,
triangular.
The B3ϩ ion is so small (radius ϭ
0.20 Å) that boron does not form
simple ionic compounds.
We can imagine that there is one electron in each of these hybrid orbitals. Each of the
three F atoms has a 2p orbital with one unpaired electron. The 2p orbitals can overlap
the three sp2 hybrid orbitals on B. Three electron pairs are shared among one B and three
F atoms:
The names of many solid figures are
based on the numbers of plane faces
they have. A regular tetrahedron is a
three-dimensional figure with four
equal-sized equilateral triangular faces
(the prefix tetra- means “four”).
318 CHAPTER 8: Molecular Structure and Covalent Bonding Theories
sp2 Hybridization occurs at the central atom whenever there are three regions of
high electron density around the central atom. AB3 molecules and ions with no
unshared pairs on the central atom have trigonal planar electronic geometry, trigonal
planar molecular geometry, and sp2 hybridization on the central atom.
TETRAHEDRAL ELECTRONIC GEOMETRY: AB4 SPECIES
(NO LONE PAIRS OF ELECTRONS ON A)
A. Experimental Facts and Lewis Formulas
Each Group IVA element has four electrons in its highest occupied energy level. The
Group IVA elements form many covalent compounds by sharing those four electrons with
four other atoms. Typical examples include CH4 (mp Ϫ182°C), CF4 (mp Ϫ184°C), CCl4
(mp Ϫ23°C), SiH4 (mp Ϫ185°C), and SiF4 (mp Ϫ90°C). All are tetrahedral, nonpolar
molecules (bond angles ϭ 109.5°). In each, the Group IVA atom is located in the center
of a regular tetrahedron. The other four atoms are located at the four corners of the tetra-
hedron.
The Group IVA atom contributes four electrons to the bonding in a tetrahedral AB4
molecule, and the other four atoms contribute one electron each. The Lewis formulas for
methane, CH4, and carbon tetrafluoride, CF4, are typical.
The ammonium ion, NH4
ϩ, and the sulfate ion, SO4
2Ϫ, are familiar examples of polyatomic
ions of this type. In each of these ions, the central atom is located at the center of
F
F
F FC
H
H
H HC
CF4, carbon tetrafluorideCH4, methane
8-7
sp2 hybrid orbitals
2p
F
F F
B
Lone pairs of electrons are not shown
for the F atoms.
a regular tetrahedron with the other atoms at the corners (HXNXH and OXSXO bond
angles ϭ 109.5°).
B. VSEPR Theory
VSEPR theory predicts that four valence shell electron pairs are directed toward the
corners of a regular tetrahedron. That shape gives the maximum separation for four electron
pairs around one atom. Thus, VSEPR theory predicts tetrahedral electronic geometry
for an AB4 molecule that has no unshared electrons on A. There are no lone pairs of electrons
on the central atom, so another atom is at each corner of the tetrahedron. VSEPR
theory predicts a tetrahedral molecular geometry for each of these molecules.
O
O
O OS
2Ϫ
ϩ
H
H
H HN
SO4
2Ϫ
, sulfate ionNH4
ϩ
, ammonium ion
8-7 Tetrahedral Electronic Geometry: AB4 Species (No Lone Pairs of Electrons on A) 319
H
C
In CH4
all H C H angles
ϭ 109.5°
H
H
H
In CF4
all F C F angles
ϭ 109.5°
F
F
C
F
F
The results that we discussed in Sections 8-5 (BeX2, CdX2, and HgX2, where X ϭ Cl,
Br, or I), and 8-6 (BX3, where X ϭ F, Cl, Br, or I), and in this section (CH4, CF4, CCl4,
SiH4, SiF4, NH4
ϩ, and SO4
2Ϫ) illustrate an important generalization:
When a molecule or polyatomic ion has no unshared pairs of valence electrons on
the central atom, the electronic geometry and the molecular geometry are the same.
Examination of bond dipoles shows that in CH4 the individual bonds are only slightly
polar, whereas in CF4 the bonds are quite polar. In CH4 the bond dipoles are directed
toward carbon, but in CF4 they are directed away from carbon. Both molecules are very
symmetrical, so the bond dipoles cancel, and both molecules are nonpolar. This is true
for all AB4 molecules in which there are no unshared electron pairs on the central element and
all four B atoms are identical.
320 CHAPTER 8: Molecular Structure and Covalent Bonding Theories
In some tetrahedral molecules, the atoms bonded to the central atom are not all the
same. Such molecules are usually polar, with the degree of polarity depending on the relative
sizes of the bond dipoles present. In CH3F or CH2F2, for example, the addition of
unequal dipoles makes the molecule polar.
The VSEPR theory also predicts that NH4
ϩ and SO4
2Ϫ ions have tetrahedral electronic
geometry. Each region of high electron density bonds the central atom to another
atom (H in NH4
ϩ, O in SO4
2Ϫ) at the corner of the tetrahedral arrangement. We describe
the molecular geometry of each of these ions as tetrahedral.
You may wonder whether square planar AB4 molecules exist. We will discuss some
examples of square planar AB4 species in Section 8-12. The bond angles in square planar
CF4
C F
EN ϭ 2.5 4.0
⌬(EN) ϭ 1.5
Net molecular
dipole ϭ 0
C
H
F
F
C
F
F
H
H
H
Net molecular
dipole ϭ 0
CH4
C H
EN ϭ 2.5 2.1
⌬(EN) ϭ 0.4
F
C
F
C
HH Net molecular
dipole > 0
Net molecular
dipole > 0
H
H
H
F
molecules are only 90°. Most AB4 molecules are tetrahedral, however, with larger bond
angles (109.5°) and greater separation of valence electron pairs around A.
C. Valence Bond Theory
According to VB theory, each Group IVA atom (C in our example) must make four equivalent
orbitals available for bonding. To do this, C forms four sp3 hybrid orbitals by mixing
the s and all three p orbitals in its outer (n ϭ 2) shell. This results in four unpaired elec-
trons.
These sp3 hybrid orbitals are directed toward the corners of a regular tetrahedron, which
has a 109.5° angle from any corner to the center to any other corner.
2p
sp3
2s
[He]C [He]C
hybridize
8-7 Tetrahedral Electronic Geometry: AB4 Species (No Lone Pairs of Electrons on A) 321
Each of the four atoms that bond to C has a half-filled atomic orbital; these can overlap
the half-filled sp3 hybrid orbitals, as is illustrated for CH4 and CF4.
C + C + C + C C
109.5°
2p
four sp3
hybrid
orbitals
2s
F
F
F
F
C
Unshared
pairs on
F atoms are
not shown
H
H
H
H
C
H
1s 2p
CF4CH4
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 10.4, Hybrid Orbitals. Be sure to
see the animation of sp3 orbital
formation on that screen. See also the
hybridization tool in Screen 10.6,
which you can use to build hybrid
orbitals.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 10.5, Sigma Bonds.
We can give the same VB description for the hybridization of the central atoms in
polyatomic ions. In NH4
ϩ and SO4
2Ϫ, the N and S atoms, respectively, form four sp3
hybrid orbitals directed toward the corners of a regular tetrahedron. Each of these sp3
hybrid orbitals overlaps with an orbital on a neighboring atom (H in NH4
ϩ, O in SO4
2Ϫ)
to form a bond.
322 CHAPTER 8: Molecular Structure and Covalent Bonding Theories
Problem-Solving Tip: Often There Is More Than One Central Atom
Many molecules contain more than one central atom, that is, there is more than one
atom that is bonded to several other atoms. We can analyze such molecules one central
atom at a time, to build up a picture of the three-dimensional aspects of the molecule.
An example is ethane, C2H6; its Lewis formula is
H
H
H C
H
H
HC
sp3 Hybridization occurs at the central atom whenever there are four regions of high
electron density around the central atom. AB4 molecules and ions with no unshared
pairs on the central atom have tetrahedral electronic geometry, tetrahedral molecular
geometry, and sp3 hybridization on the central atom.
Problem-Solving Tip: Use Periodic Relationships
Because F, Cl, Br, and I are all in the same group of the periodic table, we know that
their compounds would be similar. We expect that the detailed descriptions we have seen
for CF4 would also apply to CCl4, CBr4, and CI4, and we do not need to go through
the entire reasoning for each one. Thus, we can say that each of the CX4 molecules
(X ϭ F, Cl, Br, or I) also has tetrahedral electronic geometry, tetrahedral molecular geometry,
sp3 hybridization on the carbon atom, zero dipole moment, and so on.
O
O
O
O
S
2–
H
H
H
H
N
H
1s
+
2p
Unshared
pairs on
O atoms are
not shown
NH4
+
SO4
2–
TETRAHEDRAL ELECTRONIC GEOMETRY: AB3U SPECIES
(ONE LONE PAIR OF ELECTRONS ON A)
We are now ready to study the structures of some simple molecules with unshared valence
electron pairs (lone pairs) on the central atom. In this and subsequent sections, we use
generalized chemical formulas in which “A” represents the central atom, “B” represents
an atom bonded to A, and “U” represents an unshared valence shell electron pair (lone
pair) on the central atom A. For instance, AB3U would represent any molecule with three
B atoms bonded to a central atom A, with one unshared valence pair on A.
8-8
Let us first consider the left-hand carbon atom. The arrangements of its regions of
high electron density allows us to locate the other C and three H atoms with respect to
that C atom (the atoms outlined in red). Then we carry out a similar analysis for the
right-hand C atom to deduce the arrangements of its neighbors (outlined in blue).
Each C atom in C2H6 has four regions of high electron density. The VSEPR theory
tells us that each C atom has tetrahedral electronic geometry; the resulting atomic
arrangement around each C atom has one C and three H atoms at the corners of this
tetrahedral arrangement. The VB interpretation is that each C atom is sp3 hybridized.
The CXC bond is formed by overlap of a half-filled sp3 hybrid orbital of one C atom
with a half-filled sp3 hybrid orbital of the other C atom. Each CXH bond is formed
by the overlap of a half-filled sp3 hybrid orbital on C with the half-filled 1s orbital of an
H atom.
A. Experimental Facts and Lewis Formulas
Each Group VA element has five electrons in its valence shell. The Group VA elements
form some covalent compounds by sharing three of those electrons with three other atoms.
Let us describe two examples: ammonia, NH3, and nitrogen trifluoride, NF3. Each is a
trigonal pyramidal, polar molecule with an unshared pair on the nitrogen atom. Each has
a nitrogen atom at the apex and the other three atoms at the corners of the triangular
base of the pyramid.
The Lewis formulas for NH3 and NF3 are
Sulfite ion, SO3
2Ϫ, is an example of a polyatomic ion of the AB3U type. It is a trigonal
pyramidal ion with an unshared pair on the sulfur atom.
B. VSEPR Theory
As in Section 8-7, VSEPR theory predicts that the four regions of high electron density
around a central atom will be directed toward the corners of a tetrahedron, because this
gives maximum separation. So N has tetrahedral electronic geometry in NH3 and NF3.
We must reemphasize the distinction between electronic geometry and molecular geometry.
Electronic geometry refers to the geometric arrangement of the regions of electron density
around the central atom. But the molecular geometry excludes the unshared pairs on the
central atom, and describes only the arrangement of atoms (i.e., nuclei) around the central
atom. We can represent the tetrahedral electronic geometry around N in NH3 or NF3 as
follows.
O
O OS
2Ϫ
F
F FNHH
H
N
NF3NH3
Some Group VA elements also form
covalent compounds by sharing all five
valence electrons (Sections 7-7 and
8-11).
324 CHAPTER 8: Molecular Structure and Covalent Bonding Theories
We then use the Lewis formula as a guide to put the bonded atoms and the lone pairs in
these tetrahedral sites around the nitrogen atom.
H
N
H
H
F
N
F
F
N
Then we describe the molecular geometry as the arrangement of the atoms. In each of these
molecules, the N atom is at the apex of a (shallow) trigonal pyramidal arrangement and
the other three atoms are at the corners of the triangular base of the pyramid. Thus, the
molecular geometry of each molecule is described as trigonal pyramidal.
8-8 Tetrahedral Electronic Geometry: AB3U Species (One Lone Pair of Electrons on A) 325
We have seen that CH4, CF4, NH3, and NF3 all have tetrahedral electronic geometry.
But CH4 and CF4 (AB4) have tetrahedral molecular geometry, whereas NH3 and NF3
(AB3U) have trigonal pyramidal molecular geometry.
In molecules or polyatomic ions that contain lone (unshared) pairs of valence electrons
on the central atom, the electronic geometry and the molecular geometry cannot
be the same.
Because this trigonal pyramidal molecular geometry is a fragment of tetrahedral electronic
geometry, we expect that the HXNXH angle would be close to the tetrahedral
value, 109.5°. In CH4 (a tetrahedral AB4 molecule), all HXCXH bond angles are observed
to be this ideal value, 109.5°. In NH3, however, the HXNXH bond angles are observed
to be less than this, 107.3°. How can we explain this deviation?
A lone pair is a pair of valence electrons that is associated with only one nucleus in
contrast to a bonded pair, which is associated with two nuclei. The known geometries of
many molecules and polyatomic ions, based on measurements of bond angles, show that
lone pairs of electrons occupy more space than bonding pairs. A lone pair has only one atom
H
N
H
H
F
N
F
F
C
H
HH
H
109.5°
exerting strong attractive forces on it, so it resides closer to the nucleus than do bonding
electrons. The relative magnitudes of the repulsive forces between pairs of electrons on
an atom are
lp/lp ϾϾ lp/bp Ͼ bp/bp
where lp refers to lone (unshared) pairs and bp refers to bonding pairs of valence shell
electrons. We are most concerned with the repulsions among the electrons in the valence
shell of the central atom of a molecule or polyatomic ion. The angles at which repulsive
forces among valence shell electron pairs are minimized are the angles at which the
bonding pairs and unshared pairs (and therefore nuclei) are found in covalently bonded
molecules and polyatomic ions. Due to lp/bp repulsions in NH3 and NF3, their bond angles
are less than the angles of 109.5° we observed in CH4 and CF4 molecules.
In NH3 the bond dipoles reinforce the
effect of the unshared pair, so NH3 is
very polar (␮ ϭ 1.47 D). In NF3 the
bond dipoles oppose the effect of the
unshared pair, so NF3 is only slightly
polar (␮ ϭ 0.23 D).
326 CHAPTER 8: Molecular Structure and Covalent Bonding Theories
The formulas are frequently written as SNH3 and SNF3 to emphasize the unshared
pairs of electrons. The unshared pairs must be considered as the polarities of these molecules
are examined; they are extremely important in chemical reactions. This is why NH3
is a base, as we saw in Section 4-2.4 and as we shall discuss more fully in Chapter 10. The
contribution of each unshared pair to polarity can be depicted as shown in the margin.
The electronegativity differences in NH3 and NF3 are nearly equal, but the resulting
nearly equal bond polarities are in opposite directions.
Thus, we have
N H
N H3.0 2.1
0.9
ϭ
⌬(EN) ϭ
EN
N F
N F3.0 4.0
1.0
ϭ
⌬(EN) ϭ
EN
We can now use this information to explain the bond angles observed in NF3 and NH3.
Because of the direction of the bond dipoles in NH3, the electron-rich end of each NXH
bond is at the central atom, N. On the other hand, the fluorine end of each bond in NF3
is the electron-rich end. As a result, the lone pair can more closely approach the N in
Tetrahedral
electronic geometry
Trigonal pyramidal
molecular geometry
In NF3, F–N–F angle = 102.1°
F
F
N
F
In NH3, H–N–H angle = 107.3°
H
H
N
H
F
F
N
F
Net molecular
dipole > 0 (small)
N
H
H
H
Net molecular
dipole > 0 (large)
A model of NH3, a trigonal
pyramidal AB3U molecule, showing
the lone pair.
H
H
H
N
107.3°
NF3 than in NH3. In NF3 the lone pair therefore exerts greater repulsion toward the
bonded pairs than in NH3. In addition, the longer NXF bond length makes the bp–bp
distance greater in NF3 than in NH3, so that the bp/bp repulsion in NF3 is less than that
in NH3. The net effect is that the bond angles are reduced more in NF3. We can represent
this situation as:
We might expect the larger F atoms
(r ϭ 0.72 Å) to repel each other more
strongly than the H atoms (r ϭ 0.37 Å),
leading to larger bond angles in NF3
than in NH3. This is not the case,
however, because the NXF bond is
longer than the NXH bond. The
NXF bond density is farther from the
N than the NXH bond density.
8-8 Tetrahedral Electronic Geometry: AB3U Species (One Lone Pair of Electrons on A) 327
With the same kind of reasoning, VSEPR theory predicts that sulfite ion, SO3
2Ϫ, has
tetrahedral electronic geometry. One of these tetrahedral locations is occupied by the
sulfur lone pair, and oxygen atoms are at the other three locations. The molecular geometry
of this ion is trigonal pyramidal, the same as for other AB3U species.
C. Valence Bond Theory
Experimental results suggest four nearly equivalent orbitals (three involved in bonding, a
fourth to accommodate the lone pair), so we again need four sp3 hybrid orbitals.
In both NH3 and NF3 the lone pair of electrons occupies one of the sp3 hybrid orbitals.
Each of the other three sp3 orbitals participates in bonding by sharing electrons with
another atom. They overlap with half-filled H 1s orbitals and F 2p orbitals in NH3 and
NF3, respectively.
2p
sp3
2s
[He]N [He]N
hybridize
Lone pair/bonded pair repulsions are
stronger than bonded pair/bonded
pair repulsions
Bonded pair/bonded pair repulsions
are weaker in NF3 than in NH3
due to the longer N––F bond
107°
N
H
H
H
102°
N
F
F F
F
F
F
N
Lone pairs on F atoms
are not shown
2p
H
H
H
N
1s
A model of sulfite ion, SO3
2Ϫ.
The sulfur atom in the sulfite ion, SO3
2Ϫ, can be described as sp3 hybridized. One of
these hybrid orbitals contains the sulfur lone pair, and the remaining three overlap with
oxygen orbitals to form bonds.
AB3U molecules and ions, each having four regions of high electron density around
the central atom, usually have tetrahedral electronic geometry, trigonal pyramidal
molecular geometry, and sp3 hybridization on the central atom.
We must remember that theory (and its application) depends on fact, not the other way
around. Sometimes the experimental facts are not consistent with the existence of hybrid
orbitals. In PH3 and AsH3, each HXPXH bond angle is 93.7°, and each HXAsXH bond
angle is 91.8°. These angles very nearly correspond to three p orbitals at 90° to each other.
Thus, there appears to be no need to use the VSEPR theory or hybridization to describe
the bonding in these molecules. In such cases, we just use the “pure” atomic orbitals rather
than hybrid orbitals to describe the bonding.
328 CHAPTER 8: Molecular Structure and Covalent Bonding Theories
Problem-Solving Tip: When Do We Not Need To Describe Hybrid
Orbitals?
Students often wonder how to recognize when hybridization does not apply, as in the
PH3 and AsH3 cases just described and in H2S (see Section 8-9). Remember that models
such as hybridization are our attempts to explain observations such as bond angles. At
the level of our studies, we will be given information about the observed molecular geometry,
such as measured bond angles. If these are near the angles of pure (unhybridized)
orbitals, then hybridization is not needed; if they are near the predicted angles for
hybridized orbitals, then hybridization should be used in our explanation. If no information
about observed molecular geometry (molecular shape, bond angles, etc.) is
supplied, you should assume that the VSEPR and hybridization approaches presented in
this chapter should be used.
TETRAHEDRAL ELECTRONIC GEOMETRY: AB2U2 SPECIES
(TWO LONE PAIRS OF ELECTRONS ON A)
A. Experimental Facts and Lewis Formulas
Each Group VIA element has six electrons in its valence shell. The Group VIA elements
form many covalent compounds by sharing a pair of electrons with each of two other
atoms. Typical examples are H2O, H2S, and Cl2O. The Lewis formulas for these molecules
are
All are angular, polar molecules. The bond angle in water, for example, is 104.5°, and the
molecule is very polar with a dipole moment of 1.85 D.
OH
H
SH
H
OCl
Cl
H2O H2S Cl2O
8-9
B. VSEPR Theory
The VSEPR theory predicts that the four electron pairs around the oxygen atom in H2O
should be 109.5° apart in a tetrahedral arrangement. The observed HXOXH bond angle
is 104.5°. The two lone (unshared) pairs strongly repel each other and the bonding pairs
of electrons. These repulsions force the bonding pairs closer together and result in the
decreased bond angle. The decrease in the HXOXH bond angle (from 109.5° to 104.5°)
is greater than the corresponding decrease in the HXNXH bond angles in ammonia
(from 109.5° to 107.3°) because of the lp/lp repulsion in H2O.
The electronegativity difference is large (1.4 units), and so the bonds are quite polar.
Additionally, the bond dipoles reinforce the effect of the two unshared pairs, so the H2O
molecule is very polar. Its dipole moment is 1.8 D. Water has unusual properties, which
can be explained in large part by its high polarity.
C. Valence Bond Theory
The bond angle in H2O (104.5°) is closer to the tetrahedral value (109.5°) than to the 90°
angle that would result from bonding by pure 2p atomic orbitals on O. Valence bond
theory therefore postulates four sp3 hybrid orbitals centered on the O atom: two to participate
in bonding and two to hold the two unshared pairs.
AB2U2 molecules and ions, each having four regions of high electron density around
the central atom, usually have tetrahedral electronic geometry, angular molecular
geometry, and sp3 hybridization on the central atom.
Hydrogen sulfide, H2S, is also an angular molecule, but the HXSXH bond angle is
92.2°. This is very close to the 90° angles between two unhybridized 3p orbitals of S. We
therefore do not propose hybrid orbitals to describe the bonding in H2S. The two H atoms
are able to exist at approximately right angles to each other when they are bonded to the
larger S atom. The bond angles in H2Se and H2Te are 91° and 89.5°, respectively.
O H
3.5 2.1
1.4
ϭ
⌬(EN) ϭ
EN OH
H
Molecular dipole;
includes effect of two
unshared electron pairs.
A model of H2O, an angular AB2U2
molecule, showing the lone pairs.
8-9 Tetrahedral Electronic Geometry: AB3U Species (One Lone Pair of Electrons on A) 329
H
Lone pairs occupy one 3p
orbital and one 3s orbital
Unhybridized p
orbitals participate
in bonding
92°
H S
Sulfur is located directly below oxygen
in Group VIA.
There are now two
lone pairs that repel
the bonded pairs
lp–lp
repulsion
is strongest
O
H
H
TETRAHEDRAL ELECTRONIC GEOMETRY: ABU3
SPECIES (THREE LONE PAIRS OF ELECTRONS ON A)
Each Group VIIA element has seven electrons in its highest occupied energy level. The
Group VIIA elements form molecules such as HXF, HXCl, ClXCl, and IXI by sharing
one of their electrons with another atom. The other atom contributes one electron to the
bonding. Lewis dot formulas for these molecules are shown in the margin. Any diatomic
molecule must be linear. Neither VSEPR theory nor VB theory adds anything to what
we already know about the molecular geometry of such molecules.
8-10
330 CHAPTER 8: Molecular Structure and Covalent Bonding Theories
Problem-Solving Tip: Some Molecules Have Two Central Atoms and
Lone Pairs
Methanol, CH3OH, is a simple molecule that has more than one central atom and two
lone pairs. It is the simplest member of a family of organic molecules called alcohols; all
alcohols contain the atom grouping CXOXH. The Lewis formula for methanol is
Again, we consider the arrangements around two central atoms in sequence. The
carbon atom (outlined in red) has four regions of high electron density. VSEPR theory
tells us that this atom has tetrahedral electronic geometry; the resulting atomic arrangement
around the C atom has four atoms (O and three H) at the corners of this tetrahedral
arrangement, so the molecular geometry about the C atom is tetrahedral. The oxygen
atom (outlined in blue) has four regions of high electron density, so it, too, has tetrahedral
electronic geometry. Thus, the CXOXH arrangement is angular and there are two
unshared pairs on O. The VB interpretation is that both atoms are sp3 hybridized. The
CXO bond is formed by overlap of a half-filled sp3 hybrid orbital on the C atom with
a half-filled sp3 hybrid orbital on the O atom. Each covalent bond to an H is formed by
the overlap of a half-filled sp3 hybrid orbital with the half-filled 1s orbital on an H atom.
Each unshared pair of electrons on O is in an sp3 hybrid orbital.
H
H
H OC H
C O
HH
H
H
H
H
H
C O
H
We represent the halogen as X.
In latter case, either halogen may be
considered the “A” atom of AB.
XHX, H XX2, X
TRIGONAL BIPYRAMIDAL ELECTRONIC GEOMETRY:
AB5, AB4U, AB3U2, AND AB2U3
A. Experimental Facts and Lewis Formulas
In Section 8-8 we saw that the Group VA elements have five electrons in their outermost
occupied shells and form some molecules by sharing only three of these electrons with
other atoms (e.g., NH3, NF3, and PCl3). Group VA elements (P, As, and Sb) beyond the
second period also form some covalent compounds by sharing all five of their valence
electrons with five other atoms (see Section 7-7). Phosphorus pentafluoride, PF5 (mp
Ϫ83°C), is such a compound. Each P atom has five valence electrons to share with five
F atoms. The Lewis formula for PF5 (see Example 7-7) is shown in the margin. PF5 molecules
are trigonal bipyramidal nonpolar molecules. A trigonal bipyramid is a six-sided
polyhedron consisting of two pyramids joined at a common triangular (trigonal) base.
B. VSEPR Theory
The VSEPR theory predicts that the five regions of high electron density around the
phosphorus atom in PF5 should be as far apart as possible. Maximum separation of five
items around a sixth item is achieved when the five items (bonding pairs) are placed at
the corners and the sixth item (P atom) is placed in the center of a trigonal bipyramid.
This is in agreement with experimental observation.
8-11
FF
F
P
FF
8-11 Trigonal Bipyramidal Electronic Geometry: AB5, AB4U, AB3U2, and AB2U3 331
The three F atoms marked e are at the corners of the common base, in the same plane
as the P atom. These are called equatorial F atoms (e). The other two F atoms, one above
and one below the plane, are called axial F atoms (a). The FXPXF bond angles are 90°
(axial to equatorial), 120° (equatorial to equatorial), and 180° (axial to axial).
The large electronegativity difference between P and F (1.9) suggests very polar bonds.
Let’s consider the bond dipoles in two groups, because there are two different kinds of
PXF bonds in PF5 molecules, axial and equatorial.
a F
e
F
F
e
P
F
e
a
F
As an exercise in geometry, in how
many different ways can five fluorine
atoms be arranged symmetrically around
a phosphorus atom? Compare the
hypothetical bond angles in such
arrangements with those in a trigonal
bipyramidal arrangement.
A trigonal bipyramid.
The two axial bond dipoles cancel each other, and the three equatorial bond dipoles cancel,
so PF5 molecules are nonpolar.
C. Valence Bond Theory
Because phosphorus is the central atom in a PF5 molecule, it must have available five halffilled
orbitals to form bonds with five F atoms. Hybridization involves one d orbital from
the vacant set of 3d orbitals along with the 3s and 3p orbitals of the P atom.
P F
F
F
P2.1 4.0
1.9
ϭ
⌬(EN) ϭ
EN
F F
F
P
axial bonds equatorial bonds
180Њ
120Њ
The two axial bonds are in a linear
arrangement, like the two bonds in
BeCl2 (see Section 8-5); the three
equatorial bonds are in a trigonal
planar arrangement, like the three
bonds in BF3 (see Section 8-6).
332 CHAPTER 8: Molecular Structure and Covalent Bonding Theories
3p
3d 3d
sp3
d
3s
P [Ne] [Ne]P
hybridize
The five sp3d hybrid orbitals on P point toward the corners of a trigonal bipyramid. Each
is overlapped by a singly occupied 2p orbital of an F atom. The resulting pairing of P and
F electrons forms five covalent bonds.
F
P
F
F
F
sp3d hybrid orbitals on P
p orbitals on F
F
sp3d Hybridization occurs at the central atom whenever there are five regions of
high electron density around the central atom. AB5 molecules and ions with no
unshared pairs on the central atom have trigonal bipyramidal electronic geometry,
trigonal bipyramidal molecular geometry, and sp3d hybridization on the central atom.
We see that sp3d hybridization uses an available d orbital in the outermost occupied
shell of the central atom. The heavier Group VA elements—P, As, and Sb—can form
five covalent bonds using this hybridization. But nitrogen, also in Group VA, cannot form
five covalent bonds, because the valence shell of N has only one s and three p orbitals (and
no d orbitals). The set of s and p orbitals in a given energy level (and therefore any set of
hybrids composed only of s and p orbitals) can accommodate a maximum of eight electrons
and participate in a maximum of four covalent bonds. The same is true of all elements
of the second period, because they have only s and p orbitals in their valence shells. No
atoms in the first and second periods can exhibit expanded valence.
D. Unshared Valence Electron Pairs in Trigonal Bipyramidal Electronic
Geometry
As we saw in Sections 8-8 and 8-9, lone pairs of electrons occupy more space than bonding
pairs, resulting in increased repulsions from lone pairs. What happens when one or more
of the five regions of high electron density on the central atom are lone pairs? Let us first
consider a molecule such as SF4, for which the Lewis formula is
The central atom, S, is bonded to four atoms and has one lone pair. This is an example
of the general formula AB4U. Sulfur has five regions of high electron density, so we know
that the electronic geometry is trigonal bipyramidal and that the bonding orbitals are sp3d
hybrids. But now a new question arises: Is the lone (unshared) pair more stable in an axial
(a) or in an equatorial (e) position? If it were in an axial position, it would be 90° from
F
S
F F
F
The P atom is said to have an expanded
valence shell (see Section 7-7).
8-11 Trigonal Bipyramidal Electronic Geometry: AB5, AB4U, AB3U2, and AB2U3 333
Trigonal bipyramidal
electronic geometry
Lone pair in
axial position
Lone pair in
equatorial position
(preferred arrangement)
vs.X
e
a
e
e
a
S
a
S
e
Reminder: The relative magnitudes of
repulsive forces are:
lp/lp ϾϾ lp/bp Ͼ bp/bp
the three closest other pairs (the pairs bonding three F atoms in equatorial positions) and
180° from the other axial pair. If it were in an equatorial position, only the two axial pairs
would be at 90°, and the other two equatorial pairs would be less crowded at 120° apart.
The lone pair would be less crowded in an equatorial position. The four F atoms then
occupy the remaining four positions. We describe the resulting arrangement of atoms as
a seesaw arrangement.
Imagine rotating the arrangement so
that the line joining the two axial
positions is the board on which the
two seesaw riders sit, and the two
bonded equatorial positions are the
pivot of the seesaw.
334 CHAPTER 8: Molecular Structure and Covalent Bonding Theories
As we saw in Sections 8-8 and 8-9, the differing magnitudes of repulsions involving
lone pairs and bonding pairs often result in observed bond angles that are slightly different
from idealized values. For instance, lp/bp repulsion in the seesaw molecule SF4 causes
distortion of the axial SXF bonds away from the lone pair, to an angle of 177°; the two
equatorial SXF bonds, ideally at 120°, move much closer together to an angle of 101.6°.
By the same reasoning, we understand why additional lone pairs also take equatorial
positions (AB3U2 with both lone pairs equatorial or AB2U3 with all three lone pairs equatorial).
These arrangements are summarized in Figure 8-2.
a
e
ee
a
SF F
FF
FS
e
F
a
e
F
F
e
a
F
or
AB4U
4 bonded atoms (B)
1 lone pair (U) in
equatorial position
Seesaw molecular geometry
Example: SF4
AB3U2
3 bonded atoms (B)
2 lone pairs (U) in
equatorial positions
T-shaped molecular geometry
Examples: ClF3, BrF3
AB2U3
2 bonded atoms (B)
3 lone pairs (U) in
equatorial positions
Linear molecular geometry
Examples: XeF2, I3
–
Figure 8-2 Arrangements of bonded atoms and lone pairs (five regions of high electron
density—trigonal pyramidal electronic geometry).
OCTAHEDRAL ELECTRONIC GEOMETRY: AB6, AB5U,
AND AB4U2
A. Experimental Facts and Lewis Formulas
The Group VIA elements below oxygen form some covalent compounds of the AB6 type
by sharing their six valence electrons with six other atoms. Sulfur hexafluoride, SF6 (mp
Ϫ51°C), an unreactive gas, is an example. Sulfur hexafluoride molecules are nonpolar octahedral
molecules. The hexafluorophosphate ion, PF6
Ϫ, is an example of a polyatomic ion
of the type AB6.
B. VSEPR Theory
In an SF6 molecule we have six valence shell electron pairs and six F atoms surrounding
one S atom. Because the valence shell of sulfur contains no lone pairs, the electronic and
molecular geometries in SF6 are identical. The maximum separation possible for six electron
pairs around one S atom is achieved when the electron pairs are at the corners and
the S atom is at the center of a regular octahedron. Thus, VSEPR theory is consistent
with the observation that SF6 molecules are octahedral.
8-12
FF
F
F
S
FF
8-12 Octahedral Electronic Geometry: AB6, AB5U, and AB4U2 335
In this octahedral molecule the FXSXF bond angles are 90° and 180°. Each SXF
bond is quite polar, but each bond dipole is canceled by an equal dipole at 180° from it.
So the large bond dipoles cancel and the SF6 molecule is nonpolar.
By similar reasoning, VSEPR theory predicts octahedral electronic geometry and octahedral
molecular geometry for the PF6
Ϫ ion, which has six valence shell electron pairs
and six F atoms surrounding one P atom.
S
F
F
F
F F
F
In a regular octahedron, each of the
eight faces is an equilateral triangle.
C. Valence Bond Theory
Sulfur atoms can use one 3s, three 3p, and two 3d orbitals to form six hybrid orbitals that
accommodate six electron pairs:
Se and Te, in the same group, form
analogous compounds. O cannot do so,
for the same reasons as discussed
earlier for N (see Section 8-11).
336 CHAPTER 8: Molecular Structure and Covalent Bonding Theories
3p
3d 3d
sp3
d2
3s
[Ne]S [Ne]S
hybridize
The six sp3d2 hybrid orbitals are directed toward the corners of a regular octahedron.
Each sp3d2 hybrid orbital is overlapped by a half-filled 2p orbital from fluorine, to form a
total of six covalent bonds.
An analogous picture could be drawn for the PF6
Ϫ ion.
sp3d2 Hybridization occurs at the central atom whenever there are six regions of
high electron density around the central atom. AB6 molecules and ions with no lone
pairs on the central atom have octahedral electronic geometry, octahedral molecular
geometry, and sp3d2 hybridization on the central atom.
S
F
F sp3d2 hybrid orbitals on S
p orbitals on F
F
F
F
F
D. Unshared Valence Electron Pairs in Octahedral Electronic Geometry
We can reason along the lines used in part D of Section 8-11 to predict the preferred
locations of lone pairs on the central atom in octahedral electronic geometry. Because of
the high symmetry of the octahedral arrangement, all six positions are equivalent, so it
does not matter in which position in the drawing we put the first lone pair. AB5U molecules
and ions are described as having square pyramidal molecular geometry. When a
second lone pair is present, the most stable arrangement has the two lone pairs in two
octahedral positions at 180° angles from each other. This leads to a square planar molecular
geometry for AB4U2 species. These arrangements are shown in Figure 8-3. Table
8-3 summarizes a great deal of information—study this table carefully.
Two lone pairs at 90° from each other
would be much more crowded.
8-12 Octahedral Electronic Geometry: AB6, AB5U, and AB4U2 337
Problem-Solving Tip: Placing Lone Pairs on the Central Atom
Remember that lone pairs occupy more space than bonded pairs, so the lone pairs are
always put in positions where they will be least crowded.
If the Lewis formula for a molecule or ion shows only one lone pair: In linear, trigonal planar,
tetrahedral, or octahedral electronic geometry, all positions are equivalent, so it doesn’t
matter where we place the lone pair. In trigonal bipyramidal electronic geometry, place
the lone pair in the equatorial position where it is least crowded, and put the bonded
atoms in the other positions.
If the Lewis formula shows two lone pairs: In trigonal planar or tetrahedral electronic
geometry, we can place the lone pairs in any two positions and the bonded atoms in the
other position(s). In trigonal bipyramidal electronic geometry, place the two lone pairs
in two equatorial positions (120° apart) where they are least crowded, and put the bonded
atoms in the other positions. In octahedral electronic geometry, place the two lone pairs
in two positions across (180°) from each other, and put the bonded atoms in the other
positions.
Figure 8-3 Arrangements of bonded atoms and lone pairs (six regions of high electron
density—octahedral electronic geometry).
AB5U
5 bonded atoms (B)
1 lone pair (U)
Square pyramidal molecular geometry
Examples: IF5, BrF5
AB4U2
4 bonded atoms (B)
2 lone pairs (U)
Square planar molecular geometry
Examples: XeF4, IF4
–
338 CHAPTER 8: Molecular Structure and Covalent Bonding Theories
TABLE 8-3 Molecular Geometry of Species with Lone Pairs (U) on the Central Atom
Regions
of High
General Electron Electronic Hybridization at Lone
Formula Density Geometry Central Atom Pairs Molecular Geometry Examples
AB2U 3 trigonal sp2 1 O3, NO2
Ϫ, SO2
planar
AB3U 4 tetrahedral sp3 1 NH3, SO3
2Ϫ
AB2U2 4 tetrahedral sp3 2 H2O, NH2
Ϫ
AB4U 5 trigonal sp3d 1 SF4
bipyramidal
Angular
Trigonal
pyramidal
Angular
Seesaw
8-12 Octahedral Electronic Geometry: AB6, AB5U, and AB4U2 339
TABLE 8-3 (continued)
Regions
of High
General Electron Electronic Hybridization at Lone
Formula Density Geometry Central Atom Pairs Molecular Geometry Examples
AB3U2 5 trigonal sp3d 2 ICl3, ClF3
bipyramidal
AB2U3 5 trigonal sp3d 3 XeF2, I3
Ϫ
bipyramidal
AB5U 6 octahedral sp3d2 1 IF5, BrF5
AB4U2 6 octahedral sp3d2 2 XeF4, IF4
Ϫ
T-shaped
Linear
Square
pyramidal
Square
planar
COMPOUNDS CONTAINING DOUBLE BONDS
In Chapter 7 we constructed Lewis formulas for some molecules and polyatomic ions that
contain double and triple bonds. We have not yet considered bonding and shapes for such
species. Let us consider ethylene (ethene), C2H4, as a specific example. Its dot formula is
Each atom has three regions of high electron density. The VSEPR theory tells us that
each C atom is at the center of a trigonal plane.
Valence bond theory pictures each doubly bonded carbon atom as sp2 hybridized, with
one electron in each sp2 hybrid orbital and one electron in the unhybridized 2p orbital.
This 2p orbital is perpendicular to the plane of the three sp2 hybrid orbitals:
Recall that sp2 hybrid orbitals are directed toward the corners of an equilateral triangle.
Figure 8-4 shows top and side views of these hybrid orbitals.
The two C atoms interact by head-on (end-to-end) overlap of sp2 hybrids pointing
toward each other to form a sigma (␴) bond and by side-on overlap of the unhybridized
2p orbitals to form a pi (␲) bond.
2p 2pz
sp2
2s
[He]C C [He]
hybridize
C C
H H
HH
S ϭ N Ϫ A
ϭ 24 Ϫ 12 ϭ 12eϪ
shared
8-13
Here each C atom is considered a
central atom. Remember that each
bonded atom counts as one region of
high electron density.
340 CHAPTER 8: Molecular Structure and Covalent Bonding Theories
Figure 8-4 (a) A top view of three sp2 hybrid orbitals (green). The remaining unhybridized
p orbital (not shown in this view) is perpendicular to the plane of the drawing. (b) A side
view of a carbon atom in a trigonal planar (sp2-hybridized) environment, showing the
remaining p orbital (tan). This p orbital is perpendicular to the plane of the three sp2 hybrid
orbitals.
(a)
C
120˚
sp2
sp2
sp2
Csp2
sp2
sp2
Two lobes
of a single
p orbital
(b)
An important use of ethylene, C2H4,
is in the manufacture of
polyethylene, a nonbreakable,
nonreactive plastic.
A sigma bond is a bond resulting from head-on overlap of atomic orbitals. The
region of electron sharing is along and cylindrically around an imaginary line connecting
the bonded atoms.
All single bonds are sigma bonds. Many kinds of pure atomic orbitals and hybridized
orbitals can be involved in sigma bond formation.
A pi bond is a bond resulting from side-on overlap of atomic orbitals. The regions
of electron sharing are on opposite sides of an imaginary line connecting the bonded atoms
and parallel to this line.
A pi bond can form only if there is also a sigma bond between the same two atoms. The
sigma and pi bonds together make a double bond (Figure 8-5). The 1s orbitals (with one
eϪ each) of four hydrogen atoms overlap the remaining four sp2 orbitals (with one eϪ each)
on the carbon atoms to form four CXH sigma bonds (Figure 8-6).
Figure 8-5 A schematic
representation of the formation of a
carbon–carbon double bond. Two
sp2-hybridized carbon atoms form a
sigma (␴) bond by overlap of two sp2
orbitals (green, hatched) and a pi (␲)
bond by overlap of properly aligned
p orbitals (tan). All orbitals are fatter
than shown here.
+
sp2
sp2
C
p
C
p
sp2
sp2
σ
sp2
sp2
CC
sp2
sp2
along
with
one π bond
C
p
sp2
sp2
sp2
sp2 C
p
sp2
sp
2
one bond
H
H H
H
CC σ
σ
σ σ
σ
π
Figure 8-6 Four CXH ␴ bonds, one CXC
␴ bond (green, hatched), and one CXC ␲ bond
(tan, hatched) in the planar C2H4 molecule.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 10.5, Sigma Bonds.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 10.7, Multiple Bonding.
A double bond consists of one sigma bond and one pi bond.
As a consequence of the sp2 hybridization of C atoms in carbon–carbon double bonds,
each carbon atom is at the center of a trigonal plane. The p orbitals that overlap to form
the ␲ bond must be parallel to each other for effective overlap to occur. This adds the
further restriction that these trigonal planes (sharing a common corner) must also be
coplanar. Thus, all four atoms attached to the doubly bonded C atoms lie in the same plane
(see Figure 8-6). Many other important organic compounds contain carbon–carbon double
bonds. Several are described in Chapter 27.
COMPOUNDS CONTAINING TRIPLE BONDS
One compound that contains a triple bond is ethyne (acetylene), C2H2. Its Lewis formula
is
The VSEPR theory predicts that the two regions of high electron density around each
carbon atom are 180° apart.
Each triple-bonded carbon atom has two regions of high electron density, so valence
bond theory postulates that each is sp hybridized (see Section 8-5). Let us designate the
px orbitals as the ones involved in hybridization. Carbon has one electron in each sp hybrid
orbital and one electron in each of the 2py and 2pz orbitals (before bonding is considered).
See Figure 8-7.
The two carbon atoms form one sigma bond by head-on overlap of the sp hybrid orbitals;
each C atom also forms a sigma bond with one H atom. The sp hybrids on each atom are
180° apart. Thus, the entire molecule must be linear.
2p 2py 2pz
sp
2s
[He]C [He]C
hybridize
S ϭ N Ϫ A
ϭ 20 Ϫ 10 ϭ 10eϪ
shared
H H HC CHC C
8-14
The three p orbitals in a set are
indistinguishable. We can label the one
involved in hybridization as “px” to
help us visualize the orientations of the
two unhybridized p orbitals on carbon.
342 CHAPTER 8: Molecular Structure and Covalent Bonding Theories
The unhybridized atomic 2py and 2pz orbitals are perpendicular to each other and to the
line through the centers of the two sp hybrid orbitals (Figure 8-8). The side-on overlap
of the 2py orbitals on the two C atoms forms one pi bond; the side-on overlap of the 2pz
orbitals forms another pi bond.
A triple bond consists of one sigma bond and two pi bonds.
Figure 8-7 Diagram of the two
linear hybridized sp orbitals (green)
of an atom. These lie in a straight
line, and the two unhybridized p
orbitals py (tan, cross-hatched) and pz
(tan, hatched) lie in the perpendicular
plane and are perpendicular to each
other.
py
sp sp
pz
180°
C
Figure 8-8 The acetylene molecule, C2H2. (a) The overlap diagram of two sp-hybridized
carbon atoms and two s orbitals from two hydrogen atoms. The hybridized sp orbitals on
each C are shown in green and the unhybridized p orbitals are shown in tan. The dashed
lines, each connecting two lobes, indicate the side-by-side overlap of the four unhybridized p
orbitals to form two ␲ bonds. There are two CXH ␴ bonds, one CXC ␴ bond (green,
hatched), and two CXC ␲ bonds (hatched and cross-hatched). This makes the net carbon–
carbon bond a triple bond. (b) The ␲ bonding orbitals (tan) are positioned with one above
and below the line of the ␴ bonds (green) and the other behind and in front of the line of
the ␴ bonds.
8-15 A Summary of Electronic and Molecular Geometries 343
Some other molecules containing triply bonded atoms are nitrogen, SNmNS,
hydrogen cyanide, HXCmNS, and propyne, CH3XCmCXH. In each case, both atoms
involved in the triple bonds are sp hybridized. In the triple bond, each atom participates
in one sigma and two pi bonds. The C atom in carbon dioxide, /ÅOUCUOv", must participate
in two pi bonds (to two different O atoms). It also participates in two sigma bonds,
so it is also sp hybridized and the molecule is linear.
A SUMMARY OF ELECTRONIC AND
MOLECULAR GEOMETRIES
We have discussed several common types of polyatomic molecules and ions, and provided
a reasonable explanation for the observed structures and polarities of these species. Table
8-4 provides a summation.
Our discussion of covalent bonding illustrates two important points:
1. Molecules and polyatomic ions have definite shapes.
2. The properties of molecules and polyatomic ions are determined to a great extent
by their shapes. Incompletely filled electron shells and unshared pairs of electrons
on the central element are very important.
Our ideas about chemical bonding have developed over many years. As experimental techniques
for determining the structures of molecules have improved, our understanding of
chemical bonding has improved also. Experimental observations on molecular geometry
support our ideas about chemical bonding. The ultimate test for any theory is this: Can
8-15
pz
one bond
(a) (b)
py pz py
a second bond
H C C H H C C Hσ
ππ
In propyne, the C atom in the CH3
group is sp3 hybridized and at the
center of a tetrahedral arrangement.
344 CHAPTER 8: Molecular Structure and Covalent Bonding Theories
TABLE 8-4 A Summary of Electronic and Molecular Geometries of Polyatomic Molecules and Ions
Regions
of High Hybridization at
Electron Central Atom Hybridized Orbital
Densitya Electronic Geometry (Angles) Orientation Examples Molecular Geometry
BeCl2 linear
HgBr2 linear
sp
2 CdI2 linear
(180°)
CO2
b linear
linear C2H2
c linear
BF3 trigonal planar
BCl3 trigonal planar
NO3
Ϫe trigonal planar
sp2
3 SO2
d,e angular (AB2U)
(120°)
NO2
Ϫd,e angular (AB2U)
C2H4
f planar (trig. planar
trigonal planar at each C)
CH4 tetrahedral
CCl4 tetrahedral
NH4
ϩ tetrahedral
SO4
2Ϫ tetrahedral
sp3
4 CHCl3 distorted tet.
(109.5°)
NH3
d pyramidal (AB3U)
SO3
2Ϫd pyramidal (AB3U)
H3Oϩd pyramidal (AB3U)
tetrahedral H2Od angular (AB2U2)
PF5 trigonal bipyramidal
SbCl5 trigonal bipyramidal
sp3d
5 SF4
d seesaw (AB4U)
(90°, 120°, 180°)
ClF3
d T-shaped (AB3U2)
XeF2
d linear (AB2U3)
I3
Ϫd linear (AB2U3)
trigonal bipyramidal
SF6 octahedral
SeF6 octahedral
PF6
Ϫ octahedral
sp3d2
6 BrF5
d square pyramidal
(90°, 180°)
(AB5U)
XeF4
d square planar
(AB4U2)
octahedral
aThe number of locations of high electron density around the central atom. A region of high electron density may be a single bond, a double bond, a triple bond, or an
unshared pair. These determine the electronic geometry, and thus the hybridization of the central atom.
bContains two double bonds.
cContains a triple bond.
dCentral atom in molecule or ion has unshared pair(s) of electrons.
eBonding involves resonance.
fContains one double bond.
A
A
A
A
A
Key Terms 345
Key Terms
Angular A term used to describe the molecular geometry of a
molecule that has two atoms bonded to a central atom and one
or more unshared pairs on the central atom (AB2U or AB2U2).
Also called V-shaped or bent.
Central atom An atom in a molecule or polyatomic ion that is
bonded to more than one other atom.
Electronic geometry The geometric arrangement of orbitals
containing the shared and unshared electron pairs surrounding
the central atom of a molecule or polyatomic ion.
Hybridization The mixing of a set of atomic orbitals on an atom
to form a new set of hybrid orbitals with the same total electron
capacity and with properties and energies intermediate
between those of the original unhybridized orbitals.
Hybrid orbitals Orbitals formed on an atom by the process of
hybridization.
Lewis formula A method of representing a molecule or formula
unit by showing atoms and only outer-shell electrons; does not
show shape.
Linear A term used to describe the electronic geometry around
a central atom that has two regions of high electron density.
Also used to describe the molecular geometry of a molecule
or polyatomic ion that has one atom in the center bonded to
two atoms on opposite sides (180°) of the central atom (AB2 or
AB2U3).
Molecular geometry The arrangement of atoms (not unshared
pairs of electrons) around a central atom of a molecule or polyatomic
ion.
Octahedral A term used to describe the electronic geometry
around a central atom that has six regions of high electron density.
Also used to describe the molecular geometry of a molecule
or polyatomic ion that has one atom in the center bonded to
six atoms at the corners of an octahedron (AB6).
Octahedron A polyhedron with eight equal-sized, equilateral
triangular faces and six apices (corners).
Overlap of orbitals The interaction of orbitals on different
atoms in the same region of space.
Pi (␲) bond A bond resulting from the side-on overlap of atomic
orbitals, in which the regions of electron sharing are on opposite
sides of and parallel to an imaginary line connecting the
bonded atoms.
Seesaw A term used to describe the molecular geometry of a
molecule or polyatomic ion that has four atoms bonded to a
central atom and one unshared pair on the central atom (AB4U).
Sigma (␴) bond A bond resulting from the head-on overlap of
atomic orbitals, in which the region of electron sharing is along
and (cylindrically) symmetrical to an imaginary line connecting
the bonded atoms.
Square planar A term used to describe molecules and polyatomic
ions that have one atom in the center and four atoms at the
corners of a square.
Square pyramidal A term used to describe the molecular geometry
of a molecule or polyatomic ion that has five atoms bonded
to a central atom and one unshared pair on the central atom
(AB5U).
Tetrahedral A term used to describe the electronic geometry
around a central atom that has four regions of high electron
density. Also used to describe the molecular geometry of a molecule
or polyatomic ion that has one atom in the center bonded
to four atoms at the corners of a tetrahedron (AB4).
Tetrahedron A polyhedron with four equal-sized, equilateral triangular
faces and four apices (corners).
Trigonal bipyramid A six-sided polyhedron with five apices
(corners), consisting of two pyramids sharing a common triangular
base.
Trigonal bipyramidal A term used to describe the electronic
geometry around a central atom that has five regions of high
electron density. Also used to describe the molecular geometry
of a molecule or polyatomic ion that has one atom in the center
bonded to five atoms at the corners of a trigonal bipyramid
(AB5).
Trigonal planar (also plane triangular) A term used to describe
the electronic geometry around a central atom that has three
regions of high electron density. Also used to describe the molecular
geometry of a molecule or polyatomic ion that has one
atom in the center bonded to three atoms at the corners of an
equilateral triangle (AB3).
Trigonal pyramidal A term used to describe the molecular
geometry of a molecule or polyatomic ion that has three atoms
bonded to a central atom and one unshared pair on the central
atom (AB3U).
T-shaped A term used to describe the molecular geometry of a
molecule or polyatomic ion that has three atoms bonded to a
central atom and two unshared pairs on the central atom
(AB3U2).
Valence bond (VB) theory Assumes that covalent bonds are
formed when atomic orbitals on different atoms overlap and
electrons are shared.
Valence shell The outermost occupied electron shell of an atom.
Valence shell electron pair repulsion (VSEPR) theory
Assumes that valence electron pairs are arranged around the
central element of a molecule or polyatomic ion so that there
is maximum separation (and minimum repulsion) among
regions of high electron density.
it correctly predict the results of experiments before they are performed? When the answer
is yes, we have confidence in the theory. When the answer is no, the theory must be modified.
Current theories of chemical bonding enable us to make predictions that are usually
accurate.
346 CHAPTER 8: Molecular Structure and Covalent Bonding Theories
Exercises
VSEPR Theory: General Concepts
*01. State in your own words the basic idea of the VSEPR the-
ory.
*02. (a) Distinguish between “lone pairs” and “bonding pairs”
of electrons. (b) Which has the greater spatial requirement?
How do we know this? (c) Indicate the order of increasing
repulsions among lone pairs and bonding pairs of electrons.
*03. Identify the central atom (or atoms) in each of the following
compounds or ions: (a) H2SO4; (b) NH3; (c) NH4
ϩ;
(d) AlCl3; (e) CH3NH2; (f) CdCl2.
*04. Identify the central atom (or atoms) in each of the following
compounds or ions: (a) HCO3
Ϫ; (b) SiO2; (c) SO3;
(d) Al(OH)4
Ϫ; (e) BeBr2; (f ) (CH3)4Pb.
*05. Distinguish between electronic geometry and molecular
geometry.
*06. Under what conditions is molecular (or ionic) geometry
identical to electronic geometry about a central atom?
*07. What two shapes can a triatomic species have? How would
the electronic geometries for the two shapes differ?
*08. How are double and triple bonds treated when the VSEPR
theory is used to predict molecular geometry? How is a
single unshared electron treated?
*09. Sketch the three different possible arrangements of the
three B atoms around the central atom A for the molecule
AB3U2. Which of these structures correctly describes the
molecular geometry? Why? What are the predicted ideal
bond angles? How would observed bond angles deviate
from these values?
*10. Sketch the three different possible arrangements of the two
B atoms around the central atom A for the molecule
AB2U3. Which of these structures correctly describes the
molecular geometry? Why?
Valence Bond Theory: General Concepts
*11. What are hybridized atomic orbitals? How is the theory of
hybridized orbitals useful?
*12. (a) What is the relationship between the number of regions
of high electron density on an atom and the number of its
pure atomic orbitals that hybridize? (b) What is the relationship
between the number of atomic orbitals that
hybridize and the number of hybrid orbitals formed?
*13. (a) What is the maximum number of bonds that an atom
can form without expanding its valence shell? (b) What
must be true of the electron configuration of an element
for it to be able to expand its valence shell? (c) Tell which
of the following elements can expand its valence shell: N,
O, F, P, S, Cl.
*14. Prepare sketches of the overlaps of the following atomic
orbitals: (a) s with s; (b) s with p along the bond axis; (c) p
with p along the bond axis (head-on overlap); (d) p with p
perpendicular to the bond axis (side-on overlap).
*15. Prepare a sketch of the cross-section (through the atomic
centers) taken between two atoms that have formed (a) a
single ␴ bond, (b) a double bond consisting of a ␴ bond
and a ␲ bond, and (c) a triple bond consisting of a ␴ bond
and two ␲ bonds.
*16. Prepare sketches of the orbitals around atoms that are
(a) sp, (b) sp2, (c) sp3, (d) sp3d, and (e) sp3d2 hybridized. Show
in the sketches any unhybridized p orbitals that might participate
in multiple bonding.
*17. What hybridization is associated with these electronic
geometries: trigonal planar; linear; tetrahedral; octahedral;
trigonal bipyramidal?
*18. What angles are associated with orbitals in the following
sets of hybrid orbitals? (a) sp; (b) sp2; (c) sp3; (d) sp3d;
(e) sp3d2. Sketch each.
*19. What types of hybridization would you predict for molecules
having the following general formulas? (a) AB4;
(b) AB2U3; (c) AB3U; (d) ABU4; (e) ABU3.
*20. Repeat Exercise 19 for (a) ABU5; (b) AB2U4; (c) AB3;
(d) AB3U2; (e) AB5.
*21. What are the primary factors on which we base a decision
on whether the bonding in a molecule is better described
in terms of simple orbital overlap or overlap involving
hybridized atomic orbitals?
Electronic and Molecular Geometry
*22. Write a Lewis formula for each of the following species.
Indicate the number of regions of high electron density
and the electronic and molecular or ionic geometries.
(a) CdCl2; (b) SnCl4; (c) BrF3; (d) SbF6
Ϫ.
*23. Write a Lewis formula for each of the following species.
Indicate the number of regions of high electron density
and the electronic and molecular or ionic geometries.
(a) BF3; (b) SO2; (c) IO3
Ϫ; (d) SiCl4; (e) SeF6.
*24. (a) What would be the ideal bond angles in each molecule
or ion in Exercise 22, ignoring lone pair effects?
(b) How do these differ, if at all, from the actual values?
Why?
*25. (a) What would be the ideal bond angles in each molecule
or ion in Exercise 23, ignoring lone pair effects? (b) Are
these values greater than, less than, or equal to the actual
values? Why?
*26. The elements in Group IIA form compounds, such as
ClXBeXCl, that are linear and, therefore, nonpolar. What
is the hybridization at the central atoms?
*27. The elements in Group IIIA form compounds, such as
AlCl3, that are planar and, therefore, nonpolar. What is the
hybridization at the central atoms?
*28. Carbon forms two common oxides, CO and CO2. It also
forms a third (very uncommon) oxide, carbon suboxide,
C3O2, which is linear. The structure has terminal oxygen
atoms on both ends. Write the Lewis formula for C3O2.
Exercises 347
How many regions of high electron density are there about
each of the three carbon atoms?
*29. Pick the member of each pair that you would expect to
have the smaller bond angles, if different, and explain why.
(a) SF2 and SO2; (b) BF3 and BCl3; (c) SiF4 and SF4;
(d) NF3 and OF2.
*30. Draw a Lewis formula, sketch the three-dimensional shape,
and name the electronic and ionic geometries for the following
polyatomic ions. (a) H3Oϩ; (b) PCl6
Ϫ; (c) PCl4
Ϫ;
(d) SbCl4
ϩ.
*31. As the name implies, the interhalogens are compounds that
contain two halogens. Write Lewis formulas and threedimensional
structures for the following. Name the
electronic and molecular geometries of each. (a) BrF3;
(b) BrF; (c) BrF5.
*32. A number of ions derived from the interhalogens are
known. Write Lewis formulas and three-dimensional
structures for the following ions. Name the electronic and
molecular geometries of each. (a) IF4
ϩ; (b) ICl4
Ϫ;
(c) BrF4
Ϫ; (d) ClF3
2Ϫ.
*33. (a) Write a Lewis formula for each of the following molecules:
BF3; NF3; ClF3. (b) Contrast the molecular
geometries of these three molecules. Account for differences
in terms of the VSEPR theory.
*34. (a) Write a Lewis formula for each of the following molecules:
SiF4; SF4; XeF4. (b) Contrast the molecular
geometries of these three molecules. Account for differences
in terms of the VSEPR theory.
*35. Write the Lewis formulas and predict the shapes of these
very reactive carbon-containing species: H3Cϩ (a carbocation);
H3CSϪ (a carbanion); and SCH2 (a carbene whose
unshared electrons are paired).
*36. Write the Lewis formulas and predict the shapes of (a) I3
Ϫ;
(b) TeCl4; (c) XeO3; (d) NOBr (N is the central atom);
(e) NO2Cl (N is the central atom); (f ) SOCl2 (S is the central
atom).
*37. Describe the shapes of these polyatomic ions: (a) BO3
3Ϫ;
(b) AsO4
3Ϫ; (c) SO3
2Ϫ; (d) NO3
Ϫ.
*38. Would you predict a nitrogen–oxygen bond to have the
same magnitude of bond polarity as a hydrogen–oxygen
bond? Explain your answer.
*39. Which of the following molecules are polar? Why?
(a) CH4; (b) CH3Br; (c) CH2Br2; (d) CHBr3; (e) CBr4.
*40. Which of the following molecules are polar? Why?
(a) CdI2; (b) BCl3; (c) AsCl3; (d) H2O; (e) SF6.
*41. Which of the following molecules are nonpolar? Justify
your answer. (a) SO3; (b) IF; (c) Cl2O; (d) NF3; (e) CHCl3.
*42. The PF2Cl3 molecule is nonpolar. Use this information to
sketch its three-dimensional shape. Justify your choice.
*43. In what two major ways does the presence of unshared pairs
of valence electrons affect the polarity of a molecule?
Describe two molecules for which the presence of unshared
pairs on the central atom helps to make the molecules
polar. Can you think of a bonding arrangement that has
unshared pairs of valence electrons on the central atom but
that is nonpolar?
*44. Is the phosphorus–chlorine bond in phosphorus trichloride
a polar bond? Is phosphorus trichloride a polar
molecule? Explain.
*45. Is the phosphorus–chlorine bond in phosphorus pentachloride
a polar bond? Is phosphorus pentachloride a
polar molecule? Explain.
*46. Write the Lewis formula for each of the following. Indicate
which bonds are polar. (See Table 6-3.) Indicate which
molecules are polar. (a) CS2; (b) AlF3; (c) H2S; (d) SnF2.
*47. Write the Lewis formula for each of the following. Indicate
which bonds are polar. (See Table 6-3.) Indicate which
molecules are polar. (a) OF2; (b) CH4; (c) H2SO4; (d)SnF4.
Valence Bond Theory
*48. Describe the orbital overlap model of covalent bonding.
*49. Briefly summarize the reasoning by which we might have
predicted that the formula of the simplest stable hydrocarbon
would be CH2, if we did not consider hybridization.
Would this species satisfy the octet rule?
*50. What is the hybridization of the central atom in each of
the following? (a) NCl3; (b) molecular AlCl3; (c) CF4;
(d) SF6; (e) IO4
Ϫ.
*51. What is the hybridization of the central atom in each of
the following? (a) IF4
Ϫ; (b) SiO4
4Ϫ; (c) AlH4
Ϫ; (d) NH4
ϩ;
(e) PCl3; (f ) ClO3
Ϫ.
*52. (a) Describe the hybridization of the central atom in each
of these covalent species. (1) CHCl3; (2) CH2Cl2; (3) NF3;
(4) PO4
3Ϫ; (5) IF6
ϩ; (6) SiF6
2Ϫ. (b) Give the shape of each
species.
*53. Describe the hybridization of the underlined atoms in
C2F6, C2F2, N2F4, and (H2N)2CO.
*54. Prepare a sketch of the molecule CH3CHUCH2 showing
orbital overlaps. Identify the type of hybridization of
atomic orbitals on each carbon atom.
*55. After comparing experimental and calculated dipole
moments, Charles A. Coulson suggested that the Cl atom
in HCl is sp hybridized. (a) Give the orbital electronic
structure for an sp hybridized Cl atom. (b) Which HCl
molecule would have a larger dipole moment—one in
which the chlorine uses pure p orbitals for bonding with
the H atom or one in which sp hybrid orbitals are used?
*56. Predict the hybridization at each carbon atom in each of
the following molecules.
(a) acetone (a common solvent)
H
H
H C
H
HO
C HC C
348 CHAPTER 8: Molecular Structure and Covalent Bonding Theories
(b) glycine (an amino acid)
(c) nitrobenzene
(d) chloroprene (used to make neoprene, a synthetic
rubber)
(e) 4-penten-1-yne
*57. Predict the hybridization at the numbered atoms (➀, ➁,
and so on) in the following molecules and predict the
approximate bond angles at those atoms.
(a) diethyl ether, an anesthetic
H
C
H H
HH C O
H
H
HHH
C C
21
H
C
H H
C HH C C
H
C C
H
C
H Cl
C
H
HH C C C
H
HH
H H
N
O
O
H
H C
O
O
C
H
HN
H
(b) caffeine, a stimulant in coffee and in many over-thecounter
medicinals1
(c) acetylsalicylic acid (aspirin)1
(d) nicotine1
(e) ephedrine, a nasal decongestant1
H
HH
H H
N
H
C
H
O H
H
H
HC
1
2
4
3
C C
CC
H
H
N
N
H
H
H
H
H H H
H
H
CH3
1
4
52
3
H
H
H
C
C
O
O
CO
HO
H
H
H
H
2
1
3
H
H
H H
O
O N
N
N
C
C
C
N
HH H
H
H
H
2
1
3
1In these kinds of structural drawings, each intersection of lines
represents a C atom.
O
1
C
2
Exercises 349
*58. How many sigma and how many pi bonds are there in each
of the following molecules?
*59. How many sigma and how many pi bonds are there in each
of the following molecules?
C
Cl
C
CN
H
H
(c)
H
CH
H
C(a)
H
H
HO
CC
(d) C C HH C C
H H
CC
Cl
C
H Cl
H
C(b)
H
C H
H
H C
H
H
C
(c)
O
O
H
C
H H
C HH C
H H H
C(a)
*65. Write the Lewis formulas and predict the hybrid orbitals
and the shapes of these polyatomic ions and covalent molecules:
(a) HgCl2; (b) BF3; (c) BF4
Ϫ; (d) SbCl5; (e) SbF6
Ϫ.
*66. (a) What is the hybridization of each C in these molecules?
(i) H2CUO; (ii) HCqN; (iii) CH3CH2CH3; (iv) ketene,
H2CUCUO. (b) Describe the shape of each molecule.
*67. The following fluorides of xenon have been well characterized:
XeF2, XeF4, and XeF6. (a) Write Lewis formulas
for these substances and decide what type of hybridization
of the Xe atomic orbitals has taken place. (b) Draw all of
the possible atomic arrangements of XeF2, and discuss your
choice of molecular geometry. (c) What shape do you predict
for XeF4?
*68. Iodine and fluorine form a series of interhalogen molecules
and ions. Among these are IF (minute quantities observed
spectroscopically), IF3, IF4
Ϫ, IF5, IF6
Ϫ, and IF7. (a) Write
Lewis formulas for each of these species. (b) Identify the
type of hybridization that the orbitals of the iodine atom
have undergone in each substance. (c) Identify the shape
of the molecule or ion.
Mixed Exercises
*69. In the pyrophosphate ion, P2O7
4Ϫ, one oxygen atom is
bonded to both phosphorus atoms.
*60. Describe the bonding in the carbide ion, C2
2Ϫ, with a
three-dimensional VB structure. Show the orbital overlap,
and label the orbitals.
*61. Write Lewis formulas for molecular oxygen and ozone.
Assuming that all of the valence electrons in the oxygen
atoms are in hybrid orbitals, what would be the hybridization
of the oxygen atoms in each substance? Prepare
sketches of the molecules.
*62. Draw a Lewis formula and a three-dimensional structure
for each of the following polycentered molecules. Indicate
hybridizations and bond angles at each carbon
atom. (a) butane, C4H10; (b) propene, H2CUCHCH3;
(c) 1-butyne, HCmCCH2CH3; (d) acetaldehyde,
CH3CHO.
*63. How many ␴ bonds and how many ␲ bonds are there in
each of the molecules of Exercise 62?
*64. (a) Describe the hybridization of N in each of these
species.
(i) (ii) (iii) (iv)
(v)
(b) Give an orbital description for each species, specifying
the location of any unshared pairs and the orbitals used
for the multiple bonds.
H2N NH2.
HC N ;HN NH;NH4
ϩ
;NH3;
(d) CH2CHCH2OCH2CH3
H
C
H
O
HH C
H H
C
(b)
Write a Lewis formula, and sketch the three-dimensional
shape of the ion. Describe the electronic and molecular
geometries with respect to the central O atom and with
respect to each P atom.
*70. Briefly discuss the bond angles in the hydroxylamine molecule
in terms of the ideal geometry and the small changes
caused by electron-pair repulsions.
*71. Repeat Exercise 70 for the nitrite ion.
*72. The methyl free radical TCH3 has bond angles of about
120°, whereas the methyl carbanion SCH3
Ϫ has bond
O N
O
O N
O
ϪϪ
O HH
H
N
350 CHAPTER 8: Molecular Structure and Covalent Bonding Theories
Molecule Electronic Molecular Hybridization
or Ion Geometry Geometry of the Sulfur Atom
SO2
SCl2
SO3
SO3
2Ϫ
SF4
SO4
2Ϫ
SF5
ϩ
SF6
angles of about 109°. What can you infer from these facts
about the repulsive force exerted by an unpaired, unshared
electron as compared with that exerted by an unshared pair
of electrons?
*73. Two Lewis structures can be written for the square planar
molecule PtCl2Br2:
Show how a difference in dipole moments can distinguish
between these two possible structures.
*74. (a) Describe the hybridization of N in NO2
ϩ and NO2
Ϫ.
(b) Predict the bond angle in each case.
*75. The skeleton and a ball-and-stick model for the nitrous
acid molecule, HNO2, are shown here. Draw the Lewis
formula. What are the hybridizations at the middle O and
N atoms?
*76. Describe the change in hybridization that occurs at the
central atom of the reactant at the left in each of the following
reactions.
(a) PF5 ϩ FϪ n PF6
Ϫ
(b) 2CO ϩ O2 n 2CO2
(c) AlI3 ϩ IϪ n AlI4
Ϫ
(d) What change in hybridization occurs in the following
reaction?
*77. Consider the following proposed Lewis formulas for ozone
(O3):
(a) Which of these correspond to a polar molecule?
(b) Which of these predict covalent bonds of equal lengths
and strengths? (c) Which of these predict a diamagnetic
molecule? (d) The properties listed in parts (a), (b), and (c)
are those observed for ozone. Which structure correctly
predicts all three? (e) Which of these contain a considerable
amount of “strain”? Explain.
O OO(iii)
O O
O
(ii)O OO O OO(i)
NH3 H3NBF3 BF3ϩ
H
N
104Њ
116Њ
1.45Å
0.98Å
O
O
Pt
Cl
Cl
Br
Br
Pt
Cl
Cl
Br
Br
and
*78. What hybridizations are predicted for the central atoms in
molecules having the formulas AB2U2 and AB3U? What
are the predicted bond angles for these molecules? The
observed bond angles for representative substances are
H2O 104.5° NH3 106.7°
H2S 92.2° PH3 93.7°
H2Se 91.0° AsH3 91.8°
He2Te 89.5° SbH3 91.3°
What would be the predicted bond angle if no hybridization
occurred? What conclusion can you draw concerning
the importance of hybridization for molecules of compounds
involving elements with higher atomic numbers?
CONCEPTUAL EXERCISES
*79. Draw and explain the difference in the three-dimensional
shapes of CH3CH2CH3 and CH3COCH3.
*80. Complete the following table.
*81. Compare the shapes of the following pairs of molecules:
(a) H2CO and CH4; (b) PbCl4 and PbO2; (c) CH4 and
PbCl4.
*82. (a) Write two Lewis formulas for CO3
2Ϫ. Do the two formulas
represent resonance forms? (b) Do the same for
HSO4
Ϫ.
*83. What advantages does the VSEPR model of chemical
bonding have compared with Lewis formulas?
*84. What evidence could you present to show that two carbon
atoms joined by a single sigma bond are able to rotate about
an axis that coincides with the bond, but two carbon atoms
bonded by a double bond cannot rotate about an axis along
the double bond?
Exercises 351
BUILDING YOUR KNOWLEDGE
*85. Sketch three-dimensional representations of the following
molecules and indicate the direction of any net dipole
for each molecule: (a) CH4; (b) CH3Cl; (c) CH2Cl2;
(d) CHCl3; (e) CCl4.
*86. Write the Lewis formula and identify the electronic geometry
and molecular geometry for each polyatomic species
in the following equations:
(a) Hϩ ϩ H2O n H3Oϩ
(b) NH3 ϩ Hϩ n NH4
ϩ
*87. Draw the Lewis formula of an ammonium ion. Describe
the formation of the ammonium ion from ammonia plus
Hϩ. Does the hybridization of orbitals on nitrogen change
during the formation of the ammonium ion? Do the bond
angles change?
*88. The following is an incomplete Lewis formula for a molecule.
This formula has all the atoms at the correct places,
but it is missing several valence electrons. Complete this
Lewis formula, including lone pairs.
*89. The following is an incomplete Lewis formula for a molecule.
This formula has all the atoms at the correct places,
but it is missing several valence electrons. Complete this
Lewis formula, including lone pairs.
O
O
H
C
H
H
HH HBr
C C C N
Cl
C
H Br
C OH C
9 Molecular Orbitals in
Chemical Bonding
OUTLINE
9-1 Molecular Orbitals
9-2 Molecular Orbital Energy Level
Diagrams
9-3 Bond Order and Bond Stability
9-4 Homonuclear Diatomic
Molecules
9-5 Heteronuclear Diatomic
Molecules
9-6 Delocalization and the Shapes of
Molecular Orbitals
W
OBJECTIVES
After you have finished studying this chapter, you should be able to
• Describe the basic concepts of molecular orbital theory
• Relate the shapes and overlap of atomic orbitals to the shapes and energies of the
resulting molecular orbitals
• Distinguish among bonding, antibonding, and nonbonding orbitals
• Apply the Aufbau Principle to find molecular orbital descriptions for homonuclear
diatomic molecules and ions
• Apply the Aufbau Principle to find molecular orbital descriptions for heteronuclear
diatomic molecules and ions with small ⌬(EN) values
• Find the bond order in diatomic molecules and ions
• Relate bond order to bond stability
• Use the MO concept of delocalization for molecules in which valence bond theory would
postulate resonance
W
e have described bonding and molecular geometry in terms of valence bond
theory. In valence bond theory, we postulate that bonds result from the
sharing of electrons in overlapping orbitals of different atoms. These orbitals
may be pure atomic orbitals or hybridized atomic orbitals of individual atoms. We describe
electrons in overlapping orbitals of different atoms as being localized in the bonds between
the two atoms involved. We then use hybridization to help account for the geometry of
a molecule.
In molecular orbital theory, we postulate that
the combination of atomic orbitals on different atoms forms molecular orbitals
(MOs), so that electrons in them belong to the molecule as a whole.
Valence bond and molecular orbital theories are alternative descriptions of chemical
bonding. They have strengths and weaknesses, so they are complementary. Valence bond
A computer representation of one of
the ␲ molecular orbitals of benzene.
In some polyatomic molecules, a
molecular orbital may extend over only
a fraction of the molecule.
theory is descriptively attractive, and it lends itself well to visualization. Molecular orbital
(MO) theory gives better descriptions of electron cloud distributions, bond energies, and
magnetic properties, but its results are not as easy to visualize.
The valence bond picture of bonding in the O2 molecule involves a double bond.
This shows no unpaired electrons, so it predicts that O2 is diamagnetic. Experiments show,
however, that O2 is paramagnetic; therefore, it has unpaired electrons. Thus, the valence
bond structure is inconsistent with experiment and cannot be accepted as a description of
the bonding. Molecular orbital theory accounts for the fact that O2 has two unpaired electrons.
This ability of MO theory to explain the paramagnetism of O2 gave it credibility
as a major theory of bonding. We shall develop some of the ideas of MO theory and apply
them to some molecules and polyatomic ions.
MOLECULAR ORBITALS
We learned in Chapter 5 that each solution to the Schrödinger equation, called a wave
function, represents an atomic orbital. The mathematical pictures of hybrid orbitals in
valence bond theory can be generated by combining the wave functions that describe two
or more atomic orbitals on a single atom. Similarly, combining wave functions that describe
atomic orbitals on separate atoms generates mathematical descriptions of molecular
orbitals.
An orbital has physical meaning only when we square its wave function to describe the
electron density. Thus, the overall sign on the wave function that describes an atomic
orbital is not important, but when we combine two orbitals, the signs of the wave functions
are important. When waves are combined, they may interact either constructively
or destructively (Figure 9-1). Likewise, when two atomic orbitals overlap, they can be in
phase (added) or out of phase (subtracted). When they overlap in phase, constructive interaction
occurs in the region between the nuclei, and a bonding orbital is produced. The
energy of the bonding orbital is always lower (more stable) than the energies of the
combining orbitals. When they overlap out of phase, destructive interaction reduces the
probability of finding electrons in the region between the nuclei, and an antibonding
orbital is produced. This is higher in energy (less stable) than the original atomic orbitals.
The overlap of two atomic orbitals always produces two MOs: one bonding and one anti-
bonding.
We can illustrate this basic principle by considering the combination of the 1s atomic
orbitals on two different atoms (Figure 9-2). When these orbitals are occupied by electrons,
the shapes of the orbitals are plots of electron density. These plots show the regions in
molecules where the probabilities of finding electrons are the greatest.
In the bonding orbital, the two 1s orbitals have reinforced each other in the region
between the two nuclei by in-phase overlap, or addition of their electron waves. In the
antibonding orbital, they have canceled each other in this region by out-of-phase overlap,
or subtraction of their electron waves. We designate both molecular orbitals as sigma (␴)
molecular orbitals (which indicates that they are cylindrically symmetrical about the
internuclear axis). We indicate with subscripts the atomic orbitals that have been combined.
The star (#) denotes an antibonding orbital. Thus, two 1s orbitals produce a ␴1s (read
“sigma-1s”) bonding orbital and a ␴#
1s (read “sigma-1s-star”) antibonding orbital. The
right-hand side of Figure 9-2 shows the relative energy levels of these orbitals. All sigma
9-1
O O
Polyatomic ions such as CO3
2Ϫ,
SO4
2Ϫ, and NH4
ϩ can be described by
the molecular orbital approach.
354 CHAPTER 9: Molecular Orbitals in Chemical Bonding
An early triumph of molecular
orbital theory was its ability to
account for the observed
paramagnetism of oxygen, O2.
According to earlier theories, O2 was
expected to be diamagnetic, that is,
to have only paired electrons.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 10.9, Molecular Orbital Theory.
antibonding orbitals have nodal planes bisecting the internuclear axis. A node, or nodal
plane, is a region in which the probability of finding electrons is zero.
Another way of viewing the relative stabilities of these orbitals follows. In a bonding
molecular orbital, the electron density is high between the two atoms, where it stabilizes
the arrangement by exerting a strong attraction for both nuclei. By contrast, an antibonding
orbital has a node (a region of zero electron density) between the nuclei; this
allows for a strong net repulsion between the nuclei, which makes the arrangement less
stable. Electrons are more stable (have lower energy) in bonding molecular orbitals than
in the individual atoms. Placing electrons in antibonding orbitals, on the other hand,
requires an increase in their energy, which makes them less stable than in the individual
atoms.
Figure 9-1 An illustration of
constructive and destructive
interference of waves. (a) If the two
identical waves shown at the left are
added, they interfere constructively
to produce the more intense wave at
the right. (b) Conversely, if they are
subtracted, it is as if the phases
(signs) of one wave were reversed
and added to the first wave. This
causes destructive interference,
resulting in the wave at the right
with zero amplitude; that is, a
straight line.
9-1 Molecular Orbitals 355
Ϫ
ϩ
ϩ
ϩ
Ϫ
ϩ
Ϫ
ϩ
Ϫ
Ϫ
(a) In-phase overlap (add) (b) Out-of-phase overlap (subtract)
Figure 9-2 Molecular orbital (MO) diagram for the combination of the 1s atomic orbitals
on two identical atoms (at the left) to form two MOs. One is a bonding orbital, ␴1s (blue),
resulting from addition of the wave functions of the 1s orbitals. The other is an antibonding
orbital, ␴1s
# (red), at higher energy resulting from subtraction of the waves that describe the
combining 1s orbitals. In all ␴-type MOs, the electron density is symmetrical about an
imaginary line connecting the two nuclei. The terms “subtraction of waves,” “out of phase,”
and “destructive interference in the region between the nuclei” all refer to the formation of
an antibonding MO. Nuclei are represented by dots.
1s 1s
and
Nodal
plane
σ1s
(antibonding)
σ1s (bonding)
Molecular orbitals
Out-of-phase
overlap (subtract)
In-phaseoverlap (add)
Atomic orbitals
Energy
Atomic
orbital
Atomic
orbital
Molecular
orbitals
σ1s
(bonding)
σ1s
(antibonding)
1s 1s
#
#
For any two sets of p orbitals on two different atoms, corresponding orbitals such as
px orbitals can overlap head-on. This gives ␴p and ␴p
# orbitals, as shown in Figure 9-3 for
the head-on overlap of 2px orbitals on the two atoms. If the remaining p orbitals overlap
(py with py and pz with pz), they must do so sideways, or side-on, forming pi (␲) molecular
orbitals. Depending on whether all p orbitals overlap, there can be as many as two ␲p and
two ␲p
# orbitals. Figure 9-4 illustrates the overlap of two corresponding 2p orbitals on
two atoms to form ␲2p and ␲2p
# molecular orbitals. There is a nodal plane along the internuclear
axis for all pi molecular orbitals. If one views a sigma molecular orbital along the
internuclear axis, it appears to be symmetrical around the axis like a pure s atomic orbital.
A similar cross-sectional view of a pi molecular orbital looks like a pure p atomic orbital,
with a node along the internuclear axis.
The number of molecular orbitals (MOs) formed is equal to the number of atomic
orbitals that are combined. When two atomic orbitals are combined, one of the
resulting MOs is at a lower energy than the original atomic orbitals; this is a bonding
orbital. The other MO is at a higher energy than the original atomic orbitals; this
is an antibonding orbital.
Figure 9-3 Production of ␴2px
and
␴#
2px
molecular orbitals by overlap of
2px orbitals on two atoms.
356 CHAPTER 9: Molecular Orbitals in Chemical Bonding
How we name the axes is arbitrary. We
designate the internuclear axis as the x
direction.
This would involve rotating Figures
9-2, 9-3, and 9-4 by 90° so that the
internuclear axes are perpendicular to
the plane of the pages.
Figure 9-4 The ␲2p and ␲#
2p
molecular orbitals from overlap of
one pair of 2p atomic orbitals (for
instance, 2py orbitals). There can be
an identical pair of molecular
orbitals at right angles to these,
formed by another pair of p orbitals
on the same two atoms (in this case,
2pz orbitals).
and
σ2px
(bonding)
Molecular orbitals
σ2p
x
(antibonding)
Atomic orbitals
(head-on overlap)
2px 2px
Energy
#Out-of-phase
overlap (subtract)
In-phaseoverlap (add)
and
Out-of-phase
overlap (subtract)
In-phaseoverlap (add)2py
or 2pz
2py
or 2pz
Energy
π2p
y
or π2p
z
(antibonding)
π2py
or π2pz
(bonding)
Molecular orbitals
Atomic orbitals
(side-on overlap)
# #
If we had chosen the z axis as the axis
of head-on overlap of the 2p orbitals
in Figure 9-3, side-on overlap of the
2px–2px and 2py–2py orbitals would
form the ␲-type molecular orbitals.
MOLECULAR ORBITAL ENERGY LEVEL DIAGRAMS
Figure 9-5 shows molecular orbital energy level diagrams for homonuclear diatomic molecules
of elements in the first and second periods. Each diagram is an extension of the
right-hand diagram in Figure 9-2, to which we have added the molecular orbitals formed
from 2s and 2p atomic orbitals.
For the diatomic species shown in Figure 9-5a, the two ␲2p orbitals are lower in energy
than the ␴2p orbital. Molecular orbital calculations indicate, however, that for O2, F2, and
hypothetical Ne2 molecules, the ␴2p orbital is lower in energy than the ␲2p orbitals (see
Figure 9-5b).
9-2
Spectroscopic data support these
orders.
9-2 Molecular Orbital Energy Level Diagrams 357
Figure 9-5 Energy level diagrams for first- and second-period homonuclear diatomic
molecules and ions (not drawn to scale). The solid lines represent the relative energies of
the indicated atomic and molecular orbitals. (a) The diagram for H2, He2, Li2, Be2, B2, C2,
and N2 molecules and their ions. (b) The diagram for O2, F2 and Ne2 molecules and their
ions.
2p 2p
π2py
σ2p
ଙ
σ2p
H2 through N2
1s 1s
2s 2s
Molecular orbitalsAtomic orbitals Atomic orbitals
2p 2p
π2py
O2 through Ne2
1s 1s
2s 2s
Molecular orbitalsAtomic orbitals Atomic orbitals
(a) (b)
ଙ
π2py
ଙ
π2pz
π2pz
ଙ
σ2s
σ2s
ଙ
σ1s
Energy(notdrawntoscale)
ଙ
σ2p
ଙ
π2py
ଙ
π2pz
π2pz
σ2p
ଙ
σ2s
σ2s
σ1sσ1s
ଙ
σ1s
“Homonuclear” means consisting only
of atoms of the same element.
“Diatomic” means consisting of two
atoms.
Diagrams such as these are used to describe the bonding in a molecule in MO terms.
Electrons occupy MOs according to the same rules developed for atomic orbitals; they
follow the Aufbau Principle, the Pauli Exclusion Principle, and Hund’s Rule. (See Section
5-17.) To obtain the molecular orbital description of the bonding in a molecule or ion,
follow these steps:
1. Draw (or select) the appropriate molecular orbital energy level diagram.
2. Determine the total number of electrons in the molecule. Note that in applying
MO theory, we will account for all electrons. This includes both the inner-shell
electrons and the valence electrons.
3. Add these electrons to the energy level diagram, putting each electron into the
lowest energy level available.
a. A maximum of two electrons can occupy any given molecular orbital, and then
only if they have opposite spin (Pauli Exclusion Principle).
b. Electrons must occupy all the orbitals of the same energy singly before pairing
begins. These unpaired electrons must have parallel spins (Hund’s Rule).
BOND ORDER AND BOND STABILITY
Now we need a way to judge the stability of a molecule once its energy level diagram has
been filled with the appropriate number of electrons. This criterion is the bond order
(bo):
Bond order ϭ
Usually the bond order corresponds to the number of bonds described by the valence
bond theory. Fractional bond orders exist in species that contain an odd number of electrons,
such as the nitrogen oxide molecule, NO (15 electrons) and the superoxide ion,
O2
Ϫ (17 electrons).
A bond order equal to zero means that the molecule has equal numbers of electrons in
bonding MOs (more stable than in separate atoms) and in antibonding MOs (less stable
than in separate atoms). Such a molecule would be no more stable than separate atoms,
so it would not exist. A bond order greater than zero means that more electrons occupy
bonding MOs (stabilizing) than antibonding MOs (destabilizing). Such a molecule would
be more stable than the separate atoms, and we predict that its existence is possible. But
such a molecule could be quite reactive.
The greater the bond order of a diatomic molecule or ion, the more stable we predict
it to be. Likewise, for a bond between two given atoms, the greater the bond order,
the shorter is the bond length and the greater is the bond energy.
The bond energy is the amount of energy necessary to break a mole of bonds (Section
15-9); therefore, bond energy is a measure of bond strength.
(number of bonding electrons) Ϫ (number of antibonding electrons)
ᎏᎏᎏᎏᎏᎏᎏᎏ
2
9-3
Electrons in bonding orbitals are often
called bonding electrons, and
electrons in antibonding orbitals are
called antibonding electrons.
358 CHAPTER 9: Molecular Orbitals in Chemical Bonding
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 10.10, Molecular Electron
Configurations.
HOMONUCLEAR DIATOMIC MOLECULES
The electron distributions for the homonuclear diatomic molecules of the first and second
periods are shown in Table 9-1 together with their bond orders, bond lengths, and bond
energies.
The Hydrogen Molecule, H2
The overlap of the 1s orbitals of two hydrogen atoms produces ␴1s and ␴1s
# molecular
orbitals. The two electrons of the molecule occupy the lower energy ␴1s orbital (Figure
9-6a).
Because the two electrons in an H2 molecule are in a bonding orbital, the bond order
is one. We conclude that the H2 molecule would be stable, and we know it is. The energy
associated with two electrons in the H2 molecule is lower than that associated with the
same two electrons in separate 1s atomic orbitals. The lower the energy of a system, the
more stable it is. As the energy of a system decreases, its stability increases.
9-4
9-4 Homonuclear Diatomic Molecules 359
Problem-Solving Tip: Working with MO Theory
MO theory is often the best model to predict the bond order, bond stability, or magnetic
properties of a molecule or ion. The procedure is as follows:
1. Draw (or select) the appropriate MO energy level diagram.
2. Count the total number of electrons in the molecule or ion.
3. Follow the Pauli Exclusion Principle and Hund’s Rule to add the electrons to the
MO diagram.
4. Calculate the bond order: Bond order ϭ ΂ ΃.
5. Use the bond order to evaluate stability.
6. Look for the presence of unpaired electrons to determine if a species is paramag-
netic.
bonding e’s Ϫ antibonding e’s
ᎏᎏᎏᎏ
2
Figure 9-6 Molecular orbital diagrams for (a) H2 and (b) He2.
Energy
ls ls
H2
␴ls
␴ls
He2He
(b)(a)
H HeH
Energy
ls ls
␴ls
␴ls
H2 bond order ϭ ϭ 1
2 Ϫ 0
ᎏ
2
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 10.11, Homonuclear Diatomic
Molecules.
The Helium Molecule (Hypothetical), He2
The energy level diagram for He2 is similar to that for H2 except that it has two more
electrons. These occupy the antibonding ␴1s
# orbital (see Figures 9-5a and 9-6b and Table
9-1), giving He2 a bond order of zero. That is, the two electrons in the bonding orbital
of He2 would be more stable than in the separate atoms. But the two electrons in the antibonding
orbital would be less stable than in the separate atoms. These effects cancel, so
the molecule would be no more stable than the separate atoms. The bond order is zero,
and the molecule would not exist. In fact, He2 is unknown.
The Boron Molecule, B2
The boron atom has the configuration 1s22s22p1. Here p electrons participate in the
bonding. Figure 9-5a and Table 9-1 show that the ␲py
and ␲pz
molecular orbitals are lower
in energy than the ␴2p for B2. Thus, the electron configuration is
␴1s
2 ␴1s
#2 ␴2s
2 ␴2s
#2 ␲2py
1 ␲2pz
1
The unpaired electrons are consistent with the observed paramagnetism of B2. Here we
illustrate Hund’s Rule in molecular orbital theory. The ␲2py
and ␲2pz
orbitals are equal in
energy and contain a total of two electrons. Accordingly, one electron occupies each orbital.
The bond order is one. Experiments verify that B2 molecules exist in the vapor state.
The Nitrogen Molecule, N2
Experimental thermodynamic data show that the N2 molecule is stable, is diamagnetic,
and has a very high bond energy, 946 kJ/mol. This is consistent with molecular orbital
theory. Each nitrogen atom has seven electrons, so the diamagnetic N2 molecule has 14
electrons.
␴1s
2 ␴1s
#2 ␴2s
2 ␴2s
#2 ␲2py
2 ␲2pz
2 ␴2p
2
Six more electrons occur in bonding orbitals than in antibonding orbitals, so the bond
order is three. We see (Table 9-1) that N2 has a very short bond length, only 1.09 Å, the
shortest of any diatomic species except H2.
The Oxygen Molecule, O2
Among the homonuclear diatomic molecules, only N2 and the very small H2 have shorter
bond lengths than O2, 1.21 Å. Recall that VB theory predicts that O2 is diamagnetic.
Experiments show, however, that it is paramagnetic, with two unpaired electrons. MO
theory predicts a structure consistent with this observation. For O2, the ␴2p orbital is lower
in energy than the ␲2py
and ␲2pz
orbitals. Each oxygen atom has eight electrons, so the
O2 molecule has 16 electrons.
␴1s
2 ␴1s
#2 ␴2s
2 ␴2s
#2 ␴2p
2 ␲2py
2 ␲2pz
2 ␲#
2py
1 ␲#
2pz
1
The two unpaired electrons reside in the degenerate antibonding orbitals, ␲#
2py
and ␲#
2pz
.
Because there are four more electrons in bonding orbitals than in antibonding orbitals,
the bond order is two (see Figure 9-5b and Table 9-1). We see why the molecule is much
more stable than two free O atoms.
He2 bond order ϭ ϭ 0
2 Ϫ 2
ᎏ
2
360 CHAPTER 9: Molecular Orbitals in Chemical Bonding
O2 bond order ϭ ϭ 2
10 Ϫ 6
ᎏ
2
N2 bond order ϭ ϭ 3
In the valence bond representation, N2
is shown as SNmNS, with a triple
bond.
10 Ϫ 4
ᎏ
2
Orbitals of equal energy are called
degenerate orbitals. Hund’s Rule for
filling degenerate orbitals was
discussed in Section 5-17.
B2 bond order ϭ ϭ 1
6 Ϫ 4
ᎏ
2
Similarly, MO theory can be used to predict the structures and stabilities of ions, as
Example 9-1 shows.
EXAMPLE 9-1 Predicting Stabilities and Bond Orders
Predict the stabilities and bond orders of the ions (a) O2
ϩ and (b) O2
Ϫ.
Plan
(a) The O2
ϩ ion is formed by removing one electron from the O2 molecule. The electrons
that are withdrawn most easily are those in the highest energy orbitals. (b) The superoxide ion,
O2
Ϫ, results from adding an electron to the O2 molecule.
Solution
(a) We remove one of the ␲#
2p electrons of O2 to find the configuration of O2
ϩ:
␴1s
2 ␴1s
#2 ␴2s
2 ␴2s
#2 ␴2p
2 ␲2py
2 ␲2pz
2 ␲#
2py
1
There are five more electrons in bonding orbitals than in antibonding orbitals, so the bond
order is 2.5. We conclude that the ion would be reasonably stable relative to other diatomic
ions, and it does exist.
In fact, the unusual ionic compound [O2
ϩ][PtF6
Ϫ] played an important role in the discovery
of the first noble gas compound, XePtF6 (Section 24-2).
(b) We add one electron to the appropriate orbital of O2 to find the configuration of O2
Ϫ.
Following Hund’s Rule, we add this electron into the ␲#
2py
orbital to form a pair:
␴1s
2 ␴1s
#2 ␴2s
2 ␴2s
#2 ␴2p
2 ␲2py
2 ␲2pz
2 ␲#
2py
2 ␲#
2pz
1
TABLE 9-1 Molecular Orbitals for First- and Second-Period (Row) Diatomic Moleculesa
H2 He2
c Li2
b Be2
c B2
b C2
b N2 O2 F2 Ne2
c
␴#
2p
__ __ __ __ __ __ __ __ __ __hg
␲#
2py
␲#
2pz
__ __ __ __ __ __ __ __ __ __ __ __ __ __ __h __h __hg __hg __hg __hg
␴2p
__ __ __ __ __ __ __hg ␲2p
y
, ␲2p
z
__hg __hg __hg __hg __hg __hg
␲2py
␲2pz
__ __ __ __ __ __ __ __ __h __h __hg __hg __hg __hg ␴2p
__hg __hg __hg
␴2s
# __ __ __ __hg __hg __hg __hg __hg __hg __hg
␴2s
__ __ __hg __hg __hg __hg __hg __hg __hg __hg
␴1s
# __ __hg __hg __hg __hg __hg __hg __hg __hg __hg
␴1s
__hg __hg __hg __hg __hg __hg __hg __hg __hg __hg
Paramagnetic? no no no no yes no no yes no no
Bond order 1 0 1 0 1 2 3 2 1 0
Observed bond length (Å) 0.74 __ 2.67 __ 1.59 1.31 1.09 1.21 1.43 __
Observed bond energy (kJ/mol) 436 __ 110 9 Ϸ270 602 945 498 155 __
aElectron distribution in molecular orbitals, bond order, bond length, and bond energy of homonuclear diatomic molecules of the first- and second-period elements. Note
that nitrogen molecules, N2, have the highest bond energies listed; they have a bond order of three. The species C2 and O2, with a bond order of two, have the next
highest bond energies.
bExists only in the vapor state at elevated temperatures.
cUnknown species.
Increasingenergy
(nottoscale)
888888888888888888888888n
There are three more bonding electrons than antibonding electrons, so the bond order is
1.5. We conclude that the ion should exist but be less stable than O2.
The known superoxides of the heavier Group IA elements—KO2, RbO2, and CsO2 —
contain the superoxide ion, O2
Ϫ. These compounds are formed by combination of the free
metals with oxygen (Section 6-8, second subsection).
You should now work Exercises 19 and 20.
The Fluorine Molecule, F2
Each fluorine atom has 9 electrons, so there are 18 electrons in F2.
␴1s
2 ␴1s
#2 ␴2s
2 ␴2s
#2 ␴2p
2 ␲2py
2 ␲2pz
2 ␲#
2py
2 ␲#
2pz
2
The bond order is one. As you know, F2 exists. The FXF bond distance is longer (1.43 Å)
than the bond distances in O2 (1.21 Å) or N2 (1.09 Å) molecules. The bond order in F2
(one) is less than that in O2 (two) or N2 (three). The bond energy of the F2 molecules is
lower than that of either O2 or N2 (see Table 9-1). As a result, F2 molecules are the most
reactive of the three.
Heavier Homonuclear Diatomic Molecules
It might appear reasonable to use the same types of molecular orbital diagrams to predict
the stability or existence of homonuclear diatomic molecules of the third and subsequent
periods. However, the heavier halogens, Cl2, Br2, and I2, which contain only sigma (single)
bonds, are the only well-characterized examples at room temperature. We would predict
from both molecular orbital theory and valence bond theory that the other (nonhalogen)
homonuclear diatomic molecules from below the second period would exhibit pi bonding
and therefore multiple bonding.
Some heavier elements, such as S2, exist as diatomic species in the vapor phase at
elevated temperatures. These species are neither common nor very stable. The instability
is related to the inability of atoms of the heavier elements to form strong pi bonds with
each other. For larger atoms, the sigma bond length is too great to allow the atomic p
orbitals on different atoms to overlap very effectively. The strength of pi bonding therefore
decreases rapidly with increasing atomic size. For example, N2 is much more stable
than P2. This is because the 3p orbitals on one P atom do not overlap side by side in a
pi-bonding manner with corresponding 3p orbitals on another P atom nearly as effectively
as do the corresponding 2p orbitals on the smaller N atoms. MO theory does not predict
multiple bonding for Cl2, Br2, or I2, each of which has a bond order of one.
HETERONUCLEAR DIATOMIC MOLECULES
Heteronuclear Diatomic Molecules of Second-Period Elements
Corresponding atomic orbitals of two different elements, such as the 2s orbitals of nitrogen
and oxygen atoms, have different energies because their nuclei have different charges and
therefore different attractions for electrons. Atomic orbitals of the more electronegative
element are lower in energy than the corresponding orbitals of the less electronegative
9-5
F2 bond order ϭ ϭ 1
10 Ϫ 8
ᎏ
2
362 CHAPTER 9: Molecular Orbitals in Chemical Bonding
element. Accordingly, a molecular orbital diagram such as Figure 9-5 is inappropriate for
heteronuclear diatomic molecules. If the two elements are similar (as in NO or CN molecules,
for example), we can modify the diagram of Figure 9-5 by skewing it slightly. Figure
9-7 shows the energy level diagram and electron configuration for nitrogen oxide, NO,
also known as nitric oxide.
The closer the energy of a molecular orbital is to the energy of one of the atomic
orbitals from which it is formed, the more of the character of that atomic orbital it shows.
Thus, as we see in Figure 9-7, the bonding MOs in the NO molecule have more oxygenlike
atomic orbital character, and the antibonding orbitals have more nitrogen-like atomic
orbital character.
In general the energy differences ⌬E1, ⌬E2, and ⌬E3 (green backgrounds in Figure 9-7)
depend on the difference in electronegativities between the two atoms. The greater these
energy differences, the more polar is the bond joining the atoms and the greater is its
ionic character. On the other hand, the energy differences reflect the degree of overlap
between atomic orbitals; the smaller these differences, the more the orbitals can overlap,
and the greater is the covalent character of the bond.
We see that NO has a total of 15 electrons, making it isoelectronic with the N2
Ϫ ion.
The distribution of electrons is therefore the same in NO as in N2
Ϫ, although we expect
the energy levels of the MOs to be different. In accord with our predictions, nitrogen
oxide is a stable molecule. It has a bond order of 2.5, a short nitrogen–oxygen bond length
of 1.15 Å, a low dipole moment of 0.15 D, and a high bond energy of 891 kJ/mol.
Note: CN is a reactive molecule, not
the stable cyanide ion, CNϪ.
9-5 Heteronuclear Diatomic Molecules 363
Figure 9-7 MO energy level
diagram for nitrogen oxide, NO, a
slightly polar heteronuclear diatomic
molecule (␮ ϭ 0.15 D). The atomic
orbitals of oxygen, the more
electronegative element, are a little
lower in energy than the
corresponding atomic orbitals of
nitrogen, the less electronegative
element. For this molecule, the
energy differences ⌬E1, ⌬E2, and
⌬E3 are not very large; the molecule
is not very polar.
2p
π2p
σ2p
2p
ଙ
σ2p
∆E3
∆E2
2s
2s
ଙ
σ1s
1s
1s
∆E1
No bonds: bonding
and antibonding
electrons cancel
NO
Molecular
orbitals
N
Atomic
orbitals
O
Atomic
orbitals
Energy
ଙ
π2pz
ଙ
π2py
ଙ
σ2s
σ2s
σ1s
π2p yz
The Hydrogen Fluoride Molecule, HF
The electronegativity difference between hydrogen (EN ϭ 2.1) and fluorine (EN ϭ 4.0)
is very large (⌬(EN) ϭ 1.9). The hydrogen fluoride molecule contains a very polar bond
(␮ ϭ 1.91 D). The bond in HF involves the 1s electron of H and an unpaired electron
from a 2p orbital of F. Figure 9-8 shows the overlap of the 1s orbital of H with a 2p orbital
of F to form ␴sp and ␴sp
# molecular orbitals. The remaining two 2p orbitals of F have no
net overlap with H orbitals. They are called nonbonding orbitals. The same is true for
the F 2s and 1s orbitals. These nonbonding orbitals retain the characteristics of the
F atomic orbitals from which they are formed. The MO diagram of HF is shown in
Figure 9-9.
Other Diatomic Species with Large ⌬(EN) Values
If the energies of the atomic orbitals of the two atoms of a diatomic molecule or ion are
quite different, the MO diagram may be unlike that known for any homonuclear species.
Its unique MO diagram is constructed by combining the Schrödinger equations for the
two atoms. Construction of the MO diagram for CO is a complex case, beyond the
coverage in this textbook.
Figure 9-8 Formation of ␴sp and
␴sp
# molecular orbitals in HF by
overlap of the 1s orbital of H with a
2p orbital of F.
364 CHAPTER 9: Molecular Orbitals in Chemical Bonding
σsp
(bonding)
and
Out-of-phase
overlap (subtract)
In-phaseoverlap (add)
Energy
σsp
(antibonding)
Molecular orbitalsAtomic orbitals
hydrogen 1s fluorine 2px
#
2s
H–F
Molecular
orbitals
H
Atomic
orbitals
F
Atomic
orbitals
σsp
ଙ
σsp
1s
∆E = large
2p
Nonbonding
orbitals
Nonbonding
orbitals
2s
1s
Bonding orbital,
electrons of
highly polar
covalent bond
Bond order = ––––– = 12 – 0
2
Energy
δ+
δ–
2p
1s
Figure 9-9 MO energy level
diagram for hydrogen fluoride, HF, a
very polar molecule
(␮ ϭ 1.91 D). ⌬E is large because
the electronegativity difference is
large.
DELOCALIZATION AND THE SHAPES OF
MOLECULAR ORBITALS
In Section 7-6 we described resonance formulas for molecules and polyatomic ions. Resonance
is said to exist when two or more equivalent Lewis formulas can be written for the
same species and a single such formula does not account for the properties of a substance.
In molecular orbital terminology, a more appropriate description involves delocalization of
electrons. The shapes of molecular orbitals for species in which electron delocalization
occurs can be predicted by combining all the contributing atomic orbitals.
The Carbonate Ion, CO3
2Ϫ
Consider the trigonal planar carbonate ion, CO3
2Ϫ, as an example. All the carbon–oxygen
bonds in the ion have the same bond length and the same energy, intermediate between
those of typical CXO and CUO bonds. Valence bond theory describes the ion in terms
of three contributing resonance structures (Figure 9-10a). No one of the three resonance
forms adequately describes the bonding.
According to valence bond theory, the C atom is described as sp2 hybridized, and it
forms one sigma bond with each of the three O atoms. This leaves one unhybridized 2p
atomic orbital on the C atom, say the 2pz orbital. This orbital is capable of overlapping
and mixing with the 2pz orbital of any of the three O atoms. The sharing of two electrons
in the resulting localized pi orbital would form a pi bond. Thus, three equivalent resonance
structures can be drawn in valence bond terms (Figure 9-10b). We emphasize that
there is no evidence for the existence of these separate resonance structures.
The MO description of the pi bonding involves the simultaneous overlap and mixing
of the carbon 2pz orbital with the 2pz orbitals of all three oxygen atoms. This forms a
delocalized bonding pi molecular orbital system extending above and below the plane of
the sigma system, as well as an antibonding pi orbital system. Electrons are said to occupy
the entire set of bonding pi MOs, as depicted in Figure 9-10c. The shape is obtained by
averaging the contributing valence bond resonance structures. The bonding in such species
as nitrate ion, NO3
Ϫ, and ozone, O3, can be described similarly.
The Benzene Molecule, C6H6
Now let us consider the benzene molecule, C6H6, whose two valence bond resonance
forms are shown in Figure 9-11a. The valence bond description involves sp2 hybridization
at each C atom. Each C atom is at the center of a trigonal plane, and the entire
molecule is known to be planar. There are sigma bonds from each C atom to the two
adjacent C atoms and to one H atom. This leaves one unhybridized 2pz orbital on each
C atom and one remaining valence electron for each. According to valence bond theory,
adjacent pairs of 2pz orbitals and the six remaining electrons occupy the regions of overlap
to form a total of three pi bonds in either of the two ways shown in Figure 9-11b.
Experimental studies of the C6H6 structure prove that it does not contain alternating
single and double carbon–carbon bonds. The usual CXC single bond length is 1.54 Å,
and the usual CUC double bond length is 1.34 Å. All six of the carbon–carbon bonds in
benzene are the same length, 1.39 Å, intermediate between those of single and double
bonds.
This is well explained by the MO theory, which predicts that the six 2pz orbitals of the
C atoms overlap and mix to form three pi-bonding and three pi-antibonding molecular
9-6
The average carbon–oxygen bond
order in the CO3
2Ϫ ion is 1ᎏ1
3
ᎏ.
9-6 Delocalization and the Shapes of Molecular Orbitals 365
There is no evidence for the existence
of either of these forms of benzene.
The MO description of benzene is far
better than the valence bond
description.
Figure 9-10 Alternative representations of the bonding in the carbonate ion, CO3
2Ϫ.
(a) Lewis formulas of the three valence bond resonance structures. (b) Representation of the
p orbital overlap in the valence bond resonance structures. In each resonance form, the p
orbitals on two atoms would overlap to form the ␲ components of the hypothetical double
bonds. Each O atom has two additional sp2 orbitals (not shown) in the plane of the nuclei.
Each of these additional sp2 orbitals contains an oxygen unshared pair. (c) In the MO
description, the electrons in the ␲-bonded region are spread out, or delocalized, over all four
atoms of the CO3
2Ϫ ion. This MO description is more consistent with the experimental
observation of equal bond lengths and energies than are the valence bond pictures in parts
(a) and (b).
OO
OC
2–
OC
2–
O
O
C
2–
O
2–
O
O
O
C
••
••
••
••
••
••
••
••
2–
O
O
O
C
••
••
••
••
••
••••
••
2–
O
O
O
C
••••
••
••
••
••
••
••
OC
O
O
2–
(a) Lewis formulas for valence bond resonance structures
(b) p-Orbital overlap in valence bond resonance
(c) Delocalized MO representation
O
O
O
2–
O
O
or
C
Figure 9-11 Representations of the bonding in the benzene molecule, C6H6. (a) Lewis
formulas of the two valence bond resonance structures. (b) The six p orbitals of the benzene
ring, shown overlapping to form the (hypothetical) double bonds of the two resonance forms
of valence bond theory. (c) In the MO description the six electrons in the pi-bonded region
are delocalized, meaning they occupy an extended pi-bonding region above and below the
plane of the six C atoms.
9-6 Delocalization and the Shapes of Molecular Orbitals 367
H H
H H
H H
CC
CC
CC
H H
H H
H H
CC
C
CC
(a) Lewis formulas for valence bond resonance structures
H H
H
HH
H
C C
C C
C C H H
H
HH
H
C C
C C
C C
H
H
H
H
H
H C C
C
C C
(b) p-orbital overlap in valence bond resonance structures
(c) Delocalized MO representation
C
368 CHAPTER 9: Molecular Orbitals in Chemical Bonding
orbitals. For instance, the most strongly bonding pi molecular orbital in the benzene
pi–MO system is that in Figure 9-11c. The six pi electrons occupy three bonding MOs
of this extended (delocalized) system. Thus, they are distributed throughout the molecule
as a whole, above and below the plane of the sigma-bonded framework. This results in
identical character for all carbon–carbon bonds in benzene. Each carbon–carbon bond
has a bond order of 1.5. The MO representation of the extended pi system is the same
as that obtained by averaging the two contributing valence bond resonance structures.
Key Terms
Antibonding orbital A molecular orbital higher in energy than
any of the atomic orbitals from which it is derived; when populated
with electrons, lends instability to a molecule or ion.
Denoted with a star (#) superscript on its symbol.
Bond energy The amount of energy necessary to break one mole
of bonds of a given kind (in the gas phase).
Bond order Half the number of electrons in bonding orbitals
minus half the number of electrons in antibonding orbitals.
Bonding orbital A molecular orbital lower in energy than any
of the atomic orbitals from which it is derived; when populated
with electrons, lends stability to a molecule or ion.
Degenerate orbitals Orbitals of the same energy.
Delocalization The formation of a set of molecular orbitals that
extend over more than two atoms; important in species that
valence bond theory describes in terms of resonance.
Heteronuclear Consisting of different elements.
Homonuclear Consisting of only one element.
Molecular orbital (MO) An orbital resulting from overlap and
mixing of atomic orbitals on different atoms. An MO belongs
to the molecule as a whole.
Molecular orbital theory A theory of chemical bonding based
on the postulated existence of molecular orbitals.
Nodal plane (node) A region in which the probability of finding
an electron is zero.
Nonbonding orbital A molecular orbital derived only from an
atomic orbital of one atom; lends neither stability nor instability
to a molecule or ion when populated with electrons.
Pi (␲) bond A bond resulting from electron occupation of a pi
molecular orbital.
Pi (␲) orbital A molecular orbital resulting from side-on overlap
of atomic orbitals.
Sigma (␴) bond A bond resulting from electron occupation of
a sigma molecular orbital.
Sigma (␴) orbital A molecular orbital resulting from head-on
overlap of two atomic orbitals.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 10.13, Molecular Orbitals and
Vision.
Exercises
MO Theory: General Concepts
*01. Describe the main differences between the valence bond
theory and the molecular orbital theory.
*02. In molecular orbital theory, what is a molecular orbital?
What two types of information can be obtained from
molecular orbital calculations? How do we use such information
to describe the bonding within a molecule?
*03. What is the relationship between the maximum number of
electrons that can be accommodated by a set of molecular
orbitals and the maximum number that can be accommodated
by the atomic orbitals from which the MOs are
formed? What is the maximum number of electrons that
one MO can hold?
*04. Answer Exercise 3 after replacing “molecular orbitals” with
“hybridized atomic orbitals.”
*05. What differences and similarities exist among (a) atomic
orbitals, (b) localized hybridized atomic orbitals according
to valence bond theory, and (c) molecular orbitals?
*06. What is the relationship between the energy of a bonding
molecular orbital and the energies of the original atomic
orbitals? What is the relationship between the energy of
an antibonding molecular orbital and the energies of the
original atomic orbitals?
*07. Compare and contrast the following three concepts:
(a) bonding orbitals; (b) antibonding orbitals; (c) nonbonding
orbitals.
*08. Describe the shapes, including the locations of the nuclei,
of ␴ and ␴# orbitals.
*09. Describe the shapes, including the locations of the nuclei,
of ␲ and ␲# orbitals.
Exercises 369
*10. State the three rules for placing electrons in molecular
orbitals.
*11. What is meant by the term “bond order”? How is the value
of the bond order calculated?
*12. Compare and illustrate the differences between (a) atomic
orbitals and molecular orbitals, (b) bonding and antibonding
molecular orbitals, (c) ␴ bonds and ␲ bonds, and
(d) localized and delocalized molecular orbitals.
*13. Is it possible for a molecule or polyatomic ion in its ground
state to have a negative bond order? Why?
Homonuclear Diatomic Species
*14. What do we mean when we say that a molecule or ion is
(a) homonuclear, (b) heteronuclear, or (c) diatomic?
*15. Use the appropriate molecular orbital energy diagram to
write the electron configuration for each of the following;
calculate the bond order of each, and predict which would
exist. (a) H2
ϩ; (b) H2; (c) H2
Ϫ; (d) H2
2Ϫ.
*16. Repeat Exercise 15 for (a) He2
ϩ; (b) He2; (c) He2
2ϩ.
*17. Repeat Exercise 15 for (a) N2; (b) Ne2; (c) C2
2Ϫ.
*18. Repeat Exercise 15 for (a) Li2; (b) Li2
ϩ; (c) O2
2Ϫ.
*19. Use the appropriate molecular orbital energy diagram to
write the electron configurations of the following molecules
and ions: (a) Be2, Be2
ϩ, Be2
Ϫ; (b) B2, B2
ϩ, B2
Ϫ.
*20. What is the bond order of each of the species in Exercise
19?
*21. Which of the species in Exercise 19 are diamagnetic and
which are paramagnetic?
*22. Use MO theory to predict relative stabilities of the species
in Exercise 19. Comment on the validity of these predictions.
What else must be considered in addition to electron
occupancy of MOs?
*23. Which homonuclear diatomic molecules or ions of the second
period have the following electron distributions in
MOs? In other words, identify X in each.
(a) X2 ␴1s
2 ␴1s
#2 ␴2s
2 ␴2s
#2 ␲2py
2 ␲2pz
2 ␴2p
2
(b) X2 ␴1s
2 ␴1s
#2 ␴2s
2 ␴2s
#2 ␴2p
2 ␲2py
2 ␲2pz
2 ␲#
2py
1 ␲#
2pz
1
(c) X2
Ϫ ␴1s
2 ␴1s
#2 ␴2s
2 ␴2s
#2 ␲2py
2 ␲2pz
2 ␴2p
2 ␲#
2py
1
*24. What is the bond order of each of the species in Exercise
23?
*25. (a) Give the MO designations for O2, O2
Ϫ, O2
2Ϫ, O2
ϩ,
and O2
2ϩ. (b) Give the bond order in each case. (c) Match
these species with the following observed bond lengths:
1.04 Å; 1.12 Å; 1.21 Å; 1.33 Å; and 1.49 Å.
*26. (a) Give the MO designations for N2, N2
Ϫ, and N2
ϩ.
(b) Give the bond order in each case. (c) Rank these three
species by increasing predicted bond length.
*27. Assuming that the ␴2p MO is lower in energy than the
␲2py
and ␲2pz
MOs for the following species, write out electron
configurations for (a) F2, F2
ϩ, F2
Ϫ; (b) C2, C2
ϩ, C2
Ϫ.
*28. (a) What is the bond order of each species in Exercise 27?
(b) Are they diamagnetic or paramagnetic? (c) What would
MO theory predict about the stabilities of these species?
Heteronuclear Diatomic Species
The following is a molecular orbital energy level diagram for a
heteronuclear diatomic molecule, XY, in which both X and Y are
from Period 2 and Y is slightly more electronegative. This diagram
may be used in answering questions in this section.
*29. Use the preceding diagram to fill in an MO diagram for
NOϪ. What is the bond order of NOϪ? Is it paramagnetic?
How would you assess its stability?
*30. Repeat Exercise 29 for NOϩ.
*31. Repeat Exercise 29 for CNϩ. Refer to the preceding diagram
but assume that the ␲2py
and ␲2pz
MOs are lower in
energy than the ␴2p MO.
*32. Compare the MO descriptions for CN, CNϪ, and CN2Ϫ.
Refer to the preceding diagram but assume that the ␲2py
and ␲2pz
MOs are lower in energy than the ␴2p MO. Which
would be most stable? Why?
*33. For each of the two species OF and OFϪ: (a) Draw MO
energy level diagrams. (b) Write out electron configurations.
(c) Determine bond orders and predict relative
stabilities. (d) Predict diamagnetism or paramagnetism.
*34. For each of the two species NF and NFϩ: (a) Draw MO
energy level diagrams. (b) Write out electron configurations.
(c) Determine bond orders and predict relative
stabilities. (d) Predict diamagnetism or paramagnetism.
Energy
2p
2p
2s
2s
ls
ls
X
Atomic
orbitals
XY
Molecular
orbitals
Y
Atomic
orbitals
␴2p
␴2s
␴1s
␴2p
␴2s
␴1s
␲2pz
␲2pz
␲2py
␲2py
*35. Considering the shapes of MO energy level diagrams for
nonpolar covalent and polar covalent molecules, what
would you predict about MO diagrams, and therefore
about overlap of atomic orbitals, for ionic compounds?
*36. To increase the strength of the bonding in the hypothetical
compound BC, would you add or subtract an electron?
Explain your answer with the aid of an MO electron struc-
ture.
Delocalization
*37. Use Lewis formulas to depict the resonance structures
of the following species from the valence bond point of
view, and then sketch MOs for the delocalized ␲ systems.
(a) NO3
Ϫ, nitrate ion; (b) HCO3
Ϫ, hydrogen carbonate ion
(H is bonded to O); (c) NO2
Ϫ, nitrite ion.
*38. Use Lewis formulas to depict the resonance structures
of the following species from the valence bond point of
view, and then sketch MOs for the delocalized ␲ systems:
(a) SO2, sulfur dioxide; (b) O3 ozone; (c) HCO2
Ϫ, formate
ion (H is bonded to C).
Mixed Exercises
*39. Draw and label the complete MO energy level diagrams
for the following species. For each, determine the bond
order, predict the stability of the species, and predict
whether the species will be paramagnetic or diamagnetic.
(a) He2
ϩ; (b) CN; (c) HeHϩ.
*40. Draw and label the complete MO energy level diagrams
for the following species. For each, determine the bond
order, predict the stability of the species, and predict
whether the species will be paramagnetic or diamagnetic.
(a) O2
2ϩ; (b) HOϪ; (c) HF.
*41. Which of these species would you expect to be paramagnetic
or diamagnetic? (a) He2
Ϫ; (b) N2; (c) NOϩ; (d) N2
2ϩ;
(e) F2
ϩ.
CONCEPTUAL EXERCISES
*42. Refer to the diagrams in Figure 9-5 as needed. Can the
bond order of a diatomic species having 20 or fewer electrons
be greater than three? Can the bond order be a value
that is not divisible by 0.5? Why?
*43. As NO ionizes to form NOϩ, does the nitrogen–oxygen
bond become stronger or weaker?
NO n NOϩ ϩ eϪ
*44. Which of the homonuclear diatomic molecules of the second
row of the periodic table (Li2 to Ne2) are predicted
by MO theory to be paramagnetic? Which ones are predicted
to have a bond order of one? Which ones are
predicted to have a bond order of two? Which one is predicted
to have the highest bond order?
BUILDING YOUR KNOWLEDGE
*45. When carbon vaporizes at extremely high temperatures,
among the species present in the vapor is the diatomic
molecule C2. Write a Lewis formula for C2. Does your
Lewis formula of C2 obey the octet rule? (C2 does not
contain a quadruple bond.) Does C2 contain a single, a
double, or a triple bond? Is it paramagnetic or diamagnetic?
Show how molecular orbital theory can be used to
predict the answers to questions left unanswered by valence
bond theory.
*46. Rationalize the following observations in terms of the stabilities
of ␴ and ␲ bonds: (a) The most common form of
nitrogen is N2, whereas the most common form of phosphorus
is P4 (see the structure in Figure 2-3); (b) The most
common forms of oxygen are O2 and (less common) O3,
whereas the most common form of sulfur is S8.
370 CHAPTER 9: Molecular Orbitals in Chemical Bonding
10Reactions in Aqueous
Solutions I: Acids,
Bases, and Salts
OUTLINE
10-1 Properties of Aqueous Solutions
of Acids and Bases
10-2 The Arrhenius Theory
10-3 The Hydronium Ion (Hydrated
Hydrogen Ion)
10-4 The Brønsted–Lowry Theory
10-5 The Autoionization of Water
10-6 Amphoterism
10-7 Strengths of Acids
10-8 Acid–Base Reactions in
Aqueous Solutions
10-9 Acidic Salts and Basic Salts
10-10 The Lewis Theory
10-11 The Preparation of Acids
I
OBJECTIVES
After you have studied this chapter, you should be able to
• Describe the Arrhenius theory of acids and bases
• Describe hydrated hydrogen ions
• Describe the Brønsted–Lowry theory of acids and bases
• List properties of aqueous solutions of acids
• List properties of aqueous solutions of bases
• Arrange binary acids in order of increasing strength
• Arrange ternary acids in order of increasing strength
• Describe the Lewis theory of acids and bases
• Complete and balance equations for acid–base reactions
• Define acidic and basic salts
• Explain amphoterism
• Describe methods for preparing acids
I
n technological societies, acids, bases, and salts are indispensable compounds. Table
4-13 lists the 16 such compounds that were included in the top 50 chemicals produced
in the United States in 1997. The production of H2SO4 (number 1) was more than
twice as great as the production of lime (number 2). Sixty-five percent of the H2SO4 is
used in the production of fertilizers.
Many acids, bases, and salts occur in nature and serve a wide variety of purposes. For
instance, your “digestive juice” contains approximately 0.10 mole of hydrochloric acid per
liter. Human blood and the aqueous components of most cells are mildly acidic. The liquid
in your car battery is approximately 40% H2SO4 by mass. Baking soda is a salt of carbonic
acid. Sodium hydroxide, a base, is used in the manufacture of soaps, paper, and many other
chemicals. “Draෆno” is solid NaOH that contains some aluminum chips. Sodium chloride
is used to season food and as a food preservative. Calcium chloride is used to melt ice on
Many foods and other household
items contain weak acids, weak
bases, and salts.
You will encounter many of these in
your laboratory work.
highways and in the emergency treatment of cardiac arrest. Several ammonium salts are
used as fertilizers. Many organic acids (carboxylic acids) and their derivatives occur in
nature. Acetic acid is present in vinegar; the sting of an ant bite is due to formic acid.
Amino acids are the building blocks of proteins, which are important materials in the
bodies of animals, including humans. Amino acids are carboxylic acids that also contain
basic groups derived from ammonia. The pleasant odors and flavors of ripe fruit are due
in large part to the presence of esters (Chapter 27), which are formed from the acids in
unripe fruit.
PROPERTIES OF AQUEOUS SOLUTIONS
OF ACIDS AND BASES
Aqueous solutions of most protonic acids (those containing acidic hydrogen atoms)
exhibit certain properties, which are properties of hydrated hydrogen ions in aqueous solu-
tion.
1. They have a sour taste. Pickles are usually preserved in vinegar, a 5% solution of
acetic acid. Many pickled condiments contain large amounts of sugar so that the
taste of acetic acid is partially masked by the sweet taste of sugar. Lemons contain
citric acid, which is responsible for their characteristic sour taste.
2. They change the colors of many indicators (highly colored dyes). Acids turn blue
litmus red, and cause bromthymol blue to change from blue to yellow.
3. Nonoxidizing acids react with metals above hydrogen in the activity series (Section
4-8, part 2) to liberate hydrogen gas, H2. (HNO3, a common oxidizing acid, reacts
with metals to produce nitrogen oxides, not H2.)
4. They react with (neutralize) metal oxides and metal hydroxides to form salts and
water (Section 4-9, part 1).
5. They react with salts of weaker or more volatile acids to form the weaker or more
volatile acid and a new salt.
6. Aqueous solutions of acids conduct an electric current because they are totally or
partly ionized.
10-1
Caution: We should never try to
identify a substance by taste. You have,
however, probably experienced the
sour taste of acetic acid in vinegar or
citric acid in foods that contain citrus
fruits.
10-1 Properties of Aqueous Solutions of Acids and Bases 373
Left, Many common household liquids are acidic, including soft drinks, vinegar, and fruit
juices. Right, Most cleaning materials are basic.
Aqueous solutions of most bases also exhibit certain properties, which are due to the
hydrated hydroxide ions present in aqueous solutions of bases.
1. They have a bitter taste.
2. They have a slippery feeling. Soaps are common examples; they are mildly basic. A
solution of household bleach feels very slippery because it is quite basic.
3. They change the colors of many indicators: litmus changes from red to blue, and
bromthymol blue changes from yellow to blue, in bases.
4. They react with (neutralize) acids to form salts and, in most cases, water.
5. Their aqueous solutions conduct an electric current because they are dissociated or
ionized.
THE ARRHENIUS THEORY
In 1680, Robert Boyle noted that acids (1) dissolve many substances, (2) change the colors
of some natural dyes (indicators), and (3) lose their characteristic properties when mixed
with alkalis (bases). By 1814, J. Gay-Lussac concluded that acids neutralize bases and that
the two classes of substances should be defined in terms of their reactions with each other.
In 1884, Svante Arrhenius (1859–1927) presented his theory of electrolytic dissociation,
which resulted in the Arrhenius theory of acid–base reactions. In his view,
an acid is a substance that contains hydrogen and produces Hϩ in aqueous solution.
A base is a substance that contains the OH (hydroxyl) group and produces
hydroxide ions, OHϪ, in aqueous solution.
Neutralization is defined as the combination of Hϩ ions with OHϪ ions to form H2O
molecules.
Hϩ(aq) ϩ OHϪ(aq) 88n H2O(ᐉ) (neutralization)
The Arrhenius theory of acid–base behavior satisfactorily explained reactions of protonic
acids with metal hydroxides (hydroxy bases). It was a significant contribution to chemical
thought and theory in the latter part of the nineteenth century. The Arrhenius model of
acids and bases, although limited in scope, led to the development of more general theories
of acid–base behavior. They will be considered in later sections.
THE HYDRONIUM ION (HYDRATED HYDROGEN ION)
Although Arrhenius described Hϩ ions in water as bare ions (protons), we now know that
they are hydrated in aqueous solution and exist as Hϩ(H2O)n, in which n is some small
integer. This is due to the attraction of the Hϩ ions, or protons, for the oxygen end (␦Ϫ)
of water molecules. Although we do not know the extent of hydration of Hϩ in most solu-
10-3
10-2
The indicator bromthymol blue is
yellow in acidic solution and blue in
basic solution.
374 CHAPTER 10: Reactions in Aqueous Solutions I: Acids, Bases, and Salts
This is an extremely important idea.
We now know that all ions are
hydrated in aqueous solution.
Review Sections 4-2, 4-9, part 1, 6-7,
and 6-8.
The most common isotope of
hydrogen, 1
1H, has no neutrons. Thus,
1
1Hϩ is a bare proton. In discussions of
acids and bases, we use the terms
“hydrogen ion,” “proton,” and “Hϩ”
interchangeably.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 4.7, Acids, and Screen 4.8,
Bases.
tions, we usually represent the hydrated hydrogen ion as the hydronium ion, H3Oϩ, or
Hϩ(H2O)n, in which n ϭ 1.
The hydrated hydrogen ion is the species that gives aqueous solutions of acids their
characteristic acidic properties.
Whether we use the designation Hϩ(aq) or H3Oϩ, we always mean the hydrated hydrogen
ion.
THE BRØNSTED–LOWRY THEORY
In 1923, J. N. Brønsted (1879–1947) and T. M. Lowry (1874–1936) independently
presented logical extensions of the Arrhenius theory. Brønsted’s contribution was more
thorough than Lowry’s, and the result is known as the Brønsted theory or the Brønsted–Lowry
theory.
An acid is defined as a proton donor (Hϩ) and a base is defined as a proton acceptor.
These definitions are sufficiently broad that any hydrogen-containing molecule or ion
capable of releasing a proton, Hϩ, is an acid, whereas any molecule or ion that can accept
a proton is a base. (In the Arrhenius theory of acids and bases, only substances that contain
the OHϪ group would be called bases.)
An acid–base reaction is the transfer of a proton from an acid to a base.
Thus, the complete ionization of hydrogen chloride, HCl, a strong acid, in water is an
acid–base reaction in which water acts as a base or proton acceptor.
Step 1: HCl(aq) 88n Hϩ(aq) ϩ ClϪ(aq) (Arrhenius description)
Step 2: H2O(ᐉ) ϩ Hϩ(aq) 88n H3Oϩ
Overall: H2O(ᐉ) ϩ HCl(aq) 88n H3Oϩ ϩ ClϪ(aq) (Brønsted–Lowry description)
10-4
H
H
ϩ
O Hϩ
H
H
ϩ
OH
The Brønsted–Lowry theory is
especially useful for reactions in
aqueous solutions. It is widely used in
medicine and in the biological
sciences.
10-4 The Brønsted–Lowry Theory 375
O
H
H
H : +
–
O
H
H :
:
+
base2acid1acid2base1
H
+ transfer
+
H :
:
:Cl
:
:
:
:Cl
:
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 17.2, Brønsted Acids and Bases.
Remember that in this text we use red
to indicate acids and blue to indicate
bases.
Hydronium ion, H3Oϩ.
The ionization of hydrogen fluoride, a weak acid, is similar, but it occurs to only a slight
extent, so we use a double arrow to indicate that it is reversible.
Various measurements (electrical
conductivity, freezing point depression,
etc.) indicate that HF is only slightly
ionized in water.
376 CHAPTER 10: Reactions in Aqueous Solutions I: Acids, Bases, and Salts
Problem-Solving Tip: Conjugate Acid-Base Pairs
We can describe Brønsted–Lowry acid–base reactions in terms of conjugate acid–base
pairs. These are two species that differ by a proton. In the preceding equation, HF (acid2)
and FϪ (base2) are one conjugate acid–base pair, and H2O (base1) and H3Oϩ(acid1) are
the other pair. The members of each conjugate pair are designated by the same numerical
subscript. In the forward reaction, HF and H2O act as acid and base, respectively. In
the reverse reaction, H3Oϩ acts as the acid, or proton donor, and FϪ acts as the base, or
proton acceptor.
–
+
base2acid1acid2base1
H2O(ᐉ) + HF(aq) H3O+ + F–(aq)
O
H
H :
:
+
H
+ transfer
H :
:
: F
:
:
:
: F
:
+O
H
H :
H H+
tra
nsfer
HF F–
–H+
+H+
H2O OH–
–H+
+H+
H3O+ H2O
–H+
+H+
Conjugate acid Conjugate base
–
H
+ , remove H+, decrease charge by
1
+H +
, add H+
, increase charge by 1
When the weak acid, HF, dissolves in water, the HF molecules give up some Hϩ ions
that can be accepted by either of two bases, FϪ or H2O. The fact that HF is only slightly
ionized tells us that FϪ is a stronger base than H2O. When the strong acid, HCl, dissolves
in water, the HCl molecules give up Hϩ ions that can be accepted by either of two bases,
ClϪ or H2O. The fact that HCl is completely ionized in dilute aqueous solution tells us
FϪ is a stronger base than H2O. H2O
is a stronger base than ClϪ. FϪ is
therefore a stronger base than ClϪ.
It makes no difference which conjugate
acid–base pair, HF and FϪ or H3Oϩ
and H2O, is assigned the subscripts 1
and 2.
The double arrow is used to indicate
that the reaction occurs in both the
forward and the reverse directions.
We use rectangles to indicate one
conjugate acid–base pair and ovals to
indicate the other pair.
that ClϪ is a weaker base than H2O. Thus, the weaker acid, HF, has the stronger conjugate
base, FϪ. The stronger acid, HCl, has the weaker conjugate base, ClϪ. We can
generalize:
The stronger the acid, the weaker is its conjugate base; the weaker the acid, the
stronger is its conjugate base.
“Strong” and “weak,” like many other adjectives, are used in a relative sense. We do not
mean to imply that the fluoride ion, FϪ, is a strong base compared with species such as
the hydroxide ion, OHϪ. We mean that relative to the anions of strong acids, which are very
weak bases, FϪ is a much stronger base.
Ammonia acts as a weak Brønsted–Lowry base, and water acts as an acid in the ionization
of aqueous ammonia.
Be careful to avoid confusing solubility
in water and extent of ionization. They
are not necessarily related. Ammonia is
very soluble in water (Ϸ15 mol/L at
25°C). In 0.10 M solution, NH3 is
only 1.3% ionized and 98.7%
nonionized.
10-4 The Brønsted–Lowry Theory 377
As we see in the reverse reaction, ammonium ion, NH4
ϩ, is the conjugate acid of NH3.
The hydroxide ion, OHϪ, is the conjugate base of water. In three dimensions, the molecular
structures are
Water acts as an acid (Hϩ donor) in its reaction with NH3, whereas it acts as a base
(Hϩ acceptor) in its reactions with HCl and with HF.
Whether water acts as an acid or as a base depends on the other species present.
N
H
H
H :
–
base1 acid2 acid1 base2
NH3(aq) + H2O(ᐉ) NH4
+(aq) + OH–(aq)
+
+ O
H
:
:
:
H
+ transfer
H + O
H
:
:
:N
H
H
H
H
+ transfer
H
++N O N
H
+
O
–
Trigonal pyramidal molecule Angular molecule Tetrahedral ion Linear ion
H
H
H
H
H
H
H
H
H
H+ transfer
H+ transfer
THE AUTOIONIZATION OF WATER
Careful measurements show that pure water ionizes ever so slightly to produce equal
numbers of hydrated hydrogen ions and hydroxide ions.
10-5
The prefix “amphi-” means “of both
kinds.” “Amphiprotism” refers to
amphoterism by accepting and
donating a proton in different
reactions (see Section 10-6).
378 CHAPTER 10: Reactions in Aqueous Solutions I: Acids, Bases, and Salts
H+ transfer
++H
O O
H
H
O
H
+
H
O
–
H O +
base1 acid2 acid1 base2
H
:: H O
H
::
:
+ –
+
H
H
H
– –
–
H+ transfer
O
H
::
:–
H O
H
:–
–
H+ transfer
H
–
H+
transfer
In simplified notation, we represent this reaction as
H2O(ᐉ) 34 Hϩ(aq) ϩ OHϪ(aq)
This autoionization (self-ionization) of water is an acid–base reaction according to the
Brønsted–Lowry theory. One H2O molecule (the acid) donates a proton to another H2O
molecule (the base). The H2O molecule that donates a proton becomes an OHϪ ion, the
conjugate base of water. The H2O molecule that accepts a proton becomes an H3Oϩ ion.
Examination of the reverse reaction (right to left) shows that H3Oϩ (an acid) donates a
proton to OHϪ (a base) to form two H2O molecules. One H2O molecule behaves as an
acid and the other acts as a base in the autoionization of water. Water is said to be
amphiprotic; that is, H2O molecules can both donate and accept protons.
As we saw in Section 4-9, part 1, H3Oϩ and OHϪ ions combine to form nonionized
water molecules when strong acids and strong soluble bases react to form soluble salts
and water. The reverse reaction, the autoionization of water, occurs only slightly, as
expected.
AMPHOTERISM
As we have seen, whether a particular substance behaves as an acid or as a base depends
on its environment. Earlier we described the amphiprotic nature of water. Amphoterism
is a more general term that describes the ability of a substance to react either as an acid
or as a base. Amphiprotic behavior describes the cases in which substances exhibit amphoterism
by accepting and by donating a proton, Hϩ. Several insoluble metal hydroxides are
amphoteric; that is, they react with acids to form salts and water, but they also dissolve in
and react with excess strong bases.
Aluminum hydroxide is a typical amphoteric metal hydroxide. Its behavior as a base is
illustrated by its reaction with nitric acid to form a normal salt. The balanced formula unit,
total ionic, and net ionic equations for this reaction are, respectively:
10-6
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 17.3, Acid–Base Properties of
Water.
Al(OH)3(s) ϩ 3HNO3(aq) 88n Al(NO3)3(aq) ϩ 3H2O(ᐉ)
Al(OH)3(s) ϩ 3[Hϩ(aq) ϩ NO3
Ϫ(aq)] 88n [Al3ϩ(aq) ϩ 3NO3
Ϫ(aq)] ϩ 3H2O(ᐉ)
Al(OH)3(s) ϩ 3Hϩ(aq) 88n Al3ϩ(aq) ϩ 3H2O(ᐉ)
When an excess of a solution of any strong base, such as NaOH, is added to solid
aluminum hydroxide, the Al(OH)3 acts as an acid and dissolves. The equation for the reaction
is usually written
Al(OH)3(s) ϩ NaOH(aq) 88n NaAl(OH)4(aq)
an acid a base sodium aluminate,
a soluble compound
The total ionic and net ionic equations are
Al(OH)3(s) ϩ [Naϩ(aq) ϩ OHϪ(aq)] 88n [Naϩ(aq) ϩ Al(OH)4
Ϫ(aq)]
Al(OH)3(s) ϩ OHϪ(aq) 88n Al(OH)4
Ϫ(aq)
All hydroxides containing small, highly
charged metal ions are insoluble in
water.
Our Daily LivesCCHEMISTRY IN USE
Everyday Salts of Ternary Acids
Y ou may have encountered some salts of ternary acids without
even being aware of them. For example, the iron in many
of your breakfast cereals and breads may have been added in
the form of iron(II) sulfate, FeSO4, or iron(II) phosphate,
Fe3(PO4)2; the calcium in these foods often comes from the
addition of calcium carbonate, CaCO3. Fruits and vegetables
keep fresh longer after an application of sodium sulfite,
Na2SO3, and sodium hydrogen sulfite, NaHSO3. Restaurants
also use these two sulfites to keep their salad bars more appetizing.
The red color of fresh meat is maintained for much
longer by the additives sodium nitrate, NaNO3, and sodium
nitrite, NaNO2. Sodium phosphate, Na3PO4, is used to prevent
metal ion flavors and to control acidity in some canned
goods.
Many salts of ternary acids are used in medicine. Lithium
carbonate, Li2CO3, has been used successfully to combat
severe jet lag. Lithium carbonate is also useful in the treatment
of mania, depression, alcoholism, and schizophrenia.
Magnesium sulfate, MgSO4, sometimes helps to prevent convulsions
during pregnancy and to reduce the solubility of
toxic barium sulfate in internally administered preparations
consumed before gastrointestinal X-ray films are taken.
Other salts of ternary acids that you may find in your home
include potassium chlorate, KClO3, in matches as an oxidizing
agent and oxygen source; sodium hypochlorite, NaClO,
in bleaches and mildew removers; and ammonium carbonate,
(NH4)2CO3, which is the primary ingredient in smelling
salts. Limestone and marble are calcium carbonate; gypsum
and plaster of Paris are primarily calcium sulfate, CaSO4.
Fireworks get their brilliant colors from salts such as barium
nitrate, Ba(NO3)2, which imparts a green color; strontium
carbonate, SrCO3, which gives a red color; and copper(II)
sulfate, CuSO4, which produces a blue color. Should your
fireworks get out of hand and accidentally start a fire, the
ammonium phosphate, (NH4)3PO4, sodium hydrogen carbonate,
NaHCO3, and potassium hydrogen carbonate,
KHCO3, in your ABC dry fire extinguisher will come in
handy.
An unexpected place to find ternary acid salts is in your
long-distance phone bills; nitrates, NO3
Ϫ, are cheaper than
day rates.
Ronald DeLorenzo
Middle Georgia College
The tips of “strike anywhere” matches contain tetraphosphorus trisulfide,
red phosphorus, and potassium chlorate. Friction converts kinetic energy
into heat, which initiates a spontaneous reaction.
P4S3(s) ϩ 8O2 88n P4O10(s) ϩ 3SO2(g)
The thermal decomposition of KClO3 provides additional oxygen for this
reaction.
Other amphoteric metal hydroxides undergo similar reactions.
Table 10-1 contains lists of the common amphoteric hydroxides. Three are hydroxides
of metalloids, As, Sb, and Si, which are located along the line that divides metals and
nonmetals in the periodic table.
STRENGTHS OF ACIDS
Binary Acids
The ease of ionization of binary protonic acids depends on both (1) the ease of breaking
HXX bonds and (2) the stability of the resulting ions in solution. Let us consider the
relative strengths of the Group VIIA hydrohalic acids. Hydrogen fluoride ionizes only
slightly in dilute aqueous solutions.
HF(aq) ϩ H2O(ᐉ) 34 H3Oϩ(aq) ϩ FϪ(aq)
HCl, HBr, and HI, however, ionize completely or nearly completely in dilute aqueous
solutions because the HXX bonds are much weaker.
HX(aq) ϩ H2O(ᐉ) 88n H3Oϩ(aq) ϩ XϪ(aq) X ϭ Cl, Br, I
The order of bond strengths for the hydrogen halides is
(strongest bonds) HF ϾϾ HCl Ͼ HBr Ͼ HI (weakest bonds)
To understand why HF is so much weaker an acid than the other hydrogen halides, let
us consider the following factors.
1. In HF the electronegativity difference is 1.9, compared with 0.9 in HCl, 0.7 in HBr,
and 0.4 in HI (Section 7-8). We might expect the very polar HXF bond in HF to
ionize easily. The fact that HF is the weakest of these acids suggests that this effect
must be of minor importance.
10-7
Generally, elements of intermediate
electronegativity form amphoteric
hydroxides. Those of high and low
electronegativity form acidic and basic
“hydroxides,” respectively.
380 CHAPTER 10: Reactions in Aqueous Solutions I: Acids, Bases, and Salts
TABLE 10-1 Amphoteric Hydroxides
Insoluble Complex Ion Formed
Metal or Amphoteric in an Excess of a
Metalloid ions Hydroxide Strong Base
Be2ϩ Be(OH)2 [Be(OH)4]2Ϫ
Al3ϩ Al(OH)3 [Al(OH)4]Ϫ
Cr3ϩ Cr(OH)3 [Cr(OH)4]Ϫ
Zn2ϩ Zn(OH)2 [Zn(OH)4]2Ϫ
Sn2ϩ Sn(OH)2 [Sn(OH)3]Ϫ
Sn4ϩ Sn(OH)4 [Sn(OH)6]2Ϫ
Pb2ϩ Pb(OH)2 [Pb(OH)4]2Ϫ
As3ϩ As(OH)3 [As(OH)4]Ϫ
Sb3ϩ Sb(OH)3 [Sb(OH)4]Ϫ
Si4ϩ Si(OH)4 SiO4
4Ϫ and SiO3
2Ϫ
Co2ϩ Co(OH)2 [Co(OH)4]2Ϫ
Cu2ϩ Cu(OH)2 [Cu(OH)4]2Ϫ
A weak acid may be very reactive. For
example, HF dissolves sand and glass.
The equation for its reaction with
sand is
SiO2(s) ϩ 4HF(g) 88n
SiF4(g) ϩ 2H2O(ᐉ)
The reaction with glass and other
silicates is similar. These reactions are
not related to acid strength; none of
the three strong hydrohalic acids—
HCl, HBr, or HI—undergoes such a
reaction.
2. The bond strength is considerably greater in HF than in the other three molecules.
This tells us that the HXF bond is harder to break than the HXCl, HXBr, and
HXI bonds.
3. The small, highly charged FϪ ion, formed when HF ionizes, causes increased
ordering of the water molecules. This increase is unfavorable to the process of ioniza-
tion.
The net result of all factors is that HF is a much weaker acid than the other hydrohalic
acids: HCl, HBr, and HI.
In dilute aqueous solutions, hydrochloric, hydrobromic, and hydroiodic acids are
completely ionized, and all show the same apparent acid strength. Water is sufficiently
basic that it does not distinguish among the acid strengths of HCl, HBr, and HI, and
therefore it is referred to as a leveling solvent for these acids. It is not possible to determine
the order of the strengths of these three acids in water because they are so nearly
completely ionized.
When these compounds dissolve in anhydrous acetic acid or other solvents less basic
than water, however, they exhibit significant differences in their acid strengths. The
observed order of acid strengths is:
HCl Ͻ HBr Ͻ HI
We observe that
the hydronium ion is the strongest acid that can exist in aqueous solution. All acids
stronger than H3Oϩ(aq) react completely with water to produce H3Oϩ(aq) and their
conjugate bases.
This is called the leveling effect of water. For example HClO4 (Table 10-2) reacts
completely with H2O to form Hϩ(aq) and ClO4
Ϫ(aq).
HClO4(aq) ϩ H2O(ᐉ) 88n H3Oϩ(aq) ϩ ClO4
Ϫ(aq)
Similar observations have been made for aqueous solutions of strong bases such as
NaOH and KOH. Both are completely dissociated in dilute aqueous solutions.
H2O
NaOH(s) 8888n Naϩ(aq) ϩ OHϪ(aq)
The hydroxide ion is the strongest base that can exist in aqueous solution. Bases
stronger than OHϪ react completely with H2O to produce OHϪ and their conjugate
acids.
When metal amides such as sodium amide, NaNH2, are placed in H2O, the amide ion,
NH2
Ϫ, reacts with H2O completely.
NH2
Ϫ(aq) ϩ H2O(ᐉ) 88n NH3(aq) ϩ OHϪ(aq)
Thus, we see that H2O is a leveling solvent for all bases stronger than OHϪ.
Acid strengths for other vertical series of binary acids vary in the same way as those of
the VIIA elements. The order of bond strengths for the VIA hydrides is
(strongest bonds) H2O ϾϾ H2S Ͼ H2Se Ͼ H2Te (weakest bonds)
Bond strength is shown by the bond
energies introduced in Chapter 7 and
tabulated in Section 15-9. The
strength of the HXF bond is due
largely to the very small size of the F
atom.
10-7 Strengths of Acids 381
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 17.10, Acid–Base Properties of
Salts.
The amide ion, NH2
Ϫ, is a stronger
base than OHϪ.
HXO bonds are much stronger than the bonds in the other Group VI hydrides. As we
might expect, the order of acid strengths for these hydrides is just the reverse of the order
of bond strengths.
(weakest acid) H2O ϽϽ H2S Ͻ H2Se Ͻ H2Te (strongest acid)
Table 10-2 displays relative acid and base strengths of a number of conjugate acid–base
pairs.
Ternary Acids
Most ternary acids are hydroxyl compounds of nonmetals (oxoacids) that ionize to produce
Hϩ(aq). The formula for nitric acid is commonly written HNO3 to emphasize the presence
of an acidic hydrogen atom, but it could also be written as HONO2, as its structure
shows (see margin).
In most ternary acids the hydroxyl oxygen is bonded to a fairly electronegative
nonmetal. In nitric acid the nitrogen draws the electrons of the NXO (hydroxyl) bond
closer to itself than would a less electronegative element such as sodium. The oxygen pulls
the electrons of the OXH bond close enough so that the hydrogen atom ionizes as Hϩ,
leaving NO3
Ϫ.
HNO3(aq) 88n Hϩ(aq) ϩ NO3
Ϫ(aq)
Let us consider the hydroxyl compounds of metals. We call these compounds “hydroxides”
because they can produce hydroxide ions in water to give basic solutions. Oxygen is
much more electronegative than most metals, such as sodium. It draws the electrons of
the sodium–oxygen bond in NaOH (a strong base) so close to itself that the bonding is
The trends in binary acid strengths
across a period (e.g., CH4 Ͻ NH3 Ͻ
H2O Ͻ HF) are not those predicted
from trends in bond energies and
electronegativity differences. The
correlations used for vertical trends
cannot be used for horizontal trends.
This is because a “horizontal”
series of compounds has different
stoichiometries and different numbers
of unshared pairs of electrons on the
central atoms.
O
N
Bond that breaks
to form H
ϩ
and NO3
Ϫ
O H
O
Hydroxyl group
382 CHAPTER 10: Reactions in Aqueous Solutions I: Acids, Bases, and Salts
TABLE 10-2 Relative Strengths of Conjugate Acid–Base Pairs
Acid Base
HClO4 ClO4
Ϫ
HI IϪ
HBr BrϪ
HCl ClϪ
HNO3 NO3
Ϫ
H3Oϩ H2O
HF FϪ
CH3COOH CH3COOϪ
HCN CNϪ
NH4
ϩ NH3
H2O OHϪ
NH3 NH2
Ϫ
100% ionized in
dilute aq. soln.
No molecules of
nonionized acid.
Negligible base
strength in water.
Equilibrium mixture
of nonionized
molecules of acid,
conjugate base, and
Hϩ(aq).
Reacts completely with H2O;
cannot exist in aqueous solution.
acid loses Hϩ
u:::::::::v
base gains Hϩ




























 


Acidstrengthincreases
Basestrengthincreases
ionic. NaOH therefore exists as Naϩ and OHϪ ions, even in the solid state, and dissociates
into Naϩ and OHϪ ions when it dissolves in H2O.
H2O
NaOH(s) 8888n Naϩ(aq) ϩ OHϪ(aq)
Returning to our consideration of ternary acids, we usually write the formula for sulfuric
acid as H2SO4 to emphasize that it is a polyprotic acid. The formula can also be written
as (HO)2SO2, however, because the structure of sulfuric acid (see margin) shows clearly
that H2SO4 contains two OXH groups bound to a sulfur atom. Because the OXH bonds
are easier to break than the SXO bonds, sulfuric acid ionizes as an acid.
Step 1: H2SO4(aq) 88n Hϩ(aq) ϩ HSO4
Ϫ(aq)
Step 2: HSO4
Ϫ(aq) 34 Hϩ(aq) ϩ SO4
2Ϫ(aq)
The first step in the ionization of H2SO4 is complete in dilute aqueous solution. The
second step is nearly complete in very dilute aqueous solutions. The first step in the ionization
of a polyprotic acid always occurs to a greater extent than the second step, because
it is easier to remove a proton from a neutral acid molecule than from a negatively charged
anion.
Sulfurous acid, H2SO3, is a polyprotic acid that contains the same elements as H2SO4.
H2SO3 is a weak acid, however, which tells us that the HXO bonds in H2SO3 are stronger
than those in H2SO4.
Comparison of the acid strengths of nitric acid, HNO3, and nitrous acid, HNO2, shows
that HNO3 is a much stronger acid than HNO2.
Acid strengths of most ternary acids containing the same central element increase
with increasing oxidation state of the central element and with increasing numbers
of oxygen atoms.
The following orders of increasing acid strength are typical.
H2SO3 Ͻ H2SO4
(strongest acids
HNO2 Ͻ HNO3 are on the right side)
HClO Ͻ HClO2 Ͻ HClO3 Ͻ HClO4
Sulfuric acid is called a polyprotic acid
because it has more than one ionizable
hydrogen atom per molecule. It is the
only common polyprotic acid that is
also a strong acid.
S
H
H
O
O
O
O
10-7 Strengths of Acids 383
For most ternary acids containing different elements in the same oxidation state
from the same group in the periodic table, acid strengths increase with increasing
electronegativity of the central element.
HOCl (or HClO) HClO2 HClO3 HClO4
Nitrous acid, HNO2.
H2SeO4 Ͻ H2SO4 H2SeO3 Ͻ H2SO3
H3PO4 Ͻ HNO3
HBrO4 Ͻ HClO4 HBrO3 Ͻ HClO3
Contrary to what we might expect, H3PO3 is a stronger acid than HNO2. Care must be
exercised to compare acids that have similar structures. For example, H3PO2, which has
two H atoms bonded to the P atom, is a stronger acid than H3PO3, which has one H
atom bonded to the P atom. H3PO3 is a stronger acid than H3PO4, which has no H atoms
bonded to the P atom.
ACID–BASE REACTIONS IN AQUEOUS SOLUTIONS
In Section 4-9 we introduced classical acid–base reactions. We defined neutralization as
the reaction of an acid with a base to form a salt and (in most cases) water. Most salts are
ionic compounds that contain a cation other than Hϩ and an anion other than OHϪ or
O2Ϫ. The common strong acids and common strong bases are listed in the margin. All other
common acids may be assumed to be weak. The other common metal hydroxides (bases)
are insoluble in water.
Arrhenius and Brønsted–Lowry acid–base neutralization reactions all have one thing
in common. They involve the reaction of an acid with a base to form a salt that contains
the cation characteristic of the base and the anion characteristic of the acid. Water is also
usually formed. This is indicated in the formula unit equation. The general form of the
net ionic equation, however, is different for different acid–base reactions. The net ionic
equations depend on the solubility and extent of ionization or dissociation of each reactant
and product.
In writing ionic equations, we always write the formulas of the predominant forms of
the compounds in, or in contact with, aqueous solution. Writing ionic equations from
formula unit equations requires a knowledge of the lists of strong acids and strong bases,
as well as of the generalizations on solubilities of inorganic compounds. Please review
carefully all of Sections 4-2 and 4-3. Study Tables 4-8 and 4-9 carefully because they
summarize much information that you are about to use again.
In Section 4-2 we examined some reactions of strong acids with strong bases to form
soluble salts. Let us illustrate one additional example. Perchloric acid, HClO4, reacts with
sodium hydroxide to produce sodium perchlorate, NaClO4, a soluble ionic salt.
HClO4(aq) ϩ NaOH(aq) 88n NaClO4(aq) ϩ H2O(ᐉ)
The total ionic equation for this reaction is
[Hϩ(aq) ϩ ClO4
Ϫ(aq)] ϩ [Naϩ(aq) ϩ OHϪ(aq)] 88n [Naϩ(aq) ϩ ClO4
Ϫ(aq)] ϩ H2O(ᐉ)
Eliminating the spectator ions, Naϩ and ClO4
Ϫ, gives the net ionic equation
Hϩ(aq) ϩ OHϪ(aq) 88n H2O(ᐉ)
This is the net ionic equation for the reaction of all strong acids with strong bases to form
soluble salts and water.
Many weak acids react with strong bases to form soluble salts and water. For example,
acetic acid, CH3COOH, reacts with sodium hydroxide, NaOH, to produce sodium acetate,
NaCH3COO.
10-8
In most ternary inorganic acids, all H
atoms are bonded to O.
384 CHAPTER 10: Reactions in Aqueous Solutions I: Acids, Bases, and Salts
See Saunders Interactive General
Chemistry CD-ROM, Screen 18.2,
Acid–Base Reactions (I).
This is considered to be the same as
H3Oϩ ϩ OHϪ n 2H2O
Common Strong Acids
Binary Ternary
HCl HClO4
HBr HClO3
HI HNO3
H2SO4
Strong Bases
LiOH
NaOH
KOH Ca(OH)2
RbOH Sr(OH)2
CsOH Ba(OH)2
CH3COOH(aq) ϩ NaOH(aq) 88n NaCH3COO(aq) ϩ H2O(ᐉ)
The total ionic equation for this reaction is
CH3COOH(aq) ϩ [Naϩ(aq) ϩ OHϪ(aq)] 88n [Naϩ(aq) ϩ CH3COOϪ(aq)] ϩ H2O(ᐉ)
Elimination of Naϩ from both sides gives the net ionic equation
CH3COOH(aq) ϩ OHϪ(aq) 88n CH3COOϪ(aq) ϩ H2O(ᐉ)
In general terms, the reaction of a weak monoprotic acid with a strong base to form a
soluble salt may be represented as
HA(aq) ϩ OHϪ(aq) 88n AϪ(aq) ϩ H2O(ᐉ) (net ionic equation)
EXAMPLE 10-1 Equations for Acid–Base Reactions
Write (a) formula unit, (b) total ionic, and (c) net ionic equations for the complete neutralization
of phosphoric acid, H3PO4, with potassium hydroxide, KOH.
Plan
(a) The salt produced in the reaction contains the cation of the base, Kϩ, and the PO4
3Ϫ anion
of the acid. The salt is K3PO4.
(b) H3PO4 is a weak acid—it is not written in ionic form. KOH is a strong base, and so it is
written in ionic form. K3PO4 is a soluble salt, and so it is written in ionic form.
(c) The spectator ions are canceled to give the net ionic equation.
Solution
(a) H3PO4(aq) ϩ 3KOH(aq) 88n K3PO4(aq) ϩ 3H2O(ᐉ)
(b) H3PO4(aq) ϩ 3[Kϩ(aq) ϩ OHϪ(aq)] 88n [3Kϩ(aq) ϩ PO4
3Ϫ(aq)] ϩ 3H2O(ᐉ)
(c) H3PO4(aq) ϩ 3OHϪ(aq) 88n PO4
3Ϫ(aq) ϩ 3H2O(ᐉ)
You should now work Exercise 58.
EXAMPLE 10-2 Equations for Acid–Base Reactions
Write (a) formula unit, (b) total ionic, and (c) net ionic equations for the neutralization of
aqueous ammonia with nitric acid.
Plan
(a) The salt produced in the reaction contains the cation of the base, NH4
ϩ, and the anion of
the acid, NO3
Ϫ. The salt is NH4NO3.
(b) HNO3 is a strong acid—we write it in ionic form. Ammonia is a weak base. NH4NO3 is
a soluble salt that is completely dissociated—we write it in ionic form.
(c) We cancel the spectator ions, NO3
Ϫ, and obtain the net ionic equation.
Solution
(a) HNO3(aq) ϩ NH3(aq) 88n NH4NO3(aq)
(b) [Hϩ(aq) ϩ NO3
Ϫ(aq)] ϩ NH3(aq) 88n [NH4
ϩ(aq) ϩ NO3
Ϫ(aq)]
(c) Hϩ(aq) ϩ NH3(aq) 88n NH4
ϩ(aq)
You should now work Exercise 60.
Monoprotic acids contain one, diprotic
acids contain two, and triprotic acids
contain three acidic (ionizable)
hydrogen atoms per formula unit.
Polyprotic acids (those that contain
more than one ionizable hydrogen
atom) are discussed in detail in
Chapter 18.
When all the Hϩ from the acid and all
the OHϪ of the base have reacted, we
say that there has been complete
neutralization.
10-8 Acid–Base Reactions in Aqueous Solutions 385
Neutralization reactions of ammonia,
NH3, form salts but do not form
water.
EXAMPLE 10-3 Preparation of Salts
Write the formula unit equation for the reaction of an acid and a base that will produce each
of the following salts: (a) Na3PO4, (b) Ca(ClO3)2, (c) MgSO4.
Plan
(a) The salt contains the ions, Naϩ and PO4
3Ϫ. Naϩ is the cation in the strong base, NaOH.
PO4
3Ϫ is the anion in the weak acid, H3PO4. The reaction of NaOH with H3PO4 should
therefore produce the desired salt plus water.
(b) The cation, Ca2ϩ, is from the strong base, Ca(OH)2. The anion, ClO3
Ϫ, is from the strong
acid, HClO3. The reaction of Ca(OH)2 with HClO3 will produce water and the desired salt.
(c) MgSO4 is the salt produced in the reaction of Mg(OH)2(s) and H2SO4(aq).
Solution
(a) H3PO4(aq) ϩ 3NaOH(aq) 88n Na3PO4(aq) ϩ 3H2O(ᐉ)
(b) 2HClO3(aq) ϩ Ca(OH)2(aq) 88n Ca(ClO3)2(aq) ϩ 2H2O(ᐉ)
(c) H2SO4(aq) ϩ Mg(OH)2(s) 88n MgSO4(aq) ϩ 2H2O(ᐉ)
You should now work Exercise 62.
ACIDIC SALTS AND BASIC SALTS
To this point we have examined acid–base reactions in which stoichiometric amounts of
Arrhenius acids and bases were mixed. Those reactions form normal salts. As the name
implies, normal salts contain no ionizable H atoms or OH groups. The complete neutralization
of phosphoric acid, H3PO4, with sodium hydroxide, NaOH, produces the normal
salt, Na3PO4. The equation for this complete neutralization is
H3PO4(aq) ϩ 3NaOH(aq) 88n Na3PO4(aq) ϩ 3H2O(ᐉ)
1 mole 3 moles sodium phosphate,
a normal salt
If less than stoichiometric amounts of bases react with polyprotic acids, the resulting
salts are known as acidic salts because they are still capable of neutralizing bases.
H3PO4(aq) ϩ NaOH(aq) 88n NaH2PO4(aq) ϩ H2O(ᐉ)
1 mole 1 mole sodium dihydrogen phosphate,
an acidic salt
H3PO4(aq) ϩ 2NaOH(aq) 88n Na2HPO4(aq) ϩ 2H2O(ᐉ)
1 mole 2 moles sodium hydrogen phosphate,
an acidic salt
The reaction of phosphoric acid, H3PO4, a weak acid, with strong bases can produce
the three salts shown in the three preceding equations, depending on the relative amounts
of acid and base used. The acidic salts, NaH2PO4 and Na2HPO4, can react further with
bases such as NaOH.
NaH2PO4(aq) ϩ 2NaOH(aq) 88n Na3PO4(aq) ϩ 2H2O(ᐉ)
Na2HPO4(aq) ϩ NaOH(aq) 88n Na3PO4(aq) ϩ H2O(ᐉ)
10-9
386 CHAPTER 10: Reactions in Aqueous Solutions I: Acids, Bases, and Salts
There are many additional examples of acidic salts. Sodium hydrogen carbonate,
NaHCO3, commonly called sodium bicarbonate, is classified as an acidic salt. It is,
however, the acidic salt of an extremely weak acid—carbonic acid, H2CO3—and solutions
of sodium bicarbonate are slightly basic, as are solutions of salts of other extremely
weak acids.
Polyhydroxy bases (bases that contain more than one OH per formula unit) react with
stoichiometric amounts of acids to form normal salts.
Al(OH)3(s) ϩ 3HCl(aq) 88n AlCl3(aq) ϩ 3H2O(ᐉ)
1 mole 3 moles aluminum chloride,
a normal salt
The reaction of polyhydroxy bases with less than stoichiometric amounts of acids forms
basic salts, that is, salts that contain unreacted OH groups. For example, the reaction of
aluminum hydroxide with hydrochloric acid can produce two different basic salts:
Al(OH)3(s) ϩ HCl(aq) 88n Al(OH)2Cl(s) ϩ H2O(ᐉ)
1 mole 1 mole aluminum dihydroxide chloride,
a basic salt
Al(OH)3(s) ϩ 2HCl(aq) 88n Al(OH)Cl2(s) ϩ 2H2O(ᐉ)
1 mole 2 moles aluminum hydroxide dichloride,
a basic salt
Aqueous solutions of basic salts are not necessarily basic, but they can neutralize acids,
such as,
Al(OH)2Cl ϩ 2HCl 88n AlCl3 ϩ 2H2O
Most basic salts are rather insoluble in water.
THE LEWIS THEORY
In 1923, Professor G. N. Lewis (1875–1946) presented the most comprehensive of the
classic acid–base theories. The Lewis definitions follow.
An acid is any species that can accept a share in an electron pair. A base is any
species that can make available, or “donate,” a share in an electron pair.
10-10
These basic aluminum salts are called
“aluminum chlorohydrate.” They are
components of some deodorants.
10-10 The Lewis Theory 387
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 17.11, Lewis Acids and Bases,
and Screen 17.12, Cationic Lewis Acids.
This is the same Lewis who made
many contributions to our
understanding of chemical bonding.
Sodium hydrogen carbonate, baking soda, is the
most familiar example of an acidic salt. It can
neutralize strong bases, but its aqueous solutions are
slightly basic, as the blue color of the indicator
bromthymol blue shows.
These definitions do not specify that an electron pair must be transferred from one atom
to another—only that an electron pair, residing originally on one atom, must be shared
between two atoms. Neutralization is defined as coordinate covalent bond formation.
This results in a covalent bond in which both electrons were furnished by one atom or
ion.
The reaction of boron trichloride with ammonia is a typical Lewis acid–base reaction.
388 CHAPTER 10: Reactions in Aqueous Solutions I: Acids, Bases, and Salts
The Lewis theory is sufficiently general that it covers all acid–base reactions that the other
theories include, plus many additional reactions such as complex formation (Chapter 25).
The autoionization of water (Section 10-5) was described in terms of Brønsted–Lowry
theory. In Lewis theory terminology, this is also an acid–base reaction. The acceptance
of a proton, Hϩ, by a base involves the formation of a coordinate covalent bond.
B
Cl Cl
Cl
+ N
H
H
H
Coordinate
covalent bond
Cl
B N
H
H
Cl
Cl
H
Bond formed
BCl3(g)
acid
+ NH3(g)
base
Cl3B : NH3
product
B
:
:
:Cl
::Cl
:
:
:Cl
:
N
H
H
H
Bond formed
+ B
:
:
:Cl
::Cl
:
:
:Cl
:
N
H
H
H:
+
Theoretically, any species that contains an unshared electron pair could act as a base.
In fact, most ions and molecules that contain unshared electron pairs undergo some reactions
by sharing their electron pairs. Conversely, many Lewis acids contain only six
electrons in the highest occupied energy level of the central element. They react by
accepting a share in an additional pair of electrons. These species are said to have an open
sextet. Many compounds of the Group IIIA elements are Lewis acids, as illustrated by
the reaction of boron trichloride with ammonia, presented earlier.
Anhydrous aluminum chloride, AlCl3, is a common Lewis acid that is used to catalyze
many organic reactions. AlCl3 acts as a Lewis acid when it dissolves in hydrochloric acid
to give a solution that contains AlCl4
Ϫ ions.
Cl
Cl
Cl ClCl Al
Cl
Ϫ
(aq)AlCl3(s) AlCl4
Ϫ
(aq)ϩ
ϩ
Ϫ
Ϫ
Cl
Cl
Cl Al
productbaseacid
Al is sp
3
hybridized
10-10 The Lewis Theory 389
Other ions and molecules behave as Lewis acids by expansion of the valence shell of the
central element. Anhydrous tin(IV) chloride is a colorless liquid that also is frequently
used as a Lewis acid catalyst. The tin atom (Group IVA) can expand its valence shell by
utilizing vacant d orbitals. It can accept shares in two additional electron pairs, as its reaction
with hydrochloric acid illustrates.
O
H
H :
–
base acid
+
+ O
H
H
H + O
H
:
:
:O
H
:
:
H
Bond formed
:
:
+ Cl–
–
Cl
Cl
Cl
Cl
Al
Ϫ
Cl
Cl
Cl
Cl
Sn
2Cl
Ϫ
(aq)SnCl4(ᐉ) SnCl6
2Ϫ
(aq)
2Ϫ
ϩ
2Cl
Ϫ
ϩ
baseacid
Cl
Cl
Cl Cl
Cl Cl
Sn
Sn is sp
3
d
2
hybridized
(octahedral)
Sn is sp
3
hybridized
(tetrahedral)
390 CHAPTER 10: Reactions in Aqueous Solutions I: Acids, Bases, and Salts
Problem-Solving Tip: Which Acid–Base Theory Should You Use?
Remember the following:
1. Arrhenius acids and bases are also Brønsted–Lowry acids and bases; the reverse is not
true.
2. Brønsted–Lowry acids and bases are also Lewis acids and bases; the reverse is not
true.
3. We usually prefer the Arrhenius or the Brønsted–Lowry theory when water or another
protonic solvent is present.
4. Although the Lewis theory can be used to explain the acidic or basic property of some
species in protonic solvents, the most important use of the Lewis theory is for acidbase
reactions in many nonaqueous solvents.
Many organic and biological reactions are acid–base reactions that do not fit within
the Arrhenius or Brønsted–Lowry theories. Experienced chemists find the Lewis theory
to be very useful because so many other chemical reactions are covered by it. The less
experienced sometimes find the theory less useful, but as their knowledge expands so does
its utility.
2–
+ 2Cl–
THE PREPARATION OF ACIDS
Binary acids may be prepared by combination of appropriate elements with hydrogen
(Section 6-7, part 2).
Small quantities of the hydrogen halides (their solutions are called hydrohalic acids)
and other volatile acids are usually prepared by adding concentrated nonvolatile acids to
the appropriate salts. (Sulfuric and phosphoric acids are classified as nonvolatile acids because
they have much higher boiling points than other common acids.) The reactions of concentrated
sulfuric acid with solid sodium fluoride and sodium chloride produce gaseous
hydrogen fluoride and hydrogen chloride, respectively.
H2SO4(ᐉ) ϩ NaF(s) 88n NaHSO4(s) ϩ HF(g)
sulfuric acid sodium fluoride sodium hydrogen hydrogen fluoride
bp ϭ 336°C sulfate bp ϭ 19.6°C
H2SO4(ᐉ) ϩ NaCl(s) 88n NaHSO4(s) ϩ HCl(g)
sodium chloride hydrogen chloride
bp ϭ Ϫ84.9°C
Because concentrated sulfuric acid is a fairly strong oxidizing agent, it cannot be used to
prepare hydrogen bromide or hydrogen iodide; instead, the free halogens are produced.
Phosphoric acid, a nonoxidizing acid, is dropped onto solid sodium bromide or sodium
iodide to produce hydrogen bromide or hydrogen iodide, as the following equations show:
heat
H3PO4(ᐉ) ϩ NaBr(s) 88n NaH2PO4(s) ϩ HBr(g)
phosphoric acid sodium bromide sodium dihydrogen hydrogen bromide
bp ϭ 213°C phosphate bp ϭ Ϫ67.0°C
heat
H3PO4(ᐉ) ϩ NaI(s) 88n NaH2PO4(s) ϩ HI(g)
sodium iodide hydrogen iodide
bp ϭ Ϫ35°C
This kind of reaction may be generalized as
ᎏ
non
a
v
c
o
i
l
d
atile
ᎏ ϩ ᎏ
vol
s
a
a
ti
l
l
t
e
o
a
f
cid
ᎏ 88n ᎏ
nonv
s
o
a
l
l
a
t
ti
o
le
f
acid
ᎏ ϩ volatile acid
Dissolving each of the gaseous hydrogen halides in water gives the corresponding hydrohalic
acid.
In Section 6-8, part 2 we saw that many nonmetal oxides, called acid anhydrides, react
with water to form ternary acids with no changes in oxidation numbers. For example,
dichlorine heptoxide, Cl2O7, forms perchloric acid when it dissolves in water.
Cl2O7(ᐉ) ϩ H2O(ᐉ) 88n 2[Hϩ(aq) ϩ ClO4
Ϫ(aq)]
Some high oxidation state transition metal oxides are acidic oxides; that is, they dissolve
in water to give solutions of ternary acids. Manganese(VII) oxide, Mn2O7, and
chromium(VI) oxide, CrO3, are the most common examples.
10-11
The volatile acid HCl can be made
by dropping concentrated H2SO4
onto solid NaCl. Gaseous HCl is
liberated. HCl(g) dissolves in the
water on a piece of filter paper. The
indicator methyl red on the paper
turns red, its color in acidic solution.
10-11 The Preparation of Acids 391
ϩ7 ϩ7
A solution of dichromic acid, H2Cr2O7, is deep red.
Mn2O7(ᐉ) ϩ H2O(ᐉ) 88n 2[Hϩ(aq) ϩ MnO4
Ϫ(aq)]
manganese(VII) oxide permanganic acid
2CrO3(s) ϩ H2O(ᐉ) 88n [2Hϩ(aq) ϩ Cr2O7
2Ϫ(aq)]
chromium(VI) oxide dichromic acid
Neither permanganic acid nor dichromic acid has been isolated in pure form. Many stable
salts of both are well known.
The halides and oxyhalides of some nonmetals hydrolyze (react with water) to produce
two acids: a (binary) hydrohalic acid and a (ternary) oxyacid of the nonmetal. Phosphorus
trihalides react with water to produce the corresponding hydrohalic acids and phosphorous
acid, a weak diprotic acid, whereas phosphorus pentahalides give phosphoric acid and
the corresponding hydrohalic acid.
PX3 ϩ 3H2O(ᐉ) 88n H3PO3(aq) ϩ 3HX(aq)
PX5 ϩ 4H2O(ᐉ) 88n H3PO4(aq) ϩ 5HX(aq)
There are no changes in oxidation numbers in these reactions. Examples include the reactions
of PCl3 and PCl5 with H2O.
PCl3(ᐉ) ϩ 3H2O(ᐉ) 88n H3PO3(aq) ϩ 3[Hϩ(aq) ϩ ClϪ(aq)]
phosphorus phosphorous acid
trichloride
PCl5(s) ϩ 4H2O(ᐉ) 88n H3PO4(aq) ϩ 5[Hϩ(aq) ϩ ClϪ(aq)]
phosphorus phosphoric acid
pentachloride
A drop of PCl3 is added to water
that contains the indicator methyl
orange. As PCl3 reacts with water to
form HCl and H3PO3, the indicator
turns red, its color in acidic solution.
392 CHAPTER 10: Reactions in Aqueous Solutions I: Acids, Bases, and Salts
ϩ7 ϩ7
ϩ6 ϩ6
ϩ3
ϩ5
ϩ3
ϩ5
ϩ3 ϩ3
ϩ5 ϩ5
Key Terms
Acid (Arrhenius or Brønsted–Lowry) A substance that produces
Hϩ(aq) ions in aqueous solution. Strong acids ionize
completely or almost completely in dilute aqueous solution;
weak acids ionize only slightly.
Acid anhydride The oxide of a nonmetal that reacts with water
to form an acid.
Acidic salt A salt that contains an ionizable hydrogen atom; does
not necessarily produce acidic solutions.
Amphiprotism The ability of a substance to exhibit amphoterism
by accepting or donating protons.
Amphoterism Ability of a substance to act as either an acid or
a base.
Anhydrous Without water.
Autoionization An ionization reaction between identical mole-
cules.
Base (Arrhenius) A substance that produces OHϪ(aq) ions in
aqueous solution. Strong bases are soluble in water and are completely
dissociated. Weak bases ionize only slightly.
Basic anhydride The oxide of a metal that reacts with water to
form a base.
Basic salt A salt containing a basic OH group.
Brønsted–Lowry acid A proton donor.
Brønsted–Lowry base A proton acceptor.
Conjugate acid–base pair In Brønsted–Lowry terminology, a
reactant and product that differ by a proton, Hϩ.
Coordinate covalent bond A covalent bond in which both
shared electrons are furnished by the same species; the bond
between a Lewis acid and a Lewis base.
Dissociation In aqueous solution, the process in which a solid
ionic compound separates into its ions.
Exercises 393
Electrolyte A substance whose aqueous solutions conduct elec-
tricity.
Formula unit equation A chemical equation in which all compounds
are represented by complete formulas.
Hydration The process by which water molecules bind to ions
or molecules in the solid state or in solution.
Hydride A binary compound of hydrogen.
Hydrolysis Reaction of a substance with water.
Hydronium ion H3Oϩ, the usual representation of the hydrated
hydrogen ion.
Ionization In aqueous solution, the process in which a molecular
compound reacts with water to form ions.
Leveling effect The effect by which all acids stronger than the
acid that is characteristic of the solvent react with the solvent
to produce that acid; a similar statement applies to bases. The
strongest acid (base) that can exist in a given solvent is the acid
(base) characteristic of that solvent.
Lewis acid Any species that can accept a share in an electron
pair to form a coordinate covalent bond.
Lewis base Any species that can make available a share in an
electron pair to form a coordinate covalent bond.
Net ionic equation The equation that results from canceling
spectator ions and eliminating brackets from a total ionic equa-
tion.
Neutralization The reaction of an acid with a base to form a
salt and (usually) water; usually, the reaction of hydrogen ions
with hydroxide ions to form water molecules.
Nonelectrolyte A substance whose aqueous solutions do not
conduct electricity.
Normal oxide A metal oxide containing the oxide ion, O2Ϫ (oxygen
in the Ϫ2 oxidation state).
Normal salt A salt containing no ionizable H atoms or OH
groups.
Open sextet Refers to species that have only six electrons in the
highest energy level of the central element (many Lewis acids).
Polyprotic acid An acid that contains more than one ionizable
hydrogen atom per formula unit.
Protonic acid An Arrhenius acid, or a Brønsted–Lowry acid.
Salt A compound that contains a cation other than Hϩ and an
anion other than OHϪ or O2Ϫ.
Spectator ions Ions in solution that do not participate in a chemical
reaction.
Strong electrolyte A substance that conducts electricity well in
dilute aqueous solution.
Ternary acid An acid that contains three elements—usually H,
O, and another nonmetal.
Ternary compound A compound that contains three different
elements.
Total ionic equation The equation for a chemical reaction written
to show the predominant form of all species in aqueous
solution or in contact with water.
Weak electrolyte A substance that conducts electricity poorly
in dilute aqueous solution.
Exercises
Basic Ideas
001. Which properties of acids did Robert Boyle observe?
002. Gay-Lussac reached an important conclusion about acids
and bases. What was it?
003. Define the following terms. You may wish to refer to
Chapter 4 to check the definitions. (a) acid; (b) neutralization;
(c) ionization; (d) dissociation; (e) salt.
The Arrhenius Theory
004. Outline Arrhenius’ ideas about acids and bases. (a) How
did he define the following terms: acid, base, neutralization?
(b) Give an example that illustrates each
term.
005. Define and illustrate the following terms clearly and concisely.
Give an example of each. (a) strong electrolyte;
(b) weak electrolyte; (c) nonelectrolyte; (d) strong acid;
(e) strong base; (f) weak acid; (g) weak base; (h) insoluble
base.
006. Distinguish between the following pairs of terms, and provide
a specific example of each. (a) strong acid and weak
acid; (b) strong base and weak base; (c) strong base and
insoluble base.
007. Write formulas and names for (a) the common strong
acids; (b) three weak acids; (c) the common strong bases;
(d) the most common weak base; (e) four soluble ionic salts;
(f ) four insoluble salts.
008. Describe an experiment for classifying compounds as
strong electrolytes, weak electrolytes, or nonelectrolytes.
Tell what would be observed for each of the following compounds
and classify each. Na2SO4; HCN; C2H5COOH;
CH3CH2OH; HF; HClO4; HCOOH; NH3.
009. Summarize the electrical properties of strong electrolytes,
weak electrolytes, and nonelectrolytes.
The Hydrated Hydrogen Ion
010. Write the formula of a hydrated hydrogen ion that contains
only one water of hydration. Give another name for
the hydrated hydrogen ion.
011. Why is the hydrated hydrogen ion important?
012. Criticize the following statement: “The hydrated hydrogen
ion should always be represented as H3Oϩ.”
394 CHAPTER 10: Reactions in Aqueous Solutions I: Acids, Bases, and Salts
Brønsted–Lowry Theory
013. State the basic ideas of the Brønsted–Lowry theory.
014. Use Brønsted–Lowry terminology to define the following
terms. Illustrate each with a specific example. (a) acid;
(b) conjugate base; (c) base; (d) conjugate acid; (e) conjugate
acid–base pair.
015. Write balanced equations that describe the ionization of
the following acids in dilute aqueous solution. Use a single
arrow (n) to represent complete, or nearly complete,
ionization and a double arrow (34) to represent a small
extent of ionization. (a) HCl; (b) CH3COOH; (c) H2S;
(d) HCN; (e) HF; (f) HClO4.
016. Use words and equations to describe how ammonia
can act as a base in (a) aqueous solution and (b) the pure
state, that is, as gaseous ammonia molecules when it reacts
with gaseous hydrogen chloride or a similar anhydrous
acid.
017. What does autoionization mean? How can the autoionization
of water be described as an acid–base reaction?
What structural features must a compound have to be able
to undergo autoionization?
018. Illustrate, with appropriate equations, the fact that these
species are bases in water: NH3; HSϪ; CH3COOϪ; O2Ϫ.
019. In terms of Brønsted–Lowry theory, state the differences
between (a) a strong and a weak base and (b) a strong and
a weak acid.
020. Give the products in the following acid–base reactions.
Identify the conjugate acid–base pairs.
(a) NH4
ϩ ϩ CNϪ
(b) HSϪ ϩ H2SO4
(c) HClO4 ϩ [H2NNH3]ϩ
(d) NH2
Ϫ ϩ H2O
021. Give the conjugate acids of H2O, OHϪ, IϪ, AsO4
3Ϫ,
NH2
Ϫ, HPO4
2Ϫ, and NO2
Ϫ.
022. Give the conjugate bases of H2O, HSϪ, HCl, PH4
ϩ, and
HOCH3.
023. Identify the Brønsted–Lowry acids and bases in these reactions
and group them into conjugate acid–base pairs.
(a) NH3 ϩ HBr 34 NH4
ϩ ϩ BrϪ
(b) NH4
ϩ ϩ HSϪ 34 NH3 ϩ H2S
(c) H3Oϩ ϩ PO4
3Ϫ 34 HPO4
2Ϫ ϩ H2O
(d) HSO3
Ϫ ϩ CNϪ 34 HCN ϩ SO3
2Ϫ
024. Identify each species in the following reactions as either
an acid or a base, in the Brønsted–Lowry sense.
(a) CNϪ ϩ H2O 34 HCN ϩ OHϪ
(b) HCO3
Ϫ ϩ H2SO4 34 HSO4
Ϫ ϩ H2CO3
(c) CH3COOH ϩ NO2
Ϫ 34 HNO2 ϩ CH3COOϪ
025. Identify each species in the following reactions as either
an acid or a base, in the Brønsted–Lowry sense.
(a) NH4
ϩ ϩ HSO3
Ϫ 34 NH3 ϩ H2SO3
(b) NH2
Ϫ ϩ H2O 34 NH3 ϩ OHϪ
(c) O2Ϫ ϩ H2O 34 OHϪ ϩ OHϪ
026. Arrange the species in the reactions of Exercise 24 as Brønsted–Lowry
conjugate pairs.
027. Arrange the species in the reactions of Exercise 25 as Brønsted–Lowry
conjugate pairs.
028. Identify each reactant and product in the following chemical
reactions as a Brønsted–Lowry acid, a Brønsted–
Lowry base, or neither. Arrange the species in each reaction
as conjugate acid–base pairs.
(a) H2CO3 ϩ H2O 34 H3Oϩ ϩ HCO3
Ϫ
(b) HSO4
Ϫ ϩ H2O 34 H3Oϩ ϩ SO4
2Ϫ
(c) H3PO4 ϩ CNϪ 34 HCN ϩ H2PO4
Ϫ
(d) HSϪ ϩ OHϪ 34 H2O ϩ S2Ϫ
029. Identify each reactant and product in the following chemical
reactions as a Brønsted–Lowry acid, a Brønsted–
Lowry base, or neither. Arrange the species in each reaction
as conjugate acid–base pairs.
(a) H2SeO4 ϩ H2O 34 H3Oϩ ϩ HSeO4
Ϫ
(b) HPO4
2Ϫ ϩ H2O 34 H3Oϩ ϩ PO4
3Ϫ
(c) NH3 ϩ HϪ 34 H2 ϩ NH2
Ϫ
(d) HCl ϩ NH3 34 NH4Cl
Properties of Aqueous Solutions of Acids and Bases
030. Write equations and designate conjugate pairs for the
stepwise reactions in water of (a) H2SO4 and (b) H2SO3.
031. List six properties of aqueous solutions of protonic acids.
032. List five properties of bases in aqueous solution. Does
aqueous ammonia exhibit these properties? Why?
033. We say that strong acids, weak acids, and weak bases ionize
in water, but strong bases dissociate in water. What is
the difference between ionization and dissociation?
034. Distinguish between solubility in water and extent of ionization
in water. Provide specific examples that illustrate
the meanings of both terms.
035. Write three general statements that describe the extents
to which acids, bases, and salts are ionized in dilute aqueous
solutions.
Amphoterism
036. Use chemical equations to illustrate the hydroxides of
beryllium, zinc, arsenic, and antimony reacting (a) as acids;
(b) as bases.
037. Draw the Lewis formula of aluminum hydroxide, and
explain the features that enable it to possess amphoteric
properties.
038. What do we mean when we say that water is amphiprotic?
(a) Can we also describe water as amphoteric? Why?
(b) Illustrate the amphiprotic nature of water by writing
two equations for reactions in which water exhibits this
property.
Strengths of Acids
039. What property is characteristic of all strong acids and
strong bases but not weak acids and weak bases?
Exercises 395
040. What does “base strength” mean? What does “acid
strength” mean?
041. Classify each of the following substances as (a) a strong
base, (b) an insoluble base, (c) a strong acid, or (d) a weak
acid: LiOH; HCl; Ba(OH)2; Cu(OH)2; H2S; H2CO3;
H2SO4; Zn(OH)2.
042. (a) What are binary protonic acids? (b) Write names and
formulas for four binary protonic acids.
043. (a) How can the order of increasing acid strength in a series
of similar binary protonic acids be explained? (b) Illustrate
your answer for the series HF, HCl, HBr, and HI. (c) What
is the order of increasing base strength of the conjugate
bases of the acids in (b)? Why? (d) Is your explanation
applicable to the series H2O, H2S, H2Se, and H2Te? Why?
044. Classify each of the hydrides NaH, BeH2, BH3, CH4,
NH3, H2O, and HF as a Brønsted–Lowry base, a Brønsted–Lowry
acid, or neither.
045. (a) Which is the stronger acid of each pair? (1) NH4
ϩ,
NH3; (2) H2O, H3Oϩ; (3) HSϪ, H2S; (4) HSO3
Ϫ, H2SO3.
(b) How are acidity and charge related?
046. Arrange the members of each group in order of decreasing
acidity: (a) H2O, H2Se, H2S; (b) HI, HCl, HF, HBr;
(c) H2S, S2Ϫ, HSϪ.
047. Illustrate the leveling effect of water by writing equations
for the reactions of HCl and HNO3 with water.
Ternary Acids
048. In what sense can we describe nitric and sulfuric acids as
hydroxyl compounds of nonmetals?
049. What are ternary acids? Write names and formulas for four
of them.
050. Write proton-transfer autoionization equations for the
following amphiprotic solvents. (a) NH3, (b) NH2OH,
(c) H2SO4.
051. Explain the order of increasing acid strength for the
following groups of acids and the order of increasing
base strength for their conjugate bases. (a) H2SO3, H2SO4;
(b) HNO2, HNO3; (c) H3PO3, H3PO4; (d) HClO, HClO2,
HClO3, HClO4.
052. (a) Write a generalization that describes the order of acid
strengths for a series of ternary acids that contain different
elements in the same oxidation state from the same
group in the periodic table. (b) Indicate the order of acid
strengths for the following: (1) HNO3, H3PO4; (2) H3PO4,
H3AsO4; (3) HClO3, H2SeO4; (4) HClO3, HBrO3, HIO3.
.*53. List the following acids in order of increasing strength:
(a) sulfuric, phosphoric, and perchloric; (b) HIO3, HIO2,
HIO, and HIO4; (c) selenous, sulfurous, and tellurous
acids; (d) hydrosulfuric, hydroselenic, and hydrotelluric
acids; (e) H2CrO4, H2CrO2, HCrO3, and H3CrO3.
Reactions of Acids and Bases
054. Why are acid–base reactions described as neutralization
reactions?
055. Distinguish among (a) formula unit equations, (b) total
ionic equations, and (c) net ionic equations. What are the
advantages and limitations of each?
056. Classify each substance as either an electrolyte or a nonelectrolyte:
NH4Cl; HI; C6H6; RaF2; Zn(CH3COO)2;
Cu(NO3)2; CH3COOH; C12H22O11 (table sugar); LiOH;
KHCO3; NaClO4; La2(SO4)3; I2.
057. Classify each substance as either a strong or a weak electrolyte,
and then list (a) the strong acids, (b) the strong
bases, (c) the weak acids, and (d) the weak bases. NaCl;
MgSO4; HCl; CH3COOH; Ba(NO3)2; H3PO4; Sr(OH)2;
HNO3; HI; Ba(OH)2; LiOH; C3H5COOH; NH3;
CH3NH2; KOH; HCN; HClO4.
For Exercises 58–60, write balanced (1) formula unit, (2) total
ionic, and (3) net ionic equations for reactions between the acid–
base pairs. Name all compounds except water. Assume complete
neutralization.
058. (a) HNO2 ϩ KOH n
(b) H2SO4 ϩ NaOH n
(c) HCl ϩ Ca(OH)2 n
(d) CH3COOH ϩ KOH n
(e) HI ϩ NaOH n
059. (a) H2CO3 ϩ Sr(OH)2 n
(b) H2SO4 ϩ Ca(OH)2 n
(c) H3PO4 ϩ Ba(OH)2 n
(d) HBr ϩ KOH n
(e) H3AsO4 ϩ KOH n
060. (a) HClO4 ϩ Ba(OH)2 n
(b) HBr ϩ NH3 n
(c) HNO3 ϩ NH3 n
(d) H2SO4 ϩ Fe(OH)3 n
(e) H2SO4 ϩ Ba(OH)2 n
061. Complete these equations by writing the formulas of the
omitted compounds.
(a) Ba(OH)2 ϩ ? n BaSO4(s) ϩ 2H2O
(b) FeO(s) ϩ ? n Fe(NO3)2(aq) ϩ H2O
(c) HCl(aq) ϩ ? n AgCl(s) ϩ ?
(d) Na2O ϩ ? n 2NaOH(aq)
(e) NaOH ϩ ? n Na2HPO4(aq) ϩ ?
(two possible answers)
062. Although many salts may be formed by a variety of
reactions, salts are usually thought of as being derived
from the reaction of an acid with a base. For each of the
salts listed here, choose the acid and base that would react
with each other to form the salt. Write the (i) formula unit,
(ii) total ionic, and (iii) net ionic equations for the formation
of each salt. (a) Pb(NO3)2; (b) AlCl3; (c) (NH4)2CO3;
(d) Ca(ClO4)2; (e) Al2(SO4)3.
063. (a) Which of the following compounds are salts? CaCO3;
Na2O; U(NO3)5; AgNO3; Sr(CH3COO)2.
(b) Write an acid–base equation that accounts for the
formation of those identified as being salts.
064. Repeat Exercise 63 for: KMnO4; NiSO3; P4O10; SnF2;
K3PO4.
396 CHAPTER 10: Reactions in Aqueous Solutions I: Acids, Bases, and Salts
Acidic and Basic Salts
065. What are polyprotic acids? Write names and formulas for
five polyprotic acids.
066. What are acidic salts? Write balanced equations to show
how the following acidic salts can be prepared from the
appropriate acid and base: NaHSO3; KHCO3; NaH2PO4;
Na2HPO4; NaHS.
067. Indicate the mole ratio of acid and base required in each
case in Exercise 66.
068. The following salts are components of fertilizers. They are
made by reacting gaseous NH3 with concentrated solutions
of acids. The heat produced by the reactions
evaporates most of the water. Write balanced formula unit
equations that show the formation of each. (a) NH4NO3;
(b) NH4H2PO4; (c) (NH4)2HPO4; (d) (NH4)3PO4;
(e) (NH4)2SO4.
The Lewis Theory
075. Define and illustrate the following terms clearly and concisely.
Write an equation to illustrate the meaning of each
term. (a) Lewis acid; (b) Lewis base; (c) neutralization
according to Lewis theory.
076. What are the advantages and limitations of the Brønsted–
Lowry theory?
077. Explain the differences between the Brønsted–Lowry and
the Lewis acid–base theories, using the formation of the
ammonium ion from ammonia and water to illustrate your
points.
078. Write a Lewis formula for each species in the following
equations. Label the acids and bases using Lewis theory
terminology.
(a) H2O ϩ H2O 34 H3Oϩ ϩ OHϪ
(b) HCl(g) ϩ H2O n H3Oϩ ϩ ClϪ
(c) NH3(g) ϩ H2O 34 NH4
ϩ ϩ OHϪ
(d) NH3(g) ϩ HCl(g) n NH4Cl(s)
079. What is the term for a single covalent bond in which both
electrons in the shared pair come from the same atom?
Identify the Lewis acid and base and the donor and acceptor
atoms in the following reaction.
080. Identify the Lewis acid and base and the donor and acceptor
atoms in each of the following reactions.
081. Iodine, I2, is much more soluble in a water solution of
potassium iodide, KI, than it is in H2O. The anion found
in the solution is I3
Ϫ. Write an equation for the reaction
that forms I3
Ϫ, indicating the Lewis acid and the Lewis
base.
082. A group of very strong acids are the fluoroacids, HmXFn.
Two such acids are formed by Lewis acid–base reactions.
(a) Identify the Lewis acid and the Lewis base.
HF ϩ SbF5 88n H(SbF6) (called a “super” acid,
hexafluoroantimonic acid)
HF ϩ BF3 88n H(BF4) (tetrafluoroboric acid)
Pt
4ϩ
Cl
Cl Cl
Cl Cl
Cl ClPtϩ
Ϫ
2Ϫ
(b) 6
H
H
ϩ
H O
H
H HOϩ
ϩ
(a)
F
H
H
H BN
F
F
H
H
H N
F
F
FBϩ
069. What are polyhydroxy bases? Write names and formulas
for five polyhydroxy bases.
070. What are basic salts? (a) Write balanced equations to show
how each of the following basic salts can be prepared from
the appropriate acid and base: Ca(OH)Cl; Al(OH)2Cl;
Al(OH)Cl2. (b) Indicate the mole ratio of acid and base
required in each case.
071. What are amphoteric metal hydroxides? (a) Are they bases?
(b) Write the names and formulas for four amphoteric
metal hydroxides.
072. Chromium(III) hydroxide and lead(II) hydroxide are typical
amphoteric hydroxides. (a) Write the formula unit,
total ionic, and net ionic equations for the complete reaction
of each hydroxide with nitric acid. (b) Write the same
kinds of equations for the reaction of each hydroxide with
an excess of potassium hydroxide solution. Reference to
Table 10-1 may be helpful.
073. Write the chemical equations for the stepwise ionization
of oxalic acid, (COOH)2, a diprotic acid.
074. Write the chemical equations for the stepwise ionization
of citric acid, C3H5O(COOH)3, a triprotic acid.
Common lawn fertilizer.
Exercises 397
(b) To which atom is the H of the product bonded? How
is the H bonded?
Preparation of Acids
083. A volatile acid such as nitric acid, HNO3, can be prepared
by adding concentrated H2SO4 to a salt of the acid. (a)
Write the chemical equation for the reaction of H2SO4
with sodium nitrate (called Chile saltpeter). (b) A dilute
aqueous solution of H2SO4 cannot be used. Why?
084. Outline a method of preparing each of the following acids
and write appropriate balanced equations for each preparation:
(a) H2S; (b) HBr; (c) CH3COOH.
085. Repeat Exercise 84 for (a) carbonic acid, (b) perchloric
acid, (c) permanganic acid, and (d) phosphoric acid (two
methods).
Mixed Exercises
086. Give the formula for an example chosen from the representative
elements for (a) an acidic oxide, (b) an amphoteric
oxide, and (c) a basic oxide.
087. Identify each of the following as (i) acidic, (ii) basic, or
(iii) amphoteric. Assume all oxides are dissolved in or are
in contact with water. Do not be intimidated by the way
in which the formula of the compound is written. (a) Cs2O;
(b) Cl2O5; (c) HCl; (d) SO2(OH)2; (e) HNO2, (f ) Al2O3;
(g) BaO; (h) H2O; (i) CO2; (j) SO2.
088. Indicate which of the following substances—(a) H2S;
(b) PO(OH)3; (c) H2CaO2; (d) ClO3(OH); (e) Sb(OH)3 —
can act as (i) an acid, (ii) a base, or (iii) both according to
the Arrhenius (classical) theory or the Brønsted–Lowry
theory. Do not be confused by the way in which the formulas
are written.
089. (a) Write equations for the reactions of HCO3
Ϫ with
H3Oϩ and HCO3
Ϫ with OHϪ, and indicate the conjugate
acid–base pairs in each case. (b) A substance such as
HCO3
Ϫ that reacts with both H3Oϩ and OHϪ is said to
be ______. (Fill in the missing word.)
090. (a) List the conjugate bases of H3PO4, NH4
ϩ, and OHϪ
and the conjugate acids of HSO4
Ϫ, PH3, and PO4
3Ϫ.
(b) Given that NO2
Ϫ is a stronger base than NO3
Ϫ, which
is the stronger acid—nitric acid, HNO3, or nitrous acid,
HNO2?
*91. A 0.1 M solution of copper(II) chloride, CuCl2, causes the
light bulb in Figure 4-2 to glow brightly. When hydrogen
sulfide, H2S, a very weak acid, is added to the solution, a
black precipitate of copper(II) sulfide, CuS, forms, and the
bulb still glows brightly. The experiment is repeated with
a 0.1 M solution of copper(II) acetate, Cu(CH3COO)2,
which also causes the bulb to glow brightly. Again, CuS
forms, but this time the bulb glows dimly. With the aid of
ionic equations, explain the difference in behavior between
the CuCl2 and Cu(CH3COO)2 solutions.
092. Referring again to Figure 4-2, explain the following results
of a conductivity experiment (use ionic equations). (a) Individual
solutions of NaOH and HCl cause the bulb to glow
brightly. When the solutions are mixed, the bulb still glows
brightly but not as brightly as before. (b) Individual solutions
of NH3 and CH3COOH cause the bulb to glow
dimly. When the solutions are mixed, the bulb glows
brightly.
093. Which statements are true? Rewrite any false statement so
that it is correct. (a) Strong acids and bases are virtually
100% ionized or dissociated in dilute aqueous solutions.
(b) The leveling effect is the seemingly identical strengths
of all acids and bases in aqueous solutions. (c) A conjugate
acid is a molecule or ion formed by the addition of a proton
to a base. (d) Amphoterism and amphiprotism are the
same in aqueous solution.
CONCEPTUAL EXERCISES
094. The following diagrams are nanoscale representations of
different acids in aqueous solution; the water molecules
are not shown. The small, dark circles are hydrogen atoms
Conductivity experiment.
398 CHAPTER 10: Reactions in Aqueous Solutions I: Acids, Bases, and Salts
or ions. The larger, lighter circles represent the anions.
(a) Which diagram best represents hydrochloric acid?
(b) Which diagram best represents acetic acid?
095. When a 0.1 M aqueous ammonia solution is tested with a
conductivity apparatus (Figure 4.2), the bulb glows dimly.
When a 0.1 M hydrochloric acid solution is tested, the
bulb glows brightly. Would you expect the bulb to glow
more brightly, stop glowing, or stay the same as water is
added to each of the solutions? Explain your reasoning.
96. On the planet Baseacidopolous, the major solvent is liquid
ammonia, not water. Ammonia autoionizes much like
water (2NH3 34 NH4
ϩ ϩ NH2
Ϫ). If instead of water,
ammonia is used as a solvent: (a) What is the formula of
the cation that would indicate that a compound is an acid?
(b) What is the formula of the anion produced if a compound
is a base? (c) Look at the way that NaCl is formed
from an acid–base reaction on earth and determine if NaCl
can be a salt on Baseacidopolous.
BUILDING YOUR KNOWLEDGE
97. One of the chemical products of muscle contraction is lactic
acid (CH3CH(OH)CO2H), a monoprotic acid whose
structure is
in water. (d) Describe the geometry around each of the
carbon atoms in lactic acid.
Prolonged exercise can temporarily overload the body’s
capacity for elimination of this substance, and the resulting
increase in lactic acid concentration in the muscles
causes pain and stiffness. (a) Lactic acid has six H atoms,
yet it acts as a monoprotic acid in an aqueous environment.
Which of the H atoms is ionizable? (b) Draw the
structural formula of the conjugate base. (c) Write a net
ionic equation that illustrates the ionization of lactic acid
98. Autoionization can occur when an ion other than an Hϩ
is transferred, as exemplified by the transfer of a ClϪ ion
from one PCl5 molecule to another. Write the equation
for this reaction. What are the shapes of the two ions that
are formed?
99. Limestone, CaCO3, is a water-insoluble material, whereas
Ca(HCO3)2 is soluble. Caves are formed when rainwater
containing dissolved CO2 passes over limestone for long
periods of time. Write a chemical equation for the acid–
base reaction.
100. Acids react with metal carbonates and hydrogen carbonates
to form carbon dioxide and water. (a) Write the balanced
equation for the reaction that occurs when baking soda,
NaHCO3, and vinegar, 5% acetic acid, are mixed. What
causes the “fizz”? (b) Lactic acid, CH3CH(OH)COOH, is
found in sour milk and in buttermilk. Many of its reactions
are very similar to those of acetic acid. Write the
balanced equation for the reaction of baking soda,
NaHCO3, with lactic acid. Explain why bread “rises” during
the baking process.
101. Some of the acid formed in tissues is excreted through the
kidneys. One of the bases removing the acid is HPO4
2Ϫ.
Write the equation for the reaction. Could ClϪ serve this
function?
(a) (b) (c) (d)
Runners risk suffering from lactic acid buildup.
11Reactions in Aqueous
Solutions II:
Calculations
OUTLINE
Aqueous Acid–Base Reactions
11-1 Calculations Involving Molarity
11-2 Titrations
11-3 The Mole Method and Molarity
11-4 Equivalent Weights and
Normality
Oxidation–Reduction Reactions
11-5 The Half-Reaction Method
11-6 Adding Hϩ, OHϪ, or H2O to
Balance Oxygen or Hydrogen
11-7 Change-in-Oxidation-Number
Method
11-8 Stoichiometry of Redox
Reactions
H
OBJECTIVES
After you have studied this chapter, you should be able to
• Perform molarity calculations
• Solve acid–base stoichiometry calculations
• Describe titration and standardization
• Use the mole method and molarity in acid–base titration reactions
• Perform calculations involving equivalent weights and normality of acid and base
solutions
• Balance oxidation–reduction equations
• Perform calculations associated with redox reactions
AQUEOUS ACID–BASE REACTIONS
H
ydrochloric acid, HCl, is called “stomach acid” because it is the main acid
(Ϸ0.10 M) in our digestive juices. When the concentration of HCl is too high
in humans, problems result. These problems may range from “heartburn” to
ulcers that can eat through the lining of the stomach wall. Snakes have very high concentrations
of HCl in their digestive juices so that they can digest whole small animals and
birds.
Automobile batteries contain 40% H2SO4 by mass. When the battery has “run down,”
the concentration of H2SO4 is significantly lower than 40%. A technician checks an automobile
battery by drawing some battery acid into a hydrometer, which indicates the density
of the solution. This density is related to the concentration of H2SO4.
There are many practical applications of acid–base chemistry in which we must know
the concentration of a solution of an acid or a base.
A concentrated NaOH solution
added to a CH3COOH solution that
contains a large amount of
phenolphthalein indicator.
Digestive juice is the acidic fluid
secreted by glands in the lining of the
stomach.
CALCULATIONS INVOLVING MOLARITY
In Sections 3-6 through 3-8 we introduced methods for expressing concentrations of solutions
and discussed some related calculations. Review of those sections will be helpful as
we learn more about acid–base reactions in solutions.
In some cases, the neutralization reaction involves one mole of an acid reacting with one
mole of a base to effect neutralization.
HCl ϩ NaOH 88n NaCl ϩ H2O
HNO3 ϩ KOH 88n KNO3 ϩ H2O
Because one mole of each acid reacts with one mole of each base in these cases, one liter
of a one-molar solution of either of these acids reacts with one liter of a one-molar solution of
either of these bases. These acids have only one acidic hydrogen per formula unit, and these
bases have one hydroxide ion per formula unit, so one formula unit of base reacts with
one formula unit of acid.
The reaction ratio is the relative numbers of moles of reactants and products shown in
the balanced equation.
EXAMPLE 11-1 Acid–Base Reactions
If 100. mL of 0.100 M HCl solution and 100. mL of 0.100 M NaOH are mixed, what is the
molarity of the salt in the resulting solution? Assume that the volumes are additive.
Plan
We first write the balanced equation for the acid–base reaction and then construct the reaction
summary that shows the amounts (moles) of HCl and NaOH. We determine the amount
of salt formed from the reaction summary. The final (total) volume is the sum of the volumes
mixed. Then we calculate the molarity of the salt.
Solution
The following tabulation shows that equal numbers of moles of HCl and NaOH are mixed
and, therefore, all of the HCl and NaOH react. The resulting solution contains only NaCl,
the salt formed by the reaction, and water.
HCl ϩ NaOH 88n NaCl ϩ H2O
Rxn ratio: Ϫ0.1001 mol HCl Ϫ0.1001 mol NaOH 1 mol 1 mol
Start: ΄0.100 L ΂ ΃΅ ΄0.100 L ΂ ΃΅ Ϫ0 mol
Ϫ0.0100 mol HCl Ϫ0.0100 mol NaOH
Change: Ϫ0.0100 mol HCl Ϫ0.0100 mol NaOH ϩ0.0100 mol
After rxn: Ϫ0.1000 mol HCl Ϫ0.1000 mol NaOH Ϫ0.0100 mol
The HCl and NaOH neutralize each other exactly, and the resulting solution contains
0.0100 mol of NaCl in 0.200 L of solution. Its molarity is
_?_ ϭ ϭ 0.0500 M NaCl
0.0100 mol NaCl
ᎏᎏ
0.200 L
mol NaCl
ᎏᎏ
L
0.100 mol
ᎏᎏ
L
0.100 mol
ᎏᎏ
L
11-1
Experiments have shown that volumes
of dilute aqueous solutions are very
nearly additive. No significant error is
introduced by making this assumption.
0.100 L of NaOH solution mixed with
0.100 L of HCl solution gives 0.200 L
of solution.
11-1 Calculations Involving Molarity 401
The amount of water produced by the
reaction is negligible.
We often express the volume of a solution in milliliters rather than in liters. Likewise,
we may express the amount of solute in millimoles (mmol) rather than in moles. Because
one milliliter is 1/1000 of a liter and one millimole is 1/1000 of a mole, molarity also may
be expressed as the number of millimoles of solute per milliliter of solution:
Molarity ϭ
For volumes and concentrations that are commonly used in laboratory experiments, solving
problems in terms of millimoles and milliliters often involves more convenient numbers
than using moles and liters. We should note also that the reaction ratio that we obtain
from any balanced chemical equation is exactly the same whether we express all quantities
in moles or in millimoles. We will work many problems in this chapter using millimoles
and milliliters. Let us see how we might solve Example 11-1 in these terms.
As in Example 11-1 we first write the balanced equation for the acid–base reaction,
and then construct the reaction summary that shows the amounts (millimoles) of NaOH
and HCl. We determine the amount of salt formed from the reaction summary. The final
(total) volume is the sum of the volumes mixed. Then we can calculate the molarity of
the salt.
The following tabulation in the solution would become:
number of millimoles of solute
ᎏᎏᎏᎏ
number of milliliters of solution
1 mol ϭ 1000 mmol
1 L ϭ 1000 mL
molarity ϭ ϭ
no. mmol
ᎏᎏ
mL
no. mol
ᎏ
L
402 CHAPTER 11: Reactions in Aqueous Solutions II: Calculations
_?_ ϭ ϭ 0.0500 M NaCl
EXAMPLE 11-2 Acid–Base Reactions
If 100. mL of 1.00 M HCl and 100. mL of 0.80 M NaOH solutions are mixed, what are the
molarities of the solutes in the resulting solution?
Plan
We proceed as we did in Example 11-1. This reaction summary shows that NaOH is the
limiting reactant and that we have excess HCl.
Solution
HCl ϩ NaOH 88n NaCl ϩ H2O
Rxn ratio: 00 1 mmol 00 1 mmol ϩ8 1 mmol 1 mmol
Start: 100. mmol 080. mmol ϩ8 0 mmol
Change: Ϫ80. mmol Ϫ80. mmol ϩ80. mmol
After rxn: 020. mmol 00 0 mmol ϩ80. mmol
10.0 mmol NaCl
ᎏᎏ
200. mL
mmol NaCl
ᎏᎏ
mL
HCl ϩ NaOH 88n NaCl ϩ H2O
Rxn ratio: ϭ 10.1 mmol HCl ϭ 10.1 mmol NaOH 1 mmol 1 mmol
Start: ΄100. mL ΂ ΃΅ ΄100. mL ΂ ΃΅ 0 mmol
ϭ 10.0 mmol HCl ϭ 10.0 mmol NaOH
Change: Ϫ10.0 mmol HCl Ϫ10.0 mmol NaOH ϩ10.0 mmol
After rxn: ϭ 10.0 mmol HCl ϭ 10.0 mmol NaOH ϭ10.0 mmol
0.100 mmol
ᎏᎏ
mL
0.100 mmol
ᎏᎏ
mL
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 5-10, Solution Concentration—
Molarity.
Because two solutes are present in the solution after reaction, we must calculate the concentrations
of both.
_?_ ϭ ϭ 0.10 M HCl
_?_ ϭ ϭ 0.40 M NaCl
Both HCl and NaCl are strong electrolytes, so the solution is 0.10 M in Hϩ(aq), (0.10 ϩ 0.40)
M ϭ 0.50 M in ClϪ, and 0.40 M in Naϩ ions.
You should now work Exercises 8 and 16.
80. mmol NaCl
ᎏᎏ
200. mL
mmol NaCl
ᎏᎏ
mL
20. mmol HCl
ᎏᎏ
200. mL
mmol HCl
ᎏᎏ
mL
11-1 Calculations Involving Molarity 403
Problem-Solving Tip: Review Limiting Reactant Calculations
To solve many of the problems in this chapter, you will need to apply the limiting reactant
concept (Section 3-3). In Example 11-1, we confirm that the two reactants are
initially present in the mole ratio required by the balanced chemical equation; they both
react completely, so there is no excess of either one. In Example 11-2, we need to determine
which reactant limits the reaction. Before you proceed, be sure you understand
how the ideas of Section 3-3 are used in these examples.
In many cases more than one mole of a base will be required to neutralize completely
one mole of an acid, or more than one mole of an acid will be required to neutralize
completely one mole of a base.
H2SO4 ϩ 2NaOH 88n Na2SO4 ϩ 2H2O
1 mol 2 mol 1 mol
2HCl ϩ Ca(OH)2 88n CaCl2 ϩ 2H2O
2 mol 1 mol 1 mol
The first equation shows that one mole of H2SO4 reacts with two moles of NaOH.
Thus, two liters of 1 M NaOH solution are required to neutralize one liter of 1 M H2SO4
solution. The second equation shows that two moles of HCl react with one mole of
Ca(OH)2. Thus, two liters of HCl solution are required to neutralize one liter of Ca(OH)2
solution of equal molarity.
EXAMPLE 11-3 Volume of Acid to Neutralize Base
What volume of 0.00300 M HCl solution would just neutralize 30.0 mL of 0.00100 M Ca(OH)2
solution?
Plan
We write the balanced equation for the reaction to determine the reaction ratio. Then we
(1) convert milliliters of Ca(OH)2 solution to millimoles of Ca(OH)2 using molarity as a unit
factor, 0.00100 mmol Ca(OH)2/1.00 mL Ca(OH)2 solution; (2) convert millimoles of Ca(OH)2
to millimoles of HCl using the unit factor, 2 mmol HCl/1 mmol Ca(OH)2 (the reaction ratio
from the balanced equation); and (3) convert millimoles of HCl to milliliters of HCl solution
using the unit factor, 1.00 mL HCl/0.00300 mmol HCl, that is, molarity inverted.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 5-13, Stoichiometry of Reactions
in Solution.
ᎏ
mL C
so
a
l
(
n
OH)2
ᎏ 88n ᎏ
mmo
p
l
re
C
s
a
e
(
n
O
t
H)2
ᎏ 88n ᎏ
mm
ne
o
e
l
d
H
ed
Cl
ᎏ 88n ᎏ
mL
ne
H
ed
C
e
l
d
(aq)
ᎏ
Solution
The balanced equation for the reaction is
2HCl ϩ Ca(OH)2 88n CaCl2 ϩ 2H2O
2 mmol 1 mmol 1 mmol 2 mmol
_?_ mL HCl ϭ
30.0 mL Ca(OH)2 ϫ ϫ ϫ
ϭ 20.0 mL HCl
You should now work Exercise 18.
In the preceding example we used the unit factor, 2 mol HCl/1 mol Ca(OH)2, to convert
moles of Ca(OH)2 to moles of HCl because the balanced equation shows that two moles
of HCl are required to neutralize one mole of Ca(OH)2. We must always write the balanced
equation and determine the reaction ratio.
1.00 mL HCl
ᎏᎏᎏ
0.00300 mmol HCl
2 mmol HCl
ᎏᎏ
1 mmol Ca(OH)2
0.00100 mmol Ca(OH)2
ᎏᎏᎏ
1.00 mL Ca(OH)2
The balanced chemical equation allows
us to construct either a mole ratio or a
millimole ratio.
or
2 mmol HCl
ᎏᎏ
1 mmol Ca(OH)2
2 mol HCl
ᎏᎏ
1 mol Ca(OH)2
404 CHAPTER 11: Reactions in Aqueous Solutions II: Calculations
Problem-Solving Tip: There Is More Than One Way to Solve
Some Problems
In many problems more than one “plan” can be followed. In Example 11-3 a particular
plan was used successfully. Many students can more easily visualize the solution by
following a plan like that in Examples 11-1 and 11-4. We suggest that you use the plan
that you find most understandable.
EXAMPLE 11-4 Acid–Base Reactions
If 100. mL of 1.00 M H2SO4 solution is mixed with 200. mL of 1.00 M KOH, what salt is
produced, and what is its molarity?
Plan
We proceed as we did in Example 11-2. We note that the reaction ratio is 1 mmol of H2SO4
to 2 mmol of KOH to 1 mmol of K2SO4.
Solution
H2SO4 ϩ 2KOH 88n K2SO4 ϩ H2O
Rxn ratio: ϭ 10.1 mmol ϭ 20.2 mmol 1 mmol
Start: ΄100. mL ΂ ΃΅ ΄200. mL ΂ ΃΅
ϭ 100. mmol ϭ 200. mmol 0 mmol
Change: Ϫ100. mmol Ϫ200. mmol ϩ100. mmol
After rxn: ϭ 10.0 mmol ϭ 20.0 mmol ϩ100. mmol
1.00 mmol
ᎏᎏ
mL
1.00 mmol
ᎏᎏ
mL
The reaction produces 100. mmol of potassium sulfate. This is contained in 300. mL of solution,
and so the concentration is
_?_ ϭ ϭ 0.333 M K2SO4
You should now work Exercises 10 and 14.
TITRATIONS
In Examples 3-23 and 11-3, we calculated the volume of one solution that is required to
react with a given volume of another solution, with the concentrations of both solutions
given. In the laboratory we often measure the volume of one solution that is required to
react with a given volume of another solution of known concentration. Then we calculate
the concentration of the first solution. The process is called titration (Figure 11-1).
11-2
100. mmol K2SO4
ᎏᎏᎏ
300. mL
mmol K2SO4
ᎏᎏ
mL
Because K2SO4 is a strong electrolyte,
this corresponds to 0.666 M Kϩ and
0.333 M SO4
2Ϫ.
11-2 Titrations 405
Figure 11-1 The titration process. (a) A typical setup for titration in a teaching laboratory.
The solution to be titrated is placed in an Erlenmeyer flask, and a few drops of indicator are
added. The buret is filled with a standard solution (or the solution to be standardized). The
volume of solution in the buret is read carefully. (b) The meniscus describes the surface
of the liquid in the buret. Aqueous solutions wet glass, so the meniscus of an aqueous
solution is always concave. The position of the bottom of the meniscus is read and recorded.
(c) The solution in the buret is added (dropwise near the end point), with stirring, to the
Erlenmeyer flask until the end point is reached. (d) The end point is signaled by the
appearance (or change) of color throughout the solution being titrated. (A very large excess of
indicator was used to make this photograph.) The volume of the liquid is read again—the
difference between the final and initial buret readings is the volume of the solution used.
(a) (b) (c) (d)
Titration is the process in which a solution of one reactant, the titrant, is carefully
added to a solution of another reactant, and the volume of titrant required for
complete reaction is measured.
How does one know when to stop a titration—that is, when is the chemical reaction
just complete? In one method, a few drops of an indicator solution are added to the solution
to be titrated. An indicator is a substance that can exist in different forms, with
different colors that depend on the concentration of Hϩ in the solution. At least one of
these forms must be very intensely colored so that even very small amounts of it can be
seen.
We can titrate an acid solution of unknown concentration by adding a standardized
solution of sodium hydroxide dropwise from a buret (see Figure 11-1). A common buret
is graduated in large intervals of 1 mL and in smaller intervals of 0.1 mL so that it is
possible to estimate the volume of a solution dispensed to within at least Ϯ0.02 mL.
(Experienced individuals can often read a buret to Ϯ0.01 mL.) The analyst tries to choose
an indicator that changes color clearly at the point at which stoichiometrically equivalent
amounts of acid and base have reacted, the equivalence point. The point at which the
indicator changes color and the titration is stopped is called the end point. Ideally, the
end point should coincide with the equivalence point. Phenolphthalein is colorless in acidic
solution and reddish violet in basic solution. In a titration in which a base is added to an
acid, phenolphthalein is often used as an indicator. The end point is signaled by the first
appearance of a faint pink coloration that persists for at least 15 seconds as the solution
is swirled.
EXAMPLE 11-5 Titration
What is the molarity of a hydrochloric acid solution if 36.7 mL of the HCl solution is required
to react with 43.2 mL of 0.236 M sodium hydroxide solution?
HCl ϩ NaOH 88n NaCl ϩ H2O
Plan
The balanced equation tells us that the reaction ratio is one millimole of HCl to one millimole
of NaOH, which gives the unit factor, 1 mmol HCl/1 mmol NaOH.
HCl ϩ NaOH 88n NaCl ϩ H2O
1 mmol 1 mmol 1 mmol 1 mmol
First we find the number of millimoles of NaOH. The reaction ratio is one millimole of HCl
to one millimole of NaOH, so the HCl solution must contain the same number of millimoles
of HCl. Then we can calculate the molarity of the HCl solution because we know its volume.
Solution
The volume of a solution (in milliliters) multiplied by its molarity gives the number of
millimoles of solute.
_?_ mmol NaOH ϭ 43.2 mL NaOH soln ϫ ϭ 10.2 mmol NaOH
Because the reaction ratio is one millimole of NaOH to one millimole of HCl, the HCl solution
must contain 10.2 millimoles of HCl.
0.236 mmol NaOH
ᎏᎏᎏ
1 mL NaOH soln
The choice of indicators will be
discussed in Section 19.4.
406 CHAPTER 11: Reactions in Aqueous Solutions II: Calculations
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 5-15, Titration Simulation.
_?_ mol HCl ϭ 10.2 mmol NaOH ϫ ϭ 10.2 mmol HCl
We know the volume of the HCl solution, so we can calculate its molarity.
ϭ ϭ 0.278 M HCl
EXAMPLE 11-6 Titration
A 43.2-mL sample of 0.236 M sodium hydroxide solution reacts completely with 36.7 mL of
a sulfuric acid solution. What is the molarity of the H2SO4 solution?
H2SO4 ϩ 2NaOH 88n Na2SO4 ϩ 2H2O
Plan
The balanced equation tells us that the reaction ratio is one millimole of H2SO4 to two
millimoles of NaOH, which gives the unit factor, 1 mmol H2SO4/2 mmol NaOH.
H2SO4 ϩ 2NaOH 88n Na2SO4 ϩ 2H2O
1 mmol 2 mmol 1 mmol 2 mmol
First we find the number of millimoles of NaOH. The reaction ratio is one millimole of H2SO4
to two millimoles of NaOH, so the number of millimoles of H2SO4 must be one half of the
number of millimoles of NaOH. Then we can calculate the molarity of the H2SO4 solution
because we know its volume.
Solution
The volume of a solution (in milliliters) multiplied by its molarity gives the number of
millimoles of solute.
_?_ mmol NaOH ϭ 43.2 mL NaOH soln ϫ ϭ 10.2 mmol NaOH
0.236 mmol NaOH
ᎏᎏᎏ
1 mL NaOH soln
10.2 mmol HCl
ᎏᎏᎏ
36.7 mL HCl soln
_?_ mmol HCl
ᎏᎏ
mL HCl soln
1 mmol HCl
ᎏᎏ
1 mmol NaOH
The indicator phenolphthalein
changes from colorless, its color in
acidic solutions, to pink, its color in
basic solutions, when the reaction in
Example 11-5 reaches completion.
Note the first appearance of a faint
pink coloration in the middle beaker;
this signals that the end point is
near.
Because the reaction ratio is two millimoles of NaOH to one millimole of H2SO4, the H2SO4
solution must contain 5.10 millimoles of H2SO4.
_?_ mmol H2SO4 ϭ 10.2 mmol NaOH ϫ ϭ 5.10 mmol H2SO4
We know the volume of the H2SO4 solution, so we can calculate its molarity.
ϭ ϭ 0.139 M H2SO4
You should now work Exercise 40.
Solutions of accurately known concentrations are called standard solutions. Often we
prepare solutions of such substances and then determine their concentrations by titration
with a standard solution.
Standardization is the process by which one determines the concentration of a solution
by measuring accurately the volume of the solution required to react with an exactly
known amount of a primary standard. The standardized solution is then known as a
secondary standard and is used in the analysis of unknowns.
The properties of an ideal primary standard include the following.
1. It must not react with or absorb the components of the atmosphere, such as water
vapor, oxygen, and carbon dioxide.
2. It must react according to one invariable reaction.
3. It must have a high percentage purity.
4. It should have a high formula weight to minimize the effect of error in weighing.
5. It must be soluble in the solvent of interest.
6. It should be nontoxic.
7. It should be readily available (inexpensive).
8. It should be environmentally friendly.
The first five of these characteristics are essential to minimize the errors involved in analytical
methods. The last three characteristics are just as important as the first five in most
analytical laboratories. Because primary standards are often costly and difficult to prepare,
secondary standards are often used in day-to-day work.
THE MOLE METHOD AND MOLARITY
Let us now describe the use of a few primary standards for acids and bases. One primary
standard for solutions of acids is sodium carbonate, Na2CO3, a solid compound.
H2SO4 ϩ Na2CO3 88n Na2SO4 ϩ CO2 ϩ H2O
1 mol 1 mol 1 mol 1 mol 1 mol
1 mol Na2CO3 ϭ 106.0 g and 1 mmol Na2CO3 ϭ 0.1060 g
Sodium carbonate is a salt. Because a base can be broadly defined as a substance that
reacts with hydrogen ions, in this reaction Na2CO3 can be thought of as a base.
11-3
5.10 mmol H2SO4
ᎏᎏᎏ
36.7 mL H2SO4 soln
_?_ mmol H2SO4
ᎏᎏ
mL H2SO4 soln
1 mmol H2SO4
ᎏᎏ
2 mmol NaOH
Notice the similarity between
Examples 11-5 and 11-6 in which 43.2
mL of 0.236 M NaOH solution is
used. In Example 11-5 the reaction
ratio is 1 mmol acid/1 mmol base,
whereas in Example 11-6 the reaction
ratio is 1 mmol acid/2 mmol base, and
so the molarity of the HCl solution
(0.278 M) is twice the molarity of the
H2SO4 solution (0.139 M).
408 CHAPTER 11: Reactions in Aqueous Solutions II: Calculations
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 5-14, Titrations.
Refer to the Brønsted–Lowry theory.
(Section 10-4).
CO2, H2O, and O2 are present in the
atmosphere. They react with many
substances.
EXAMPLE 11-7 Standardization of an Acid Solution
Calculate the molarity of a solution of H2SO4 if 40.0 mL of the solution neutralizes 0.364
gram of Na2CO3.
Plan
We know from the balanced equation that 1 mol of H2SO4 reacts with 1 mol of Na2CO3,
106.0 g. This provides the unit factors that convert 0.364 g of Na2CO3 to the corresponding
number of moles of H2SO4, from which we can calculate molarity.
ᎏ
g
a
N
va
a
il
2
a
C
b
O
le
3
ᎏ 88n ᎏ
mo
p
l
r
N
es
a
e
2
n
C
t
O3
ᎏ 88n ᎏ
mol
u
H
se
2
d
SO4
ᎏ 88n ᎏ
o
m
f
o
H
la
2
r
S
i
O
ty
4
ᎏ
Solution
_?_ mol H2SO4 ϭ 0.364 g Na2CO3 ϫ ϫ
ϭ 0.00343 mol H2SO4 (present in 40.0 mL of solution)
Now we calculate the molarity of the H2SO4 solution
ϭ ϭ 0.0858 M H2SO4
You should now work Exercise 28.
Most inorganic bases are metal hydroxides, all of which are solids. Even in the solid
state, however, most inorganic bases react rapidly with CO2 (an acid anhydride) from the
atmosphere. Most metal hydroxides also absorb H2O from the air. These properties make
it very difficult to accurately weigh out samples of pure metal hydroxides. Chemists obtain
solutions of bases of accurately known concentration by standardizing the solutions against
an acidic salt, potassium hydrogen phthalate, KC6H4(COO)(COOH). This is produced
by neutralization of one of the two ionizable hydrogens of an organic acid, phthalic acid.
This acidic salt, known simply as KHP, has one acidic hydrogen (highlighted) that reacts
with bases. KHP is easily obtained in a high state of purity, and is soluble in water. It is
used as a primary standard for bases.
EXAMPLE 11-8 Standardization of Base Solution
A 20.00-mL sample of a solution of NaOH reacts with 0.3641 gram of KHP. Calculate the
molarity of the NaOH solution.
C
C
C
C
C
C
C
C
O
O
O
O
H
H
H
H
H
H
HC
C
C
C
C
C
C
C
O
Ϫ
K
ϩ
O
O
O
H
H
H
H
ϩ ϩKOH H2O
C6H4(COOH)2
phthalic acid
KC6H4(COO)(COOH)
potassium hydrogen
phthalate (KHP)
0.00343 mol H2SO4
ᎏᎏᎏ
0.0400 L
_?_ mol H2SO4
ᎏᎏ
L
1 mol H2SO4
ᎏᎏ
1 mol Na2CO3
1 mol Na2CO3
ᎏᎏ
106.0 g Na2CO3
Sodium carbonate is often used as a
primary standard for acids.
11-3 The Mole Method and Molarity 409
The “ph” in phthalate is silent.
Phthalate is pronounced “thalate.”
The P in KHP stands for the phthalate
ion, C6H4(COO)2
2Ϫ, not phosphorus.
Plan
We first write the balanced equation for the reaction between NaOH and KHP. We then calculate
the number of moles of NaOH in 20.00 mL of solution from the amount of KHP that
reacts with it. Then we can calculate the molarity of the NaOH solution.
ᎏ
a
g
va
K
il
P
ab
H
le
ᎏ 88n ᎏ
m
av
o
a
l
il
K
ab
H
le
P
ᎏ 88n ᎏ
m
r
o
e
l
q
N
ui
a
r
O
ed
H
ᎏ 88n ᎏ
o
m
f
o
N
la
a
r
O
it
H
y
ᎏ
Solution
NaOH ϩ KHP 88n NaKP ϩ H2O
1 mol 1 mol 1 mol 1 mol
We see that NaOH and KHP react in a 1Ϻ1 mole ratio. One mole of KHP is 204.2 g.
_?_ mol NaOH ϭ 0.3641 g KHP ϫ ϫ ϭ 0.001783 mol NaOH
Then we calculate the molarity of the NaOH solution.
ϭ ϭ 0.08915 M NaOH
You should now work Exercise 30.
Impure samples of acids can be titrated with standard solutions of bases. The results
can be used to determine percentage purity of the samples.
EXAMPLE 11-9 Determination of Percent Acid
Oxalic acid, (COOH)2, is used to remove rust stains and some ink stains from fabrics. A
0.1743-gram sample of impure oxalic acid required 39.82 mL of 0.08915 M NaOH solution
for complete neutralization. No acidic impurities were present. Calculate the percentage purity
of the (COOH)2.
Plan
We write the balanced equation for the reaction and calculate the number of moles of NaOH
in the standard solution. Then we calculate the mass of (COOH)2 in the sample, which gives
us the information we need to calculate percentage purity.
Solution
The equation for the complete neutralization of (COOH)2 with NaOH is
2NaOH ϩ (COOH)2 88n Na2(COO)2 ϩ 2H2O
2 mol 1 mol 1 mol 2 mol
Two moles of NaOH neutralizes completely one mole of (COOH)2. The number of moles of
NaOH that react is the volume times the molarity of the solution.
_?_ mol NaOH ϭ 0.03982 L ϫ ϭ 0.003550 mol NaOH
Now we calculate the mass of (COOH)2 that reacts with 0.003550 mol NaOH.
0.08915 mol NaOH
ᎏᎏᎏ
L
0.001783 mol NaOH
ᎏᎏᎏ
0.02000 L
_?_ mol NaOH
ᎏᎏ
L
1 mol NaOH
ᎏᎏ
1 mol KHP
1 mol KHP
ᎏᎏ
204.2 g KHP
Very pure KHP is readily available.
410 CHAPTER 11: Reactions in Aqueous Solutions II: Calculations
Each molecule of (COOH)2 contains
two acidic H’s.
C
O
O HH O C
O
1 mol ϭ 90.04 g
_?_ g (COOH)2 ϭ 0.003550 mol NaOH ϫ ϫ
ϭ 0.1598 g (COOH)2
The 0.1743-g sample contained 0.1598 g of (COOH)2, so its percentage purity was
% purity ϭ ϫ 100% ϭ 91.68% pure (COOH)2
You should now work Exercise 36.
EQUIVALENT WEIGHTS AND NORMALITY
Because one mole of an acid does not necessarily neutralize one mole of a base, some
chemists prefer a method of expressing concentration other than molarity to retain a oneto-one
relationship. Concentrations of solutions of acids and bases are frequently expressed
as normality (N). The normality of a solution is defined as the number of equivalent
weights, or simply equivalents (eq), of solute per liter of solution. Normality may be represented
symbolically as
Normality ϭ ϭ
By definition there are 1000 milliequivalent weights (meq) in one equivalent weight of
an acid or base. Normality may also be represented as
Normality ϭ ϭ
In acid–base reactions, one equivalent weight, or equivalent (eq), of an acid is
defined as the mass of the acid (expressed in grams) that could furnish 6.022 ϫ 1023
hydrogen ions (1 mol) or that could react with 6.022 ϫ 1023 hydroxide ions (1 mol). One
mole of an acid contains 6.022 ϫ 1023 formula units of the acid. Consider hydrochloric
acid as a typical monoprotic acid.
H2O
HCl 8888n Hϩ(aq) ϩ ClϪ(aq)
1 mol 1 mol 1 mol
36.46 g 1.008 g 35.45 g
6.022 ϫ 1023 6.022 ϫ 1023 6.022 ϫ 1023
formula units formula units formula units
We see that one mole of HCl can produce 6.022 ϫ 1023 Hϩ ions, and so one mole of HCl
is one equivalent. The same is true for all monoprotic acids.
no. meq
ᎏ
mL
number of milliequivalent weights of solute
ᎏᎏᎏᎏᎏ
milliliter of solution
no. eq
ᎏ
L
number of equivalent weights of solute
ᎏᎏᎏᎏᎏ
liter of solution
11-4
0.1598 g (COOH)2
ᎏᎏᎏ
0.1743 g sample
90.04 g (COOH)2
ᎏᎏᎏ
1 mol (COOH)2
1 mol (COOH)2
ᎏᎏ
2 mol NaOH
Any calculation that can be carried out
with equivalent weights and normality
can also be done by the mole method
using molarity. The methods of this
section are widely used, however, in
health-related fields and in many
industrial laboratories.
An equivalent weight is often referred
to simply as an equivalent (eq).
11-4 Equivalent Weights and Normality 411
A milliequivalent weight is often
referred to simply as a milliequivalent
(meq).
Sulfuric acid is a diprotic acid. One molecule of H2SO4 can furnish 2Hϩ ions.
H2O
H2SO4 8888n 2Hϩ(aq) ϩ SO4
2Ϫ(aq)
1 mol 2 mol 1 mol
98.08 g 2(1.008 g) 96.06 g
6.022 ϫ 1023 2(6.022 ϫ 1023) 6.022 ϫ 1023
formula units formula units formula units
This equation shows that one mole of H2SO4 can produce 2(6.022 ϫ 1023) Hϩ; therefore,
one mole of H2SO4 is two equivalent weights in all reactions in which both acidic
hydrogen atoms react.
One equivalent weight of a base is defined as the mass of the base (expressed in
grams) that will furnish 6.022 ϫ 1023 hydroxide ions or the mass of the base that will react
with 6.022 ϫ 1023 hydrogen ions.
The equivalent weight of an acid is obtained by dividing its formula weight in grams
either by the number of acidic hydrogens that could be furnished by one formula unit of
the acid or by the number of hydroxide ions with which one formula unit of the acid reacts.
The equivalent weight of a base is obtained by dividing its formula weight in grams either
by the number of hydroxide ions furnished by one formula unit or by the number of
hydrogen ions with which one formula unit of the base reacts. Equivalent weights of some
common acids and bases are given in Table 11-1.
EXAMPLE 11-10 Concentration of a Solution
Calculate the normality of a solution of 4.202 grams of HNO3 in 600. mL of solution.
Plan
We convert grams of HNO3 to moles of HNO3 and then to equivalents of HNO3, which lets
us calculate the normality.
88n 88n ϭ N HNO3
eq HNO3
ᎏᎏ
L
mol HNO3
ᎏᎏ
L
g HNO3
ᎏᎏ
L
One mole of H2SO4 is two
equivalent weights of H2SO4.
TABLE 11-1 Equivalent Weights* of Some Acids and Bases
Acids Bases
Symbolic Symbolic
representation One eq representation One eq
ϭ ϭ 63.02 g HNO3 ϭ ϭ 40.00 g NaOH
ϭ ϭ 60.03 g CH3COOH ϭ ϭ 17.04 g NH3
ϭ ϭ 204.2 g KHP ϭ ϭ 37.05 g Ca(OH)2
ϭ ϭ 49.04 g H2SO4 ϭ ϭ 85.68 g Ba(OH)2
*Complete neutralization is assumed.
171.36 g
ᎏ
2
Ba(OH)2
ᎏᎏ
2
98.08 g
ᎏ
2
H2SO4
ᎏ
2
74.10 g
ᎏ
2
Ca(OH)2
ᎏᎏ
2
204.2 g
ᎏ
1
KHP
ᎏ
1
17.04 g
ᎏ
1
NH3
ᎏ
1
60.03 g
ᎏ
1
CH3COOH
ᎏᎏ
1
40.00 g
ᎏ
1
NaOH
ᎏ
1
63.02 g
ᎏ
1
HNO3
ᎏ
1
Solution
N ϭ
_?_ ϭ ϫ ϫ ϭ 0.111 N HNO3
MHNO3
Because normality is equal to molarity times the number of equivalents per mole of
solute, a solution’s normality is always equal to or greater than its molarity.
Normality ϭ molarity ϫ or N ϭ M ϫ
EXAMPLE 11-11 Concentration of a Solution
Calculate (a) the molarity and (b) the normality of a solution that contains 9.50 grams of barium
hydroxide in 2000. mL of solution.
Plan
(a) We use the same kind of logic we used in Example 11-10.
(b) Because each mole of Ba(OH)2 produces 2 moles of OHϪ ions, 1 mole of Ba(OH)2 is 2
equivalents. Thus,
N ϭ M ϫ or M ϭ
Solution
(a) _?_ ϭ ϫ ϭ 0.0277 M Ba(OH)2
(b) _?_ ϭ ϫ ϭ 0.0554 N Ba(OH)2
You should now work Exercises 42 through 44.
From the definitions of one equivalent of an acid and of a base, we see that one equivalent
of an acid reacts with one equivalent of any base. It is not true that one mole of any acid
reacts with one mole of any base in any specific chemical reaction that goes to completion.
As a consequence of the definition of equivalents, 1 eq acid Х 1 eq base. We may
write the following for all acid–base reactions that go to completion.
Number of equivalents of acid ϭ number of equivalents of base
The product of the volume of a solution, in liters, and its normality is equal to the
number of equivalents of solute contained in the solution. For a solution of an acid,
Lacid ϫ Nacid ϭ Lacid ϫ ϭ eq acid
eq acid
ᎏ
Lacid
2 eq Ba(OH)2
ᎏᎏ
1 mol Ba(OH)2
0.0277 mol Ba(OH)2
ᎏᎏᎏ
L
eq Ba(OH)2
ᎏᎏ
L
1 mol Ba (OH)2
ᎏᎏᎏ
171.36 g Ba(OH)2
9.50 g Ba(OH)2
ᎏᎏ
2.00 L
mol Ba(OH)2
ᎏᎏ
L
N
ᎏᎏ
2 eq/mol
2 eq
ᎏ
mol
no. eq
ᎏ
mol
no. eq
ᎏ
mol
1 eq HNO3
ᎏᎏ
mol HNO3
1 mol HNO3
ᎏᎏ
63.02 g HNO3
4.202 g HNO3
ᎏᎏ
0.600 L
eq HNO3
ᎏᎏ
L
no. eq HNO3
ᎏᎏ
L
Because each formula unit of Ba(OH)2
contains two OHϪ ions,
1 mol Ba(OH)2 ϭ 2 eq Ba(OH)2
Thus, molarity is one half of normality
for Ba(OH)2 solutions.
11-4 Equivalent Weights and Normality 413



















Remember that the product of volume
and concentration equals the amount
of solute.
The notation Х is read “is equivalent
to.”
Similar relationships can be written for a solution of a base. Because 1 eq of acid always
reacts with 1 eq of base, we may write
Number of equivalents of acid ϭ number of equivalents of base
so
Lacid ϫ Nacid ϭ Lbase ϫ Nbase or mLacid ϫ Nacid ϭ mLbase ϫ Nbase
EXAMPLE 11-12 Volume Required for Neutralization
What volume of 0.100 N HNO3 solution is required to neutralize completely 50.0 mL of a
0.150 N solution of Ba(OH)2?
Plan
We know three of the four variables in the relationship
mLacid ϫ Nacid ϭ mLbase ϫ Nbase, and so we solve for mLacid.
Solution
_?_ mLacid ϭ ϭ ϭ 75.0 mL of HNO3 solution
You should now work Exercise 48.
In Example 11-13 let us again solve Example 11-7, this time using normality rather
than molarity. The balanced equation for the reaction of H2SO4 with Na2CO3, interpreted
in terms of equivalent weights, is
H2SO4 ϩ Na2CO3 88n Na2SO4 ϩ CO2 ϩ H2O
1 mol 1 mol 1 mol 1 mol 1 mol
2 eq 2 eq 2 eq 2 eq 2 eq
98.08 g 106.0 g
So, 1 eq Na2CO3 ϭ 53.0 g
EXAMPLE 11-13 Standardization of Acid Solution
Calculate the normality of a solution of H2SO4 if 40.0 mL of the solution reacts completely
with 0.364 gram of Na2CO3.
Plan
We refer to the balanced equation. We are given the mass of Na2CO3, so we convert grams
of Na2CO3 to equivalents of Na2CO3, then to equivalents of H2SO4, which lets us calculate
the normality of the H2SO4 solution.
ᎏ
g
p
N
re
a
s
2
e
C
n
O
t
3
ᎏ 88n ᎏ
eq
p
N
re
a
s
2
e
C
nt
O3
ᎏ 88n ᎏ
eq
ne
H
ed
2S
e
O
d
4
ᎏ 88n ᎏ
eq H
L
2SO4
ᎏ
50.0 mL ϫ 0.150 N
ᎏᎏᎏ
0.100 N
mLbase ϫ Nbase
ᎏᎏ
Nacid
The conversion factors needed to
convert liters to milliliters on each side
of this equation will cancel.
414 CHAPTER 11: Reactions in Aqueous Solutions II: Calculations
By definition, there must be equal
numbers of equivalents of all reactants
and products in a balanced chemical
equation.
Solution
First we calculate the number of equivalents of Na2CO3 in the sample.
no. eq Na2CO3 ϭ 0.364 g Na2CO3 ϫ ϭ 6.87 ϫ 10Ϫ3 eq Na2CO3
Because no. eq H2SO4 ϭ no. eq Na2CO3, we can write
LH2SO4
ϫ NH2SO4
ϭ 6.87 ϫ 10Ϫ3 eq H2SO4
NH2SO4
ϭ ϭ ϭ 0.172 N H2SO4
You should now work Exercise 50.
OXIDATION–REDUCTION REACTIONS
Our rules for assigning oxidation numbers are constructed so that in all redox reactions
the total increase in oxidation numbers must equal the total decrease in oxidation
numbers.
This equivalence provides the basis for balancing redox equations. Although there is no
single “best method” for balancing all redox equations, two methods are particularly useful:
(1) the half-reaction method, which is used extensively in electrochemistry (Chapter 21)
6.87 ϫ 10Ϫ3 eq H2SO4
ᎏᎏᎏ
0.040 L
6.87 ϫ 10Ϫ3 eq H2SO4
ᎏᎏᎏ
LH2SO4
1 eq Na2CO3
ᎏᎏ
53.0 g Na2CO3
Automatic titrators are used in modern
analytical laboratories. Such titrators rely
on electrical properties of the solutions.
Methyl red indicator changes from yellow
to red at the end point of this titration.
11-4 Equivalent Weights and Normality 415
The starting values in this example are
the same as those in Example 11-7.
The normality of this H2SO4 solution
is twice the molarity obtained in
Example 11-7 because 1 mol of H2SO4
is 2 eq.
and (2) the change-in-oxidation-number method. Many redox equations can be balanced
by simple inspection, but you should master one of these systematic methods so it can be
used to balance difficult equations.
All balanced equations must satisfy two criteria.
1. There must be mass balance. That is, the same number of atoms of each kind
must appear in reactants and products.
2. There must be charge balance. The sums of actual charges on the left and right
sides of the equation must equal each other. In a balanced formula unit equation,
the total charge on each side will be equal to zero. In a balanced net ionic equation,
the total charge on each side might not be zero, but it still must be equal
on the two sides of the equation.
THE HALF-REACTION METHOD
In the half-reaction method we separate and completely balance equations describing
oxidation and reduction half-reactions. Then we equalize the numbers of electrons gained
and lost in each. Finally, we add the resulting half-reactions to give the overall balanced
equation. The general procedure follows.
1. Write as much of the overall unbalanced equation as possible, omitting spectator
ions.
2. Construct unbalanced oxidation and reduction half-reactions (these are usually
incomplete as well as unbalanced). Show complete formulas for polyatomic ions and
molecules.
3. Balance by inspection all elements in each half-reaction, except H and O. Then use
the chart in Section 11-6 to balance H and O in each half-reaction.
4. Balance the charge in each half-reaction by adding electrons as “products” or “reac-
tants.”
5. Balance the electron transfer by multiplying the balanced half-reactions by appropriate
integers.
6. Add the resulting half-reactions and eliminate any common terms.
EXAMPLE 11-14 Balancing Redox Equations (HR method)
A useful analytical procedure involves the oxidation of iodide ions to free iodine. The free
iodine is then titrated with a standard solution of sodium thiosulfate, Na2S2O3. Iodine oxidizes
S2O3
2Ϫ ions to tetrathionate ions, S4O6
2Ϫ, and is reduced to IϪ ions. Write the balanced net
ionic equation for this reaction.
Plan
We are given the formulas for two reactants and two products. We use these to write as much
of the equations as possible. We construct and balance the appropriate half-reactions using the
rules just described. Then we add the half-reactions and eliminate common terms.
11-5
Most redox equations can be balanced
by both methods, but in some
instances one may be easier to use than
the other.
416 CHAPTER 11: Reactions in Aqueous Solutions II: Calculations
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 21-3, Balancing Equations for
Redox Reactions.
For brevity, we will refer to this as the
HR method. It is sometimes referred
to as the ion–electron method.
Solution
I2 ϩ S2O3
2Ϫ 88n IϪ ϩ S4O6
2Ϫ
I2 88n IϪ (red. half-reaction)
I2 88n 2IϪ
I2 ϩ 2eϪ 88n 2IϪ (balanced red. half-reaction)
S2O3
2Ϫ 88n S4O6
2Ϫ (ox. half-reaction)
2S2O3
2Ϫ 88n S4O6
2Ϫ
2S2O3
2Ϫ 88n S4O6
2Ϫ ϩ 2eϪ (balanced ox. half-reaction)
Each balanced half-reaction involves a transfer of two electrons. We add these half-reactions
and cancel the electrons.
I2 ϩ 2eϪ 88n 2IϪ
2S2O3
2Ϫ 88n S4O6
2Ϫ ϩ 2eϪ
I2(s) ϩ 2S2O3
2Ϫ(aq) 88n 2IϪ(aq) ϩ S4O6
2Ϫ(aq)
ADDING Hϩ, OHϪ, OR H2O TO BALANCE OXYGEN
OR HYDROGEN
Frequently we need more oxygen or hydrogen to complete the mass balance for a reaction
or half-reaction in aqueous solution. We must be careful, however, not to introduce
other changes in oxidation number or to use species that could not actually be present in
the solution. We cannot add H2 or O2 to equations because these species are not present
in aqueous solutions. Acidic solutions do not contain significant concentrations of OHϪ
ions. Basic solutions do not contain significant concentrations of Hϩ ions.
In acidic solution: We add only Hϩ or H2O (not OHϪ in acidic solution).
In basic solution: We add only OHϪ or H2O (not Hϩ in basic solution).
The following chart shows how to balance hydrogen and oxygen.
11-6
Each I2 gains 2eϪ. I2 is reduced; it is
the oxidizing agent.
Each S2O3
2Ϫ ion loses an eϪ. S2O3
2Ϫ
is oxidized; it is the reducing agent.
11-6 Adding H؉, OH؊, or H2O to Balance Oxygen or Hydrogen 417
In Acidic Solution:
In Basic Solution:
To balance O: To balance H:and
88n
then
Add H2O Add Hϩ
To balance O: To balance H:
For each O needed, For each H needed,
(1) add two OHϪ to side needing O (1) add one H2O to side needing H
and and
(2) add one H2O to other side (2) add one OHϪ to other side
and
88n
then
When balancing redox equations, we often find it convenient to omit the spectator
ions (Section 4-3) so that we can focus on the oxidation and reduction processes. We use
the methods presented in this chapter to balance the net ionic equation. If necessary we
add the spectator ions and combine species to write the balanced formula unit equation.
Examples 11-15 and 11-16 illustrate this approach.
EXAMPLE 11-15 Net Ionic Equations
Permanganate ions oxidize iron(II) to iron(III) in sulfuric acid solution. Permanganate ions are
reduced to manganese(II) ions. Write the balanced net ionic equation for this reaction.
Plan
We use the given information to write as much of the equation as possible. Then we follow
steps 2 through 6 in Section 11-5. The reaction occurs in H2SO4 solution; we can add Hϩ and
H2O as needed to balance H and O in the half-reactions (Step 3).
Solution
Fe2ϩ ϩ MnO4
Ϫ 88n Fe3ϩ ϩ Mn2ϩ
Fe2ϩ 88n Fe3ϩ (ox. half-reaction)
Fe2ϩ 88n Fe3ϩ ϩ 1eϪ (balanced ox. half-reaction)
MnO4
Ϫ 88n Mn2ϩ (red. half-reaction)
MnO4
Ϫ ϩ 8Hϩ 88n Mn2ϩ ϩ 4H2O
MnO4
Ϫ ϩ 8Hϩ ϩ 5eϪ 88n Mn2ϩ ϩ 4H2O (balanced red. half-reaction)
The oxidation half-reaction involves one electron, and the reduction half-reaction involves
five electrons. Now we balance the electron transfer and then add the two equations term
by term. This gives the balanced net ionic equation.
5(Fe2ϩ 88n Fe3ϩ ϩ 1eϪ)
1(MnO4
Ϫ ϩ 8Hϩ ϩ 5eϪ 88n Mn2ϩ ϩ 4H2O)
5Fe2ϩ(aq) ϩ MnO4
Ϫ(aq) ϩ 8Hϩ(aq) 88n 5Fe3ϩ(aq) ϩ Mn2ϩ(aq) ϩ 4H2O(ᐉ)
EXAMPLE 11-16 Total Ionic and Formula Unit Equations
Write the balanced total ionic and the formula unit equations for the reaction in Example
11-15, given that the reactants were KMnO4, FeSO4, and H2SO4.
Plan
The Kϩ is the cationic spectator ion, and the anionic spectator ion is SO4
2Ϫ. The Fe3ϩ ion
will need to occur twice in the product Fe2(SO4)3, so there must be an even number of Fe
atoms. So the net ionic equation is multiplied by two. It now becomes:
10Fe2ϩ(aq) ϩ 2MnO4
Ϫ(aq) ϩ 16Hϩ(aq) 88n 10Fe3ϩ(aq) ϩ 2Mn2ϩ(aq) ϩ 8H2O(ᐉ)
Based on the 10Fe2ϩ and the 16Hϩ, we add 18SO4
2Ϫ to the reactant side of the equation; we
must also add them to the product side to keep the equation balanced. Based on the 2MnO4
Ϫ,
we add 2Kϩ to each side of the equation.
418 CHAPTER 11: Reactions in Aqueous Solutions II: Calculations
In Section 4-3, we first wrote the
formula unit equation. We separated
any ionized or dissociated species into
ions to obtain the total ionic equation.
Then we eliminated the spectator ions
to obtain the net ionic equation. In
Examples 11-15 and 11-16, we reverse
the procedure.
Solution
Total ionic equation
10[Fe2ϩ(aq) ϩ SO4
2Ϫ(aq)] ϩ 2[Kϩ(aq) ϩ MnO4
Ϫ(aq)] ϩ 8[2Hϩ(aq) ϩ SO4
2Ϫ(aq)] 88n
5[2Fe3ϩ(aq) ϩ 3SO4
2Ϫ(aq)] ϩ 2[Mn2ϩ(aq) ϩ SO4
2Ϫ(aq)] ϩ 8H2O(ᐉ)
ϩ [2 Kϩ(aq) ϩ SO4
2Ϫ(aq)]
Balanced formula unit equation
10FeSO4(aq) ϩ 2KMnO4(aq) ϩ 8H2SO4(aq) 88n
5Fe2(SO4)3(aq) ϩ 2MnSO4(aq) ϩ K2SO4(aq) ϩ 8H2O(ᐉ)
You should now work Exercise 62.
Bleaches sold under trade names such as Clorox and Purex are 5% solutions of sodium
hypochlorite. The hypochlorite ion is a very strong oxidizing agent in basic solution. It
oxidizes many stains to colorless substances.
EXAMPLE 11-17 Balancing Redox Equations (HR method)
In basic solution, hypochlorite ions, ClOϪ, oxidize chromite ions, CrO2
Ϫ, to chromate ions,
CrO4
2Ϫ, and are reduced to chloride ions. Write the balanced net ionic equation for this
reaction.
Plan
We are given the formulas for two reactants and two products; we write as much of the equations
as possible. The reaction occurs in basic solution; we can add OHϪ and H2O as needed.
We construct and balance the appropriate half-reactions, equalize the electron transfer, add the
half-reactions, and eliminate common terms.
Solution
CrO2
Ϫ ϩ ClOϪ 88n CrO4
2Ϫ ϩ ClϪ
CrO2
Ϫ 88n CrO4
2Ϫ (ox. half-rxn)
CrO2
Ϫ ϩ 4OHϪ 88n CrO4
2Ϫ ϩ 2H2O
CrO2
Ϫ ϩ 4OHϪ 88n CrO4
2Ϫ ϩ 2H2O ϩ 3eϪ (balanced ox. half-rxn)
ClOϪ 88n ClϪ (red. half-rxn)
ClOϪ ϩ H2O 88n ClϪ ϩ 2OHϪ
ClOϪ ϩ H2O ϩ 2eϪ 88n ClϪ ϩ 2OHϪ (balanced red. half-rxn)
11-6 Adding H؉, OH؊, or H2O to Balance Oxygen or Hydrogen 419
These common household
chemicals, ammonia and bleach,
should never be mixed because they
react to form chloramine (NH2Cl), a
very poisonous volatile compound.
NH3(aq) ϩ ClOϪ(aq) 88n
NH2Cl(aq) ϩ OHϪ(aq)
The oxidation half-reaction involves three electrons, and the reduction half-reaction involves
two electrons. We balance the electron transfer and add the half-reactions term by term.
2(CrO2
Ϫ ϩ 4OHϪ 88n CrO4
2Ϫ ϩ 2H2O ϩ 3eϪ)
3(ClOϪ ϩ H2O ϩ 2eϪ 88n ClϪ ϩ 2OHϪ)
2CrO2
Ϫ ϩ 8OHϪ ϩ 3ClOϪ ϩ 3H2O 88n 2CrO4
2Ϫ ϩ 4H2O ϩ 3ClϪ ϩ 6OHϪ
We see 6 OHϪ and 3 H2O that can be eliminated from both sides to give the balanced net
ionic equation.
2CrO2
Ϫ(aq) ϩ 2OHϪ(aq) ϩ 3ClOϪ(aq) 88n 2CrO4
2Ϫ(aq) ϩ H2O(ᐉ) ϩ 3ClϪ(aq)
You should now work Exercise 58.
For brevity, we will refer to this as the
CON method.
420 CHAPTER 11: Reactions in Aqueous Solutions II: Calculations
Problem-Solving Tip: Converting Ionic to Formula Unit Equations
We learned in Section 4-3 how to convert the formula unit equation to the net ionic
equation. To do this, we convert the formulas for all strong electrolytes into their ions, and
then cancel spectator ions from both sides of the equation. In Example 11-16 we reverse
this procedure. To balance this excess charge, we must add negatively charged spectator
ions to combine with the positively charged reactants, and we add positively charged
spectator ions to combine with the negatively charged reactants. Any spectator ions added
to the reactant side of the equation must also be added to the product side. Then we
combine species to give complete formula units. Now we can write total ionic and formula
unit equations for Exercise 11-17 if we know what spectator ions are present. Just for
practice, consider the spectator ions to be Naϩ.
CHANGE-IN-OXIDATION-NUMBER METHOD
The next few examples illustrate this method, which is based on equal total increases and
decreases in oxidation numbers. The general procedure follows.
1. Write as much of the overall unbalanced equation as possible.
2. Assign oxidation numbers (Section 4-4) to find the elements that undergo changes
in oxidation numbers.
3. a. Draw a bracket to connect atoms of the element that is oxidized. Show the
increase in oxidation number per atom. Draw a bracket to connect atoms of the
element that is reduced. Show the decrease in oxidation number per atom.
b. Determine the factors that will make the total increase and decrease in oxidation
numbers equal.
11-7
These common household
chemicals, vinegar and bleach,
should never be mixed because they
react to form chlorine, a very
poisonous gas.
2Hϩ(aq) ϩ ClOϪ(aq) ϩ ClϪ(aq) 88n
Cl2(g) ϩ H2O(ᐉ)
4. Insert coefficients into the equation to make the total increase and decrease in oxidation
numbers equal.
5. Balance the other atoms by inspection. The chart in Section 11-6 will help with the
balancing of H’s and O’s in aqueous solutions.
EXAMPLE 11-18 Balancing Redox Equations (CON method)
Iron(II) chloride reacts with tin(IV) chloride to form iron(III) chloride and tin(II) chloride.
Balance the formula unit equation, and identify the oxidizing and reducing agents.
Plan
We follow the five-step procedure, one step at a time.
Solution
The unbalanced formula unit equation and oxidation numbers (steps 1 and 2) are
FeCl2 ϩ SnCl4 88n SnCl2 ϩ FeCl3
The oxidation number of Fe increases from ϩ2 to ϩ3. Fe2ϩ is the reducing agent; it is oxidized.
The oxidation number of Sn decreases from ϩ4 to ϩ2. Sn4ϩ is the oxidizing agent; it is reduced.
FeCl2 ϩ SnCl4 88n SnCl2 ϩ FeCl3 (Step 3a)
Ϫ2
ϩ1
We make the total increase and decrease in oxidation numbers equal (step 3b).
Oxidation Numbers Change/Atom Equalizing Changes Gives
Fe ϭ ϩ2 88n Fe ϭ ϩ3 ϩ1 2(ϩ1) ϭ ϩ2
Sn ϭ ϩ4 88n Sn ϭ ϩ2 Ϫ2 1(Ϫ2) ϭ Ϫ2
2(ϩ1) ϭ ϩ2 (total increase) 1(Ϫ2) ϭ Ϫ2 (total decrease)
We need 2 Fe2ϩ on each side of the equation (step 4). All elements are now balanced, so Step
5 is not required.
2FeCl2 ϩ SnCl4 88n SnCl2 ϩ 2FeCl3
EXAMPLE 11-19 Balancing Redox Equations (CON method)
Copper is a widely used metal. Before it is welded (brazed), copper is cleaned by dipping it
into nitric acid. HNO3 oxidizes Cu to Cu2ϩ ions and is reduced to NO. The other product is
H2O. Write the balanced net ionic and formula unit equations for the reaction. Excess HNO3
is present.
Plan
In writing ionic equations, we recall that strong acids, strong bases, and most soluble salts are
strong electrolytes. Then we apply our five-step procedure for redox equations.
Solution
We write the unbalanced net ionic equation and assign oxidation numbers. HNO3 is a strong
acid.
11-7 Change-in-Oxidation-Number Method 421
ϩ2 Ϫ1 ϩ4 Ϫ1 ϩ2 Ϫ1 ϩ3 Ϫ1
ϩ2 ϩ4 ϩ2 ϩ3
8n
888n
888888
888888
888888888888888888888888888
888888888888
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 21-3, Balancing Equations for
Redox Reactions.
ϩ5 0 ϩ2 ϩ2
8n
888n
888888
888888
8888888888888888888888888
8888888888
0 ϩ1 0ϩ3
Hϩ(aq) ϩ NO3
Ϫ(aq) ϩ Cu(s) 88n Cu2ϩ(aq) ϩ NO(g) ϩ H2O(ᐉ)
We see that copper is oxidized; it is the reducing agent. Nitrate ions are reduced; they are the
oxidizing agent.
Hϩ ϩ NO3
Ϫ ϩ Cu 88n Cu2ϩ ϩ NO ϩ H2O
ϩ2
Ϫ3
We make the total increase and decrease in oxidation numbers equal.
Oxidation Numbers Change/Atom Equalizing Changes Gives
Cu ϭ 0 88n Cu ϭ ϩ2 ϩ2 3(ϩ2) ϭ ϩ6
N ϭ ϩ5 88n N ϭ ϩ2 Ϫ3 2(Ϫ3) ϭ Ϫ6
Now we balance the redox part of the reaction.
Hϩ ϩ 2NO3
Ϫ ϩ 3Cu 88n 3Cu2ϩ ϩ 2NO ϩ H2O
There are 6 O on the left in NO3
Ϫ ions. A coefficient of 4 before H2O balances O and gives
8 H on the right. So we need 8 Hϩ ions on the left to balance the net ionic equation.
8Hϩ(aq) ϩ 2NO3
Ϫ(aq) ϩ 3Cu(s) 88n 3Cu2ϩ(aq) ϩ 2NO(g) ϩ 4H2O(ᐉ)
This solution contains excess HNO3, so NO3
Ϫ is the only anion present in significant concentration.
Therefore, we add six more NO3
Ϫ ions to each side to give the total ionic equation.
8[Hϩ(aq) ϩ NO3
Ϫ(aq)] ϩ 3Cu(s) 88n 3[Cu2ϩ(aq) ϩ 2NO3
Ϫ(aq)] ϩ 2NO(g) ϩ 4H2O(ᐉ)
Now we can write the balanced formula unit equation.
8HNO3(aq) ϩ 3Cu(s) 88n 3Cu(NO3)2(aq) ϩ 2NO(g) ϩ 4H2O(ᐉ)
You should now work Exercise 52.
EXAMPLE 11-20 Balancing Redox Equations (CON method)
Draෆno drain cleaner is solid sodium hydroxide that contains some aluminum chips. When
Draෆno is added to water, the NaOH dissolves rapidly with the evolution of a lot of heat. The
Al reduces H2O in the basic solution to produce [Al(OH)4]Ϫ ions and H2 gas, which gives the
bubbling action. Write the balanced net ionic and formula unit equations for this reaction.
Plan
We are given formulas for reactants and products. Recall that NaOH is a strong soluble base
(OHϪ and H2O can be added to either side as needed). We apply our five-step procedure.
Solution
We write the unbalanced net ionic equation and assign oxidation numbers.
OHϪ(aq) ϩ Al(s) ϩ H2O(ᐉ) 88n [Al(OH)4]Ϫ(aq) ϩ H2(g)
Copper is cleaned by dipping it into
nitric acid.
422 CHAPTER 11: Reactions in Aqueous Solutions II: Calculations
ϩ5ϩ1 Ϫ2 ϩ2ϩ20 Ϫ2 ϩ1 Ϫ2
The Draෆno reaction.
Aluminum is oxidized; it is the reducing agent. H2O is reduced; it is the oxidizing agent.
OHϪ ϩ H2O ϩ Al 88n [Al(OH)4]Ϫ ϩ H2
ϩ3
Ϫ1
We make the total increase and decrease in oxidation numbers equal.
Oxidation Numbers Change/Atom Equalizing Changes Gives
Al ϭ 0 88n Al ϭ ϩ3 ϩ3 1(ϩ3) ϭ ϩ3
H ϭ ϩ1 88n H ϭ 0 Ϫ1 3(Ϫ1) ϭ Ϫ3
Each change must be multiplied by two because there are 2 H in each H2.
2(ϩ3) ϭ ϩ6 (total increase) 2(Ϫ3) ϭ Ϫ6 (total decrease)
Now we balance the redox part of the equation. We need 2 Al on each side. Because only one
H in each H2O molecule is reduced (the other H is in OHϪ), we show 6 H2O on the left and
3 H2 on the right.
OHϪ ϩ 6H2O ϩ 2Al 88n 2[Al(OH)4]Ϫ ϩ 3H2
The net charge on the right is 2Ϫ, and so we need 2 OHϪ on the left to balance the net ionic
equation.
2OHϪ(aq) ϩ 6H2O(ᐉ) ϩ 2Al(s) 88n 2[Al(OH)4]Ϫ(aq) ϩ 3H2(g)
This reaction occurs in excess NaOH solution. We need 2 Naϩ(aq) on each side to balance
the negative charges.
2NaOH(aq) ϩ 6H2O(ᐉ) ϩ 2Al(s) 88n 2Na[Al(OH)4](aq) ϩ 3H2(g)
You should now work Exercise 58.
EXAMPLE 11-21 Balancing Redox Equations (CON method)
The breathalyzer detects the presence of ethanol (ethyl alcohol) in the breath of persons
suspected of drunken driving. It utilizes the oxidation of ethanol to acetaldehyde by dichromate
ions in acidic solution. The Cr2O7
2Ϫ(aq) ion is orange (see Example 11-23). The Cr3ϩ(aq)
ion is green. The appearance of a green color signals alcohol in the breath that exceeds the
legal limit. Balance the net ionic equation for this reaction.
Hϩ(aq) ϩ Cr2O7
2Ϫ(aq) ϩ C2H5OH(ᐉ) 88n Cr3ϩ(aq) ϩ C2H4O(ᐉ) ϩ H2O(ᐉ)
Plan
We are given the unbalanced equation, which includes Hϩ. This tells us that the reaction occurs
in acidic solution. We apply our five-step procedure.
Solution
We first assign oxidation numbers to the elements that change.
Hϩ ϩ C2H5OH ϩ Cr2O7
2Ϫ 88n Cr3ϩ ϩ C2H4O ϩ H2O(ᐉ)
Many breathalyzers have been
developed over the years. This
model has proved to be effective.
11-7 Change-in-Oxidation-Number Method 423
8n
888n
888888
888888
88888888888888888888888888888
8888888888
0ϩ1 0ϩ3
Ϫ2 ϩ6 ϩ3 Ϫ1
We see that ethanol is oxidized; it is the reducing agent. Cr2O7
2Ϫ ions are reduced; they are
the oxidizing agent.
Hϩ ϩ C2H5OH ϩ Cr2O7
2Ϫ 88n Cr3ϩ ϩ C2H4O ϩ H2O
Ϫ3
ϩ1
Oxidation Numbers Change/Atom Equalizing Changes Gives
Cr ϭ ϩ6 88n Cr ϭ ϩ3 Ϫ3 1(Ϫ3) ϭ Ϫ3
C ϭ Ϫ2 88n C ϭ Ϫ1 ϩ1 3(ϩ1) ϭ ϩ3
Each change must be multiplied by two because there are 2 Cr in each Cr2O7
2Ϫ and 2 C in
C2H5OH.
2(Ϫ3) ϭ Ϫ6 (total decrease) 2(ϩ3) ϭ ϩ6 (total increase)
We need 2 Cr and 6 C on each side of the equation to balance the redox part.
Hϩ ϩ 3C2H5OH ϩ Cr2O7
2Ϫ 88n 2Cr3ϩ ϩ 3C2H4O ϩ H2O
Now we balance H and O using our chart. There are 10 O on the left and only 4 O on the
right. So we add 6 more H2O molecules on the right.
Hϩ ϩ 3C2H5OH ϩ Cr2O7
2Ϫ 88n 2Cr3ϩ ϩ 3C2H4O ϩ 7H2O
Now there are 26 H on the right and only 19 on the left. So we add 7 more Hϩ ions on the
left to give the balanced net ionic equation.
8Hϩ(aq) ϩ 3C2H5OH(aq) ϩ Cr2O7
2Ϫ(aq) 88n 2Cr3ϩ(aq) ϩ 3C2H4O(aq) ϩ 7H2O(ᐉ)
You should now work Exercise 54.
Every balanced equation must have both mass balance and charge balance. Once the redox
part of an equation has been balanced, we must next count either atoms or charges. After
we balanced the redox part in Example 11-21, we had
Hϩ ϩ 3C2H5OH ϩ Cr2O7
2Ϫ 88n 2Cr3ϩ ϩ 3C2H4O ϩ H2O
The net charge on the left side is (1 ϩ 2Ϫ) ϭ 1Ϫ. On the right, it is 2(3ϩ) ϭ 6ϩ. Because
Hϩ is the only charged species whose coefficient isn’t known, we add 7 more Hϩ to give a net
charge of 6ϩ on both sides.
8Hϩ ϩ 3C2H5OH ϩ Cr2O7
2Ϫ 88n 2Cr3ϩ ϩ 3C2H4O ϩ H2O
Now we have 10 O on the left and only 4 O on the right. We add six more H2O molecules
to give the balanced net ionic equation.
8Hϩ(aq) ϩ 3C2H5OH(ᐉ) ϩ Cr2O7
2Ϫ(aq) 88n 2Cr3ϩ(aq) ϩ 3C2H4O(ᐉ) ϩ 7H2O(ᐉ)
How can you tell whether to balance atoms or charges first? Look at the equation after
you have balanced the redox part. Decide which is simpler, and do that. In the preceding
equation, it is easier to balance charges than to balance atoms.
424 CHAPTER 11: Reactions in Aqueous Solutions II: Calculations
Ϫ2 ϩ6 ϩ3 Ϫ1
8n
888n
888888
888888
888888888888888888888888888888888
8888888888888888
STOICHIOMETRY OF REDOX REACTIONS
One method of analyzing samples quantitatively for the presence of oxidizable or reducible
substances is by redox titration. In such analyses, the concentration of a solution is determined
by allowing it to react with a carefully measured amount of a standard solution of
an oxidizing or reducing agent.
As in other kinds of chemical reactions, we must pay particular attention to the mole
ratio in which oxidizing agents and reducing agents react.
Potassium permanganate, KMnO4, is a strong oxidizing agent. Through the years it
has been the “workhorse” of redox titrations. For example, in acidic solution, KMnO4
reacts with iron(II) sulfate, FeSO4, according to the balanced equation in the following
example. A strong acid, such as H2SO4, is used in such titrations (Example 11-15).
A word about terminology. The reaction involves MnO4
Ϫ ions and Fe2ϩ ions in acidic
solution. The source of MnO4
Ϫ ions usually is the soluble ionic compound KMnO4. We
often refer to “permanganate solutions.” Such solutions also contain cations—in this case,
Kϩ. Likewise, we often refer to “iron(II) solutions” without specifying what the anion is.
EXAMPLE 11-22 Redox Titration
What volume of 0.0200 M KMnO4 solution is required to oxidize 40.0 mL of 0.100 M FeSO4
in sulfuric acid solution (Figure 11-2)?
Plan
The balanced equation in Example 11-15 gives the reaction ratio, 1 mol MnO4
Ϫ/5 mol Fe2ϩ.
Then we calculate the number of moles of Fe2ϩ to be titrated, which lets us find the number
of moles of MnO4
Ϫ required and the volume in which this number of moles of KMnO4 is
contained.
11-8
Because it has an intense purple color,
KMnO4 acts as its own indicator. One
drop of 0.020 M KMnO4 solution
imparts a pink color to a liter of pure
water. When KMnO4 solution is added
to a solution of a reducing agent, the
end point in the titration is taken as
the point at which a pale pink color
appears in the solution being titrated
and persists for at least 30 seconds.
11-8 Stoichiometry of Redox Reactions 425
One mole of KMnO4 contains one
mole of MnO4
Ϫ ions. The number of
moles of KMnO4 is therefore always
equal to the number of moles of
MnO4
Ϫ ions required in a reaction.
Similarly, one mole of FeSO4 contains
1 mole of Fe2ϩ ions.
Figure 11-2 (a) Nearly colorless FeSO4 solution is titrated with deep-purple KMnO4.
(b) The end point is the point at which the solution becomes pink, owing to a very small
excess of KMnO4. Here a considerable excess of KMnO4 was added so that the pink color
could be reproduced photographically.
(a) (b)
Solution
The reaction ratio is
MnO4
Ϫ(aq) ϩ 8Hϩ(aq) ϩ 5Fe2ϩ(aq) 88n 5Fe3ϩ(aq) ϩ Mn2ϩ(aq) ϩ 4H2O(ᐉ)
rxn ratio: 1 mol 5 mol
The number of moles of Fe2ϩ to be titrated is
_?_mol Fe2ϩ ϭ 40.0 mL ϫ ϭ 4.00 ϫ 10Ϫ3 mol Fe2ϩ
We use the balanced equation to find the number of moles of MnO4
Ϫ required.
_?_mol MnO4
Ϫ ϭ 4.00 ϫ 10Ϫ3 mol Fe2ϩ ϫ ϭ 8.00 ϫ 10Ϫ4 mol MnO4
Ϫ
Each formula unit of KMnO4 contains one MnO4
Ϫ ion, and so
1 mol KMnO4 Х 1 mol MnO4
Ϫ
The volume of 0.0200 M KMnO4 solution that contains 8.00 ϫ 10Ϫ4 mol of KMnO4 is
_?_mL KMnO4 soln ϭ 8.00 ϫ 10Ϫ4 mol KMnO4 ϫ
ϭ 40.0 mL KMnO4 soln
You should now work Exercises 66 and 68.
Potassium dichromate, K2Cr2O7, is another frequently used oxidizing agent. However,
an indicator must be used when reducing agents are titrated with dichromate solutions.
K2Cr2O7 is orange, and its reduction product, Cr3ϩ, is green.
Consider the oxidation of sulfite ions, SO3
2Ϫ, to sulfate ions, SO4
2Ϫ, by Cr2O7
2Ϫ ions
in the presence of a strong acid such as sulfuric acid. We shall balance the equation by
the half-reaction method.
Cr2O7
2Ϫ 88n Cr3ϩ (red. half-rxn)
Cr2O7
2Ϫ 88n 2Cr3ϩ
14Hϩ ϩ Cr2O7
2Ϫ 88n 2Cr3ϩ ϩ 7H2O
6eϪ ϩ 14Hϩ ϩ Cr2O7
2Ϫ 88n 2Cr3ϩ ϩ 7H2O (balanced red. half-rxn)
SO3
2Ϫ 88n SO4
2Ϫ (ox. half-rxn)
SO3
2Ϫ ϩ H2O 88n SO4
2Ϫ ϩ 2Hϩ
SO3
2Ϫ ϩ H2O 88n SO4
2Ϫ ϩ 2Hϩ ϩ 2eϪ (balanced ox. half-rxn)
We now equalize the electron transfer, add the balanced half-reactions, and eliminate
common terms.
(6eϪ ϩ 14Hϩ ϩ Cr2O7
2Ϫ 88n 2Cr3ϩ ϩ 7H2O) (reduction)
3(SO3
2Ϫ ϩ H2O 88n SO4
2Ϫ ϩ 2Hϩ ϩ 2eϪ) (oxidation)
8Hϩ(aq) ϩ Cr2O7
2Ϫ(aq) ϩ 3SO3
2Ϫ(aq) 88n 2Cr3ϩ(aq) ϩ 3SO4
2Ϫ(aq) ϩ 4H2O(ᐉ)
The balanced equation tells us that the reaction ratio is 3 mol SO3
2Ϫ/mol Cr2O7
2Ϫ or
1 mol Cr2O7
2Ϫ/3 mol SO3
2Ϫ. Potassium dichromate is the usual source of Cr2O7
2Ϫ ions,
1000 mL KMnO4 soln
ᎏᎏᎏ
0.0200 mol KMnO4
1 mol MnO4
Ϫ
ᎏᎏ
5 mol Fe2ϩ
0.100 mol Fe2ϩ
ᎏᎏ
1000 mL
426 CHAPTER 11: Reactions in Aqueous Solutions II: Calculations
and Na2SO3 is the usual source of SO3
2Ϫ ions. Thus, the preceding reaction ratio could
also be expressed as 1 mol K2Cr2O7/3 mol Na2SO3.
EXAMPLE 11-23 Redox Titration
A 20.00-mL sample of Na2SO3 was titrated with 36.30 mL of 0.05130 M K2Cr2O7 solution
in the presence of H2SO4. Calculate the molarity of the Na2SO3 solution.
Plan
We can calculate the number of millimoles of Cr2O7
2Ϫ in the standard solution. Then we refer
to the balanced equation in the preceding discussion, which gives us the reaction ratio, 3 mmol
SO3
2Ϫ/1 mmol Cr2O7
2Ϫ. The reaction ratio lets us calculate the number of millimoles of
SO3
2Ϫ (Na2SO3) that reacted and the molarity of the solution.
mL Cr2O7
2Ϫ soln 88n mmol Cr2O7
2Ϫ 88n mmol SO3
2Ϫ 88n M SO3
2Ϫ soln
Solution
From the preceding discussion we know the balanced equation and the reaction ratio.
3SO3
2Ϫ ϩ Cr2O7
2Ϫ ϩ 8Hϩ 88n 3SO4
2Ϫ ϩ 2Cr3ϩ ϩ 4H2O
3 mmol 1 mmol
The number of millimoles of Cr2O7
2Ϫ used is
_?_mmol Cr2O7
2Ϫ ϭ 36.30 mL ϫ ϭ 1.862 mmol Cr2O7
2Ϫ
The number of millimoles of SO3
2Ϫ that reacted with 1.862 mmol of Cr2O7
2Ϫ is
_?_mmol SO3
2Ϫ ϭ 1.862 mmol Cr2O7
2Ϫ ϫ ϭ 5.586 mmol SO3
2Ϫ
The Na2SO3 solution contained 5.586 mmol of SO3
2Ϫ (or 5.586 mmol of Na2SO3). Its molarity
is
_?_ ϭ ϭ 0.2793 M Na2SO3
You should now work Exercise 70.
5.586 mmol Na2SO3
ᎏᎏᎏ
20.00 mL
mmol Na2SO3
ᎏᎏ
mL
3 mmol SO3
2Ϫ
ᎏᎏ
1 mmol Cr2O7
2Ϫ
0.05130 mmol Cr2O7
2Ϫ
ᎏᎏᎏ
mL
Key Terms 427
Key Terms
Buret A piece of volumetric glassware, usually graduated in
0.1-mL intervals, that is used in titrations to deliver solutions
in a quantitative (dropwise) manner.
End point The point at which an indicator changes color and a
titration is stopped.
Equivalence point The point at which chemically equivalent
amounts of reactants have reacted.
Equivalent weight in acid–base reactions The mass of an acid
or base that furnishes or reacts with 6.022 ϫ 1023 H3Oϩ or
OHϪ ions.
Half-reaction Either the oxidation part or the reduction part of
a redox reaction.
Indicator For acid–base titrations, an organic compound that
exhibits its different colors in solutions of different acidities;
used to determine the point at which the reaction between two
solutes is complete.
Millimole 1/1000 mole.
Molarity (M) The number of moles of solute per liter of solution
or the number of millimoles of solute per milliliter of
solution.
Normality (N) The number of equivalent weights (equivalents)
of solute per liter of solution.
Oxidation An algebraic increase in oxidation number; may correspond
to a loss of electrons.
Cr2(SO4)3 is green in acidic solution.
K2Cr2O7 is orange in acidic
solution.
428 CHAPTER 11: Reactions in Aqueous Solutions II: Calculations
Exercises
Molarity
*01. Why can we describe molarity as a “method of convenience”
for expressing concentrations of solutions?
*02. Why is the molarity of a solution the same number whether
we describe it in mol/L or in mmol/mL?
*03. Calculate the molarities of solutions that contain the following
masses of solute in the indicated volumes: (a) 45 g
of H3AsO4 in 500. mL of solution; (b) 8.3 g of (COOH)2
in 600. mL of solution; (c) 8.25 g of (COOH)2и2H2O in
750. mL of solution.
*04. What is the molarity of a solution made by dissolving
132.6 g of magnesium sulfate in sufficient water to produce
a total of 3.500 L?
*05. There is 75.0 g of iron(II) nitrate present in 750. mL of a
solution. Calculate the molarity of that solution.
*06. Calculate the molarity of a solution that is 39.77% H2SO4
by mass. The specific gravity of the solution is 1.305.
*07. Calculate the molarity of a solution that is 19.0% HNO3
by mass. The specific gravity of the solution is 1.11.
*08. If 150. mL of 4.32 M HCl solution is added to 300. mL
of 2.16 M NaOH solution, the resulting solution will be
______ molar in NaCl.
*09. What is the molarity of the salt solution produced when
500. mL of 3.00 M HCl and 500. mL of 3.00 M LiOH are
mixed? (Assume that the volumes are additive.) Give the
name and formula of the salt formed.
*10. Potassium iodide is sometimes used as a sodium chloride
replacement for those people who cannot tolerate table
salt. Calculate the molarity of potassium iodide solution
produced when 25.0 mL of 9.00 M HI and 25.0 mL of
9.00 M KOH are mixed.
*11. Calculate the molarity of the salt solution produced when
3.60 mL of 0.00100 M Mg(OH)2 and 3.60 mL of 0.00100
M H2SO4 are mixed. Give the name and formula of the
salt formed.
*12. What is the concentration of the salt solution produced
when 32.5 mL of 2.00 M H2SO4 and 32.5 mL of 4.00 M
NaOH are mixed? Give the name and formula of the salt
formed.
*13. What is the salt concentration produced if we mix 8.00 mL
of 3.00 M HCl with 3.00 mL of 4.00 M Ba(OH)2? Give
the name and formula of the salt formed.
*14. What is the concentration of barium iodide produced by
mixing 5.00 mL of 0.125 M Ba(OH)2 with 12.0 mL of
0.0650 M HI?
*15. What is the concentration of the ammonium chloride produced
when 44.0 mL of 12.0 M HCl and 37.0 mL of 8.00
M NH3 are mixed?
*16. If 225 mL of 3.68 M H3PO4 solution is added to 775 mL
of 3.68 M NaOH solution, the resulting solution will be
______ molar in Na3PO4 and ______ molar in ______.
*17. If 400 mL of 0.200 M HCl solution is added to 800. mL
of 0.0400 M Ba(OH)2 solution, the resulting solution will
be ______ molar in BaCl2 and ______ molar in ______.
*18. What volume of 0.0150 M acetic acid solution would completely
neutralize 18.7 mL of 0.0105 M Ba(OH)2 solution?
*19. What volume of 0.300 M potassium hydroxide solution
would completely neutralize 35.0 mL of 0.100 M H2SO4
solution?
*20. A vinegar solution is 5.11% acetic acid. Its density is 1.007
g/mL. What is its molarity?
*21. A household ammonia solution is 5.03% ammonia. Its density
is 0.979 g/mL. What is its molarity?
*22. (a) What volumes of 1.50 M NaOH and 3.00 M H3PO4
solutions would be required to form 1.00 mol of Na3PO4?
(b) What volumes of the solutions would be required to
form 1.00 mol of Na2HPO4?
Oxidation–reduction reaction A reaction in which oxidation
and reduction occur; also called redox reaction.
Oxidizing agent The substance that oxidizes another substance
and is reduced.
Primary standard A substance of a known high degree of purity
that undergoes one invariable reaction with the other reactant
of interest.
Redox reaction An oxidation-reduction reaction.
Redox titration The quantitative analysis of the amount or concentration
of an oxidizing or reducing agent in a sample by
observing its reaction with a known amount or concentration
of a reducing or oxidizing agent.
Reducing agent The substance that reduces another substance
and is oxidized.
Reduction An algebraic decrease in oxidation number; may correspond
to a gain of electrons.
Secondary standard A solution that has been titrated against
a primary standard. A standard solution in a secondary
standard.
Standard solution A solution of accurately known concentra-
tion.
Standardization The process by which the concentration of a
solution is accurately determined by titrating it against an accurately
known amount of a primary standard.
Titration The process by which the volume of a standard solution
required to react with a specific amount of a substance is
determined.
Exercises 429
Standardization and Acid–Base Titrations:
Mole Method
*23. Define and illustrate the following terms clearly and
concisely: (a) standard solution; (b) titration; (c) primary
standard; (d) secondary standard.
*24. Describe the preparation of a standard solution of
NaOH, a compound that absorbs both CO2 and H2O from
the air.
*25. Distinguish between the net ionic equation and the formula
unit equation.
*26. (a) What is potassium hydrogen phthalate, KHP? (b) For
what is it used?
*27. Why can sodium carbonate be used as a primary standard
for solutions of acids?
*28. Calculate the molarity of a solution of HNO3 if 35.72 mL
of the solution neutralizes 0.302 g of Na2CO3.
*29. If 41.38 mL of a sulfuric acid solution reacts completely
with 0.3545 g of Na2CO3, what is the molarity of the sulfuric
acid solution?
*30. A solution of sodium hydroxide is standardized against
potassium hydrogen phthalate. From the following data,
calculate the molarity of the NaOH solution.
mass of KHP 0.6536
buret reading before titration 0.23 mL
buret reading after titration 31.26 mL
*31. Calculate the molarity of a KOH solution if 40.68 mL of
the KOH solution reacted with 0.4084 g of potassium
hydrogen phthalate, KHP.
*32. Calcium carbonate tablets can be used as an antacid and a
source of dietary calcium. A bottle of generic antacid tablets
states that each tablet contains 500. mg calcium carbonate.
What volume of 6.0 M HNO3 could be neutralized by the
calcium carbonate in one tablet?
*33. What volume of 18.0 M H2SO4 is required to react with
100. mL of 6.00 M NaOH to produce a Na2SO4 solution?
What volume of water must be added to the resulting solution
to obtain a 1.25 M Na2SO4 solution?
*34. (a) What are the properties of an ideal primary standard?
(b) What is the importance of each property?
*35. The secondary standard solution of NaOH of Exercise 30
was used to titrate a solution of unknown concentration of
HCl. A 30.00-mL sample of the HCl solution required
34.21 mL of the NaOH solution for complete neutralization.
What is the molarity of the HCl solution?
*36. An impure sample of (COOH)2и2H2O that had a mass of
2.00 g was dissolved in water and titrated with standard
NaOH solution. The titration required 38.32 mL of 0.198
M NaOH solution. Calculate the percent (COOH)2и2H2O
in the sample. Assume that the sample contains no acidic
impurities.
*37. A 50.0-mL sample of 0.0500 M Ca(OH)2 is added to 10.0
mL of 0.200 M HNO3. (a) Is the resulting solution acidic
or basic? (b) How many moles of excess acid or base are
present? (c) How many additional mL of 0.0500 M
Ca(OH)2 or 0.200 M HNO3 would be required to completely
neutralize the solution?
*38. An antacid tablet containing calcium carbonate as an active
ingredient required 24.5 mL of 0.0932 M HCl for complete
neutralization. What mass of CaCO3 did the tablet
contain?
*39. Butyric acid, whose empirical formula is C2H4O, is the acid
responsible for the odor of rancid butter. The acid has one
ionizable hydrogen per molecule. A 1.000-g sample of
butyric acid is neutralized by 54.42 mL of 0.2088 M NaOH
solution. What are (a) the molecular weight and (b) the
molecular formula of butyric acid?
*40. What is the molarity of a solution of sodium hydroxide,
NaOH, if 36.2 mL of this solution is required to react with
25.0 mL of 0.0513 M nitric acid solution according to the
following reaction?
HNO3 ϩ NaOH 88n NaNO3 ϩ H2O
*41. What is the molarity of a solution of sodium hydroxide,
NaOH, if 36.9 mL of this solution is required to react with
35.2 mL of 0.101 M hydrochloric acid solution according
to the following reaction?
HCl ϩ NaOH 88n NaCl ϩ H2O
Standardization and Acid–Base Titrations:
Equivalent Weight Method
In answering Exercises 42–51, assume that the acids and bases will
be completely neutralized.
*42. What is the normality of each of the following acid or base
solutions? (a) 0.35 M HCl; (b) 0.35 M H2SO4; (c) 0.35 M
H3PO4; (d) 0.35 M NaOH.
*43. What is the normality of each of the following acid or base
solutions? (a) 0.105 M Ca(OH)2; (b) 0.105 M Al(OH)3;
(c) 0.105 M HNO3; (d) 0.105 M H2Se.
*44. What is the normality of a solution that contains 7.08 g of
H3PO4 in 185 mL of solution?
*45. Calculate the molarity and the normality of a solution that
was prepared by dissolving 24.2 g of barium hydroxide in
enough water to make 4000. mL of solution.
*46. Calculate the molarity and the normality of a solution that
contains 19.6 g of arsenic acid, H3AsO4, in enough water
to make 500. mL of solution.
*47. What are the molarity and normality of a sulfuric acid solution
that is 19.6% H2SO4 by mass? The density of the
solution is 1.14 g/mL.
*48. A 25.0-mL sample of 0.206 normal nitric acid solution
required 35.2 mL of barium hydroxide solution for neutralization.
Calculate the molarity of the barium hydroxide
solution.
*49. Vinegar is an aqueous solution of acetic acid, CH3COOH.
Suppose you titrate a 25.00-mL sample of vinegar with
430 CHAPTER 11: Reactions in Aqueous Solutions II: Calculations
17.62 mL of a standardized 0.1060 N solution of NaOH.
(a) What is the normality of acetic acid in this vinegar?
(b) What is the mass of acetic acid contained in 1.000 L of
vinegar?
*50. Calculate the normality and molarity of an HCl solution
if 43.1 mL of the solution reacts with 0.318 g of Na2CO3.
2HCl ϩ Na2CO3 88n 2NaCl ϩ CO2 ϩ H2O
*51. Calculate the normality and molarity of an H2SO4 solution
if 40.0 mL of the solution reacts with 0.424 g of
Na2CO3.
H2SO4 ϩ Na2CO3 88n Na2SO4 ϩ CO2 ϩ H2O
Balancing Redox Equations
In Exercises 52 and 53, write balanced formula unit equations for
the reactions described by words.
*52. (a) Iron reacts with hydrochloric acid to form aqueous
iron(II) chloride and gaseous hydrogen. (b) Chromium
reacts with sulfuric acid to form aqueous chromium(III)
sulfate and gaseous hydrogen. (c) Tin reacts with concentrated
nitric acid to form tin(IV) oxide, nitrogen dioxide,
and water.
*53. (a) Carbon reacts with hot concentrated nitric acid to form
carbon dioxide, nitrogen dioxide, and water. (b) Sodium
reacts with water to form aqueous sodium hydroxide and
gaseous hydrogen. (c) Zinc reacts with sodium hydroxide
solution to form aqueous sodium tetrahydroxozincate and
gaseous hydrogen. (The tetrahydroxozincate ion is
[Zn(OH)4]2Ϫ.)
*54. Balance the following equations. For each equation tell
what is oxidized, what is reduced, what is the oxidizing
agent, and what is the reducing agent.
(a) Zn(s) ϩ HClO4(aq) 88n Zn(ClO4)2(aq) ϩ H2(g)
(b) K(s) ϩ H2O(ᐉ) 88n KOH(aq) ϩ H2(g)
heat
(c) NaClO3(s) 8888n NaCl(s) ϩ O2(g)
*55. Balance the following equations. For each equation tell
what is oxidized, what is reduced, what is the oxidizing
agent, and what is the reducing agent.
heat
(a) Cu(NO3)2(s) 888n CuO(s) ϩ NO2(g) ϩ O2(g)
(b) Hg2Cl2(s) ϩ NH3(aq) 88n
Hg(ᐉ) ϩ HgNH2Cl(s) ϩ NH4
ϩ(aq) ϩ ClϪ(aq)
(c) Ba(s) ϩ H2O(ᐉ) 88n Ba(OH)2(aq) ϩ H2(g)
*56. Balance the following equations. For each equation tell
what is oxidized, what is reduced, what is the oxidizing
agent, and what is the reducing agent.
(a) MnO4
Ϫ(aq) ϩ Hϩ(aq) ϩ BrϪ(aq) 88n
Mn2ϩ(aq) ϩ Br2(ᐉ) ϩ H2O(ᐉ)
(b) Cr2O7
2Ϫ(aq) ϩ Hϩ(aq) ϩ IϪ(aq) 88n
Cr3ϩ(aq) ϩ I2(s) ϩ H2O(ᐉ)
(c) MnO4
Ϫ(aq) ϩ SO3
2Ϫ(aq) ϩ Hϩ(aq) 88n
Mn2ϩ(aq) ϩ SO4
2Ϫ(aq) ϩ H2O(ᐉ)
(d) Cr2O7
2Ϫ(aq) ϩ Fe2ϩ(aq) ϩ Hϩ(aq) 88n
Cr3ϩ(aq) ϩ Fe3ϩ(aq) ϩ H2O(ᐉ)
*57. Balance the following ionic equations. For each equation
tell what is oxidized, what is reduced, what is the oxidizing
agent, and what is the reducing agent.
(a) C2H4(g) ϩ MnO4
Ϫ(aq) ϩ Hϩ(aq) n
CO2(g) ϩ Mn2ϩ(aq) ϩ H2O(ᐉ)
(b) H2S(aq) ϩ Hϩ(aq) ϩ Cr2O7
2Ϫ(aq) n
Cr3ϩ(aq) ϩ S(s) ϩ H2O(ᐉ)
(c) ClO3
Ϫ(aq) ϩ H2O(ᐉ) ϩ I2(s) n
IO3
Ϫ(aq) ϩ ClϪ(aq) ϩ Hϩ(aq)
(d) Cu(s) ϩ Hϩ(aq) ϩ SO4
2Ϫ(aq) n
Cu2ϩ(aq) ϩ H2O(ᐉ) ϩ SO2(g)
*58. Balance the following ionic equations. For each equation
tell what is oxidized, what is reduced, what is the oxidizing
agent, and what is the reducing agent.
(a) Al(s) ϩ NO3
Ϫ(aq) ϩ OHϪ(aq) ϩ H2O n
Al(OH)4
Ϫ(aq) ϩ NH3(g)
(b) NO2(g) ϩ OHϪ(aq) n
NO3
Ϫ(aq) ϩ NO2
Ϫ(aq) ϩ H2O(ᐉ)
(c) MnO4
Ϫ(aq) ϩ H2O(ᐉ) ϩ NO2
Ϫ(aq) n
MnO2(s) ϩ NO3
Ϫ(aq) ϩ OHϪ(aq)
*59. Balance the following ionic equations. For each equation
tell what is oxidized, what is reduced, what is the oxidizing
agent, and what is the reducing agent.
(a) Cr(OH)4
Ϫ(aq) ϩ OHϪ(aq) ϩ H2O2(aq) n
CrO4
2Ϫ(aq) ϩ H2O(ᐉ)
(b) MnO2(s) ϩ Hϩ(aq) ϩ NO2
Ϫ(aq) n
NO3
Ϫ(aq) ϩ Mn2ϩ(aq) ϩ H2O(ᐉ)
(c) Sn(OH)3
Ϫ(aq) ϩ Bi(OH)3(s) ϩ OHϪ(aq) n
Sn(OH)6
2Ϫ(aq) ϩ Bi(s)
(d) CrO4
2Ϫ(aq) ϩ H2O(ᐉ) ϩ HSnO2
Ϫ(aq) n
CrO2
Ϫ(aq) ϩ OHϪ(aq) ϩ HSnO3
Ϫ(aq)
*60. Balance the following ionic equations for reactions in acidic
solution. Hϩ or H2O (but not OHϪ) may be added as nec-
essary.
(a) Fe2ϩ(aq) ϩ MnO4
Ϫ(aq) n Fe3ϩ(aq) ϩ Mn2ϩ(aq)
(b) Br2(ᐉ) ϩ SO2(g) n BrϪ(aq) ϩ SO4
2Ϫ(aq)
(c) Cu(s) ϩ NO3
Ϫ(aq) n Cu2ϩ(aq) ϩ NO2(g)
(d) PbO2(s) ϩ ClϪ(aq) n PbCl2(s) ϩ Cl2(g)
(e) Zn(s) ϩ NO3
Ϫ(aq) n Zn2ϩ(aq) ϩ N2(g)
*61. Balance the following ionic equations for reactions in acidic
solution. Hϩ or H2O (but not OHϪ) may be added as nec-
essary.
(a) P4(s) ϩ NO3
Ϫ(aq) n H3PO4(aq) ϩ NO(g)
(b) H2O2(aq) ϩ MnO4
Ϫ(aq) n Mn2ϩ(aq) ϩ O2(g)
(c) HgS(s) ϩ ClϪ(aq) ϩ NO3
Ϫ(aq) n
HgCl4
2Ϫ(aq) ϩ NO2(g) ϩ S(s)
(d) HBrO(aq) n BrϪ(aq) ϩ O2(g)
*62. Write the balanced net ionic equations for the reactions
given. Then, using the reactants shown in parentheses convert
each balanced net ionic equation to a balanced formula
unit equation.
(a) MnO4
Ϫ ϩ C2O4
2Ϫ ϩ Hϩ n Mn2ϩ ϩ CO2 ϩ H2O
(KMnO4, HCl, and K2C2O4)
(b) Zn ϩ NO3
Ϫ ϩ Hϩ n Zn2ϩ ϩ NH4
ϩ ϩ H2O
(Zn(s) and HNO3)
Exercises 431
*63. Write the balanced net ionic equations for the reactions
given. Then, using the reactants shown in parentheses convert
each balanced net ionic equation to a balanced formula
unit equation.
(a) I2 ϩ S2O3
2Ϫ n IϪ ϩ S4O6
2Ϫ
(I2 and Na2S2O3)
(b) IO3
Ϫ ϩ N2H4 ϩ ClϪ ϩ Hϩ n N2 ϩ ICl2
Ϫ ϩ H2O
(NaIO3 ϩ N2H4, and HCl)
*64. Write the balanced net ionic equations for the reactions
given. Then, using the reactants shown in parentheses convert
each balanced net ionic equation to a balanced
formula unit equation.
(a) Zn ϩ Cu2ϩ n Cu ϩ Zn2ϩ (Zn and CuSO4)
(b) Cr ϩ Hϩ n Cr3ϩ ϩ H2 (Cr and H2SO4)
*65. Write the balanced net ionic equations for the reactions
given. Then, using the reactants shown in parentheses convert
each balanced net ionic equation to a balanced formula
unit equation.
(a) Cl2 ϩ OHϪ n ClO3
Ϫ ϩ ClϪ ϩ H2O
(Cl2 and hot NaOH)
(b) Pb ϩ Hϩ ϩ BrϪ n PbBr2(s) ϩ H2 (Pb(s) and HBr)
Redox Titrations: Mole Method and Molarity
*66. What volume of 0.150 M KMnO4 would be required to
oxidize 25.0 mL of 0.100 M FeSO4 in acidic solution? Refer
to Example 11-22.
*67. What volume of 0.150 M K2Cr2O7 would be required to
oxidize 70.0 mL of 0.100 M Na2SO3 in acidic solution?
The products include Cr3ϩ and SO4
2Ϫ ions. Refer to
Example 11-23.
*68. What volume of 0.200 M KMnO4 would be required
to oxidize 40.0 mL of 0.100 M KI in acidic solution?
Products include Mn2ϩ and I2.
*69. What volume of 0.200 M K2Cr2O7 would be required
to oxidize 50.0 mL of 0.150 M KI in acidic solution?
Products include Cr3ϩ and I2.
*70. (a) A solution of sodium thiosulfate, Na2S2O3, is 0.1442 M.
40.00 mL of this solution reacts with 26.36 mL of I2 solution.
Calculate the molarity of the I2 solution.
2Na2S2O3 ϩ I2 88n Na2S4O6 ϩ 2NaI
(b) 25.32 mL of the I2 solution is required to titrate a
sample containing As2O3. Calculate the mass of As2O3
(197.8 g/mol) in the sample.
As2O3 ϩ 5H2O ϩ 2I2 88n 2 H3AsO4 ϩ 4HI
*71. Copper(II) ions, Cu2ϩ, can be determined by the net reac-
tion
2Cu2ϩ ϩ 2IϪ ϩ 2S2O3
2Ϫ 88n 2CuI(s) ϩ S4O6
2Ϫ
A 2.075-g sample containing CuSO4 and excess KI is
titrated with 32.55 mL of 0.1214 M solution of Na2S2O3.
What is the percent CuSO4 (159.6 g/mol) in the sample?
*72. What volume of 3.0 M nitrate ion solution would be
required to react with 25. mL of 0.80 M sulfide ion solution?
(Hint: The equation is not balanced.)
NO3
Ϫ ϩ S2Ϫ n NO ϩ S (acidic solution)
*73. The iron in a 5.675-g sample containing some Fe2O3 is
reduced to Fe2ϩ. The Fe2ϩ is titrated with 12.02 mL of
0.1467 M K2Cr2O7 in an acid solution.
6Fe2ϩ ϩ Cr2O7
2Ϫ ϩ 14Hϩ 88n
6Fe3ϩ ϩ 2Cr3ϩ ϩ 7H2O
Find (a) the mass of Fe and (b) the percentage of Fe in the
sample.
*74. Calculate the molarity of a solution that contains 12.6 g of
KMnO4 in 500. mL of solution to be used in the reaction
that produces MnO4
2Ϫ ions as the reduction product.
*75. A 0.783-g sample of an ore of iron is dissolved in acid and
converted to Fe(II). The sample is oxidized by 38.50 mL
of 0.161 M ceric sulfate, Ce(SO4)2, solution; the cerium(IV)
ion, Ce4ϩ, is reduced to Ce3ϩ ion. (a) Write a balanced
equation for the reaction. (b) What is the percent iron in
the ore?
Mixed Exercises
*76. Calculate the molarity of a hydrochloric acid solution if
32.75 mL of it reacts with 0.4011 g of sodium carbonate.
*77. Calculate the molarity and the normality of a sulfuric acid
solution if 32.75 mL of it reacts with 0.4011 g of sodium
carbonate.
*78. Find the number of mmols of HCl that reacts with 25.5
mL of 0.110 M NaOH. What volume of 0.303 M HCl is
needed to furnish this amount of HCl?
*79. What is the composition of the final solution when 25.5
mL of 0.110 M NaOH and 25.5 mL of 0.205 M HCl solutions
are mixed?
*80. What volume of 0.1123 M HCl is needed to neutralize
completely 1.58 g of Ca(OH)2?
*81. What mass of NaOH is needed to neutralize 45.50 mL of
0.1036 M HCl? If the NaOH is available as a 0.1333 M
aqueous solution, what volume will be required?
*82. What volume of 0.246 M H2SO4 solution would be
required to neutralize completely 39.4 mL of 0.302 M
KOH solution?
*83. What volume of 0.388 N H2SO4 solution would be
required to neutralize completely 37.4 mL of 0.302 N
KOH solution?
*84. What volume of 0.1945 normal sodium hydroxide would
be required to neutralize completely 38.38 mL of 0.1023
normal H2SO4 solution?
*85. Benzoic acid, C6H5COOH, is sometimes used as a primary
standard for the standardization of solutions of bases. A
1.862-g sample of the acid is neutralized by 33.00 mL of
NaOH solution. What is the molarity of the base solution?
C6H5COOH(s) ϩ NaOH(aq) 88n
C6H5COONa(aq) ϩ H2O(ᐉ)
432 CHAPTER 11: Reactions in Aqueous Solutions II: Calculations
*86. Find the volume of 0.250 M HI solution required to titrate
(a) 25.0 mL of 0.100 M NaOH
(b) 5.03 g of AgNO3 (Agϩ ϩ IϪ n AgI(s))
(c) 0.621 g CuSO4 (2Cu2ϩ ϩ 4IϪ n 2CuI(s) ϩ I2(s))
CONCEPTUAL EXERCISES
*87. Describe how you could prepare 1.00 L of 1.00 ϫ 10Ϫ6 M
NaCl solution by using a balance that can only measure
masses to 0.01 g.
*88. Ascorbic acid (vitamin C), along with many other reputed
properties, acts as an antioxidant. The following equation
illustrates its antioxidant properties.
H2C6H6O6 88n C6H6O6 ϩ H2
neutralize 4750 L of solution containing 1.50 g of
Mg(OH)2 per liter.
*92. Silver nitrate and calcium chloride solutions produce a heavy,
white precipitate when mixed. Chemical analysis indicates
that the precipitate is silver chloride. What mass of silver
chloride would be produced if 45 mL of 6.0 M silver nitrate
is mixed with 45 mL of 6.0 M calcium chloride?
*93. A 0.500-g sample of a crystalline monoprotic acid was dissolved
in sufficient water to produce 100. mL of solution.
Neutralization of the resulting solution required 75.0 mL
of 0.150 M NaOH. How many moles of the acid were
present in the initial acid solution?
*94. The typical concentration of HCl in stomach acid (digestive
juice) is a concentration of about 8.0 ϫ 10Ϫ2 M. One
experiences “acid stomach” when the stomach contents
reach about 1.0 ϫ 10Ϫ1 M HCl. One antacid tablet contains
334 mg of active ingredient, NaAl(OH)2CO3. Assume
that you have acid stomach and that your stomach contains
800. mL of 1.0 ϫ 10Ϫ1 M HCl. Calculate the number of
mmol of HCl in the stomach and the number of mmol of
HCl that the tablet can neutralize. Which is greater? (The
neutralization reaction produces NaCl, AlCl3, CO2, and
H2O.)
*95. Refer to Exercises 20 and 21. Notice that the percent by
mass of solute is nearly the same for both solutions. How
many moles of solute are present per liter of each solution?
Are the moles of solute per liter also nearly equal? Why
or why not?
*96. The etching of glass by hydrofluoric acid may be represented
by the simplified reaction of silica with HF.
SiO2(s) ϩ HF(aq) 88n H2SiF6(aq) ϩ H2O(ᐉ)
This is an acid–base reaction in which a weak acid is used
to produce an even weaker acid. Is it also an oxidation–
reduction reaction? Balance the equation.
*97. Write a Lewis formula for the anion SiF6
2Ϫ that would be
produced from the weak acid H2SiF6. Use the VSEPR theory
to predict the shape of SiF6
2Ϫ.
*98. Baking soda, NaHCO3, used to be a common remedy for
“acid stomach.” What weight of baking soda would be
required to neutralize 85 mL of digestive juice, corresponding
in acidity to 0.17 M HCl?
*99. Oxalic acid, a poisonous compound, is found in certain vegetables
such as spinach and rhubarb, but in concentrations
well below toxic limits. The manufacturers of a spinach
juice concentration routinely test their product using an
oxalic acid analysis to avoid any problems from an unexpectedly
high concentration of this chemical. A titration
with potassium permanganate is used for the oxalic acid
assay, according to the following net equation.
5H2C2O4 ϩ 2MnO4
Ϫ ϩ 6Hϩ 88n
10CO2 ϩ 2Mn2ϩ ϩ 8H2O
Calculate the molarity of an oxalic acid solution requiring
23.2 mL of 0.127 M permanganate for a 25.0 mL portion
of the solution.
What is an antioxidant? Assign oxidation numbers. Is
vitamin C oxidized or reduced in this reaction?
BUILDING YOUR KNOWLEDGE
*89. For the formation of 1.00 mol of water, which reaction uses
the most nitric acid?
(a) 3Cu(s) ϩ 8HNO3(aq) n
3Cu(NO3)2(aq) ϩ 2NO(g) ϩ 4H2O(ᐉ)
(b) Al2O3(s) ϩ 6HNO3(aq) n 2Al(NO3)3(aq) ϩ 3H2O(ᐉ)
(c) 4Zn(s) ϩ 10HNO3(aq) n
4Zn(NO3)2(aq) ϩ NH4NO3(aq) ϩ 3H2O(ᐉ)
*90. Limonite is an ore of iron that contains 2Fe2O3и3H2O. A
0.5166-g sample of limonite is dissolved in acid and treated
so that all the iron is converted to ferrous ions, Fe2ϩ. This
sample requires 42.96 mL of 0.02130 M sodium dichromate
solution, Na2Cr2O7, for titration. Fe2ϩ is oxidized to
Fe3ϩ, and Cr2O7
2Ϫ is reduced to Cr3ϩ. What is the percent
iron in the limonite? If your answer had been over
100% limonite, what conclusion could you make, presuming
that the analytical data are correct?
*91. One of the troublesome products of a water treatment plant
in some areas of the country is Mg(OH)2, a gelatinous precipitate
formed during water softening. A suggestion was
made that instead of shoveling the precipitate out of the
pool during cleaning, the Mg(OH)2 could be neutralized
with hydrochloric acid to produce a soluble compound,
MgCl2. Then the pool could be flushed out with fresh
water. Calculate the volume of 12.0 M HCl necessary to
12Gases and the
Kinetic–Molecular
Theory
OUTLINE
12-1 Comparison of Solids, Liquids,
and Gases
12-2 Composition of the Atmosphere
and Some Common Properties
of Gases
12-3 Pressure
12-4 Boyle’s Law: The Volume–
Pressure Relationship
12-5 Charles’s Law: The Volume–
Temperature Relationship; The
Absolute Temperature Scale
12-6 Standard Temperature and
Pressure
12-7 The Combined Gas Law
Equation
12-8 Avogadro’s Law and the
Standard Molar Volume
12-9 Summary of Gas Laws: The
Ideal Gas Equation
12-10 Determination of Molecular
Weights and Molecular
Formulas of Gaseous
Substances
12-11 Dalton’s Law of Partial
Pressures
12-12 Mass–Volume Relationships in
Reactions Involving Gases
12-13 The Kinetic–Molecular
Theory
12-14 Diffusion and Effusion of
Gases
12-15 Real Gases: Deviations from
Ideality
OBJECTIVES
After you have studied this chapter, you should be able to
• List the properties of gases and compare gases, liquids, and solids
• Describe how pressure is measured
• Use and understand the absolute (Kelvin) temperature scale
• Describe the relationships among pressure, volume, temperature, and amount of gas
(Boyle’s Law, Charles’s Law, Avogadro’s Law, and the Combined Gas Law), and the
limitations of each
• Use Boyle’s Law, Charles’s Law, Avogadro’s Law, and the Combined Gas Law, as
appropriate, to calculate changes in pressure, volume, temperature, and amount of gas
• Calculate gas densities and the standard molar volume
• Use the ideal gas equation to do pressure, volume, temperature, and mole calculations
as related to gas samples
• Determine molecular weights and formulas of gaseous substances from measured
properties of gases
• Describe how mixtures of gases behave and predict their properties (Dalton’s Law of
Partial Pressures)
• Carry out calculations about the gases involved in chemical reactions
• Apply the kinetic–molecular theory of gases and describe how this theory is consistent
with the observed gas laws
• Describe molecular motion, diffusion, and effusion of gases
• Describe the molecular features that are responsible for nonideal behavior of real gases
and explain when this nonideal behavior is important
Immersed in water, this green plant
oxidizes the water to form bubbles
of gaseous oxygen, O2.
COMPARISON OF SOLIDS, LIQUIDS, AND GASES
Matter exists in three physical states: solids, liquids, and gases. In the solid state H2O is
known as ice, in the liquid state it is called water, and in the gaseous state it is known as
steam or water vapor. Most, but not all, substances can exist in all three states. Most solids
change to liquids and most liquids change to gases as they are heated. Liquids and gases
are known as fluids because they flow freely. Solids and liquids are referred to as
condensed states because they have much higher densities than gases. Table 12-1 displays
the densities of a few common substances in different physical states.
As the data in Table 12-1 indicate, solids and liquids are many times denser than gases.
The molecules must be very far apart in gases and much closer together in liquids and
solids. For example, the volume of one mole of liquid water is about 18 milliliters, whereas
one mole of steam occupies about 30,600 milliliters at 100°C and atmospheric pressure.
Gases are easily compressed, and they completely fill any container in which they are
present. This tells us that the molecules in a gas are far apart relative to their sizes and
that interactions among them are weak. The possibilities for interaction among gaseous
molecules would be minimal (because they are so far apart) were it not for their rapid
motion.
All substances that are gases at room temperature may be liquefied by cooling and
compressing them. Volatile liquids are easily converted to gases at room temperature or
slightly above. The term vapor refers to a gas that is formed by evaporation of a liquid
or sublimation of a solid. We often use this term when some of the liquid or solid remains
in contact with the gas.
COMPOSITION OF THE ATMOSPHERE AND
SOME COMMON PROPERTIES OF GASES
Many important chemical substances are gases at ambient conditions. The earth’s atmosphere
is a mixture of gases and particles of liquids and solids (Table 12-2). The major
gaseous components are N2 (bp Ϫ195.79°C) and O2 (bp Ϫ182.98°C), with smaller concentrations
of other gases. All gases are miscible; that is, they mix completely unless they react
with one another.
Several scientists, notably Torricelli (1643), Boyle (1660), Charles (1787), and Graham
(1831), laid an experimental foundation on which our present understanding of gases is
based. For example, their investigations showed that
12-2
12-1
Some compounds decompose before
melting or boiling.
Ice is less dense than liquid water. This
behavior is quite unusual; most
substances are denser in the solid state
than in the liquid state.
12-2 Composition of the Atmosphere and Some Common Properties of Gases 435
TABLE 12-1 Densities and Molar Volumes of Three Substances at Atmospheric Pressure*
Substance Solid Liquid (20°C) Gas (100°C)
Density Molar volume Density Molar volume Density Molar volume
(g/mL) (mL/mol) (g/mL) (mL/mol) (g/mL) (mL/mol)
water (H2O) 0.917 (0°C) 19.6 0.998 18.0 0.000588 30,600
benzene (C6H6) 0.899 (0°C) 86.9 0.876 89.2 0.00255 30,600
carbon tetrachloride (CCl4) 1.70 (Ϫ25°C) 90.5 1.59 96.8 0.00503 30,600
*The molar volume of a substance is the volume occupied by one mole of that substance.
Volatile liquids evaporate readily. They
have low boiling points.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 1.3, States of Matter.
TABLE 12-2 Composition
of Dry Air
Gas % by Volume
N2 78.09
O2 20.94
Ar 0.93
CO2 0.03*
He, Ne, Kr, Xe 0.002
CH4 0.00015*
H2 0.00005
All others combined† Ͻ 0.00004
*Variable.
†Atmospheric moisture varies.
1. Gases can be compressed into smaller volumes; that is, their densities can be
increased by applying increased pressure.
2. Gases exert pressure on their surroundings; in turn, pressure must be exerted to
confine gases.
3. Gases expand without limits, and so gas samples completely and uniformly occupy
the volume of any container.
4. Gases diffuse into one another, and so samples of gas placed in the same container
mix completely. Conversely, different gases in a mixture do not separate on standing.
5. The amounts and properties of gases are described in terms of temperature, pressure,
the volume occupied, and the number of molecules present. For example, a
sample of gas occupies a greater volume when hot than it does when cold at the
same pressure, but the number of molecules does not change.
PRESSURE
Pressure is defined as force per unit area—for example, pounds per square inch (lb/in.2),
commonly known as psi. Pressure may be expressed in many different units, as we shall
see. The mercury barometer is a simple device for measuring atmospheric pressures.
Figure 12-1a illustrates the “heart” of the mercury barometer. A glass tube (about 800
mm long) is sealed at one end, filled with mercury, and then carefully inverted into a dish
of mercury without air being allowed to enter. The mercury in the tube falls to the level
at which the pressure of the air on the surface of the mercury in the dish equals the gravitational
pull downward on the mercury in the tube. The air pressure is measured in terms
12-3
Investigating four variables at once
is difficult. In Sections 12-4 through
12-8 we shall see how to study these
variables two at a time. Section 12-9
will consolidate these descriptions into
a single relationship, the ideal gas
equation.
436 CHAPTER 12: Gases and the Kinetic–Molecular Theory
Diffusion of bromine vapor in air. Some liquid bromine (dark reddish brown) was placed in
the small inner bottle. As the liquid evaporated, the resulting reddish brown gas diffused.
Br2 (gas)
Br2 (liquid)
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 12.2, Properties of Gases:
Pressure.
of the height of the mercury column, that is, the vertical distance between the surface of
the mercury in the open dish and that inside the closed tube. The pressure exerted by the
atmosphere is equal to the pressure exerted by the column of mercury.
Mercury barometers are simple and well known, so gas pressures are frequently
expressed in terms of millimeters of mercury (mm Hg, or just mm). In recent years the
unit torr has been used to indicate pressure; it is defined as 1 torr ϭ 1 mm Hg.
A mercury manometer consists of a glass U-tube partially filled with mercury. One
arm is open to the atmosphere, and the other is connected to a container of gas (see Figure
12-1b,c).
Atmospheric pressure varies with atmospheric conditions and distance above sea level.
The atmospheric pressure decreases with increasing elevation because there is a decreasing
mass of air above it. Approximately one half of the matter in the atmosphere is less than
20,000 feet above sea level. Thus, atmospheric pressure is only about one half as great at
The unit torr was named for
Evangelista Torricelli (1608–1647),
who invented the mercury barometer.
12-3 Pressure 437
Figure 12-1 Some laboratory devices for measuring pressure. (a) Schematic diagram of a
closed-end barometer. At the level of the lower mercury surface, the pressure both inside
and outside the tube must be equal to that of the atmosphere. There is no air inside
the tube, so the pressure is exerted only by the mercury column h mm high. Hence, the
atmospheric pressure must equal the pressure exerted by h mm Hg, or h torr. (b) The twoarm
mercury barometer is called a manometer. In this sample, the pressure of the gas
inside the flask is greater than the external atmospheric pressure. At the level of the lower
mercury surface, the total pressure on the mercury in the left arm must equal the total
pressure on the mercury in the right arm. The pressure exerted by the gas is equal to the
external pressure plus the pressure exerted by the mercury column of height h mm, or
Pgas (in torr) ϭ Patm (in torr) ϩ h torr. (c) When the gas pressure measured by the
manometer is less than the external atmospheric pressure, the pressure exerted by the
atmosphere is equal to the gas pressure plus the pressure exerted by the mercury column, or
Patm ϭ Pgas ϩ h. We can rearrange this to write Pgas (in torr) ϭ Patm (in torr) Ϫ h torr.
h (mm) h h
Atmospheric
pressure, Patm
Pressure
unknown
The pressure of the gas
in the flask is less than
atmospheric pressure.
(c)
Pgas
(in torr) = Patm
(in torr) – h torr
Atmospheric
pressure, Patm
Pressure
unknown
The pressure of the gas
in the flask is greater than
atmospheric pressure.
(b)
Pgas
(in torr) = Patm
(in torr) + h torr
Atmospheric
pressure
Mercury surface
(a)
Vacuum
(text continues on page 440)
438 CHAPTER 12: Gases and the Kinetic–Molecular Theory
The EnvironmentCCHEMISTRY IN USE
The Greenhouse Effect
During the twentieth century, the great increase in our use
of fossil fuels caused a significant rise in the concentration of
carbon dioxide, CO2, in the atmosphere. Scientists believe
that the concentration of atmospheric CO2 could double by
early in the 21st century, compared with its level just before
the Industrial Revolution. During the last 100 to 200 years,
the CO2 concentration has increased by 25%. The curve in
Figure (a) shows the recent steady rise in atmospheric CO2
concentration.
Energy from the sun reaches the earth in the form of light.
Neither CO2 nor H2O vapor absorbs the visible light in sunlight,
so they do not prevent it from reaching the surface of
the earth. The energy given off by the earth in the form of
lower-energy infrared (heat) radiation, however, is readily
absorbed by both CO2 and H2O (as it is by the glass or plastic
of greenhouses). Thus, some of the heat the earth must
lose to stay in thermal equilibrium can become trapped in
the atmosphere, causing the temperature to rise (Figure b).
This phenomenon, called the greenhouse effect, has been
the subject of much discussion among scientists and the topic
of many articles in the popular press. The anticipated rise in
average global temperature by the year 2050 due to increased
CO2 concentration is predicted to be 2 to 5°C.
An increase of 2 to 5°C may not seem like much. However,
this is thought to be enough to cause a dramatic change
in climate, transforming now productive land into desert and
altering the habitats of many animals and plants beyond their
ability to adapt. Another drastic consequence of even this
small temperature rise would be the partial melting of the
polar ice caps. The resulting rise in sea level, though only a
few feet, would mean that water would inundate coastal cities
such as Los Angeles, New York, and Houston, and low-lying
coastal areas such as southern Florida and Louisiana. On a
global scale, the effects would be devastating.
The earth’s forests and jungles play a crucial role in maintaining
the balance of gases in the atmosphere, removing CO2
and supplying O2. The massive destruction, for economic
reasons, of heavily forested areas such as the Amazon rain
forest in South America is cited as another long-term contributor
to global environmental problems. Worldwide, more
than 3 million square miles of once-forested land is now barren
for some reason; at least 60% of this land is now unused.
Environmental scientists estimate that if even one quarter of
this land could be reforested, the vegetation would absorb
1.1 billion tons of CO2 annually.
Some scientists are more skeptical than others about the
role of human-produced CO2 in climate changes and, indeed,
(a) A plot of the monthly average CO2 concentration in
parts per million (ppm), measured at Mauna Loa
Observatory, Hawaii, far from significant sources of CO2
from human activities. Annual fluctuations occur because
plants in the Northern Hemisphere absorb CO2 in the
spring and release it as they decay in the fall.
Seasonal
fluctuations
Carbondioxideconcentration(partspermillion)
348
344
340
336
332
328
324
320
316
312
Year
352
356
Mean
Summer (low)
Winter (high)
360
60 64 68 72 76 80 84 88 92
364
96
12-3 Pressure 439
about whether global warming is a significant phenomenon
or simply another of the recognized warm–cold cycles that
have occurred throughout the earth’s history. Such skeptics
point out an unexplained increase in atmospheric CO2 during
an extended period in the seventeenth century, and an
even higher and more prolonged peak about 130,000 years
ago. Even the most skeptical observers, however, seem to
agree that responsible stewardship of the planet requires that
we do something in a reasoned fashion to reduce production
of greenhouse gases, primarily CO2, and that this will involve
decreasing our dependence on energy from fossil fuels.
Despite the technical and political problems of waste disposal,
an essentially all-electric economy based on nuclear power
may someday be a usable solution.
Much CO2 is eventually absorbed by the vast amount of
water in the oceans, where the carbonate–bicarbonate buffer
system almost entirely counteracts any adverse effects of
ocean water acidity. Ironically, there is also evidence to suggest
that other types of air pollution in the form of particulate
matter may partially counteract the greenhouse effect. The
particles reflect visible (sun) radiation rather than absorbing
it, blocking some light from entering the atmosphere. It
seems foolish, however, to depend on one form of pollution
to help rescue us from the effects of another! Real solutions
to current environmental problems such as the greenhouse
effect are not subject to quick fixes, but depend on long-term
cooperative international efforts that are based on the firm
knowledge resulting from scientific research.
Tropical rain forests are important in maintaining the
balance of CO2 and O2 in the earth’s atmosphere. In recent
years a portion of the South American forests (by far the
world’s largest) larger than France has been destroyed,
either by flooding caused by hydroelectric dams or by
clearing of forest land for agricultural or ranching use.
Such destruction continues at a rate of more than 20,000
square kilometers per year. If current trends continue,
many of the world’s rain forests will be severely reduced or
even obliterated in the next few years. The fundamental
question—“What are the long-term consequences of the
destruction of tropical rain forests?”—remains unanswered.
(b) The greenhouse effect. Visible light passes through
atmospheric H2O and CO2, but heat radiated from the
surface of the earth is absorbed by these gases.
Sunlight
Surface of earth
Heat Heat
Atmospheric CO2, H2O vapor
20,000 feet as it is at sea level. Mountain climbers and pilots use portable barometers to
determine their altitudes (Figure 12-2). At sea level, at a latitude of 45°, the average atmospheric
pressure supports a column of mercury 760 mm high in a simple mercury barometer
when the mercury is at 0°C. This average sea-level pressure of 760 mm Hg is called one
atmosphere of pressure.
1 atmosphere (atm) ϭ 760 mm Hg at 0°C ϭ 760 torr
The SI unit of pressure is the pascal (Pa), defined as the pressure exerted by a force
of one newton acting on an area of one square meter. By definition, one newton (N) is
the force required to give a mass of one kilogram an acceleration of one meter per second
per second. Symbolically we represent one newton as
1 N ϭ so 1 Pa ϭ ϭ
One atmosphere of pressure ϭ 1.01325 ϫ 105 Pa, or 101.325 kPa.
1 kg
ᎏ
mиs2
1 N
ᎏ
m2
1 kgиm
ᎏ
s2
Figure 12-2 Some commercial
pressure-measuring devices. (a) A
commercial mercury barometer.
(b) Portable barometers. This type
is called an aneroid (“not wet”)
barometer. Some of the air has been
removed from the airtight box,
which is made of thin, flexible
metal. When the pressure of the
atmosphere changes, the remaining
air in the box expands or contracts
(Boyle’s Law), moving the flexible
box surface and an attached pointer
along a scale. (c) A tire gauge. This
kind of gauge registers “relative”
pressure, that is, the difference
between internal pressure and the
external atmospheric pressure. For
instance, when the gauge reads 30
psi (pounds per square inch), the
total gas pressure in the tire is
30 psi ϩ 1 atm, or about 45 psi. In
engineering terminology, this is
termed “psig” (g ϭ gauge).
(b)
(c)(a)
Acceleration is the change in velocity
(m/s) per unit time (s), m/s2.
BOYLE’S LAW: THE VOLUME–PRESSURE RELATIONSHIP
Early experiments on the behavior of gases were carried out by Robert Boyle (1627–1691)
in the seventeenth century. In a typical experiment (Figure 12-3), a sample of a gas was
trapped in a U-tube and allowed to come to constant temperature. Then its volume and
the difference in the heights of the two mercury columns were recorded. This difference
in height plus the pressure of the atmosphere represents the pressure on the gas. Addition
of more mercury to the tube increases the pressure by changing the height of the
mercury column. As a result, the gas volume decreases. The results of several such experiments
are tabulated in Figure 12-4a.
Boyle showed that for a given sample of gas at constant temperature, the product of
pressure and volume, P ϫ V, was always the same number.
At a given temperature, the product of pressure and volume of a definite mass of
gas is constant.
PV ϭ k (constant n, T )
This relationship is Boyle’s Law. The value of k depends on the amount (number of
moles, n) of gas present and on the temperature, T. Units for k are determined by the
units used to express the volume (V) and pressure (P).
12-4
Figure 12-3 A representation of Boyle’s experiment. A sample of air is trapped in a tube in
such a way that the pressure on the air can be changed and the volume of the air measured.
Patm is the atmospheric pressure, measured with a barometer. P1 ϭ h1 ϩ Patm, P2 ϭ h2 ϩ Patm.
12-4 Boyle’s Law: The Volume–Pressure Relationship 441
Volume
scale
Volume
scale
After addition of mercury
Pressure, P1
Volume, V1, of
entrapped air
Patm
h1
Patm
Pressure, P2
Volume, V2, of
entrapped air
h2
When the volume of a gas is plotted against its pressure at constant temperature, the
resulting curve is one branch of a hyperbola. Figure 12-4b is a graphic illustration of this
inverse relationship. When volume is plotted versus the reciprocal of the pressure, 1/P, a
straight line results (Figure 12-4c). In 1662, Boyle summarized the results of his experiments
on various samples of gases in an alternative statement of Boyle’s Law:
At constant temperature the volume, V, occupied by a definite mass of a gas is
inversely proportional to the applied pressure, P.
V ϰ ᎏ
P
1
ᎏ or V ϭ k ΂ᎏ
P
1
ᎏ΃ (constant n, T )
At normal temperatures and pressure, most gases obey Boyle’s Law rather well. We call
this ideal behavior. Deviations from ideality are discussed in Section 12-15.
Let us think about a fixed mass of gas at constant temperature, but at two different
conditions of pressure and volume (see Figure 12-3). For the first condition we can write
P1V1 ϭ k (constant n, T )
and for the second condition we can write
P2V2 ϭ k (constant n, T )
Because the right-hand sides of these two equations are the same, the left-hand sides must
be equal, or
P1V1 ϭ P2V2 (for a given amount of a gas at constant temperature)
This form of Boyle’s Law is useful for calculations involving pressure and volume changes,
as the following examples demonstrate.
The symbol ϰ reads “is proportional
to.” A proportionality is converted into
an equality by introducing a
proportionality constant, k.
442 CHAPTER 12: Gases and the Kinetic–Molecular Theory
0 10 20 30 40 50
0
10
20
30
40
50
Volume
Pressure(b)
0.00 0.05 0.10 0.15 0.20 0.25
0
10
20
30
40
50
Volume
1/Pressure(c)
0.30
Temperature
constant
Temperature
constant
5.0
P V P ϫ V 1/P
10.0
15.0
17.0
20.0
22.0
30.0
40.0
(a)
40.0
20.0
13.3
11.8
10.0
9.10
6.70
5.00
200
200
200
201
200
200
201
200
0.20
0.10
0.0667
0.0588
0.0500
0.0455
0.0333
0.0250
Figure 12-4 (a) Some typical data from an experiment such as that shown in Figure 12-3.
Measured values of P and V are presented in the first two columns, on an arbitrary scale.
(b, c) Graphical representations of Boyle’s Law, using the data of part (a). (b) V versus P.
(c) V versus 1/P.
EXAMPLE 12-1 Boyle’s Law Calculation
A sample of gas occupies 12 L under a pressure of 1.2 atm. What would its volume be if the
pressure were increased to 2.4 atm?
Plan
We know the volume at one pressure and want to find the volume at another pressure (constant
temperature). This suggests that we use Boyle’s Law. We tabulate what is known and what is
asked for, and then solve the Boyle’s Law equation for the unknown quantity, V2.
Solution
We have
V1 ϭ 12 L P1 ϭ 1.2 atm
V2 ϭ _?_ P2 ϭ 2.4 atm
Solving Boyle’s Law, P1V1 ϭ P2V2, for V2 and substituting gives
V2 ϭ ϭ ϭ 6.0 L
(1.2 atm)(12 L)
ᎏᎏ
2.4 atm
P1V1
ᎏ
P2
Pressure and volume are inversely
proportional. Doubling the pressure
halves the volume of a sample of gas at
constant temperature.
It is often helpful to tabulate what is
given and what is asked for in a
problem.
12-4 Boyle’s Law: The Volume–Pressure Relationship 443
EXAMPLE 12-2 Boyle’s Law Calculation
A sample of oxygen occupies 10.0 L under a pressure of 790. torr (105 kPa). At what pressure
would it occupy 13.4 L if the temperature did not change?
Plan
We know the pressure at one volume and wish to find the pressure at another volume (at
constant temperature). We can solve Boyle’s Law for the second pressure and substitute.
Solution
We have P1 ϭ 790. torr; V1 ϭ 10.0 L; P2 ϭ _?_; V2 ϭ 13.4 L. Solving Boyle’s Law, P1V1 ϭ P2V2,
for P2 and substituting yields
P2 ϭ ϭ ϭ 590. torr ΂ϫ ϭ 78.6 kPa ΃
You should now work Exercises 16 and 17.
101.3 kPa
ᎏᎏ
760. torr
(790. torr)(10.0 L)
ᎏᎏᎏ
13.4 L
P1V1
ᎏ
V2
Problem-Solving Tip: Units in Boyle’s Law Calculations
Which units for volume and pressure are appropriate for Boyle’s Law calculations? Boyle’s
Law in the form P1V1 ϭ P2V2 can be written as V1/V2 ϭ P2/P1. This involves a ratio of
volumes, so they can be expressed in any volume units—liters, milliliters, cubic feet—
as long as the same units are used for both volumes. Likewise, because Boyle’s Law
involves a ratio of pressures, you can use any units for pressures—atmospheres, torr,
pascals—as long as the same units are used for both pressures.
CHARLES’S LAW: THE VOLUME–TEMPERATURE
RELATIONSHIP; THE ABSOLUTE TEMPERATURE SCALE
In his pressure–volume studies on gases, Robert Boyle noticed that heating a sample of
gas caused some volume change, but he did not follow up on this observation. About 1800,
two French scientists—Jacques Charles (1746–1823) and Joseph Gay-Lussac (1778–
1850), pioneer balloonists at the time—began studying the expansion of gases with
increasing temperature. Their studies showed that the rate of expansion with increased
temperature was constant and was the same for all the gases they studied as long as the
pressure remained constant. The implications of their discovery were not fully recognized
until nearly a century later. Then scientists used this behavior of gases as the basis of a
new temperature scale, the absolute temperature scale.
The change of volume with temperature, at constant pressure, is illustrated in Figure
12-5. From the table of typical data in Figure 12-5b, we see that volume (V, mL) increases
as temperature (t,°C) increases, but the quantitative relationship is not yet obvious. These
data are plotted in Figure 12-5c (line A), together with similar data for the same gas sample
at different pressures (lines B and C).
Lord Kelvin, a British physicist, noticed that an extension of the different temperature–
volume lines back to zero volume (dashed line) yields a common intercept at Ϫ273.15°C
on the temperature axis. Kelvin named this temperature absolute zero. The degrees are
the same size over the entire scale, so 0°C becomes 273.15 degrees above absolute zero.
In honor of Lord Kelvin’s work, this scale is called the Kelvin temperature scale. As pointed
out in Section 1-12, the relationship between the Celsius and Kelvin temperature scales
is K ϭ °C ϩ 273.15°.
If we convert temperatures (°C) to absolute temperatures (K), the green scale in Figure
12-5c, the volume–temperature relationship becomes obvious. This relationship is known
as Charles’s Law.
At constant pressure, the volume occupied by a definite mass of a gas is directly
proportional to its absolute temperature.
12-5
An artist’s representation of Jacques
Charles’s first ascent in a hydrogen
balloon at the Tuileries, Paris,
December 1, 1783.
Lord Kelvin (1824–1907) was born
William Thompson. At the age
of ten he was admitted to Glasgow
University. Because its new appliance
was based on Kelvin’s theories, a
refrigerator company named its
product the Kelvinator.
Recall that temperatures on the Kelvin
scale are expressed in kelvins (not
degrees Kelvin) and represented by K,
not °K.
Absolute zero may be thought of as
the limit of thermal contraction for an
ideal gas.
444 CHAPTER 12: Gases and the Kinetic–Molecular Theory
Problem-Solving Tip: Use What You Can Predict About the Answer
In Example 12-1 the calculated volume decrease is consistent with the increase in pressure.
We can use this reasoning in another method for solving that problem, that is, by
setting up a “Boyle’s Law factor” to change the volume in the direction required by the
pressure change. We reason that the pressure increases from 1.2 atm to 2.4 atm, so the
volume decreases by the factor (1.2 atm/2.4 atm). The solution then becomes
_?_L ϭ 12 L ϫ (Boyle’s Law factor that would decrease the volume)
ϭ 12 L ϫ ΂ ΃ϭ 6.0 L
Now solve Example 12-2 using a Boyle’s Law factor.
1.2 atm
ᎏ
2.4 atm
Figure 12-5 An experiment
showing that the volume of an ideal
gas increases as the temperature is
increased at constant pressure. (a) A
mercury plug of constant weight,
plus atmospheric pressure, maintains
a constant pressure on the trapped
air. (b) Some representative volume–
temperature data at constant
pressure. The relationship becomes
clear when t (°C) is converted to T
(K) by adding 273°C. (c) A graph in
which volume is plotted versus
temperature on two different scales.
Lines A, B, and C represent the
same mass of the same ideal gas
at different pressures. Line A
represents the data tabulated in part
(b). Graph D shows the behavior of
a gas that condenses to form a liquid
(in this case, at 50°C) as it is cooled.
We can express Charles’s Law in mathematical terms as
V ϰ T or V ϭ kT (constant n, P)
Rearranging the expression gives V/T ϭ k, a concise statement of Charles’s Law. As the
temperature increases, the volume must increase proportionally. If we let subscripts 1 and
2 represent values for the same sample of gas at two different temperatures, we obtain
ϭ (for a definite mass of gas at constant pressure)
which is the more useful form of Charles’s Law. This relationship is valid only when temperature,
T, is expressed on an absolute (usually the Kelvin) scale.
V2
ᎏ
T2
V1
ᎏ
T1
Ϫ300 Ϫ200 Ϫ100 0 100 200 300 400 t (°C)
0 73 173 273 373 473 573 673 T (K)
10
20
30
40
0
Pressure constant
(c)
V(mL)
B A
C
D
Ϫ273
Trapped
air sample
Ice
water
Moveable mercury plug,
constant mass
Boiling
water
Glass
tubeWater
bath
Patm Patm Patm
0ЊC
(ice water)
20ЊC
(room
temperature)
100ЊC
(boiling
water)
(a)
27
100
227
427
20.0
24.9
33.3
46.7
300
373
500
700
t (ЊC) V (mL) T (K)
(b)
54
127
327 40.0
26.7
21.8
600
400
327
When balloons filled with air are cooled in liquid nitrogen (bp Ϫ196°C), each shrinks to a
small fraction of its original volume. Because the boiling points of the other components of
air, except He and Ne, are higher than Ϫ196°C, they condense to form liquids. When the
balloons are removed from the liquid nitrogen, the liquids vaporize to form gases again. As
the air warms to room temperature, the balloons expand to their original volume (Charles’s
Law).
(a)
(b)
(c)
EXAMPLE 12-3 Charles’s Law Calculation
A sample of nitrogen occupies 117 mL at 100.°C. At what temperature in °C would it occupy
234 mL if the pressure did not change?
Plan
We know the volume of the sample at one temperature and wish to know its temperature corresponding
to a second volume (constant pressure). We can solve Charles’s Law for the second
temperature. We must remember to carry out calculations with all temperatures expressed on
the Kelvin scale, converting to or from Celsius as necessary.
Solution
V1 ϭ 117 mL V2 ϭ 234 mL
T1 ϭ 100.°C ϩ 273° ϭ 373 K T2 ϭ _?_
ϭ and T2 ϭ ϭ ϭ 746 K
°C ϭ 746 K Ϫ 273° ϭ 473°C
The temperature doubles on the Kelvin scale, from 373 K to 746 K, so the volume doubles.
You should now work Exercise 24.
(234 mL)(373 K)
ᎏᎏ
(117 mL)
V2T1
ᎏ
V1
V2
ᎏ
T2
V1
ᎏ
T1
446 CHAPTER 12: Gases and the Kinetic–Molecular Theory
Problem-Solving Tip: Be Careful of Units in Charles’s Law
Calculations
Which units for volume and temperature are appropriate for Charles’s Law calculations?
The equation ᎏ
V
T1
1
ᎏ ϭ ᎏ
V
T2
2
ᎏ can be written as ᎏ
V
V
1
2
ᎏ ϭ ᎏ
T
T
1
2
ᎏ. This involves a ratio of volumes,
so they can be expressed in any volume units—liters, milliliters, cubic feet—as long as
the same units are used for both volumes. But the relationship does not apply at all unless
the temperatures are both expressed on an absolute scale. Remember to express all
temperatures in kelvins for Charles’s Law calculations.
STANDARD TEMPERATURE AND PRESSURE
We have seen that both temperature and pressure affect the volumes (and therefore the
densities) of gases. It is often convenient to choose some “standard” temperature and pressure
as a reference point for discussing gases. Standard temperature and pressure (STP)
are, by international agreement, exactly 0°C (273.15 K) and one atmosphere of pressure
(760. torr).
THE COMBINED GAS LAW EQUATION
Boyle’s Law relates the pressures and volumes of a sample of gas at constant temperature,
P1V1 ϭ P2V2. Charles’s Law relates the volumes and temperatures at constant pressure,
12-7
12-6
V1/T1 ϭ V2/T2. Combination of Boyle’s Law and Charles’s Law into a single expression
gives the combined gas law equation.
ϭ (constant amount of gas)
When any five of the variables in the equation are known, the sixth variable can be
calculated.
EXAMPLE 12-4 Combined Gas Law Calculation
A sample of neon occupies 105 liters at 27°C under a pressure of 985 torr. What volume would
it occupy at standard temperature and pressure (STP)?
Plan
A sample of gas is changing in all three quantities P, V, and T. This suggests that we use the
combined gas law equation. We tabulate what is known and what is asked for, solve the combined
gas law equation for the unknown quantity, V2, and substitute known values.
Solution
V1 ϭ 105 L P1 ϭ 985 torr T1 ϭ 27°C ϩ 273° ϭ 300. K
V2 ϭ _?_ P2 ϭ 760. torr T2 ϭ 273 K
Solving for V2,
ϭ so V2 ϭ ϭ ϭ 124 L
Alternatively, we can multiply the original volume by a Boyle’s Law factor and a Charles’s Law
factor. As the pressure decreases from 985 torr to 760. torr, the volume increases, so the Boyle’s
Law factor is 985 torr/760. torr. As the temperature decreases from 300. K to 273 K, the volume
decreases, so the Charles’s Law factor is 273 K/300. K. Multiplication of the original volume
by these factors gives the same result.
_?_L ϭ 105 L ϫ ϫ ϭ 124 L
The temperature decrease (from 300. K to 273 K) alone would give only a small decrease
in the volume of neon, by a factor of 273 K/300. K, or 0.910. The pressure decrease (from
985 torr to 760. torr) alone would result in a greater increase in the volume, by a factor
of 985 torr/760. torr, or 1.30. The result of the two changes is that the volume increases
from 105 liters to 124 liters.
EXAMPLE 12-5 Combined Gas Law Calculation
A sample of gas occupies 12.0 liters at 240.°C under a pressure of 80.0 kPa. At what temperature
would the gas occupy 15.0 liters if the pressure were increased to 107 kPa?
Plan
The approach is the same as for Example 12-4 except that the unknown quantity is now the
temperature, T2.
273 K
ᎏ
300. K
985 torr
ᎏᎏ
760. torr
(985 torr)(105 L)(273 K)
ᎏᎏᎏ
(760. torr)(300. K)
P1V1T2
ᎏ
P2T1
P2V2
ᎏ
T2
P1V1
ᎏ
T1
P2V2
ᎏ
T2
P1V1
ᎏ
T1
12-7 The Combined Gas Law Equation 447
Notice that the combined gas law
equation becomes
1. P1V1 ϭ P2V2 (Boyle’s Law)
when T is constant;
2. ϭ (Charles’s Law)
when P is constant; and
3. ϭ when V is constant.
P2
ᎏ
T2
P1
ᎏ
T1
V2
ᎏ
T2
V1
ᎏ
T1
Solution
V1 ϭ 12.0 L P1 ϭ 80.0 kPa T1 ϭ 240.°C ϩ 273° ϭ 513 K
V2 ϭ 15.0 L P2 ϭ 107 kPa T2 ϭ _?_
We solve the combined gas law equation for T2.
ϭ so T2 ϭ ϭ ϭ 858 K
K ϭ °C ϩ 273° so °C ϭ 858 K Ϫ 273° ϭ 585°C
You should now work Exercises 32 and 33.
(107 kPa)(15.0 L)(513 K)
ᎏᎏᎏ
(80.0 kPa)(12.0 L)
P2V2T1
ᎏ
P1V1
P2V2
ᎏ
T2
P1V1
ᎏ
T1
448 CHAPTER 12: Gases and the Kinetic–Molecular Theory
Problem-Solving Tip: Units in Combined Gas Law Calculations
The combined gas law equation is derived by combining Boyle’s and Charles’s Laws, so
the comments in earlier Problem-Solving Tips also apply to this equation. Remember
to express all temperatures in kelvins. Volumes can be expressed in any units as long as
both are in the same units. Similarly, any pressure units can be used, so long as both are
in the same units. Example 12-4 uses torr for both pressures; Example 12-5 uses kPa for
both pressures.
AVOGADRO’S LAW AND THE STANDARD MOLAR
VOLUME
In 1811, Amedeo Avogadro postulated that
at the same temperature and pressure, equal volumes of all gases contain the same
number of molecules.
Many experiments have demonstrated that Avogadro’s hypothesis is accurate to within
about Ϯ2%, and the statement is now known as Avogadro’s Law.
Avogadro’s Law can also be stated as follows.
At constant temperature and pressure, the volume, V, occupied by a gas sample is
directly proportional to the number of moles, n, of gas.
V ϰ n or V ϭ kn or ϭ k (constant P, T )
For two samples of gas at the same temperature and pressure, the relation between volumes
and numbers of moles can be represented as
ϭ (constant T, P)
V2
ᎏ
n2
V1
ᎏ
n1
V
ᎏ
n
12-8
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 12.3, Gas Laws.
The volume occupied by a mole of gas at standard temperature and pressure, STP, is
referred to as the standard molar volume. It is nearly constant for all gases (Table 12-3).
The standard molar volume of an ideal gas is taken to be 22.414 liters per mole
at STP.
Gas densities depend on pressure and temperature; however, the number of moles of
gas in a given sample does not change with temperature or pressure. Pressure changes
affect volumes of gases according to Boyle’s Law, and temperature changes affect volumes
of gases according to Charles’s Law. We can use these laws to convert gas densities at
various temperatures and pressures to standard temperature and pressure. Table 12-3 gives
the experimentally determined densities of several gases at standard temperature and
pressure.
EXAMPLE 12-6 Molecular Weight, Density
One (1.00) mole of a gas occupies 27.0 liters, and its density is 1.41 g/L at a particular temperature
and pressure. What is its molecular weight? What is the density of the gas at STP?
Plan
We can use dimensional analysis to convert the density, 1.41 g/L, to molecular weight, g/mol.
To calculate the density at STP, we recall that the volume occupied by one mole would be
22.4 L.
Solution
We multiply the density under the original conditions by the unit factor 27.0 L/1.00 mol to
generate the appropriate units, g/mol.
ϭ ϫ ϭ 38.1 g/mol
At STP, 1.00 mol of the gas, 38.1 g, would occupy 22.4 L, and its density would be
Density ϭ ϫ ϭ 1.70 g/L at STP
You should now work Exercises 40 and 42.
1 mol
ᎏ
22.4 L
38.1 g
ᎏ
1 mol
27.0 L
ᎏ
mol
1.41 g
ᎏ
L
_?_g
ᎏ
mol
The volume percentages given in
Table 12-2 are also equal to mole
percentages.
Density is defined as mass per unit
volume.
12-8 Avogadro’s Law and the Standard Molar Volume 449
TABLE 12-3 Standard Molar Volumes and Densities of Some Gases
Standard Molar Density at
Gas Formula (g/mol) Volume (L/mol) STP (g/L)
hydrogen H2 2.02 22.428 0.090
helium He 4.003 22.426 0.178
neon Ne 20.18 22.425 0.900
nitrogen N2 28.01 22.404 1.250
oxygen O2 32.00 22.394 1.429
argon Ar 39.95 22.393 1.784
carbon dioxide CO2 44.01 22.256 1.977
ammonia NH3 17.03 22.094 0.771
chlorine Cl2 70.91 22.063 3.214
Deviations in standard molar volume
indicate that gases do not behave
ideally.
SUMMARY OF GAS LAWS: THE IDEAL GAS EQUATION
Let us summarize what we have learned about gases. Any sample of gas can be described
in terms of its pressure, temperature (in kelvins), volume, and the number of moles, n,
present. Any three of these variables determine the fourth. The gas laws we have studied
give several relationships among these variables. An ideal gas is one that exactly obeys
these gas laws. Many real gases show slight deviations from ideality, but at normal temperatures
and pressures the deviations are usually small enough to be ignored. We will do so
for the present and discuss deviations later.
We can summarize the behavior of ideal gases as follows.
Boyle’s Law V ϰ (at constant T and n)
Charles’s Law V ϰ T (at constant P and n)
Avogadro’s Law V ϰ n (at constant T and P)
Summarizing V ϰ (no restrictions)
As before, a proportionality can be written as an equality by introducing a proportionality
constant, for which we’ll use the symbol R. This gives
V ϭ R΂ ΃ or, rearranging, PV ϭ nRT
This relationship is called the ideal gas equation or the ideal gas law. The numerical value
of R, the universal gas constant, depends on the choices of the units for P, V, and T. One
mole of an ideal gas occupies 22.414 liters at 1.0000 atmosphere and 273.15 K (STP).
Solving the ideal gas law for R gives
R ϭ ϭ ϭ 0.082057
In working problems, we often round R to 0.0821 Lиatm/molиK. We can express R in
other units, as shown inside the back cover of this text.
EXAMPLE 12-7 Units of R
R can have any energy units per mole per kelvin. Calculate R in terms of joules per mole per
kelvin and in SI units of kPaиdm3/molиK.
Plan
We apply dimensional analysis to convert to the required units.
Solution
Appendix C shows that 1 Lиatm ϭ 101.325 joules.
R ϭ ϫ ϭ 8.3144 J/molиK
Now evaluate R in SI units. One atmosphere pressure is 101.325 kilopascals, and the molar
volume at STP is 22.414 dm3.
101.325 J
ᎏᎏ
1 Lиatm
0.082057 Lиatm
ᎏᎏ
molиK
Lиatm
ᎏ
molиK
(1.0000 atm)(22.414 L)
ᎏᎏᎏ
(1.0000 mol)(273.15 K)
PV
ᎏ
nT
nT
ᎏ
P
nT
ᎏ
P
1
ᎏ
P
12-9
This equation takes into account the
values of n, T, P, and V. Restrictions
that apply to the individual gas laws
are therefore not needed for the ideal
gas equation.
450 CHAPTER 12: Gases and the Kinetic–Molecular Theory
Recall that 1 dm3 ϭ 1 L.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 12.4, The Ideal Gas Law.
R ϭ ϭ ϭ 8.3145
You should now work Exercise 43.
We can now express R, the universal gas constant, to four digits in three different sets of units.
R ϭ 0.08206 ᎏ
m
L
o
иa
l
t
и
m
K
ᎏ ϭ ᎏ
8
m
.3
o
1
lи
4
K
J
ᎏ ϭ 8.314 ᎏ
k
m
Pa
o
и
l
d
иK
m3
ᎏ
The usefulness of the ideal gas equation is that it relates the four variables, P, V, n,
and T, that describe a sample of gas at one set of conditions. If any three of these variables
are known, the fourth can be calculated.
EXAMPLE 12-8 Ideal Gas Equation
What pressure, in atm, is exerted by 54.0 grams of Xe in a 1.00-liter flask at 20.°C?
Plan
We list the variables with the proper units. Then we solve the ideal gas equation for P and
substitute values.
Solution
V ϭ 1.00 L n ϭ 54.0 g Xe ϫ ϭ 0.411 mol
T ϭ 20.°C ϩ 273° ϭ 293 K P ϭ _?_
Solving PV ϭ nRT for P and substituting gives
P ϭ ϭ ϭ 9.89 atm
EXAMPLE 12-9 Ideal Gas Equation
What is the volume of a gas balloon filled with 4.00 moles of He when the atmospheric pressure
is 748 torr and the temperature is 30.°C?
Plan
We first list the variables with the proper units. Then we solve the ideal gas equation for V
and substitute known values.
Solution
P ϭ 748 torr ϫ ϭ 0.984 atm n ϭ 4.00 mol
T ϭ 30.°C ϩ 273° ϭ 303 K V ϭ _?_
Solving PV ϭ nRT for V and substituting gives
1 atm
ᎏ
760 torr
(0.411 mol)΂ᎏ
0.08
m
21
ol
L
иK
иatm
ᎏ΃(293 K)
ᎏᎏᎏᎏ
1.00 L
nRT
ᎏ
V
1 mol
ᎏᎏ
131.3 g Xe
kPaиdm3
ᎏᎏ
molиK
101.325 kPa ϫ 22.414 dm3
ᎏᎏᎏ
1 mol ϫ 273.15 K
PV
ᎏ
nT
12-9 Summary of Gas Laws: The Ideal Gas Equation 451
V ϭ ϭ ϭ 101 L
You should now work Exercise 46.
You may wonder why pressures are given in torr or mm Hg and temperatures in °C.
This is because pressures are often measured with mercury barometers, and temperatures
are measured with Celsius thermometers.
EXAMPLE 12-10 Ideal Gas Equation
A helium-filled weather balloon has a volume of 7240 cubic feet. How many grams of helium
would be required to inflate this balloon to a pressure of 745 torr at 21°C? (1 ft3 ϭ 28.3 L)
Plan
We use the ideal gas equation to find n, the number of moles required, and then convert to
grams. We must convert each quantity to one of the units stated for R. (R ϭ 0.0821
Lиatm/molиK)
Solution
P ϭ 745 torr ϫ ϭ 0.980 atm T ϭ 21°C ϩ 273° ϭ 294 K
V ϭ 7240 ft3 ϫ ϭ 2.05 ϫ 105 L n ϭ _?_
Solving PV ϭ nRT for n and substituting gives
n ϭ ϭ ϭ 8.32 ϫ 103 mol He
_?_g He ϭ (8.32 ϫ 103 mol He)΂4.00 ΃ϭ 3.33 ϫ 104 g He
You should now work Exercise 47.
g
ᎏ
mol
(0.980 atm)(2.05 ϫ 105 L)
ᎏᎏᎏ
΂0.0821 ᎏ
m
Lи
o
a
l
t
и
m
K
ᎏ΃(294 K)
PV
ᎏ
RT
28.3 L
ᎏ
1 ft3
1 atm
ᎏ
760 torr
(4.00 mol)΂0.0821 ᎏ
m
Lи
o
a
l
t
и
m
K
ᎏ΃(303 K)
ᎏᎏᎏᎏ
0.984 atm
nRT
ᎏ
P
A helium-filled weather balloon.
452 CHAPTER 12: Gases and the Kinetic–Molecular Theory
Problem-Solving Tip: Watch Out for Units in Ideal Gas Law
Calculations
The units of R that are appropriate for ideal gas law calculations are those that involve
units of volume, pressure, moles, and temperature. When you use the value R ϭ 0.0821
Lиatm/molиK, remember to express all quantities in a calculation in these units. Pressures
should be expressed in atmospheres, volumes in liters, temperature in kelvins, and
amount of gas in moles. In Examples 12-9 and 12-10 we converted pressures from torr
to atm. In Example 12-10 the volume was converted from ft3 to L.
Summary of the ideal gas laws
1. The individual gas laws are usually used to calculate the changes in conditions for
a sample of gas (subscripts can be thought of as “before” and “after”).
Boyle’s Law P1V1 ϭ P2V2 (for a given amount of a gas
at constant temperature)
Charles’s Law ᎏ
V
T1
1
ᎏ ϭ ᎏ
V
T2
2
ᎏ (for a given amount of a gas
at constant pressure)
Combined Gas Law ᎏ
P
T
1V
1
1
ᎏ ϭ ᎏ
P
T
2V
2
2
ᎏ (for a given amount of a gas)
Avogadro’s Law ᎏ
V
n1
1
ᎏ ϭ ᎏ
V
n2
2
ᎏ (for gas samples at the same
temperature and pressure)
2. The ideal gas equation is used to calculate one of the four variables P, V, n, and
T, which describe a sample of gas at any single set of conditions.
PV ϭ nRT
The ideal gas equation can also be used to calculate the densities of gases.
EXAMPLE 12-11 Ideal Gas Equation
Nitric acid, a very important industrial chemical, is made by dissolving the gas nitrogen dioxide,
NO2, in water. Calculate the density of NO2 gas, in g/L, at 1.24 atm and 50.°C.
Plan
We use the ideal gas equation to find the number of moles, n, in any volume, V, at the specified
pressure and temperature. Then we convert moles to grams. Because we want to express
density in g/L, we choose a volume of one liter.
Solution
V ϭ 1.00 L n ϭ _?_
T ϭ 50.°C ϩ 273° ϭ 323 K P ϭ 1.24 atm
Solving PV ϭ nRT for n and substituting gives
n ϭ ϭ ϭ 0.0468 mol
So there is 0.0468 mol NO2/L at the specified P and T. Converting this to grams of NO2 per
liter, we obtain
Density ϭ ϭ ϫ ϭ 2.15 g/L
You should now work Exercise 40.
46.0 g NO2
ᎏᎏ
mol NO2
0.0468 mol NO2
ᎏᎏ
L
_?_g
ᎏ
L
(1.24 atm)(1.00 L)
ᎏᎏᎏ
΂0.0821 ᎏ
m
Lи
o
a
l
t
и
m
K
ᎏ΃(323 K)
PV
ᎏ
RT
12-9 Summary of Gas Laws: The Ideal Gas Equation 453
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 12.5, Gas Density.
DETERMINATION OF MOLECULAR WEIGHTS AND
MOLECULAR FORMULAS OF GASEOUS SUBSTANCES
In Section 2-10 we distinguished between simplest and molecular formulas of compounds.
We showed how simplest formulas can be calculated from percent compositions of
compounds. The molecular weight must be known to determine the molecular formula
of a compound. For compounds that are gases at convenient temperatures and pressures,
the ideal gas law provides a basis for determining molecular weights.
EXAMPLE 12-12 Molecular Weight
A 0.109-gram sample of a pure gaseous compound occupies 112 mL at 100.°C and 750. torr.
What is the molecular weight of the compound?
Plan
We first use the ideal gas law, PV ϭ nRT, to find the number of moles of gas. Then, knowing
the mass of that number of moles of gas, we calculate the mass of one mole, the molecular
weight.
Solution
V ϭ 0.112 L T ϭ 100.°C ϩ 273° ϭ 373 K P ϭ 750. torr ϫ ϭ 0.987 atm
n ϭ ϭ ϭ 0.00361 mol
The mass of 0.00361 mole of this gas is 0.109 g, so the mass of one mole is
ϭ ϭ 30.2 g/mol
EXAMPLE 12-13 Molecular Weight
A 120.-mL flask contained 0.345 gram of a gaseous compound at 100.°C and 1.00 atm pressure.
What is the molecular weight of the compound?
Plan
We use the ideal gas law, PV ϭ nRT, to determine the number of moles of gas that filled
the flask. Then, knowing the mass of this number of moles, we can calculate the mass of
one mole.
Solution
V ϭ 0.120 L P ϭ 1.00 atm T ϭ 100.°C ϩ 273° ϭ 373 K
n ϭ ϭ ϭ 0.00392 mol
(1.00 atm)(0.120 L)
ᎏᎏᎏ
΂0.0821 ᎏ
m
L
o
иa
l
t
и
m
K
ᎏ΃(373 K)
PV
ᎏ
RT
0.109 g
ᎏᎏ
0.00361 mol
_?_g
ᎏ
mol
(0.987 atm)(0.112 L)
ᎏᎏᎏ
΂0.0821 ᎏ
m
Lи
o
a
l
t
и
m
K
ᎏ΃(373 K)
PV
ᎏ
RT
1 atm
ᎏᎏ
760. torr
12-10
The molecular weight of the gas is
30.2 amu. The gas could be ethane,
C2H6, MW ϭ 30.1 amu. Can you
think of other possibilities? Could the
gas have been NO, CH3OH, O2, or
CH3NH2?
454 CHAPTER 12: Gases and the Kinetic–Molecular Theory
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 12.6, Using Gas Laws:
Determining Molar Mass.
The mass of 0.00392 mol of gas is 0.345 g, so the mass of one mole is
ϭ ϭ 88.0 g/mol
You should now work Exercises 51 and 54.
Let’s carry the calculation one step further in the next example.
EXAMPLE 12-14 Molecular Formula
Additional analysis of the gaseous compound in Example 12-13 showed that it contained 54.5%
carbon, 9.10% hydrogen, and 36.4% oxygen by mass. What is its molecular formula?
Plan
We first find the simplest formula for the compound as we did in Section 2-9 (Examples 2-12
and 2-13). Then we use the molecular weight that we determined in Example 12-13 to find
the molecular formula. To find the molecular formula, we reason as in Example 2-16. We use
the experimentally known molecular weight to find the ratio
n ϭ
The molecular weight is n times the simplest-formula weight, so the molecular formula is n
times the simplest formula.
Solution
Relative Number Smallest WholeRelative
Mass of Atoms Divide by Number Ratio
Element of Element (divide mass by AW) Smallest Number of Atoms
C 54.5 ᎏ
5
1
4
2
.
.
5
0
ᎏ ϭ 4.54 ᎏ
4
2
.
.
5
2
4
8
ᎏ ϭ 1.99 2
H 9.10 ᎏ
9
1
.
.
1
0
0
1
ᎏ ϭ 9.01 ᎏ
9
2
.
.
0
2
1
8
ᎏ ϭ 3.95 4 C2H4O
O 36.4 ᎏ
3
1
6
6
.
.
4
0
ᎏ ϭ 2.28 ᎏ
2
2
.
.
2
2
8
8
ᎏ ϭ 1.00 1
The simplest formula is C2H4O and the simplest-formula weight is 44.0 amu.
Division of the molecular weight by the simplest-formula weight gives
ϭ ϭ 2
The molecular formula is therefore 2 ϫ (C2H4O) ϭ C4H8O2.
The gas could be either ethyl acetate or butyric acid. Both have the formula C4H8O2. They
have very different odors, however. Ethyl acetate has the odor of nail polish remover. Butyric
acid has the foul odor of rancid butter.
You should now work Exercise 52.
88.0 amu
ᎏᎏ
44.0 amu
molecular weight
ᎏᎏᎏ
simplest-formula weight
molecular weight
ᎏᎏᎏ
simplest-formula weight
0.345 g
ᎏᎏ
0.00392 mol
_?_g
ᎏ
mol
12-10 Determination of Molecular Weights and Molecular Formulas of Gaseous Substances 455
H
H
H
C
C C H
C
O
O
O
H H
H H
H
H
H
C
H
H
C
H
H
C C
O H
ethyl acetate
butyric acid
DALTON’S LAW OF PARTIAL PRESSURES
Many gas samples, including our atmosphere, are mixtures that consist of different kinds
of gases. The total number of moles in a mixture of gases is
ntotal ϭ nA ϩ nB ϩ nC ϩ и и и
where nA, nB, and so on represent the number of moles of each kind of gas present. Rearranging
the ideal gas equation, PtotalV ϭ ntotalRT, for the total pressure, Ptotal, and then
substituting for ntotal gives
Ptotal ϭ ϭ
Multiplying out the right-hand side gives
Ptotal ϭ ϩ ϩ ϩ и и и
Now nART/V is the partial pressure PA that the nA moles of gas A alone would exert in
the container at temperature T; similarly, nBRT/V ϭ PB, and so on. Substituting these into
the equation for Ptotal, we obtain Dalton’s Law of Partial Pressures (Figure 12-6).
Ptotal ϭ PA ϩ PB ϩ PC ϩ и и и (constant V, T)
The total pressure exerted by a mixture of ideal gases is the sum of the partial pressures
of those gases.
Dalton’s Law is useful in describing real gaseous mixtures at moderate pressures because
it allows us to relate total measured pressures to the composition of mixtures.
EXAMPLE 12-15 Mixture of Gases
A 10.0-liter flask contains 0.200 mole of methane, 0.300 mole of hydrogen, and 0.400 mole of
nitrogen at 25°C. (a) What is the pressure, in atmospheres, inside the flask? (b) What is the
partial pressure of each component of the mixture of gases?
nCRT
ᎏ
V
nBRT
ᎏ
V
nART
ᎏ
V
(nA ϩ nB ϩ nC ϩ и и и) RT
ᎏᎏᎏ
V
ntotalRT
ᎏ
V
12-11
John Dalton was the first to notice
this effect. He did so in 1807 while
studying the compositions of moist and
dry air. The pressure that each gas
exerts in a mixture is called its partial
pressure. No way has been devised to
measure the pressure of an individual
gas in a mixture; it must be calculated
from other quantities.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 12.8, Gas Mixtures and Partial
Pressures.
Figure 12-6 An illustration of Dalton’s Law. When the two gases A and B are mixed in
the same container at the same temperature, they exert a total pressure equal to the sum of
their partial pressures.
Gas A Gas B
Mixture
A + B
+
PA PB Mix Ptotal = PA + PB
Plan
(a) We are given the number of moles of each component. The ideal gas law is then used to
calculate the total pressure from the total number of moles. (b) The partial pressure of each
gas in the mixture can be calculated by substituting the number of moles of each gas individually
into PV ϭ nRT.
Solution
(a) n ϭ 0.200 mol CH4 ϩ 0.300 mol H2 ϩ 0.400 mol N2 ϭ 0.900 mol of gas
(a) V ϭ 10.0 L T ϭ 25°C ϩ 273° ϭ 298 K
Solving PV ϭ nRT for P gives P ϭ nRT/V. Substitution gives
P ϭ ϭ 2.20 atm
(b) Now we find the partial pressures. For CH4, n ϭ 0.200 mol, and the values for V and T
are the same as above.
PCH4
ϭ ϭ ϭ 0.489 atm
Similar calculations for the partial pressures of hydrogen and nitrogen give
PH2
ϭ 0.734 atm and PN2
ϭ 0.979 atm
As a check, we use Dalton’s Law: Ptotal ϭ PA ϩ PB ϩ PC ϩ и и и . Addition of the partial pressures
in this mixture gives the total pressure.
Ptotal ϭ PCH4
ϩ PH2
ϩ PN2
ϭ (0.489 ϩ 0.734 ϩ 0.979) atm ϭ 2.20 atm
You should now work Exercises 58 and 59.
(0.200 mol)΂0.0821 ᎏ
m
Lи
o
a
l
t
и
m
K
ᎏ΃(298 K)
ᎏᎏᎏᎏ
10.0 L
(nCH4
)RT
ᎏᎏ
V
(0.900 mol)΂0.0821 ᎏ
m
Lи
o
a
l
t
и
m
K
ᎏ΃(298 K)
ᎏᎏᎏᎏ
10.0 L
12-11 Dalton’s Law of Partial Pressures 457
Problem-Solving Tip: Amounts of Gases in Mixtures Can Be
Expressed in Various Units
In Example 12-15 we were given the number of moles of each gas. Sometimes the amount
of a gas is expressed in other units that can be converted to number of moles. For instance,
if we know the formula weight (or the formula), we can convert a given mass of gas to
number of moles.
We can describe the composition of any mixture in terms of the mole fraction of each
component. The mole fraction, XA, of component A in a mixture is defined as
XA ϭ
Like any other fraction, mole fraction is a dimensionless quantity. For each component
in a mixture, the mole fraction is
no. mol A
ᎏᎏᎏᎏ
total no. mol of all components
XA ϭ ,
XB ϭ , and so on
For a gaseous mixture, we can relate the mole fraction of each component to its partial
pressure as follows. From the ideal gas equation, the number of moles of each component
can be written as
nA ϭ PAV/RT, nB ϭ PBV/RT, and so on
and the total number of moles is
ntotal ϭ PtotalV/RT
Substituting into the definition of XA,
XA ϭ ϭ
The quantities V, R, and T cancel to give
XA ϭ ; similarly, XB ϭ ; and so on
We can rearrange these equations to give another statement of Dalton’s Law of Partial
Pressures.
PA ϭ XA ϫ Ptotal; PB ϭ XB ϫ Ptotal; and so on
The partial pressure of each gas is equal to its mole fraction in the gaseous mixture
times the total pressure of the mixture.
EXAMPLE 12-16 Mole Fraction, Partial Pressure
Calculate the mole fractions of the three gases in Example 12-15.
Plan
One way to solve this problem is to use the numbers of moles given in the problem. Alternatively
we could use the partial pressures and the total pressure from Example 12-15.
Solution
Using the moles given in Example 12-15,
XCH4
ϭ ϭ ϭ 0.222
XH2
ϭ ϭ ϭ 0.333
XN2
ϭ ϭ ϭ 0.444
0.400 mol
ᎏᎏ
0.900 mol
nN2
ᎏ
ntotal
0.300 mol
ᎏᎏ
0.900 mol
nH2
ᎏ
ntotal
0.200 mol
ᎏᎏ
0.900 mol
nCH4
ᎏ
ntotal
PB
ᎏ
Ptotal
PA
ᎏ
Ptotal
PAV/RT
ᎏᎏ
PtotalV/RT
nA
ᎏᎏ
nA ϩ nB ϩ и и и
no. mol B
ᎏᎏᎏᎏ
no. mol A ϩ no. mol B ϩ и и и
no. mol A
ᎏᎏᎏᎏ
no. mol A ϩ no. mol B ϩ и и и
The sum of all mole fractions in a
mixture is equal to 1.
XA ϩ XB ϩ и и и ϭ 1 for any mixture
We can use this relationship to check
mole fraction calculations or to find a
remaining mole fraction if we know all
the others.
In Example 12-16 we see that, for a
gas mixture, relative numbers of moles
of components are the same as relative
pressures of the components.
Using the partial and total pressures calculated in Example 12-15,
XCH4
ϭ ϭ ϭ 0.222
XH2
ϭ ϭ ϭ 0.334
XN2
ϭ ϭ ϭ 0.445
You should now work Exercise 60.
EXAMPLE 12-17 Partial Pressure, Mole Fraction
The mole fraction of oxygen in the atmosphere is 0.2094. Calculate the partial pressure of O2
in air when the atmospheric pressure is 760. torr.
Plan
The partial pressure of each gas in a mixture is equal to its mole fraction in the mixture times
the total pressure of the mixture.
Solution
PO2
ϭ XO2
ϫ Ptotal
ϭ 0.2094 ϫ 760. torr ϭ 159 torr
Dalton’s Law can be used in combination with other gas laws, as the following example
shows.
EXAMPLE 12-18 Mixture of Gases
Two tanks are connected by a closed valve. Each tank is filled with gas as shown, and both
tanks are held at the same temperature. We open the valve and allow the gases to mix.
(a) After the gases mix, what is the partial pressure of each gas, and what is the total pressure?
(b) What is the mole fraction of each gas in the mixture?
0.979 atm
ᎏᎏ
2.20 atm
PN2
ᎏ
Ptotal
0.734 atm
ᎏᎏ
2.20 atm
PH2
ᎏ
Ptotal
0.489
ᎏ
2.20 atm
PCH4
ᎏ
Ptotal
The difference between the two
calculated results is due to rounding.
12-11 Dalton’s Law of Partial Pressures 459
Plan
(a) Each gas expands to fill the available volume, 5.00 liters plus 3.00 liters or a total volume
of 8.00 liters. We can use Boyle’s Law to calculate the partial pressure that each gas would
exert after it expands to fill 8.00 L. The total pressure is equal to the sum of the partial pressures
of the two gases. (b) The mole fractions can be calculated from the ratio of the partial
pressure of each gas to the total pressure.
Tank A Tank B
5.00 L of O2
24.0 atm
3.00 L of N2
32.0 atm
Solution
(a) For O2,
P1V1 ϭ P2V2 or P2,O2
ϭ ᎏ
P
V
1V
2
1
ᎏ ϭ ϭ 15.0 atm
For N2,
P1V1 ϭ P2V2 or P2,N2
ϭ ᎏ
P
V
1V
2
1
ᎏ ϭ ϭ 12.0 atm
The total pressure is the sum of the partial pressures.
Ptotal ϭ P2,O2
ϩ P2,N2
ϭ 15.0 atm ϩ 12.0 atm ϭ 27.0 atm
(b) XO2
ϭ ϭ ϭ 0.556
XN2
ϭ ϭ ϭ 0.444
As a check, the sum of the mole fractions is 1.
You should now work Exercise 62.
Some gases can be collected over water. Figure 12-7 illustrates the collection of a sample
of hydrogen by displacement of water. A gas produced in a reaction displaces the denser
water from the inverted water-filled container. The pressure on the gas inside the collection
container can be made equal to atmospheric pressure by raising or lowering the
container until the water level inside is the same as that outside.
One complication arises, however. A gas in contact with water soon becomes saturated
with water vapor. The pressure inside the container is the sum of the partial pressure of
the gas itself plus the partial pressure exerted by the water vapor in the gas mixture (the
vapor pressure of water). Every liquid shows a characteristic vapor pressure that varies
only with temperature, and not with the volume of vapor present, so long as both liquid
12.0 atm
ᎏ
27.0 atm
P2,N2
ᎏ
Ptotal
15.0 atm
ᎏ
27.0 atm
P2,O2
ᎏ
Ptotal
32.0 atm ϫ 3.00 L
ᎏᎏᎏ
8.00 L
24.0 atm ϫ 5.00 L
ᎏᎏᎏ
8.00 L
Notice that this problem has been
solved without calculating the number
of moles of either gas.
Gases that are soluble in water or that
react with water cannot be collected by
this method. Other liquids can be
used.
The partial pressure exerted by the
vapor above a liquid is called the vapor
pressure of that liquid. A more
extensive table of the vapor pressure of
water appears in Appendix E.
460 CHAPTER 12: Gases and the Kinetic–Molecular Theory
Figure 12-7 Apparatus for
preparing hydrogen from zinc and
sulfuric acid.
Zn(s) ϩ 2Hϩ(aq) 88n
Zn2ϩ(aq) ϩ H2(g)
The hydrogen is collected by
displacement of water.
and vapor are present. Table 12-4 displays the vapor pressure of water near room temper-
ature.
The relevant point here is that a gas collected over water is “moist”; that is, it is saturated
with water vapor. Measuring the atmospheric pressure at which the gas is collected,
we can write
Patm ϭ Pgas ϩ PH2O or Pgas ϭ Patm Ϫ PH2O
Example 12-19 provides a detailed illustration.
EXAMPLE 12-19 Gas Collected Over Water
Hydrogen was collected over water (Figure 12-7) at 21°C on a day when the atmospheric pressure
was 748 torr. The volume of the gas sample collected was 300. mL. (a) How many moles
of H2 were present? (b) How many moles of water vapor were present in the moist gas mixture?
(c) What is the mole fraction of hydrogen in the moist gas mixture? (d) What would be the
mass of the gas sample if it were dry?
Plan
(a) The vapor pressure of H2O, PH2O ϭ 19 torr at 21°C, is obtained from Table 12-4. Applying
Dalton’s Law, PH2
ϭ Patm Ϫ PH2O. We then use the partial pressure of H2 in the ideal gas equation
to find the number of moles of H2 present. (b) The partial pressure of water vapor (the
vapor pressure of water at the stated temperature) is used in the ideal gas equation to find the
number of moles of water vapor present. (c) The mole fraction of H2 is the ratio of its partial
pressure to the total pressure. (d) The number of moles found in part (a) can be converted to
mass of H2.
Solution
(a) PH2
ϭ Patm Ϫ PH2O ϭ (748 Ϫ 19) torr ϭ 729 torr ϫ ϭ 0.959 atm
We also know
V ϭ 300. mL ϭ 0.300 L and T ϭ 21°C ϩ 273° ϭ 294 K
Solving the ideal gas equation for nH2
gives
nH2
ϭ ϭ ϭ 1.19 ϫ 10Ϫ2 mol H2
(b) PH2O ϭ 19 torr ϫ ϭ 0.025 atm
V and T have the same values as in part (a).
nH2O ϭ ϭ ϭ 3.1 ϫ 10Ϫ4 mol H2O vapor
(c) XH2
ϭ ϭ ϭ 0.974
(d) _?_g H2 ϭ 1.19 ϫ 10Ϫ2 mol ϫ ϭ 2.40 ϫ 10Ϫ2 g H2
You should now work Exercise 64.
2.02 g
ᎏ
1 mol
729 torr
ᎏ
748 torr
PH2
ᎏ
Ptotal
(0.025 atm)(0.300 L)
ᎏᎏᎏ
΂0.0821 ᎏ
m
Lи
o
a
l
t
и
m
K
ᎏ΃(294 K)
PH2OV
ᎏ
RT
1 atm
ᎏ
760 torr
(0.959 atm)(0.300 L)
ᎏᎏᎏ
΂0.0821 ᎏ
m
Lи
o
a
l
t
и
m
K
ᎏ΃(294 K)
PH2
V
ᎏ
RT
1 atm
ᎏ
760 torr
Remember that each gas occupies the
total volume of the container.
At STP, this dry hydrogen would
occupy 267 mL. Can you calculate
this?
12-11 Dalton’s Law of Partial Pressures 461
TABLE 12-4 Vapor Pressure
of Water
Near Room
Temperature
Vapor Pressure
Temperature of Water
(°C) (torr)
19 16.48
20 17.54
21 18.65
22 19.83
23 21.07
24 22.38
25 23.76
26 25.21
27 26.74
28 28.35
MASS–VOLUME RELATIONSHIPS IN REACTIONS
INVOLVING GASES
Many chemical reactions produce gases. For instance, the combustion of hydrocarbon in
excess oxygen at high temperatures produces both carbon dioxide and water as gases, as
illustrated for octane.
2C8H18(g) ϩ 25O2(g) 88n 16CO2(g) ϩ 18H2O(g)
The N2 gas produced by the very rapid decomposition of sodium azide, NaN3(s), inflates
air bags used as safety devices in automobiles.
We know that one mole of gas, measured at STP, occupies 22.4 liters; we can use the
ideal gas equation to find the volume of a mole of gas at any other conditions. This information
can be utilized in stoichiometry calculations (Section 3-2).
Small amounts of oxygen can be produced in the laboratory by heating solid potassium
chlorate, KClO3, in the presence of a catalyst, manganese(IV) oxide, MnO2. Solid potassium
chloride, KCl, is also produced. (CAUTION: Heating KClO3 can be dangerous.)
MnO2
2KClO3(s) 8888n 2KCl(s) ϩ 3O2(g)
heat
2 mol 2 mol 3 mol
2(122.6 g) 2(74.6 g) 3(22.4 LSTP)
Unit factors can be constructed using any two of these quantities.
EXAMPLE 12-20 Gas Volume in a Chemical Reaction
What volume of O2 (STP) can be produced by heating 112 grams of KClO3?
Plan
The preceding equation shows that two moles of KClO3 produce three moles of O2. We
construct appropriate unit factors from the balanced equation and the standard molar volume
of oxygen to solve the problem.
Solution
_?_LSTP O2 ϭ 112 g KClO3 ϫ ϫ ϫ
ϭ 30.7 LSTP O2
This calculation shows that the thermal decomposition of 112 grams of KClO3 produces 30.7
liters of oxygen measured at standard conditions.
You should now work Exercise 76.
EXAMPLE 12-21 Gas Volume in a Chemical Reaction
A 1.80-gram mixture of potassium chlorate, KClO3, and potassium chloride, KCl, was heated
until all of the KClO3 had decomposed. After being dried, the liberated oxygen occupied 405
mL at 25°C when the barometric pressure was 745 torr. (a) How many moles of O2 were
produced? (b) What percentage of the mixture was KClO3?
22.4 LSTP O2
ᎏᎏ
1 mol O2
3 mol O2
ᎏᎏ
2 mol KClO3
1 mol KClO3
ᎏᎏ
122.6 g KClO3
12-12
The nitrogen gas formed in the
rapid reaction
2NaN3(s) 88n 2Na(s) ϩ 3N2(g)
fills an automobile air bag during a
collision. The air bag fills within
1/20th of a second after a front
collision.
462 CHAPTER 12: Gases and the Kinetic–Molecular Theory
Production of a gas by a reaction.
2NaOH(g) ϩ 2Al(s) ϩ 6H2O(ᐉ) 88n
2Na[Al(OH)4](aq) ϩ 3H2(g)
This reaction is used in some solid
drain cleaners.
Plan
(a) The number of moles of O2 produced can be calculated from the ideal gas equation. (b)
Then we use the balanced chemical equation to relate the known number of moles of O2
formed and the mass of KClO3 that decomposed to produce it.
Solution
(a) V ϭ 405 mL ϭ 0.405 L; P ϭ 745 torr ϫ ϭ 0.980 atm
(a) T ϭ 25°C ϩ 273° ϭ 298 K
Solving the ideal gas equation for n and evaluating gives
n ϭ ϭ ϭ 0.0162 mol O2
(b) _?_ g KClO3 ϭ 0.0162 mol O2 ϫ ϫ ϭ 1.32 g KClO3
The sample contained 1.32 grams of KClO3. The percent of KClO3 in the sample is
% KClO3 ϭ ϫ 100% ϭ ϫ 100% ϭ 73.3% KClO3
You should now work Exercise 78.
Our study of stochiometry has shown that substances react in definite mole and mass
proportions. Using previously discussed gas laws, we can show that gases also react in
simple, definite proportions by volume. For example, one volume of hydrogen always
combines (reacts) with one volume of chlorine to form two volumes of hydrogen chloride,
if all volumes are measured at the same temperature and pressure
H2(g) ϩ Cl2(g) 88n 2HCl(g)
1 volume ϩ 1 volume 88n 2 volumes
Volumes may be expressed in any units as long as the same unit is used for all. Gay-Lussac
summarized several experimental observations on combining volumes of gases. The
summary is known as Gay-Lussac’s Law of Combining Volumes:
At constant temperature and pressure, the volumes of reacting gases can be expressed
as a ratio of simple whole numbers.
The ratio is obtained from the coefficients in the balanced equation for the reaction.
Clearly, the law applies only to gaseous substances at the same temperature and pressure.
No generalizations can be made about the volumes of solids and liquids as they undergo
chemical reactions. Consider the following examples, based on experimental observations
at constant temperature and pressure. Hundreds more could be cited.
1.32 g
ᎏ
1.80 g
g KClO3
ᎏᎏ
g sample
122.6 g KClO3
ᎏᎏ
1 mol KClO3
2 mol KClO3
ᎏᎏ
3 mol O2
(0.980 atm)(0.405 L)
ᎏᎏᎏ
΂0.0821 ᎏ
m
Lи
o
a
l
t
и
m
K
ᎏ (298 K)΃
PV
ᎏ
RT
1 atm
ᎏ
760 torr
12-12 Mass-Volume Relationships in Reactions Involving Gases 463
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 12.7, Gas Laws and Chemical
Reactions.
1. One volume of nitrogen can react with three volumes of hydrogen to form two
volumes of ammonia
N2(g) ϩ 3H2(g) 88n 2NH3(g)
1 volume ϩ 3 volumes 88n 2 volumes
2. One volume of methane reacts with (burns in) two volumes of oxygen to give one
volume of carbon dioxide and two volumes of steam
CH4(g) ϩ 2O2(g) 88n CO2(g) ϩ 2H2O(g)
1 volume ϩ 2 volumes 88n 1 volume ϩ 2 volumes
3. Sulfur (a solid) reacts with one volume of oxygen to form one volume of sulfur
dioxide
S(s) ϩ O2(g) 88n SO2(g)
1 volume 88n 1 volume
4. Four volumes of ammonia burn in five volumes of oxygen to produce four volumes
of nitric oxide and six volumes of steam
4NH3(g) ϩ 5O2(g) 88n 4 NO(g) ϩ 6H2O(g)
4 volumes 5 volumes 88n 4 volumes 6 volumes
THE KINETIC–MOLECULAR THEORY
As early as 1738, Daniel Bernoulli (1700–1782) envisioned gaseous molecules in ceaseless
motion striking the walls of their container and thereby exerting pressure. In 1857, Rudolf
Clausius (1822–1888) published a theory that attempted to explain various experimental
observations that had been summarized by Boyle’s, Dalton’s, Charles’s, and Avogadro’s
laws. The basic assumptions of the kinetic–molecular theory for an ideal gas follow.
1. Gases consist of discrete molecules. The individual molecules are very small and
are very far apart relative to their own sizes.
2. The gas molecules are in continuous, random, straight-line motion with varying
velocities (see Figure 12-8).
3. The collisions between gas molecules and with the walls of the container are
elastic; the total energy is conserved during a collision; that is, there is no net
energy gain or loss.
4. Between collisions, the molecules exert no attractive or repulsive forces on one
another; instead, each molecule travels in a straight line with a constant velocity.
Kinetic energy is the energy a body possesses by virtue of its motion. It is ᎏ1
2
ᎏmu2, where
m, the body’s mass, can be expressed in grams and u, its velocity, can be expressed in
meters per second (m/s). The assumptions of the kinetic–molecular theory can be used
to relate temperature and molecular kinetic energy (see the Enrichment section, pages
467–469).
12-13
1. The observation that gases can be
easily compressed indicates that the
molecules are far apart. At ordinary
temperatures and pressures, the gas
molecules themselves occupy an
insignificant fraction of the total
volume of the container.
2. Near temperatures and pressures at
which a gas liquefies, the gas does
not behave ideally (Section 12-15)
and attractions or repulsions among
gas molecules are significant.
3. At any given instant, only a small
fraction of the molecules are
involved in collisions.
464 CHAPTER 12: Gases and the Kinetic–Molecular Theory
Figure 12-8 A representation of
molecular motion. Gaseous
molecules, in constant motion,
undergo collisions with one another
and with the walls of the container.
The average kinetic energy of gaseous molecules is directly proportional to the
absolute temperature of the sample. The average kinetic energies of molecules of
different gases are equal at a given temperature.
For instance, in samples of H2, He, CO2, and SO2 at the same temperature, all the molecules
have the same average kinetic energies. But the lighter molecules, H2 and He, have
much higher average velocities than do the heavier molecules, CO2 and SO2, at the same
temperature.
We can summarize this very important result from the kinetic–molecular theory.
Average molecular KE ϭ KෆEෆ ϰ T
or
Average molecular speed ϭ uෆ ϰ
Ί๶๶
Molecular kinetic energies of gases increase with increasing temperature and decrease
with decreasing temperature. We have referred only to the average kinetic energy; in a
given sample, some molecules may be moving quite rapidly while others are moving more
slowly. Figure 12-9 shows the distribution of speeds of gaseous molecules at two temper-
atures.
The kinetic–molecular theory satisfactorily explains most of the observed behavior of
gases in terms of molecular behavior. Let’s look at the gas laws in terms of the kinetic–
molecular theory.
T
ᎏᎏ
molecular weight
A bar over a quantity denotes an
average of that quantity.
12-13 The Kinetic–Molecular Theory 465
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 12.11, Distribution of Molecular
Speeds.
Figure 12-9 The Maxwellian distribution function for molecular speeds. This graph shows
the relative numbers of O2 molecules having a given speed at 25°C and at 1000°C. At 25°C,
most O2 molecules have speeds between 200 and 600 m/s (450–1350 miles per hour). Some
of the molecules have very high speeds, so the distribution curve never reaches the
horizontal axis.
Fractionofmolecules
ataparticularspeed
Very few
molecules
have very
low speeds
At 25°C more molecules are moving
at about 400 m/s than at any other speed
At 1000°C many,
many more molecules
are moving at
1600 m/s than at 25°C
O2 at 25°C
O2 at 1000°C
O2 at 25°C O2 at 1000°C
Molecular speed (m/s)
0 200 400 600 800 1000 1200 1400 1600 1800 2000
Boyle’s Law
The pressure exerted by a gas on the walls of its container is caused by gas molecules
striking the walls. Clearly, pressure depends on two factors: (1) the number of molecules
striking the walls per unit time and (2) how vigorously the molecules strike the walls. If
the temperature is held constant, the average speed and the force of the collisions remain
the same. But halving the volume of a sample of gas doubles the pressure because twice
as many molecules strike a given area on the walls per unit time. Likewise, doubling the
volume of a sample of gas halves the pressure because only half as many gas molecules
strike a given area on the walls per unit time (Figure 12-10).
Dalton’s Law
In a gas sample the molecules are very far apart and do not attract one another significantly.
Each kind of gas molecule acts independently of the presence of the other kind.
The molecules of each gas thus collide with the walls with a frequency and vigor that do
not change even if other molecules are present (Figure 12-11). As a result, each gas exerts
a partial pressure that is independent of the presence of the other gas, and the total pressure
is due to the sum of all the molecule–wall collisions.
Figure 12-10 A molecular
interpretation of Boyle’s Law—
the change in pressure of a
gas with changes in
volume (at constant
temperature). The
entire apparatus is
enclosed in a vacuum.
In the smaller volume,
more molecules strike
the walls per unit time to
give a higher pressure.
Piston
10 g
Gas
sample
3.0 L 1.5 L
20 g
Figure 12-11 A molecular
interpretation of Dalton’s Law. The
molecules act independently, so each
gas exerts its own partial pressure
due to its molecular collisions with
the walls.
20 g
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 12.10, Gas Laws and the
Kinetic–Molecular Theory.
Charles’s Law
Recall that average kinetic energy is directly proportional to the absolute temperature.
Doubling the absolute temperature of a sample of gas doubles the average kinetic energy
of the gaseous molecules, and the increased force of the collisions of molecules with the
walls doubles the volume at constant pressure. Similarly, halving the absolute temperature
decreases kinetic energy to half its original value; at constant pressure, the volume
decreases by half because of the reduced vigor of the collision of gaseous molecules with
the container walls (Figure 12-12).
12-13 The Kinetic–Molecular Theory 467
Piston
10 g
Gas
sample
3.0 L 1.5 L
10 g
600 K 300 K
Figure 12-12 A molecular interpretation of Charles’s Law—the change in volume of a gas
with changes in temperature (at constant pressure). At the lower temperature, molecules
strike the walls less often and less vigorously. Thus, the volume must be less to maintain the
same pressure.
Kinetic–Molecular Theory, the Ideal Gas Equation, and Molecular
Speeds
In 1738, Daniel Bernoulli derived Boyle’s Law from Newton’s laws of motion applied to
gas molecules. This derivation was the basis for an extensive mathematical development of
the kinetic–molecular theory more than a century later by Clausius, Maxwell, Boltzmann,
and others. Although we do not need to study the detailed mathematical presentation of
this theory, we can gain some insight into its concepts from the reasoning behind Bernoulli’s
derivation. Here we present that reasoning based on proportionality arguments.
In the kinetic–molecular theory pressure is viewed as the result of collisions of gas molecules
with the walls of the container. As each molecule strikes a wall, it exerts a small impulse.
The pressure is the total force thus exerted on the walls divided by the area of the walls.
The total force on the walls (and thus the pressure) is proportional to two factors: (1) the
E n r i c h m e n t
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 12.9, Kinetic–Molecular Theory
of Gases.
impulse exerted by each collision and (2) the rate of collisions (number of collisions in a
given time interval).
P ϰ (impulse per collision) ϫ (rate of collisions)
Let us represent the mass of an individual molecule by m and its speed by u. The heavier
the molecule is (greater m) and the faster it is moving (greater u), the harder it pushes on
the wall when it collides. The impulse due to each molecule is proportional to its momentum,
mu.
Impulse per collision ϰ mu
The rate of collisions, in turn, is proportional to two factors. First, the rate of collision
must be proportional to the molecular speed; the faster the molecules move, the more often
they reach the wall to collide. Second, this collision rate must be proportional to the number
of molecules per unit volume, N/V. The greater the number of molecules, N, in a given
volume, the more molecules collide in a given time interval.
Rate of collisions ϰ (molecular speed) ϫ (molecules per unit volume)
or
Rate of collisions ϰ (u) ϫ ΂ ΃
We can introduce these proportionalities into the one describing pressure, to conclude that
P ϰ (mu) ϫ u ϫ or P ϰ or PV ϰ Nmu2
At any instant not all molecules are moving at the same speed, u. We should reason in terms
of the average behavior of the molecules, and express the quantity u2 in average terms as
uෆ2ෆ, the mean-square speed.
PV ϰ Nmuෆ2ෆ
Not all molecules collide with the walls at right angles, so we must average (using calculus)
over all the trajectories. This gives a proportionality constant of ᎏ1
3
ᎏ, and
PV ϭ ᎏ1
3
ᎏNmuෆ2ෆ
This describes the quantity PV (pressure ϫ volume) in terms of molecular quantities—
number of molecules, molecular masses, and molecular speeds. The number of molecules,
N, is given by the number of moles, n, times Avogadro’s number, NAv, or N ϭ nNAv. Making
this substitution, we obtain
PV ϭ ᎏ1
3
ᎏnNAvmuෆ2ෆ
The ideal gas equation describes (pressure ϫ volume) in terms of measurable quantities—
number of moles and absolute temperature.
PV ϭ nRT
So we see that the ideas of the kinetic–molecular theory lead to an equation of the same
form as the macroscopic ideal gas equation. Thus, the molecular picture of the theory is
consistent with the ideal gas equation and gives support to the theory. Equating the righthand
sides of these last two equations and canceling n gives
ᎏ1
3
ᎏNAvmuෆ2ෆ ϭ RT
Nmu2
ᎏ
V
N
ᎏ
V
N
ᎏ
V
Recall that momentum is mass ϫ
speed.
468 CHAPTER 12: Gases and the Kinetic–Molecular Theory
(Enrichment, continued)
uෆ2ෆ is the average of the squares of the
molecular speeds. It is proportional to
the square of the average speed, but
the two quantities are not equal.
This equation can also be written as
ᎏ1
3
ᎏNAv ϫ (2 ϫ ᎏ1
2
ᎏ muෆ2ෆ) ϭ RT
From physics we know that the kinetic energy of a particle of mass m moving at speed u is
ᎏ1
2
ᎏmu2. So we can write
ᎏ2
3
ᎏ NAv ϫ (avg KE per molecule) ϭ RT
or
NAv ϫ (avg KE per molecule) ϭ ᎏ3
2
ᎏRT
This equation shows that the absolute temperature is directly proportional to the average
molecular kinetic energy, as postulated by the kinetic–molecular theory. Because there are
NAv molecules in a mole, the left-hand side of this equation is equal to the total kinetic
energy of a mole of molecules.
Total kinetic energy per mole of gas ϭ ᎏ3
2
ᎏRT
With this interpretation, the total molecular–kinetic energy of a mole of gas depends only
on the temperature, and not on the mass of the molecules or the gas density.
We can also obtain some useful equations for molecular speeds from the previous
reasoning. Solving the equation
ᎏ1
3
ᎏNAvmuෆ2ෆ ϭ RT
for root-mean-square speed, urms ϭ ͙uෆෆ2ෆෆ, we obtain
urms ϭ
Ί๶
We recall that m is the mass of a single molecule. So NAvm is the mass of Avogadro’s number
of molecules, or one mole of substance; this is equal to the molecular weight, M, of the gas.
urms ϭ
Ί๶
3RT
ᎏ
M
3RT
ᎏ
NAvm
12-13 The Kinetic–Molecular Theory 469
EXAMPLE 12-22 Molecular Speed
Calculate the root-mean-square speed of H2 molecules in meters per second at 20°C. Recall
that
1 J ϭ 1 ᎏ
kg
s
и
2
m2
ᎏ
Plan
We substitute the appropriate values into the equation relating urms to temperature and molecular
weight. Remember that R must be expressed in the appropriate units.
R ϭ 8.314 ᎏ
mo
J
lиK
ᎏ ϭ 8.314 ᎏ
m
k
o
g
lи
и
K
m
и
2
s2
ᎏ
Solution
urms ϭ
Ί๶ϭ
Ί๶
urms ϭ ͙3ෆ.6ෆ2ෆ ϫෆ 1ෆ0ෆ6ෆmෆ2/ෆs2ෆ ϭ 1.90 ϫ 103 m/s (about 4250 mph)
You should now work Exercise 86.
DIFFUSION AND EFFUSION OF GASES
Because gas molecules are in constant, rapid, random motion, they diffuse quickly
throughout any container (Figure 12-13). For example, if hydrogen sulfide (the smell of
rotten eggs) is released in a large room, the odor can eventually be detected throughout
the room. If a mixture of gases is placed in a container with thin porous walls, the molecules
effuse through the walls. Because they move faster, lighter gas molecules effuse
through the tiny openings of porous materials faster than heavier molecules (Figure
12-14).
Although they are the most abundant elements in the universe, hydrogen and helium
occur as gases only in trace amounts in our atmosphere. This is due to the high average
molecular speeds resulting from their low molecular weights. At temperatures in our
atmosphere, these molecules reach speeds exceeding the escape velocity required for them
to break out of the earth’s gravitational pull and diffuse into interplanetary space. Thus,
most of the gaseous hydrogen and helium that were probably present in large concentrations
in the earth’s early atmosphere have long since diffused away. The same is true for
the abundance of these gases on other small planets in our solar system, especially those
with higher average temperatures than ours (Mercury and Venus). The Mariner 10 spacecraft
in 1974 revealed measurable amounts of He in the atmosphere of Mercury; the source
of this helium is unknown. Massive bodies such as stars (including our own sun) are mainly
H and He.
12-14
3 ϫ 8.314 ᎏ
m
k
o
g
lи
и
K
m
и
2
s2
ᎏ ϫ 293 K
ᎏᎏᎏᎏ
2.016 ᎏ
m
g
ol
ᎏ ϫ ᎏ
10
1
0
k
0
g
g
ᎏ
3RT
ᎏ
M
Scientists use the word “effusion” to
describe the escape of a gas through a
tiny hole, and the word “diffusion” to
describe movement of a gas into a
space or the mixing of one gas with
another. The distinction made by
scientists is somewhat sharper than
that usually found in a dictionary.
470 CHAPTER 12: Gases and the Kinetic–Molecular Theory
Figure 12-13 A representation of diffusion of gases. The space between the molecules
allows for ease of mixing one gas with another. Collisions of molecules with the walls of the
container are responsible for the pressure of the gas.
O2
+ N2
( – atm each)1
2
Equal
volumes
Stopcock
closed
O2
(1 atm) N2
(1 atm)
(a)
Stopcock
open
(b)
O2
+ N2
( – atm each)1
2
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 12.12, Applications of the
Kinetic–Molecular Theory.
REAL GASES: DEVIATIONS FROM IDEALITY
Until now our discussions have dealt with ideal behavior of gases. By this we mean that
the identity of a gas does not affect how it behaves, and the same equations should work
equally well for all gases. Under ordinary conditions most real gases do behave ideally;
their P and V are predicted by the ideal gas laws, so they do obey the postulates of the
kinetic–molecular theory. According to the kinetic–molecular model, (1) all but a negligible
volume of a gas sample is empty space, and (2) the molecules of ideal gases do not
attract one another because they are so far apart relative to their own sizes.
Under some conditions, however, most gases can have pressures and/or volumes that
are not accurately predicted by the ideal gas laws. This tells us that they are not behaving
entirely as postulated by the kinetic–molecular theory.
Nonideal gas behavior (deviation from the predictions of the ideal gas laws) is most
significant at high pressures and/or low temperatures, that is, near the conditions under
which the gas liquefies.
Johannes van der Waals (1837–1923) studied deviations of real gases from ideal
behavior. In 1867, he empirically adjusted the ideal gas equation
PidealVideal ϭ nRT
to take into account two complicating factors.
12-15
NH3 gas (left) and HCl gas (right)
escape from concentrated aqueous
solutions. The white smoke (solid
NH4Cl) shows where the gases mix
and react.
NH3(g) ϩ HCl(g) 88n NH4Cl(s)
12-15 Real Gases: Deviations from Ideality 471
(a) (b) (c)
A
B
Bell jar
Porous cup
Hydrogen gas
Air
Figure 12-14 Effusion of gases. (a) A molecular interpretation of effusion. Molecules are
in constant motion; occasionally they strike the opening and escape. (b) Latex balloons were
filled with the same volume of He (yellow), N2 (blue), and O2 (red). Lighter molecules, such
as He, effuse through the tiny pores of the latex balloons more rapidly than does N2 or O2.
The silver party balloon is made of a metal-coated polymer with pores that are too small to
allow rapid He effusion. (c) If a bell jar full of hydrogen is brought down over a porous cup
full of air, rapidly moving hydrogen diffuses into the cup faster than the oxygen and
nitrogen in the air can effuse out of the cup. This causes an increase in pressure in the cup
sufficient to produce bubbles in the water in the beaker.
1. According to the kinetic–molecular theory, the molecules are so small, relative to
the total volume of the gas, that each molecule can move through virtually the entire
measured volume of the container, Vmeasured (Figure 12-15a). But under high pressures,
a gas is compressed so that the volume of the molecules themselves becomes
a significant fraction of the total volume occupied by the gas. As a result, the available
volume, Vavailable, for any molecule to move in is less than the measured volume
by an amount that depends on the volume excluded by the presence of the other
molecules (Figure 12-15b). To account for this, we subtract a correction factor, nb.
Videally available ϭ Vmeasured Ϫ nb
The factor nb corrects for the volume occupied by the molecules themselves. Larger
molecules have greater values of b, and the greater the number of molecules in a
sample (higher n), the larger is the volume correction. The correction term becomes
negligibly small, however, when the volume is large.
2. The kinetic–molecular theory describes pressure as resulting from molecular collisions
with the walls of the container; this theory assumes that attractive forces
between molecules are insignificant. For any real gas, the molecules can attract one
another. But at higher temperatures, the potential energy due to intermolecular
attractions is negligibly small compared with the high kinetic energy due to the
rapid motion of the molecules and to the great distances between them. When the
temperature is quite low (low kinetic energy), the molecules move so slowly that
the potential energy due to even small attractive forces does become important. This
perturbation becomes even more important when the molecules are very close
together (at high pressure). As a result, the molecules deviate from their straightline
paths and take longer to reach the walls, so fewer collisions take place in a given
time interval. Furthermore for a molecule about to collide with the wall, the attraction
by its neighbors causes the collision to be less energetic than it would otherwise
be (Figure 12-16). As a consequence, the pressure that the gas exerts, Pmeasured, is
less than the pressure it would exert if attractions were truly negligible, Pideally exerted.
To correct for this, we subtract a correction factor, n2a/V2, from the ideal pressure.
Figure 12-15 A molecular
interpretation of deviations from
ideal behavior. (a) A sample of
gas at a low temperature. Each
sphere represents a molecule.
Because of their low kinetic
energies, attractive forces
between molecules can now
cause a few molecules to
“stick together.” (b) A sample of
gas under high pressure. The
molecules are quite close together.
The free volume is now a much
smaller fraction of the total volume.
472 CHAPTER 12: Gases and the Kinetic–Molecular Theory
(a) Low temperature (b) High pressure
Figure 12-16 A gas molecule
strikes the walls of a container with
diminished force. The attractive
forces between a molecule and its
neighbors are significant.
Pmeasured ϭ Pideally exerted Ϫ
or
Pideally exerted ϭ Pmeasured ϩ
In this correction term, large values of a indicate strong attractive forces. When
more molecules are present (greater n) and when the molecules are close together
(smaller V2 in the denominator), the correction term becomes larger. The correction
term becomes negligibly small, however, when the volume is large.
When we substitute these two expressions for corrections into the ideal gas equation,
we obtain the equation
΂Pmeasured ϩ ΃ (Vmeasured Ϫ nb) ϭ nRT
or
΂P ϩ ΃(V Ϫ nb) ϭ nRT
This is the van der Waals equation. In this equation, P, V, T, and n represent the measured
values of pressure, volume, temperature (expressed on the absolute scale), and number of
moles, respectively, just as in the ideal gas equation. The quantities a and b are experimentally
derived constants that differ for different gases (Table 12-5). When a and b are
both zero, the van der Waals equation reduces to the ideal gas equation.
We can understand the relative values of a and b in Table 12-5 in terms of molecular
properties. Note that a for helium is very small. This is the case for all noble gases and
many other nonpolar molecules, because only very weak attractive forces, called dispersion
forces, exist between them. Dispersion forces result from short-lived electrical
dipoles produced by the attraction of one atom’s nucleus for an adjacent atom’s electrons.
These forces exist for all molecules but are especially important for nonpolar molecules,
which would never liquefy if dispersion forces did not exist. Polar molecules such as
ammonia, NH3, have permanent charge separations (dipoles), so they exhibit greater forces
of attraction for one another. This explains the high value of a for ammonia. Dispersion
forces and permanent dipole forces of attraction are discussed in more detail in Chapter 13.
Larger molecules have greater values of b. For instance, H2, a first-row diatomic molecule,
has a greater b value than the first-row monatomic He. The b value for CO2, which
contains three second-row atoms, is greater than that for N2, which contains only two
second-row atoms.
The following example illustrates the deviation of methane, CH4, from ideal gas
behavior under high pressure.
EXAMPLE 12-23 van der Waals Equation
Calculate the pressure exerted by 1.00 mole of methane, CH4, in a 500.-mL vessel at 25.0°C
assuming (a) ideal behavior and (b) nonideal behavior.
n2a
ᎏ
V2
n2a
ᎏᎏ
V2
measured
n2a
ᎏᎏ
V2
measured
n2a
ᎏᎏ
V2
measured
The van der Waals equation, like the
ideal gas equation, is known as an
equation of state, that is, an equation
that describes a state of matter.
12-15 Real Gases: Deviations from Ideality 473
TABLE 12-5 van der Waals
Constants
a b
Gas (L2иatm/mol2) (L/mol)
H2 0.244 0.0266
He 0.034 0.0237
N2 1.39 0.0391
NH3 4.17 0.0371
CO2 3.59 0.0427
CH4 2.25 0.0428
Plan
(a) Ideal gases obey the ideal gas equation. We can solve this equation for P.
(b) To describe methane as a nonideal gas, we use the van der Waals equation and solve
for P.
Solution
(a) Using the ideal gas equation to describe ideal gas behavior,
PV ϭ nRT
P ϭ ϭ ϭ 48.9 atm
(b) Using the van der Waals equation to describe nonideal gas behavior,
΂P ϩ ΃(V Ϫ nb) ϭ nRT
For CH4, a ϭ 2.25 L2иatm/mol2 and b ϭ 0.0428 L/mol (see Table 12-5).
΄P ϩ ΅΄0.500 L Ϫ (1.00 mol)΂0.0428 ΃΅
ϭ (1.00 mol)΂ ΃(298 K)
Combining terms and canceling units, we get
P ϩ 9.00 atm ϭ ᎏ
24
0
.5
.45
L
7
иa
L
tm
ᎏ ϭ 53.6 atm
P ϭ 44.6 atm
You should now work Exercises 96 and 97.
Repeating the calculations of Example 12-23 with the volume twenty times higher
(V ϭ 10.0 L) gives ideal and nonideal pressures, respectively, of 2.45 and 2.44 atm, a difference
of only 0.4%.
Many other equations have been developed to describe the behavior of real gases. Each
of these contains quantities that must be empirically derived for each gas.
0.0821 Lиatm
ᎏᎏ
molиK
L
ᎏ
mol
(1.00 mol)2(2.25 L2иatm/mol2)
ᎏᎏᎏᎏ
(0.500 L)2
n2a
ᎏ
V 2
(1.00 mol)΂ᎏ
0.08
m
21
ol
L
иK
иatm
ᎏ΃(298 K)
ᎏᎏᎏᎏ
0.500 L
nRT
ᎏ
V
The pressure is 4.3 atm (8.8%) less
than that calculated from the ideal
gas law. A significant error would be
introduced by assuming ideal behavior
at this high pressure.
474 CHAPTER 12: Gases and the Kinetic–Molecular Theory
Key Terms
Absolute zero The zero point on the absolute temperature scale;
Ϫ273.15°C or 0 K; theoretically, the temperature at which molecular
motion is a minimum.
Atmosphere (atm) A unit of pressure; the pressure that will support
a column of mercury 760 mm high at 0°C; 760 torr.
Avogadro’s Law At the same temperature and pressure,
equal volumes of all gases contain the same number of
molecules.
Barometer A device for measuring atmospheric pressure. See
Figures 12-1 and 12-2. The liquid is usually mercury.
Boyle’s Law At constant temperature, the volume occupied by
a given mass of a gas is inversely proportional to the applied
pressure.
Charles’s Law At constant pressure, the volume occupied by a
definite mass of a gas is directly proportional to its absolute
temperature.
Condensed states The solid and liquid states.
Dalton’s Law of Partial Pressures The total pressure exerted
by a mixture of gases is the sum of the partial pressures of the
individual gases.
Exercises 475
Diffusion The movement of a substance (e.g., a gas) into a space
or the mixing of one substance (e.g., a gas) with another.
Dispersion Forces Weak, short-range attractive forces between
short-lived temporary dipoles.
Effusion The escape of a gas through a tiny hole or a thin porous
wall.
Equation of state An equation that describes the behavior of
matter in a given state; for example, the van der Waals equation
describes the behavior of the gaseous state.
Fluids Substances that flow freely; gases and liquids.
Gay-Lussac’s Law of Combining Volumes At constant temperature
and pressure, the volumes of reacting gases (and any
gaseous products) can be expressed as ratios of small whole
numbers.
Ideal gas A hypothetical gas that obeys exactly all postulates of
the kinetic–molecular theory.
Ideal Gas Equation The product of the pressure and volume of
an ideal gas is directly proportional to the number of moles of
the gas and the absolute temperature.
Kinetic–molecular theory A theory that attempts to explain
macroscopic observations on gases in microscopic or molecular
terms.
Manometer A two-armed barometer. See Figure 12-1.
Mole fraction The number of moles of a component of a mixture
divided by the total number of moles in the mixture.
Partial pressure The pressure exerted by one gas in a mixture
of gases.
Pascal (Pa) The SI unit of pressure; it is defined as the pressure
exerted by a force of one newton acting on an area of one square
meter.
Pressure Force per unit area.
Real gases Gases that deviate from ideal gas behavior.
Root-mean-square speed, urms The square root of the
mean-square speed, ͙uෆෆ2ෆෆ. This is equal to
Ί๶for an ideal
gas. The root-mean-square speed is slightly different from the
average speed, but the two quantities are proportional.
Standard molar volume The volume occupied by one mole of
an ideal gas under standard conditions; 22.414 liters.
Standard temperature and pressure (STP) Standard temperature
0°C (273.15 K), and standard pressure, one atmosphere,
are standard conditions for gases.
Torr A unit of pressure; the pressure that will support a column
of mercury 1 mm high at 0°C.
Universal gas constant R, the proportionality constant in the
ideal gas equation, PV ϭ nRT.
van der Waals equation An equation of state that extends the
ideal gas law to real gases by inclusion of two empirically determined
parameters, which are different for different gases.
Vapor A gas formed by boiling or evaporation of a liquid or sublimation
of a solid; a term commonly used when some of the
liquid or solid remains in contact with the gas.
Vapor pressure The pressure exerted by a vapor in equilibrium
with its liquid or solid.
3RT
ᎏ
M
Exercises
You may assume ideal gas behavior unless otherwise
indicated.
Basic Ideas
*001. Define pressure. Give a precise scientific definition—one
that can be understood by someone without any scientific
training.
*002. State whether each property is characteristic of all gases,
some gases, or no gas: (a) transparent to light; (b) colorless;
(c) unable to pass through filter paper; (d) more
difficult to compress than liquid water; (e) odorless;
(f ) settles on standing.
*003. Describe the mercury barometer. How does it work?
*004. What is a manometer? How does it work?
*005. Express a pressure of 685 torr in the following units:
(a) mm Hg; (b) atm; (c) Pa; (d) kPa.
*006. A typical laboratory atmospheric pressure reading is 755
torr. Convert this value to (a) psi, (b) cm Hg, (c) inches
Hg, (d) kPa, (e) atm, and (f ) ft H2O.
*007. Complete the following table.
atm torr Pa kPa
Standard atmosphere 1
Partial pressure of
nitrogen in the
atmosphere 593
A tank of compressed
hydrogen 1.61 ϫ 105
Atmospheric pressure
at the summit
of Mt. Everest 33.7
*008. State whether each of the following samples of matter is
a gas. If the information is insufficient for you to decide,
write “insufficient information.”
(a) A material is in a steel tank at 100. atm pressure.
When the tank is opened to the atmosphere, the material
immediately expands, increasing its volume many-fold.
(b) A material, on being emitted from an industrial
476 CHAPTER 12: Gases and the Kinetic–Molecular Theory
smokestack, rises about 10 m into the air. Viewed against
a clear sky, it has a white appearance.
(c) 1.0 mL of material weighs 8.2 g.
(d) When a material is released from a point 30 ft below
the level of a lake at sea level (equivalent in pressure to
about 76 cm of mercury), it rises rapidly to the surface,
at the same time doubling its volume.
(e) A material is transparent and pale green in color.
(f ) One cubic meter of a material contains as many molecules
as 1 m3 of air at the same temperature and pressure.
00*9. The densities of mercury and corn oil are 13.5 g/mL and
0.92 g/mL, respectively. If corn oil were used in a barometer,
what would be the height of the column, in meters,
at standard atmospheric pressure? (The vapor pressure of
the oil is negligible.)
*010. Steel tanks for storage of gases are capable of withstanding
pressures greater than 125 atm. Express this pressure
in psi.
*011. Automobile tires are normally inflated to a pressure of 28
psi as measured by a tire gauge. (a) Express this pressure
in atmospheres. (b) Assuming standard atmospheric pressure,
calculate the internal pressure of the tire.
Boyle’s Law: The Pressure–Volume Relationship
*012. (a) On what kinds of observations (measurements) is
Boyle’s Law based? State the law. (b) Use the statement
of Boyle’s Law to derive a simple mathematical expression
for Boyle’s Law.
*013. Could the words “a fixed number of moles” be substituted
for “a definite mass” in the statement of Boyle’s
Law? Explain.
*014. A gas sample contained in a cylinder equipped with a
moveable piston occupied 300. mL at a pressure of 2.00
atm. What would be the final pressure if the volume were
increased to 500. mL at constant temperature?
*015. A balloon that contains 1.50 liters of air at 1.00 atm is
taken under water to a depth at which the pressure is 3.00
atm. Calculate the new volume of the balloon. Assume
that the temperature remains constant.
*016. A 50.-L sample of gas collected in the upper atmosphere
at a pressure of 18.3 torr is compressed into a 150.-mL
container at the same temperature. (a) What is the new
pressure, in atmospheres? (b) To what volume would the
original sample have had to be compressed to exert a pressure
of 10.0 atm?
*017. A sample of krypton gas occupies 75.0 mL at 0.400 atm.
If the temperature remained constant, what volume would
the krypton occupy at (a) 4.00 atm, (b) 0.00400 atm,
(c) 765 torr, (d) 4.00 torr, and (e) 3.5 ϫ 10Ϫ2 torr?
0*18. A cylinder containing 15 L of helium gas at a pressure of
165 atm is to be used to fill toy balloons to a pressure of
1.1 atm. Each inflated balloon has a volume of 2.0 L.
What is the maximum number of balloons that can be
inflated? (Remember that 15 L of helium at 1.1 atm will
remain in the “exhausted” cylinder.)
*019. (a) Can an absolute temperature scale based on Fahrenheit
rather than Celsius degrees be developed? Why?
(b) Can an absolute temperature scale that is based on a
“degree” twice as large as a Celsius degree be developed?
Why?
*020. (a) What does “absolute temperature scale” mean?
(b) Describe the experiments that led to the evolution of
the absolute temperature scale. What is the relationship
between the Celsius and Kelvin temperature scales?
(c) What does “absolute zero” mean?
*021. Complete the table by making the required temperature
conversions. Pay attention to significant figures.
Temperature
K °C
Normal boiling point
of water 100
Reference for
thermodynamic data 298.15
Dry ice becomes a gas at
atmospheric pressure Ϫ78.5
The center of the sun
(estimated) 1.53 ϫ 107
Charles’s Law: The Volume–Temperature
Relationship
*022. (a) Why is a plot of volume versus temperature at constant
pressure a straight line (see Figure 12-5)? (b) On
what kind of observations (measurements) is Charles’s
Law based? State the law.
*023. A gas occupies a volume of 31.0 L at 17.0°C. If the gas
temperature rises to 34.0°C at constant pressure,
(a) would you expect the volume to double to 62.0 L?
Explain. Calculate the new volume (b) at 34.0°C, (c) at
400. K, and (d) at 0.00°C.
*024. Several balloons are inflated with helium to a volume of
0.75 L at 27°C. One of the balloons was found several
hours later; the temperature had dropped to 22°C. What
would be the volume of the balloon when found, if no
helium has escaped?
*025. Which of the following statements are true? Which are
false? Why is each true or false? Assume constant pressure
in each case.
(a) If a sample of gas is heated from 100°C to 200°C, the
volume will double.
(b) If a sample of gas is heated from 0.°C to 273°C, the
volume will double.
(c) If a sample of gas is cooled from 1273°C to 500.°C,
the volume will decrease by a factor of 2.
Exercises 477
(d) If a sample of gas is cooled from 1000.°C to 200.°C,
the volume will decrease by a factor of 5.
(e) If a sample of gas is heated from 473°C to 1219°C,
the volume will increase by a factor of 2.
0*26. The device shown here is a gas thermometer. (a) At the
ice point, the gas volume is 1.400 L. What would be the
new volume if the gas temperature were raised from the
ice point to 8.0°C? (b) Assume the cross-sectional area of
the graduated arm is 1.0 cm2. What would be the difference
in height if the gas temperature changed from 0°C
to 8.0°C? (c) What modifications could be made to
increase the sensitivity of the thermometer?
*033. A 247-mL sample of a gas exerts a pressure of 3.13 atm
at 16.0°C. What volume would it occupy at 100.°C and
1.00 atm?
*034. Show how Boyle’s and Charles’s gas laws can be obtained
from the combined gas law equation.
STP, Standard Molar Volume, and Gas Densities
*035. How many molecules of an ideal gas are contained in a
2.00-L flask at STP?
*036. (a) What is Avogadro’s Law? What does it mean?
(b) What does “standard molar volume” mean? (c) Are
there conditions other than STP at which 1 mole of an
ideal gas would occupy 22.4 L? Explain.
*037. Sodium vapor has been detected recently as a major component
of the thin atmosphere of Mercury using a
ground-based telescope and a spectrometer. Its concentration
is estimated to be about 1.0 ϫ 105 atoms per cm3.
(a) Express this in moles per liter. (b) The maximum temperature
of the atmosphere was measured by Mariner 10
to be about 970.°C. What is the approximate partial pressure
of sodium vapor at that temperature?
*027. A weather balloon is filled to the volume of 150. L on a
day when the temperature is 19°C. If no gases escaped,
what would be the volume of the weather balloon after it
rises to an altitude where the temperature is Ϫ8°C?
*028. Calculate the volume of an ideal gas at the temperatures
of dry ice (Ϫ78.5°C), liquid N2 (Ϫ195.8°C), and liquid
He (Ϫ268.9°C) if it occupies 5.00 L at 25.0°C. Assume
constant pressure. Plot your results, and extrapolate to
zero volume. At what temperature would zero volume be
theoretically reached?
The Combined Gas Law
*029. Classify the relationship between the variables (a) P and
V, (b) V and T, and (c) P and T as either (i) directly
proportional or (ii) inversely proportional.
*030. Prepare sketches of plots of (a) P vs. V, (b) P vs. 1/V,
(c) V vs. T, and (d) P vs. T for an ideal gas.
*031. A sample of gas occupies 400. mL at STP. Under what
pressure would this sample occupy 200. mL if the temperature
were increased to 819°C?
*032. A 280.-mL sample of neon exerts a pressure of 660. torr
at 26°C. At what temperature in °C would it exert a pressure
of 940. torr in a volume of 440. mL?
Temperature
scale
Leveling
bulb (adjust
height until
the liquid
levels are
equal)
*038. The limit of sensitivity for the analysis of carbon monoxide,
CO, in air is 1 ppb (ppb ϭ parts per billion) by
volume. What is the smallest number of CO molecules
that can be detected in 10. L of air at STP?
*039. A lab instructor asked a student to calculate the number
of moles of gas contained in a 320.-mL bulb in the laboratory.
The student determined that the pressure and
temperature of the gas were 732 torr and 23.3°C. The
student calculated the following:
_?_mol ϭ 0.320 L ϫ ϭ 0.0143 mol
Was the student correct? Why?
*040. Ethylene dibromide (EDB) was formerly used as a fumigant
for fruits and grains, but now it is banned because
it is a potential health hazard. EDB is a liquid that boils
at 109°C. Its molecular weight is 188 g/mol. Calculate
the density of its vapor at 180°C and 1.00 atm.
1 mol
ᎏ
22.4 L
The surface of the planet Mercury.
478 CHAPTER 12: Gases and the Kinetic–Molecular Theory
0*41. A 503-mL flask contains 0.0243 mol of an ideal gas at a
given temperature and pressure. Another flask contains
0.0388 mol of the gas at the same temperature and pressure.
What is the volume of the second flask?
*042. A laboratory technician forgot what the color coding on
some commercial cylinders of gas meant, but remembered
that each of two specific tanks contained one of the
following gases: He, Ne, Ar, or Kr. Measurements at STP
made on samples of the gases from the two cylinders
showed the gas densities to be 3.74 g/L and 0.178 g/L.
(a) Determine by calculation which of these gases was
present in each tank. (b) Could this determination be
made if the densities had been at a temperature and pressure
different from STP?
The Ideal Gas Equation
*043. Calculate R in Lиatm/molиK, in kPaиdm3/molиK, in
J/molиK, and in kJ/molиK.
*044. (a) What is an ideal gas? (b) What is the ideal gas equation?
(c) Outline the logic used to obtain the ideal gas
equation. (d) What is R? How is it obtained?
*045. (a) A chemist is preparing to carry out a reaction at high
pressure that requires 36.0 mol of hydrogen gas. The
chemist pumps the hydrogen into a 12.3-L rigid steel container
at 25°C. To what pressure (in atmospheres) must
the hydrogen be compressed? (b) What would be the density
of the high-pressure hydrogen?
*046. Calculate the pressure needed to contain 2.44 mol of an
ideal gas at 45°C in a volume of 3.70 L.
*047. (a) How many molecules are in a 1.00-L container of
gaseous oxygen if the pressure is 2.50 ϫ 10Ϫ9 torr and the
temperature is 1225 K? (b) How many grams of oxygen
are in the container?
0*48. A barge containing 580. tons of liquid chlorine was
involved in an accident. (a) What volume would this
amount of chlorine occupy if it were all converted to a
gas at 750. torr and 18°C? (b) Assume that the chlorine
is confined to a width of 0.500 mile and an average depth
of 60. ft. What would be the length, in feet, of this chlorine
“cloud”?
Molecular Weights and Formulas for Gaseous
Compounds
*049. Analysis of a volatile liquid shows that it contains 37.23%
carbon, 7.81% hydrogen, and 54.96% chlorine by mass.
At 150.°C and 1.00 atm, 500. mL of the vapor has a mass
of 0.922 g. (a) What is the molecular weight of the compound?
(b) What is its molecular formula?
0*50. A student was given a container of ethane, C2H6, that had
been closed at STP. By making appropriate measurements,
the student found that the mass of the sample of
ethane was 0.244 g and the volume of the container was
185 mL. Use the student’s data to calculate the molecular
weight of ethane. What percent error is obtained?
Suggest some possible sources of the error.
*051. Calculate the molecular weight of a gaseous element if
0.480 g of the gas occupies 367 mL at 365 torr and 45°C.
Suggest the identity of the element.
*052. A cylinder was found in a storeroom of a manufacturing
plant. The label on the cylinder was gone and no one
remembered what the cylinder held. A 0.00500-gram
sample was found to occupy 4.13 mL at 23°C and 745
torr. The sample was also found to be composed of only
carbon and hydrogen. Identify the gas.
*053. A sample of porous rock was brought back from the planet
Farout on the other side of the galaxy. Trapped in the
rock was a carbon–oxygen gas. The unknown gas was
extracted and evaluated. A volume of 3.70 mL of the gas
was collected under the conditions of STP. The mass of
the gas sample was determined to be 0.00726 grams.
Additional analysis proved that there was only one compound
present in the sample. What was the most probable
identity of the gas based on these analyses?
0*54. A highly volatile liquid was allowed to vaporize completely
into a 250.-mL flask immersed in boiling water.
From the following data, calculate the molecular weight
(in amu per molecule) of the liquid. Mass of empty
flask ϭ 65.347 g; mass of flask filled with water at room
temperature ϭ 327.4 g; mass of flask and condensed
liquid ϭ 65.739 g; atmospheric pressure ϭ 743.3 torr;
temperature of boiling water ϭ 99.8°C; density of water
at room temperature ϭ 0.997 g/mL.
*055. A pure gas contains 85.63% carbon and 14.37% hydrogen
by mass. Its density is 2.50 g/L at STP. What is its
molecular formula?
Gas Mixtures and Dalton’s Law
*056. (a) What are partial pressures of gases? (b) State Dalton’s
Law. Express it symbolically.
*057. A sample of oxygen of mass 30.0 g is confined in a vessel
at 0°C and 1000. torr. Then 8.00 g of hydrogen is
pumped into the vessel at constant temperature. What
will be the final pressure in the vessel (assuming only mixing
with no reaction)?
*058. A gaseous mixture contains 5.23 g of chloroform, CHCl3,
and 1.66 g of methane, CH4. What pressure is exerted by
the mixture inside a 50.0-mL metal container at 275°C?
What pressure is contributed by the CHCl3?
*059. A cyclopropane–oxygen mixture can be used as an anesthetic.
If the partial pressures of cyclopropane and oxygen
are 140. torr and 560. torr, respectively, what is the ratio
of the number of moles of cyclopropane to the number
of moles of oxygen in this mixture? What is the corresponding
ratio of molecules?
*060. What is the mole fraction of each gas in a mixture having
the partial pressures of 0.267 atm of He, 0.317 atm
of Ar, and 0.277 atm of Xe?
Exercises 479
0*61. Assume that unpolluted air has the composition shown in
Table 12-2. (a) Calculate the number of molecules of N2,
of O2, and of Ar in 1.00 L of air at 21°C and 1.00 atm.
(b) Calculate the mole fractions of N2, O2, and Ar in the
air.
*062. Individual samples of O2, N2, and He are present in three
2.25-L vessels. Each exerts a pressure of 1.50 atm. (a) If
all three gases are forced into the same 1.00-L container
with no change in temperature, what will be the resulting
pressure? (b) What is the partial pressure of O2 in the
mixture? (c) What are the partial pressures of N2 and He?
*063. Hydrogen was collected over water at 20°C and 755 torr.
The volume of this gas sample was 45.3 mL. What volume
would the dry hydrogen occupy at STP?
*064. A nitrogen sample occupies 447 mL at STP. If the same
sample were collected over water at 25°C and 750. torr,
what would be the volume of the gas sample?
0*65. A study of climbers who reached the summit of Mt. Everest
without supplemental oxygen revealed that the partial
pressures of O2 and CO2 in their lungs were 35 torr and
7.5 torr, respectively. The barometric pressure at the summit
was 253 torr. Assume that the lung gases are saturated
with moisture at a body temperature of 37°C. Calculate
the partial pressure of inert gas (mostly nitrogen) in the
climbers’ lungs.
*066. A 4.00-L flask containing He at 6.00 atm is connected to
a 3.00-L flask containing N2 at 3.00 atm and the gases
are allowed to mix. (a) Find the partial pressures of each
gas after they are allowed to mix. (b) Find the total pressure
of the mixture. (c) What is the mole fraction of
helium?
*067. A 3.46-liter sample of a gas was collected over water on
a day when the temperature was 21°C and the barometric
pressure was 718 torr. The dry sample of gas had a
mass of 4.20 g. What is the molecular weight of the gas?
At 21°C the vapor pressure of water is 18.65 torr.
Stoichiometry in Reactions Involving Gases
*068. During a collision, automobile air bags are inflated by the
N2 gas formed by the explosive decomposition of sodium
azide, NaN3.
2NaN3 88n 2Na ϩ 3N2
What mass of sodium azide would be needed to inflate a
30.0-L bag to a pressure of 1.40 atm at 25°C?
*069. Assuming the volumes of all gases in the reaction are measured
at the same temperature and pressure, calculate the
volume of water vapor obtainable by the explosive reaction
of a mixture of 440. mL of hydrogen gas and 325
mL of oxygen gas.
0*70. One liter of sulfur vapor, S8(g), at 600°C and 1.00 atm is
burned in excess pure oxygen to give sulfur dioxide gas,
SO2, measured at the same temperature and pressure.
What mass of SO2 gas is obtained?
*071. Calculate the volume of methane, CH4, measured at
300. K and 815 torr, that can be produced by the bacterial
breakdown of 1.00 kg of a simple sugar.
C6H12O6 88n 3CH4 ϩ 3CO2
0*72. A common laboratory preparation of oxygen is
MnO2
2KClO3(s) 8888n 2KCl(s) ϩ 3O2(g)
heat
If you were designing an experiment to generate four bottles
(each containing 250. mL) of O2 at 25°C and 755
torr and allowing for 25% waste, what mass of potassium
chlorate would be required?
*073. Many campers use small propane stoves to cook meals.
What volume of air (see Table 12-2) will be required to
burn 11.5 L of propane, C3H8? Assume all gas volumes
are measured at the same temperature and pressure.
C3H8(g) ϩ 5O2(g) 88n 3CO2(g) ϩ 4H2O(g)
0*74. If 2.00 L of nitrogen and 5.00 L of hydrogen were allowed
to react, how many liters of NH3(g) could form? Assume
all gases are at the same temperature and pressure, and
that the limiting reactant is used up.
N2(g) ϩ 3H2(g) 88n 2NH3(g)
0*75. We burn 15.00 L of ammonia in 20.00 L of oxygen at
500.°C. What volume of nitric oxide, NO, gas can form?
What volume of steam, H2O(g), is formed? Assume all
gases are at the same temperature and pressure, and that
the limiting reactant is used up.
4NH3(g) ϩ 5O2(g) 88n 4NO(g) ϩ 6H2O(g)
*076. What mass of KNO3 would have to be decomposed to
produce 21.1 L of oxygen measured at STP?
heat
2KNO3(s) 888n 2KNO2(s) ϩ O2(g)
*077. Refer to Exercise 76. An impure sample of KNO3 that
had a mass of 50.3 g was heated until all of the KNO3
had decomposed. The liberated oxygen occupied 4.22 L
at STP. What percentage of the sample was KNO3?
Assume that no impurities decompose to produce
oxygen.
0*78. Heating a 5.913-g sample of an ore containing a metal
sulfide, in the presence of excess oxygen, produces 1.177
L of dry SO2, measured at 35.0°C and 755 torr. Calculate
the percentage by mass of sulfur in the ore.
0*79. The following reactions occur in a gas mask (a selfcontained
breathing apparatus) sometimes used by underground
miners. The H2O and CO2 come from exhaled
air, and O2 is inhaled as it is produced. KO2 is potassium
superoxide. The CO2 is converted to the solid salt
480 CHAPTER 12: Gases and the Kinetic–Molecular Theory
KHCO3, potassium hydrogen carbonate, so that CO2 is
not inhaled in significant amounts.
4KO2(s) ϩ 2H2O(ᐉ) 88n 4KOH(s) ϩ 3O2(g)
CO2(g) ϩ KOH(s) 88n KHCO3(s)
(a) What volume of O2, measured at STP, is produced by
the complete reaction of 1.00 g of KO2? (b) What is this
volume at body temperature, 37°C, and 1.00 atm?
(c) What mass of KOH is produced in part (a)? (d) What
volume of CO2, measured at STP, will react with the mass
of KOH of part (c)? (e) What is the volume of CO2 in
part (d) measured at 37°C and 1.00 atm?
0*80. Let us represent gasoline as octane, C8H18. When hydrocarbon
fuels burn in the presence of sufficient oxygen,
CO2 is formed.
Reaction A: 2C8H18 ϩ 25O2 88n 16CO2 ϩ 18H2O
But when the supply of oxygen is limited, the poisonous
gas carbon monoxide, CO, is formed.
Reaction B: 2C8H18 ϩ 17O2 88n 16CO ϩ 18H2O
Any automobile engine, no matter how well tuned, burns
its fuel by some combination of these two reactions. Suppose
an automobile engine is running at idle speed in a
closed garage with air volume 97.5 m3. This engine burns
95.0% of its fuel by reaction A, and the remainder by
reaction B. (a) How many liters of octane, density 0.702
g/mL, must be burned for the CO to reach a concentration
of 2.00 g/m3? (b) If the engine running at idle speed
burns fuel at the rate of 1.00 gal/h (0.0631 L/min), how
long does it take to reach the CO concentration in (a)?
The Kinetic—Molecular Theory and
Molecular Speeds
*081. Outline the kinetic–molecular theory.
*082. The radius of a typical molecule of a gas is 2.00 Å.
(a) Find the volume of a molecule assuming it to be spherical.
For a sphere, V ϭ 4/3 ␲r3. (b) Calculate the volume
actually occupied by 1.00 mol of these molecules. (c) If
1.0 mol of this gas occupies 22.4 L, find the fraction of
the volume actually occupied by the molecules. (d) Comment
on your answer to part (c) in view of the first
statement summarizing the kinetic–molecular theory of
an ideal gas.
*083. How does the kinetic–molecular theory explain
(a) Boyle’s Law? (b) Dalton’s Law? (c) Charles’s Law?
*084. SiH4 molecules are heavier than CH4 molecules; yet,
according to kinetic–molecular theory, the average
kinetic energies of the two gases at the same temperature
are equal. How can this be?
0*85. At 22°C, Cl2 molecules have some rms speed (which we
need not calculate). At what temperature would the rms
speed of F2 molecules be the same?
0*86. (a) How do average speeds of gaseous molecules vary with
temperature? (b) Calculate the ratio of the rms speed of
N2 molecules at 100.°C to the rms speed of the same molecules
at 0.0°C.
*087. How do the average kinetic energies and average speeds
of each gas in a mixture compare?
*088. (a) If you heat a gaseous sample in a fixed volume container,
the pressure increases. Use the kinetic–molecular
theory to explain the increased pressure. (b) If the volume
of a gaseous sample is reduced at constant temperature,
the pressure increases. Use the kinetic–molecular theory
to explain the increase in pressure.
Real Gases and Deviations from Ideality
*089. What is the van der Waals equation? How does it differ
from the ideal gas equation?
*090. Which of the following gases would be expected to behave
most nearly ideally under the same conditions? H2, F2,
HF. Which one would be expected to deviate from ideal
behavior the most? Explain both answers.
*091. Does the effect of intermolecular attraction on the properties
of a gas become more significant or less significant
if (a) the gas is compressed to a smaller volume at constant
temperature? (b) more gas is forced into the same
volume at the same temperature? (c) the temperature of
the gas is raised at constant pressure?
*092. Does the effect of molecular volume on the properties of
a gas become more significant or less significant if (a) the
gas is compressed to a smaller volume at constant temperature?
(b) more gas is forced into the same volume at
the same temperature? (c) the temperature of the gas is
raised at constant pressure?
*093. A sample of gas has a molar volume of 10.1 L at a pressure
of 745 torr and a temperature of Ϫ138°C. Is the gas
behaving ideally?
*094. Calculate the compressibility factor, (Preal)(Vreal)/RT,
for a 1.00-mol sample of NH3 under the following
conditions: in a 500.-mL vessel at Ϫ10.0°C it exerts a
pressure of 30.0 atm. What would be the ideal pressure
for 1.00 mol of NH3 at Ϫ10.0°C in a 500.-mL vessel?
Compare this with the real pressure and account for the
difference.
*095. (a) How do “real” and “ideal” gases differ? (b) Under what
kinds of conditions are deviations from ideality most
important? Why?
*096. Find the pressure of a sample of carbon tetrachloride,
CCl4, if 1.00 mol occupies 35.0 L at 77.0°C (slightly
above its normal boiling point). Assume that CCl4 obeys
(a) the ideal gas law; (b) the van der Waals equation.
The van der Waals constants for CCl4 are a ϭ 20.39
L2иatm/mol2 and b ϭ 0.1383 L/mol.
*097. Repeat the calculations of Exercise 96 using a 3.10-mol
gas sample confined to 6.15 L at 135°C.
Exercises 481
Mixed Exercises
*098. A student is to perform a laboratory experiment that
requires the evolution and collection of 75 mL of dry oxygen
gas at one atmosphere and 25°C. What is the
minimum mass of water required to generate the oxygen
by electrolysis of water?
*099. A tilting McLeod gauge is used to measure very low pressures
of gases in glass vacuum lines in the laboratory. It
operates by compressing a large volume of gas at low pressure
to a much smaller volume so that the pressure is
more easily measured. What is the pressure of a gas in a
vacuum line if a 53.3-mL volume of the gas, when compressed
to 0.133 mL, supports a 16.9-mm column of
mercury?
Relative humidity
ϭ
Often this quantity is multiplied by 100 to give the percent
relative humidity. Suppose the percent relative
humidity is 80.0% at 91.4°F (33.0°C) in a house with volume
245 m3. Then an air conditioner is turned on. Due
to the condensation of water vapor on the cold coils of
the air conditioner, water vapor is also removed from the
air as it cools. After the air temperature has reached
77.0°F (25.0°C), the percent relative humidity is measured
to be 15.0%. (a) What mass of water has been
removed from the air in the house? (Reminder: Take into
account the difference in saturated water vapor pressure
at the two temperatures.) (b) What volume would this liquid
water occupy at 25°C? (Density of liquid water at
25.0°C ϭ 0.997 g/cm3.)
*105. A 450.-mL flask contains 0.500 g of nitrogen gas at a pressure
of 744 torr. Are these data sufficient to allow you to
calculate the temperature of the gas? If not, what is missing?
If so, what is the temperature in °C?
*106. Use both the ideal gas law and the van der Waals equation
to calculate the pressure exerted by a 10.0-mol
sample of ammonia in a 60.0-L container at 100.°C. By
what percentage do the two results differ?
*107. What volume of hydrogen fluoride at 743 torr and 24°C
will be released by the reaction of 38.3 g of xenon difluoride
with a stoichiometric amount of water? The
unbalanced equation is
XeF2(s) ϩ H2O(ᐉ) 88n Xe(g) ϩ O2(g) ϩ HF(g)
What volumes of oxygen and xenon will be released under
these conditions?
*108. Cyanogen is 46.2% carbon and 53.8% nitrogen by mass.
At a temperature of 25°C and a pressure of 750. torr,
1.00 g of cyanogen gas occupies 0.476 L. Determine
the empirical formula and the molecular formula of
cyanogen.
*109. Incandescent light bulbs contain noble gases, such as
argon, so that the filament will last longer. The approximate
volume of a 100.-watt bulb is 130. cm3, and the
bulb contains 0.125 g of argon. How many grams of argon
would be contained in a 150.-watt bulb under the same
pressure and temperature conditions if the volume of the
larger wattage bulb is 180. cm3?
CONCEPTUAL EXERCISES
*110. An attempt was made to collect carbon dioxide, isolated
from the decomposition of a carbonate-containing mineral,
by first bubbling the gas through pure liquid acetic
acid. The experiment yielded 500. mL of a gaseous mixture
of acetic acid and carbon dioxide at 1.00 atm and
actual partial pressure of H2O vapor
ᎏᎏᎏᎏᎏ
partial pressure of H2O vapor if sat’d
*100. Imagine that you live in a cabin with an interior volume
of 150. m3. On a cold morning your indoor air temperature
is 10.°C, but by the afternoon the sun has warmed
the cabin air to 18°C. The cabin is not sealed; therefore,
the pressure inside is the same as it is outdoors. Assume
that the pressure remains constant during the day. How
many cubic meters of air would have been forced out of
the cabin by the sun’s warming? How many liters?
*101. A particular tank can safely hold gas up to a pressure of
44.3 atm. When the tank contains 38.1 g of N2 at 25°C,
the gas exerts a pressure of 10.1 atm. What is the highest
temperature to which the gas sample can be heated
safely?
*102. Find the molecular weight of Freon-12 (a chlorofluorocarbon)
if 8.29 L of vapor at 200.°C and 790. torr has a
mass of 26.8 g.
*103. A flask of unknown volume was filled with air to a pressure
of 3.25 atm. This flask was then attached to an
evacuated flask with a known volume of 5.00 L, and the
air was allowed to expand into the flask. The final pressure
of the air (in both flasks) was 2.40 atm. Calculate the
volume of the first flask.
*104. Relative humidity is the ratio of the pressure of water
vapor in the air to the pressure of water vapor in air that
is saturated with water vapor at the same temperature.
A McLeod gauge.
482 CHAPTER 12: Gases and the Kinetic–Molecular Theory
16.0°C. The vapor pressure of pure acetic acid at 16.0°C
is 400. torr. What should be the total mass of the collected
sample?
*111. The ideal gas equation deals with one set of volume, temperature,
and pressure conditions; the combined gas law
equation deals with two sets of conditions. Derive the
combined gas law equation from the ideal gas equation.
*112. Redraw Figure 12-12 so that it not only depicts the
decrease in space between molecules as the system is
cooled from 600. K to 300. K but also emphasizes the
change in kinetic energy.
*113. Suppose you were asked to supply a particular mass of a
specified gas in a container of fixed volume at a specified
pressure and temperature. Is it likely that you could fulfill
the request? Explain.
*114. The gas molecules in the box undergo a reaction at constant
temperature and pressure.
example earlier in this textbook where percent was used
instead of a fraction. Write an equation that relates mole
fraction and mole percent.
*117. Explain why gases at high pressure or low temperature
do not act as ideal gases.
BUILDING YOUR KNOWLEDGE
*118. A 5.00-L reaction vessel contains hydrogen at a partial
pressure of 0.588 atm and oxygen gas at a partial pressure
of 0.302 atm. Which element is the limiting reactant
in the following reaction?
2H2(g) ϩ O2(g) 88n 2H2O(g)
*119. Suppose the gas mixture in Exercise 118 is ignited and
the reaction produces the theoretical yield of the product.
What would be the partial pressure of each substance
present in the final mixture?
*120. A 0.422-g sample of pentane, C5H12, is placed in a
4.00-L reaction vessel with excess O2. The mixture is then
ignited and the sample burns with complete combustion.
What will be the partial pressures of CO2 and of H2O(g)
in the reaction vessel if the final temperature is 300.°C?
*121. When magnesium carbonate, MgCO3, is heated to a high
temperature, it decomposes.
heat
MgCO3(s) 888n MgO(s) ϩ CO2(g)
A 20.29-gram sample of impure magnesium carbonate is
completely decomposed at 1000.°C in a previously evacuated
2.00-L reaction vessel. After the reaction was
complete, the solid residue (consisting only of MgO and
the original impurities) had a mass of 15.90 grams.
Assume that no other constituent of the sample produced
a gas and that the volume of any solid was negligible compared
with the gas volume. (a) How many grams of CO2
were produced? (b) What was the pressure of the CO2
produced? (c) What percent of the original sample was
magnesium carbonate?
*122. One natural source of atmospheric carbon dioxide is precipitation
reactions such as the precipitation of silicates
in the oceans.
Mg2ϩ(aq) ϩ SiO2(dispersed) ϩ 2HCO3
Ϫ(aq) 88n
MgSiO3(s) ϩ 2CO2(g) ϩ H2O(ᐉ)
How many grams of magnesium silicate would be precipitated
during the formation of 100. L of carbon dioxide
at 30.°C and 775 torr?
*123. Table 12-2 states that dry air is 20.94% (by volume) oxygen.
What is the partial pressure of oxygen under the
conditions of STP? Hint: For gaseous samples the mole
ratios are equal to the volume ratios.
If the initial volume is 1.8 L and the final volume is 0.9
L, which of the following boxes could be the products of
the reaction? Explain your reasoning.
(a) (b) (c)
(d) (e)
*115. Use Dalton’s Law of Partial Pressures to illustrate the
changes that take place in the partial pressure of compound
A and the total pressure as gas B is pumped into
a closed system containing A.
*116. Mole fraction is one of the few quantities that is expressed
as a fraction. Frequently, percent is used instead. Cite an
13Liquids and Solids
OUTLINE
13-1 Kinetic–Molecular Description
of Liquids and Solids
13-2 Intermolecular Attractions and
Phase Changes
The Liquid State
13-3 Viscosity
13-4 Surface Tension
13-5 Capillary Action
13-6 Evaporation
13-7 Vapor Pressure
13-8 Boiling Points and Distillation
13-9 Heat Transfer Involving Liquids
The Solid State
13-10 Melting Point
13-11 Heat Transfer Involving Solids
13-12 Sublimation and the Vapor
Pressure of Solids
13-13 Phase Diagrams (P versus T)
13-14 Amorphous Solids and
Crystalline Solids
13-15 Structures of Crystals
13-16 Bonding in Solids
13-17 Band Theory of Metals
OBJECTIVES
After you have studied this chapter, you should be able to
• Describe the properties of liquids and solids and how they differ from gases
• Understand the kinetic–molecular description of liquids and solids, and show how this
description differs from that for gases
• Use the terminology of phase changes
• Understand various kinds of intermolecular attractions and how they are related to
physical properties such as vapor pressure, viscosity, melting point, and boiling point
• Describe evaporation, condensation, and boiling in molecular terms
• Calculate the heat transfer involved in warming or cooling without change of phase
• Calculate the heat transfer involved in phase changes
• Describe melting, solidification, sublimation, and deposition in molecular terms
• Interpret P versus T phase diagrams
• Describe the regular structure of crystalline solids
• Describe various types of solids
• Relate the properties of different types of solids to the bonding or interactions among
particles in these solids
• Visualize some common simple arrangements of atoms in solids
• Carry out calculations relating atomic arrangement, density, unit cell size, and ionic or
atomic radii in some simple crystalline arrangements
• Describe the bonding in metals
• Explain why some substances are conductors, some are insulators, and others are
semiconductors
The shapes of soap bubbles are due
to surface tension, an important
physical property of liquids. White
light striking the bubbles gives
brightly colored interference
patterns.
TT
he molecules of most gases are so widely separated at ordinary temperatures and
pressures that they do not interact with one another significantly. The physical
properties of gases are reasonably well described by the simple relationships in
Chapter 12. In liquids and solids, the so-called condensed phases, the particles are close
together so they interact strongly. Although the properties of liquids and solids can be
described, they cannot be adequately explained by simple mathematical relationships. Table
13-1 and Figure 13-1 summarize some of the characteristics of gases, liquids, and solids.
Figure 13-1 Representations of the kinetic–molecular description of the three phases of
matter.
CHAPTER 13: Liquids and Solids 485
TABLE 13-1 Some Characteristics of Solids, Liquids, and Gases
Solids
1. Have definite shape (resist
deformation)
2. Are nearly incompressible
3. Usually have higher density than
liquids
4. Are not fluid
5. Diffuse only very slowly through
solids
6. Have an ordered arrangement of
particles that are very close together;
particles have vibrational motion only
Liquids
1. Have no definite shape (assume
shapes of containers)
2. Have definite volume (are only very
slightly compressible)
3. Have high density
4. Are fluid
5. Diffuse through other liquids
6. Consist of disordered clusters of
particles that are quite close together;
particles have random motion in three
dimensions
Gases
1. Have no definite shape (fill containers
completely)
2. Are compressible
3. Have low density
4. Are fluid
5. Diffuse rapidly
6. Consist of extremely disordered
particles with much empty space
between them; particles have rapid,
random motion in three dimensions
Heat
Cool
Heat or
decrease
pressure
Cool or
increase
pressure
Solid
Liquid
Gas
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 13.2, Phases of Matter; the
Kinetic–Molecular Theory. This screen
contains an animated version of Figure
13-1.
KINETIC–MOLECULAR DESCRIPTION
OF LIQUIDS AND SOLIDS
The properties listed in Table 13-1 can be qualitatively explained in terms of the kinetic–
molecular theory of Chapter 12. We saw in Section 12-13 that the average kinetic energy
of a collection of gas molecules decreases as the temperature is lowered. As a sample of
gas is cooled and compressed, the rapid, random motion of gaseous molecules decreases.
The molecules approach one another, and the intermolecular attractions increase. Eventually,
these increasing intermolecular attractions overcome the reduced kinetic energies.
At this point condensation (liquefaction) occurs. The temperatures and pressures required
for condensation vary from gas to gas, because different kinds of molecules have different
attractive forces.
In the liquid state the forces of attraction among particles are great enough that disordered
clustering occurs. The particles are so close together that very little of the volume
occupied by a liquid is empty space. As a result, it is very hard to compress a liquid. Particles
in liquids have sufficient energy of motion to overcome partially the attractive forces
among them. They are able to slide past one another so that liquids assume the shapes of
their containers up to the volume of the liquid.
Liquids diffuse into other liquids with which they are miscible. For example, when a
drop of red food coloring is added to a glass of water, the water becomes red throughout
after diffusion is complete. The natural diffusion rate is slow at normal temperatures.
Because the average separations among particles in liquids are far less than those in gases,
the densities of liquids are much higher than the densities of gases (Table 12-1).
Cooling a liquid lowers its molecular kinetic energy and causes its molecules to slow
down even more. If the temperature is lowered sufficiently, at ordinary pressures, stronger
but shorter-range attractive interactions overcome the reduced kinetic energies of the
molecules to cause solidification. The temperature required for crystallization at a given pressure
depends on the nature of short-range interactions among the particles and is
characteristic of each substance.
Most solids have ordered arrangements of particles with a very restricted range of
motion. Particles in the solid state cannot move freely past one another so they only vibrate
about fixed positions. Consequently, solids have definite shapes and volumes. Because the
particles are so close together, solids are nearly incompressible and are very dense relative
to gases. Solid particles do not diffuse readily into other solids. However, analysis of
13-1
Figure 13-2 A representation of
diffusion in solids. When blocks of
two different metals are clamped
together for a long time, a few
atoms of each metal diffuse into the
other metal.
486 CHAPTER 13: Liquids and Solids
Lead Copper
Intermolecular attractions are those
between different molecules or ions.
Intramolecular attractions are those
between atoms within a single
molecule or ion.
The miscibility of two liquids refers
to their ability to mix and produce a
homogeneous solution.
Solidification and crystallization refer to
the process in which a liquid changes
to a solid.
two blocks of different solids, such as copper and lead, that have been pressed together
for a period of years shows that each block contains some atoms of the other element.
This demonstrates that solids do diffuse, but very slowly (Figure 13-2).
INTERMOLECULAR ATTRACTIONS AND PHASE
CHANGES
We have seen (Section 12-15) how the presence of strong attractive forces between gas
molecules can cause gas behavior to become nonideal when the molecules get close
together. In liquids and solids the molecules are much closer together than in gases. As a
result, properties of liquids, such as boiling point, vapor pressure, viscosity, and heat of
vaporization, depend markedly on the strengths of the intermolecular attractive forces.
These forces are also directly related to the properties of solids, such as melting point and
heat of fusion. Let us preface our study of these condensed phases with a discussion of
the types of attractive forces that can exist between molecules and ions.
Intermolecular forces refer to the forces between individual particles (atoms, molecules,
ions) of a substance. These forces are quite weak relative to intramolecular forces, that is,
covalent and ionic bonds within compounds. For example, 920 kJ of energy is required to
decompose one mole of water vapor into H and O atoms. This reflects the strength of
intramolecular forces (covalent bonds).
Only 40.7 kJ is required to convert one mole of liquid water into steam at 100°C.
H2O(ᐉ) 88n H2O(g) (absorbs 40.7 kJ/mol)
This reflects the strength of the intermolecular forces of attraction between the water
molecules, mainly hydrogen bonding.
If it were not for the existence of intermolecular attractions, condensed phases (liquids
and solids) could not exist. These are the forces that hold the particles close to one another
in liquids and solids. As we shall see, the effects of these attractions on melting points of
solids parallel those on boiling points of liquids. High boiling points are associated with
compounds that have strong intermolecular attractions. Let us consider the effects of the
general types of forces that exist among ionic, covalent, and monatomic species.
Ion–Ion Interactions
According to Coulomb’s Law, the force of attraction between two oppositely charged ions
is directly proportional to the charges on the ions, qϩ and qϪ, and inversely proportional
to the square of the distance between them, d.
F ϰ
Energy has the units of force ϫ distance, F ϫ d, so the energy of attraction between two
oppositely charged ions is directly proportional to the charges on the ions and inversely
proportional to the distance of separation.
E ϰ
qϩqϪ
ᎏ
d
qϩqϪ
ᎏ
d2
O H(g) 2H (g) O (g)H ϩ (absorbs 920 kJ/mol)
13-2
It is important to be able to tell
whether a substance is ionic, nonpolar
covalent, or polar covalent. You should
review the discussion of bonding in
Chapters 7 and 8.
13-2 Intermolecular Attractions and Phase Changes 487
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 13.3, Intermolecular Forces (1).
When oppositely charged ions are
close together, d (the denominator)
is small, so F, the attractive force
between them, is large.
Ionic compounds such as NaCl, CaBr2, and K2SO4 exist as extended arrays of discrete
ions in the solid state. As we shall see in Section 13-16, the oppositely charged ions in
these arrays are quite close together. As a result of these small distances, d, the energies
of attraction in these solids are substantial. Most ionic bonding is strong, and as a result
most ionic compounds have relatively high melting points (Table 13-2). At high enough
temperatures, ionic solids melt as the added heat energy overcomes the potential energy
associated with the attraction of oppositely charged ions. The ions in the resulting liquid
samples are free to move about, which accounts for the excellent electrical conductivity
of molten ionic compounds.
For most substances, the liquid is less dense than the solid, but H2O is one of the rare
exceptions. Melting a solid nearly always produces greater average separations among the
particles. This means that the forces (and energies) of attractions among the ions in an
ionic liquid are less than in the solid state because the average d is greater in the melt.
However, these energies of attraction are still much greater in magnitude than the energies
of attraction among neutral species (molecules or atoms).
The product qϩqϪ increases as the charges on ions increase. Ionic substances containing
multiply charged ions, such as Al3ϩ, Mg2ϩ, O2Ϫ, and S2Ϫ ions, usually have higher melting
and boiling points than ionic compounds containing only singly charged ions, such as
Naϩ, Kϩ, FϪ, and ClϪ. For a series of ions of similar charges, the closer approach of
smaller ions results in stronger interionic attractive forces and higher melting points
(compare NaF, NaCl, and NaBr in Table 13-2).
Dipole–Dipole Interactions
Permanent dipole–dipole interactions occur between polar covalent molecules
because of the attraction of the ␦ϩ atoms of one molecule to the ␦Ϫ atoms of another
molecule (Section 7-9).
Electrostatic forces between two ions decrease by the factor 1/d2 as their separation,
d, increases. But dipole–dipole forces vary as 1/d4. Because of the higher power of d in
the denominator, 1/d4 diminishes with increasing d much more rapidly than does 1/d2.
As a result, dipole forces are effective only over very short distances. Furthermore, for
dipole–dipole forces, qϩ and qϪ represent only “partial charges,” so these forces are weaker
than ion–ion forces. Average dipole–dipole interaction energies are approximately 4 kJ
per mole of bonds. They are much weaker than ionic and covalent bonds, which have
typical energies of about 400 kJ per mole of bonds. Substances in which permanent dipole–
dipole interactions affect physical properties include bromine fluoride, BrF, and sulfur
Ionic bonding may be thought of as
both inter- and intramolecular bonding.
488 CHAPTER 13: Liquids and Solids
TABLE 13-2 Melting Points of Some Ionic Compounds
Compound mp (°C) Compound mp (°C) Compound mp (°C)
NaF 993 CaF2 1423 MgO 2800
NaCl 801 Na2S 1180 CaO 2580
NaBr 747 K2S 840 BaO 1923
KCl 770
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 13.4, Intermolecular Forces (2),
especially the subsection “Dipole–
Dipole Forces.”
dioxide, SO2. Dipole–dipole interactions are illustrated in Figure 13-3. All dipole–dipole
interactions, including hydrogen bonding (discussed in the following section), are somewhat
directional. An increase in temperature causes an increase in translational, rotational,
and vibrational motion of molecules. This produces more random orientations of molecules
relative to one another. Consequently, dipole–dipole interactions become less
important as temperature increases. All these factors make compounds having only dipole–
dipole interactions more volatile than ionic compounds.
Hydrogen Bonding
Hydrogen bonds are a special case of very strong dipole–dipole interaction. They are
not really chemical bonds in the formal sense.
Strong hydrogen bonding occurs among polar covalent molecules containing H and
one of the three small, highly electronegative elements—F, O, or N.
Like ordinary dipole–dipole interactions, hydrogen bonds result from the attractions
between ␦ϩ atoms of one molecule, in this case H atoms, and the ␦Ϫ atoms of another
molecule. The small sizes of the F, O, and N atoms, combined with their high electronegativities,
concentrate the electrons of these molecules around these ␦Ϫ atoms. This
causes an H atom bonded to one of these highly electronegative elements to become quite
positive. The ␦ϩ H atom is attracted to a lone pair of electrons on an F, O, or N atom
other than the atom to which it is covalently bonded (Figure 13-4). The molecule that
contains the hydrogen-bonding ␦ϩ H atom is often referred to as the hydrogen-bond donor;
the ␦Ϫ atom to which it is attracted is called the hydrogen-bond acceptor.
Recently, careful studies of light absorption and magnetic properties in solution and of
the arrangements of molecules in solids have led to the conclusion that the same kind of
attraction occurs (although more weakly) when H is bonded to carbon. In some cases,
very weak CXH,O “hydrogen bonds” exist. Similar observations suggest the existence
Figure 13-3 Dipole–dipole
interactions among polar molecules.
(a) Bromine fluoride, BrF. (b) Sulfur
dioxide, SO2. Each polar molecule is
shaded with regions of highest
negative charge (␦Ϫ) darkest and
regions of highest positive charge
(␦ϩ) lightest. Attractive forces are
shown as XAX, and repulsive forces
are shown as XRX. Molecules tend
to arrange themselves to maximize
attractions by bringing regions of
opposite charge together while
minimizing repulsions by separating
regions of like charge.
13-2 Intermolecular Attractions and Phase Changes 489
A
R
R
R
A
A
R
R
R
R
R
A
A
A
A A
A
A A
AAAA
A
A
A
A
A
A
A
shown as
Br
(a) (b)
F
␦ϩ
␦ϩ
␦ϩ ␦ϩ
␦ϩ
␦ϩ
␦ϩ
␦ϩ
␦ϩ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
␦ϩ
␦ϩ
␦ϩ
␦ϩ
␦ϩ
␦ϩ
␦ϩ
␦ϩ
␦ϩ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ
␦Ϫ ␦Ϫ ␦Ϫ␦Ϫ
␦Ϫ
shown
asO O
S
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 13.6, Hydrogen Bonding.
of weak hydrogen bonds to chlorine atoms, such as OXH,Cl. However, most chemists
usually restrict usage of the term “hydrogen bonding” to compounds in which H is covalently
bonded to F, O, or N, and we will do likewise throughout this text.
Typical hydrogen-bond energies are in the range 15 to 20 kJ/mol, which is four to five
times greater than the energies of other dipole–dipole interactions. As a result, hydrogen
bonds exert a considerable influence on the properties of substances. Hydrogen bonding
is responsible for the unusually high melting and boiling points of compounds such as
water, methyl alcohol, and ammonia compared with other compounds of similar molecular
weight and molecular geometry (Figure 13-5). Hydrogen bonding between amino
acid subunits is very important in establishing the three-dimensional structures of proteins.
Figure 13-4 Hydrogen bonding
(indicated by dashed lines) in (a)
water, H2O; (b) methyl alcohol,
CH3OH; and (c) ammonia, NH3.
Hydrogen bonding is a special case
of very strong dipole interaction.
490 CHAPTER 13: Liquids and Solids
(a) (b) (c)
Figure 13-5 Boiling points of some hydrides as a function of molecular weight. The
unusually high boiling points of NH3, H2O, and HF compared with those of other hydrides
of the same groups are due to hydrogen bonding. The electronegativity difference between
H and C is small, and there are no unshared pairs on C; thus, CH4 is not hydrogen bonded.
Increasing molecular weight corresponds to increasing number of electrons; this makes the
electron clouds easier to deform and causes increased dispersion forces, accounting for the
increase in boiling points for the nonhydrogen-bonded members of each series.
GeH4
SnH4
Molecular weight
Boilingtemperature(°C)
20 40 60 80 100 120 140
Ϫ200
Ϫ150
Ϫ100
Ϫ50
0
50
100
150
Group VA
NH3
Group VIA
H2O
Group IVA
H2S
H2Se
H2Te
SiH4
CH4
Group VIIA
HF
HBrHCl
HI
PH3
AsH3
SbH3
Dispersion Forces
Dispersion forces are weak attractive forces that are important only over extremely short
distances because they vary as 1/d7. They are present between all types of molecules in condensed
phases but are weak for small molecules. Dispersion forces are the only kind of
intermolecular forces present among symmetrical nonpolar substances such as SO3, CO2,
O2, N2, Br2, H2, and monatomic species such as the noble gases. Without dispersion
forces, such substances could not condense to form liquids or solidify to form solids. Condensation
of some substances occurs only at very low temperatures and/or high pressures.
Dispersion forces result from the attraction of the positively charged nucleus of one
atom for the electron cloud of an atom in nearby molecules. This induces temporary dipoles
in neighboring atoms or molecules. As electron clouds become larger and more diffuse,
they are attracted less strongly by their own (positively charged) nuclei. Thus, they are
more easily distorted, or polarized, by adjacent nuclei.
Polarizability increases with increasing numbers of electrons and therefore with
increasing sizes of molecules. Therefore, dispersion forces are generally stronger for
molecules that have more electrons or are larger.
Dispersion forces are depicted in Figure 13-6. They exist in all substances.
Figure 13-5 shows that polar covalent compounds with hydrogen bonding (H2O, HF,
NH3) boil at higher temperatures than analogous polar compounds without hydrogen
bonding (H2S, HCl, PH3). Symmetrical, nonpolar compounds (CH4, SiH4) of compaDispersion
forces are often called
London forces, after the German-born
physicist Fritz London (1900–1954).
He initially postulated their existence
in 1930, on the basis of quantum
theory.
Although the term “van der
Waals forces” usually refers to all
intermolecular attractions, it is also
often used interchangeably with
“dispersion forces,” as are the terms
“London forces” and “dipole-induced
dipole forces.”
13-2 Intermolecular Attractions and Phase Changes 491
Figure 13-6 An illustration of how a temporary dipole can be induced in an atom. (a) An
isolated argon atom, with spherical charge distribution (no dipole). (b) When a cation
approaches the argon atom, the outer portion of the electron cloud is weakly attracted by
the ion’s positive charge. This induces a weak temporary dipole in the argon atom. (c) A
temporary dipole can also be induced if the argon atom is approached by an anion. (d) The
approach of a molecule with a permanent dipole (for instance, HF) could also temporarily
polarize the argon atom. (e) Even in pure argon, the close approach of one argon atom to
another results in temporary dipole formation in both atoms as each atom’s electron cloud is
attracted by the nucleus of the other atom or is repelled by the other atom’s electron cloud.
The resulting temporary dipoles cause weak attractions among the argon atoms. Molecules
are even more easily polarized than isolated atoms.
(a) (b) (c)
(d) (e)
δ+ δ– A + –A
A
A
A
A
δ+δ–
δ+δ–
δ–
δ+ δ+δ–
δ–
δ+
δ+δ– δ–
δ+
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 13.5, Intermolecular Forces (3).
rable molecular weight boil at lower temperatures. In the absence of hydrogen bonding,
boiling points of analogous substances (CH4, SiH4, GeH4, SnH4) increase fairly regularly
with increasing number of electrons and molecular size (molecular weight). This is due
to increasing effectiveness of dispersion forces of attraction in the larger molecules and
occurs even in the case of some polar covalent molecules. The increasing effectiveness of
dispersion forces, for example, accounts for the increase in boiling points in the sequences
HCl Ͻ HBr Ͻ HI and H2S Ͻ H2Se Ͻ H2Te, which involve nonhydrogen-bonded polar
covalent molecules. The differences in electronegativities between hydrogen and other
nonmetals decrease in these sequences, and the increasing dispersion forces override the
decreasing permanent dipole–dipole forces. The permanent dipole–dipole interactions
therefore have very little effect on the boiling point trend of these compounds.
Let us compare the magnitudes of the various contributions to the total energy of interactions
in some simple molecules. Table 13-3 shows the permanent dipole moments and
the energy contributions for five simple molecules. The contribution from dispersion
forces is substantial in all cases. The permanent dipole–dipole energy is greatest for
substances in which hydrogen bonding occurs. The variations of these total energies of
interaction are closely related to molar heats of vaporization. As we shall see in Section
13-9, the heat of vaporization measures the amount of energy required to overcome the
attractive forces that hold the molecules together in a liquid.
The properties of a liquid or a solid are often the result of many forces. The properties
of an ionic compound are determined mainly by the very strong ion–ion interactions,
even though other forces may also be present. In a polar covalent compound that contains
NXH, OXH, or FXH bonds, strong hydrogen bonding is often the strongest force
present. If hydrogen bonding is absent in a polar covalent compound, dispersion forces
are likely to be the most important forces. In a slightly polar or nonpolar covalent
compound or a monatomic nonmetal, the dispersion forces, though weak, are still the
strongest ones present, so they determine the forces. For large molecules, even the very
weak dispersion forces can total up to a considerable interactive force.
THE LIQUID STATE
We shall briefly describe several properties of the liquid state. These properties vary
markedly among various liquids, depending on the nature and strength of the attractive
forces among the particles (atoms, molecules, ions) making up the liquid.
The covalent bonding within a
polyatomic ion such as NH4
ϩ or
SO4
2Ϫ is very strong, but the forces
that hold the entire substance together
are ionic. Thus a compound that
contains a polyatomic ion is an ionic
compound (see Section 7-10).
492 CHAPTER 13: Liquids and Solids
TABLE 13-3 Approximate Contributions to the Total Energy of Interaction
Between Molecules, in kJ/mol
Permanent Permanent Molar Heat of
Dipole Moment Dipole–Dipole Dispersion Total Vaporization
Molecule (D) Energy Energy Energy (kJ/mol)
Ar 0 0 8.5 8.5 6.7
CO 0.1 Ϸ0 8.7 8.7 8.0
HCl 1.03 3.3 17.8 21 16.2
NH3 1.47 13* 16.3 29 27.4
H2O 1.85 36* 10.9 47 40.7
*Hydrogen-bonded.
Honey is a very viscous liquid.
VISCOSITY
Viscosity is the resistance to flow of a liquid. Honey has a high viscosity at room temperature,
and freely flowing gasoline has a low viscosity. The viscosity of a liquid can be
measured with a viscometer such as the one in Figure 13-7.
For a liquid to flow, the molecules must be able to slide past one another. In general,
the stronger the intermolecular forces of attraction, the more viscous the liquid is.
Substances that have a great ability to form hydrogen bonds, especially involving several
hydrogen-bonding sites per molecule, such as glycerine (margin), usually have high viscosities.
Increasing the size and surface area of molecules generally results in increased
viscosity, due to the increased dispersion forces. For example, the shorter-chain hydrocarbon
pentane (a free-flowing liquid at room temperature) is less viscous than dodecane
(an oily liquid at room temperature). The longer the molecules are, the more they can
get “tangled up” with one another, and the harder it is for them to flow.
pentane, C5H12
viscosity ϭ 0.215
centipoise at 25°C
dodecane, C12H26
viscosity ϭ 1.38
centipoise at 25°C
As temperature increases and the molecules move more rapidly, their kinetic energies
are better able to overcome intermolecular attractions. Thus, viscosity decreases with
increasing temperature, as long as no changes in composition occur.
SURFACE TENSION
Molecules below the surface of a liquid are influenced by intermolecular attractions from
all directions. Those on the surface are attracted only toward the interior (Figure 13-8);
these attractions pull the surface layer toward the center. The most stable situation is one
13-4
CH H
H
H
C
H
H
C
H
H
C
H
H
C
H
H
C
H
H
C
H
H
C
H
H
C
H
H
C
H
H
C
H
H
C
H
H
CH H
H
H
C
H
H
C
H
H
C
H
H
C
H
H
13-3
Viscosity ϭ 945 centipoise at 25°C
C
H
C C
OH
H H
H
OH
H
OH
glycerine
13-4 Surface Tension 493
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 13.11, Properties of Liquids (4):
Surface Tension/Capillary Action/
Viscosity.
The poise is the unit used to express
viscosity. The viscosity of water at
25°C is 0.89 centipoise.
stronger
higher
attractive mn
viscosity
forces
increasing lower
mn
temperature viscosity
΂ ΃ ΂ ΃
΂ ΃΂ ΃
Figure 13-7 The Ostwald viscometer, a device used
to measure viscosity of liquids. The time it takes for a
known volume of a liquid to flow through a small neck
of known size is measured. Liquids with low viscosities
flow rapidly.
Figure 13-8 A molecular-level
view of the attractive forces
experienced by molecules at and
below the surface of a liquid.
Calibration
marks
in which the surface area is minimal. For a given volume, a sphere has the least possible
surface area, so drops of liquid tend to assume spherical shapes. Surface tension is a
measure of the inward forces that must be overcome to expand the surface area of a liquid.
CAPILLARY ACTION
All forces holding a liquid together are called cohesive forces. The forces of attraction
between a liquid and another surface are adhesive forces. The partial positive charges
on the H atoms of water hydrogen bond strongly to the partial negative charges on the
oxygen atoms at the surface of the glass. As a result, water adheres to glass, or is said to
wet glass. As water creeps up the side of the glass tube, its favorable area of contact with
the glass increases. The surface of the water, its meniscus, has a concave shape (Figure
13-9). On the other hand, mercury does not wet glass because its cohesive forces (due to
dispersion forces) are much stronger than its attraction to glass. Thus, its meniscus is
convex. Capillary action occurs when one end of a capillary tube, a glass tube with a
small bore (inside diameter), is immersed in a liquid. If adhesive forces exceed cohesive
forces, the liquid creeps up the sides of the tube until a balance is reached between adhesive
forces and the weight of liquid. The smaller the bore, the higher the liquid climbs.
Capillary action helps plant roots take up water and dissolved nutrients from the soil and
transmit them up the stems. The roots, like glass, exhibit strong adhesive forces for water.
Osmotic pressure (Section 14-15) also plays a major role in this process.
13-5
(Right) The shape of a soap bubble is due to
the inward force (surface tension) that acts
to minimize the surface area.
The surface tension of water
supports this water strider. The
nonpolar surfaces of its feet also
help to repel the water.
Droplets of mercury lying on a glass
surface. The small droplets are
almost spherical, whereas the larger
droplets are flattened due to the
effects of gravity. This shows that
surface tension has more influence
on the shape of the small (lighter)
droplets.
Figure 13-9 The meniscus, as
observed in glass tubes with water
and with mercury.
Coating glass with a silicone polymer greatly reduces the adhesion of water to the
glass. The left side of each glass has been treated with Rain-X, which contains a
silicone polymer. Water on the treated side forms droplets that are easily swept away.
EVAPORATION
Evaporation, or vaporization, is the process by which molecules on the surface of a liquid
break away and go into the gas phase (Figure 13-10). Kinetic energies of molecules in
liquids depend on temperature in the same way as they do in gases. The distribution of
kinetic energies among liquid molecules at two different temperatures is shown in Figure
13-11. To break away, the molecules must possess at least some minimum kinetic energy.
Figure 13-11 shows that at a higher temperature, a greater fraction of molecules possess
at least that minimum energy. The rate of evaporation increases as temperature increases.
Only the higher-energy molecules can escape from the liquid phase. The average molecular
kinetic energy of the molecules remaining in the liquid state is thereby lowered,
resulting in a lower temperature in the liquid. The liquid would then be cooler than its
surroundings, so it absorbs heat from its surroundings. The cooling of your body by evaporation
of perspiration is a familiar example of the cooling of the surroundings by
evaporation of a liquid. This is called “cooling by evaporation.”
13-6
Figure 13-11 Distribution of
kinetic energies of molecules in a
liquid at different temperatures. At
the lower temperature, a smaller
fraction of the molecules have the
energy required to escape from the
liquid, so evaporation is slower and
the equilibrium vapor pressure
(Section 13-7) is lower.
13-6 Evaporation 495
Breeze
Molecules
of the air
(a) (b) (c)
Figure 13-10 (a) Liquid continuously evaporates from an open vessel. (b) Equilibrium
between liquid and vapor is established in a closed container in which molecules return to
the liquid at the same rate as they leave it. (c) A bottle in which liquid–vapor equilibrium
has been established. Note that droplets have condensed.
Minimum kinetic energy required
for escape of molecules from
surface of liquid
Kinetic energy
Fractionofmoleculesat
givenkineticenergy
Higher temperature
Lower temperature
The dew on this spider web was
formed by condensation of water
vapor from the air.
A molecule in the vapor may strike the liquid surface and be captured there. This
process, the reverse of evaporation, is called condensation. As evaporation occurs in a
closed container, the volume of liquid decreases and the number of gas molecules above
the surface increases. Because more gas phase molecules can collide with the surface, the
rate of condensation increases. The system composed of the liquid and gas molecules of
the same substance eventually achieves a dynamic equilibrium in which the rate of evaporation
equals the rate of condensation in the closed container.
evaporation
liquid 3:::::::::4 vapor
condensation
The two opposing rates are not zero, but are equal to each other—hence we call this
“dynamic,” rather than “static,” equilibrium. Even though evaporation and condensation
are both continuously occurring, no net change occurs because the rates are equal.
However, if the vessel were left open to the air, this equilibrium could not be reached.
Molecules would diffuse away, and slight air currents would also sweep some gas molecules
away from the liquid surface. This would allow more evaporation to occur to replace
the lost vapor molecules. Consequently, a liquid can eventually evaporate entirely if it is
left uncovered. This situation illustrates LeChatelier’s Principle:
A system at equilibrium, or changing toward equilibrium, responds in the way that
tends to relieve or “undo” any stress placed on it.
In this example the stress is the removal of molecules in the vapor phase. The response
is the continued evaporation of the liquid.
VAPOR PRESSURE
Vapor molecules cannot escape when vaporization of a liquid occurs in a closed container.
As more molecules leave the liquid, more gaseous molecules collide with the walls of the
container, with one another, and with the liquid surface, so more condensation occurs.
This is responsible for the formation of liquid droplets that adhere to the sides of the
vessel above a liquid surface and for the eventual establishment of equilibrium between
liquid and vapor (see Figure 13-10b and c).
The partial pressure of vapor molecules above the surface of a liquid at equilibrium
at a given temperature is the vapor pressure (vp) of the liquid at that temperature.
Because the rate of evaporation increases with increasing temperature, vapor pressures
of liquids always increase as temperature increases.
13-7
As an analogy, suppose that 30
students per minute leave a classroom,
moving into the closed hallway
outside, and 30 students per minute
enter it. The total number of students
in the room would remain constant, as
would the total number of students
outside the room.
496 CHAPTER 13: Liquids and Solids
TABLE 13-4 Vapor Pressures (in torr) of Some Liquids
0°C 25°C 50°C 75°C 100°C 125°C
water 4.6 23.8 92.5 300 760 1741
benzene 27.1 94.4 271 644 1360
methyl alcohol 29.7 122 404 1126
diethyl ether 185 470 1325 2680 4859
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 13.9, Properties of Liquids (2):
Vapor Pressure; this screen contains an
animation of the vaporization process.
As long as some liquid remains in
contact with the vapor, the pressure
does not depend on the volume or
surface area of the liquid.
This is one of the guiding principles
that allows us to understand chemical
equilibrium. It is discussed further in
Chapter 17.
Easily vaporized liquids are called volatile liquids, and they have relatively high vapor
pressures. The most volatile liquid in Table 13-4 is diethyl ether. Water is the least volatile.
Vapor pressures can be measured with manometers (Figure 13-12).
Stronger cohesive forces tend to hold molecules in the liquid state. Methyl alcohol
molecules are strongly linked by hydrogen bonding, whereas diethyl ether molecules are
not, so methyl alcohol has a lower vapor pressure than diethyl ether. The very strong
hydrogen bonding in water accounts for its unusually low vapor pressure (see Table
13-4). Dispersion forces generally increase with increasing molecular size, so substances
composed of larger molecules have lower vapor pressures.
Figure 13-12 A representation of
the measurement of vapor pressure
of a liquid at a given temperature.
The container is evacuated before
the liquid is added. At the instant
the liquid is added to the container,
there are no molecules in the gas
phase so the pressure is zero. Some
of the liquid then vaporizes until
equilibrium is established. The
difference in heights of the mercury
column is a measure of the vapor
pressure of the liquid at that
temperature.
Vapor pressure
at temperature
of measurement
Time
Equilibrium:
liquid and vapor
Ptotal = Pvapor
Initial: liquid only
Figure 13-13 Plots of the vapor
pressures of the liquids in Table
13-4. The normal boiling point of a
liquid is the temperature at which
its vapor pressure is equal to one
atmosphere. Normal boiling points
are: water, 100°C; benzene; 80.1°C;
methyl alcohol, 65.0°C; and diethyl
ether, 34.6°C. Notice that the
increase in vapor pressure is not
linear with temperature.
Methyl
alcohol
Diethyl
ether
Benzene
Water
T (ЊC)
2000
1500
1000
760
500
Vaporpressure(torr)
0 25 50 75 100
1 atm
We can understand the order of vapor pressures of the four liquids cited in Table
13-4 and Figure 13-13 by considering the strengths of their intermolecular attractions.
Water has the lowest vapor pressure (strongest cohesive forces) because each molecule has
two hydrogen atoms to act as hydrogen-bond donors and each molecule can accept
hydrogen bonds from two other molecules. Methyl alcohol has only one potential
hydrogen-bond donor, so its average cohesive forces are weaker than those in water and
its vapor pressure is higher. In benzene and diethyl ether, the hydrogen atoms are all
bonded to carbon, so strong hydrogen bonds are not possible. Electrons can move easily
throughout the delocalized ␲-bonding orbitals of benzene, however, so benzene is quite
polarizable and exhibits significant dispersion forces. In addition, the hydrogen atoms of
benzene are more positive than most hydrogens that are bonded to carbon. The H atoms
of benzene are attracted to the electron-rich ␲-bonding regions of nearby molecules. The
accumulation of these forces gives benzene rather strong cohesive forces, resulting in a
lower vapor pressure than we might expect for a hydrocarbon. The diethyl ether molecule
is only slightly polar, resulting in weak dipole–dipole forces and a high vapor pressure.
BOILING POINTS AND DISTILLATION
When heat energy is added to a liquid, it increases the kinetic energy of the molecules,
and the temperature of the liquid increases. Heating a liquid always increases its vapor
pressure. When a liquid is heated to a sufficiently high temperature under a given applied
(usually atmospheric) pressure, bubbles of vapor begin to form below the surface. If the
vapor pressure inside the bubbles is less than the applied pressure on the surface of the
liquid, the bubbles collapse as soon as they form. If the temperature is raised sufficiently,
the vapor pressure is high enough that the bubbles can persist, rise to the surface, and
burst, releasing the vapor into the air. This process is called boiling and is different from
evaporation. The boiling point of a liquid is the temperature at which its vapor pressure
equals the external pressure. The normal boiling point is the temperature at which the
vapor pressure of a liquid is equal to exactly one atmosphere (760 torr). The vapor pressure
of water is 760 torr at 100°C, its normal boiling point. As heat energy is added to a
pure liquid at its boiling point, the temperature remains constant, because the energy is
used to overcome the cohesive forces in the liquid to form vapor.
If the applied pressure is lower than 760 torr, say on the top of a mountain, water boils
below 100°C. The chemical reactions involved in cooking food occur more slowly at the
lower temperature, so it takes longer to cook food in boiling water at high altitudes than
at sea level. A pressure cooker cooks food rapidly because water boils at higher temperatures
under increased pressures. The higher temperature of the boiling water increases
the rate of cooking.
Different liquids have different cohesive forces, so they have different vapor pressures
and boil at different temperatures. A mixture of liquids with sufficiently different boiling
points can often be separated into its components by distillation. In this process the
mixture is heated slowly until the temperature reaches the point at which the most volatile
liquid boils off. If this component is a liquid under ordinary conditions, it is subsequently
recondensed in a water-cooled condensing column (Figure 13-14) and collected as a distillate.
After enough heat has been added to vaporize all of the most volatile liquid, the
temperature again rises slowly until the boiling point of the next substance is reached, and
the process continues. Any nonvolatile substances dissolved in the liquid do not boil, but
remain in the distilling flask. Impure water can be purified and separated from its dissolved
salts by distillation. Compounds with similar boiling points, especially those that interact
13-8
As water is being heated, but before it
boils, small bubbles may appear in the
container. This is not boiling, but
rather the formation of bubbles of
dissolved gases such as CO2 and O2
whose solubilities in water decrease
with increasing temperature.
498 CHAPTER 13: Liquids and Solids
stronger lower
attractive mn vapor
forces pressure
higher
increasing
mn vapor
temperature
pressure
΂ ΃ ΂ ΃
΂ ΃΂ ΃
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 13.10, Properties of Liquids (3):
Boiling Point.
Figure 13-14 A laboratory setup
for distillation. During distillation
of an impure liquid, nonvolatile
substances remain in the distilling
flask. The liquid is vaporized and
condensed before being collected in
the receiving flask. If any of the
substances are flammable, using an
open flame would be dangerous; in
such a case, another source of heat,
such as an electric heater, should be
used.
very strongly with one another, are not effectively separated by simple distillation but
require a modification called fractional distillation (Section 14-10).
HEAT TRANSFER INVOLVING LIQUIDS
Heat must be added to a liquid to raise its temperature (Section 1-13). The specific heat
(J/gи°C) or molar heat capacity (J/molи°C) of a liquid is the amount of heat that must
be added to the stated mass of liquid to raise its temperature by one degree Celsius. If
heat is added to a liquid under constant pressure, the temperature rises until its boiling
point is reached. Then the temperature remains constant until enough heat has been added
to boil away all the liquid. The molar heat (or enthalpy) of vaporization (⌬Hvap) of a
liquid is the amount of heat that must be added to one mole of the liquid at its boiling
point to convert it to vapor with no change in temperature. Heats of vaporization can also
be expressed as energy per gram. For example, the heat of vaporization for water at its
boiling point is 40.7 kJ/mol, or 2.26 ϫ 103 J/g.
ϭ ϫ ϫ ϭ 2.26 ϫ 103 J/g
Like many other properties of liquids, heats of vaporization reflect the strengths of
intermolecular forces. Heats of vaporization generally increase as boiling points and intermolecular
forces increase and as vapor pressures decrease. Table 13-5 illustrates this. The
high heats of vaporization of water, ethylene glycol, and ethyl alcohol are due mainly to
the strong hydrogen-bonding interactions in these liquids (see Section 13-2). The very
high value for water makes it very effective as a coolant and, in the form of steam, as a
source of heat.
Liquids can evaporate even below their boiling points. The water in perspiration is an
effective coolant for our bodies. Each gram of water that evaporates absorbs 2.41 kJ of
heat from the body. We feel even cooler in a breeze because perspiration evaporates faster,
so heat is removed more rapidly.
1 mol
ᎏ
18.0 g
1000 J
ᎏ
kJ
40.7 kJ
ᎏ
mol
_?_ J
ᎏ
g
13-9
The specific heat and heat capacity of
a substance change somewhat with its
temperature. For most substances, this
variation is small enough to ignore.
Molar heats of vaporization (also called
molar enthalpies of vaporization) are
often expressed in kilojoules rather
than joules. The units of heat
of vaporization do not include
temperature. This is because boiling
occurs with no change in temperature.
13-9 Heat Transfer Involving Liquids 499
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 13.8, Properties of Liquids (1):
Enthalpy of Vaporization.
The heat of vaporization of water is
higher at 37°C (normal body
temperature) than at 100°C (2.41 kJ/g
compared to 2.26 kJ/g).
TABLE 13-5 Heats of Vaporization, Boiling Points, and Vapor Pressures
of Some Common Liquids
Heat of
Vaporization
at
Vapor Boiling Boiling Point
Pressure Point
Liquid (torr at 20°C) at 1 atm (°C) J/g kJ/mol
water, H2O 17.5 100. 2260 40.7
ethyl alcohol, CH3CH2OH 43.9 78.3 855 39.3
benzene, C6H6 74.6 80.1 395 30.8
diethyl ether, CH3CH2OCH2CH3 442. 34.6 351 26.0
carbon tetrachloride, CCl4 85.6 76.8 213 32.8
ethylene glycol, CH2OHCH2OH 0.1 197.3 984 58.9
Condensation is the reverse of evaporation. The amount of heat that must be removed
from a vapor to condense it (without change in temperature) is called the heat of conden-
sation.
evaporation
liquid ϩ heat 3:::::::::4 vapor
condensation
The heat of condensation of a liquid is equal in magnitude to the heat of vaporization. It
is released by the vapor during condensation.
Because 2.26 kJ must be absorbed to vaporize one gram of water at 100°C, that same
amount of heat must be released to the environment when one gram of steam at 100°C
condenses to form liquid water at 100°C. In steam-heated radiators, steam condenses and
releases 2.26 kJ of heat per gram as its molecules collide with the cooler radiator walls
and condense there. The metallic walls conduct heat well. They transfer the heat to the
air in contact with the outside walls of the radiator. The heats of condensation and vaporization
of non–hydrogen-bonded liquids, such as benzene, have smaller magnitudes than
those of hydrogen-bonded liquids (see Table 13-5). They are therefore much less effective
as heating and cooling agents.
EXAMPLE 13-1 Heat of Vaporization
Calculate the amount of heat, in joules, required to convert 180. grams of water at 10.0°C to
steam at 105.0°C.
Plan
The total amount of heat absorbed is the sum of the amounts required to (1) warm the liquid
water from 10.0°C to 100.0°C, (2) convert the liquid water to steam at 100.0°C, and (3) warm
the steam from 100.0°C to 105.0°C.
Because of the large amount of heat
released by steam as it condenses,
burns caused by steam at 100°C are
much more severe than burns caused
by liquid water at 100°C.
500 CHAPTER 13: Liquids and Solids
180. g H2O(ᐉ)
at 10.0ЊC
180. g H2O(ᐉ)
at 100.0ЊC
Step 1:
warm the liquid
(temp. change)
Step 2:
boil the liquid
(phase change)
180. g H2O(g)
at 100.0ЊC
180. g H2O(g)
at 105.0ЊC
Step 3:
warm the steam
(temp. change)
Steps 1 and 3 involve the specific heats of water and steam, 4.18 J/gи°C and 2.03 J/gи°C, respectively
(Appendix E), whereas step 2 involves the heat of vaporization of water (2.26 ϫ 103 J/g).
Solution
1. _?_ J ϭ 180. g ϫ ᎏ
4
g
.
и
1
°
8
C
J
ᎏ ϫ (100.0°C Ϫ 10.0°C) ϭ 6.77 ϫ 104 J ϭ 0.677 ϫ 105 J
2. _?_ J ϭ 180. g ϫ ᎏ
2.26 ϫ
g
103 J
ᎏ ϭ 4.07 ϫ 105 J
3. _?_ J ϭ 180. g ϫ ᎏ
2
g
.
и
0
°
3
C
J
ᎏ ϫ (105.0°C Ϫ 100.0°C) ϭ 1.8 ϫ 103 J ϭ 0.018 ϫ 105 J
Total amount of heat absorbed ϭ 4.76 ϫ 105 J
You should now work Exercises 50 and 51.
Distillation is not an economical way to purify large quantities of water for public water
supplies. The high heat of vaporization of water makes it too expensive to vaporize large
volumes of water.
Steps 1 and 3 of this example involve
warming with no phase change. Such
calculations were introduced in Section
1-13.
Step 1: Temperature change only
Step 2: Phase change only
Step 3: Temperature change only
EXAMPLE 13-2 Heat of Vaporization
Compare the amount of “cooling” experienced by an individual who drinks 400. mL of ice
water (0.0°C) with the amount of “cooling” experienced by an individual who “sweats out”
400. mL of water. Assume that the sweat is essentially pure water and that all of it evaporates.
The density of water is very nearly 1.00 g/mL at both 0.0°C and 37.0°C, average body temperature.
The heat of vaporization of water is 2.41 kJ/g at 37.0°C.
Plan
In the case of drinking ice water, the body is cooled by the amount of heat required to raise
the temperature of 400. mL (400. g) of water from 0.0°C to 37.0°C. The amount of heat lost
by perspiration is equal to the amount of heat required to vaporize 400. g of water at 37.0°C.
Solution
Raising the temperature of 400. g of water from 0.0°C to 37.0°C requires
_?_ J ϭ (400. g)(4.18 J/gи°C)(37.0°C) ϭ 6.19 ϫ 104 J, or 61.9 kJ
Evaporating (i.e., “sweating out”) 400. mL of water at 37°C requires
_?_ J ϭ (400. g)(2.41 ϫ 103 J/g) ϭ 9.64 ϫ 105 J, or 964 kJ
Thus, we see that “sweating out” 400. mL of water removes 964 kJ of heat from one’s body,
whereas drinking 400. mL of ice water cools it by only 61.9 kJ. Stated differently, sweating
removes (964/61.9) ϭ 15.6 times more heat than drinking ice water!
You should now work Exercise 55.
For health reasons, it is important
to replace the water lost by
perspiration.
Problem-Solving Tip: Temperature Change or Phase Change?
A problem such as Example 13-1 can be broken down into steps so that each involves
either a temperature change or a phase change, but not both. A temperature change calculation
uses the specific heat of the substance (steps 1 and 3 of Example 13-1); remember
that each different phase has its own specific heat. A phase change always takes place
with no change in temperature, so that calculation does not involve temperature (step 2
of Example 13-1).
The Clausius–Clapeyron Equation
We have seen (Figure 13-13) that vapor pressure increases with increasing temperature. Let
us now discuss the quantitative expression of this relationship.
When the temperature of a liquid is changed from T1 to T2, the vapor pressure of the
liquid changes from P1 to P2. These changes are related to the molar heat of vaporization,
⌬Hvap, for the liquid by the Clausius–Clapeyron equation.
ln ΂ ΃ϭ ΂ Ϫ ΃
Although ⌬Hvap changes somewhat with temperature, it is usually adequate to use the value
tabulated at the normal boiling point of the liquid (Appendix E) unless more precise values
are available. The units of R must be consistent with those of ⌬Hvap.
1
ᎏ
T2
1
ᎏ
T1
⌬Hvap
ᎏ
R
P2
ᎏ
P1
E n r i c h m e n t
The Clausius–Clapeyron equation is
used for three types of calculations:
(1) to predict the vapor pressure of a
liquid at a specified temperature, as in
Example 13-3; (2) to determine the
temperature at which a liquid has a
specified vapor pressure; and (3) to
calculate ⌬Hvap from measurement
of vapor pressures at different
temperatures.
EXAMPLE 13-3 Vapor Pressure Versus Temperature
The normal boiling point of ethanol, C2H5OH, is 78.3°C, and its molar heat of vaporization
is 39.3 kJ/mol (Appendix E). What would be the vapor pressure, in torr, of ethanol at
50.0°C?
Plan
The normal boiling point of a liquid is the temperature at which its vapor pressure is 760
torr, so we designate this as one of the conditions (subscript 1). We wish to find the vapor
pressure at another temperature (subscript 2), and we know the molar heat of vaporization.
We use the Clausius–Clapeyron equation to solve for P2.
Solution
P1 ϭ 760 torr at T1 ϭ 78.3°C ϩ 273.2 ϭ 351.5 K
P2 ϭ _?_ at T2 ϭ 50.0°C ϩ 273.2 ϭ 323.2 K
⌬Hvap ϭ 39.3 kJ/mol or 3.93 ϫ 104 J/mol
We solve for P2.
ln ΂ ΃ϭ ΂ Ϫ ΃
ln ΂ ΃ϭ Ϫ1.18 so ΂ ΃ϭ eϪ1.18 ϭ 0.307
P2 ϭ 0.307(760 torr) ϭ 233 torr (lower vapor pressure
at lower temperature)
You should now work Exercise 58.
P2
ᎏ
760 torr
P2
ᎏ
760 torr
1
ᎏ
323.2 K
1
ᎏ
351.5 K
3.93 ϫ 104 J/mol
ᎏᎏ
΂8.314 ᎏ
mo
J
lиK
ᎏ΃
P2
ᎏ
760 torr
502 CHAPTER 13: Liquids and Solids
(Enrichment, continued)
TABLE 13-6 General Effects of Intermolecular Attractions on Physical
Properties of Liquids
Volatile Liquids Nonvolatile Liquids
(weak intermolecular (strong intermolecular
Property attractions) attractions)
cohesive forces low high
viscosity low high
surface tension low high
specific heat low high
vapor pressure high low
rate of evaporation high low
boiling point low high
heat of vaporization low high
We have described many properties of liquids and discussed how they depend on intermolecular
forces of attraction. The general effects of these attractions on the physical
properties of liquids are summarized in Table 13-6. “High” and “low” are relative terms.
Table 13-6 is intended to show only very general trends. Example 13-4 illustrates the use
of intermolecular attractions to predict boiling points.
EXAMPLE 13-4 Boiling Points Versus Intermolecular Forces
Predict the order of increasing boiling points for the following: H2S; H2O; CH4; H2; KBr.
Plan
We analyze the polarity and size of each substance to determine the kinds of intermolecular
forces that are present. In general, the stronger the intermolecular forces, the higher is the
boiling point of the substance.
Solution
KBr is ionic, so it boils at the highest temperature. Water exhibits hydrogen bonding and boils
at the next highest temperature. Hydrogen sulfide is the only other polar covalent substance
in the list, so it boils below H2O but above the other two substances. Both CH4 and H2 are
nonpolar. The larger CH4 molecule is more easily polarized than the very small H2 molecule,
so dispersion forces are stronger in CH4. Thus, CH4 boils at a higher temperature than H2.
H2 Ͻ CH4 Ͻ H2S Ͻ H2O Ͻ KBr
88888888888888888888888888888888n
increasing boiling points
You should now work Exercise 24.
THE SOLID STATE
MELTING POINT
The melting point (freezing point) of a substance is the temperature at which its solid
and liquid phases coexist in equilibrium.
melting
solid 3::::4 liquid
freezing
The melting point of a solid is the same as the freezing point of its liquid. It is the temperature
at which the rate of melting of a solid is the same as the rate of freezing of its liquid
under a given applied pressure.
The normal melting point of a substance is its melting point at one atmosphere pressure.
Changes in pressure have very small effects on melting points; they have large effects
on boiling points.
HEAT TRANSFER INVOLVING SOLIDS
When heat is added to a solid below its melting point, its temperature rises. After enough
heat has been added to bring the solid to its melting point, additional heat is required to
convert the solid to liquid. During this melting process, the temperature remains constant
13-11
13-10
13-11 Heat Transfer Involving Solids 503
at the melting point until all of the substance has melted. After melting is complete, the
continued addition of heat results in an increase in the temperature of the liquid, until
the boiling point is reached. This is illustrated graphically in the first three segments of
the heating curve in Figure 13-15.
The molar heat (or enthalpy) of fusion (⌬Hfus; kJ/mol) is the amount of heat required
to melt one mole of a solid at its melting point. Heats of fusion can also be expressed on
a per gram basis. The heat of fusion depends on the intermolecular forces of attraction in
the solid state. These forces “hold the molecules together” as a solid. Heats of fusion are
usually higher for substances with higher melting points. Values for some common
compounds are shown in Table 13-7. Appendix E has more values.
The heat (or enthalpy) of solidification of a liquid is equal in magnitude to the heat
of fusion. It represents removal of a sufficient amount of heat from a given amount (1 mol
or 1 g) of liquid to solidify the liquid at its freezing point. For water,
6.02 kJ/mol
or 334 J/g absorbed
ice 3::::::::::::::4 water (at 0°C)
6.02 kJ/mol
or 334 J/g released
Melting is always endothermic. The
term “fusion” means “melting.”
Heat added
B
ATemperature
Gas is
cooled
EXOTHERMIC
PROCESSES
(LIBERATE HEAT)
ENDOTHERMIC
PROCESSES
(ABSORB HEAT)
Liquid is
cooled
Solid is
cooled
Freezing
Condensing
Gas is
warmed
Solid is
warmed
Melting
Boiling
Liquid is
warmed
Figure 13-15 A typical heating curve at constant pressure. When heat energy is added to
a solid below its melting point, the temperature of the solid rises until its melting point is
reached (point A). In this region of the plot, the slope is rather steep because of the low
specific heats of solids [e.g., 2.09 J/gи°C for H2O(s)]. If the solid is heated at its melting
point (A), its temperature remains constant until the solid has melted, because the melting
process requires energy. The length of this horizontal line is proportional to the heat of
fusion of the substance—the higher the heat of fusion, the longer the line. When all of the
solid has melted, heating the liquid raises its temperature until its boiling point is reached
(point B). The slope of this line is less steep than that for warming the solid, because the
specific heat of the liquid phase [e.g., 4.18 J/gи°C for H2O(ᐉ)] is usually greater than that of
the corresponding solid. If heat is added to the liquid at its boiling point (B), the added heat
energy is absorbed as the liquid boils. This horizontal line is longer than the previous one,
because the heat of vaporization of a substance is always higher than its heat of fusion.
When all of the liquid has been converted to a gas (vapor), the addition of more heat
raises the temperature of the gas. This segment of the plot has a steep slope because of
the relatively low specific heat of the gas phase [e.g., 2.03 J/gи°C for H2O(g)]. Each step
in the process can be reversed by removing the same amount of heat.
EXAMPLE 13-5 Heat of Fusion
The molar heat of fusion, ⌬Hfus, of Na is 2.6 kJ/mol at its melting point, 97.5°C. How much
heat must be absorbed by 5.0 g of solid Na at 97.5°C to melt it?
Plan
Melting takes place at a constant temperature. The molar heat of fusion tells us that every mole
of Na, 23 grams, absorbs 2.6 kJ of heat at 97.5°C during the melting process. We want to know
the amount of heat that 5.0 grams would absorb. We use the appropriate unit factors,
constructed from the atomic weight and ⌬Hfus, to find the amount of heat absorbed.
Solution
_?_ kJ ϭ 5.0 g Na ϫ ϫ ϭ 0.57 kJ
2.6 kJ
ᎏᎏ
1 mol Na
1 mol Na
ᎏᎏ
23 g Na
13-11 Heat Transfer Involving Solids 505
TABLE 13-7 Some Melting Points and Heats of Fusion
Heat of Fusion
Melting Point
Substance (°C) J/g kJ/mol
methane, CH4 Ϫ182 58.6 0.92
ethyl alcohol, CH3CH2OH Ϫ117 109 5.02
water, H2O 0 334 6.02
naphthalene, C10H8 80.2 147 18.8
silver nitrate, AgNO3 209 67.8 11.5
aluminum, Al 658 395 10.6
sodium chloride, NaCl 801 519 30.3
Transitions among the three states of matter. The transitions shown in blue are endothermic
(absorb heat); those shown in red are exothermic (release heat).
Gas
LiquidSolid
Melting
Freezing
Condensation
Evaporation
DepositionSublimation
EXAMPLE 13-6 Heat of Fusion
Calculate the amount of heat that must be absorbed by 50.0 grams of ice at Ϫ12.0°C to convert
it to water at 20.0°C. Refer to Appendix E.
Plan
We must determine the amount of heat absorbed during three steps: (1) warming 50.0 g of ice
from Ϫ12.0°C to its melting point, 0.0°C (we use the specific heat of ice, 2.09 J/gи°C);
(2) melting the ice with no change in temperature (we use the heat of fusion of ice at 0.0°C,
334 J/g; and (3) warming the resulting liquid from 0.0°C to 20.0°C (we use the specific heat
of water, 4.18 J/gи°C).
Solution
1. 50.0 g ϫ ᎏ
2
g
.
и
0
°
9
C
J
ᎏ ϫ [0.0 Ϫ (Ϫ12.0)]°C ϭ 1.25 ϫ 103 J ϭ 0.125 ϫ 104 J
2. 50.0 g ϫ ᎏ
33
g
4 J
ᎏ ϭ 1.67 ϫ 104 J
3. 50.0 g ϫ ᎏ
4
g
.
и
1
°
8
C
J
ᎏ ϫ (20.0 Ϫ 0.0)°C ϭ 4.18 ϫ 103 J ϭ 0.418 ϫ 104 J
Total amount of heat absorbed ϭ 2.21 ϫ 104 J ϭ 22.1 kJ
Note that most of the heat was absorbed in step 2, melting the ice.
You should now work Exercise 56.
SUBLIMATION AND THE VAPOR PRESSURE OF SOLIDS
Some solids, such as iodine and carbon dioxide, vaporize at atmospheric pressure without
passing through the liquid state. This process is known as sublimation. Solids exhibit
vapor pressures just as liquids do, but they generally have much lower vapor pressures.
Solids with high vapor pressures sublime easily. The characteristic odor of a common
household solid, para-dichlorobenzene (moth repellent), is due to sublimation. The reverse
process, by which a vapor solidifies without passing through the liquid phase, is called
deposition.
sublimation
solid 3:::::::4 gas
deposition
Some impure solids can be purified by sublimation and subsequent deposition of the
vapor (as a solid) onto a cooler surface. Purification of iodine by sublimation is illustrated
in Figure 13-16.
13-12
50.0 g H2O(ᐉ)
at 0.0ЊC
50.0 g H2O(ᐉ)
at 20ЊC
Step 3:
warm the liquid
(temp. change)
50.0 g H2O(s)
at Ϫ12.0ЊC
50.0 g H2O(s)
at 0.0ЊC
Step 1:
warm the ice
(temp. change)
Step 2:
melt the ice
(phase change)
Ice is very efficient for cooling because
considerable heat is required to melt a
given mass of it. However, ⌬Hvap is
generally much greater than ⌬Hfusion,
so evaporative cooling is preferable
when possible.
506 CHAPTER 13: Liquids and Solids
Figure 13-16 Sublimation can be
used to purify volatile solids. The
high vapor pressure of the solid
substance causes it to sublime
when heated. Crystals of purified
substance are formed when the
vapor is deposited to form solid on
the cooler (upper) portion of the
apparatus. Iodine, I2, sublimes
readily. I2 vapor is purple.
PHASE DIAGRAMS (P VERSUS T )
We have discussed the general properties of the three phases of matter. Now we can
describe phase diagrams that show the equilibrium pressure–temperature relationships
among the different phases of a given pure substance in a closed system. Our discussion
of phase diagrams applies only to closed systems (e.g., a sample in a sealed container), in
which matter does not escape into the surroundings. This limitation is especially important
when the vapor phase is involved. Figure 13-17 shows a portion of the phase diagrams
for water and carbon dioxide. The curves are not drawn to scale. The distortion allows
us to describe the changes of state over wide ranges of pressure or temperature using one
diagram.
The curved line from A to C in Figure 13-17a is a vapor pressure curve obtained experimentally
by measuring the vapor pressures of water at various temperatures (Table 13-8).
Points along this curve represent the temperature–pressure combinations for which liquid
and gas (vapor) coexist in equilibrium. At points above AC, the stable form of water is
liquid; below the curve, it is vapor.
13-13
13-13 Phase Diagrams (P versus T ) 507
TABLE 13-8 Points on the Vapor Pressure Curve for Water
temperature (°C) Ϫ10 0 20 30 50 70 90 95 100 101
vapor pressure (torr) 2.1 4.6 17.5 31.8 92.5 234 526 634 760 788
0.01 374
2.1
4.6
355
218 atm
CB
A
D
Gas
Liquid
water, H2O
Temperature (°C)
Pressure(torr)
(a)
B
1
5
10
73 atm
–78 –57 310
C
Gas
Liquid
carbon dioxide, CO2
Solid
5.2 atm
Temperature (°C)
Pressure(atm)
(b)
Figure 13-17 Phase diagrams (not to scale). (a) Diagram for water. For water and a
few other substances for which the solid is less dense than the liquid, the solid–liquid
equilibrium line (AB) has negative slope, that is, up and to the left. (b) Diagram for carbon
dioxide, a substance for which the solid is denser than the liquid. Note that the solid–liquid
equilibrium line has positive slope, that is, up and to the right. This is true for most
substances.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 13.17, Phase Diagrams.
Line AB represents the liquid–solid equilibrium conditions. We see that it has a negative
slope. Water is one of the very few substances for which this is the case. The negative
slope (up and to the left) indicates that increasing the pressure sufficiently on the surface
of ice causes it to melt. This is because ice is less dense than liquid water in the vicinity of
the liquid–solid equilibrium. The network of hydrogen bonding in ice is more extensive
than that in liquid water and requires a greater separation of H2O molecules. This causes
ice to float in liquid water. Almost all other solids are denser than their corresponding
liquids; they would have positive slopes associated with line AB. The stable form of water
at points to the left of AB is solid (ice). Thus AB is called a melting curve.
There is only one point, A, at which all three phases of a substance—solid, liquid, and
gas—can coexist at equilibrium. This is called the triple point. For water it occurs at 4.6
torr and 0.01°C.
At pressures below the triple-point pressure, the liquid phase does not exist; rather, the
substance goes directly from solid to gas (sublimes) or the reverse happens (crystals deposit
from the gas). At pressures and temperatures along AD, the sublimation curve, solid and
vapor are in equilibrium.
Consider CO2 (Figure 13-17b). The triple point is at 5.2 atmospheres and Ϫ57°C.
This pressure is above normal atmospheric pressure, so liquid CO2 cannot exist at atmospheric
pressure. Dry ice (solid CO2) sublimes and does not melt at atmospheric pressure.
The critical temperature is the temperature above which a gas cannot be liquefied,
that is, the temperature above which the liquid and gas do not exist as distinct phases. A
substance at a temperature above its critical temperature is called a supercritical fluid. The
critical pressure is the pressure required to liquefy a gas (vapor) at its critical temperature.
The combination of critical temperature and critical pressure is called the critical
point (C in Figure 13-17). For H2O, the critical point is at 374°C and 218 atmospheres;
for CO2, it is at 31°C and 73 atmospheres. There is no such upper limit to the solid–
liquid line, however, as emphasized by the arrowhead at the top of that line.
To illustrate the use of a phase diagram in determining the physical state or states of
a system under different sets of pressures and temperatures, let’s consider a sample of
water at point E in Figure 13-18a (355 torr and Ϫ10°C). At this point all the water is in
the form of ice, H2O(s). Suppose that we hold the pressure constant and gradually increase
the temperature—in other words, trace a path from left to right along EG. At the temperature
at which EG intersects AB, the melting curve, some of the ice melts. If we stopped
here, equilibrium between solid and liquid water would eventually be established, and both
phases would be present. If we added more heat, all the solid would melt with no temperature
change. Remember that all phase changes of pure substances occur at constant
temperature.
Once the solid is completely melted, additional heat causes the temperature to rise.
Eventually, at point F (355 torr and 80°C), some of the liquid begins to boil; liquid, H2O(ᐉ),
and vapor, H2O(g), are in equilibrium. Adding more heat at constant pressure vaporizes
the rest of the water with no temperature change. Adding still more heat warms the vapor
(gas) from F to G. Complete vaporization would also occur if, at point F and before all
the liquid had vaporized, the temperature were held constant and the pressure were
Benzene is denser as a solid than as a
liquid, so the solid sinks in the liquid
(left). This is the behavior shown by
nearly all known substances except
water (right).
508 CHAPTER 13: Liquids and Solids
Camphor, which is used in inhalers, has a high vapor pressure. When stored in a bottle,
camphor sublimes and then deposits elsewhere in the bottle.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 13.7, The Weird Properties of
Water.
The CO2 in common fire
extinguishers is liquid. As you can see
from Figure 13-17b, the liquid must be
at some pressure greater than 10 atm
for temperatures above 0°C. It is
ordinarily at about 65 atm (more than
900 lb/in.2), so these cylinders must be
handled with care.
Phase diagrams are obtained by
combining the results of heating curves
measured experimentally at different
pressures.
decreased to, say, 234 torr at point H. If we wished to hold the pressure at 234 torr and
condense some of the vapor, it would be necessary to cool the vapor to 70°C, point I,
which lies on the vapor pressure curve, AC. To state this in another way, the vapor pressure
of water at 70°C is 234 torr.
Suppose we move back to solid at point E (355 torr and Ϫ10°C). If we now hold the
temperature at Ϫ10°C and reduce the pressure, we move vertically down along EJ. At a
pressure of 2.1 torr we reach the sublimation curve, at which point the solid passes directly
into the gas phase (sublimes) until all the ice has sublimed. An important application of
this phenomenon is in the freeze-drying of foods. In this process a water-containing food
is cooled below the freezing point of water to form ice, which is then removed as a vapor
by decreasing the pressure.
Let us clarify the nature of the fluid phases (liquid and gas) and of the critical point by
describing two different ways that a gas can be liquefied. A sample at point W in the phase
diagram of Figure 13-18b is in the vapor (gas) phase, below its critical temperature.
Suppose we compress the sample at constant T from point W to point Z. We can identify
a definite pressure (the intersection of line WZ with the vapor pressure curve AC)
where the transition from gas to liquid takes place. If we go around the critical point by
the path WXYZ, however, no such clear-cut transition takes place. By this second path,
the density and other properties of the sample vary in a continuous manner; there is no
definite point at which we can say that the sample changes from gas to liquid.
A weighted wire cuts through a
block of ice. The ice melts under the
high pressure of the wire, and then
refreezes behind the wire.
A fluid below its critical temperature
may properly be identified as a
liquid or as a gas. Above the critical
temperature, we should use the term
“fluid.”
–10 0.01 70 80 374
2.1
4.6
355
218 atm
G
H
CB
E
Solid
A
J
D
Gas
F
Liquid
Temperature (°C)
Pressure(torr)
(a)
I
TC
A
Solid
Gas
Liquid
Z Y
XW
CPC
(b)
P
T
Figure 13-18 Some interpretations of phase diagrams. (a) The phase diagram of water.
Phase relationships at various points in this diagram are described in the text. (b) Two paths
by which a gas can be liquefied. (1) Below the critical temperature. Compressing the sample
at constant temperature is represented by the vertical line WZ. Where this line crosses the
vapor pressure curve AC, the gas liquefies; at that set of conditions, two distinct phases, gas
and liquid, are present in equilibrium with each other. These two phases have different
properties, for example, different densities. Raising the pressure further results in a
completely liquid sample at point Z. (2) Above the critical temperature. Suppose that we
instead first warm the gas at constant pressure from W to X, a temperature above its critical
temperature. Then, holding the temperature constant, we increase the pressure to point Y.
Along this path, the sample increases smoothly in density, with no sharp transition between
phases. From Y, we then decrease the temperature to reach final point Z, where the sample
is clearly a liquid.
AMORPHOUS SOLIDS AND CRYSTALLINE SOLIDS
We have already seen that solids have definite shapes and volumes, are not very compressible,
are dense, and diffuse only very slowly into other solids. They are generally
characterized by compact, ordered arrangements of particles that vibrate about fixed positions
in their structures.
Some noncrystalline solids, called amorphous solids, have no well-defined, ordered
structure. Examples include rubber, some kinds of plastics, and amorphous sulfur.
Some amorphous solids are called glasses because, like liquids, they flow, although very
slowly. The irregular structures of glasses are intermediate between those of freely flowing
liquids and those of crystalline solids; there is only short-range order. Crystalline solids
such as ice and sodium chloride have well-defined, sharp melting temperatures. Particles
in amorphous solids are irregularly arranged, so intermolecular forces among their particles
vary in strength within a sample. Melting occurs at different temperatures for various
portions of the same sample as the intermolecular forces are overcome. Unlike crystalline
solids, glasses and other amorphous solids do not exhibit sharp melting points, but soften
over a temperature range.
The shattering of a crystalline solid produces fragments having the same (or related)
interfacial angles and structural characteristics as the original sample. The shattering of a
cube of rock salt produces several smaller cubes of rock salt. This cleaving occurs preferentially
along crystal lattice planes between which the interionic or intermolecular forces
of attraction are weakest. Amorphous solids with irregular structures, such as glasses,
shatter irregularly to yield pieces with curved edges and irregular angles.
13-14
The regular external shape of a crystal is the
result of regular internal arrangements of
atoms, molecules, or ions. Crystals of the
ionic solid sodium chloride, NaCl, from a
kitchen saltshaker have the same shape as the
large crystal shown here.
510 CHAPTER 13: Liquids and Solids
E n r i c h m e n t
X-Ray Diffraction
Atoms, molecules, and ions are much too small to be seen with the eye. The arrangements
of particles in crystalline solids are determined indirectly by X-ray diffraction (scattering).
In 1912, the German physicist Max von Laue (1879–1960) showed that any crystal could
serve as a three-dimensional diffraction grating for incident electromagnetic radiation with
wavelengths approximating the internuclear separations of atoms in the crystal. Such radiation
is in the X-ray region of the electromagnetic spectrum. Using an apparatus such as
that shown in Figure 13-19, a monochromatic (single-wavelength) X-ray beam is defined
by a system of slits and directed onto a crystal. The crystal is rotated to vary the angle of
incidence ␪. At various angles, strong beams of deflected X-rays hit a photographic plate.
Upon development, the plate shows a set of symmetrically arranged spots due to deflected
X-rays. Different crystals produce different arrangements of spots.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 13.12, Solid Structures (1):
Crystalline and Amorphous Solids.
One test for the purity of a crystalline
solid is the sharpness of its melting
point. Impurities disrupt the
intermolecular forces and cause
melting to occur over a considerable
temperature range.
The lattice planes are planes within
the crystal containing ordered
arrangements of particles.
William and Lawrence Bragg are the
only father and son to receive the
Nobel Prize, which they shared in
physics in 1915.
Figure 13-19 (a) X-ray diffraction
by crystals (schematic). (b) A
photograph of the X-ray diffraction
pattern from a crystal of the enzyme
histidine decarboxylase (MW Ϸ
37,000 amu). The crystal was rotated
so that many different lattice planes
with different spacings were moved
in succession into diffracting
position (see Figure 13-20).
13-14 Amorphous Solids and Crystalline Solids 511
In 1913, the English scientists William (1862–1942) and Lawrence (1890–1971) Bragg
found that diffraction photographs are more easily interpreted by considering the crystal as
a reflection grating rather than a diffraction grating. The analysis of the spots is somewhat
complicated, but an experienced crystallographer can determine the separations between
atoms within identical layers and the distances between layers of atoms. The more electrons
an atom has, the more strongly it scatters X-rays, so it is also possible to determine the
identities of individual atoms using this technique.
Figure 13-20 illustrates the determination of spacings between layers of atoms. The Xray
beam strikes parallel layers of atoms in the crystal at an angle ␪. Those rays colliding
with atoms in the first layer are reflected at the same angle ␪. Those passing through the
first layer may be reflected from the second layer, third layer, and so forth. A reflected beam
results only if all rays are in phase.
For the waves to be in phase (interact constructively), the difference in path length must
be equal to the wavelength, ␭, times an integer, n. This leads to the condition known as the
Bragg equation.
n␭ ϭ 2d sin ␪ or sin ␪ ϭ
It tells us that for X-rays of a given wavelength ␭, atoms in planes separated by distances d
give rise to reflections at angles of incidence ␪. The reflection angles increases with
increasing order, n ϭ 1, 2, 3, . . . .
n␭
ᎏ
2d
(a) Crystal
Slit system
X-ray tube
Photographic film
or recording device
(ϩ)
(Ϫ)
␪
(b)
d
␪
␪
Figure 13-20 Reflection of a
monochromatic beam of X-rays by
two lattice planes (layers of atoms)
of a crystal.
STRUCTURES OF CRYSTALS
All crystals contain regularly repeating arrangements of atoms, molecules, or ions. They
are analogous (but in three dimensions) to a wallpaper pattern (Figure 13-21). Once we
discover the pattern of a wallpaper, we can repeat it in two dimensions to cover a wall.
To describe such a repeating pattern we must specify two things: (1) the size and shape
of the repeating unit and (2) the contents of this unit. In the wallpaper pattern of Figure
13-21a, two different choices of the repeating unit are outlined. Repeating unit A contains
one complete cat; unit B, with the same area, contains parts of four different cats, but
these still add up to one complete cat. From whichever unit we choose, we can obtain the
entire pattern by repeatedly translating the contents of that unit in two dimensions.
In a crystal the repeating unit is three-dimensional; its contents consist of atoms, molecules,
or ions. The smallest unit of volume of a crystal that shows all the characteristics
of the crystal’s pattern is a unit cell. We note that the unit cell is just the fundamental
box that describes the arrangement. The unit cell is described by the lengths of its edges—
a, b, c (which are related to the spacings between layers, d)—and the angles between the
edges—␣, ␤, ␥ (Figure 13-22). Unit cells are stacked in three dimensions to build a lattice,
the three-dimensional arrangement corresponding to the crystal. It can be proven that unit
cells must fit into one of the seven crystal systems (Table 13-9). Each crystal system is
13-15
Figure 13-21 Patterns that repeat
in two dimensions. Such patterns
might be used to make wallpaper.
We must imagine that the pattern
extends indefinitely (to the end of
the wall). In each pattern two of the
many possible choices of unit cells
are outlined. Once we identify a unit
cell and its contents, repetition by
translating this unit generates the
entire pattern. In (a) the unit cell
contains only one cat. In (b) each
cell contains two cats related to one
another by a 180° rotation. Any
crystal is an analogous pattern in
which the contents of the threedimensional
unit cell consist of
atoms, molecules, or ions. The
pattern extends in three dimensions
to the boundaries of the crystal,
usually including many thousands of
unit cells.
512 CHAPTER 13: Liquids and Solids
A A
B
(a) (b)
B
TABLE 13-9 The Unit Cell Relationships for the Seven Crystal Systems*
Unit Cell
System Lengths Angles Example (common name)
cubic a ϭ b ϭ c ␣ ϭ ␤ ϭ ␥ ϭ 90° NaCl (rock salt)
tetragonal a ϭ b c ␣ ϭ ␤ ϭ ␥ ϭ 90° TiO2 (rutile)
orthorhombic a b c ␣ ϭ ␤ ϭ ␥ ϭ 90° MgSO4и7H2O (epsomite)
monoclinic a b c ␣ ϭ ␥ ϭ 90°; ␤ 90° CaSO4и2H2O (gypsum)
triclinic a b c ␣ ␤ ␥ 90° K2Cr2O7 (potassium dichromate)
hexagonal a ϭ b c ␣ ϭ ␤ ϭ 90°; ␥ ϭ 120° SiO2 (silica)
rhombohedral a ϭ b ϭ c ␣ ϭ ␤ ϭ ␥ 90° CaCO3 (calcite)
*In these definitions, the sign means “is not necessarily equal to.”
c
b
a
α β
γ
Figure 13-22 A representation of a
unit cell.
distinguished by the relations between the unit cell lengths and angles and by the symmetry
of the resulting three-dimensional patterns. Crystals have the same symmetry as their
constituent unit cells because all crystals are repetitive multiples of such cells.
Let us replace each repeat unit in the crystal by a point (called a lattice point) placed at
the same place in the unit. All such points have the same environment and are indistinguishable
from one another. The resulting three-dimensional array of points is called a
lattice. It is a simple but complete description of the way in which a crystal structure is
built up.
The unit cells shown in Figure 13-23a are the simple, or primitive, unit cells corresponding
to the seven crystal systems listed in Table 13-9. Each of these unit cells
corresponds to one lattice point. As a two-dimensional representation of the reasoning
behind this statement, look at the unit cell marked “B” in Figure 13-21a. Each corner of
the unit cell is a lattice point, and can be imagined to represent one cat. The cat at each
corner is shared among four unit cells (remember—we are working in two dimensions
here). The unit cell has four corners, and in the corners of the unit cell are enough pieces
to make one complete cat. Thus, unit cell B contains one cat, the same as the alternative
unit cell choice marked “A.” Now imagine that each lattice point in a three-dimensional
crystal represents an object (a molecule, an atom, and so on). Such an object at a corner
(Figure 13-24a) is shared by the eight unit cells that meet at that corner. Each unit cell
has eight corners, so it contains eight “pieces” of the object, so it contains 8(ᎏ1
8
ᎏ) ϭ 1 object.
Similarly, an object on an edge, but not at a corner, is shared by four unit cells (Figure
13-24b), and an object on a face is shared by two unit cells (Figure 13-24c).
Each unit cell contains atoms, molecules, or ions in a definite arrangement. Often the
unit cell contents are related by some additional symmetry. (For instance, the unit cell in
Figure 13-21b contains two cats, related to one another by a rotation of 180°.) Different
Figure 13-23 Shapes of unit cells
for the seven crystal systems and a
representative mineral of each
system.
13-15 Structures of Crystals 513
Fluorite
Cubic
Chalcopyrite
Tetragonal
Aragonite
Orthorhombic
Calcite
Trigonal
Emerald
Hexagonal
Azurite
Monoclinic
Rhodonite
Triclinic
substances that crystallize in the same type of lattice with the same atomic arrangement
are said to be isomorphous. A single substance that can crystallize in more than one
arrangement is said to be polymorphous.
In a simple, or primitive, lattice, only the eight corners of the unit cell are equivalent.
In other types of crystals, objects equivalent to those forming the outline of the unit cell
may occupy extra positions within the unit cell. (In this context, “equivalent” means that
the same atoms, molecules, or ions appear in identical environments and orientations at the
eight corners of the cell and, when applicable, at other locations in the unit cell.) This
results in additional lattices besides the simple ones in Figure 13-23. Two of these are
shown in Figure 13-25b, c. A body-centered lattice has equivalent points at the eight unit
cell corners and at the center of the unit cell (see Figure 13-25). Iron, chromium, and
many other metals crystallize in a body-centered cubic (bcc) arrangement. The unit cell
of such a metal contains 8(ᎏ1
8
ᎏ) ϭ 1 atom at the corners of the cell plus one atom at the center
of the cell (and therefore entirely in this cell); this makes a total of two atoms per unit
cell. A face-centered structure involves the eight points at the corners and six more equivalent
points, one in the middle of each of the six square faces of the cell. A metal (calcium
and silver are cubic examples) that crystallizes in this arrangement has 8(ᎏ1
8
ᎏ) ϭ 1 atom at
the corners plus 6(ᎏ1
2
ᎏ) ϭ 3 more in the faces, for a total of four atoms per unit cell. In more
complicated crystals, each lattice site may represent several atoms or an entire molecule.
We have discussed some simple structures that are easy to visualize. More complex
compounds crystallize in structures with unit cells that can be more difficult to describe.
Experimental determination of the crystal structures of such solids is correspondingly
more complex. Modern computer-controlled instrumentation can collect and analyze the
large amounts of X-ray diffraction data used in such studies. This now allows analysis of
structures ranging from simple metals to complex biological molecules such as proteins
and nucleic acids. Most of our knowledge about the three-dimensional arrangements of
atoms depends on crystal structure studies.
A crystal of one form of manganese
metal has Mn atoms at the corners of a
simple cubic unit cell that is 6.30° Å
on edge (Example 13-7).
514 CHAPTER 13: Liquids and Solids
᭤ Figure 13-25 Unit cells for (a) simple cubic, (b) body-centered cubic, and (c) face-centered
cubic. The spheres in each figure represent identical atoms or ions; different colors are
shown only to help you visualize the spheres in the center of the cube in body-centered
cubic (b) and in face-centered cubic (c) forms.
Figure 13-24 Representation of the sharing of an object (an atom, ion, or molecule)
among unit cells. The fraction of each sphere that “belongs” to a single unit cell is shown in
red. (a) The sharing of an object at a corner by eight unit cells. (b) The sharing of an object
on an edge by four unit cells. (c) The sharing of an object in a face by two unit cells. (d) A
representation of a unit cell that illustrates the portions of atoms presented in more detail in
Figure 13-28. The green ion at each corner is shared by eight unit cells, as in part (a). The
gray ion at each edge is shared by four unit cells, as in part (b). The green ion in each face
is shared by two unit cells, as in part (c).
(a) (b) (c) (d)
Face Corner
Edge
Each object in a face is shared between
two unit cells, so it is counted ᎏ1
2
ᎏ in
each; there are six faces in each unit
cell.
13-15 Structures of Crystals 515
Simple cubic
(a)
Body-centered
cubic
(b)
Face-centered
cubic
(c)
BONDING IN SOLIDS
We classify crystalline solids into categories according to the types of particles in the crystal
and the bonding or interactions among them. The four categories are (1) metallic solids,
(2) ionic solids, (3) molecular solids, and (4) covalent solids. Table 13-10 summarizes these
categories of solids and their typical properties.
Metallic Solids
Metals crystallize as solids in which metal ions may be thought to occupy the lattice sites
and are embedded in a cloud of delocalized valence electrons. Nearly all metals crystallize
in one of three types of lattices: (1) body-centered cubic (bcc), (2) face-centered cubic
(fcc; also called cubic close-packed), and (3) hexagonal close-packed. The latter two types
are called close-packed structures because the particles (in this case metal atoms) are packed
together as closely as possible. The differences between the two close-packed structures
are illustrated in Figures 13-26 and 13-27. Let spheres of equal size represent identical
metal atoms, or any other particles, that form close-packed structures. Consider a layer
of spheres packed in a plane, A, as closely as possible (Figure 13-27a). An identical plane
of spheres, B, is placed in the depressions of plane A. If the third plane is placed with its
spheres directly above those in plane A, the ABA arrangement results. This is the hexagonal
close-packed structure (Figure 13-27a). The extended pattern of arrangement of
planes is ABABAB . . . . If the third layer is placed in the alternate set of depressions in
the second layer so that spheres in the first and third layers are not directly above and
below each other, the cubic close-packed structure, ABCABCABC . . . , results (Figure
13-27b). In close-packed structures each sphere has a coordination number of 12, that is, 12
nearest neighbors. In ideal close-packed structures 74% of a given volume is due to spheres
and 26% is empty space. The body-centered cubic structure is less efficient in packing;
each sphere has only eight nearest neighbors, and there is more empty space.
13-16
The term “coordination number”
is used in crystallography in a
somewhat different sense from that in
coordination chemistry (Section 25-3).
Here it refers to the number of nearest
neighbors.
516 CHAPTER 13: Liquids and Solids
TABLE 13-10 Characteristics of Types of Solids
Metallic Ionic Molecular Covalent
Particles of Metal ions in “electron Anions, cations Molecules (or atoms) Atoms
unit cell cloud”
Strongest Metallic bonds Electrostatic Dispersion, dipole–dipole, Covalent bonds
interparticle attraction between and/or hydrogen bonds
forces cations and eϪ’s)
Properties Soft to very hard; good Hard; brittle; poor Soft; poor thermal and Very hard; poor
thermal and thermal and electrical conductors; thermal and electrical
electrical conductors; electrical conductors; low melting points conductors;* high
wide range of high melting points (Ϫ272 to 400°C) melting points
melting points (400 to 3000°C) (1200 to 4000°C)
(Ϫ39 to 3400°C)
Examples Li, K, Ca, Cu, Cr, Ni NaCl, CaBr2, K2SO4 CH4 (methane), P4, O2, C (diamond), SiO2
(metals) (typical salts) Ar, CO2, H2O, S8 (quartz)
*Exceptions: Diamond is a good conductor of heat; graphite is soft and conducts electricity well.
13-16 Bonding in Solids 517
Figure 13-26 (a) Spheres in the same plane, packed as closely as possible. Each sphere
touches six others. (b) Spheres in two planes, packed as closely as possible. All spheres
represent identical atoms or ions; different colors are shown only to help you visualize the
layers. Real crystals have many more than two planes. Each sphere touches six others in its
own layer, three in the layer below it, and three in the layer above it; that is, it contacts a
total of 12 other spheres (has a coordination number of 12).
(a) (b)
Figure 13-27 There are two crystal structures in which atoms are packed together as
compactly as possible. The diagrams show the structures expanded to clarify the difference
between them. (a) In the hexagonal close-packed structure, the first and third layers are
oriented in the same direction, so that each atom in the third layer (A) lies directly above an
atom in the first layer (A). (b) In the cubic close-packed structure, the first and third layers
are oriented in opposite directions, so that no atom in the third layer (C) is directly above
an atom in either of the first two layers (A and B). In both cases, every atom is surrounded
by 12 other atoms if the structure is extended indefinitely, so each atom has a coordination
number of 12. Although it is not obvious from this figure, the cubic close-packed structure
is face-centered cubic. To see this, we would have to include additional atoms and tilt the
resulting cluster of atoms.
A
B
C
CA
B
A
B
A
A
(a)
Hexagonal
close-packed
crystal structure
Cubic
close-packed
(face-centered)
crystal structure
Expanded view
(b)
518 CHAPTER 13: Liquids and Solids
Research & TechnologyCCHEMISTRY IN USE
Semiconductors
A semiconductor is an element or a compound with filled
bands that are only slightly below, but do not overlap with,
empty bands. The difference between an insulator and a semiconductor
is only the size of the energy gap, and there is no
sharp distinction between them. An intrinsic semiconductor
(i.e., a semiconductor in its pure form) is a much poorer conductor
of electricity than a metal because, for conduction to
occur in a semiconductor, electrons must be excited from
bonding orbitals in the filled valence band into the empty conduction
band. Figure (a) shows how this happens. An electron
that is given an excitation energy greater than or equal to the
band gap (Eg) enters the conduction band and leaves behind
a positively charged hole (hϩ, the absence of a bonding electron)
in the valence band. Both the electron and the hole
reside in delocalized orbitals, and both can move in an electric
field, much as electrons move in a metal. (Holes migrate
when an electron in a nearby orbital moves to fill in the hole,
thereby creating a new hole in the nearby orbital.) Electrons
and holes move in opposite directions in an electric field.
Silicon, a semiconductor of great importance in electronics,
has a band gap of 1.94 ϫ 10Ϫ22 kJ, or 1.21 electron volts
(eV). This is the energy needed to create one electron and
one hole or, put another way, the energy needed to break one
SiXSi bond. This energy can be supplied either thermally or
by using light with a photon energy greater than the band
gap. To excite one mole of electrons from the valence band
to the conduction band, an energy of
ϫ ϭ 117 kJ/mol
1.94 ϫ 10Ϫ22 kJ
ᎏᎏ
electron
6.022 ϫ 1023 electrons
ᎏᎏᎏ
mol
is required. For silicon, a large amount of energy is required,
so there are very few mobile electrons and holes (about one
electron in a trillion—i.e., 1 in 1012 —is excited thermally at
room temperature); the conductivity of pure silicon is
therefore about 1011 times lower than that of highly
conductive metals such as silver. The number of electrons
excited thermally is proportional to eϪEg/2RT. Increasing the
temperature or decreasing the band gap energy leads to
higher conductivity for an intrinsic semiconductor. Insulators
such as diamond and silicon dioxide (quartz), which have
very large values of Eg, have conductivities 1015 to 1020 times
lower than most metals.
The electrical conductivity of a semiconductor can be
greatly increased by doping with impurities. For example,
silicon, a Group IVA element, can be doped by adding small
amounts of a Group VA element, such as phosphorus, or a
Group IIIA element, such as boron. Figure (b) shows the
effect of substituting phosphorus for silicon in the crystal
structure (silicon has the same structure as diamond, Figure
13-31a). There are exactly enough valence band orbitals to
accommodate four of the valence electrons from the phosphorus
atom. However, a phosphorus atom has one more
electron (and one more proton in its nucleus) than does silicon.
The fifth electron enters a higher energy orbital that is
localized in the lattice near the phosphorus atom; the energy
of this orbital, called a donor level, is just below the conduction
band, within the energy gap. An electron in this
orbital can easily become delocalized when a small amount of
thermal energy promotes it into the conduction band.
Because the phosphorus-doped silicon contains mobile, negatively
charged carriers (electrons), it is said to be doped
n-type. Doping the silicon crystal
with boron produces a related, but
opposite, effect. Each boron atom contributes
only three valance electrons to
bonding orbitals in the valence band,
and therefore a hole is localized near
each boron atom. Thermal energy
is enough to separate the negatively
SiSiSiSi
SiSiSiSi
SiSiSiSi
Valence band
Conduction band
Thermal or
light energy
Eg
Si Si Si Si Si
Si Si Si Si Si
Si Si Si Si Si
Si Si Si Si Si
eϪ
hϩ
Ϫ
ϩ
(a) Generation of an electron–hole pair in
silicon, an intrinsic semiconductor. The electron
(eϪ) and hole (hϩ) have opposite charges, and
so move in opposite directions in an electric
field.
13-16 Bonding in Solids 519
charged boron atom from the hole, delocalizing the latter. In
this case the charge carriers are the holes, which are positive,
and the crystal is doped p-type. In both p- and n-type doping,
an extremely small concentration of dopants (as little as
one part per billion) is enough to cause a significant increase
in conductivity. For this reason, great pains are taken to purify
the semiconductors used in electronic devices.
Even in a doped semiconductor, mobile electrons and
holes are both present, although one carrier type is predominant.
For example, in a sample of silicon doped with arsenic
(n-type doping), the concentrations of mobile electrons are
slightly less than the concentration of arsenic atoms (usually
expressed in terms of atoms/cm3), and the concentrations of
mobile holes are extremely low. Interestingly, the concentrations
of electrons and holes always follow an equilibrium
expression that is entirely analogous to that for the autodissociation
of water into Hϩ and OHϪ ions (Chapter 18); that
is,
[eϪ][hϩ] ϭ Keq
where the equilibrium constant Keq depends only on the identity
of the semiconductor and the absolute temperature. For
silicon at room temperature, Keq ϭ 4.9 ϫ 1019 carriers2/cm6.
Doped semiconductors are extremely important in electronic
applications. A p–n junction is formed by joining pand
n-type semiconductors. At the junction, free electrons
and holes combine, annihilating each other and leaving positively
and negatively charged dopant atoms on opposite
sides. The unequal charge distribution on the two sides of
the junction causes an electric field to develop and gives rise
to current rectification (electrons can flow, with a small
applied voltage, only from the n side to the p side of the junction;
holes flow only in the reverse direction). Devices such
as diodes and transistors, which form the bases of most analog
and digital electronic circuits, are composed of p–n
junctions.
Professor Thomas A. Mallouk
Pennsylvania State University
SiSiSiSi
SiSiSiSi
SiSiSi
Valence band
Conduction band
Thermal
excitation
Donor
level
Si Si Si Si Si
Si Si Si Si Si
Si Si Si Si Si
Si Si Si Si Si
Pϩ
eϪ
Ϫ
The colors of semiconductors are determined by the band gap energy Eg.
Only photons with energy greater than Eg can be absorbed. From the
Planck radiation formula (E ϭ hv) and ␭v ϭ c, we calculate that the
wavelength, ␭, of an absorbed photon must be less than hc/ Eg. Gallium
arsenide (GaAs; Eg ϭ 1.4 eV) absorbs photons of wavelengths shorter
than 890 nm, which is in the near infrared region. Because it absorbs all
wavelengths of visible light, gallium arsenide appears black to the eye.
Iron oxide (Fe2O3; Eg ϭ 2.2 eV) absorbs light of wavelengths shorter
than 570 nm; it absorbs both yellow and blue light, and therefore
appears red. Cadmium sulfide (CdS; Eg ϭ 2.6 eV), which absorbs blue
light (␭ Յ 470 nm), appears yellow. Strontium titanate (SrTiO3; Eg ϭ
3.2 eV) absorbs only in the ultraviolet (␭ Յ 390 nm). It appears white
to the eye because visible light of all colors is reflected by the fine particles.
(b) n-type doping of silicon by phosphorus. The
extra valence electron from a phosphorus atom
is thermally excited into the conduction band,
leaving a fixed positive charge on the
phosphorus atom.
EXAMPLE 13-7 Nearest Neighbors
In the simple cubic form of manganese there are Mn atoms at the corners of a simple cubic
unit cell that is 6.30 Å on edge. (a) What is the shortest distance between centers of neighboring
Mn atoms? (b) How many nearest neighbors does each atom have?
520 CHAPTER 13: Liquids and Solids
Problem-Solving Tip: The Locations of the Nearest Neighbors
in Cubic Crystals
The distance from an atom to one of its nearest neighbors in any crystal structure depends
on the arrangement of atoms and on the size of the unit cell. For structures (such as
metals) that contain only one kind of atom, the nearest neighbors of each atom can be
visualized as follows. (Recall that for a cubic structure, the unit cell edge is a). In a simple
cubic structure, the nearest neighbors are along the cell edge (i). In a face-centered cubic
structure, the nearest neighbors are along the face diagonal (ii). In a body-centered cubic
structure, they are along the body diagonal (iii).
The relationships just described hold only for structures composed of a single kind of
atom. For other structures, the relationships are more complex.
Cell
edge, a
Cell edge = a
= 2 atom radii
(i)
Cell
edge, a
Face diagonal = √2 a
= 4 atom radii
(ii)
Cell
edge, a
Body diagonal = √3 a
= 4 atom radii
(iii)
Plan
We visualize the simple cubic cell.
13-16 Bonding in Solids 521
Solution
(a) One face of the cubic unit cell is shown in the left-hand drawing, with the atoms touching.
The centers of the nearest neighbor atoms are separated by one unit cell edge, at the distance
6.30 Å. (b) A three-dimensional representation of eight unit cells is also shown. In that drawing
the atoms are shown smaller for clarity. Some atoms are represented with different colors to
aid in visualizing the arrangement, but all atoms are identical. Consider the atom shown in red
at the center (at the intersection of the eight unit cells). Its nearest neighbors in all of the unit
cell directions are shown as light red atoms. As we can see, there are six nearest neighbors.
The same would be true of any atom in the structure.
EXAMPLE 13-8 Nearest Neighbors
Silver crystals are face-centered cubic, with a cell edge of 4.086 Å. (a) What is the distance
between centers of the two closest Ag atoms? (b) What is the atomic radius of silver in this
crystal? (c) How many nearest neighbors does each atom have?
Plan
We reason as in Example 13-7, except that now the two atoms closest to each other are those
along the face diagonal.
One face of the simple
cubic unit cell
Several unit cells
(atoms shown smaller for clarity)
6.30 Å
6.30 Å
6.30 Å
One face of four adjacent unit cells
(x-y plane)
One face of the face-centered
cubic unit cell
4.086 Å
4.086 Å
h
Solution
(a) One face of the face-centered cubic unit cell is shown in the left-hand drawing, with the
atoms touching. The nearest neighbor atoms are the ones along the diagonal of the face of the
cube. We may visualize the face as consisting of two right isosceles triangles sharing a common
hypotenuse, h, and having sides of length a ϭ 4.086 Å. The hypotenuse is equal to twice the
center-to-center distance. The hypotenuse can be calculated from the Pythagorean theorem,
h2 ϭ a2 ϩ a2. The length of the hypotenuse equals the square root of the sum of the squares
of the sides.
h ϭ ͙a2ෆ ϩෆ aෆ2ෆ ϭ ͙2ෆa2ෆ ϭ ͙2ෆ(4ෆ.0ෆ8ෆ6ෆ Åෆ)2ෆ ϭ 5.778 Å
The distance between centers of adjacent silver atoms is one half of h, so
Distance ϭ ϭ 2.889 Å
(b) The hypotenuse of the unit cell face is four times the radius of the silver atom.
Atom radius ϭ ϭ 1.444 Å
(c) To see the number of nearest neighbors, we expand the left-hand drawing to include several
unit cells, as shown in the right-hand drawing. Suppose that this is the x–y plane. The atom
shown in red has four nearest neighbors in this plane. There are four more such neighbors in
the x–z plane (perpendicular to the x–y plane), and four additional neighbors in the y–z plane
(also perpendicular to the x–y plane). This gives a total of 12 nearest neighbors.
EXAMPLE 13-9 Density and Cell Volume
From data in Example 13-8, calculate the density of metallic silver.
Plan
We first determine the mass of a unit cell, that is, the mass of four atoms of silver. The density
of the unit cell, and therefore of silver, is its mass divided by its volume.
Solution
_?_ g Ag per unit cell ϭ ϫ ϫ
ϭ 7.165 ϫ 10Ϫ22 g Ag/unit cell
Vunit cell ϭ (4.086 Å)3 ϭ 68.22 Å3 ϫ ΂ ΃
3
ϭ 6.822 ϫ 10Ϫ23 cm3/unit cell
Density ϭ ϭ 10.50 g/cm3
You should now work Exercises 90 and 92.
Data obtained from crystal structures and observed densities give us information from
which we can calculate the value of Avogadro’s number. The next example illustrates these
calculations.
7.165 ϫ 10Ϫ22 g Ag/unit cell
ᎏᎏᎏᎏ
6.822 ϫ 10Ϫ23 cm3/unit cell
10Ϫ8 cm
ᎏ
Å
107.87 g Ag
ᎏᎏ
1 mol Ag
1 mol Ag
ᎏᎏᎏ
6.022 ϫ 1023 Ag atoms
4 Ag atoms
ᎏᎏ
unit cell
5.778 Å
ᎏ
4
5.778 Å
ᎏ
2
A handbook gives the density of silver
as 10.5 g/cm3 at 20°C.
522 CHAPTER 13: Liquids and Solids
EXAMPLE 13-10 Density, Cell Volume, and Avogadro’s Number
Titanium crystallizes in a body-centered cubic unit cell with an edge length of 3.306 Å. The
density of titanium is 4.401 g/cm3. Use these data to calculate Avogadro’s number.
Plan
We relate the density and the volume of the unit cell to find the total mass contained in one
unit cell. Knowing the number of atoms per unit cell, we can then find the mass of one atom.
Comparing this to the known atomic weight, which is the mass of one mole (Avogadro’s number)
of atoms, we can evaluate Avogadro’s number.
Solution
We first determine the volume of the unit cell.
Vcell ϭ (3.306 Å)3 ϭ 36.13 Å3
We now convert Å3 to cm3.
_?_cm3 ϭ 36.13 Å3 ϫ ΂ ΃
3
ϭ 3.613 ϫ 10Ϫ23 cm3
The mass of the unit cell is its volume times the observed density.
Mass of unit cell ϭ 3.613 ϫ 10Ϫ23 cm3 ϫ ϭ 1.590 ϫ 10Ϫ22 g
The bcc unit cell contains 8(ᎏ1
8
ᎏ) ϩ 1 ϭ 2 Ti atoms, so this represents the mass of two Ti atoms.
The mass of a single Ti atom is
Mass of atom ϭ ϭ 7.950 ϫ 10Ϫ23 g/atom
From the known atomic weight of Ti (47.88), we know that the mass of one mole of Ti is
47.88 g/mol. Avogadro’s number represents the number of atoms per mole, and can be calculated
as
NAv ϭ ϫ ϭ 6.023 ϫ 1023 atoms/mol
You should now work Exercise 96.
Ionic Solids
Most salts crystallize as ionic solids with ions occupying the unit cell. Sodium chloride (Figure
13-28) is an example. Many other salts crystallize in the sodium chloride (face-centered cubic)
arrangement. Examples are the halides of Liϩ, Kϩ, and Rbϩ, and M2ϩX2Ϫ oxides and sulfides
such as MgO, CaO, CaS, and MnO. Two other common ionic structures are those of cesium
chloride, CsCl (simple cubic lattice), and zincblende, ZnS (face-centered cubic lattice), shown
in Figure 13-29. Salts that are isomorphous with the CsCl structure include CsBr, CsI, NH4Cl,
TlCl, TlBr, and TlI. The sulfides of Be2ϩ, Cd2ϩ, and Hg2ϩ, together with CuBr, CuI, AgI,
and ZnO, are isomorphous with the zincblende structure (Figure 13-29c).
The ions in an ionic solid can vibrate only about their fixed positions, so ionic solids are
poor electrical and thermal conductors. Liquid (molten) ionic compounds are excellent conductors,
however, because their ions are freely mobile.
1 atom
ᎏᎏ
7.950 ϫ 10Ϫ23 g
47.88 g
ᎏ
mol
1.590 ϫ 10Ϫ22 g
ᎏᎏ
2 atoms
4.401 g
ᎏ
cm3
10Ϫ8 cm
ᎏ
Å
13-16 Bonding in Solids 523
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 13.13, Solid Structures (2): Ionic
Solids.
Figure 13-28 Some
representations of the crystal
structure of sodium chloride,
NaCl. Sodium ions are shown
in gray and chloride ions are
shown in green. (a) One unit
cell of the crystal structure
of sodium chloride. (b) A
representation of the unit cell of
sodium chloride that indicates
the relative sizes of the Naϩ
and ClϪ ions as well as how
ions are shared between unit
cells. Particles at the corners,
edges, and faces of unit cells
are shared by other unit cells.
Remember that there is an
additional Naϩ ion at the center
of the cube. (c) A cross-section
of the structure of NaCl,
showing the repeating pattern
of its unit cell at the right.
The dashed lines outline an
alterative choice of the unit cell.
The entire pattern is generated
by repeating either unit cell
(and its contents) in all three
directions. Several such choices
of unit cells are usually possible.
(d) A stereoview of the sodium
chloride structure, extending
over several unit cells.
524 CHAPTER 13: Liquids and Solids
Figure 13-29 Crystal structures of some ionic compounds of the MX type. The gray
circles represent cations. One unit cell of each structure is shown. (a) The structure
of cesium chloride, CsCl, is simple cubic. It is not body-centered, because the point
at the center of the cell (Csϩ, gray) is not the same as the point at a corner of the cell
(ClϪ, green). (b) Sodium chloride, NaCl, is face-centered cubic. (c) Zincblende, ZnS, is
face-centered cubic, with four Zn2ϩ (gray) and four S2Ϫ (yellow) ions per unit cell. The
Zn2ϩ ions are related by the same translations as the S2Ϫ ions.
(a) (c)(b)
Ϫ Ϫ Ϫ
Ϫ Ϫ Ϫ
Ϫ Ϫ Ϫ Ϫ Ϫ
Ϫ Ϫ Ϫ Ϫ Ϫ
Ϫ Ϫ Ϫ
Ϫ ϪϪ Ϫ
ϩ ϩ ϩ
ϩ ϩ ϩ
ϩ ϩ ϩ ϩ ϩ
ϩ ϩ ϩ ϩ
ϩϩ
ϩϩϩ
ϩϩ
Naϩ
ClϪ
(d)
(c) Zincblende
ZnS: ϭ Zn2ϩ, ϭ S2Ϫ
(b) Sodium chloride
NaCl: ϭ Naϩ, ϭ ClϪ
(a) Cesium chloride
CsCl: ϭ Csϩ, ϭ ClϪ
In certain types of solids, including ionic crystals, particles different from those at the corners
of the unit cell occupy extra positions within the unit cell. For example, the face-centered cubic
unit cell of sodium chloride can be visualized as having chloride ions at the corners and middles
of the faces; sodium ions are on the edges between the chloride ions and in the center (see
Figure 13-28). Thus, a unit cell of NaCl contains the following.
eight at six in middles
ClϪ: ϩ ϭ 1 ϩ 3 ϭ 4 ClϪ ions/unit cell
corners of faces
8(ᎏ1
8
ᎏ) ϩ 6(ᎏ1
2
ᎏ)
twelve on
Naϩ: ϩ one in center ϭ 3 ϩ 1 ϭ 4 Naϩ ions/unit cell
edges
12(ᎏ1
4
ᎏ) ϩ 1
The unit cell contains equal numbers of Naϩ and ClϪ ions, as required by its chemical
formula. Alternatively, we could translate the unit cell by half its length in any axial direction
within the lattice, and visualize the unit cell in which sodium and chloride ions
have exchanged positions. Such an exchange is not always possible. You should confirm
that this alternative description also gives four chloride ions and four sodium ions per
unit cell.
Ionic radii such as those in Figure 6-1 and Table 14-1 are obtained from X-ray crystallographic
determinations of unit cell dimensions, assuming that adjacent ions are in
contact with each other.
EXAMPLE 13-11 Ionic Radii from Crystal Data
Lithium bromide, LiBr, crystallizes in the NaCl face-centered cubic structure with a unit cell
edge length of a ϭ b ϭ c ϭ 5.501 Å. Assume that the BrϪ ions at the corners of the unit cell
are in contact with those at the centers of the faces. Determine the ionic radius of the BrϪ ion.
One face of the unit cell is depicted in Figure 13-30.
Plan
We may visualize the face as consisting of two right isosceles triangles sharing a common
hypotenuse, h, and having sides of length a ϭ 5.501 Å. The hypotenuse is equal to four times
the radius of the bromide ion, h ϭ 4rBrϪ.
Solution
The hypotenuse can be calculated from the Pythagorean theorem, h2 ϭ a2 ϩ a2. The length
of the hypotenuse equals the square root of the sum of the squares of the sides.
h ϭ ͙a2ෆ ϩෆ aෆ2ෆ ϭ ͙2ෆa2ෆ ϭ ͙2ෆ(5ෆ.5ෆ0ෆ1ෆ Åෆ)2ෆ ϭ 7.780 Å
The radius of the bromide ion is one fourth of h, so
rBrϪ ϭ ϭ 1.945 Å
EXAMPLE 13-12 Ionic Radii from Crystal Data
Refer to Example 13-11. Calculate the ionic radius of Liϩ in LiBr, assuming anion–cation
contact along an edge of the unit cell.
7.780 Å
ᎏ
4
Figure 13-30 One face of the facecentered
cubic unit cell of lithium
bromide (Example 13-11).
13-16 Bonding in Solids 525
5.501 Å
BrϪ
Liϩ
5.501Å
h
a
a
Plan
The edge length, a ϭ 5.501 Å, is twice the radius of the BrϪ ion plus twice the radius of the
Liϩ ion. We know from Example 13-11 that the radius for the BrϪ ion is 1.945 Å.
Solution
5.501 Å ϭ 2 rBrϪ ϩ 2 rLiϩ
2 rLiϩ ϭ 5.501 Å Ϫ 2(1.945 Å) ϭ 1.611 Å
rLiϩ ϭ 0.806 Å
You should now work Exercise 88.
The value of 1.945 Å for the BrϪ radius calculated in Example 13-11 is a little different
from the value of 1.82 Å given in Figure 6-1; the Liϩ value of 0.806 Å from Example
13-12 also differs somewhat from the value of 0.90 Å given in Figure 6-1. We should
remember that the tabulated value in Figure 6-1 is the average value obtained from a
number of crystal structures of compounds containing the specified ion. Calculations of
ionic radii usually assume that anion–anion contact exists, but this assumption is not always
true. Calculated radii therefore vary from structure to structure, and we should not place
too much emphasis on a value of an ionic radius obtained from any single structure determination.
We now see that there is some difficulty in determining precise values of ionic
radii. Similar difficulties can arise in the determination of atomic radii from molecular
and covalent solids or of metallic radii from solid metals.
Molecular Solids
The lattice positions that describe unit cells of molecular solids represent molecules or
monatomic elements (sometimes referred to as monatomic molecules). Figure 13-31 shows
the unit cells of two simple molecular crystals. Although the bonds within molecules are
covalent and strong, the forces of attraction between molecules are much weaker. They
range from hydrogen bonds and weaker dipole–dipole interactions in polar molecules such
as H2O and SO2 to very weak dispersion forces in symmetrical, nonpolar molecules such
as CH4, CO2, and O2 and monatomic elements, such as the noble gases. Because of the
relatively weak intermolecular forces of attraction, molecules can be easily displaced. Thus,
molecular solids are usually soft substances with low melting points. Because electrons do
not move from one molecule to another under ordinary conditions, molecular solids are
poor electrical conductors and good insulators.
Covalent Solids
Covalent solids (or “network solids”) can be considered giant molecules that consist of
covalently bonded atoms in an extended, rigid crystalline network. Diamond (one crystalline
form of carbon) and quartz are examples of covalent solids (Figure 13-32). Because
of their rigid, strongly bonded structures, most covalent solids are very hard and melt at
high temperatures. Because electrons are localized in covalent bonds, they are not freely
Dispersion forces are also present
among polar molecules.
526 CHAPTER 13: Liquids and Solids
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 13.14, Solid Structures (3):
Molecular Solids.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 13.15, Solid Structures (4):
Network Solids.
mobile. As a result, covalent solids are usually poor thermal and electrical conductors at
ordinary temperatures. (Diamond, however, is a good conductor of heat; jewelers use this
property to distinguish diamonds from imitations.)
An important exception to these generalizations about properties is graphite, an
allotropic form of carbon. It has the layer structure shown in Figure 13-32c. The overlap
of an extended ␲-electron network in each plane makes graphite an excellent conductor.
The very weak attraction between layers allows these layers to slide over one another
easily. Graphite is used as a lubricant, as an additive for motor oil, and in pencil “lead”
(combined with clay and other fillers to control hardness).
It is interesting to note that the
allotropes of carbon include one very
hard substance and one very soft
substance. They differ only in the
arrangement and bonding of the C
atoms.
13-16 Bonding in Solids 527
Figure 13-31 The packing
arrangement in a molecular crystal
depends on the shape of the
molecule as well as on the
electrostatic interactions of any
regions of excess positive and
negative charge in the molecules.
The arrangements in some
molecular crystals are shown here:
(a) carbon dioxide, CO2; (b)
benzene, C6H6.
(a)
(b)
BAND THEORY OF METALS
As described in the previous section, most metals crystallize in close-packed structures.
The ability of metals to conduct electricity and heat must result from strong electronic
interactions of an atom with its 8 to 12 nearest neighbors. This might be surprising at
first if we recall that each Group IA and Group IIA metal atom has only one or two
valence electrons available for bonding. This is too few to participate in bonds localized
between it and each of its nearest neighbors.
Bonding in metals is called metallic bonding. It results from the electrical attractions
among positively charged metal ions and mobile, delocalized electrons belonging to the
crystal as a whole. The properties associated with metals—metallic luster, high thermal
and electrical conductivity, and so on—can be explained by the band theory of metals,
which we now describe.
The overlap interaction of two atomic orbitals, say the 3s orbitals of two sodium atoms,
produces two molecular orbitals, one bonding orbital and one antibonding orbital (Chapter
9). If N atomic orbitals interact, N molecular orbitals are formed. In a single metallic
13-17
Figure 13-32 Portions of the atomic arrangements in three covalent solids. (a) Diamond.
Each C is bonded tetrahedrally to four others through sp3-sp3 ␴-bonds. (1.54 Å).
(b) Graphite. C atoms are linked in planes by sp2-sp2 ␴-bonds (1.42 Å). The crystal is soft,
owing to the weakness of the attractions between planes (3.40 Å). Electrons move freely
through the ␲-bonding network in these planes, but they do not jump between planes easily.
(c) Quartz (SiO2). Each Si atom (gray) is bonded tetrahedrally to four O atoms (red).
528 CHAPTER 13: Liquids and Solids
Metals can be formed into many
shapes because of their malleability
and ductility.
(a) Diamond (b) Graphite (c) A natural quartz crystal
crystal containing one mole of sodium atoms, for example, the interaction (overlap) of
6.022 ϫ 1023 3s atomic orbitals produces 6.022 ϫ 1023 molecular orbitals. Atoms interact
more strongly with nearby atoms than with those farther away. The energy that separates
bonding and antibonding molecular orbitals resulting from two given atomic orbitals
decreases as the overlap between the atomic orbitals decreases (Chapter 9). The interactions
among the mole of Na atoms result in a series of very closely spaced molecular
orbitals (formally ␴3s and ␴3s
#). These constitute a nearly continuous band of orbitals that
belongs to the crystal as a whole. One mole of Na atoms contributes 6.022 ϫ 1023 valence
electrons (Figure 13-33a), so the 6.022 ϫ 1023 orbitals in the band are half-filled.
The ability of metallic Na to conduct electricity is due to the ability of any of the
highest energy electrons in the “3s” band to jump to a slightly higher-energy vacant orbital
in the same band when an electric field is applied. The resulting net flow of electrons
through the crystal is in the direction of the applied field.
The empty 3p atomic orbitals of the Na atoms also interact to form a wide band of 3
ϫ 6.022 ϫ 1023 orbitals. The 3s and 3p atomic orbitals are quite close in energy, so the
fanned-out bands of molecular orbitals overlap, as shown in Figure 13-33b. The two overlapping
bands contain 4 ϫ 6.022 ϫ 1023 orbitals and only 6.022 ϫ 1023 electrons. Because
each orbital can hold two electrons, the resulting combination of bands is only one-eighth
full.
Overlap of “3s” and “3p” bands is not necessary to explain the ability of Na or of any
other Group IA metal to conduct electricity. It can do so utilizing only the half-filled “3s”
band. In the Group IIA metals, however, such overlap is important. Consider a crystal of
magnesium as an example. The 3s atomic orbital of an isolated Mg atom is filled with two
electrons. Thus, without this overlap, the “3s” band in a crystal of Mg is also filled. Mg
is a good conductor at room temperature because the highest energy electrons are able
to move readily into vacant orbitals in the “3p” band (Figure 13-34).
According to band theory, the highest energy electrons of metallic crystals occupy either
a partially filled band or a filled band that overlaps an empty band. A band within which
The alkali metals are those of Group
IA; the alkaline earth metals are those
of Group IIA.
13-17 Band Theory of Metals 529
Figure 13-33 (a) The band of orbitals resulting from interaction of the 3s orbitals in a
crystal of sodium. (b) Overlapping of a half-filled “3s” band (black) with an empty “3p” band
(red) of NaN crystal.
3s
Half-filled
band of
N molecular
orbitals
Empty band of
3N molecular
orbitals from 3p
orbitals
Bands overlap
Half-filled band of N
molecular orbitals
from 3s orbitals
NaN NaNNa NaNa2 Na3 Na8
Energy
Energy
(a) (b)
3p
3s
}
(or into which) electrons move to allow electrical conduction is called a conduction band.
The electrical conductivity of a metal decreases as temperature increases. The increase in
temperature causes thermal agitation of the metal ions. This impedes the flow of electrons
when an electric field is applied.
Crystalline nonmetals, such as diamond and phosphorus, are insulators—they do not
conduct electricity. The reason for this is that their highest energy electrons occupy filled
bands of molecular orbitals that are separated from the lowest empty band (conduction
band) by an energy difference called the band gap. In an insulator, this band gap is an
energy difference that is too large for electrons to jump to get to the conduction band
(Figure 13-35).
Elements that are semiconductors have filled bands that are only slightly below, but
do not overlap with, empty bands. They do not conduct electricity at low temperatures,
but a small increase in temperature is sufficient to excite some of the highest energy electrons
into the empty conduction band.
Let us now summarize some of the physical properties of metals in terms of the band
theory of metallic bonding.
1. We have just accounted for the ability of metals to conduct electricity.
2. Metals are also conductors of heat. They can absorb heat as electrons become thermally
excited to low-lying vacant orbitals in a conduction band. The reverse process
accompanies the release of heat.
3. Metals have a lustrous appearance because the mobile electrons can absorb a wide
range of wavelengths of radiant energy as they jump to higher energy levels. Then
they emit photons of visible light and fall back to lower levels within the conduction
band.
Various samples of elemental silicon.
The circle at the lower right is a
disk of ultrapure silicon on which
many electronic circuits have been
etched.
530 CHAPTER 13: Liquids and Solids
Figure 13-34 Overlapping of a filled “3s” band (blue) with an empty “3p” band of MgN
crystal. The higher-energy electrons are able to move into the “3p” band (red) as a result of
this overlap.
Figure 13-35 Distinction among metals, insulators, and semiconductors. In each case an
unshaded area represents a conduction band.
Energy
Metals Insulators Semiconductors
Partially
filled
band
Overlapping
bands
or
Empty band
Filled band
Forbidden zone
(large energy gap)
Small energy gap
Empty band
Energy
3s
3p
Mg MgN
4. Metals are malleable or ductile (or both). A crystal of a metal is easily deformed when
a mechanical stress is applied to it. All of the metal ions are identical, and they are
imbedded in a “sea of electrons.” As bonds are broken, new ones are readily formed
with adjacent metal ions. The features of the arrangement remain unchanged, and
the environment of each metal ion is the same as before the deformation occurred
(Figure 13-36). The breakage of bonds involves the promotion of electrons to
higher-energy levels. The formation of bonds is accompanied by the return of the
electrons to the original energy levels.
A crystal can be cleaved into smaller
crystals that have the same
appearance as the larger crystal.
Here a worker is cleaving a large
crystal of sodium chloride.
Key Terms 531
A malleable substance can be rolled
or pounded into sheets. A ductile
substance can be drawn into wires.
+ + + + + + + +
+ + + + + + + +
+ + + + + + + +
+ + + + + + + +
+ + + + + + + +
+ + + + + + + +
+ + + + + + + +
+ + + + + + + +
+ + + + + + + +
+ + + + + + + +
Force
Force
(a)
+– +– +– +– +–
+ – + – + – + – + –
+– +– +– +– +–
+ – + – + – + – + –
+– +– +– +– +–
+– +– +– +– +–
+ – + – + – + – + –
+– +– +– +– +–
+– +– +– +– +–
+ – + – + – + – + –
Force
Force
(b)
Figure 13-36 (a) In a metal, the positively charged metal ions are immersed in a
delocalized “cloud of electrons.” When the metal is distorted (e.g., rolled into sheets or
drawn into wires), the environment around the metal atoms is essentially unchanged, and no
new repulsive forces occur. This explains why metal sheets and wires remain intact. (b) By
contrast, when an ionic crystal is subjected to a force that causes it to slip along a plane, the
increased repulsive forces between like-charged ions cause the crystal to break.
Key Terms
Adhesive force Force of attraction between a liquid and another
surface.
Allotropes Different forms of the same element in the same
physical state.
Amorphous solid A noncrystalline solid with no well-defined,
ordered structure.
Band A series of very closely spaced, nearly continuous molecular
orbitals that belong to the crystal as a whole.
Band gap An energy separation between an insulator’s highest
filled electron energy band and the next higher-energy vacant
band.
Band theory of metals A theory that accounts for the bonding
and properties of metallic solids.
Boiling point The temperature at which the vapor pressure of
a liquid is equal to the external pressure; also the condensation
point.
Capillary action The drawing of a liquid up the inside of a smallbore
tube when adhesive forces exceed cohesive forces, or the
depression of the surface of the liquid when cohesive forces
exceed adhesive forces.
Cohesive forces All the forces of attraction among particles of
a liquid.
532 CHAPTER 13: Liquids and Solids
Condensation Liquefaction of vapor.
Condensed phases The liquid and solid phases; phases in which
particles interact strongly.
Conduction band A partially filled band or a band of vacant
energy levels just higher in energy than a filled band; a band
within which, or into which, electrons must be promoted to
allow electrical conduction to occur in a solid.
Coordination number In describing crystals, the number of
nearest neighbors of an atom or ion.
Critical point The combination of critical temperature and critical
pressure of a substance.
Critical pressure The pressure required to liquefy a gas (vapor)
at its critical temperature.
Critical temperature The temperature above which a gas cannot
be liquefied; the temperature above which a substance
cannot exhibit distinct gas and liquid phases.
Crystal lattice The pattern of arrangement of particles in a crys-
tal.
Crystalline solid A solid characterized by a regular, ordered
arrangement of particles.
Deposition The direct solidification of a vapor by cooling; the
reverse of sublimation.
Dipole–dipole interactions Interactions between polar molecules,
that is, between molecules with permanent dipoles.
Dipole-induced dipole interaction See Dispersion forces.
Dispersion forces Very weak and very short-range attractive
forces between short-lived temporary (induced) dipoles; also
called London forces.
Distillation The separation of a liquid mixture into its components
on the basis of differences in boiling points.
Dynamic equilibrium A situation in which two (or more)
processes occur at the same rate so that no net change occurs.
Evaporation Vaporization of a liquid below its boiling point.
Heat of condensation The amount of heat that must be
removed from a specific amount of a vapor at its condensation
point to condense the vapor with no change in temperature;
usually expressed in J/g or kJ/mol; in the latter case it is called
the molar heat of condensation.
Heat of fusion The amount of heat required to melt a specific
amount of a solid at its melting point with no change in temperature;
usually expressed in J/g or kJ/mol; in the latter case
it is called the molar heat of fusion.
Heat of solidification The amount of heat that must be removed
from a specific amount of a liquid at its freezing point to freeze
it with no change in temperature; usually expressed in J/g or
kJ/mol; in the latter case it is called the molar heat of solidifica-
tion.
Heat of vaporization The amount of heat required to vaporize
a specific amount of a liquid at its boiling point with no change
in temperature; usually expressed in J/g or kJ/mol; in the latter
case it is called the molar heat of vaporization.
Hydrogen bond A fairly strong dipole–dipole interaction (but
still considerably weaker than covalent or ionic bonds) between
molecules containing hydrogen directly bonded to a small,
highly electronegative atom, such as N, O, or F.
Insulator A poor conductor of electricity and heat.
Intermolecular forces Forces between individual particles
(atoms, molecules, ions) of a substance.
Intramolecular forces Forces between atoms (or ions) within
molecules (or formula units).
Isomorphous Refers to crystals having the same atomic arrange-
ment.
LeChatelier’s Principle A system at equilibrium, or striving to
attain equilibrium, responds in such a way as to counteract any
stress placed upon it.
London forces See Dispersion forces.
Melting point The temperature at which liquid and solid coexist
in equilibrium; also the freezing point.
Meniscus The upper surface of a liquid in a cylindrical container.
Metallic bonding Bonding within metals due to the electrical
attraction of positively charged metal ions for mobile electrons
that belong to the crystal as a whole.
Molar heat capacity The amount of heat necessary to raise the
temperature of one mole of a substance one degree Celsius with
no change in state; usually expressed in kJ/molи°C. See Specific
heat.
Molar heat of condensation The amount of heat that must be
removed from one mole of a vapor at its condensation point to
condense the vapor with no change in temperature; usually
expressed in kJ/mol. See Heat of condensation.
Molar heat of fusion The amount of heat required to melt one
mole of a solid at its melting point with no change in temperature;
usually expressed in kJ/mol. See Heat of fusion.
Molar heat of vaporization The amount of heat required to
vaporize one mole of a liquid at its boiling point with no change
in temperature; usually expressed in kJ/mol. See Heat of vapor-
ization.
Normal boiling point The temperature at which the vapor pressure
of a liquid is equal to one atmosphere pressure.
Normal melting point The melting (freezing) point at one
atmosphere pressure.
Phase diagram A diagram that shows equilibrium temperature–
pressure relationships for different phases of a substance.
Polymorphous Refers to substances that crystallize in more than
one crystalline arrangement.
Semiconductor A substance that does not conduct electricity
well at low temperatures but that does at higher tempera-
tures.
Specific heat The amount of heat necessary to raise the temperature
of a specific amount of a substance one degree Celsius
with no change in state; usually expressed in J/gи°C. See Molar
heat capacity.
Sublimation The direct vaporization of a solid by heating without
passing through the liquid state.
Supercritical fluid A substance at a temperature above its critical
temperature.
Exercises 533
Surface tension The result of inward intermolecular forces of
attraction among liquid particles that must be overcome to
expand the surface area.
Triple point The point on a phase diagram that corresponds to
the only pressure and temperature at which three phases (usually
solid, liquid, and gas) of a substance can coexist at
equilibrium.
Unit cell The smallest repeating unit showing all the structural
characteristics of a crystal.
Vapor pressure The partial pressure of a vapor in equilibrium
with its parent liquid or solid.
Viscosity The tendency of a liquid to resist flow; the inverse of
its fluidity.
Volatility The ease with which a liquid vaporizes.
Exercises
General Concepts
*001. What causes dispersion forces? What factors determine
the strengths of dispersion forces between molecules?
*002. What is hydrogen bonding? Under what conditions can
strong hydrogen bonds be formed?
*003. Which of the following substances have permanent
dipole–dipole forces? (a) SiH4, (b) molecular MgCl2,
(c) NBr3, (d) F2O.
*004. Which of the following substances have permanent
dipole–dipole forces? (a) Molecular AlCl3, (b) PCl5,
(c) NO, (d) SeF4.
*005. For which of the substances in Exercise 3 are dispersion
forces the only important forces in determining boiling
points?
*006. For which of the substances in Exercise 4 are dispersion
forces the only important forces in determining boiling
points?
*007. For each of the following pairs of compounds, predict
which compound would exhibit stronger hydrogen bonding.
Justify your prediction. It may help to write a Lewis
formula for each. (a) water, H2O, or hydrogen sulfide,
H2S; (b) dichloromethane, CH2Cl2 or fluoroamine,
NH2F; (c) acetone, C3H6O (contains a CUO double
bond) or ethyl alcohol, C2H6O (contains one CXO single
bond).
*008. Hydrogen bonding is a very strong dipole–dipole interaction.
Why is hydrogen bonding so strong in comparison
with other dipole–dipole interactions?
*009. What is the effect of pressure on sublimation?
*010. Which of the following substances exhibit strong
hydrogen bonding in the liquid and solid states?
(a) CH3OH (methyl alcohol), (b) PH3, (c) CH4,
(d) CH2Cl2, (e) CH3NH2.
*011. Which of the following substances exhibit strong hydrogen
bonding in the liquid and solid states? (a) H2S,
(b) NH3, (c) SiH4, (d) HF, (e) HCl.
*012. The molecular weights of SiH4 and PH3 are nearly the
same. Account for the fact that the melting and boiling
points of PH3 (Ϫ133°C and Ϫ88°C) are higher than those
of SiH4 (Ϫ185°C and Ϫ112°C).
0*13. Imagine replacing one H atom of a methane molecule,
CH4, with another atom or group of atoms. Account for
the order in the normal boiling points of the resulting
compounds: CH4(Ϫ161°C), CH3Br (3.59°C), CH3F
(Ϫ78°C), CH3OH (65°C).
*014. For each of the following pairs of compounds, predict
which would exhibit stronger hydrogen bonding. Justify
your prediction. It may help to write a Lewis formula for
each. (a) ammonia, NH3, or phosphine, PH3; (b) ethylene,
C2H4, or hydrazine, N2H4; (c) hydrogen fluoride,
HF, or hydrogen chloride, HCl.
*015. Describe the intermolecular forces that are present in
each of the following compounds. Which kind of force
would have the greatest influence on the properties
of each compound? (a) bromine pentafluoride, BrF5,
(b) acetone, C3H6O (contains a central CUO double
bond); (c) formaldehyde, H2CO.
*016. Describe the intermolecular forces that are present in
each of the following compounds. Which kind of force
would have the greatest influence on the properties of
each compound? (a) ethyl alcohol, C2H6O (contains one
CXO single bond); (b) phosphine, PH3; (c) sulfur hexafluoride,
SF6.
*017. Give the correct names for these changes in state:
(a) Crystals of para-dichlorobenzene, used as a moth
repellent, gradually become vapor without passing
through the liquid phase. (b) As you enter a warm room
from the outdoors on a cold winter day, your eyeglasses
become fogged with a film of moisture. (c) On the same
(windy) winter day, a pan of water is left outdoors. Some
of it turns to vapor, the rest to ice.
*018. The normal boiling point of trichlorofluoromethane,
CCl3F, is 24°C, and its freezing point is Ϫ111°C. Complete
these sentences by supplying the proper terms that
describe a state of matter or a change in state. (a) At
standard temperature and pressure, CCl3F is a ________.
(b) In an arctic winter at Ϫ40°C and 1 atm pressure,
CCl3F is a ________. If it is cooled to Ϫ120°C, the
molecules arrange themselves in an orderly lattice, the
CCl3F ________ and becomes a ________. (c) If
534 CHAPTER 13: Liquids and Solids
crystalline CCl3F is held at a temperature of Ϫ120°C
while a stream of helium gas is blown over it, the
crystals will gradually disappear by the process of
________. If liquid CCl3F is boiled at atmospheric
pressure, it is converted to a ________ at a temperature
of ________.
*019. Why does HF have a lower boiling point and lower heat
of vaporization than H2O, even though their molecular
weights are nearly the same and the hydrogen bonds
between molecules of HF are stronger?
0*20. Many carboxylic acids form dimers in which two molecules
“stick together.” These dimers result from the
formation of two hydrogen bonds between the two molecules.
Use acetic acid to draw a likely structure for this
kind of hydrogen-bonded dimer.
The Liquid State
*021. Use the kinetic–molecular theory to describe the behavior
of liquids with changing temperature. Why are liquids
more dense than gases?
*022. Distinguish between evaporation and boiling. Use the
kinetic–molecular theory to explain the dependence of
rate of evaporation on temperature.
*023. Support or criticize the statement that liquids with high
normal boiling points have low vapor pressures. Give
examples of three common liquids that have relatively
high vapor pressures at 25°C and three that have low
vapor pressures at 25°C.
*024. Within each group, assign each of the boiling points
to the appropriate substance on the basis of intermolecular
forces. (a) Ne, Ar, Kr: Ϫ246°C, Ϫ186°C, Ϫ152°C;
(b) NH3 H2O, HF: Ϫ33°C, 20°C, 100°C.
*025. Within each group, assign each of the boiling points
to the respective substances on the basis of intermolecular
forces. (a) N2, HCN, C2H6: Ϫ196°C, Ϫ89°C, 26°C;
(b) H2, HCl, Cl2: Ϫ35°C, Ϫ259°C, Ϫ85°C.
*026. (a) What is the definition of the normal boiling point?
(b) Why is it necessary to specify the atmospheric pressure
over a liquid when measuring a boiling point?
*027. What factors determine how viscous a liquid is? How does
viscosity change with increasing temperature?
*028. What is the surface tension of a liquid? What causes this
property? How does surface tension change with increasing
temperature?
*029. What happens inside a capillary tube when a liquid “wets”
the tube? What happens when a liquid does not “wet” the
tube?
O HC
O
CH3
*030. What are some of the similarities of the molecular-level
descriptions of the viscosity, surface tension, vapor pressure
and the rate of evaporation of a liquid?
*031. Dispersion forces are extremely weak in comparison to
the other intermolecular attractions. Explain why this
is so.
*032. Choose from each pair the substance that, in the liquid
state, would have the greater vapor pressure at a given
temperature. Base your choice on predicted strengths of
intermolecular forces. (a) BiBr3 or BiCl3, (b) CO or CO2,
(c) N2 or NO, (d) CH3COOH or HCOOCH3.
*033. Repeat Exercise 32 for (a) C6H6 or C6Cl6, (b) F2CUO
or CH3OH, (c) He or H2.
*034. The temperatures at which the vapor pressures of the following
liquids are all 100 torr are given. Predict the order
of increasing boiling points of the liquids. Butane, C4H10,
Ϫ44.2°C; 1-butanol, C4H10O, 70.1°C; diethyl ether,
C4H10O, Ϫ11.5°C.
*035. Plot a vapor pressure curve for GaCl3 from the following
vapor pressures. Determine the boiling point of GaCl3
under a pressure of 250 torr from the plot:
t (°C) 91 108 118 132 153 176 200
vp (torr) 20 40 60 100 200 400 760
*036. Plot a vapor pressure curve for Cl2O7 from the following
vapor pressures. Determine the boiling point of Cl2O7
under a pressure of 125 torr from the plot:
t (°C) Ϫ24 Ϫ13 Ϫ2 10 29 45 62 79
vp (torr) 5 10 20 40 100 200 400 760
*037. The vapor pressure of liquid bromine at room temperature
is 168 torr. Suppose that bromine is introduced drop
by drop into a closed system containing air at 775 torr
and room temperature. (The volume of liquid bromine is
negligible compared to the gas volume.) If the bromine
is added until no more vaporizes and a few drops of liquid
are present in the flask, what would be the total
pressure? What would be the total pressure if the volume
of this closed system were decreased to one half its original
value at the same temperature?
*038. A closed flask contains water at 75.0°C. The total pressure
of the air-and-water-vapor mixture is 633.5 torr. The
vapor pressure of water at this temperature is given in
Appendix E as 289.1 torr. What is the partial pressure of
the air in the flask?
0*39. ⌬Hvap is usually greater than ⌬Hfusion for a substance,
yet the nature of interactions that must be overcome in
the vaporization and fusion processes are similar. Why
is ⌬Hvap greater?
0*40. The heat of vaporization of water at 100°C is 2.26 kJ/g;
at 37°C (body temperature) it is 2.41 kJ/g.
Exercises 535
(a) Convert the latter value to standard molar heat of
vaporization, ⌬H°vap, at 37°C.
(b) Why is the heat of vaporization greater at 37°C than
at 100°C?
*041. Plot a vapor pressure curve for C2Cl2F4 from the following
vapor pressures. Determine the boiling point of
C2Cl2F4 under a pressure of 300 torr from the plot:
t (°C) Ϫ95.4 Ϫ72.3 Ϫ53.7 Ϫ39.1 Ϫ12.0 3.5
vp (torr) 1 10 40 100 400 760
*042. Plot a vapor pressure curve for C2H4F2 from the following
vapor pressures. From the plot, determine the boiling
point of C2H4F2 under a pressure of 200 torr.
t (°C) Ϫ77.2 Ϫ51.2 Ϫ31.1 Ϫ15.0 14.8 31.7
vp (torr) 1 10 40 100 400 760
Phase Changes and Associated Heat Transfer
The following values will be useful in some exercises in this section:
Specific heat of ice 2.09 J/gи°C
Heat of fusion of ice at 0°C 334 J/g
Specific heat of liquid H2O 4.18 J/gи°C
Heat of vaporization of liquid H2O at 100°C 2.26 ϫ 103 J/g
Specific heat of steam 2.03 J/gи°C
*043. What amount of heat energy, in joules, must be removed
to condense 23.2 g of water vapor at 125.0°C to liquid at
22.8°C?
*044. Is the equilibrium that is established between two physical
states of matter an example of static or dynamic
equilibrium? Explain your answer.
*045. Which of the following changes of state are exothermic?
(a) fusion, (b) liquefaction, (c) sublimation, (d) deposition.
Explain.
*046. Suppose that heat was added to a 21.8-g sample of solid
zinc at the rate of 9.84 J/s. After the temperature reached
the normal melting point of zinc, 420.°C, it remained
constant for 3.60 minutes. Calculate ⌬H°fusion at 420.°C,
in J/mol, for zinc.
*047. The specific heat of silver is 0.237 J/gи°C. Its melting
point is 961°C. Its heat of fusion is 11 J/g. How much
heat is needed to change 8.72 g of silver from solid at
25°C to liquid at 961°C?
*048. The heat of fusion of thallium is 21 J/g, and its heat of
vaporization is 795 J/g. The melting and boiling points
are 304°C and 1457°C. The specific heat of liquid thallium
is 0.13 J/gи°C. How much heat is needed to change
243 g of solid thallium at 304°C to vapor at 1457°C and
1 atm?
*049. Calculate the amount of heat required to convert 75.0 g
of ice at 0°C to liquid water at 100.°C.
*050. Calculate the amount of heat required to convert 75.0 g
of ice at Ϫ15.0°C to steam at 125.0°C.
*051. Use data in Appendix E to calculate the amount of heat
required to warm 175 g of mercury from 25°C to its boiling
point and then to vaporize it.
*052. If 275 g of liquid water at 100.°C and 475 g of water at
30.0°C are mixed in an insulated container, what is the
final temperature?
*053. If 25.0 g of ice at Ϫ10.0°C and 25.0 g of liquid water at
100.°C are mixed in an insulated container, what will the
final temperature be?
*054. If 175 g of liquid water at 0°C and 17.5 g of steam at
110.°C are mixed in an insulated container, what will the
final temperature be?
*055. Water can be cooled in hot climates by the evaporation
of water from the surfaces of canvas bags. What mass of
water can be cooled from 35.0°C to 20.0°C by the evaporation
of one gram of water? Assume that ⌬Hvap does
not change with temperature.
*056. (a) How much heat must be removed to prepare 15.0 g
of ice at 0.0°C from 15.0 g of water at 25.0°C? (b) Calculate
the mass of water at 100.0°C that could be cooled
to 23.5°C by the same amount of heat as that calculated
in part (a).
Clausius–Clapeyron Equation
*057. Toluene, C6H5CH3, is a liquid used in the manufacture
of TNT. Its normal boiling point is 111.0°C, and its molar
heat of vaporization is 35.9 kJ/mol. What would be the
vapor pressure (torr) of toluene at 75.00°C?
*058. At their normal boiling points, the heat of vaporization
of water (100°C) is 40,656 J/mol and that of heavy water
(101.41°C) is 41,606 J/mol. Use these data to calculate
the vapor pressure of each liquid at 80.00°C.
*059. (a) Use the Clausius–Clapeyron equation to calculate the
temperature (°C) at which pure water would boil at a pressure
of 400.0 torr. (b) Compare this result with the
temperature read from Figure 13-13. (c) Compare the
results of (a) and (b) with a value obtained from Appendix
E.
0*60. Show that the Clausius–Clapeyron equation can be written
as
ln P ϭ ᎏ
Ϫ⌬
R
H
T
vap
ᎏ ϩ B
where B is a constant that has different values for different
substances. This is an equation for a straight line.
(a) What is the expression for the slope of this line?
536 CHAPTER 13: Liquids and Solids
(b) Using the following vapor pressure data, plot ln P vs.
1/T for ethyl acetate, CH3COOC2H5, a common organic
solvent used in nail polish removers.
t (°C) Ϫ43.4. Ϫ23.5. Ϫ13.5. Ϫ3.0. ϩ9.1.
vp (torr) 1. 5. 10. 20. 40.
t (°C) 16.6. 27.0. 42.0. 59.3.
vp (torr) 60. 100. 200. 400.
(c) From the plot, estimate ⌬Hvap for ethyl acetate.
(d) From the plot, estimate the normal boiling point of
ethyl acetate.
0*61. Repeat Exercise 60(b) and 60(c) for mercury, using the
following data for liquid mercury. Then compare this
value with the one in Appendix E.
t (°C) 126.2 184.0. 228.8. 261.7. 323.0.
vp (torr) 1 10. 40. 100. 400.
*062. Isopropyl alcohol, C3H8O, is marketed as “rubbing alcohol.”
Its vapor pressure is 100. torr at 39.5°C and 400. torr
at 67.8°C. Estimate the molar heat of vaporization of isopropyl
alcohol.
*063. Using data from Exercise 62, predict the normal boiling
point of isopropyl alcohol.
*064. Boiling mercury is often used in diffusion pumps to attain
a very high vacuum; pressures down to 10Ϫ10 atm can be
readily attained with such a system. Mercury vapor is very
toxic to inhale, however. The normal boiling point of liquid
mercury is 357°C. What would be the vapor pressure
of mercury at 25°C?
Phase Diagrams
*065. How many phases exist at a triple point? Describe what
would happen if a small amount of heat were added under
constant-volume conditions to a sample of water at the
triple point. Assume a negligible volume change during
fusion.
*066. What is the critical point? Will a substance always be a
liquid below the critical temperature? Why or why not?
Refer to the phase diagram of CO2 in Figure 13-17b to answer
Exercises 67–70.
*067. What phase of CO2 exists at 2 atm pressure and a temperature
of Ϫ90°C? Ϫ60°C? 0°C?
*068. What phases of CO2 are present (a) at a temperature of
Ϫ78°C and a pressure of 1 atm? (b) at Ϫ57°C and a pressure
of 5.2 atm?
*069. List the phases that would be observed if a sample of CO2
at 8 atm pressure were heated from Ϫ80°C to 40°C.
*070. How does the melting point of CO2 change with pressure?
What does this indicate about the relative density
of solid CO2 versus liquid CO2?
0*71. You are given the following data for ethanol, C2H5OH.
Normal melting point Ϫ117°C
Normal boiling point 78.0°C
Critical temperature 243°C
Critical pressure 63.0 atm
Assume that the triple point is slightly lower in temperature
than the melting point and that the vapor pressure
at the triple point is about 10Ϫ5 torr. (a) Sketch a phase
diagram for ethanol. (b) Ethanol at 1 atm and 140.°C is
compressed to 70. atm. Are two phases present at any
time during this process? (c) Ethanol at 1 atm and 270.°C
is compressed to 70. atm. Are two phases present at any
time during this process?
0*72. You are given the following data for butane, C4H10.
Normal melting point Ϫ138°C
Normal boiling point 0°C
Critical temperature 152°C
Critical pressure 38 atm
Assume that the triple point is slightly lower in temperature
than the melting point and that the vapor pressure
at the triple point is 3 ϫ 10Ϫ5 torr. (a) Sketch a phase diagram
for butane. (b) Butane at 1 atm and 140°C is
compressed to 40 atm. Are two phases present at any time
during this process? (c) Butane at 1 atm and 200°C is
compressed to 40 atm. Are two phases present at any time
during this process?
Temperature (°C)
Pressure(atm)
40 60 80 100 120 140 160 180 200
to F
Liquid
Rhombic
Vapor
Monoclinic
D
G
C
E
B
A10– 6
10– 5
10– 4
10– 3
10– 2
10– 1
1
101
102
103
104
Phase Diagram for Sulfur
Exercises 537
Exercises 73 and 74 refer to the phase diagram for sulfur on page
536. (The vertical axis is on a logarithmic scale.) Sulfur has two
solid forms, monoclinic and rhombic.
0*73. (a) How many triple points are there for sulfur? (b) Indicate
the approximate pressure and temperature at each
triple point. (c) Which phases are in equilibrium at each
triple point?
0*74. Which physical states should be present at equilibrium
under the following conditions? (a) 10Ϫ1 atm and 110°C,
(b) 10Ϫ5 atm and 80°C, (c) 5 ϫ 103 atm and 160°C,
(d) 10Ϫ1 atm and 80°C, (e) 10Ϫ5 atm and 140°C, (f) 1
atm and 140°C.
The Solid State
*075. Comment on the following statement: “The only perfectly
ordered state of matter is the crystalline state.”
*076. Ice floats in water. Why? Would you expect solid mercury
to float in liquid mercury at its freezing point?
Explain.
*077. Distinguish among and compare the characteristics of
molecular, covalent, ionic, and metallic solids. Give two
examples of each kind of solid.
*078. Classify each of the following substances, in the solid
state, as molecular, ionic, covalent (network), or metallic
solids:
Electrical
Melting Boiling Conductor
Point Point
(°C) (°C) Solid Liquid
MoF6 17.5 (at 406 torr) 35 no no
BN 3000 (sublimes) — no no
Se8 217 684 poor poor
Pt 1769 3827 yes yes
RbI 642 1300 no yes
*079. Classify each of the following substances, in the solid
state, as molecular, ionic, covalent (network), or metallic
solids:
Electrical
Melting Boiling Conductor
Point Point
(°C) (°C) Solid Liquid
CeCl3 848 1727 no yes
Ti 1675 3260 yes yes
TiCl4 Ϫ25 136 no no
NO3F Ϫ175 Ϫ45.9 no no
B 2300 2550 no no
*080. Based only on their formulas, classify each of the following
in the solid state as a molecular, ionic, covalent
(network), or metallic solid: (a) SO2F, (b) MgF2, (c) W,
(d) Pb, (e) PF5.
*081. Based only on their formulas, classify each of the following
in the solid state as a molecular, ionic, covalent
(network), or metallic solid: (a) Au, (b) NO2, (c) CaF2,
(d) SF4, (e) Cdiamond.
*082. Arrange the following solids in order of increasing melting
points and account for the order: NaF, MgF2, AlF3.
*083. Arrange the following solids in order of increasing melting
points and account for the order: MgO, CaO, SrO,
BaO.
*084. Distinguish among and sketch simple cubic, bodycentered
cubic (bcc), and face-centered cubic (fcc) lattices.
Use CsCl, sodium, and nickel as examples of solids existing
in simple cubic, bcc, and fcc lattices, respectively.
*085. Describe a unit cell as precisely as you can.
Unit Cell Data: Atomic and Ionic Sizes
*086. Determine the number of ions of each type present in
each unit cell shown in Figure 13-29.
*087. Refer to Figure 13-29(a). (a) If the unit cell edge is represented
as a, what is the distance (center to center) from
Csϩ to its nearest neighbor? (b) How many equidistant
nearest neighbors does each Csϩ ion have? What are the
identities of these nearest neighbors? (c) What is the distance
(center to center), in terms of a, from a Csϩ ion to
the nearest Csϩ ion? (d) How many equidistant nearest
neighbors does each ClϪ ion have? What are their iden-
tities?
*088. Refer to Figure 13-28. (a) If the unit cell edge is represented
as a, what is the distance (center to center) from
Naϩ to its nearest neighbor? (b) How many equidistant
nearest neighbors does each Naϩ ion have? What are the
identities of these nearest neighbors? (c) What is the distance
(center to center), in terms of a, from an Naϩ ion
to the nearest Naϩ ion? (d) How many equidistant nearest
neighbors does each ClϪ ion have? What are their
identities?
*089. Polonium crystallizes in a simple cubic unit cell with an
edge length of 3.36 Å. (a) What is the mass of the unit
cell? (b) What is the volume of the unit cell? (c) What is
the theoretical density of Po?
*090. Calculate the density of Na metal. The length of the
body-centered cubic unit cell is 4.24 Å.
*091. Tungsten has a density of 19.3 g/cm3 and crystallizes in
a cubic lattice whose unit cell edge length is 3.16 Å.
Which type of cubic unit cell is it?
*092. The atomic radius of iridium is 1.36 Å. The unit cell of
iridium is a face-centered cube. Calculate the density of
iridium.
538 CHAPTER 13: Liquids and Solids
*093. A certain metal has a specific gravity of 10.200 at 25°C.
It crystallizes in a body-centered cubic arrangement with
a unit cell edge length of 3.147 Å. Determine the atomic
weight and identify the metal.
*094. The structure of diamond is shown below, with each
sphere representing a carbon atom. (a) How many carbon
atoms are there per unit cell in the diamond
structure? (b) Verify, by extending the drawing if necessary,
that each carbon atom has four nearest neighbors.
What is the arrangement of these nearest neighbors?
(c) What is the distance (center to center) from any carbon
atom to its nearest neighbor, expressed in terms of
a, the unit cell edge? (d) The observed unit cell edge
length in diamond is 3.567 Å. What is the CXC single
bond length in diamond? (e) Calculate the density of
diamond.
0*99. (a) Write the Bragg equation. Identify each symbol.
(b) X-rays from a palladium source (␭ ϭ 0.576 Å) were
reflected by a sample of copper at an angle of 9.40°. This
reflection corresponds to the unit cell length (d ϭ a) with
n ϭ 2 in the Bragg equation. Calculate the length of the
copper unit cell.
*100. The spacing between successive planes of platinum atoms
parallel to the cubic unit cell face is 2.256 Å. When
X-radiation emitted by copper strikes a crystal of platinum
metal, the minimum diffraction angle of X-rays is
19.98°. What is the wavelength of the Cu radiation?
*101. Gold crystallizes in an fcc structure. When X-radiation
of 0.70926 Å wavelength from molybdenum is used to
determine the structure of metallic gold, the minimum
diffraction angle of X-rays by the gold is 8.683°. Calculate
the spacing between parallel layers of gold atoms.
Metallic Bonding and Semiconductors
*102. In general, metallic solids are ductile and malleable,
whereas ionic salts are brittle and shatter readily (although
they are hard). Explain this observation.
*103. What single factor accounts for the ability of metals to
conduct both heat and electricity in the solid state? Why
are ionic solids poor conductors of heat and electricity
even though they are composed of charged particles?
*104. Compare the temperature dependence of electrical conductivity
of a metal with that of a typical metalloid.
Explain the difference.
Mixed Exercises
*105. Benzene, C6H6, boils at 80.1°C. How much energy, in
joules, would be required to change 500.0 g of liquid benzene
at 21.5°C to a vapor at its boiling point? (The specific
heat of liquid benzene is 1.74 J/gи°C and its heat of vaporization
is 395 J/g.)
*106. The three major components of air are N2 (bp Ϫ196°C),
O2 (bp Ϫ183°C), and Ar (bp Ϫ186°C). Suppose we have
a sample of liquid air at Ϫ200°C. In what order will these
gases evaporate as the temperature is raised?
*107. A 10.0-g sample of liquid ethanol, C2H5OH, absorbs 3.42
ϫ 103 J of heat at its normal boiling point, 78.0°C. The
molar enthalpy of vaporization of ethanol, ⌬Hvap, is 39.3
kJ/mol.
(a) What volume of C2H5OH vapor is produced? The
volume is measured at 78.0°C and 1.00 atm pressure.
(b) What mass of C2H5OH remains in the liquid state?
*108. What is the pressure predicted by the ideal gas law
for one mole of steam in 31.0 L at 100°C? What is
the pressure predicted by the van der Waals equation
(Section 12-15) given that a ϭ 5.464 L2иatm/mol2 and
b ϭ 0.03049 L/mol? What is the percent difference
between these values? Does steam deviate from ideality
significantly at 100°C? Why?
*095. The crystal structure of CO2 is cubic, with a cell edge
length of 5.540 Å. A diagram of the cell is shown in Figure
13-31a. (a) What is the number of molecules of CO2
per unit cell? (b) Is this structure face-centered cubic?
How can you tell? (c) What is the density of solid CO2
at this temperature?
*096. A Group IVA element with a density of 11.35 g/cm3 crystallizes
in a face-centered cubic lattice whose unit cell
edge length is 4.95 Å. Calculate its atomic weight. What
is the element?
*097. Crystalline silicon has the same structure as diamond,
with a unit cell edge length of 5.430 Å. (a) What is the
SiXSi distance in this crystal? (b) Calculate the density
of crystalline silicon.
0*98. (a) What types of electromagnetic radiation are suitable
for diffraction studies of crystals? (b) Describe the X-ray
diffraction experiment. (c) What must be the relationship
between the wavelength of incident radiation and the
spacing of the particles in a crystal for diffraction to
occur?
a = 3.567
Exercises 539
*109. The boiling points of HCl, HBr, and HI increase with
increasing molecular weight. Yet the melting and boiling
points of the sodium halides, NaCl, NaBr, and NaI,
decrease with increasing formula weight. Explain why the
trends are opposite.
*110. The structures for three molecules having the formula
C2H2Cl2 are
the planet Mars at a temperature of Ϫ100°C, they will
gradually disappear by the process of ________ and form
a ________.
*117. The van der Waals constants (Section 12-15) are a ϭ
19.01 L2иatm/mol2, b ϭ 0.1460 L/mol for pentane, and
a ϭ 18.05 L2иatm/mol2, b ϭ 0.1417 L/mol for isopentane.
Describe the intermolecular forces present in each of
these compounds and predict which has the lowest boiling
point.
*111. Are the following statements true or false? Indicate why
if a statement is false.
(a) The vapor pressure of a liquid will decrease if the
volume of liquid decreases.
(b) The normal boiling point of a liquid is the temperature
at which the external pressure equals the vapor
pressure of the liquid.
(c) The vapor pressures of liquids in a similar series tend
to increase with increasing molecular weight.
*112. Are the following statements true or false? Indicate why
if a statement is false.
(a) The equilibrium vapor pressure of a liquid is independent
of the volume occupied by the vapor above the
liquid.
(b) The normal boiling point of a liquid changes with
changing atmospheric pressure.
(c) The vapor pressure of a liquid will increase if the mass
of liquid is increased.
*113. The following are vapor pressures at 20°C. Predict the
order of increasing normal boiling points of the liquids,
acetone, 185 torr; ethanol, 44 torr; carbon disulfide, CS2,
309 torr.
*114. Refer to Exercise 113. What is the expected order of
increasing molar heats of vaporization, ⌬Hvap, of these
liquids at their boiling points? Account for the orders.
*115. Refer to the sulfur phase diagram on page 536. (a) Can
rhombic sulfur be sublimed? If so, under what conditions?
(b) Can monoclinic sulfur be sublimed? If so, under what
conditions? (c) Describe what happens if rhombic sulfur
is slowly heated from 80°C to 140°C at constant 1-atm
pressure. (d) What happens if rhombic sulfur is heated
from 80°C to 140°C under constant pressure of 5 ϫ 10Ϫ6
atm?
*116. The normal boiling point of ammonia, NH3, is Ϫ33°C,
and its freezing point is Ϫ78°C. Fill in the blanks. (a) At
STP (0°C, 1 atm pressure), NH3 is a ________. (b) If the
temperature drops to Ϫ40°C, the ammonia will ________
and become a ________. (c) If the temperature drops further
to Ϫ80°C and the molecules arrange themselves in
an orderly pattern, the ammonia will ________ and
become a ________. (d) If crystals of ammonia are left on
CH2C
H
CH3
CH2CH2CH2CH3 CH3 CH3 CH3
isopentanepentane
CC
Cl
HH
Cl
CC
Cl H
H Cl
CC
Cl H
HCl
(a) Basing your reasoning on intermolecular forces, why
would you expect a for pentane to be greater? (b) Basing
your reasoning on molecular size, why would you expect
b for pentane to be greater?
CONCEPTUAL EXERCISES
*118. Iodine sublimes at room temperature and pressure; water
does not. Explain the differences you would expect to
observe at room temperature if 5.0 grams of iodine crystals
were sealed in a 10.-mL container and 5.0 mL of
water were sealed in a similar container.
*119. Referring to the phase diagram for carbon dioxide shown
in Figure 13-17b for approximate values, draw a heating
curve similar to that in Figure 13-15 for carbon dioxide
at 1 atmosphere pressure. Draw a second heating curve
for carbon dioxide at 5 atm pressure. Estimate the transition
temperatures.
*120. A popular misconception is that “hot water freezes more
quickly than cold water.” In an experiment two 100.-mL
samples of water, in identical containers, were placed far
apart in a freezer at Ϫ25°C. One sample had an initial
temperature of 78°C, while the other was at 24°C. The
second sample took 151 minutes to freeze, and the
warmer sample took 166 minutes. The warmer sample
took more time but not much. How can you explain their
taking about the same length of time to freeze?
*121. Consider the portions of the heating curves shown here.
Which compound has the highest specific heat capacity?
Heat Added
Temperature
Compound A
Compound B
540 CHAPTER 13: Liquids and Solids
BUILDING YOUR KNOWLEDGE
*122. More than 150 years ago Pierre Dulong and A. T. Petit
discovered a rule of thumb that the heat capacity of one
mole of a pure solid element is about 6.0 calories per °C
(i.e., about 25 J/°C). A 100.2-g sample of an unknown
metal at 99.9°C is placed in 50.6 g of water at 24.8°C.
The temperature is 36.6°C when the system comes to
equilibrium. Assume all heat lost by the metal is absorbed
by the water. What is the likely identity of this metal?
*123. In a flask containing dry air and some liquid silicon tetrachloride,
SiCl4, the total pressure was 988 torr at 225°C.
Halving the volume of the flask increased the pressure to
1742 torr at constant temperature. What is the vapor
pressure of SiCl4 in this flask at 225°C?
*124. A friend comes to you with this problem: “I looked up
the vapor pressure of water in a table; it is 26.7 torr at
300 K and 92,826 torr at 600 K. That means that the
vapor pressure increases by a factor of 3477 when the
absolute temperature doubles over this temperature
range. But I thought the pressure was proportional to the
absolute temperature, P ϭ nRT/V. The pressure doesn’t
just double. Why?” How would you help the friend?
*125. Using as many as six drawings (frames) for each, depict
the changes that occur at the molecular level during each
of the following physical changes: (a) melting an ice cube,
(b) sublimation of an ice cube below 4.6°C and 4.6 torr,
and (c) evaporation of a droplet of water at room temperature
and pressure.
*126. Write the Lewis formula of each member of each of the
following pairs. Then use VSEPR theory to predict the
geometry about each central atom, and describe any features
that lead you to decide which member of each pair
would have the lower boiling point. (a) CH3COOH and
HCOOCH3, (b) NHF2 and BH2Cl, (c) CH3CH2OH and
CH3OCH3.
*127. At its normal melting point of 271.3°C, solid bismuth has
a density of 9.73 g/cm3 and liquid bismuth has a density
of 10.05 g/cm3. A mixture of liquid and solid bismuth is
in equilibrium at 271.3°C. If the pressure were increased
from 1 atm to 10 atm, would more solid bismuth melt or
would more liquid bismuth freeze? What unusual property
does bismuth share with water?
*128. The doping of silicon with boron to an atomic concentration
of 0.0010% boron atoms vs. silicon atoms
increases its conductivity by a factor of 103 at room temperature.
How many atoms of boron would be needed to
dope 12.5 g of silicon? What mass of boron is this?
14Solutions
OUTLINE
The Dissolution Process
14-1 Spontaneity of the Dissolution
Process
14-2 Dissolution of Solids in Liquids
14-3 Dissolution of Liquids in
Liquids (Miscibility)
14-4 Dissolution of Gases in Liquids
14-5 Rates of Dissolution and
Saturation
14-6 Effect of Temperature on
Solubility
14-7 Effect of Pressure on Solubility
14-8 Molality and Mole Fraction
Colligative Properties of Solutions
14-9 Lowering of Vapor Pressure
and Raoult’s Law
14-10 Fractional Distillation
14-11 Boiling Point Elevation
14-12 Freezing Point Depression
14-13 Determination of Molecular
Weight by Freezing Point
Depression or Boiling Point
Elevation
14-14 Colligative Properties and
Dissociation of Electrolytes
14-15 Osmotic Pressure
Colloids
14-16 The Tyndall Effect
14-17 The Adsorption Phenomenon
14-18 Hydrophilic and Hydrophobic
Colloids
A
OBJECTIVES
After you have studied this chapter, you should be able to
• Describe the factors that favor the dissolution process
• Describe the dissolution of solids in liquids, liquids in liquids, and gases in liquids
• Describe how temperature and pressure affect solubility
• Express concentrations of solutions in terms of molality and mole fractions
• Describe the four colligative properties of solutions and some of their applications
• Carry out calculations involving the four colligative properties of solutions: lowering of
vapor pressure (Raoult’s Law), boiling point elevation, freezing point depression and
osmotic pressure
• Use colligative properties to determine molecular weights of compounds
• Describe dissociation and ionization of compounds, and the associated effects on
colligative properties
• Recognize and describe colloids: the Tyndall effect, the adsorption phenomenon,
hydrophilic and hydrophobic colloids
A
solution is defined as a homogeneous mixture of substances in which no settling occurs.
A solution consists of a solvent and one or more solutes, whose proportions vary from
one solution to another. By contrast, a pure substance has fixed composition. The
solvent is the medium in which the solutes are dissolved. The fundamental units of solutes
are usually ions or molecules.
Spreading a soluble salt such as
calcium chloride on icy roadways or
sidewalks forms a solution with a
lower freezing point than pure
water, causing the ice or snow to
melt at a lower temperature.
Solutions include different combinations in which a solid, liquid, or gas acts as either solvent
or solute. Usually the solvent is a liquid. For instance, sea water is an aqueous solution of many
salts and some gases such as carbon dioxide and oxygen. Carbonated water is a saturated solution
of carbon dioxide in water. Solutions are common in nature and are extremely important
in all life processes, in all scientific areas, and in many industrial processes. The body fluids of
all forms of life are solutions. Variations in concentrations of our bodily fluids, especially those
of blood and urine, give physicians valuable clues about a person’s health. Solutions in which
the solvent is not a liquid are also common. Air is a solution of gases with variable composition.
Dental fillings are solid amalgams, or solutions of liquid mercury dissolved in solid
metals. Alloys are solid solutions of solids dissolved in a metal.
It is usually obvious which of the components of a solution is the solvent and which is (are)
the solute(s): The solvent is usually the most abundant species present. In a cup of instant
coffee, the coffee and any added sugar are considered solutes, and the hot water is the solvent.
If we mix 10 grams of alcohol with 90 grams of water, alcohol is the solute. If we mix 10 grams
of water with 90 grams of alcohol, water is the solute. But which is the solute and which is the
solvent in a solution of 50 grams of water and 50 grams of alcohol? In such cases, the terminology
is arbitrary and, in fact, unimportant.
THE DISSOLUTION PROCESS
SPONTANEITY OF THE DISSOLUTION PROCESS
In Section 4-2, part 5, we listed the solubility guidelines for aqueous solutions. Now we
investigate the major factors that influence solubility.
A substance may dissolve with or without reaction with the solvent. For example, when
metallic sodium reacts with water, there is the evolution of bubbles of hydrogen and a
great deal of heat. A chemical change occurs in which H2 and soluble ionic sodium
hydroxide, NaOH, are produced.
2Na(s) ϩ 2H2O 88n 2Naϩ(aq) ϩ 2OHϪ(aq) ϩ H2(g)
If the resulting solution is evaporated to dryness, solid sodium hydroxide, NaOH, is
obtained rather than metallic sodium. This, along with the production of bubbles of
hydrogen, is evidence of a reaction with the solvent. Reactions that involve oxidation state
changes are usually considered as chemical reactions and not as dissolution.
Solid sodium chloride, NaCl, on the other hand, dissolves in water with no evidence
of chemical reaction.
H2O
NaCl(s) 8888n Naϩ(aq) ϩ ClϪ(aq)
Evaporation of the water from the sodium chloride solution yields the original NaCl. In
this chapter we focus on dissolution processes of this type, in which no irreversible reaction
occurs between components.
The ease of dissolution of a solute depends on two factors: (1) the change in energy
and (2) the change in disorder (called entropy change) that accompanies the process. In
the next chapter we will study both of these factors in detail for many kinds of physical
and chemical changes. For now, we point out that a process is favored by (1) a decrease in
the energy of the system, which corresponds to an exothermic process, and (2) an increase in
the disorder, or randomness, of the system.
14-1
Many naturally occurring fluids
contain particulate matter suspended
in a solution. For example, blood
contains a solution (plasma) with
suspended blood cells. Sea water
contains dissolved substances as well
as suspended solids.
14-1 Spontaneity of the Dissolution Process 543
Ionic solutes that do not react with the
solvent undergo solvation. This is a
kind of reaction in which molecules
of solvent are attached in oriented
clusters to the solute particles.
Let us look at the first of these factors. If a solution gets hotter as a substance dissolves,
energy is being released in the form of heat. The energy change that accompanies a dissolution
process is called the heat of solution, ⌬Hsolution. It depends mainly on how strongly
solute and solvent particles interact. A negative value of ⌬Hsolution designates the release
of heat. More negative (less positive) values of ⌬Hsolution favor the dissolution process.
In a pure liquid all the intermolecular forces are between like molecules; when the
liquid and a solute are mixed, each molecule then interacts with molecules (or ions) unlike
it as well as with like molecules. The relative strengths of these interactions help to determine
the extent of solubility of a solute in a solvent. The main interactions that affect the
dissolution of a solute in a solvent follow.
a. Weak solute–solute attractions favor solubility.
b. Weak solvent–solvent attractions favor solubility.
c. Strong solvent–solute attractions favor solubility.
Figure 14-1 illustrates the interplay of these factors. The intermolecular or interionic
attractions among solute particles in the pure solute must be overcome (step a) to dissolve
the solute. This part of the process requires an input of energy (endothermic). Separating
the solvent molecules from one another (step b) to “make room” for the solute particles
also requires the input of energy (endothermic). Energy is released, however, as the solute
particles and solvent molecules interact in the solution (step c, exothermic). The overall
dissolution process is exothermic (and favored) if the amount of heat absorbed in hypothetical
steps a and b is less than the amount of energy released in step c. The process is
endothermic (and disfavored) if the amount of energy absorbed in steps a and b is greater
than the amount of heat released in step c.
Many solids do dissolve in liquids by endothermic processes, however. The reason such
processes can occur is that the endothermicity can be outweighed by a large increase in
disorder of the solute during the dissolution process. The solute particles are highly
ordered in a solid crystal, but are free to move about randomly in liquid solutions. Likewise,
the degree of disorder in the solvent increases as the solution is formed, because
solvent molecules are then in a more random environment. They are surrounded by a
mixture of solvent and solute particles.
We can consider the energy changes
separately, even though the actual
process cannot be carried out in these
separate steps.
544 CHAPTER 14: Solutions
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 14.3, The Solution Process.
Figure 14-1 A diagram
representing the changes in energy
content associated with the
hypothetical three-step sequence in a
dissolution process—in this case, for
a solid solute dissolving in a liquid
solvent. (Similar considerations
would apply to other combinations.)
An exothermic process is depicted
here. The amount of energy
absorbed in steps a and b is less than
the amount of energy released in
step c, so the energy of the solution
is favorable. In an endothermic
process (not shown), the heat
content of the solution would be
higher than that of the original
solvent plus solute. Thus, the
amount of energy absorbed in steps
a and b would be greater than the
amount of heat released in step c, so
energy (heat of solution) would be
unfavorable.
Solvent
Solute
Potentialenergy
Step
a
Step
b
Solvent Expanded
solute
Expanded
solvent
Expanded
solute
+
+ Step
c
Net energy change
⌬Hsolution (negative value;
exothermic)
Solution
Most dissolving processes are accompanied by an overall increase in disorder. Thus,
the disorder factor is usually favorable to solubility. The determining factor, then, is whether
the heat of solution (energy) also favors dissolution or, if it does not, whether it is small
enough to be outweighed by the favorable effects of the increasing disorder. In gases, for
instance, the molecules are so far apart that intermolecular forces are quite weak. Thus,
when gases are mixed, changes in the intermolecular forces are very slight. So the very
favorable increase in disorder that accompanies mixing is always more important than
possible changes in intermolecular attractions (energy). Hence, gases that do not react
with one another can always be mixed in any proportion.
The most common types of solutions are those in which the solvent is a liquid. In the
next several sections we consider liquid solutions in more detail.
DISSOLUTION OF SOLIDS IN LIQUIDS
The ability of a solid to go into solution depends most strongly on its crystal lattice energy,
or the strength of attractions among the particles making up the solid. The crystal lattice
energy is defined as the energy change accompanying the formation of one mole of
formula units in the crystalline state from constituent particles in the gaseous state. This
process is always exothermic; that is, crystal lattice energies are always negative. For an
ionic solid, the process is written as
Mϩ(g) ϩ XϪ(g) 88n MX(s) ϩ energy
The amount of energy involved in this process depends on the attraction between ions in
the solid. When these attractions are strong, a large amount of energy is released as the
solid forms, and so the solid is very stable.
The reverse of the crystal formation reaction is the separation of the crystal into ions.
MX(s) ϩ energy 88n Mϩ(g) ϩ XϪ(g)
This process can be considered the hypothetical first step (step a in Figure 14-1) in forming
a solution of a solid in a liquid. It is always endothermic. The smaller the magnitude of
the crystal lattice energy (a measure of the solute–solute interactions), the more readily
dissolution occurs. Less energy must be supplied to start the dissolution process.
If the solvent is water, the energy that must be supplied to expand the solvent (step b
in Figure 14-1) includes that required to break up some of the hydrogen bonding between
water molecules.
The third major factor contributing to the heat of solution is the extent to which solvent
molecules interact with particles of the solid. The process in which solvent molecules
surround and interact with solute ions or molecules is called solvation. When the solvent
is water, the more specific term is hydration. Hydration energy (equal to the sum of
steps b and c in Figure 14-1) is defined as the energy change involved in the (exothermic)
hydration of one mole of gaseous ions.
Mnϩ(g) ϩ xH2O 88n M(OH2)x
nϩ ϩ energy (for cation)
XyϪ(g) ϩ rH2O 88n X(H2O)r
yϪ ϩ energy (for anion)
Hydration is usually highly exothermic for ionic or polar covalent compounds, because
the polar water molecules interact very strongly with ions and polar molecules. In fact,
the only solutes that are appreciably soluble in water either undergo dissociation or ionization
or are able to form hydrogen bonds with water.
14-2
One of the few exceptions is the
dissolution of NaF. The water
molecules become more ordered
around the small FϪ ions. This is
due to the strong hydrogen bonding
between H2O molecules and FϪ ions.
The amount of heat released on
mixing, however, outweighs the
disadvantage of this ordering.
FϪ
H
O H OH
H
14-2 Dissolution of Solids in Liquids 545
Hydration energy is also referred to as
the heat of hydration.
A very negative crystal lattice energy
indicates very strong attractions within
the solid.
The overall heat of solution for a solid dissolving in a liquid is equal to the heat of
solvation minus the crystal lattic energy.
⌬Hsolution ϭ (heat of solvation) Ϫ (crystal lattice energy)
Remember that both terms on the right are always negative.
Nonpolar solids such as naphthalene, C10H8, do not dissolve appreciably in polar
solvents such as water because the two substances do not attract each other significantly.
This is true despite the fact that crystal lattice energies of solids consisting of nonpolar
molecules are much less negative (smaller in magnitude) than those of ionic solids. Naphthalene
dissolves readily in nonpolar solvents such as benzene because there are no strong
attractive forces between solute molecules or between solvent molecules. In such cases,
the increase in disorder controls the process. These facts help explain the observation that
“like dissolves like.”
The statement “like dissolves like” means that polar solvents dissolve ionic and polar
molecular solutes, and nonpolar solvents dissolve nonpolar molecular liquids.
Consider what happens when a piece of sodium chloride, a typical ionic solid, is placed
in water. The ␦ϩ ends of water molecules attract the negative chloride ions on the surface
of the solid NaCl, as shown in Figure 14-2. Likewise, the ␦Ϫ ends of H2O molecules (O
atoms) orient themselves toward the Naϩ ions and solvate them. These attractions help
to overcome the forces holding the ions in the crystal, and NaCl dissolves in the H2O.
H2O
NaCl(s) 8888n Naϩ(aq) ϩ ClϪ(aq)
A cube of sugar is slowly lifted
through a solution.
546 CHAPTER 14: Solutions
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 14.4, Energetics of Solution
Formation.
Figure 14-2 The role of electrostatic attractions in the dissolution of NaCl in water. The
␦ϩ H of the polar H2O molecule helps to attract ClϪ away from the crystal. Likewise, Naϩ
is attracted by the ␦Ϫ O. Once they are separated from the crystal, both kinds of ions are
surrounded by water molecules, to complete the hydration process.
Cl–
H
Cl–
–
O
+ H H +
Cl–
Na+
Na+
H
+
+
O–
O–
H
+
H
+
Cl–
Cl–Na+
Na+
Na+
Cl–
Cl–
When we write Naϩ(aq) and ClϪ(aq), we refer to hydrated ions. The number of H2O
molecules attached to an ion differs with different ions. Sodium ions are thought to be
hexahydrated; that is, Naϩ(aq) probably represents [Na(OH2)6]ϩ. Most cations in aqueous
solution are surrounded by four to nine H2O molecules, with six being the most common.
Generally, larger cations can accommodate more H2O molecules than smaller cations.
Many solids that are appreciably soluble in water are ionic compounds. Magnitudes of
crystal lattice energies generally increase with increasing charge and decreasing size of
ions. That is, the size of the lattice energy increases as the ionic charge densities increase
and, therefore, as the strength of electrostatic attractions within the crystal increases.
Hydration energies vary in the same order (Table 14-1). As we indicated earlier, crystal
lattice energies and hydration energies are generally much smaller in magnitude for molecular
solids than for ionic solids.
Hydration and the effects of attractions in a crystal oppose each other in the dissolution
process. Hydration energies and lattice energies are usually of about the same
magnitude for low-charge species, so they often nearly cancel each other. As a result, the
dissolution process is slightly endothermic for many ionic substances. Ammonium nitrate,
NH4NO3, is an example of a salt that dissolves endothermically. This property is used in
the “instant cold packs” used to treat sprains and other minor injuries. Ammonium nitrate
and water are packaged in a plastic bag in which they are kept separate by a partition that
is easily broken when squeezed. As the NH4NO3 dissolves in the H2O, the mixture absorbs
heat from its surroundings and the bag becomes cold to the touch.
Some ionic solids dissolve with the release of heat. Examples are anhydrous sodium
sulfate, Na2SO4; calcium acetate, Ca(CH3COO)2; calcium chloride, CaCl2; and lithium
sulfate hydrate, Li2SO4иH2O.
As the charge-to-size ratio (charge density) increases for ions in ionic solids, the magnitude
of the crystal lattice energy usually increases more than the hydration energy. This
makes dissolution of solids that contain highly charged ions—such as aluminum fluoride,
AlF3; magnesium oxide, MgO; and chromium(III) oxide, Cr2O3—very endothermic. As
a result, these compounds are not very soluble in water.
For simplicity, we often omit the (aq)
designations from dissolved ions.
Remember that all ions are hydrated
in aqueous solution, whether this is
indicated or not.
14-2 Dissolution of Solids in Liquids 547
TABLE 14-1 Ionic Radii, Charge/Radius Ratios, and Hydration Energies
for Some Cations
Ionic Charge/Radius Hydration Energy
Ion Radius (Å) Ratio (kJ/mol)
Kϩ 1.52 0.66 Ϫ351
Naϩ 1.16 0.86 Ϫ435
Liϩ 0.90 1.11 Ϫ544
Ca2ϩ 1.14 1.75 Ϫ1650
Fe2ϩ 0.76 2.63 Ϫ1980
Zn2ϩ 0.74 2.70 Ϫ2100
Cu2ϩ 0.72 2.78 Ϫ2160
Fe3ϩ 0.64 4.69 Ϫ4340
Cr3ϩ 0.62 4.84 Ϫ4370
Al3ϩ 0.68 4.41 Ϫ4750
Solid ammonium nitrate, NH4NO3,
dissolves in water in a very
endothermic process, absorbing
heat from its surroundings. It is used
in instant cold packs for early
treatment of injuries, such as sprains
and bruises, to minimize swelling.
The charge/radius ratio is the ionic
charge divided by the ionic radius in
angstroms. This is a measure of the
charge density around the ion. A
negative value for heat of hydration
indicates that heat is released during
hydration.
Review the sizes of ions in Figure 6-1
carefully.
DISSOLUTION OF LIQUIDS IN LIQUIDS (MISCIBILITY)
In science, miscibility is used to describe the ability of one liquid to dissolve in another.
The three kinds of attractive interactions (solute–solute, solvent–solvent, and solvent–
solute) must be considered for liquid–liquid solutions just as they were for solid–liquid
solutions. Because solute–solute attractions are usually much weaker for liquid solutes
than for solids, this factor is less important and so the mixing process is often exothermic
for miscible liquids. Polar liquids tend to interact strongly with and dissolve readily in
other polar liquids. Methanol, CH3OH; ethanol, CH3CH2OH; acetonitrile, CH3CN; and
sulfuric acid, H2SO4, are all polar liquids that are soluble in most polar solvents (such as
water). The hydrogen bonding between methanol and water molecules and the dipolar
interaction between acetonitrile and water molecules are depicted in Figure 14-3.
Because hydrogen bonding is so strong between sulfuric acid, H2SO4, and water, a
large amount of heat is released when concentrated H2SO4 is diluted with water (Figure
14-4). This can cause the solution to boil and spatter. If the major component of the
mixture is water, this heat can be absorbed with less increase in temperature because of
the unusually high specific heat of H2O. For this reason, sulfuric acid (as well as other mineral
acids) is always diluted by adding the acid slowly and carefully to water. Water should never be
added to the acid. If spattering does occur when the acid is added to water, it is mainly water
that spatters, not the corrosive concentrated acid.
Nonpolar liquids that do not react with the solvent generally are not very soluble in
polar liquids because of the mismatch of forces of interaction. Nonpolar liquids are,
however, usually quite soluble in other nonpolar liquids. Between nonpolar molecules
(whether alike or different) there are only dispersion forces, which are weak and easily
overcome. As a result, when two nonpolar liquids are mixed, their molecules just “slide
between” one another.
14-3
Hydrogen bonding and dipolar
interactions were discussed in Section
13-2.
548 CHAPTER 14: Solutions
When water is added to concentrated
acid, the danger is due more to the
spattering of the acid itself than to the
steam from boiling water.
The nonpolar molecules in oil do
not attract polar water molecules, so
oil and water are immiscible. The
polar water molecules attract one
another strongly—they “squeeze
out” the nonpolar molecules in the
oil. Oil is less dense than water, so it
floats on water.
Figure 14-3 (a) Hydrogen bonding in methanol–water solution. (b) Dipolar interaction in
acetonitrile–water solution. Each polar molecule is labeled with regions of highest negative
charge (␦Ϫ) and regions of highest positive charge (␦ϩ). Attractive forces are shown.
Molecules tend to arrange themselves to maximize attractions by bringing regions of
opposite charge together while minimizing repulsions by separating regions of like charge.
H
(b)
C
δ–
δ+
C
H
H
N
δ+
δ–
••
••
O
H
H
H
H
δ–
δ+
O
••
(a)
H
H
H
H
C
O
• •
••
H
H
δ–
δ+ δ–
• •
••
O
δ–
δ+
δ+
δ+
••
••
H
H
O
H
H
H
H
C
O
• •
••
δ–
δ+
••
••
DISSOLUTION OF GASES IN LIQUIDS
Based on Section 13-2 and the foregoing discussion, we expect that polar gases are most
soluble in polar solvents and nonpolar gases are most soluble in nonpolar liquids. Although
carbon dioxide and oxygen are nonpolar gases, they do dissolve slightly in water. CO2 is
somewhat more soluble because it reacts with water to some extent to form carbonic acid,
H2CO3. This in turn ionizes slightly in two steps to give hydrogen ions, bicarbonate ions,
and carbonate ions.
CO2(g) ϩ H2O(ᐉ) 34 H2CO3(aq) carbonic acid (exists only in solution)
H2CO3(aq) 34 Hϩ(aq) ϩ HCO3
Ϫ(aq)
HCO3
Ϫ(aq) 34 Hϩ(aq) ϩ CO3
2Ϫ(aq)
Approximately 1.45 grams of CO2 (0.0329 mole) dissolves in a liter of water at 25°C and
one atmosphere pressure.
Oxygen, O2, is less soluble in water than CO2, but it does dissolve to a noticeable extent
due to dispersion forces (induced dipoles, Section 13-2). Only about 0.041 gram of O2
(1.3 ϫ 10Ϫ3 mole) dissolves in a liter of water at 25°C and 1 atm pressure. This is sufficient
to support aquatic life.
The hydrogen halides, HF, HCl, HBr, and HI, are all polar covalent gases. In the gas
phase the interactions among the widely separated molecules are not very strong, so solute–
solute attractions are minimal, and the dissolution processes in water are exothermic. The
14-4
Figure 14-4 The heat released by
pouring 50 mL of sulfuric acid,
H2SO4, into 50 mL of water
increases the temperature by 100°C
(from 21°C to 121°C)!
14-4 Dissolution of Gases in Liquids 549
Carbon dioxide is called an acid
anhydride, that is, an “acid without
water.” As noted in Section 6-8, part 2,
many other oxides of nonmetals, such
as N2O5, SO3, and P4O10, are also acid
anhydrides.
Aqueous HCl, HBr, and HI are strong
acids (Sections 4-2, part 2, and Section
10-7). Aqueous HF is a weak acid.
resulting solutions, called hydrohalic acids, contain predominantly ionized HX (X ϭ Cl,
Br, I). The ionization involves protonation of a water molecule by HX to form a hydrated
hydrogen ion and halide ion XϪ (also hydrated). HCl is used as an example.
HF is only slightly ionized in aqueous solution because of the strong covalent bond.
In addition, the polar bond between H and the small F atoms in HF causes very strong
hydrogen bonding between H2O and the largely intact HF molecules.
The only gases that dissolve appreciably in water are those that are capable of
hydrogen bonding (such as HF), those that ionize (such as HCl, HBr, and HI), and
those that react with water (such as CO2).
RATES OF DISSOLUTION AND SATURATION
At a given temperature, the rate of dissolution of a solid increases if large crystals are
ground to a powder. Grinding increases the surface area, which in turn increases the
number of solute ions or molecules in contact with the solvent. When a solid is placed in
water, some of its particles solvate and dissolve. The rate of this process slows as time
passes because the surface area of the crystals gets smaller and smaller. At the same time,
the number of solute particles in solution increases, so they collide with the solid more
14-5
O
H
H H F O
H
HH Fas well as
␦ϩ ␦ϩ
␦Ϫ␦ϩ␦Ϫ␦ϩ ␦Ϫ␦ϩ␦Ϫ␦ϩ
O
H
HH Cl O
H
H
HϩϪϩ Cl
H
ϩ transfer ϩ
Oxygen gas is sufficiently soluble in
water to support a wide variety of
aquatic life.
550 CHAPTER 14: Solutions
A mortar and pestle are used for
grinding solids.
A saturated solution of copper(II)
sulfate, CuSO4, in water. As H2O
evaporates, blue CuSO4и5H2O
crystals form. They are in dynamic
equilibrium with the saturated
solution.
A solution that contains less than the amount of solute
necessary for saturation is said to be unsaturated.
Concentrationofsolute
Supersaturated
Saturated
Unsaturated
(at constant T, P)
0
frequently. Some of these collisions result in recrystallization. The rates of the two
opposing processes become equal after some time. The solid and dissolved ions are then
in equilibrium with each other.
Such a solution is said to be saturated. Saturation occurs at very low concentrations of
dissolved species for slightly soluble substances and at high concentrations for very soluble
substances. When imperfect crystals are placed in saturated solutions of their ions, surface
defects on the crystals are slowly “patched” with no net increase in mass of the solid.
Often, after some time has passed, we see fewer but larger crystals. These observations
provide evidence of the dynamic nature of the solubility equilibrium. After equilibrium is
established, no more solid dissolves without the simultaneous crystallization of an equal
mass of dissolved ions.
The solubilities of many solids increase at higher temperatures. Supersaturated solutions
contain higher-than-saturated concentrations of solute. They can sometimes be
prepared by saturating a solution at a high temperature. The saturated solution is cooled
slowly, without agitation, to a temperature at which the solute is less soluble. At this point,
the resulting supersaturated solution is metastable (temporarily stable). This may be thought
of as a state of pseudoequilibrium in which the system is at a higher energy than in its
most stable state. In such a case, the solute has not yet become sufficiently organized for
crystallization to begin. A supersaturated solution produces crystals rapidly if it is slightly
disturbed or if it is “seeded” with a dust particle or a tiny crystal. Under such conditions
enough solid crystallizes to leave a saturated solution (Figure 14-5).
solid dissolved particles
dissolution
crystallization
Dynamic equilibria occur in all
saturated solutions; for instance, there
is a continuous exchange of oxygen
molecules across the surface of water
in an open container. This is fortunate
for fish, which “breathe” dissolved
oxygen.
14-5 Rates of Dissolution and Saturation 551
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 14.2, Solubility.
A tiny crystal of sodium acetate,
NaCH3COO, was added to a clear,
colorless, supersaturated solution of
NaCH3COO. This photo shows
solid NaCH3COO just beginning to
crystallize in a very rapid process.
Figure 14-5 Another method of seeding a supersaturated solution is by pouring it very
slowly onto a seed crystal. A supersaturated sodium acetate solution was used in these
photographs.
EFFECT OF TEMPERATURE ON SOLUBILITY
In Section 13-6 we introduced LeChatelier’s Principle, which states that when a stress is
applied to a system at equilibrium, the system responds in a way that best relieves the stress. Recall
that exothermic processes release heat and endothermic processes absorb heat.
Exothermic: reactants 88n products ϩ heat
Endothermic: reactants ϩ heat 88n products
Many ionic solids dissolve by endothermic processes. Their solubilities in water usually
increase as heat is added and the temperature increases. For example, KCl dissolves
endothermically.
H2O
KCl(s) ϩ 17.2 kJ 8888n Kϩ(aq) ϩ ClϪ(aq)
Figure 14-6 shows that the solubility of KCl increases as the temperature increases because
more heat is available to increase the dissolving process. Raising the temperature (adding
heat) causes a stress on the solubility equilibrium. This stress favors the process that
consumes heat. In this case, more KCl dissolves.
14-6
Solid iodine, I2, dissolves to a
limited extent in water to give an
orange solution. This aqueous
solution does not mix with nonpolar
carbon tetrachloride, CCl4 (left).
Iodine is much more soluble in the
nonpolar carbon tetrachloride. After
the funnel is shaken and the liquids
are allowed to separate (right), the
upper aqueous phase is lighter
orange and the lower CCl4 layer is
much more highly colored. This is
because iodine is much more soluble
in the nonpolar carbon tetrachloride
than in water; much of the iodine
dissolves preferentially in the lower
(CCl4) phase. The design of the
separatory funnel allows the lower
(denser) layer to be drained off.
Fresh CCl4 could be added and the
process repeated. This method of
separation is called extraction. It takes
advantage of the different solubilities
of a solute in two immiscible liquids.
552 CHAPTER 14: Solutions
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 14.6, Factors Affecting Solubility
(2): Temperature and LeChatelier’s
Principle.
Figure 14-6 A graph that illustrates the effect of temperature on the solubilities of some
salts. Some compounds exist either as nonhydrated crystalline substances or as hydrated
crystals. Hydrated and nonhydrated crystal forms of the same compounds often have
different solubilities because of the different total forces of attraction in the solids. The
discontinuities in the solubility curves for CaCl2 and Na2SO4 are due to transitions between
hydrated and nonhydrated crystal forms.
200
180
160
140
120
100
80
60
40
20
0
0 10 20 30 40 50 60 70 80 90 100
Temperature in degrees Celsius
Solubilityingramsper100g(water)
KNO3
CaCl2
Na2SO4
K2SO4
NaCl
KBr
KCl
KI
CaCl2
и6H 2
O
CaCl 2
и4H
2O
Na 2
S
O
4и
1
0H
2O
Some solids, such as anhydrous Na2SO4, and many liquids and gases dissolve by
exothermic processes. Their solubilities usually decrease as temperature increases. The
solubility of O2 in water decreases (by 34%) from 0.041 gram per liter of water at 25°C
to 0.027 gram per liter at 50°C. Raising the temperature of rivers and lakes by dumping
heated waste water from industrial plants and nuclear power plants is called thermal pollution.
A slight increase in the temperature of the water causes a small but significant
decrease in the concentration of dissolved oxygen. As a result, the water can no longer
support the marine life it ordinarily could.
The dissolution of anhydrous
calcium chloride, CaCl2, in water is
quite exothermic. This dissolution
process is utilized in commercial
instant hot packs for quick treatment
of injuries requiring heat.
14-6 Effect of Temperature on Solubility 553
Calcium acetate, Ca(CH3COO)2, is more soluble in
cold water than in hot water. When a cold,
concentrated solution of calcium acetate is heated,
solid calcium acetate precipitates.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 14.5, Factors Affecting Solubility
(1): Henry’s Law and Gas Pressure.
EFFECT OF PRESSURE ON SOLUBILITY
Changing the pressure has no appreciable effect on the solubilities of either solids or
liquids in liquids. The solubilities of gases in all solvents increase, however, as the partial
pressures of the gases increase (Figure 14-7). Carbonated water is a saturated solution of
carbon dioxide in water under pressure. When a can or bottle of a carbonated beverage
is opened, the pressure on the surface of the beverage is reduced to atmospheric pressure,
and much of the CO2 bubbles out of solution. If the container is left open, the beverage
becomes “flat” because the released CO2 escapes.
Henry’s Law applies to gases that do not react with the solvent in which they dissolve
(or, in some cases, gases that react incompletely). It is usually stated as follows.
The pressure of a gas above the surface of a solution is proportional to the concentration
of the gas in the solution. Henry’s Law can be represented symbolically as
Pgas ϭ kCgas
Pgas is the pressure of the gas above the solution, and k is a constant for a particular gas
and solvent at a particular temperature. Cgas represents the concentration of dissolved gas;
it is usually expressed either as molarity (Section 3-6) or as mole fraction (Section 14-8).
The relationship is valid at low concentrations and low pressures.
MOLALITY AND MOLE FRACTION
We saw in Section 3-6 that concentrations of solutions are often expressed as percent by
mass of solute or as molarity. Discussion of many physical properties of solutions is often
made easier by expressing concentrations either in molality units or as mole fractions
(Sections 14-9 to 14-14).
14-8
14-7
Figure 14-7 An illustration of
Henry’s Law. The solubility of a gas
(that does not react completely with
the solvent) increases with increasing
pressure of the gas above the
solution.
554 CHAPTER 14: Solutions
(a) (b)
Increase
pressure
Carbonated beverages can be used to
illustrate Henry’s Law. When the
bottle is opened, the equilibrium is
disturbed and bubbles of CO2 form
within the liquid and rise to the
surface. After some time, an
equilibrium between dissolved CO2
and atmospheric CO2 is reached.
Molality
The molality, m, of a solute in solution is the number of moles of solute per kilogram
of solvent.
molality ϭ
EXAMPLE 14-1 Molality
What is the molality of a solution that contains 128 g of CH3OH in 108 g of water?
Plan
We convert the amount of solute (CH3OH) to moles, express the amount of solvent (water)
in kilograms, and apply the definition of molality.
Solution
ϭ ϫ ϭ
ϭ 37.0 m CH3OH
You should now work Exercise 28.
EXAMPLE 14-2 Molality
How many grams of H2O must be used to dissolve 50.0 grams of sucrose to prepare a 1.25 m
solution of sucrose, C12H22O11?
Plan
We convert the amount of solute (C12H22O11) to moles, solve the molality expression for kilograms
of solvent (water), and then express the result in grams.
Solution
_?_ mol C12H22O11 ϭ 50.0 g C12H22O11 ϫ ϭ 0.146 mol C12H22O11
molality of solution ϭ
Rearranging gives
kg H2O ϭ ϭ
ϭ 0.117 kg H2O ϭ 117 g H2O
0.146 mol C12H22O11
ᎏᎏᎏᎏ
1.25 mol C12H22O11/kg H2O
mol C12H22O11
ᎏᎏᎏ
molality of solution
mol C12H22O11
ᎏᎏ
kg H2O
1 mol C12H22O11
ᎏᎏ
342 g C12H22O11
37.0 mol CH3OH
ᎏᎏᎏ
kg H2O
1 mol CH3OH
ᎏᎏ
32.0 g CH3OH
128 g CH3OH
ᎏᎏ
0.108 kg H2O
_?_ mol CH3OH
ᎏᎏ
kg H2O
g CH3OH mol CH3OH
g H2O kg H2O
molality
number of moles solute
ᎏᎏᎏᎏ
number of kilograms solvent
Molality is based on the amount of
solvent not solution.
14-8 Molality and Mole Fraction 555
In other examples later in this chapter
we will calculate several properties of
the solution in Example 14-2.
Each beaker holds the amount of a
crystalline ionic compound that will
dissolve in 100. grams of water at
100°C. The compounds are (top row,
left to right) 39 grams of sodium
chloride (NaCl, white), 102 grams of
potassium dichromate (K2Cr2O7,
red-orange), 341 grams of nickel
sulfate hexahydrate (NiSO4и6H2O,
green); (bottom row, left to right)
79 grams of potassium chromate
(K2CrO4, yellow), 191 grams of
cobalt(II) chloride hexahydrate
(CoCl2и6H2O, dark red), and
203 grams of copper sulfate
pentahydrate (CuSO4и5H2O, blue).
Mole Fraction
Recall that in Chapter 12 the mole fractions, XA and XB, of each component in a mixture
containing components A and B were defined as
XA ϭ and XB ϭ
Mole fraction is a dimensionless quantity, that is, it has no units.
EXAMPLE 14-3 Mole Fraction
What are the mole fractions of CH3OH and H2O in the solution described in Example 14-1?
It contains 128 grams of CH3OH and 108 grams of H2O.
Plan
We express the amount of both components in moles, and then apply the definition of mole
fraction.
Solution
_?_ mol CH3OH ϭ 128 g CH3OH ϫ ϭ 4.00 mol CH3OH
_?_ mol H2O ϭ 108 g H2O ϫ ϭ 6.00 mol H2O
Now we calculate the mole fraction of each component.
XCH3OH ϭ ϭ ϭ 0.400
XH2O ϭ ϭ ϭ 0.600
You should now work Exercise 30.
COLLIGATIVE PROPERTIES OF SOLUTIONS
Physical properties of solutions that depend on the number, but not the kind, of solute particles
in a given amount of solvent are called colligative properties. There are four
important colligative properties of a solution that are directly proportional to the number
of solute particles present. They are (1) vapor pressure lowering, (2) boiling point elevation,
(3) freezing point depression, and (4) osmotic pressure. These properties of a solution
depend on the total concentration of all solute particles, regardless of their ionic or molecular
nature, charge, or size. For most of this chapter, we will consider nonelectrolyte solutes
(Section 4-2, part 1); these substances dissolve to give one mole of dissolved particles for
each mole of solute. In Section 14-14 we will learn to modify our predictions of colligative
properties to account for ion formation in electrolyte solutions.
6.00 mol
ᎏᎏ
(4.00 ϩ 6.00) mol
no. mol H2O
ᎏᎏᎏᎏ
no. mol CH3OH ϩ no. mol H2O
4.00 mol
ᎏᎏ
(4.00 ϩ 6.00) mol
no. mol CH3OH
ᎏᎏᎏᎏ
no. mol CH3OH ϩ no. mol H2O
1 mol H2O
ᎏᎏ
18.0 g H2O
1 mol CH3OH
ᎏᎏ
32.0 g CH3OH
no. mol B
ᎏᎏᎏ
no. mol A ϩ no. mol B
no. mol A
ᎏᎏᎏ
no. mol A ϩ no. mol. B
In any mixture the sum of the mole
fractions must be 1:
0.400 ϩ 0.600 ϭ 1
556 CHAPTER 14: Solutions
Colligative means “tied together.”
LOWERING OF VAPOR PRESSURE AND RAOULT’S LAW
Many experiments have shown that a solution containing a nonvolatile liquid or a solid as
a solute always has a lower vapor pressure than the pure solvent (Figure 14-8). The vapor
pressure of a liquid depends on the ease with which the molecules are able to escape from
the surface of the liquid. When a solute is dissolved in a liquid, some of the total volume
of the solution is occupied by solute molecules, and so there are fewer solvent molecules
per unit area at the surface. As a result, solvent molecules vaporize at a slower rate than if
no solute were present. The increase in disorder that accompanies evaporation is also a
significant factor. Because a solution is already more disordered (“mixed up”) than a pure
solvent, the evaporation of the pure solvent involves a larger increase in disorder, and is
thus more favorable. Hence, the pure solvent exhibits a higher vapor pressure than does
the solution. The lowering of the vapor pressure of the solution is a colligative property.
It is a function of the number, and not the kind, of solute particles in solution. We emphasize
that solutions of gases or low-boiling (volatile) liquid solutes can have higher total
vapor pressures than the pure solvents, so this discussion does not apply to them.
The lowering of the vapor pressure of a solvent due to the presence of nonvolatile,
nonionizing solutes is summarized by Raoult’s Law.
14-9
A vapor is a gas formed by the boiling
or evaporation of a liquid or
sublimation of a solid. The vapor
pressure of a liquid is the pressure
(partial pressure) exerted by a vapor in
equilibrium with its liquid (see Section
13-7).
14-9 Lowering of Vapor Pressure and Raoult’s Law 557
Figure 14-8 Lowering of vapor pressure. If no air is present in the apparatus, the pressure
above each liquid is due to water vapor. This pressure is less over the solution of sugar and
water.
Water Sugar
solution
Manometer
The vapor pressure of a solvent in an ideal solution is directly proportional to the
mole fraction of the solvent in the solution.
The relationship can be expressed mathematically as
Psolvent ϭ XsolventP0
solvent
where Xsolvent represents the mole fraction of the solvent in a solution, P0
solvent is the
vapor pressure of the pure solvent, and Psolvent is the vapor pressure of the solvent in the
solution (see Figure 14-9). If the solute is nonvolatile, the vapor pressure of the solution
is entirely due to the vapor pressure of the solvent, Psolution ϭ Psolvent.
The lowering of the vapor pressure, ⌬Psolvent, is defined as
⌬Psolvent ϭ P0
solvent Ϫ Psolvent
Thus,
⌬Psolvent ϭ P0
solvent Ϫ (XsolventP0
solvent) ϭ (1 Ϫ Xsolvent)P0
solvent
Now Xsolvent ϩ Xsolute ϭ 1, so 1 Ϫ Xsolvent ϭ Xsolute. We can express the lowering of the
vapor pressure in terms of the mole fraction of solute.
⌬Psolvent ϭ XsoluteP0
solvent
Solutions that obey this relationship exactly are called ideal solutions. The vapor pressures
of many solutions, however, do not behave ideally.
EXAMPLE 14-4 Vapor Pressure of a Solution of Nonvolatile Solute
Sucrose is a nonvolatile, nonionizing solute in water. Determine the vapor pressure lowering,
at 25°C, of the 1.25 m sucrose solution in Example 14-2. Assume that the solution behaves
ideally. The vapor pressure of pure water at 25°C is 23.8 torr (Appendix E).
Plan
The solution in Example 14-2 was made by dissolving 50.0 grams of sucrose (0.146 mol) in
117 grams of water (6.50 mol). We calculate the mole fraction of solute in the solution. Then
we apply Raoult’s Law to find the vapor pressure lowering, ⌬Psolvent.
Solution
Xsucrose ϭ ϭ 0.0220
Applying Raoult’s Law in terms of the vapor pressure lowering,
⌬Psolvent ϭ (Xsolute)(P0
solvent) ϭ (0.0220)(23.8 torr) ϭ 0.524 torr
You should now work Exercise 38.
When a solution consists of two components that are very similar, each component
behaves essentially as it would if it were pure. For example, the two liquids heptane, C7H16,
and octane, C8H18, are so similar that each heptane molecule experiences nearly the same
intermolecular forces whether it is near another heptane molecule or near an octane mole-
0.146 mol
ᎏᎏᎏ
0.146 mol ϩ 6.50 mol
Figure 14-9 Raoult’s Law for an
ideal solution of a solute in a volatile
liquid. The vapor pressure exerted
by the liquid is proportional to its
mole fraction in the solution.
558 CHAPTER 14: Solutions
The vapor pressure of water in the
solution is (23.8 Ϫ 0.524) torr ϭ
23.3 torr. We could calculate this
vapor pressure directly from the mole
fraction of the solvent (water) in the
solution, using the relationship
Psolvent ϭ XsolventP0
solvent.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 14.7, Colligative Properties (1):
Vapor Pressure and Raoult’s Law.
Vaporpressureofsolvent
Xsolvent
P0
solvent
Xsolute
0 0
0
0
1
1
cule, and similarly for each octane molecule. The properties of such a solution can be
predicted from a knowledge of its composition and the properties of each component.
Such a solution is very nearly ideal.
Consider an ideal solution of two volatile components, A and B. The vapor pressure
of each component above the solution is proportional to its mole fraction in the solution.
PA ϭ XAP0
A and PB ϭ XBP0
B
The total vapor pressure of the solution is, by Dalton’s Law of Partial Pressures (Section
12-11), equal to the sum of the vapor pressures of the two components.
Ptotal ϭ PA ϩ PB or Ptotal ϭ XAP0
A ϩ XBP0
B
This is shown graphically in Figure 14-10. We can use these relationships to predict the
vapor pressures of an ideal solution, as Example 14-5 illustrates.
EXAMPLE 14-5 Vapor Pressure of a Solution of Volatile Components
At 40°C, the vapor pressure of pure heptane is 92.0 torr and the vapor pressure of pure octane
is 31.0 torr. Consider a solution that contains 1.00 mole of heptane and 4.00 moles of octane.
Calculate the vapor pressure of each component and the total vapor pressure above the solution.
Plan
We first calculate the mole fraction of each component in the liquid solution. Then we apply
Raoult’s Law to each of the two volatile components. The total vapor pressure is the sum of
the vapor pressures of the components.
Solution
We first calculate the mole fraction of each component in the liquid solution.
Xheptane ϭ ϭ 0.200
Xoctane ϭ 1 Ϫ Xheptane ϭ 0.800
Then, applying Raoult’s Law for volatile components,
Pheptane ϭ XheptaneP0
heptane ϭ (0.200)(92.0 torr) ϭ 18.4 torr
Poctane ϭ XoctaneP0
octane ϭ (0.800)(31.0 torr) ϭ 24.8 torr
Ptotal ϭ Pheptane ϩ Poctane ϭ 18.4 torr ϩ 24.8 torr ϭ 43.2 torr
You should now work Exercise 40.
The vapor in equilibrium with a liquid solution of two or more volatile components
has a higher mole fraction of the more volatile component than does the liquid solution.
EXAMPLE 14-6 Composition of Vapor
Calculate the mole fractions of heptane and octane in the vapor that is in equilibrium with the
solution in Example 14-5.
Plan
We learned in Section 12-11 that the mole fraction of a component in a gaseous mixture equals
the ratio of its partial pressure to the total pressure. In Example 14-5 we calculated the partial
pressure of each component in the vapor and the total vapor pressure.
1.00 mol heptane
ᎏᎏᎏᎏᎏ
(1.00 mol heptane) ϩ (4.00 mol octane)
If component B were nonvolatile,
then P0
B would be zero, and this
description would be the same as
that given earlier for a solution of a
nonvolatile, nonionizing solute in a
volatile solvent, Ptotal ϭ Psolvent.
14-9 Lowering of Vapor Pressure and Raoult’s Law 559
Figure 14-10 Raoult’s Law for
an ideal solution of two volatile
components. The left-hand side
of the plot corresponds to pure B
(XA ϭ 0, XB ϭ 1), and the right-hand
side corresponds to pure A (XA ϭ 1,
XB ϭ 0). Of these hypothetical
liquids, B is more volatile than A
(P0
B Ͼ P0
A).
Vaporpressure
P0
B
0
1 0
0
0 1
XB
XA
P0
A
Ptotal = PA + PB
PB
PA
Solution
In the vapor
Xheptane ϭ ϭ ϭ 0.426
Xoctane ϭ ϭ ϭ 0.574
or
Xoctane ϭ 1.000 Ϫ 0.426 ϭ 0.574
You should now work Exercise 42.
Many dilute solutions behave ideally. Some solutions do not behave ideally over the
entire concentration range. For some solutions, the observed vapor pressure is greater
than that predicted by Raoult’s Law (Figure 14-11a). This kind of deviation, known as a
positive deviation, is due to differences in polarity of the two components. On the molecular
level, the two substances do not mix entirely randomly, so there is self-association of
each component with local regions enriched in one type of molecule or the other. In a
region enriched in A molecules, substance A acts as though its mole fraction were greater
than it is in the solution as a whole, and the vapor pressure due to A is greater than if the
solution were ideal. A similar description applies to component B. The total vapor pressure
is then greater than it would be if the solution were behaving ideally. A solution of
acetone and carbon disulfide is an example of a solution that shows a positive deviation
from Raoult’s Law.
Another, more common type of deviation occurs when the total vapor pressure is less
than that predicted (Figure 14-11b). This is called a negative deviation. Such an effect is
due to unusually strong attractions (such as hydrogen bonding) between unlike molecules.
24.8 torr
ᎏᎏ
43.2 torr
Poctane
ᎏ
Ptotal
18.4 torr
ᎏᎏ
43.2 torr
Pheptane
ᎏ
Ptotal
Heptane (pure vapor pressure ϭ
92.0 torr at 40°C) is a more volatile
liquid than octane (pure vapor
pressure ϭ 31.0 torr at 40°C). Its mole
fraction in the vapor, 0.426, is higher
than its mole fraction in the liquid,
0.200.
560 CHAPTER 14: Solutions
Figure 14-11 Deviations from Raoult’s Law for two volatile components. (a) Positive
deviation. (b) Negative deviation.
Vaporpressure
P0
B
P0
A
XA
XB
0
0
1 0
0
1
(a)
Vaporpressure
P0
B
P0
A
XA
XB
0
0
1 0
0
1
(b)
O
C
C
CH3 CH3
SS
acetone
carbon disulfide
As a result, unlike species hold one another especially tightly in the liquid phase, so fewer
molecules escape to the vapor phase. The observed vapor pressure of each component is
thus less than ideally predicted. An acetone–chloroform solution and an ethanol–water
solution are two examples that show negative deviations from Raoult’s Law.
FRACTIONAL DISTILLATION
In Section 13-8 we described simple distillation as a process in which a liquid solution can
be separated into volatile and nonvolatile components. But separation of volatile components
is not very efficient by this method. Consider the simple distillation of a liquid
solution consisting of two volatile components. If the temperature is slowly raised, the
solution begins to boil when the sum of the vapor pressures of the components reaches
the applied pressure on the surface of the solution. Both components exert vapor pressures,
so both are carried away as a vapor. The resulting distillate is richer than the original
liquid in the more volatile component (Example 14-6).
As a mixture of volatile liquids is distilled, the compositions of both the liquid and the
vapor, as well as the boiling point of the solution, change continuously. At constant pressure,
we can represent these quantities in a boiling point diagram, Figure 14-12. In such
a diagram the lower curve represents the boiling point of a liquid mixture with the indicated
composition. The upper curve represents the composition of the vapor in equilibrium
14-10
chloroform CHCl3
ethanol CH3CH2OH
14-10 Fractional Distillation 561
Figure 14-12 A boiling point diagram for a solution of two volatile liquids, A and B. The
lower curve represents the boiling point of a liquid mixture with the indicated composition.
The upper curve represents the composition of the vapor in equilibrium with the boiling
liquid mixture at the indicated temperature. Pure liquid A boils at a lower temperature than
pure liquid B; hence, A is the more volatile liquid in this illustration. Suppose we begin with
an ideal equimolar mixture (XA ϭ XB ϭ 0.5) of liquids A and B. The point P represents the
temperature at which this solution boils, T1. The vapor that is present at this equilibrium is
indicated by point Q (XA Ϸ 0.8). Condensation of that vapor at temperature T2 gives a liquid
of the same composition (point R). At this point we have described one step of simple
distillation. The boiling liquid at point R is in equilibrium with the vapor of composition
indicated by point S (XA Ͼ 0.95), and so on.
XA
XB
XB = 0.5
XA = 0.5
0
01
1
Temperature
bp pure B
bp pure A
T1
T2
Vapor
Liquid
Q
S
P
R
T
The applied pressure is usually
atmospheric pressure.
with the boiling liquid mixture at the indicated temperature. The intercepts at the two
vertical axes show the boiling points of the two pure liquids. The distillation of the two
liquids is described in the legend to Figure 14-12.
From the boiling point diagram in Figure 14-12, we see that two or more volatile
liquids cannot be completely separated from each other by a single distillation step. The
vapor collected at any boiling temperature is always enriched in the more volatile component
(A); however, at any temperature the vapor still contains both components. A series
of simple distillations would provide distillates increasingly richer in the more volatile
component, but the repeated distillations would be very tedious.
Repeated distillations may be avoided by using fractional distillation. A fractionating
column is inserted above the solution and attached to the condenser, as shown in Figure
14-13. The column is constructed so that it has a large surface area or is packed with
many small glass beads or another material with a large surface area. These provide surfaces
on which condensation can occur. Contact between the vapor and the packing favors
condensation of the less volatile component. The column is cooler at the top than at the
bottom. By the time the vapor reaches the top of the column, practically all of the less
volatile component has condensed and fallen back down the column. The more volatile
component goes into the condenser, where it is liquefied and delivered as a highly enriched
distillate into the collection flask. The longer the column or the greater the packing, the
more efficient is the separation.
Distillation under vacuum lowers the
applied pressure. This allows boiling
at lower temperatures than under
atmospheric pressure. This technique
allows distillation of some substances
that would decompose at higher
temperatures.
562 CHAPTER 14: Solutions
Many fractions, such as gasoline,
kerosene, fuel oil, paraffin, and asphalt,
are separated from crude oil by
fractional distillation.
Figure 14-13 A fractional
distillation apparatus. The vapor
phase rising in the column is in
equilibrium with the liquid phase
that has condensed out and is
flowing slowly back down the
column.
Fractional distillation is used for
separations in many industrial
processes. In this Pennsylvania
plant, atmospheric air is liquefied
by cooling and compression and
then is separated by distillation in
towers. This plant produces more
than 1000 tons daily of gases from
air (nitrogen, oxygen, and argon).
BOILING POINT ELEVATION
Recall that the boiling point of a liquid is the temperature at which its vapor pressure
equals the applied pressure on its surface (see Section 13-8). For liquids in open containers,
this is atmospheric pressure. We have seen that the vapor pressure of a solvent at a given
temperature is lowered by the presence in it of a nonvolatile solute. Such a solution must
be heated to a higher temperature than the pure solvent to cause the vapor pressure of
the solvent to equal atmospheric pressure (Figure 14-14). In accord with Raoult’s Law,
the elevation of the boiling point of a solvent caused by the presence of a nonvolatile,
nonionized solute is proportional to the number of moles of solute dissolved in a given
mass of solvent. Mathematically, this is expressed as
⌬Tb ϭ Kbm
The term ⌬Tb represents the elevation of the boiling point of the solvent, that is, the
boiling point of the solution minus the boiling point of the pure solvent. The m is the
molality of the solute, and Kb is a proportionality constant called the molal boiling point
elevation constant. This constant is different for different solvents and does not depend
on the solute (Table 14-2).
Kb corresponds to the change in boiling point produced by a one-molal ideal solution
of a nonvolatile nonelectrolyte. The units of Kb are °C/m.
14-11
When the solute is nonvolatile, only
the solvent distills from the solution.
14-11 Boiling Point Elevation 563
⌬Tb ϭ Tb(soln) Ϫ Tb(solvent). The boiling
points of solutions that contain
nonvolatile solutes are always higher
than the boiling points of the pure
solvents. So ⌬Tb is always positive.
Figure 14-14 Because a nonvolatile solute lowers the vapor pressure of a solvent, the
boiling point of a solution is higher and the freezing point lower than the corresponding
points for the pure solvent. The magnitude of the boiling point elevation, ⌬Tb, is less than
the magnitude of the freezing point depression, ⌬Tf.
Solid
Liquid
Gas
⌬Psolvent
at bp of
pure solvent
Solutio
n
Pure solven
t
760
Vaporpressureofsolvent(torr)
Tf
solution
Tb
solution
Tf
pure
solvent
Tb
pure
solvent
⌬Tf ⌬Tb
Temperature (°C)
Elevations of boiling points and depressions of freezing points, which will be discussed
later, are usually quite small for solutions of typical concentrations. They can be measured,
however, with specially constructed differential thermometers that measure small temperature
changes accurately to the nearest 0.001°C.
EXAMPLE 14-7 Boiling Point Elevation
Predict the boiling point of the 1.25 m sucrose solution in Example 14-2.
Plan
We first find the increase in boiling point from the relationship ⌬Tb ϭ Kbm. The boiling point
is higher by this amount than the normal boiling point of pure water.
Solution
From Table 14-2, Kb for H2O ϭ 0.512°C/m, so
⌬Tb ϭ (0.512°C/m)(1.25 m) ϭ 0.640°C
The solution would boil at a temperature that is 0.640°C higher than pure water would boil.
The normal boiling point of pure water is exactly 100°C, so at 1.00 atm this solution is predicted
to boil at 100°C ϩ 0.640°C ϭ 100.640°C.
You should now work Exercise 46.
FREEZING POINT DEPRESSION
Molecules of liquids move more slowly and approach one another more closely as the
temperature is lowered. The freezing point of a liquid is the temperature at which the
forces of attraction among molecules are just great enough to overcome their kinetic energies
and thus cause a phase change from the liquid to the solid state. Strictly speaking,
the freezing (melting) point of a substance is the temperature at which the liquid and solid
phases are in equilibrium. When a dilute solution freezes, it is the solvent that begins to
solidify first, leaving the solute in a more concentrated solution. Solvent molecules in a
solution are somewhat more separated from one another (because of solute particles) than
they are in the pure solvent. Consequently, the temperature of a solution must be lowered
below the freezing point of the pure solvent to freeze it.
14-12
When a solution freezes, the solvent
solidifies as the pure substance. For
this photo a dye has been added. As
the solute freezes along the wall of
the test tube, the dye concentration
increases near the center.
564 CHAPTER 14: Solutions
TABLE 14-2 Some Properties of Common Solvents
Solvent bp (pure) Kb (°C/m) fp (pure) Kf (°C/m)
water 100* 0.512 0* 1.86
benzene 80.1 2.53 5.48 5.12
acetic acid 118.1 3.07 16.6 3.90
nitrobenzene 210.88 5.24 5.7 7.00
phenol 182 3.56 43 7.40
camphor 207.42 5.61 178.40 40.0
*Exact values.
The freezing point depressions of solutions of nonelectrolytes have been found to be
equal to the molality of the solute times a proportionality constant called the molal
freezing point depression constant, Kf.
⌬Tf ϭ Kf m
The values of Kf for a few solvents are given in Table 14-2. Each is numerically equal
to the freezing point depression of a one-molal ideal solution of a nonelectrolyte in that
solvent.
EXAMPLE 14-8 Freezing Point Depression
When 15.0 grams of ethyl alcohol, C2H5OH, is dissolved in 750. grams of formic acid, the
freezing point of the solution is 7.20°C. The freezing point of pure formic acid is 8.40°C. Evaluate
Kf for formic acid.
Plan
The molality and the depression of the freezing point are calculated first. Then we solve the
equation ⌬Tf ϭ Kf m for Kf and substitute values for m and ⌬Tf.
Kf ϭ
Solution
_?_ ϭ ϫ ϭ 0.435 m
⌬Tf ϭ (Tf[formic acid]) Ϫ (Tf[solution]) ϭ 8.40°C Ϫ 7.20°C ϭ 1.20°C (depression)
Then Kf ϭ ϭ ϭ 2.76°C/m for formic acid.
You should now work Exercise 48.
EXAMPLE 14-9 Freezing Point Depression
Calculate the freezing point of the 1.25 m sucrose solution in Example 14-2.
Plan
We first find the decrease in freezing point from the relationship ⌬Tf ϭ Kf m. The temperature
at which the solution freezes is lower than the freezing point of pure water by this amount.
Solution
From Table 14-2, Kf for H2O ϭ 1.86°C/m, so
⌬Tf ϭ (1.86°C/m)(1.25 m) ϭ 2.32°C
The solution freezes at a temperature 2.32°C below the freezing point of pure water, or
Tf(solution) ϭ 0.00°C Ϫ 2.32°C ϭ Ϫ2.32°C
You should now work Exercise 56.
1.20°C
ᎏ
0.435 m
⌬Tf
ᎏ
m
1 mol C2H5OH
ᎏᎏ
46.0 g C2H5OH
15.0 g C2H5OH
ᎏᎏᎏ
0.750 kg formic acid
mol C2H5OH
ᎏᎏ
kg formic acid
⌬Tf
ᎏ
m
⌬Tf is the depression of freezing point.
It is defined as
⌬Tf ϭ Tf(solvent) Ϫ Tf(soln)
so it is always positive.
14-12 Freezing Point Depression 565
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 14.8, Colligative Properties (2):
Boiling Point and Freezing Point.
Lime, CaO, is added to molten iron
ore during the manufacture of pig
iron. It lowers the melting point of
the mixture. The metallurgy of iron
is discussed in more detail in
Chapter 22.
You may be familiar with several examples of the effects we have studied. Sea water
does not freeze on some days when fresh water does, because sea water contains higher
concentrations of solutes, mostly ionic solutes. Spreading soluble salts such as sodium
chloride, NaCl, or calcium chloride, CaCl2, on an icy road lowers the freezing point of
the ice, causing the ice to melt.
A familiar application is the addition of “permanent” antifreeze, mostly ethylene glycol,
HOCH2CH2OH, to the water in an automobile radiator. Because the boiling point of the
solution is elevated, addition of a solute as a winter antifreeze also helps protect against
loss of the coolant by summer “boil-over.” The amounts by which the freezing and boiling
points change depend on the concentration of the ethylene glycol solution. The addition
of too much ethylene glycol is counterproductive, however. The freezing point of pure
ethylene glycol is about Ϫ12°C. A solution that is mostly ethylene glycol would have a
somewhat lower freezing point due to the presence of water as a solute. Suppose you graph
the freezing point depression of water below 0°C as ethylene glycol is added, and also
graph the freezing point depression of ethylene glycol below Ϫ12°C as water is added.
These two curves would intersect at some temperature, indicating the limit of lowering
that can occur. (At these high concentrations, the solutions do not behave ideally, so the
temperatures could not be accurately predicted by the equations we have introduced in
this chapter, but the main ideas still apply.) Most antifreeze labels recommend a 50Ϻ50
mixture by volume (fp Ϫ34°F, bp 265°F with a 15-psi pressure cap on the radiator), and
cite the limit of possible protection with a 70Ϻ30 mixture by volume of antifreezeϺwater
(fp Ϫ84°F, bp 276°F with a 15-psi pressure cap).
DETERMINATION OF MOLECULAR WEIGHT BY
FREEZING POINT DEPRESSION OR
BOILING POINT ELEVATION
The colligative properties of freezing point depression and, to a lesser extent, boiling point
elevation are useful in the determination of molecular weights of solutes. The solutes must
be nonvolatile in the temperature range of the investigation if boiling point elevations are
to be determined. We will restrict our discussion of determination of molecular weight
to nonelectrolytes.
EXAMPLE 14-10 Molecular Weight from a Colligative Property
A 1.20-gram sample of an unknown covalent compound is dissolved in 50.0 grams of benzene.
The solution freezes at 4.92°C. Calculate the molecular weight of the compound.
Plan
To calculate the molecular weight of the unknown compound, we find the number of moles
that is represented by the 1.20 grams of unknown compound. We first use the freezing point
data to find the molality of the solution. The molality relates the number of moles of solute
and the mass of solvent (known), so this allows us to calculate the number of moles of unknown.
14-13
The total concentration of all dissolved
solute species determines the
colligative properties. As we will
emphasize in Section 14-14, we must
take into account the extent of ion
formation in solutions of ionic solutes.
566 CHAPTER 14: Solutions
Ethylene glycol, HOCH2CH2OH, is the major component of “permanent” antifreeze. It
depresses the freezing point of water in an automobile radiator and also raises its boiling
point. The solution remains in the liquid phase over a wider temperature range than does
pure water. This protects against both freezing and boil-over.
Solution
From Table 14-2, the freezing point of pure benzene is 5.48°C and Kf is 5.12°C/m.
⌬Tf ϭ 5.48°C Ϫ 4.92°C ϭ 0.56°C
m ϭ ϭ ϭ 0.11 m
The molality is the number of moles of solute per kilogram of benzene, so the number of moles
of solute in 50.0 g (0.0500 kg) of benzene can be calculated.
0.11 m ϭ
_?_ mol solute ϭ (0.11 m)(0.0500 kg) ϭ 0.0055 mol solute
mass of 1.0 mol ϭ ϭ ϭ 2.2 ϫ 102 g/mol
molecular weight ϭ 2.2 ϫ 102 amu
You should now work Exercise 61.
EXAMPLE 14-11 Molecular Weight from a Colligative Property
Either camphor (C10H16O, molecular weight ϭ 152 g/mol) or naphthalene (C10H8, molecular
weight ϭ 128 g/mol) can be used to make mothballs. A 5.2-gram sample of mothballs was
dissolved in 100. grams of ethyl alcohol, and the resulting solution had a boiling point of
78.90°C. Were the mothballs made of camphor or naphthalene? Pure ethyl alcohol has a boiling
point of 78.41°C; its Kb ϭ 1.22°C/m.
Plan
We can distinguish between the two possibilities by determining the molecular weight of the
unknown solute. We do this by the method shown in Example 14-10, except that now we use
the observed boiling point data.
Solution
The observed boiling point elevation is
⌬Tb ϭ Tb(solution) Ϫ Tb(solvent) ϭ (78.90 Ϫ 78.41)°C ϭ 0.49°C
Using ⌬Tb ϭ 0.49°C and Kb ϭ 1.22°C/m, we can find the molality of the solution.
molality ϭ ϭ ϭ 0.40 m
The number of moles of solute in the 100. g (0.1000 kg) of solvent used is
΂0.40 ΃(0.100 kg solvent) ϭ 0.040 mol solute
The molecular weight of the solute is its mass divided by the number of moles.
ϭ ϭ 130 g/mol
The value 130 g/mol for the molecular weight indicates that naphthalene was used to make
these mothballs.
You should now work Exercise 59.
5.2 g
ᎏᎏ
0.040 mol
_?_ g
ᎏ
mol
mol solute
ᎏᎏ
kg solvent
0.49°C
ᎏᎏ
1.22°C/m
⌬Tb
ᎏ
Kb
1.20 g solute
ᎏᎏ
0.0055 mol solute
no. of g solute
ᎏᎏ
no. of mol solute
_?_ mol solute
ᎏᎏᎏ
0.0500 kg benzene
0.56°C
ᎏᎏ
5.12°C/m
⌬Tf
ᎏ
Kf
14-13 Determination of Molecular Weight by Freezing Point Depression or Boiling Point Elevation 567
COLLIGATIVE PROPERTIES AND DISSOCIATION
OF ELECTROLYTES
As we have emphasized, colligative properties depend on the number of solute particles in
a given mass of solvent. A 0.100 molal aqueous solution of a covalent compound that does
not ionize gives a freezing point depression of 0.186°C. If dissociation were complete,
0.100 m KBr would have an effective molality of 0.200 m (i.e., 0.100 m Kϩ ϩ 0.100 m BrϪ).
So we might predict that a 0.100 molal solution of this 1Ϻ1 strong electrolyte would have
a freezing point depression of 2 ϫ 0.186°C, or 0.372°C. In fact, the observed depression is
only 0.349°C. This value for ⌬Tf is about 6% less than we would expect for an effective
molarity of 0.200 m.
In an ionic solution the solute particles are not randomly distributed. Rather, each positive
ion has more negative than positive ions near it. The resulting electrical interactions
cause the solution to behave nonideally. Some of the ions undergo association in solution
(Figure 14-15). At any given instant, some Kϩ and BrϪ ions collide and “stick
together.” During the brief time that they are in contact, they behave as a single particle.
This tends to reduce the effective molality. The freezing point depression (⌬Tf) is therefore
reduced (as well as the boiling point elevation (⌬Tb) and the lowering of vapor
pressure).
A (more concentrated) 1.00 m solution of KBr might be expected to have a freezing
point depression of 2 ϫ 1.86°C ϭ 3.72°C, but the observed depression is only 3.29°C.
This value for ⌬Tf is about 11% less than we would expect. We see a greater deviation
from the depression predicted (ignoring ionic association) in the more concentrated solution.
This is because the ions are closer together and collide more often in the more
concentrated solution. Consequently, the ionic association is greater.
One measure of the extent of dissociation (or ionization) of an electrolyte in water is
the van’t Hoff factor, i, for the solution. This is the ratio of the actual colligative property
to the value that would be observed if no dissociation occurred.
i ϭ ϭ ϭ
The ideal, or limiting, value of i for a solution of KBr would be 2, and the value for a 2Ϻ1
electrolyte such as Na2SO4 would be 3; these values would apply to infinitely dilute solutions
in which no appreciable ion association occurs. For 0.10 m and 1.0 m solutions of
KBr, i is less than 2.
For 0.10 m: i ϭ ᎏ
0
0
.
.
3
1
4
8
9
6
°
°
C
C
ᎏ ϭ 1.88 For 1.0 m: i ϭ ᎏ
3
1
.
.
2
8
9
6
°
°
C
C
ᎏ ϭ 1.77
Table 14-3 lists actual and ideal values of i for solutions of some strong electrolytes, based
on measurements of freezing point depressions.
Many weak electrolytes are quite soluble in water, but they ionize only slightly. The
percent ionization and i value for a weak electrolyte in solution can also be determined
from freezing point depression data (Example 14-12).
meffective
ᎏ
mstated
Kf meffective
ᎏᎏ
Kf mstated
⌬Tf(actual)
ᎏᎏ
⌬Tf(if nonelectrolyte)
14-14
⌬Tf ϭ Kf m ϭ (1.86°C/m)(0.100 m)
ϭ 0.186°C
568 CHAPTER 14: Solutions
Ionic solutions are elegantly described
by the Debye–Hückel theory, which is
beyond the scope of this text.
Weak acids and weak bases (Section
4-2) are weak electrolytes.
Figure 14-15 Diagrammatic
representation of the various species
thought to be present in a solution
of KBr in water. This would explain
unexpected values for its colligative
properties, such as freezing point
depression.
BrϪ
BrϪ
BrϪ
Kϩ
Associated ions
Associated ions
Kϩ
BrϪ
BrϪ
Kϩ
ϩ
ϩ
ϩ ϩ
ϩ
ϩ
ϩ
ϩϩ
ϩ
ϩ ϩ
ϩ
ϩ
ϩ
ϩ
ϩ
ϩ ϩ
ϩ ϩ ϩ
ϩϩ
ϩ
ϩ ϩ
ϩ
ϩϩϩϩ
ϩ
ϩ
ϩ
ϩ
ϩ
ϩ
ϩ
ϩ
ϩ
ϩ
ϩ
ϩ
ϩϩ
ϩ
ϩ
ϩ
ϩ
ϩ
ϩ ϩ
ϩ
ϩ
ϩ
ϩϩ
ϩϩ
Ϫ
Ϫ
Ϫ
Ϫ
Ϫ
Ϫ
ϪϪ
Ϫ
Ϫ
ϪϪ
Ϫ Ϫ
ϪϪ
Ϫ
Ϫ
Ϫ
ϪϪ
Ϫ
Ϫ
Ϫ
Ϫ
Ϫ
Ϫ
Ϫ
Ϫ
Ϫ
Problem-Solving Tip: Selection of a van’t Hoff Factor
Use an ideal value for the van’t Hoff factor unless the question clearly indicates to do
otherwise, as in the following example and in some of the end-of-chapter exercises. For
a strong electrolyte dissolved in water, the ideal value for its van’t Hoff factor is listed
in Table 14-3. For nonelectrolytes dissolved in water or any solute dissolved in common
nonaqueous solvents, the van’t Hoff factor is considered to be 1. For weak electrolytes
dissolved in water, the van’t Hoff factor is a little greater than 1.
TABLE 14-3 Actual and Ideal van’t Hoff Factors, i, for Aqueous Solutions of
Nonelectrolytes and Strong Electrolytes
i for i for
Compound 1.00 m Solution 0.100 m Solution
nonelectrolytes 1.00 (ideal) 1.00 (ideal)
sucrose, C12H22O11 1.00 1.00
If 2 ions in solution/formula unit 2.00 (ideal) 2.00 (ideal)
KBr 1.77 1.88
NaCl 1.83 1.87
If 3 ions in solution/formula unit 3.00 (ideal) 3.00 (ideal)
K2CO3 2.39 2.45
K2CrO4 1.95 2.39
If 4 ions in solution/formula unit 4.00 (ideal) 4.00 (ideal)
K3[Fe(CN)6] — 2.85
Percent Ionization and i Value from Freezing Point Depression Data
EXAMPLE 14-12 Colligative Property and Weak Electrolytes
Lactic acid, C2H4(OH)COOH, is found in sour milk. It is also formed in muscles during
intense physical activity and is responsible for the pain felt during strenuous exercise. It is
a weak monoprotic acid and therefore a weak electrolyte. The freezing point of a 0.0100 m
aqueous solution of lactic acid is Ϫ0.0206°C. Calculate (a) the i value and (b) the percent
ionization in the solution.
Plan for (a)
To evaluate the van’t Hoff factor, i, we first calculate meffective from the observed freezing
point depression and Kf for water; we then compare meffective and mstated to find i.
Solution for (a)
meffective ϭ ϭ ϭ 0.0111 m
i ϭ ϭ ϭ 1.11
0.0111 m
ᎏᎏ
0.0100 m
meffective
ᎏ
mstated
0.0206°C
ᎏᎏ
1.86°C/m
⌬Tf
ᎏ
Kf
E n r i c h m e n t
570 CHAPTER 14: Solutions
Plan for (b)
The percent ionization is given by
% ionization ϭ ϫ 100% (where moriginal ϭ mstated ϭ 0.0100 m)
The freezing point depression is caused by the meffective, the total concentration of all dissolved
species—in this case, the sum of the concentrations of HA, Hϩ, and AϪ. We know the value
of meffective from part (a). Thus, we need to construct an expression for the effective molality
in terms of the amount of lactic acid that ionizes. We represent the molality of lactic acid
that ionizes as an unknown, x, and write the concentrations of all species in terms of this
unknown.
Solution for (b)
In many calculations, it is helpful to write down (1) the values, or symbols for the values,
of initial concentrations; (2) changes in concentrations due to reaction; and (3) final concentrations,
as shown here. The coefficients of the equation are all ones, so the reaction ratio
must be 1Ϻ1Ϻ1.
Let x ϭ molality of lactic acid that ionizes; then
x ϭ molality of Hϩ and lactate ions that have been formed
HA 88n Hϩ ϩ AϪ
Start 0.0100 m 0 0
Change Ϫx m ϩx m ϩx m
Final (0.0100 Ϫ x) m x m x m
The meffective is equal to the sum of the molalities of all the solute particles.
meffective ϭ mHA ϩ mHϩ ϩ mAϪ
ϭ (0.0100 Ϫ x) m ϩ x m ϩ x m ϭ (0.0100 ϩ x) m
This must equal the value for meffective calculated earlier, 0.0111 m.
0.0111 m ϭ (0.0100 ϩ x) m
x ϭ 0.0011 m ϭ molality of the acid that ionizes
We can now calculate the percent ionization.
% ionization ϭ ϫ 100% ϭ ϫ 100% ϭ 11%
This experiment shows that in 0.0100 m solutions, only 11% of the lactic acid has been
converted into Hϩ and C2H4(OH)COOϪ ions. The remainder, 89%, exists as nonionized
molecules.
You should now work Exercises 78 and 80.
0.0011 m
ᎏᎏ
0.0100 m
mionized
ᎏ
moriginal
mionized
ᎏ
moriginal
OSMOTIC PRESSURE
Osmosis is the spontaneous process by which the solvent molecules pass through a semipermeable
membrane from a solution of lower concentration of solute into a solution of
higher concentration of solute. A semipermeable membrane (e.g., cellophane) separates
14-15
(Enrichment, continued)
To simplify the notation, we denote
the weak acid as HA and its anion as
AϪ. The reaction summary used here
to analyze the extent of reaction was
introduced in Chapter 11.
Osmosis is one of the main ways in
which water molecules move into and
out of living cells. The membranes and
cell walls in living organisms allow
solvent to pass through. Some of these
also selectively permit passage of ions
and other small solute particles.
two solutions. Solvent molecules may pass through the membrane in either direction, but
the rate at which they pass into the more concentrated solution is found to be greater
than the rate in the opposite direction. The initial difference between the two rates is
directly proportional to the difference in concentration between the two solutions. Solvent
particles continue to pass through the membrane (Figure 14-16a). The column of liquid
continues to rise until the hydrostatic pressure due to the weight of the solution in the
column is sufficient to force solvent molecules back through the membrane at the same
rate at which they enter from the dilute side. The pressure exerted under this condition
is called the osmotic pressure of the solution.
Figure 14-16 (a) Laboratory
apparatus for demonstrating osmosis.
The picture at the right gives some
details of the process, which is
analogous to the transfer of solvent
into a solution through the space
above them (b). In (a) the solute
particles cannot pass through the
semipermeable membrane. In (b) the
solute particles cannot pass through
the vapor phase because they are
nonvolatile.
14-15 Osmotic Pressure 571
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 14.9, Colligative Properties (3):
Osmosis.
Thistle tube
Level at start
Sugar solution
Rubber band
WaterMembrane
Solution level
after osmosis
has progressed
for a time
Sugar
molecules
cannot pass
through
membrane
Some H2O
molecules
pass out of
the solution
Water molecules pass
in through membrane
(a)
(b)
Start Later
Start Later
Solution Solvent
Solution
Osmotic pressure depends on the number, and not the kind, of solute particles in
solution; it is therefore a colligative property.
The osmotic pressure of a given aqueous solution can be measured with an apparatus
such as that depicted in Figure 14-16a. The solution of interest is placed inside an inverted
glass (thistle) tube that has a membrane firmly fastened across the bottom. This part of
the thistle tube and its membrane are then immersed in a container of pure water. As time
passes, the height of the solution in the neck rises until the pressure it exerts just counterbalances
the osmotic pressure.
Alternatively, we can view osmotic pressure as the external pressure exactly sufficient
to prevent osmosis. The pressure required (Figure 14-17) is equal to the osmotic pressure
of the solution.
Like molecules of an ideal gas, solute particles are widely separated in very dilute solutions
and do not interact significantly with one another. For very dilute solutions, osmotic
pressure, ␲, is found to follow the equation
␲ ϭ ᎏ
nR
V
T
ᎏ
In this equation n is the number of moles of solute in volume, V, (in liters) of the solution.
The other quantities have the same meaning as in the ideal gas law. The term n/V
is a concentration term. In terms of molarity, M,
␲ ϭ MRT
Osmotic pressure increases with increasing temperature because T affects the number of
solvent–membrane collisions per unit time. It also increases with increasing molarity
because M affects the difference in the numbers of solvent molecules hitting the membrane
from the two sides, and because a higher M leads to a stronger drive to equalize the
concentration difference by dilution and to increase disorder in the solution. For dilute
aqueous solutions, the molarity is approximately equal to the molality (because the density
of the solution is nearly 1 kg/L), so
␲ ϭ mRT (dilute aqueous solutions)
Osmotic pressures represent very significant forces. For example, a 1.00 molal solution
of a nonelectrolyte in water at 0°C produces an equilibrium osmotic pressure of approximately
22.4 atmospheres (Ϸ330 psi).
EXAMPLE 14-13 Osmotic Pressure Calculation
What osmotic pressure would the 1.25 m sucrose solution in Example 14-2 exhibit at 25°C?
The density of this solution is 1.34 g/mL.
Plan
We note that the approximation M Ϸ m is not very good for this solution, because the density
of this solution is quite different from 1 g/mL or kg/L. Thus, we must first find the molarity
of sucrose, and then use the relationship ␲ ϭ MRT.
Solution
Recall from Example 14-2 that there is 50.0 g of sucrose (0.146 mol) in 117 g of H2O which
gives 167 g of solution. The volume of this solution is
vol solution ϭ 167 g ϫ ᎏ
1
1
.3
m
4
L
g
ᎏ ϭ 125 mL, or 0.125 L
The greater the number of solute
particles, the greater the height to
which the column rises, and the
greater the osmotic pressure.
572 CHAPTER 14: Solutions
For a solution of an electrolyte,
␲ ϭ meffectiveRT.
Figure 14-17 The pressure that is
just sufficient to prevent solvent flow
from the pure solvent side through
the semipermeable membrane to the
solution side is a measure of the
osmotic pressure of the solution.
SolventSolution
Equal
levels
Pressure =
Membrane
␲
Thus, the molarity of sucrose in the solution is
Msucrose ϭ ᎏ
0
0
.1
.1
4
2
6
5
m
L
ol
ᎏ ϭ 1.17 mol/L
Now we can calculate the osmotic pressure.
␲ ϭ MRT ϭ (1.17 mol/L)΂0.0821 ᎏ
m
Lи
o
a
l
t
и
m
K
ᎏ΃(298 K) ϭ 28.6 atm
You should now work Exercise 84.
Let’s compare the calculated values of the four colligative properties for the 1.25 m
sucrose solution.
vapor pressure lowering ϭ 0.524 torr (Example 14-4)
boiling point elevation ϭ 0.640°C (Example 14-7)
freezing point depression ϭ 2.32°C (Example 14-9)
osmotic pressure ϭ 28.6 atm (Example 14-13)
The first of these is so small that it would be hard to measure precisely. Even this small
lowering of the vapor pressure is sufficient to raise the boiling point by an amount that
could be measured, although with difficulty. The freezing point depression is greater, but
still could not be measured very precisely without a special apparatus. The osmotic pressure,
on the other hand, is so large that it could be measured much more precisely. Thus,
osmotic pressure is often the most easily measured of the four colligative properties, especially
when very dilute solutions are used.
The use of measurements of osmotic pressure for the determination of molecular
weights has several advantages. Even very dilute solutions give easily measurable osmotic
pressures. This method therefore is useful in determination of the molecular weights of
(1) very expensive substances, (2) substances that can be prepared only in very small
amounts, and (3) substances of very high molecular weight that are not very soluble.
Because high-molecular-weight materials are often difficult, and in some cases impossible,
to obtain in a high state of purity, determinations of their molecular weights are not as
accurate as we might like. Nonetheless, osmotic pressures provide a very useful method
of estimating molecular weights.
EXAMPLE 14-14 Molecular Weight from Osmotic Pressure
Pepsin is an enzyme present in the human digestive tract. A solution of a 0.500-gram sample
of purified pepsin in 30.0 mL of aqueous solution exhibits an osmotic pressure of 8.92 torr at
27.0°C. Estimate the molecular weight of pepsin.
Plan
As we did in earlier molecular weight determinations (Section 14-13), we must first find n, the
number of moles that 0.500 grams of pepsin represents. We start with the equation ␲ ϭ MRT.
The molarity of pepsin is equal to the number of moles of pepsin per liter of solution, n/V. We
substitute this for M and solve for n.
Solution
␲ ϭ MRT ϭ ΂ᎏ
V
n
ᎏ΃RT or n ϭ ᎏ
␲
RT
V
ᎏ
An illustration of osmosis. When a
carrot is soaked in a concentrated
salt solution, water flows out of the
plant cells by osmosis. A carrot
soaked overnight in salt solution
(left) has lost much water and
become limp. A carrot soaked
overnight in pure water (right)
is little affected.
14-15 Osmotic Pressure 573
An enzyme is a protein that acts as a
biological catalyst. Pepsin catalyzes the
metabolic cleavage of amino acid
chains (called peptide chains) in other
proteins.
We convert 8.92 torr to atmospheres to be consistent with the units of R.
n ϭ ᎏ
␲
RT
V
ᎏ ϭ ϭ 1.43 ϫ 10Ϫ5 mol pepsin
Thus, 0.500 g of pepsin is 1.43 ϫ 10Ϫ5 mol. We now estimate its molecular weight.
_?_ g/mol ϭ ᎏ
1.43
0
ϫ
.5
1
0
0
0
Ϫ
g
5 mol
ᎏ ϭ 3.50 ϫ 104 g/mol
The molecular weight of pepsin is approximately 35,000 amu. This is typical for medium-sized
proteins.
You should now work Exercise 90.
΂8.92 torr ϫ ᎏ
76
1
0
at
t
m
orr
ᎏ΃(0.0300 L)
ᎏᎏᎏᎏ
΂0.0821 ᎏ
m
Lи
o
a
l
t
и
m
K
ᎏ΃(300 K)
Living cells contain solutions. When
living cells are put in contact with
solutions having different total solute
concentrations, the resulting osmotic
pressure can cause solvent to flow
into (left) or out of (right) the cells.
The freezing point depression of this
very dilute solution would be only
about 0.0009°C, which would be
difficult to measure accurately. The
osmotic pressure of 8.92 torr, on the
other hand, is easily measured.
Problem-Solving Tip: Units in Osmotic Pressure Calculations
Strictly speaking, the equation for osmotic pressure is presented in terms of molarity,
␲ ϭ MRT. Osmotic pressure, ␲, has the units of pressure (atmospheres); M (mol/L); and
T (kelvins). Therefore, the appropriate value of R is 0.0821 ᎏ
m
Lи
o
a
l
t
и
m
K
ᎏ. We can balance the
units in this equation as follows.
atm ϭ ΂ᎏ
m
L
ol
ᎏ΃΂ᎏ
m
Lи
o
a
l
t
и
m
K
ᎏ΃(K)
When we use the approximation that M Ϸ m, we might think that the units do not
balance. For very dilute aqueous solutions, we can think of M as being numerically equal
to m, but having the units mol/L, as shown above.
Cells expand in solution of
lower solute concentration
(a hypotonic solution)
Normal cells in isotonic solution Cells shrink in solution of
greater solute concentration
(a hypertonic solution)
COLLOIDS
A solution is a homogeneous mixture in which no settling occurs and in which solute
particles are at the molecular or ionic state of subdivision. This represents one extreme
of mixtures. The other extreme is a suspension, a clearly heterogeneous mixture in which
solute-like particles settle out after mixing with a solvent-like phase. Such a situation results
when a handful of sand is stirred into water. Colloids (colloidal dispersions) represent
an intermediate kind of mixture in which the solute-like particles, or dispersed phase,
are suspended in the solvent-like phase, or dispersing medium. The particles of the
dispersed phase are so small that settling is negligible. They are large enough, however,
to make the mixture appear cloudy or even opaque, because light is scattered as it passes
through the colloid.
Table 14-4 indicates that all combinations of solids, liquids, and gases can form colloids
except mixtures of nonreacting gases (all of which are homogeneous and, therefore, true
solutions). Whether a given mixture forms a solution, a colloidal dispersion, or a suspension
depends on the size of the solute-like particles (Table 14-5), as well as solubility and
miscibility.
Freshly made wines are often cloudy
because of colloidal particles.
Removing these colloidal particles
clarifies the wine.
14-15 Osmotic Pressure 575
TABLE 14-5 Approximate Sizes of Dispersed Particles
Mixture Example Approximate Particle Size
suspension sand in water larger than 10,000 Å
colloidal dispersion starch in water 10–10,000 Å
solution sugar in water 1–10 Å
TABLE 14-4 Types of Colloids
Dispersed Dispersing
(solute-like) (solvent-like) Common
Phase Medium Name Examples
solid in solid solid sol Many alloys (e.g., steel and
duralumin), some colored
gems, reinforced rubber,
porcelain, pigmented plastics
liquid in solid solid emulsion Cheese, butter, jellies
gas in solid solid foam Sponge, rubber, pumice,
Styrofoam
solid in liquid sols and gels Milk of magnesia, paints,
mud, puddings
liquid in liquid emulsion Milk, face cream, salad
dressings, mayonnaise
gas in liquid foam Shaving cream, whipped
cream, foam on beer
solid in gas solid aerosol Smoke, airborne viruses and
particulate matter, auto
exhaust
liquid in gas liquid aerosol Fog, mist, aerosol spray,
clouds
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 14.10, Colloids.
576 CHAPTER 14: Solutions
Our Daily LivesCCHEMISTRY IN USE
Water Purification and Hemodialysis
Semipermeable membranes play important roles in the normal
functioning of many living systems. In addition, they are
used in a wide variety of industrial and medical applications.
Membranes with different permeability characteristics have
been developed for many different purposes. One of these is
the purification of water by reverse osmosis.
Suppose we place a semipermeable membrane between a
saline (salt) solution and pure water. If the saline solution is
pressurized under a greater pressure than its osmotic pressure,
the direction of flow can be reversed. That is, the net
flow of water molecules will be from the saline solution
through the membrane into the pure water. This process is
called reverse osmosis. The membrane usually consists of
cellulose acetate or hollow fibers of a material structurally
similar to nylon. This method has been used for the purification
of brackish (mildly saline) water. It has the economic
advantages of low cost, ease of apparatus construction, and
simplicity of operation. Because this method of water purification
requires no heat, it has a great advantage over
distillation.
The city of Sarasota, Florida, has built a large reverse
osmosis plant to purify drinking water. It processes more than
4 million gallons of water per day from local wells. Total dissolved
solids are reduced in concentration from 1744 parts
per million (ppm) (0.1744% by mass) to 90 ppm. This water
is mixed with additional well water purified by an ion
exchange system. The final product is more than 10 million
gallons of water per day containing less than 500 ppm of total
dissolved solids, the standard for drinking water set by the
World Health Organization. The Kuwaiti and Saudi water
purification plants that were of strategic concern in the Persian
Gulf war use reverse osmosis in one of their primary
stages.
Human kidneys carry out many important functions. One
of the most crucial is the removal of metabolic waste prodThe
reverse osmosis method
of water purification.
A reverse osmosis unit used to provide all the fresh water
(82,500 gallons per day) for the steamship Norway.
Product-water
Membrane
High
pressure
pump
Impure
water
Membrane
support
Solution more
concentrated
in impurities
14-15 Osmotic Pressure 577
ucts (e.g., creatinine, urea, and uric acid) from the blood without
removal of substances needed by the body (e.g., glucose,
electrolytes, and amino acids). The process by which this is
accomplished in the kidney involves dialysis, a phenomenon
in which the membrane allows transfer of both solvent molecules
and certain solute molecules and ions, usually small
ones. Many patients whose kidneys have failed can have this
dialysis performed by an artificial kidney machine. In this
mechanical procedure, called hemodialysis, the blood is withdrawn
from the body and passed in contact with a
semipermeable membrane.
The membrane separates the blood from a dialyzing solution,
or dialysate, that is similar to blood plasma in its
concentration of needed substances (e.g., electrolytes and
amino acids) but contains none of the waste products. Because
the concentrations of undesirable substances are thus higher
in the blood than in the dialysate, they flow preferentially out
of the blood and are washed away. The concentrations of
needed substances are the same on both sides of the membrane,
so these substances are maintained at the proper
concentrations in the blood. The small pore size of the membrane
prevents passage of blood cells. However, Naϩ and ClϪ
ions and some small molecules do pass through the membrane.
A patient with total kidney failure may require up to
four hemodialysis sessions per week, at 3 to 4 hours per session.
To help hold down the cost of such treatment, the
dialysate solution is later purified by a combination of filtration,
distillation, and reverse osmosis and is then reused.
Blood in
Hollow
membrane
tubes
Jacket
Blood out
Dialysate
out
Dialysate in
A schematic diagram of the hollow fiber (or capillary)
dialyzer, the most commonly used artificial kidney. The
blood flows through many small tubes constructed of
semipermeable membrane; these tubes are bathed in the
dialyzing solution.
A portable dialysis unit.
THE TYNDALL EFFECT
The scattering of light by colloidal particles is called the Tyndall effect (Figure 14-18).
Particles cannot scatter light if they are too small. Solute particles in solutions are below
this limit. The maximum dimension of colloidal particles is about 10,000 Å.
The scattering of light from automobile headlights by fogs and mists is an example of
the Tyndall effect, as is the scattering of a light beam in a laser show by dust particles in
the air in a darkened room.
THE ADSORPTION PHENOMENON
Much of the chemistry of everyday life is the chemistry of colloids, as one can tell from
the examples in Table 14-4. Because colloidal particles are so finely divided, they have
tremendously high total surface area in relation to their volume. It is not surprising, therefore,
that an understanding of colloidal behavior requires an understanding of surface
phenomena.
Atoms on the surface of a colloidal particle are bonded only to other atoms of the
particle on and below the surface. These atoms interact with whatever comes in contact
with the surface. Colloidal particles often adsorb ions or other charged particles, as well
as gases and liquids. The process of adsorption involves adhesion of any such species onto
the surfaces of particles. For example, a bright red sol (solid dispersed in liquid) is formed
by mixing hot water with a concentrated aqueous solution of iron(III) chloride (Figure
14-19).
2x[Fe3ϩ(aq) ϩ 3ClϪ(aq)] ϩ x(3 ϩ y)H2O 88n [Fe2O3иyH2O]x(s) ϩ 6x[Hϩ ϩ ClϪ]
yellow solution bright red sol
Each colloidal particle of this sol is a cluster of many formula units of hydrated Fe2O3.
Each attracts positively charged Fe3ϩ ions to its surface. Because each particle is then
surrounded by a shell of positively charged ions, the particles repel one another and cannot
combine to the extent necessary to cause precipitation.
14-17
14-16
Figure 14-18 The dispersion of a beam of light by colloidal particles is called the Tyndall
effect. The presence of colloidal particles is easily detected with the aid of a light beam.
578 CHAPTER 14: Solutions
Light source
Beam of light not visible
Colloidal gold
A gold sol
Light beam
visible
NaCl solution
Light beam
not visible
Tobacco smoke
An aerosol
Light beam
visible
HYDROPHILIC AND HYDROPHOBIC COLLOIDS
Colloids are classified as hydrophilic (“water loving”) or hydrophobic (“water hating”)
based on the surface characteristics of the dispersed particles.
Hydrophilic Colloids
Proteins such as the oxygen-carrier hemoglobin form hydrophilic sols when they are
suspended in saline aqueous body fluids such as blood plasma. Such proteins are macromolecules
(giant molecules) that fold and twist in an aqueous environment so that polar
groups are exposed to the fluid, whereas nonpolar groups are encased (Figure 14-20).
Protoplasm and human cells are examples of gels, which are special types of sols in which
the solid particles (in this case mainly proteins and carbohydrates) join together in a semirigid
network structure that encloses the dispersing medium. Other examples of gels are
gelatin, jellies, and gelatinous precipitates such as Al(OH)3 and Fe(OH)3.
Hydrophobic Colloids
Hydrophobic colloids cannot exist in polar solvents without the presence of emulsifying
agents, or emulsifiers. These agents coat the particles of the dispersed phase to prevent
their coagulation into a separate phase. Milk and mayonnaise are examples of hydrophobic
14-18
Fe(OH)3 is a gelatinous precipitate
(a gel).
14-18 Hydrophilic and Hydrophobic Colloids 579
Some edible colloids.
Figure 14-20 Examples of
hydrophilic groups at the surface of
a giant molecule (macromolecule)
that help keep the macromolecule
suspended in water.
Figure 14-19 Stabilization of a colloid (Fe2O3 sol) by electrostatic forces. Each colloidal
particle of this red sol is a cluster of many formula units of hydrated Fe2O3. Each attracts
positively charged Fe3ϩ ions to its surface. (Fe3ϩ ions fit readily into the crystal structure, so
they are preferentially adsorbed rather than the ClϪ ions.) Each particle is then surrounded
by a shell of positively charged ions, so the particles repel one another and cannot combine
to the extent necessary to cause actual precipitation. The suspended particles scatter light,
making the path of the light beam through the suspension visible.
[Fe2O3иyH2O]x
[Fe2O3иyH2O]x
Fe3ϩ
Fe3ϩ
Fe3ϩ
Fe3ϩ
Fe3ϩ
Fe3ϩ
Fe3ϩ
Fe3ϩ
Fe3ϩ
Fe3ϩ
Hϩ
Hϩ
Hϩ
Hϩ
ClϪ
ClϪ
ClϪ ClϪ
ClϪ
ClϪ
ClϪ
ClϪ ClϪ
ClϪ
ClϪ
ClϪ
ClϪ ClϪ ClϪ
Hϩ
Hydrogen bond
N+
H
N
C
O
Protein
colloids (milk fat in milk, vegetable oil in mayonnaise) that stay suspended with the aid
of emulsifying agents (casein in milk and egg yolk in mayonnaise).
Consider the mixture resulting from vigorous shaking of salad oil (nonpolar) and vinegar
(polar). Droplets of hydrophobic oil are temporarily suspended in the water. In a short
time, however, the very polar water molecules, which attract one another strongly, squeeze
out the nonpolar oil molecules. The oil then coalesces and floats to the top. If we add an
emulsifying agent, such as egg yolk, and shake or beat the mixture, a stable emulsion
(mayonnaise) results.
Oil and grease are mostly long-chain hydrocarbons that are very nearly nonpolar. Our
most common solvent is water, a polar substance that does not dissolve nonpolar
substances. To use water to wash soiled fabrics, greasy dishes, or our bodies, we must
enable the water to suspend and remove nonpolar substances. Soaps and detergents are
emulsifying agents that accomplish this. Their function is controlled by the intermolecular
interactions that result from their structures.
Solid soaps are usually sodium salts of long-chain organic acids called fatty acids. They
have a polar “head” and a nonpolar “hydrocarbon tail.” Sodium stearate, a typical soap,
is shown here.
The stearate ion is typical of the anions in soaps. It has a polar carboxylate head,
O
B
XCXOϪ, and a long nonpolar tail, CH3(CH2)16X. The head of the stearate ion is
compatible with (“soluble in”) water, whereas the hydrocarbon tail is compatible with
(“soluble in”) oil and grease. Groups of such ions can be dispersed in water because they
form micelles (Figure 14-21a). Their “water-insoluble” tails are in the interior of a micelle
and their polar heads on the outside where they can interact with the polar water molecules.
When sodium stearate is stirred into water, the result is not a true solution. Instead
it contains negatively charged micelles of stearate ions, surrounded by the positively
“Dry” cleaning, on the other hand,
does not involve water. The solvents
that are used in dry cleaning dissolve
grease to form true solutions.
580 CHAPTER 14: Solutions
CH
H
H
C
H
H
C
H
H
C
H
H
C
H
H
C
H
H
C
H
H
C
H
H
C
H
H
C
H
H
C
H
H
C
H
H
C
H
H
C
H
H
C
H
H
C
H
H
C
H
O
O
H
C Naϩ
Ϫ
hydrocarbon tail (soluble in oil) polar head
(soluble in H2O)
sodium stearate (a soap)
(Right) Powdered sulfur floats on
pure water because of the high
surface tension of water.
(Left) When a drop of detergent
solution is added to the water, its
surface tension is lowered and sulfur
sinks. This lowering of surface
tension enhances the cleaning action
of detergent solutions.
Sodium stearate is also a major
component of some stick deodorants.
charged Naϩ ions. The result is a suspension of micelles in water. These micelles are large
enough to scatter light, so a soap–water mixture appears cloudy. Oil and grease “dissolve”
in soapy water because the nonpolar oil and grease are taken into the nonpolar interior
of micelles (Figure 14-21b). Micelles form a true emulsion in water, so the oil and grease
can be washed away. Sodium stearate is called a surfactant (meaning “surface-active
agent”) or wetting agent because it has the ability to suspend and wash away oil and grease.
Other soaps and detergents behave similarly.
“Hard” water contains Fe3ϩ, Ca2ϩ, and/or Mg2ϩ ions, all of which displace Naϩ from
soaps to form precipitates. This removes the soap from the water and puts an undesirable
coating on the bathtub or on the fabric being laundered. Synthetic detergents are soaplike
emulsifiers that contain sulfonate, XSO3
Ϫ, or sulfate, XOSO3
Ϫ, instead of carboxylate
groups, XCOOϪ. They do not precipitate the ions of hard water, so they can be used in
hard water as soap substitutes without forming undesirable scum.
Phosphates were added to commercial detergents for various purposes. They complexed
the metal ions that contribute to water hardness and kept them dissolved, controlled acidity,
and influenced micelle formation. The use of detergents containing phosphates is now
discouraged because they cause eutrophication in rivers and streams that receive sewage.
Figure 14-21 (a) A representation
of a micelle. The nonpolar tails
“dissolve” in one another in the
center of the cluster and the polar
heads on the outside interact
favorably with the polar water
molecules. (b) Attachment of soap or
detergent molecules to a droplet of
oily dirt to suspend it in water.
14-18 Hydrophilic and Hydrophobic Colloids 581
Oil droplet (dirt)
suspended in water
Long-chain-hydrocarbon
portion of detergent
(oil-soluble)
Water-soluble portion
of detergent, –COO–
or –SO4
– (polar part)
(a) (b)
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 14.11, Surfactants.
Phosphates and nonbiodegradable
detergents are responsible for the
devastation of plant life in and
along this river.
This is a condition (not related to colloids) in which an overgrowth of vegetation is caused
by the high concentration of phosphorus, a plant nutrient. This overgrowth and the subsequent
decay of the dead plants lead to decreased dissolved O2 in the water, which causes
the gradual elimination of marine life. There is also a foaming problem associated with
branched alkylbenzenesulfonate (ABS) detergents in streams and in pipes, tanks, and
pumps of sewage treatment plants. Such detergents are not biodegradable; that is, they
cannot be broken down by bacteria.
Currently used linear-chain alkylbenzenesulfonate (LAS) detergents are biodegradable and
do not cause such foaming.
CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2 SO3
Ϫ
Naϩ
sodium lauryl benzenesulfonate
a linear alkylbenzenesulfonate (LAS)—a biodegradable detergent
CH3CHCH2CHCH2CHCH2CH
CH3 CH3 CH3 CH3
SO3
Ϫ
Naϩ
a sodium branched alkylbenzenesulfonate (ABS)—a nonbiodegradable detergent
The two detergents shown here each
have a C12H25 tail but the branched
one is not biodegradable.
582 CHAPTER 14: Solutions
Our Daily LivesCCHEMISTRY IN USE
Why Does Red Wine Go with Red Meat?
Choosing the appropriate wine to go with dinner is a problem
for some diners. Experts, however, have offered a simple
rule for generations, “serve red wine with red meat and white
wine with fish and chicken.” Are these cuisine choices just
traditions, or are there fundamental reasons for them?
Red wine is usually served with red meat because of a
desirable matching of the chemicals found in each. The most
influential ingredient in red meat is fat; it gives red meats
their desirable flavor. As you chew a piece of red meat, the
fat from the meat coats your tongue and palate, which desensitizes
your taste buds. As a result, your second bite of red
meat is less tasty than the first. Your steak would taste better
if you washed your mouth between mouthfuls; there is an
easy way to wash away the fat deposits.
Red wine contains a surfactant that cleanses your mouth,
removing fat deposits, re-exposing your taste buds, and allowing
you to savor the next bite of red meat almost as well as
the first bite. The tannic acid (also called tannin) in red wine
provides a soap-like action. Like soap, tannic acid consists of
both a nonpolar complex hydrocarbon part as well as a polar
one. The polar part of tannic acid dissolves in polar saliva,
while the nonpolar part dissolves in the fat film that coats
your palate. When you sip red wine, a suspension of micelles
forms in the saliva. This micelle emulsion has the fat molecules
in its interior; the fat is washed away by swallowing the
red wine.
White wines go poorly with red meats because they lack
the tannic acid needed to cleanse the palate. In fact, the presence
or absence of tannic acid distinguishes red wines from
white wines. Grapes fermented with their skins produce red
wines; grapes fermented without their skins produce white
wines.
Because fish and chicken have less fat than red meats,
they can be enjoyed without surfactants to cleanse the palate.
Also, tannic acid has a strong flavor that can overpower the
delicate flavor of many fish. The absence of tannic acid in
white wines gives them a lighter flavor than red wines, and
many people prefer this lighter flavor with their fish or
chicken dinners.
Ronald DeLorenzo
Middle Georgia College
Key Terms 583
Key Terms
The following terms were defined at the end of Chapter 3: concentration,
dilution, molarity, percent by mass, solute,
solution, and solvent. The following terms were defined at the
end of Chapter 13: condensation, condensed phases, evaporation,
phase diagram, and vapor pressure.
Adsorption Adhesion of species onto surfaces of particles.
Associated ions Short-lived species formed by the collision of
dissolved ions of opposite charge.
Biodegradability The ability of a substance to be broken down
into simpler substances by bacteria.
Boiling point elevation The increase in the boiling point of a
solvent caused by dissolution of a nonvolatile solute.
Boiling point elevation constant, Kb A constant that corresponds
to the change (increase) in boiling point produced by a
one-molal ideal solution of a nonvolatile nonelectrolyte.
Colligative properties Physical properties of solutions that depend
on the number but not the kind of solute particles present.
Colloid A heterogeneous mixture in which solute-like particles
do not settle out; also called colloidal dispersion.
Crystal lattice energy The energy change when one mole of
formula units of a crystalline solid is formed from its ions,
atoms, or molecules in the gas phase; always negative.
Detergent A soap-like emulsifier that contains a sulfonate,
XSO3
Ϫ, or sulfate, XOSO3
Ϫ, group instead of a carboxylate,
XCOOϪ, group.
Dispersed phase The solute-like species in a colloid.
Dispersing medium The solvent-like phase in a colloid.
Dispersion See Colloid.
Distillation The process in which components of a mixture are
separated by boiling away the more volatile liquid.
Effective molality The sum of the molalities of all solute particles
in solution.
Emulsifier See Emulsifying agent.
Emulsifying agent A substance that coats the particles of a dispersed
phase and prevents coagulation of colloidal particles; an
emulsifier.
Emulsion A colloidal dispersion of a liquid in a liquid.
Eutrophication The undesirable overgrowth of vegetation
caused by high concentrations of plant nutrients in bodies of
water.
Foam A colloidal dispersion of a gas in a liquid.
Fractional distillation The process in which a fractionating column
is used in a distillation apparatus to separate components
of a liquid mixture that have different boiling points.
Freezing point depression The decrease in the freezing point
of a solvent caused by the presence of a solute.
Freezing point depression constant, Kf A constant that corresponds
to the change in freezing point produced by a
one-molal ideal solution of a nonvolatile nonelectrolyte.
Gel A colloidal dispersion of a solid in a liquid; a semirigid sol.
Hard water Water containing Fe3ϩ, Ca2ϩ, or Mg2ϩ ions, which
form precipitates with soaps.
Heat of solution (molar) The amount of heat absorbed in the
formation of a solution that contains one mole of solute; the
value is positive if heat is absorbed (endothermic) and negative
if heat is released (exothermic).
Henry’s Law The pressure of the gas above a solution is proportional
to the concentration of the gas in the solution.
Hydration The interaction (surrounding) of solute particles with
water molecules.
Hydration energy (molar) of an ion The energy change accompanying
the hydration of a mole of gaseous ions.
Hydrophilic colloids Colloidal particles that attract water mol-
ecules.
Hydrophobic colloids Colloidal particles that repel water mol-
ecules.
Ideal solution A solution that obeys Raoult’s Law exactly.
Liquid aerosol A colloidal dispersion of a liquid in a gas.
Micelle A cluster of a large number of soap or detergent molecules
or ions, assembled with their hydrophobic tails directed
toward the center and their hydrophilic heads directed outward.
Miscibility The ability of one liquid to mix with (dissolve in)
another liquid.
Molality (m) Concentration expressed as number of moles of
solute per kilogram of solvent.
Mole fraction of a component in solution The number of
moles of the component divided by the total number of moles
of all components.
Osmosis The process by which solvent molecules pass through
a semipermeable membrane from a dilute solution into a more
concentrated solution.
Osmotic pressure The hydrostatic pressure produced on the
surface of a semipermeable membrane by osmosis.
Percent ionization of weak electrolytes The percent of the
weak electrolyte that ionizes in a solution of a given concen-
tration.
Raoult’s Law The vapor pressure of a solvent in an ideal solution
is directly proportional to the mole fraction of the solvent
in the solution.
Reverse osmosis The forced flow of solvent molecules through
a semipermeable membrane from a concentrated solution into
a dilute solution. This is accomplished by application of hydrostatic
pressure on the concentrated side greater than the osmotic
pressure that is opposing it.
Saturated solution A solution in which no more solute will dissolve
at a given temperature.
Semipermeable membrane A thin partition between two solutions
through which certain molecules can pass but others
cannot.
Soap An emulsifier that can disperse nonpolar substances in
water; the sodium salt of a long-chain organic acid; consists of
a long hydrocarbon chain attached to a carboxylate group,
XCO2
ϪNaϩ.
Sol A colloidal dispersion of a solid in a liquid.
Solid aerosol A colloidal dispersion of a solid in a gas.
584 CHAPTER 14: Solutions
Solid emulsion A colloidal dispersion of a liquid in a solid.
Solid foam A colloidal dispersion of a gas in a solid.
Solid sol A colloidal dispersion of a solid in a solid.
Solvation The process by which solvent molecules surround and
interact with solute ions or molecules.
Supersaturated solution A (metastable) solution that contains
a higher-than-saturation concentration of solute; slight disturbance
or seeding causes crystallization of excess solute.
Surfactant A “surface-active agent”; a substance that has the ability
to emulsify and wash away oil and grease in an aqueous sus-
pension.
Thermal pollution Introduction of heated waste water into natural
waters.
Tyndall effect The scattering of light by colloidal particles.
van’t Hoff factor, i A number that indicates the extent of dissociation
or ionization of a solute; equal to the actual colligative
property divided by the colligative property calculated assuming
no ionization or dissociation.
Exercises
General Concepts: The Dissolving Process
*001. Support or criticize the statement “Solutions and mixtures
are the same thing.”
*002. Give an example of a solution that contains each of the
following: (a) a solid dissolved in a liquid; (b) a gas dissolved
in a gas; (c) a gas dissolved in a liquid; (d) a liquid
dissolved in a liquid; (e) a solid dissolved in a solid. Identify
the solvent and the solute in each case.
*003. There are no true solutions in which the solvent is gaseous
and the solute is either liquid or solid. Why?
*004. Explain why (a) solute–solute, (b) solvent–solvent, and
(c) solute–solvent interactions are important in determining
the extent to which a solute dissolves in a solvent.
*005. Define and distinguish between dissolution, solvation,
and hydration.
*006. The amount of heat released or absorbed in the dissolution
process is important in determining whether the
dissolution process is spontaneous, meaning, whether it
can occur. What is the other important factor? How does
it influence solubility?
*007. An old saying is that “oil and water don’t mix.” Explain,
on a molecular basis, why this saying is true.
*008. Two liquids, A and B, do not react chemically and are
completely miscible. What would be observed as one is
poured into the other? What would be observed in the
case of two completely immiscible liquids and in the case
of two partially miscible liquids?
*009. Consider the following solutions. In each case, predict
whether the solubility of the solute should be high or low.
Justify your answers. (a) KCl in hexane, C6H14; (b) CaCl2
in H2O; (c) C6H14 in H2O; (d) CCl4 in C6H14; (e) C6H14
in CCl4.
*010. Consider the following solutions. In each case predict
whether the solubility of the solute should be high or low.
Justify your answers. (a) HCl in H2O; (b) HF in H2O;
(c) Al2O3 in H2O; (d) S8 in H2O; (e) NaNO3 in hexane,
C6H14.
*011. For those solutions in Exercise 9 that can be prepared in
“reasonable” concentrations, classify the solutes as nonelectrolytes,
weak electrolytes, or strong electrolytes.
*012. For those solutions in Exercise 10 that can be prepared
in “reasonable” concentrations, classify the solutes as nonelectrolytes,
weak electrolytes, or strong electrolytes.
*013. Both methanol, CH3OH, and ethanol, CH3CH2OH,
are completely miscible with water at room temperature
because of strong solvent–solute intermolecular
attractions. Predict the trend in solubility
in water for 1-propanol, CH3CH2CH2OH;
1-butanol, CH3CH2CH2CH2OH; and 1-pentanol,
CH3CH2CH2CH2CH2OH.
*014. (a) Does the solubility of a solid in a liquid exhibit an
appreciable dependence on pressure? (b) Is the same true
for the solubility of a liquid in a liquid? Why?
*015. Describe a technique for determining whether or not a
solution contains an electrolyte.
*016. A reagent bottle in the storeroom is labeled as containing
a saturated sodium chloride solution. How can one
determine whether or not the solution is saturated?
0*17. A handbook lists the value of the Henry’s Law constant
as 3.02 ϫ 104 atm for ethane, C2H6, dissolved in water at
25°C. The absence of concentration units on k means that
the constant is meant to be used with concentration
expressed as a mole fraction. Calculate the mole fraction
of ethane in water at an ethane pressure of 0.15 atm.
0*18. The mole fraction of methane, CH4, dissolved in water
can be calculated from the Henry’s Law constants of 4.13
ϫ 104 atm at 25°C and 5.77 ϫ 104 atm at 50.°C. Calculate
the solubility of methane at these temperatures for a
methane pressure of 10. atm above the solution. Does the
solubility increase or decrease with increasing temperature?
(See Exercise 17 for interpretation of units.)
*019. Choose the ionic compound from each pair for which the
crystal lattice energy should be the most negative. Justify
your choice. (a) LiF or LiBr; (b) KF or CaF2; (c) FeF2 or
FeF3; (d) NaF or KF.
*020. Choose the ion from each pair that should be more
strongly hydrated in aqueous solution. Justify your
choice. (a) Naϩ or Rbϩ; (b) ClϪ or BrϪ; (c) Fe3ϩ or Fe2ϩ;
(d) Naϩ or Mg2ϩ.
0*21. The crystal lattice energy for LiBr(s) is Ϫ818.6 kJ/mol
at 25°C. The hydration energy of the ions of LiBr is
Ϫ867.4 kJ/mol at 25°C (for infinite dilution). (a) What is
the heat of solution of LiBr(s) at 25°C (for infinite
Exercises 585
dilution)? (b) The hydration energy of Liϩ(g) is
Ϫ544 kJ/mol at 25°C. What is the hydration energy for
BrϪ(g) at 25°C?
*022. Why is the dissolving of many ionic solids in water an
endothermic process, whereas the mixing of most miscible
liquids is an exothermic process?
Concentrations of Solutions
*023. Under what conditions are the molarity and molality of
a solution nearly the same? Which concentration unit is
more useful when measuring volume with burets, pipets,
and volumetric flasks in the laboratory? Why?
*024. Many handbooks list solubilities in units of (g solute/
100. g H2O). How would you convert from this unit to
mass percent?
*025. A 60.0-mL sample of diethyl ether, (C2H5)2O, is dissolved
in enough methanol, CH3OH, to make 300.0 mL of solution.
The density of the ether is 0.714 g/mL. What is the
molarity of this solution?
0*26. Describe how to prepare 1.000 L of 0.215 m NaCl. The
density of this solution is 1.01 g/mL.
0*27. Urea, (NH2)2CO, is a product of metabolism of proteins.
An aqueous solution is 32.0% urea by mass and has a density
of 1.087 g/mL. Calculate the molality of urea in the
solution.
*028. Calculate the molality of a solution that contains 71.5 g
of benzoic acid, C6H5COOH, in 325 mL of ethanol,
C2H5OH. The density of ethanol is 0.789 g/mL.
*029. Sodium fluoride has a solubility of 4.22 g in 100.0 g of
water at 18°C. Express the solute concentration in terms
of (a) mass percent, (b) mole fraction, and (c) molality.
*030. What are the mole fractions of ethanol, C2H5OH, and
water in a solution prepared by mixing 70.0 g of ethanol
with 30.0 g of water?
*031. What are the mole fractions of ethanol, C2H5OH, and
water in a solution prepared by mixing 70.0 mL of ethanol
with 30.0 mL of water at 25°C? The density of ethanol
is 0.789 g/mL, and that of water is 1.00 g/mL.
*032. The density of an aqueous solution containing 12.50 g
K2SO4 in 100.00 g solution is 1.083 g/mL. Calculate the
concentration of this solution in terms of molarity, molality,
percent of K2SO4, and mole fraction of solvent.
0*33. A piece of jewelry is marked “14 carat gold,” meaning
that on a mass basis the jewelry is 14/24 pure gold. What
is the molality of this alloy—considering the other metal
as the solvent?
0*34. A solution that is 24.0% fructose, C6H12O6, in water
has a density of 1.10 g/mL at 20°C. (a) What is the
molality of fructose in this solution? (b) At a higher
temperature, the density is lower. Would the molality
be less than, greater than, or the same as the molality at
20°C? Explain.
*035. The density of a sulfuric acid solution taken from a car
battery is 1.225 g/cm3. This corresponds to a 3.75 M solution.
Express the concentration of this solution in terms
of molality, mole fraction of H2SO4, and percentage of
water by mass.
Raoult’s Law and Vapor Pressure
*036. In your own words, explain briefly why the vapor pressure
of a solvent is lowered by dissolving a nonvolatile
solute in it.
*037. (a) Calculate the vapor pressure lowering associated with
dissolving 35.5 g of table sugar, C12H22O11, in 400. g of
water at 25.0°C. (b) What is the vapor pressure of the
solution? Assume that the solution is ideal. The vapor
pressure of pure water at 25°C is 23.76 torr. (c) What is
the vapor pressure of the solution at 100.°C?
*038. Calculate (a) the lowering of the vapor pressure and
(b) the vapor pressure of a solution prepared by dissolving
35.5 g of naphthalene, C10H8 (a nonvolatile
nonelectrolyte), in 150.0 g of benzene, C6H6, at 20.°C.
Assume that the solution is ideal. The vapor pressure of
pure benzene is 74.6 torr at 20.°C.
*039. At Ϫ100.°C ethane, CH3CH3, and propane, CH3CH2CH3,
are liquids. At that temperature, the vapor pressure of
pure ethane is 394 torr and that of pure propane is
22 torr. What is the vapor pressure at Ϫ100.°C over
a solution containing equal molar amounts of these
substances?
*040. Using Raoult’s Law, predict the partial pressures in
the vapor above a solution containing 0.250 mol
acetone (P0 ϭ 345 torr) and 0.300 mol chloroform
(P0 ϭ 295 torr).
*041. What is the composition of the vapor above the solution
described in Exercise 39?
*042. What is the composition of the vapor above the solution
described in Exercise 40?
*043. Use the following vapor pressure diagram to estimate
(a) the partial pressure of chloroform, (b) the partial pressure
of acetone, and (c) the total vapor pressure of a
solution in which the mole fraction of CHCl3 is 0.30,
assuming ideal behavior.
300
200
100
0
0 0.2 0.4 0.6 0.8 1.0
Pchloroform
Pacetone
Ptotal
Mole fraction CHCl3
P(torr)
586 CHAPTER 14: Solutions
*044. Answer Exercise 43 for the real solution of acetone and
chloroform.
*045. A solution is prepared by mixing 60.0 g of dichloromethane,
CH2Cl2, and 30.0 g of dibromomethane,
CH2Br2, at 0°C. The vapor pressure at 0°C of pure
CH2Cl2 is 0.175 atm, and that of CH2Br2 is 0.0150 atm.
(a) Assuming ideal behavior, calculate the total vapor pressure
of the solution. (b) Calculate the mole fractions of
CH2Cl2 and CH2Br2 in the vapor above the liquid.
Assume that both the vapor and the solution behave
ideally.
Boiling Point Elevation and Freezing Point
Depression: Solutions of Nonelectrolytes
*046. Calculate the boiling point of a 2.15 m aqueous solution
of ethylene glycol, a nonvolatile nonelectrolyte.
*052. Lemon juice is a complex solution that does not freeze in
a home freezer at temperatures as low as Ϫ11°C. At what
temperature will the lemon juice freeze if its effective
molality is the equivalent of a 10.48 m glucose solution?
0*53. The “proof ” of an alcoholic beverage is twice the volume
percent of ethyl alcohol, C2H5OH, in water. The density
of ethyl alcohol is 0.789 g/mL and that of water is
1.00 g/mL. A bottle of 100-proof rum is left outside on
a cold winter day. (a) Will the rum freeze if the temperature
drops to Ϫ18°C? (b) Rum is used in cooking and
baking. At what temperature does 100-proof rum boil?
*054. A 3.0-g sample of a nonelectrolyte was isolated from beef
fat. The molecular weight of the compound was determined
to be 137 amu. The sample was dissolved in
250.0 mL of ethyl alcohol. At what temperature should
the solution boil? For ethyl alcohol: boiling point ϭ
78.41°C; Kb ϭ 1.22°C/m, density ϭ 0.789 g/mL.
*055. The normal boiling point of benzene is 80.1°C. A
0.77-gram sample of a nonvolatile compound with the
molar mass of 185 g/mol is dissolved in 2.75 g of benzene.
What is the expected boiling point of this solution?
*056. Pure copper melts at 1083°C. Its molal freezing point
depression constant is 23°C/m. What will be the melting
point of a brass made of 10% Zn and 90% Cu by mass?
*057. How many grams of the nonelectrolyte sucrose,
C12H22O11, should be dissolved in 575 g of water to produce
a solution that freezes at Ϫ1.75°C?
*058. What mass of naphthalene, C10H8, should be dissolved
in 375 g of nitrobenzene, C6H5NO2, to produce a solution
that boils at 214.20°C? See Table 14-2.
*059. A solution was made by dissolving 3.75 g of a nonvolatile
solute in 108.7 g of acetone. The solution boiled at
56.58°C. The boiling point of pure acetone is 55.95°C,
and Kb ϭ 1.71°C/m. Calculate the molecular weight of
the solute.
*060. The molecular weight of an organic compound was determined
by measuring the freezing point depression of a
benzene solution. A 0.500-g sample was dissolved in
50.0 g of benzene, and the resulting depression was
0.42°C. What is the approximate molecular weight?
The compound gave the following elemental analysis:
40.0% C, 6.67% H, 53.3% O by mass. Determine the
formula and exact molecular weight of the substance.
*061. When 0.154 g of sulfur is finely ground and melted with
4.38 g of camphor, the freezing point of the camphor is
lowered by 5.47°C. What is the molecular weight of sulfur?
What is its molecular formula?
0*62. (a) Suppose we dissolve a 6.00-g sample of a mixture of
naphthalene, C10H8, and anthracene, C14H10, in 360. g
of benzene. The solution is observed to freeze at 4.85°C.
Find the percent composition (by mass) of the sample. (b)
At what temperature should the solution boil? Assume
that naphthalene and anthracene are nonvolatile non-
electrolytes.
Ethylene glycol solutions.
*047. A solution is prepared by dissolving 6.41 g of ordinary
sugar (sucrose, C12H22O11, 342 g/mol) in 32.0 g of water.
Calculate the boiling point of the solution. Sucrose is a
nonvolatile nonelectrolyte.
*048. What is the freezing point of the solution described in
Exercise 46?
*049. What is the freezing point of the solution described in
Exercise 47?
*050. Refer to Table 14-2. Suppose you had a 0.150 m solution
of a nonvolatile nonelectrolyte in each of the solvents
listed there. Which one should have (a) the greatest freezing
point depression, (b) the lowest freezing point, (c) the
greatest boiling point elevation, and (d) the highest boiling
point?
*051. A 5.00% solution of dextrose, C6H12O6, in water is
referred to as DW5. This solution can be used as a source
of nourishment when introduced by intravenous injection.
Calculate the freezing point and the boiling point
of a DW5 solution.
Exercises 587
Boiling Point Elevation and Freezing Point
Depression: Solutions of Electrolytes
*063. What is ion association in solution? Can you suggest why
the term “ion pairing” is sometimes used to describe this
phenomenon?
*064. You have separate 0.10 M aqueous solutions of the following
salts: LiNO3, Ca(NO3)2, and Al(NO3)3. In which
one would you expect to find the highest particle concentration?
Which solution would you expect to conduct
electricity most strongly? Explain your reasoning.
*065. What is the significance of the van’t Hoff factor, i?
*066. What is the value of the van’t Hoff factor, i, for the
following strong electrolytes at infinite dilution?
(a) Na2SO4; (b) KOH; (c) Al2(SO4)3; (d) SrSO4.
*067. Compare the number of solute particles that are present
in solutions of equal concentrations of strong electrolytes,
weak electrolytes, and nonelectrolytes.
*068. Four beakers contain 0.010 m aqueous solutions of
CH3OH, KClO3, CaCl2, and CH3COOH, respectively.
Without calculating the actual freezing points of each of
these solutions, arrange them from lowest to highest
freezing point.
0*69. A 0.050 m aqueous solution of K3[Fe(CN)6] has a freezing
point of Ϫ0.2800°C. Calculate the total concentration
of solute particles in this solution and interpret your
results.
0*70. A solution is made by dissolving 1.00 gram each of NaCl,
NaBr, and NaI in 100. g of water. What is the vapor pressure
above this solution at 100°C? Assume complete
dissociation of the three salts.
*071. The ice fish lives under the polar ice cap where the water
temperature is Ϫ4°C. This fish does not freeze at that
temperature due to the solutes in its blood. The solute
concentration of the fish’s blood can be related to a
sodium chloride solution that would have the same freezing
point. What is the minimum concentration of a
sodium chloride solution that would not freeze at Ϫ4°C?
*072. Which solution freezes at a lower temperature, 0.1 m
sodium sulfate or 0.1 m calcium sulfate? Explain.
*073. Solutions are produced by dissolving 25.0 grams of
sodium sulfate in 0.500 kg of water and by dissolving
25.0 grams of calcium sulfate in 0.500 kg of water. Which
solution will freeze at a lower temperature? Explain.
*074. A series of 1.1 m aqueous solutions is produced. Predict
which solution of each pair would boil at the higher temperature.
(a) NaCl or LiCl; (b) LiCl or Li2SO4; (c) Li2SO4
or HCl; (d) HCl or C6H12O6; (e) C6H12O6 or CH3OH;
(f ) CH3OH or CH3COOH; (g) CH3COOH or NaCl.
*075. Identify which of the following solutions, each prepared
by dissolving 65.0 g of the solute in 150. mL of water,
will display the highest boiling point and explain your
choice. (a) NaCl or LiCl; (b) LiCl or Li2SO4; (c) Li2SO4
or HCl; (d) HCl or C6H12O6; (e) C6H12O6 or CH3OH;
(f ) CH3OH or CH3COOH; (g) CH3COOH or NaCl.
*076. Synthetic ocean water can be produced by dissolving
36.0 g of table salt in 1000. mL of water. What is the
boiling point of this solution? (Assume that the density
of the water is 0.998 g/mL and that table salt is pure
sodium chloride.)
*077. Formic acid, HCOOH, ionizes slightly in water.
HCOOH(aq) 34 Hϩ(aq) ϩ HCOOϪ(aq)
A 0.0100 m formic acid solution freezes at Ϫ0.0209°C.
Calculate the percent ionization of HCOOH in this
solution.
*078. A 0.100 m acetic acid solution in water freezes at
Ϫ0.1884°C. Calculate the percent ionization of
CH3COOH in this solution.
0*79. In a home ice cream freezer, we lower the freezing point
of the water bath surrounding the ice cream can by
dissolving NaCl in water to make a brine solution. A
15.0% brine solution is observed to freeze at Ϫ10.888°C.
What is the van’t Hoff factor, i, for this solution?
*080. CsCl dissolves in water according to
H2O
CsCl(s) 8888n Csϩ(aq) ϩ ClϪ(aq)
A 0.121 m solution of CsCl freezes at Ϫ0.403°C. Calculate
i and the apparent percent dissociation of CsCl in
this solution.
Osmotic Pressure
*081. What are osmosis and osmotic pressure?
0*82. Show numerically that the molality and molarity of 1.00
ϫ 10Ϫ4 M aqueous sodium chloride are nearly equal.
Why is this true? Would this be true if another solvent,
say acetonitrile, CH3CN, replaced water? Why or why
not? The density of CH3CN is 0.786 g/mL at 20.0°C.
*083. Show how the expression ␲ ϭ MRT, where ␲ is osmotic
pressure, is similar to the ideal gas law. Rationalize qualitatively
why this should be so.
*084. What is the osmotic pressure associated with a 0.0111 M
aqueous solution of a nonvolatile nonelectrolyte solute at
75°C?
*085. The osmotic pressure of an aqueous solution of a nonvolatile
nonelectrolyte solute is 1.17 atm at 0.0°C. What
is the molarity of the solution?
*086. Calculate the freezing point depression and boiling point
elevation associated with the solution in Exercise 85.
*087. Estimate the osmotic pressure associated with 24.5 g of
an enzyme of molecular weight 4.21 ϫ 106 dissolved in
1740. mL of ethyl acetate solution at 38.0°C.
*088. Calculate the osmotic pressure at 25°C of 0.10 m K2CrO4
in water, taking ion association into account. Refer to
Table 14-3.
*089. Estimate the osmotic pressure at 25°C of 0.10 m K2CrO4
in water, assuming no ion association.
588 CHAPTER 14: Solutions
0*90. Many biological compounds are isolated and purified in
very small amounts. We dissolve 11.0 mg of a biological
macromolecule with molecular weight of 2.00 ϫ 104 in
10.0 g of water. (a) Calculate the freezing point of the
solution. (b) Calculate the osmotic pressure of the solution
at 25°C. (c) Suppose we are trying to use freezing
point measurements to determine the molecular weight of
this substance and that we make an error of only 0.001°C
in the temperature measurement. What percent error
would this cause in the calculated molecular weight?
(d) Suppose we could measure the osmotic pressure with
an error of only 0.1 torr (not a very difficult experiment).
What percent error would this cause in the calculated
molecular weight?
Colloids
*091. How does a colloidal dispersion differ from a true solu-
tion?
*092. Distinguish among (a) sol, (b) gel, (c) emulsion, (d) foam,
(e) solid sol, (f) solid emulsion, (g) solid foam, (h) solid
aerosol, and (i) liquid aerosol. Try to give an example of
each that is not listed in Table 14-4.
*093. What is the Tyndall effect, and how is it caused?
*094. Distinguish between hydrophilic and hydrophobic col-
loids.
*095. What is an emulsifier?
*096. Distinguish between soaps and detergents. How do they
interact with hard water? Write an equation to show the
interaction between a soap and hard water that contains
Ca2ϩ ions.
*097. What is the disadvantage of branched alkylbenzenesulfonate
(ABS) detergents compared with linear alkylbenzenesulfonate
(LAS) detergents?
Mixed Exercises
0*98. The heat of solution (for infinite dilution) of KF(s) is
Ϫ17.7 kJ/mol at 25°C. The crystal lattice energy, ⌬Hxtal,
is Ϫ825.9 kJ/mol at 25°C. What is the hydration energy
of KF for infinite dilution at 25°C? [Here we refer to the
sum of the hydration energies of Kϩ(g) and FϪ(g).]
*099. Dry air contains 20.94% O2 by volume. The solubility of
O2 in water at 25°C is 0.041 gram O2 per liter of water.
How many liters of water would dissolve the O2 in one
liter of dry air at 25°C and 1.00 atm?
*100. (a) The freezing point of a 1.00% aqueous solution of
acetic acid, CH3COOH, is Ϫ0.310°C. What is the
approximate formula weight of acetic acid in water?
(b) A 1.00% solution of acetic acid in benzene has a
freezing point depression of 0.441°C. What is the formula
weight of acetic acid in this solvent? Explain the
difference.
*101. An aqueous ammonium chloride solution contains 6.50
mass % NH4Cl. The density of the solution is 1.0201 g/mL.
Express the concentration of this solution in molarity,
molality, and mole fraction of solute.
*102. Starch contains CXC, CXH, CXO, and OXH bonds.
Hydrocarbons contain only CXC and CXH bonds. Both
starch and hydrocarbon oils can form colloidal dispersions
in water. (a) Which dispersion is classified as
hydrophobic? (b) Which is hydrophilic? (c) Which
dispersion would be easier to make and maintain?
*103. Suppose we put some one-celled microorganisms in various
aqueous NaCl solutions. We observe that the cells
remain unperturbed in 0.7% NaCl, whereas they shrink
in more concentrated solutions and expand in more dilute
solutions. Assume that 0.7% NaCl behaves as an ideal 1Ϻ1
electrolyte. Calculate the osmotic pressure of the aqueous
fluid within the cells at 25°C.
*104. A sample of a drug (C21H23O5N, molecular weight ϭ
369 g/mol) mixed with lactose (a sugar, C12H22O11,
molecular weight ϭ 342 g/mol) was analyzed by osmotic
pressure to determine the amount of sugar present. If
100. mL of solution containing 1.00 g of the drug–sugar
mixture has an osmotic pressure of 527 torr at 25°C, what
is the percent sugar present?
*105. A solution containing 4.22 g of a nonelectrolyte polymer
per liter of benzene solution has an osmotic pressure of
0.646 torr at 20.0°C. (a) Calculate the molecular weight
of the polymer. (b) Assume that the density of the dilute
solution is the same as that of benzene, 0.879 g/mL. What
would be the freezing point depression for this solution?
(c) Why are boiling point elevations and freezing point
depressions difficult to use to measure molecular weights
of polymers?
*106. On what basis would you choose the components to prepare
an ideal solution of a molecular solute? Which of
the following combinations would you expect to act most
nearly ideally? (a) CH4(ᐉ) and CH3OH(ᐉ); (b) CH3OH(ᐉ)
and NaCl(s); (c) CH4(ᐉ) and CH3CH3(ᐉ).
*107. Physiological saline (normal saline) is a 0.90% NaCl
solution. This solution is isotonic with human blood.
Calculate the freezing point and boiling point of physiological
saline.
CONCEPTUAL EXERCISES
*108. In the first five sections of this chapter the term “dissolution”
was used to describe the process by which a solute
is dispersed by a solvent to form a solution. A popular
dictionary defines dissolution as “decomposition into
fragments or parts.” Using either of the definitions, compare
the following two uses of the term. “It was ruled that
there must be dissolution of the estate.” “The ease of dissolution
of a solute depends on two factors. . . .”
*109. Consider two nonelectrolytes A and B; A has a higher
molecular weight than B, and both are soluble in solvent
Exercises 589
C. A solution is made by dissolving x grams of A in 100
grams of C; another solution is made by dissolving the
same number of grams, x, of B in 100 grams of C. Assume
that the two solutions have the same density. (a) Which
solution has the higher molality? (b) Which solution has
the higher mole fraction? (c) Which solution has the
higher percent by mass? (d) Which solution has the
higher molarity?
*110. Can the actual value for the van’t Hoff factor, i, ever be
larger than the ideal value? Could the actual value ever
be less than 1?
*111. When the van’t Hoff factor, i, is included, the boiling
point elevation equation becomes ⌬Tb ϭ i Kbm. The van’t
Hoff factor can be inserted in a similar fashion in the
freezing point depression equation and the osmotic
pressure equation. The van’t Hoff factor, however,
cannot be inserted in the same way into the vapor
pressure-lowering equation. Show that ⌬Psolvent ϭ
iXsoluteP0
solvent is not equivalent to ⌬Psolvent ϭ
{(i x molessolute)/[(i x molessolute) ϩ (molessolvent)]}P0
solvent.
Which equation for ⌬Psolvent is correct?
*112. The two solutions shown were both prepared by dissolving
194 g of K2CrO4 (1.00 mol) in a 1.00-L volumetric
flask. One was diluted to the 1-L mark on the neck of the
flask, and the other was diluted by adding 1.00 L of water.
Which solution, the one on the left or the one on the
right, is: (a) more concentrated; (b) a 1.00 m solution;
(c) a 1.00 M solution?
*113. Would you expect lowering the freezing point or elevating
the boiling point to be the better method to obtain
the approximate molecular weight of an unknown?
Explain your choice.
*114. Rock candy consists of crystals of sugar on a string or
stick. Propose a method of making rock candy, and explain
each step.
*115. Concentrations expressed in units of parts per million and
parts per billion often have no meaning for people until
they relate these small and large numbers to their own
experiences. How many seconds are equal to 1 ppm of a
year?
BUILDING YOUR KNOWLEDGE
*116. DDT is a toxin still found in the fatty tissues of some animals.
DDT was transported into our lakes and streams as
runoff from agricultural operations where it was originally
used several years ago as an insecticide. In the lakes
and streams it did not dissolve to any great extent; it
collected in the lake and stream bottoms. It entered
the bodies of animals via fatty tissues in their diet;
microorganisms collected the DDT, the fish ate the
microorganisms, and so on. Fortunately, much of the once
large quantities of DDT in lakes and streams has biodegraded.
Based on this information, what can you conclude
regarding the intermolecular forces present in DDT?
*117. Draw Figure 14-1, but instead of using colored circles to
represent the solvent and solute molecules, use Lewis formulas
to represent water as the solvent and acetone,
CH3COCH3, as the solute. Use dashed lines to show
hydrogen bonds. Twelve water molecules and two acetone
molecules should be sufficient to illustrate the
interaction between these two kinds of molecules.
*118. A sugar maple tree grows to a height of about 45 feet,
and its roots are in contact with water in the soil. What
must be the concentration of the sugar in its sap if osmotic
pressure is responsible for forcing the sap to the top of
the tree at 10°C? The density of mercury is 13.6 g/cm3,
and the density of the sap can be considered to be
1.10 g/cm3.
*119. Many metal ions become hydrated in solution by forming
coordinate covalent bonds with the unshared pair of
electrons from the water molecules to form “AB6” ions.
Because of their sizes, these hydrated ions are unable to
pass through the semipermeable membrane described in
Section 14-15, whereas water as a trimer, (H2O)3, or
a tetramer, (H2O)4, can pass through. Anions tend to
be less hydrated. Using the VSEPR theory, prepare
three-dimensional drawings of Cu(H2O)6
2ϩ and a possible
(H2O)3 that show their relative shapes and sizes.
15 Chemical
Thermodynamics
OUTLINE
Heat Changes and Thermochemistry
15-1 The First Law of
Thermodynamics
15-2 Some Thermodynamic Terms
15-3 Enthalpy Changes
15-4 Calorimetry
15-5 Thermochemical Equations
15-6 Standard States and Standard
Enthalpy Changes
15-7 Standard Molar Enthalpies of
Formation, ⌬Hf
0
15-8 Hess’s Law
15-9 Bond Energies
15-10 Changes in Internal Energy,
⌬E
15-11 Relationship of ⌬H and ⌬E
Spontaneity of Physical and
Chemical Changes
15-12 The Two Aspects of
Spontaneity
15-13 The Second Law of
Thermodynamics
15-14 Entropy, S
15-15 Free Energy Change, ⌬G, and
Spontaneity
15-16 The Temperature Dependence
of Spontaneity
OBJECTIVES
After you have finished studying this chapter, you should be able to
• Understand the terminology of thermodynamics, and the meaning of the signs of changes
• Use the concept of state functions
• Carry out calculations of calorimetry to determine changes in energy and enthalpy
• Use Hess’s Law to find the enthalpy change, ⌬H, for a reaction by combining
thermochemical equations with known ⌬H values
• Use Hess’s Law to find the enthalpy change, ⌬H, for a reaction by using tabulated
values of standard molar enthalpies of formation
• Use Hess’s Law to find the enthalpy of formation given ⌬H for a reaction and the
known enthalpies of formation of the other substances in the reaction
• Use the First Law of Thermodynamics to relate heat, work, and energy changes
• Relate the work done on or by a system to changes in its volume
• Use bond energies to estimate heats of reaction for gas phase reactions; use ⌬H values
for gas phase reactions to find bond energies
• Understand what is meant by a product-favored process; by a reactant-favored process
• Understand the relationship of entropy to the order or disorder of a system
• Understand how the spontaneity of a process is related to entropy changes—the Second
Law of Thermodynamics
• Use tabulated values of absolute entropies to calculate the entropy change, ⌬S
• Calculate changes in Gibbs free energy, ⌬G, (a) from values of ⌬H and ⌬S and (b)
from tabulated values of standard molar free energies of formation; know when to use
each type of calculation
• Use ⌬G to predict whether a process is product-favored at constant T and P
• Understand how changes in temperature can affect the spontaneity of a process
• Predict the temperature range of spontaneity of a chemical or physical process
The launch of the space shuttle
requires about 2 ؋ 1010 kilojoules
of energy. About one sixth of this
comes from the combustion of
hydrogen, H2, in oxygen, O2. The
rest comes from the explosive
decomposition of ammonium
perchlorate, NH4ClO4, in
solid-fuel rockets.
Some forms of energy are potential,
kinetic, electrical, nuclear, heat, and
light.
592 CHAPTER 15: Chemical Thermodynamics
EE
nergy is very important in every aspect of our daily lives. The food we eat supplies
the energy to sustain life with all of its activities and concerns. The availability of
relatively inexpensive energy is an important factor in our technological society.
This is seen in the costs of fuel, heating and cooling our homes and workplaces, and the
electricity to power our lights, appliances, and computers. It is also seen in the costs of
the goods and services we purchase, because a substantial part of the cost of production
is for energy in one form or another. We must understand the storage and use of energy
on a scientific basis to learn how to decrease our dependence on consumable oil and natural
gas as our main energy sources. Such understanding has profound ramifications, ranging
from our daily lifestyles to international relations.
The concept of energy is at the very heart of science. All physical and chemical processes
are accompanied by the transfer of energy. Because energy cannot be created or destroyed,
we must understand how to do the “accounting” of energy transfers from one body or
one substance to another or from one form of energy to another.
In thermodynamics we study the energy changes that accompany physical and chemical
processes. Usually these energy changes involve heat—hence the “thermo-” part of
the term. In this chapter we study the two main aspects of thermodynamics. The first is
thermochemistry. This practical subject is concerned with how we observe, measure, and
predict energy changes for both physical changes and chemical reactions. The second part
of the chapter addresses a more fundamental aspect of thermodynamics. There we will
learn to use energy changes to tell whether or not a given process can occur under specified
conditions to give predominantly products (or reactants) and how to make a process
more (or less) favorable.
HEAT CHANGES AND THERMOCHEMISTRY
THE FIRST LAW OF THERMODYNAMICS
We can define energy as follows.
Energy is the capacity to do work or to transfer heat.
We classify energy into two general types: kinetic and potential. Kinetic energy is the
energy of motion. The kinetic energy of an object is equal to one half its mass, m, times
the square of its velocity, v.
Ekinetic ϭ ᎏ1
2
ᎏ mv2
The heavier a hammer is and the more rapidly it moves, the greater its kinetic energy and
the more work it can accomplish.
Potential energy is the energy that a system possesses by virtue of its position or
composition. The work that we do to lift an object is stored in the object as energy; we
describe this as potential energy. If we drop a hammer, its potential energy is converted
into kinetic energy as it falls, and it could do work on something it hits—for example,
drive a nail or break a piece of glass. Similarly, an electron in an atom has potential energy
because of the electrostatic force on it that is due to the positively charged nucleus and
the other electrons in that atom and surrounding atoms. Energy can take many other
forms: electrical energy, radiant energy (light), nuclear energy, and chemical energy. At
15-1
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 6.4, Energy, and Screen 6.5,
Forms of Energy.
As matter falls from a higher to a
lower level, its gravitational potential
energy is converted into kinetic
energy. A hydroelectric power plant
converts the kinetic energy of falling
water into electrical (potential)
energy.
the atomic or molecular level, we can think of each of these as either kinetic or potential
energy.
The chemical energy in a fuel or food comes from potential energy stored in atoms
due to their arrangements in the molecules. This stored chemical energy can be released
when compounds undergo chemical changes, such as those that occur in combustion and
metabolism. Reactions that release energy in the form of heat are called exothermic reac-
tions.
Combustion reactions of fossil fuels are familiar examples of exothermic reactions.
Hydrocarbons—including methane, the main component of natural gas, and octane, a
minor component of gasoline—undergo combustion with an excess of O2 to yield CO2
and H2O. These reactions release heat energy. The amounts of heat energy released at
constant pressure are shown for the reactions of one mole of methane and of two moles
of octane.
CH4(g) ϩ 2O2(g) 88n CO2(g) ϩ 2H2O(ᐉ) ϩ 890 kJ
2C8H18(ᐉ) ϩ 25O2(g) 88n 16CO2(g) ϩ 18H2O(ᐉ) ϩ 1.090 ϫ 104 kJ
In such reactions, the total energy of the products is lower than that of the reactants by
the amount of energy released, most of which is heat. Some initial activation (e.g., by heat)
is needed to get these reactions started. This is shown for CH4 in Figure 15-1. This activation
energy plus 890 kJ is released as one mole of CO2(g) and two moles of H2O(ᐉ) are
formed. A process that absorbs energy from its surroundings is called endothermic. One
such process is shown in Figure 15-2.
Energy changes accompany physical changes, too (Chapter 13). For example, the
melting of one mole of ice at 0°C at constant pressure must be accompanied by the absorption
of 6.02 kJ of energy.
H2O(s) ϩ 6.02 kJ 88n H2O(ᐉ)
This tells us that the total energy of the water is raised by 6.02 kJ in the form of heat
during the phase change.
A hydrocarbon is a binary compound
of only hydrogen and carbon.
Hydrocarbons may be gaseous, liquid,
or solid. All burn.
The amount of heat shown in such an
equation always refers to the reaction
for the number of moles of reactants
and products specified by the
coefficients. We call this one mole of
reaction. It is important to specify the
physical states of all substances,
because different physical states have
different energy contents.
15-1 The First Law of Thermodynamics 593
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 6.12, Energy Changes in
Chemical Processes.
Figure 15-1 The difference
between the potential energy of the
reactants—one mole of CH4(g) and
two moles of O2(g)—and that of
the products—one mole of CO2(g)
and two moles of H2O(ᐉ)—is the
amount of heat evolved in this
exothermic reaction at constant
pressure. For this reaction, it is
890 kJ/mol of reaction. In this
chapter, we see how to measure the
heat absorbed or released and how
to calculate it from other known
heat changes. Some initial activation,
for example by heat, is needed to get
the reaction started. In the absence
of such activation energy, a mixture
of CH4 and O2 can be kept at room
temperature for a long time without
reacting. For an endothermic reaction,
the final level is higher than the
initial level.
CH4(g) ϩ 2O2(g)
PotentialEnergy
CO2(g) ϩ 2H2O(ᐉ)
Reactants Products
Progress of reaction
Heat evolved ϭ 890 kJ
Amount of energy
needed to activate
the reaction
Some important ideas about energy are summarized in the First Law of Thermody-
namics.
The total amount of energy in the universe is constant.
The Law of Conservation of Energy is just another statement of the First Law of Ther-
modynamics.
Energy is neither created nor destroyed in ordinary chemical reactions and physical
changes.
SOME THERMODYNAMIC TERMS
The substances involved in the chemical and physical changes that we are studying are
called the system. Everything in the system’s environment constitutes its surroundings.
The universe is the system plus its surroundings. The system may be thought of as the
part of the universe under investigation. The First Law of Thermodynamics tells us that
energy is neither created nor destroyed; it is only transferred between the system and its
surroundings.
15-2
In Chapter 1 we pointed out the
equivalence of matter and energy.
The word “energy” is understood to
include the energy equivalent of
all matter in the universe. Stated
differently, the total amount of mass
and energy in the universe is constant.
594 CHAPTER 15: Chemical Thermodynamics
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 6.13, The First Law of
Thermodynamics.
Figure 15-2 An endothermic process. (a) When solid hydrated barium hydroxide,
Ba(OH)2и8H2O, and excess solid ammonium nitrate, NH4NO3, are mixed, a reaction occurs.
Ba(OH)2и8H2O(s) ϩ 2NH4NO3(s) 88n Ba(NO3)2(s) ϩ 2NH3(g) ϩ 10H2O(ᐉ)
The excess ammonium nitrate dissolves in the water produced in the reaction. (b) The
dissolution process is very endothermic. If the flask is placed on a wet wooden block, the
water freezes and attaches the block to the flask.
(a) (b)
The thermodynamic state of a system is defined by a set of conditions that completely
specifies all the properties of the system. This set commonly includes the temperature,
pressure, composition (identity and number of moles of each component), and physical
state (gas, liquid, or solid) of each part of the system. Once the state has been specified,
all other properties—both physical and chemical—are fixed.
The properties of a system—such as P, V, T—are called state functions. The value
of a state function depends only on the state of the system and not on the way in which
the system came to be in that state. A change in a state function describes a difference
between the two states. It is independent of the process or pathway by which the change
occurs.
For instance, consider a sample of one mole of pure liquid water at 30°C and 1 atm
pressure. If at some later time the temperature of the sample is 22°C at the same pressure,
then it is in a different thermodynamic state. We can tell that the net temperature
change is Ϫ8°C. It does not matter whether (1) the cooling took place directly (either
slowly or rapidly) from 30°C to 22°C, or (2) the sample was first heated to 36°C, then
cooled to 10°C, and finally warmed to 22°C, or (3) any other conceivable path was followed
from the initial state to the final state. The change in other properties (e.g., the pressure)
of the sample is likewise independent of path.
The most important use of state functions in thermodynamics is to describe changes.
We describe the difference in any quantity, X, as
⌬X ϭ Xfinal Ϫ Xinitial
When X increases, the final value is greater than the initial value, so ⌬X is positive; a
decrease in X makes ⌬X a negative value.
You can consider a state function as analogous to a bank account. With a bank account,
at any time you can measure the amount of money in your account (your balance) in
convenient terms—dollars and cents. Changes in this balance can occur for several reasons,
such as deposit of your paycheck, writing of checks, or service charges assessed by the
bank. In our analogy these transactions are not state functions, but they do cause changes
in the state function (the balance in the account). You can think of the bank balance on a
vertical scale; a deposit of $150 changes the balance by ϩ$150, no matter what it was at
the start, just as a withdrawal of $150 would change the balance by Ϫ$150. Similarly, we
shall see that the energy of a system is a state function that can be changed—for instance,
by an energy “deposit” of heat absorbed or work done on the system, or by an energy
“withdrawal” of heat given off or work done by the system.
We can describe differences between levels of a state function, regardless of where the
zero level is located. In the case of a bank balance, the “natural” zero level is obviously
the point at which we open the account, before any deposits or withdrawals. In contrast,
the zero levels on most temperature scales are set arbitrarily. When we say that the temperature
of an ice–water mixture is “zero degrees Celsius,” we are not saying that the mixture
contains no temperature! We have simply chosen to describe this point on the temperature
scale by the number zero; conditions of higher temperature are described by positive
temperature values, and those of lower temperature have negative values, “below zero.”
The phrase “15 degrees cooler” has the same meaning anywhere on the scale. Many of
the scales that we use in thermodynamics are arbitrarily defined in this way. Arbitrary
scales are useful when we are interested only in changes in the quantity being described.
Any property of a system that depends only on the values of its state functions is also
a state function. For instance, the volume of a given sample of water depends only on
temperature, pressure, and physical state; volume is a state function. We shall encounter
other thermodynamic state functions.
State functions are represented by
capital letters. Here P refers to
pressure, V to volume, and T to
absolute temperature.
15-2 Some Thermodynamic Terms 595
Initial balance
⌬ ϭ Ϫ$150
Final balance
Here is a graphical representation of a
$150 decrease in your bank balance.
We express the change in your bank
balance as ⌬$ ϭ $final Ϫ $initial. Your
final balance is less than your initial
balance, so the result is negative,
indicating a decrease. There are many
ways to get this same net change—one
large withdrawal or some combination
of deposits, withdrawals, interest
earned, and service charges. All of the
⌬ values we will see in this chapter can
be thought of in this way.
Increasingaccountbalance
88888888888888888888888888n
88888888888n888888888
ENTHALPY CHANGES
Most chemical reactions and physical changes occur at constant (usually atmospheric) pres-
sure.
The quantity of heat transferred into or out of a system as it undergoes a chemical
or physical change at constant pressure, qp, is defined as the enthalpy change, ⌬H,
of the process.
An enthalpy change is sometimes loosely referred to as a heat change or a heat of reaction.
The enthalpy change is equal to the enthalpy or “heat content,” H, of the substances
produced minus the enthalpy of the substances consumed.
⌬H ϭ Hfinal Ϫ Hinitial or ⌬H ϭ Hsubstances produced Ϫ Hsubstances consumed
It is impossible to know the absolute enthalpy (heat content) of a system. Enthalpy is a
state function, however, and it is the change in enthalpy in which we are interested; this can
be measured for many processes. In the next several sections we focus on chemical reactions
and the enthalpy changes that occur in these processes. We first discuss the
experimental determination of enthalpy changes.
CALORIMETRY
We can determine the energy change associated with a chemical or physical process by
using an experimental technique called calorimetry. This technique is based on observing
the temperature change when a system absorbs or releases energy in the form of heat.
The experiment is carried out in a device called a calorimeter, in which the temperature
change of a known amount of substance (often water) of known specific heat is measured.
The temperature change is caused by the absorption or release of heat by the chemical
or physical process under study. A review of calculations involved with heat transfer
(Sections 1-13, 13-9, and 13-11) may be helpful for understanding this section.
A “coffee-cup” calorimeter (Figure 15-3) is often used in laboratory classes to measure
“heats of reaction” at constant pressure, qp, in aqueous solutions. Reactions are chosen so
that there are no gaseous reactants or products. Thus, all reactants and products remain
in the vessel throughout the experiment. Such a calorimeter could be used to measure the
amount of heat absorbed or released when a reaction takes place in aqueous solution. We
can consider the reactants and products as the system and the calorimeter plus the solution
(mostly water) as the surroundings. For an exothermic reaction, the amount of heat
evolved by the reaction can be calculated from the amount by which it causes the temperature
of the calorimeter and the solution to rise. The heat can be visualized as divided
into two parts.
΂ ΃ϭ ΂ ΃ϩ ΂ ΃
The heat capacity of a calorimeter is determined by adding a known amount of heat
and measuring the rise in temperature of the calorimeter and of the solution it contains.
This heat capacity of a calorimeter is sometimes called its calorimeter constant.
amount of heat
ᎏᎏᎏ
gained by solution
amount of heat gained
ᎏᎏᎏ
by calorimeter
amount of heat
ᎏᎏᎏ
released by reaction
15-4
15-3
We use q to represent the amount of
heat absorbed by the system. The
subscript p indicates a constantpressure
process.
596 CHAPTER 15: Chemical Thermodynamics
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 6.10, Heat Transfer Between
Substances, and Screen 6.18,
Measuring Heats of Reaction:
Calorimetry.
The amount of heat absorbed by a
calorimeter is sometimes expressed as
the heat capacity of the calorimeter, in
joules per degree.
The polystyrene insulation of the
coffee-cup calorimeter ensures that
little or no heat escapes from or enters
the container.
EXAMPLE 15-1 Heat Capacity of a Calorimeter
We add 3.358 kJ of heat to a calorimeter that contains 50.00 g of water. The temperature of
the water and the calorimeter, originally at 22.34°C, increases to 36.74°C. Calculate the heat
capacity of the calorimeter in J/°C. The specific heat of water is 4.184 J/gи°C.
Plan
We first calculate the amount of heat gained by the water in the calorimeter. The rest of the
heat must have been gained by the calorimeter, so we can determine the heat capacity of the
calorimeter.
Solution
50.00 g H2O(ᐉ) at 22.34°C 88n 50.00 g H2O(ᐉ) at 36.74°C
The temperature change is (36.74 Ϫ 22.34)°C ϭ 14.40°C.
__? J ϭ 50.00 g ϫ ᎏ
4
g
.1
и
8
°
4
C
J
ᎏ ϫ 14.40°C ϭ 3.012 ϫ 103 J
The total amount of heat added was 3.358 kJ or 3.358 ϫ 103 J. The difference between these
heat values is the amount of heat absorbed by the calorimeter.
__? J ϭ 3.358 ϫ 103 J Ϫ 3.012 ϫ 103 J ϭ 0.346 ϫ 103 J, or 346 J absorbed by calorimeter
To obtain the heat capacity of the calorimeter, we divide the amount of heat absorbed by the
calorimeter, 346 J, by its temperature change.
__? ᎏ
°
J
C
ᎏ ϭ ᎏ
14
3
.
4
4
6
0°
J
C
ᎏ ϭ 24.0 J/°C
The calorimeter absorbs 24.0 J of heat for each degree Celsius increase in its temperature.
You should now work Exercise 58.
Figure 15-3 A coffee-cup calorimeter. The
stirring rod is moved up and down to ensure
thorough mixing and uniform heating of the
solution during reaction. The polystyrene walls
and top provide insulation so that very little heat
escapes. This kind of calorimeter measures qp,
the heat transfer due to a reaction occurring at
constant pressure.
15-4 Calorimetry 597
Thermometer
Polystyrene
cover
Polystyrene
cups
Glass
stirring rod
Water
Beaker
One way to add heat is to use an
electric heater.
EXAMPLE 15-2 Heat Measurements Using a Calorimeter
A 50.0-mL sample of 0.400 M copper(II) sulfate solution at 23.35°C is mixed with 50.0 mL of
0.600 M sodium hydroxide solution, also at 23.35°C, in the coffee-cup calorimeter of Example
15-1. After the reaction occurs, the temperature of the resulting mixture is measured to be
25.23°C. The density of the final solution is 1.02 g/mL. Calculate the amount of heat evolved.
Assume that the specific heat of the solution is the same as that of pure water, 4.184 J/gи°C.
CuSO4(aq) ϩ 2NaOH(aq) 88n Cu(OH)2(s) ϩ Na2SO4(aq)
Plan
The amount of heat released by the reaction is absorbed by the calorimeter and by the solution.
To find the amount of heat absorbed by the solution, we must know the mass of solution;
to find that, we assume that the volume of the reaction mixture is the sum of volumes of the
original solutions.
Solution
The mass of solution is
__? g soln ϭ (50.0 ϩ 50.0) mL ϫ ᎏ
1.02
m
g
L
soln
ᎏ ϭ 102 g soln
The amount of heat absorbed by the calorimeter plus the amount absorbed by the solution is
amount of heat amount of heat
absorbed by calorimeter absorbed by solution
__? J ϭ ᎏ
24
°
.
C
0 J
ᎏ ϫ (25.23 Ϫ 23.35)°C ϩ 102 g ϫ ᎏ
4
g
.
и
1
°
8
C
J
ᎏ ϫ (25.23 Ϫ 23.35)°C
ϭ 45 J ϩ 801 J ϭ 846 J absorbed by solution plus calorimeter
Thus, the reaction must have liberated 846 J, or 0.846 kJ, of heat.
You should now work Exercise 62(a).
THERMOCHEMICAL EQUATIONS
A balanced chemical equation, together with its value of ⌬H, is called a thermochemical
equation. For example,
C2H5OH(ᐉ) ϩ 3O2(g) 88n 2CO2(g) ϩ 3H2O(ᐉ) ϩ 1367 kJ
1 mol 3 mol 2 mol 3 mol
is a thermochemical equation that describes the combustion (burning) of one mole of
liquid ethanol at a particular temperature and pressure. The coefficients in such a description
must be interpreted as numbers of moles. Thus, 1367 kJ of heat is released when one
mole of C2H5OH(ᐉ) reacts with three moles of O2(g) to give two moles of CO2(g) and
three moles of H2O(ᐉ). We can refer to this amount of reaction as one mole of reaction,
which we abbreviate “mol rxn.” This interpretation allows us to write various unit factors
as desired.
, , and so on
2 mol CO2(g)
ᎏᎏ
1 mol rxn
1 mol C2H5OH(ᐉ)
ᎏᎏᎏ
1 mol rxn
15-5
When dilute aqueous solutions are mixed,
their volumes are very nearly additive.
598 CHAPTER 15: Chemical Thermodynamics




































See the Saunders Interactive
General Chemistry CD-ROM,
Screen 6.14, Enthalpy Change and ⌬H.
The heat released by the reaction of
HCl(aq) with NaOH(aq) causes the
temperature of the solution to rise.
We can also write the thermochemical equation as
C2H5OH(ᐉ) ϩ 3O2(g) 88n 2CO2(g) ϩ 3H2O(ᐉ) ⌬H ϭ Ϫ1367 kJ/mol rxn
The negative sign indicates that this is an exothermic reaction (i.e., it gives off heat).
We always interpret ⌬H as the enthalpy change for the reaction as written; that is,
as (enthalpy change)/(mole of reaction), where the denominator means “for the
number of moles of each substance shown in the balanced equation.”
We can then use several unit factors to interpret this thermochemical equation.
ϭ ϭ
ϭ ϭ
The reverse reaction would require the absorption of 1367 kJ under the same conditions;
1367 kJ ϩ 2CO2(g) ϩ 3H2O(ᐉ) 88n C2H5OH(ᐉ) ϩ 3O2(g)
That is, it is endothermic, with ⌬H ϭ ϩ1367 kJ.
2CO2(g) ϩ 3H2O(ᐉ) 88n C2H5OH(ᐉ) ϩ 3O2(g) ⌬H ϭ ϩ1367 kJ/mol rxn
It is important to remember the following conventions regarding thermochemical equa-
tions:
1. The coefficients in a balanced thermochemical equation refer to the numbers of
moles of reactants and products involved. In the thermodynamic interpretation
of equations we never interpret the coefficients as numbers of molecules. Thus, it
is acceptable to write coefficients as fractions rather than as integers, when neces-
sary.
2. The numerical value of ⌬H (or any other thermodynamic change) refers to the
number of moles of substances specified by the equation. This amount of change
of substances is called one mole of reaction, so we can express ⌬H in units of
energy/mol rxn. For brevity, the units of ⌬H are sometimes written kJ/mol or
even just kJ. No matter what units are used, be sure that you interpret the thermodynamic
change per mole of reaction for the balanced chemical equation to which
it refers. If a different amount of material is involved in the reaction, then the
⌬H (or other change) must be scaled accordingly.
3. The physical states of all species are important and must be specified. Heat is
given off or absorbed when phase changes occur, so different amounts of heat
could be involved in a reaction depending on the phases of reactants and prod-
ucts.
4. The value of ⌬H usually does not change significantly with moderate changes in
temperature.
1367 kJ given off
ᎏᎏᎏ
3 mol H2O(ᐉ) formed
1367 kJ given off
ᎏᎏᎏ
2 mol CO2(g) formed
1367 kJ given off
ᎏᎏᎏ
3 mol O2(g) consumed
1367 kJ given off
ᎏᎏᎏᎏ
mol C2H5OH(ᐉ) consumed
1367 kJ given off
ᎏᎏ
mol of reaction
15-5 Thermochemical Equations 599
C2H5OH(ᐉ) ϩ 3O2(g)
⌬H ϭ Ϫ1367 kJ
2CO2(g) ϩ 3H2O(ᐉ)
Before we know what a value of ⌬H
means, we must know the balanced
chemical equation to which it refers.
Increasingenthalpy
88888888888888888888888888n
8888888888888888888888nSee the Saunders Interactive
General Chemistry CD-ROM,
Screen 6.15, Enthalpy Changes for
Chemical Reactions.
EXAMPLE 15-3 Thermochemical Equations
When 2.61 grams of dimethyl ether, CH3OCH3, is burned at constant pressure, 82.5 kJ of
heat is given off. Find ⌬H for the reaction
CH3OCH3(ᐉ) ϩ 3O2(g) 88n 2CO2(g) ϩ 3H2O(ᐉ)
Plan
We scale the amount of heat given off in the experiment to correspond to the amount of
CH3OCH3 shown in the balanced equation.
Solution
ϭ ϫ ϫ
ϭ 1450 kJ/mol rxn
Because heat is given off, we know that the reaction is exothermic and the value of ⌬H is negative,
so
⌬H ϭ Ϫ1450 kJ/mol rxn
You should now work Exercise 19.
EXAMPLE 15-4 Thermochemical Equations
Write the thermochemical equation for the reaction in Example 15-2.
Plan
We must determine how much reaction occurred—that is, how many moles of reactants were
consumed. We first multiply the volume, in liters, of each solution by its concentration in
mol/L (molarity) to determine the number of moles of each reactant mixed. Then we identify
the limiting reactant. We scale the amount of heat released in the experiment to correspond
to the number of moles of that reactant shown in the balanced equation.
Solution
Using the data from Example 15-2,
__? mol CuSO4 ϭ 0.0500 L ϫ ϭ 0.0200 mol CuSO4
__? mol NaOH ϭ 0.0500 L ϫ ϭ 0.0300 mol NaOH
We determine which is the limiting reactant (see Section 3-3).
Required Ratio Available Ratio
ϭ ϭ
More CuSO4 is available than is required to react with the NaOH. Thus, 0.846 kJ of heat was
given off during the consumption of 0.0300 mol of NaOH. The amount of heat given off per
“mole of reaction” is
ϭ ϫ ϭ
56.4 kJ given off
ᎏᎏ
mol rxn
2 mol NaOH
ᎏᎏ
mol rxn
0.846 kJ given off
ᎏᎏᎏ
0.0300 mol NaOH
__? kJ released
ᎏᎏ
mol rxn
0.667 mol CuSO4
ᎏᎏ
1.00 mol NaOH
0.0200 mol CuSO4
ᎏᎏᎏ
0.0300 mol NaOH
0.50 mol CuSO4
ᎏᎏ
1.00 mol NaOH
1 mol CuSO4
ᎏᎏ
2 mol NaOH
0.600 mol NaOH
ᎏᎏ
1.00 L
0.400 mol CuSO4
ᎏᎏ
1.00 L
1 mol CH3OCH3
ᎏᎏ
mol rxn
46.0 g CH3OCH3
ᎏᎏᎏ
mol CH3OCH3
82.5 kJ given off
ᎏᎏᎏ
2.61 g CH3OCH3
? kJ given off
ᎏᎏ
mol rxn
NaOH is the limiting reactant.
600 CHAPTER 15: Chemical Thermodynamics
Thus, when the reaction occurs to the extent indicated by the balanced chemical equation, 56.4 kJ
is released. Remembering that exothermic reactions have negative values of ⌬Hrxn, we write
CuSO4(aq) ϩ 2NaOH(aq) 88n Cu(OH)2(s) ϩ Na2SO4(aq) ⌬Hrxn ϭ Ϫ56.4 kJ/mol rxn
You should now work Exercise 62(b).
EXAMPLE 15-5 Amount of Heat Versus Extent of Reaction
When aluminum metal is exposed to atmospheric oxygen (as in aluminum doors and windows),
it is oxidized to form aluminum oxide. How much heat is released by the complete oxidation
of 24.2 grams of aluminum at 25°C and 1 atm? The thermochemical equation is
4Al(s) ϩ 3O2(g) 88n 2Al2O3(s) ⌬H ϭ Ϫ3352 kJ/mol rxn
Plan
The thermochemical equation tells us that 3352 kJ of heat is released for every mole of reaction,
that is, for every 4 moles of Al that reacts. We convert 24.2 g of Al to moles, and then
calculate the number of kilojoules corresponding to that number of moles of Al, using the unit
factors
ᎏ
Ϫ
m
3
o
3
l
5
r
2
xn
kJ
ᎏ and ᎏ
1
4
m
m
o
o
l
l
r
A
xn
l
ᎏ
Solution
For 24.2 g Al,
__? kJ ϭ 24.2 g Al ϫ ᎏ
2
1
7
m
.0
o
g
l A
A
l
l
ᎏ ϫ ᎏ
1
4
m
m
o
o
l
l
r
A
xn
l
ᎏ ϫ ᎏ
Ϫ
m
3
o
3
l
5
r
2
xn
kJ
ᎏ ϭ Ϫ751 kJ
This tells us that 751 kJ of heat is released to the surroundings during the oxidation of 24.2
grams of aluminum.
You should now work Exercises 14 and 15.
Heat is released, so this is an
exothermic reaction.
15-6 Standard States and Standard Enthalpy Changes 601
STANDARD STATES AND STANDARD
ENTHALPY CHANGES
The thermodynamic standard state of a substance is its most stable pure form under
standard pressure (one atmosphere)* and at some specific temperature (25°C or 298 K
unless otherwise specified). Examples of elements in their standard states at 25°C are
hydrogen, gaseous diatomic molecules, H2(g); mercury, a silver-colored liquid metal,
15-6
Problem-Solving Tip: Mole of Reaction
Remember that a thermochemical equation can imply different numbers of moles of
different reactants or products. In Example 15-5 one mole of reaction also corresponds
to 3 moles of O2(g) and to 2 moles of Al2O3(s).
*IUPAC has changed the standard pressure from 1 atm to 1 bar. Because 1 bar is equal to 0.987 atm, the differences
in thermodynamic calculations are negligible except in work of very high precision. Many tables of thermodynamic
data are still based on a standard pressure of 1 atm, so we will use it in this book.
A temperature of 25°C is 77°F.
This is slightly above typical room
temperature. Notice that these
thermodynamic “standard conditions”
are not the same as the “standard
temperature and pressure (STP)” that
we used in gas calculations involving
standard molar volume (Chapter 12).
The sign tells us that heat was released,
but it would be grammatical nonsense
to say in words that “Ϫ751 kJ of heat
was released.” As an analogy, suppose
you give your friend $5. Your ⌬$ is
Ϫ$5, but in describing the transaction
you would not say “I gave her minus
five dollars,” but rather “I gave her five
dollars.”
If the substance exists in several
different forms, we take the form that
is most stable at 25°C and 1 atm as the
standard state.
602 CHAPTER 15: Chemical Thermodynamics
Hg(ᐉ); sodium, a silvery white solid metal, Na(s); and carbon, a grayish black solid called
graphite, C(graphite). We use C(graphite) instead of C(s) to distinguish it from other solid
forms of carbon, such as C(diamond). The reaction C(diamond) n C(graphite) would be
exothermic by 1.897 kJ/mol rxn; C(graphite) is thus more stable than C(diamond). Examples
of standard states of compounds include ethanol (ethyl alcohol or grain alcohol), a
liquid, C2H5OH(ᐉ); water, a liquid, H2O(ᐉ); calcium carbonate, a solid, CaCO3(s); and
carbon dioxide, a gas, CO2(g). Keep in mind the following conventions for thermochemical
standard states.
1. For a pure substance in the liquid or solid phase, the standard state is the pure
liquid or solid.
2. For a gas, the standard state is the gas at a pressure of one atmosphere; in a mixture
of gases, its partial pressure must be one atmosphere.
3. For a substance in solution, the standard state refers to one-molar concentration.
For ease of comparison and tabulation, we often refer to thermochemical or thermodynamic
changes “at standard states” or, more simply, to a standard change. To indicate a
change at standard pressure, we add a superscript zero. If some temperature other than
standard temperature of 25°C (298 K) is specified, we indicate it with a subscript; if no
subscript appears, a temperature of 25°C (298 K) is implied.
The standard enthalpy change, ⌬H 0
rxn, for reaction
reactants 88n products
refers to the ⌬H when the specified number of moles of reactants, all at standard
states, are converted completely to the specified number of moles of products, all at
standard states.
We allow a reaction to take place, with changes in temperature or pressure if necessary;
when the reaction is complete, we return the products to the same conditions of temperature
and pressure that we started with, keeping track of energy or enthalpy changes as we do
so. When we describe a process as taking place “at constant T and P,” we mean that the
initial and final conditions are the same. Because we are dealing with changes in state
functions, the net change is the same as the change we would have obtained hypothetically
with T and P actually held constant.
STANDARD MOLAR ENTHALPIES OF FORMATION, ⌬Hf
0
It is not possible to determine the total enthalpy content of a substance on an absolute
scale. We need to describe only changes in this state function, however, so we can define
an arbitrary scale as follows.
The standard molar enthalpy of formation, ⌬H f
0
, of a substance is the enthalpy
change for the reaction in which one mole of the substance in a specified state is
formed from its elements in their standard states. By convention, the ⌬H f
0
value for
any element in its standard state is zero.
15-7See the Saunders Interactive
General Chemistry CD-ROM,
Screen 6.17, Standard Enthalpy of
Formation.
We can think of ⌬H0
f as the enthalpy
content of each substance, in its
standard state, relative to the enthalpy
content of the elements, in their
standard states. This is why ⌬H0
f for an
element in its standard state is zero.
This is sometimes referred to as the
standard heat of reaction.
Standard molar enthalpy of formation is often called standard molar heat of formation
or, more simply, heat of formation. The superscript zero in ⌬H f
0
signifies standard
pressure, 1 atmosphere. Negative values for ⌬H f
0
describe exothermic formation reactions,
whereas positive values for ⌬Hf
0
describe endothermic formation reactions.
The enthalpy change for a balanced equation that gives a compound from its elements
does not necessarily give a molar enthalpy of formation for the compound. Consider the
following exothermic reaction at standard conditions.
H2(g) ϩ Br2(ᐉ) 88n 2HBr(g) ⌬H0
rxn ϭ Ϫ72.8 kJ/mol rxn
We see that two moles of HBr(g) are formed in the reaction as written. Half as much
energy, 36.4 kJ, is liberated when one mole of HBr(g) is produced from its constituent
elements in their standard states. For HBr(g), ⌬H f
0
ϭ Ϫ36.4 kJ/mol. This can be shown
by dividing all coefficients in the balanced equation by 2.
ᎏ1
2
ᎏH2(g) ϩ ᎏ1
2
ᎏBr2(ᐉ) 88n HBr(g) ⌬H 0
rxn ϭ Ϫ36.4 kJ/mol rxn
⌬H0
f HBr(g) ϭ Ϫ36.4 kJ/mol HBr(g)
Standard heats of formation of some common substances are tabulated in Table 15-1.
Appendix K contains a larger listing.
When referring to a thermodynamic quantity for a substance, we often omit the description
of the substance from the units. Units for tabulated ⌬Hf
0
values are given as “kJ/mol”;
we must interpret this as “per mole of the substance in the specified state.” For instance,
for HBr(g) the tabulated ⌬H f
0
value of Ϫ36.4 kJ/mol should be interpreted as .
EXAMPLE 15-6 Interpretation of ⌬Hf
0
The standard molar enthalpy of formation of ethanol, C2H5OH(ᐉ), is Ϫ277.7 kJ/mol. Write
the thermochemical equation for the reaction for which ⌬H0
rxn ϭ Ϫ277.7 kJ/mol rxn.
Ϫ36.4 kJ
ᎏᎏ
mol HBr(g)
The coefficients ᎏ1
2
ᎏ preceding H2(g) and
Br2(ᐉ) do not imply half a molecule of
each. In thermochemical equations, the
coefficients always refer to the number
of moles under consideration.
15-7 Standard Molar Enthalpies of Formation, ⌬Hf
0 603
TABLE 15-1 Selected Standard Molar Enthalpies of Formation at 298 K
Substance ⌬Hf
0
(kJ/mol) Substance ⌬Hf
0
(kJ/mol)
Br2(ᐉ) 0 HgS(s) red Ϫ58.2
Br2(g) 30.91 H2(g) 0
C(diamond) 1.897 HBr(g) Ϫ36.4
C(graphite) 0 H2O(ᐉ) Ϫ285.8
CH4(g) Ϫ74.81 H2O(g) Ϫ241.8
C2H4(g) 52.26 NO(g) 90.25
C6H6(ᐉ) 49.03 Na(s) 0
C2H5OH(ᐉ) Ϫ277.7 NaCl(s) Ϫ411.0
CO(g) Ϫ110.5 O2(g) 0
CO2(g) Ϫ393.5 SO2(g) Ϫ296.8
CaO(s) Ϫ635.5 SiH4(g) 34.0
CaCO3(s) Ϫ1207.0 SiCl4(g) Ϫ657.0
Cl2(g) 0 SiO2(s) Ϫ910.9
The ⌬H0
f values of Br2(g) and
C(diamond) are not equal to 0 at 298 K.
The standard states of these elements
are Br2(ᐉ) and C(graphite),
respectively.
Plan
The definition of ⌬H0
f of a substance refers to a reaction in which one mole of the substance is
formed. We put one mole of C2H5OH(ᐉ) on the right side of the chemical equation and put
the appropriate elements in their standard states on the left. We balance the equation without
changing the coefficient of the product, even if we must use fractional coefficients on the left.
Solution
2C(graphite) ϩ 3H2(g) ϩ ᎏ1
2
ᎏO2(g) 88n C2H5OH(ᐉ) ⌬H ϭ Ϫ277.7 kJ/mol rxn
You should now work Exercise 26.
As an analogy, consider traveling from
Kansas City (elevation 884 ft above sea
level) to Denver (elevation 5280 ft).
The change in elevation is (5280 Ϫ
884) ft ϭ 4396 ft, regardless of the
route taken.
604 CHAPTER 15: Chemical Thermodynamics
HESS’S LAW
In 1840, G. H. Hess (1802–1850) published his law of heat summation, which he derived
on the basis of numerous thermochemical observations.
The enthalpy change for a reaction is the same whether it occurs by one step or by
any series of steps.
Enthalpy is a state function. Its change is therefore independent of the pathway by which
a reaction occurs. We do not need to know whether the reaction does, or even can, occur
15-8
Problem-Solving Tip: How Do We Interpret Fractional Coefficients?
Remember that we always interpret the coefficients in thermochemical equations as
numbers of moles of reactants or products. The ᎏ1
2
ᎏO2(g) in the answer to Example 15-6
refers to ᎏ1
2
ᎏ mole of O2 molecules, or
ᎏ1
2
ᎏ mol O2 ϫ ᎏ
3
m
2.0
ol
g
O
O
2
2
ᎏ ϭ 16.0 g O2
It is important to realize that this is not the same as one mole of O atoms (though that
would also weigh 16.0 g).
Similarly, the fractional coefficients in
ᎏ1
2
ᎏH2(g) ϩ ᎏ1
2
ᎏBr2(ᐉ) 88n HBr(g)
refer to
ᎏ1
2
ᎏ mol H2 ϫ ϭ 1.0 g H2
and
ᎏ1
2
ᎏ mol Br2 ϫ ϭ 79.9 g Br2
respectively.
159.8 g Br2
ᎏᎏ
mol Br2
2.0 g H2
ᎏ
mol H2
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 6.16, Hess’s Law.
by the series of steps used in the calculation. The steps must (if only “on paper”) result
in the overall reaction. Hess’s Law lets us calculate enthalpy changes for reactions for
which the changes could be measured only with difficulty, if at all. In general terms, Hess’s
Law of heat summation may be represented as
⌬H0
rxn ϭ ⌬Ha
0 ϩ ⌬H b
0
ϩ ⌬H c
0 ϩ и и и
Here a, b, c, . . . refer to balanced thermochemical equations that can be summed to give
the equation for the desired reaction.
Consider the following reaction.
C(graphite) ϩ ᎏ1
2
ᎏO2(g) 88n CO(g) ⌬H0
rxn ϭ __?
The enthalpy change for this reaction cannot be measured directly. Even though CO(g)
is the predominant product of the reaction of graphite with a limited amount of O2(g),
some CO2(g) is always produced as well. The following reactions do go to completion
with excess O2(g); therefore, ⌬H0 values have been measured experimentally for them.
[Pure CO(g) is readily available.]
C(graphite) ϩ O2(g) 88n CO2(g) ⌬H0
rxn ϭ Ϫ393.5 kJ/mol rxn (1)
CO(g) ϩ ᎏ1
2
ᎏO2(g) 88n CO2(g) ⌬H0
rxn ϭ Ϫ283.0 kJ/mol rxn (2)
We can “work backward” to find out how to combine these two known equations to obtain
the desired equation. We want one mole of CO on the right, so we reverse equation (2)
[designated below as (Ϫ2)]; heat is then absorbed instead of released, so we must change
the sign of its ⌬H0 value. Then we add it to equation (1), canceling equal numbers of
moles of the same species on each side. This gives the equation for the reaction we want.
Adding the corresponding enthalpy changes gives the enthalpy change we seek.
⌬H0
C(graphite) ϩ O2(g) 88n CO2(g) Ϫ393.5 kJ/mol rxn) (1)
CO2(g) 88n CO(g) ϩ ᎏ1
2
ᎏO2(g) Ϫ(Ϫ283.0 kJ/mol rxn) (Ϫ2)
C(graphite) ϩ ᎏ1
2
ᎏO2(g) 88n CO(g) ⌬H0
rxn ϭ Ϫ110.5 kJ/mol rxn)
This equation shows the formation of one mole of CO(g) in its standard state from
the elements in their standard states. In this way, we determine that ⌬H0
f for CO(g) is
Ϫ110.5 kJ/mol.
EXAMPLE 15-7 Combining Thermochemical Equations: Hess’s Law
Use the thermochemical equations shown here to determine ⌬H0
rxn at 25°C for the following
reaction.
C(graphite) ϩ 2H2(g) 88n CH4(g)
⌬H0
C(graphite) ϩ O2(g) 88n CO2(g) Ϫ393.5 kJ/mol rxn (1)
H2(g) ϩ ᎏ1
2
ᎏO2(g) 88n H2O(ᐉ) Ϫ285.8 kJ/mol rxn (2)
CH4(g) ϩ 2O2(g) 88n CO2(g) ϩ 2H2O(ᐉ) Ϫ890.3 kJ/mol rxn (3)
You are familiar with the addition and
subtraction of algebraic equations.
This method of combining
thermochemical equations is
analogous.
15-8 Hess’s Law 605
These are combustion reactions, for
which ⌬H0
rxn values can be readily
determined from calorimetry
experiments.
88n8
88888888888888n
C(graphite) ϩ O2(g)
Ϫ110.5 kJ
CO(g) ϩ ᎏ1
2
ᎏO2(g)
Ϫ393.5 kJ
Ϫ283.0 kJ
CO2(g)
Above is a schematic representation of
the enthalpy changes for the reaction
C(graphite) ϩ ᎏ1
2
ᎏO2(g) n CO(g). The
⌬H value for each step is based on the
number of moles of each substance
indicated.
888n8888
Plan
(i) We want one mole of C(graphite) as reactant, so we write down equation (1).
(ii) We want two moles of H2(g) as reactants, so we multiply equation (2) by 2 [designated
below as 2 ϫ (2)].
(iii) We want one mole of CH4(g) as product, so we reverse equation (3) to give (Ϫ3).
(iv) We do the same operations on each ⌬H0 value.
(v) Then we add these equations term by term. The result is the desired thermochemical
equation, with all unwanted substances canceling. The sum of the ⌬H0 values is the
⌬H0 for the desired reaction.
Solution
⌬H0
C(graphite) ϩ O2(g) 88n CO2(g) Ϫ393.5 kJ/mol rxn) (1)
2H2(g) ϩ O2(g) 88n 2H2O(ᐉ) 2(Ϫ285.8 kJ/mol rxn) 2 ϫ (2)
CO2(g) ϩ 2H2O(ᐉ) 88n CH4(g) ϩ 2O2(g) ϩ890.3 kJ/mol rxn) (Ϫ3)
C(graphite) ϩ 2H2(g) 88n CH4(g) ⌬H0
rxn ϭ Ϫ74.8 kJ/mol rxn)
CH4(g) cannot be formed directly from C(graphite) and H2(g), so its ⌬H 0
f value cannot be
measured directly. The result of this example tells us that this value is Ϫ74.8 kJ/mol.
EXAMPLE 15-8 Combining Thermochemical Equations: Hess’s Law
Given the following thermochemical equations, calculate the heat of reaction at 298 K for the
reaction of ethylene with water to form ethanol.
C2H4(g) ϩ H2O(ᐉ) 88n C2H5OH(ᐉ)
⌬H0
C2H5OH(ᐉ) ϩ 3O2(g) 88n 2CO2(g) ϩ 3H2O(ᐉ) Ϫ1367 kJ/mol rxn (1)
C2H4(g) ϩ 3O2(g) 88n 2CO2(g) ϩ 2H2O(ᐉ) Ϫ1411 kJ/mol rxn (2)
Plan
We reverse equation (1) to give (Ϫ1); when the equation is reversed, the sign of ⌬H0 is
changed because the reverse of an exothermic reaction is endothermic. Then we add it to
equation (2).
Solution
⌬H0
2CO2(g) ϩ 3H2O(ᐉ) 88n C2H5OH(ᐉ) ϩ 3O2(g) ϩ1367 kJ/mol rxn (Ϫ1)
C2H4(g) ϩ 3O2(g) 88n 2CO2(g) ϩ 2H2O(ᐉ) Ϫ1411 kJ/mol rxn (2)
C2H4(g) ϩ H2O(ᐉ) 88n C2H5OH(ᐉ) ⌬H0
rxn ϭ Ϫ44 kJ/mol rxn
You should now work Exercises 30 and 32.
We have used a series of reactions
for which ⌬H0 values can be easily
measured to calculate ⌬H0 for a
reaction that cannot be carried out.
606 CHAPTER 15: Chemical Thermodynamics
Another interpretation of Hess’s Law lets us use tables of ⌬H0
f values to calculate the
enthalpy change for a reaction. Let us consider again the reaction of Example 15-8.
C2H4(g) ϩ H2O(ᐉ) 88n C2H5OH(ᐉ)
A table of ⌬H0
f values (Appendix K) gives ⌬H0
f C2H5OH(ᐉ) ϭ Ϫ277.7 kJ/mol,
⌬H0
f C2H4(g) ϭ 52.3 kJ/mol, and ⌬H0
f H2O(ᐉ) ϭ Ϫ285.8 kJ/mol. We may express this information
in the form of the following thermochemical equations.
⌬H0
2C(graphite) ϩ 3H2(g) ϩ ᎏ1
2
ᎏO2(g) 88n C2H5OH(ᐉ) Ϫ277.7 kJ/mol rxn (1)
2C(graphite) ϩ 2H2(g) 88n C2H4(g) 52.3 kJ/mol rxn (2)
H2(g) ϩ ᎏ1
2
ᎏO2(g) 88n H2O(ᐉ) Ϫ285.8 kJ/mol rxn (3)
We may generate the equation for the desired net reaction by adding equation (1) to the
reverse of equations (2) and (3). The value of ⌬H0 for the desired reaction is then the
sum of the corresponding ⌬H0 values.
The capital Greek letter sigma (⌺) is
read “the sum of.” The ⌺ n means that
the ⌬H0
f value of each product and
reactant must be multiplied by its
coefficient, n, in the balanced equation.
The resulting values are then added.
15-8 Hess’s Law 607
Problem-Solving Tip: ⌬Hf
0 Refers to Specific Reaction
The ⌬H0 for the reaction in Example 15-8 is Ϫ44 kJ for each mole of C2H5OH(ᐉ)
formed. This reaction, however, does not involve formation of C2H5OH(ᐉ) from its
constituent elements; therefore, ⌬H0
rxn is not ⌬H0
f for C2H5OH(ᐉ). We have seen the
reaction for ⌬H0
f of C2H5OH(ᐉ) in Example 15-6.
Similarly, the ⌬H0
rxn for
CO(g) ϩ ᎏ1
2
ᎏH2(g) 88n CO2(g)
is not ⌬H0
f for CO2(g).
We see that ⌬H0 for this reaction is given by
⌬H0
rxn ϭ ⌬H0
(1) ϩ ⌬H0
(Ϫ2) ϩ ⌬H0
(Ϫ3)
or by
product reactants
⌬H0
rxn ϭ ⌬H0
f C2H5OH(ᐉ) Ϫ [⌬H0
f C2H4(g) ϩ ⌬H0
f H2O(ᐉ)]
In general terms this is a very useful form of Hess’s Law.
⌬H0
rxn ϭ ͚ n ⌬H0
f products Ϫ ͚ n ⌬H0
f reactants
The standard enthalpy change of a reaction is equal to the sum of the standard molar
enthalpies of formation of the products, each multiplied by its coefficient, n, in the
balanced equation, minus the corresponding sum of the standard molar enthalpies of
formation of the reactants.
⌬H0
2C(graphite) ϩ 3H2(g) ϩ ᎏ1
2
ᎏO2(g) 88n C2H5OH(ᐉ) Ϫ277.7 kJ/mol rxn (1)
C2H4(g) 88n 2C(graphite) ϩ 2H2(g) Ϫ52.3 kJ/mol rxn (Ϫ2)
H2O(ᐉ) 88n H2(g) ϩ ᎏ1
2
ᎏO2(g) ϩ285.8 kJ/mol rxn (Ϫ3)
net rxn: C2H4(g) ϩ H2O(ᐉ) 88n C2H5OH(ᐉ) ⌬H0
rxn ϭ Ϫ44.2 kJ/mol rxn
b
o
m8
g
In effect this form of Hess’s Law supposes that the reaction occurs by converting reactants
to the elements in their standard states, then converting these to products (Figure
15-4). Few, if any, reactions actually occur by such a pathway. Nevertheless, the ⌬H0 for
this hypothetical pathway for reactants n products would be the same as that for any other
pathway—including the one by which the reaction actually occurs.
EXAMPLE 15-9 Using ⌬Hf
0 Values: Hess’s Law
Calculate ⌬H0
rxn for the following reaction at 298 K.
SiH4(g) ϩ 2O2(g) 88n SiO2(s) ϩ 2H2O(ᐉ)
Plan
We apply Hess’s Law in the form ⌬H0
rxn ϭ ͚ n ⌬H0
f products Ϫ ͚ n ⌬H0
f reactants, so we use the
⌬Hf
0
values tabulated in Appendix K.
Solution
We can first list the ⌬H0
f values we obtain from Appendix K:
SiH4(g) O2(g) SiO2(g) H2O(ᐉ)
⌬H0
f, kJ/mol: 34.3 0 Ϫ910.9 Ϫ285.8
⌬H0
rxn ϭ ͚ n ⌬H0
f products Ϫ ͚ n ⌬H0
f reactants
⌬H0
rxn ϭ [⌬H0
f SiO2(s) ϩ 2 ⌬H0
f H2O(ᐉ)] Ϫ [⌬H0
f SiH4(g) ϩ 2 ⌬H0
f O2(g)]
⌬H0
rxn ϭ ΄ ϫ ϩ ϫ ΅
Ϫ ΄ ϫ ϩ ϫ ΅
⌬H0
rxn ϭ Ϫ1515.7 kJ/mol rxn
You should now work Exercise 36.
Each term in the sums on the right-hand side of the solution in Example 15-9 has the
units
ϫ or
For brevity, we shall omit units in the intermediate steps of calculations of this type,
and just assign the proper units to the answer. Be sure that you understand how
these units arise.
Suppose we measure ⌬H0
rxn at 298 K and know all but one of the ⌬H0
f values for reactants
and products. We can then calculate the unknown ⌬H0
f value.
kJ
ᎏ
mol rxn
kJ
ᎏᎏ
mol substance
mol substance
ᎏᎏ
mol rxn
0 kJ
ᎏᎏ
mol O2(g)
2 mol O2(g)
ᎏᎏ
mol rxn
ϩ34.3 kJ
ᎏᎏ
mol SiH4(g)
1 mol SiH4(g)
ᎏᎏ
mol rxn
Ϫ285.8 kJ
ᎏᎏ
mol H2O(ᐉ)
2 mol H2O(ᐉ)
ᎏᎏ
mol rxn
Ϫ910.9 kJ
ᎏᎏ
mol SiO2(s)
1 mol SiO2(s)
ᎏᎏ
mol rxn
O2(g) is an element in its standard
state, so its ⌬H0
f is zero.
608 CHAPTER 15: Chemical Thermodynamics
Figure 15-4 A schematic
representation of Hess’s Law. The
red arrow represents the direct path
from reactants to products. The
series of blue arrows is a path
(hypothetical) in which reactants are
converted to elements, and they in
turn are converted to products—all
in their standard states.
Ϫ⌺n⌬H0
f reactants
⌺n⌬H0
⌬H0
f products
rxn
Elements
Reactants
Products
EXAMPLE 15-10 Using ⌬Hf
0 Values: Hess’s Law
Use the following information to determine ⌬Hf
0
for PbO(s, yellow).
PbO(s, yellow) ϩ CO(g) 88n Pb(s) ϩ CO2(g) ⌬H0
rxn ϭ Ϫ65.69 kJ
⌬H0
f for CO2(g) ϭ Ϫ393.5 kJ/mol and ⌬H0
f for CO(g) ϭ Ϫ110.5 kJ/mol
Plan
We again use Hess’s Law in the form ⌬H0
rxn ϭ ͚ n ⌬H0
f products Ϫ ͚ n ⌬H0
f reactants. The standard
state of lead is Pb(s), so ⌬H0
f Pb(s) ϭ 0 kJ/mol. Now we are given ⌬H0
rxn and the ⌬Hf
0
values for all substances except PbO(s, yellow). We can solve for this unknown.
Solution
We list the known ⌬H0
f values:
PbO(s, yellow) CO(g) Pb(s) CO2(g),
⌬H0
f, kJ/mol: ⌬H0
f PbO2(s, yellow) Ϫ110.5 0 Ϫ393.5
⌬H0
rxn ϭ ͚ n ⌬H0
f products Ϫ ͚ n ⌬H0
f reactants
⌬H0
rxn ϭ ⌬H0
f Pb(s) ϩ ⌬H0
f CO2(g) Ϫ [⌬H0
f PbO(s, yellow) ϩ ⌬H0
f CO(g)]
Ϫ65.69 ϭ 0 ϩ (Ϫ393.5) Ϫ [⌬H0
f PbO(s, yellow) ϩ (Ϫ110.5)]
Rearranging to solve for ⌬H0
f PbO(s, yellow), we have
⌬H0
f PbO(s, yellow) ϭ 65.69 Ϫ 393.5 ϩ 110.5 ϭ Ϫ217.3 kJ/mol of PbO
You should now work Exercise 42.
For all practical purposes, the bond
energy is the same as bond enthalpy.
Tabulated values of average bond
energies are actually average bond
enthalpies. We use the term “bond
energy” rather than “bond enthalpy”
because it is common practice to
do so.
15-9 Bond Energies 609
BOND ENERGIES
Chemical reactions involve the breaking and making of chemical bonds. Energy is always
required to break a chemical bond. Often this energy is supplied in the form of heat.
The bond energy (B.E.) is the amount of energy necessary to break one mole of
bonds in a gaseous covalent substance to form products in the gaseous state at
constant temperature and pressure.
The greater the bond energy, the more stable (stronger) the bond is, and the harder it is
to break. Thus bond energy is a measure of bond strengths.
15-9
We will consult Appendix K, only after
working the problem, to check the
answer.
Problem-Solving Tip: Remember the Values of ⌬Hf
0 for Elements
In Example 15-10, we were not given the value of ⌬H f
0
for Pb(s). We should know
without reference to tables that ⌬H f
0
for an element in its most stable form is exactly 0
kJ/mol. But the element must be in its most stable form. Thus, ⌬H f
0
for O2(g) is zero,
because ordinary oxygen is gaseous and diatomic. We would not assume that ⌬H f
0
would
be zero for oxygen atoms, O(g), or for ozone, O3(g). Similarly, ⌬H f
0
is zero for Cl2(g)
and for Br2(ᐉ), but not for Br2(g). Recall that bromine is one of the few elements that is
liquid at room temperature and 1 atm pressure.
Consider the following reaction.
H2(g) 88n 2H(g) ⌬H0
rxn ϭ ⌬HHXH ϭ ϩ436 kJ/mol HXH bonds
The bond energy of the hydrogen–hydrogen bond is 436 kJ/mol of bonds. In other words,
436 kJ of energy must be absorbed for every mole of HXH bonds that are broken. This
endothermic reaction (⌬H0
rxn is positive) can be written
H2(g) ϩ 436 kJ 88n 2H(g)
Some average bond energies are listed in Tables 15-2 and 15-3. We see from Table
15-3 that for any combination of elements, a triple bond is stronger than a double bond,
which in turn is stronger than a single bond. Bond energies for double and triple bonds
are not simply two or three times those for the corresponding single bonds. A single bond
is a ␴ bond, whereas double and triple bonds involve a combination of ␴ and ␲ bonding.
The bond energy measures the difficulty of overcoming the orbital overlap, and we should
not expect the strength of a ␲ bond to be the same as that of a ␴ bond between the same
two atoms.
We should keep in mind that each of the values listed is the average bond energy from
a variety of compounds. The average CXH bond energy is 413 kJ/mol of bonds. Average
We have discussed these changes in
terms of absorption or release of heat.
Another way of breaking bonds is by
absorption of light energy (Chapter 5).
Bond energies can be determined from
the energies of the photons that cause
bond dissociation.
610 CHAPTER 15: Chemical Thermodynamics
TABLE 15-2 Some Average Single Bond Energies (kJ/mol of bonds)
H C N O F Si P S Cl Br I
436 413 391 463 565 318 322 347 432 366 299 H
346 305 358 485 272 339 285 213 C
163 201 283 192 N
146 452 335 218 201 201 O
155 565 490 284 253 249 278 F
222 293 381 310 234 Si
201 326 184 P
226 255 S
242 216 208 Cl
193 175 Br
151 I
TABLE 15-3 Comparison of Some Average
Single and Multiple Bond
Energies (kJ/mol of bonds)
Single Bonds Double Bonds Triple Bonds
CXC 346 CUC 602 CmC 835
NXN 163 NUN 418 NmN 945
OXO 146 OUO 498
CXN 305 CUN 615 CmN 887
CXO 358 CUO 732* CmO 1072
*Except in CO2, where it is 799 kJ/mol.
CXH bond energies differ slightly from compound to compound, as in CH4, CH3Cl,
CH3NO2, and so on. Nevertheless, they are sufficiently constant to be useful in estimating
thermodynamic data that are not readily available by another approach. Values of ⌬H0
rxn
estimated in this way are not as reliable as those obtained from ⌬H0
f values for the
substances involved in the reaction.
A special case of Hess’s Law involves the use of bond energies to estimate heats of reaction.
Consider the enthalpy diagrams in Figure 15-5. In general terms, ⌬H0
rxn is related
to the bond energies of the reactants and products in gas phase reactions by the following
version of Hess’s Law.
⌬H0
rxn ϭ ͚ B.E.reactants Ϫ ͚ B.E.products in gas phase reactions only
The net enthalpy change of a reaction is the amount of energy required to break all
the bonds in reactant molecules minus the amount of energy required to break all the
bonds in product molecules. Stated in another way, the amount of energy released when
a bond is formed is equal to the amount absorbed when the same bond is broken. The
heat of reaction for a gas phase reaction can be described as the amount of energy released
in forming all the bonds in the products minus the amount of energy released in forming
all the bonds in the reactants (see Figure 15-5). This heat of reaction can be estimated
using the average bond energies in Tables 15-2 and 15-3.
The definition of bond energies is limited to the bond-breaking process only, and does
not include any provision for changes of state. Thus, it is valid only for substances in the
gaseous state. The calculations of this section therefore apply only when all substances in
the reaction are gases. If liquids or solids were involved, then additional information such
as heats of vaporization and fusion would be needed to account for phase changes.
Figure 15-5 A schematic representation of the relationship between bond energies and
⌬Hrxn for gas phase reactions. (a) For a general reaction (exothermic). (b) For the gas phase
reaction
H2(g) ϩ Br2(g) 88n 2HBr(g)
As usual for such diagrams, the value shown for each change refers to the number of moles
of substances or bonds indicated in the diagram.
⌬Hrxn ϭ Ϫ103 kJ
⌬HBrϪBr ϭ 193 kJ
Ϫ2⌬HHϪBr ϭ Ϫ732 kJ
⌬HHϪH ϭ 436 kJ
2HBr
2H ϩ 2Br
2H ϩ Br2
H2 ϩ Br2
Atoms (g)
Reactants (g)
(a) (b)
⌬Hrxn
Energy
released in
bond
formation
Energy
required to
break bonds
Products (g)
Note that this equation involves bond
energies of reactants minus bond
energies of products.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 9.10, Bond Energy and ⌬H0
rxn.
EXAMPLE 15-11 Bond Energies
Use the bond energies listed in Table 15-2 to estimate the heat of reaction at 298 K for the
following reaction.
N2(g) ϩ 3H2(g) 88n 2NH3(g)
Plan
Each NH3 molecule contains three NXH bonds, so two moles of NH3 contain six moles of
NXH bonds. Three moles of H2 contain a total of three moles of HXH bonds, and one mole
of N2 contains one mole of NmN bonds. Using the bond energy form of Hess’s Law,
Solution
⌬H0
rxn ϭ [⌬HNmN ϩ 3⌬HHXH] Ϫ [6⌬HNXH]
ϭ 945 ϩ 3(436) Ϫ 6(391) ϭ Ϫ93 kJ/mol rxn
You should now work Exercise 46.
EXAMPLE 15-12 Bond Energies
Use the bond energies listed in Table 15-2 to estimate the heat of reaction at 298 K for the
following reaction.
Plan
Two moles of CXC bonds and seven moles of CXH bonds are the same before and after reaction,
so we do not need to include them in the bond energy calculation. The only reactant
bonds that are broken are one mole of CXH bonds and one mole of ClXCl bonds. On the
product side, the only new bonds formed are one mole of CXCl bonds and one mole of HXCl
bonds. We need to take into account only the bonds that are different on the two sides of the
equation. As before, we add and subtract the appropriate bond energies, using values from
Table 15-2.
Solution
⌬H0
rxn ϭ [⌬HCXH ϩ ⌬HClXCl] Ϫ [⌬HCXCl ϩ ⌬HHXCl]
ϭ [413 ϩ 242] Ϫ [339 ϩ 432] ϭ Ϫ116 kJ/mol rxn
You should now work Exercises 48 and 50.
CHANGES IN INTERNAL ENERGY, ⌬E
The internal energy, E, of a specific amount of a substance represents all the energy
contained within the substance. It includes such forms as kinetic energies of the molecules;
energies of attraction and repulsion among subatomic particles, atoms, ions, or
molecules; and other forms of energy. The internal energy of a collection of molecules is
15-10
CH H
H
H
C Cl
Cl2(g) C3H7Cl(g) HCl(g)C3H8(g)
Cl
H
H
C
H
H
CH
H
H
C
H
H
C
H
H
ϩ
ϩ
H ClCl ϩ
ϩ
We would get the same value for
⌬H0
rxn if we used the full bond energy
form of Hess’s Law and assumed that
all bonds in reactants were broken and
then all bonds in products were
formed. In such a calculation the bond
energies for the unchanged bonds
would cancel. Why? Try it!
612 CHAPTER 15: Chemical Thermodynamics
N
H
H H
For each term in the sum, the units
are
ᎏ
m
m
ol
ol
bo
rx
n
n
ds
ᎏ ϫ ᎏ
mol
k
b
J
onds
ᎏ
Internal energy is a state function, so it
is represented by a capital letter.
a state function. The difference between the internal energy of the products and the
internal energy of the reactants of a chemical reaction or physical change, ⌬E, is given by
the equation
⌬E ϭ Efinal Ϫ Einitial ϭ Eproducts Ϫ Ereactants ϭ q ϩ w
The terms q and w represent heat and work, respectively. These are two ways in which
energy can flow into or out of a system. Work involves a change of energy in which a
body is moved through a distance, d, against some force, f; that is, w ϭ fd.
⌬E ϭ (amount of heat absorbed by system) ϩ (amount of work done on system)
The following conventions apply to the signs of q and w.
q is positive: Heat is absorbed by the system from the surroundings.
q is negative: Heat is released by the system to the surroundings.
w is positive: Work is done on the system by the surroundings.
w is negative: Work is done by the system on the surroundings.
Whenever a given amount of energy is added to or removed from a system, either as
heat or as work, the energy of the system changes by that same amount. Thus the equation
⌬E ϭ q ϩ w is another way of expressing the First Law of Thermodynamics (see
Section 15-1).
The only type of work involved in most chemical and physical changes is pressure–
volume work. From dimensional analysis we can see that the product of pressure and
volume is work. Pressure is the force exerted per unit area, where area is distance squared,
d2; volume is distance cubed, d3. Thus, the product of pressure and volume is force times
distance, which is work. An example of a physical change (a phase change) in which the
system expands and thus does work as it absorbs heat is shown in Figure 15-6. Even if
the weight of the book had not been present, the expanding system pushing against the
atmosphere would have done work for the expansion.
When energy is released by a reacting system, ⌬E is negative; energy can be written
as a product in the equation for the reaction. When the system absorbs energy from the
surroundings, ⌬E is positive; energy can be written as a reactant in the equation.
For example, the complete combustion of CH4 at constant volume at 25°C releases
energy.
Figure 15-6 A system that absorbs
heat and does work. (a) Some
powdered dry ice (solid CO2) is
placed into a flexible bag, which is
then sealed. (b) As the dry ice
absorbs heat from the surroundings,
some solid CO2 sublimes to form
gaseous CO2. The larger volume of
the gas causes the bag to expand.
The expanding gas does the work of
raising a book that has been placed
on the bag. Work would be done by
the same amount of expansion, even
if the book were not present, as the
bag pushes against the surrounding
atmosphere. The heat absorbed by
such a process at constant pressure,
qp, is equal to ⌬H for the process.
15-10 Changes in Internal Energy, ⌬E 613
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 6.13, The First Law of
Thermodynamics.
Sign conventions for q and w.
SYSTEM
q > 0
q < 0
w > 0
w < 0
SURROUNDINGS
Heat
released
by system
Work done
by system
Heat
absorbed
by system
Work done
on system
(b)(a)
ᎏ
d
F
2
ᎏ ϫ d3 ϭ Fd ϭ w
h
h
6
666
P V
CH4(g) ϩ 2O2(g) 88n CO2(g) ϩ 2H2O(ᐉ) ϩ 887 kJ
indicates release of energy
We can write the change in energy that accompanies this reaction as
CH4(g) ϩ 2O2(g) 88n CO2(g) ϩ 2H2O(ᐉ) ⌬E ϭ Ϫ887 kJ/mol rxn
As discussed in Section 15-2, the negative sign indicates a decrease in energy of the system,
or a release of energy by the system.
The reverse of this reaction absorbs energy. It can be written as
CO2(g) ϩ 2H2O(ᐉ) ϩ 887 kJ 88n CH4(g) ϩ 2O2(g)
indicates absorption of energy
or
CO2(g) ϩ 2H2O(ᐉ) 88n CH4(g) ϩ 2O2(g) ⌬E ϭ ϩ887 kJ/mol rxn
If the latter reaction could be forced to occur, the system would have to absorb 887 kJ of
energy per mole of reaction from its surroundings.
When a gas is produced against constant external pressure, such as in an open vessel
at atmospheric pressure, the gas does work as it expands against the pressure of the atmosphere.
If no heat is absorbed during the expansion, the result is a decrease in the internal
energy of the system. On the other hand, when a gas is consumed in a process, the atmosphere
does work on the reacting system.
Let us illustrate the latter case. Consider the complete reaction of a 2Ϻ1 mole ratio of
H2 and O2 to produce steam at some constant temperature above 100°C and at one atmosphere
pressure (Figure 15-7).
2H2(g) ϩ O2(g) 88n 2H2O(g) ϩ heat
Assume that the constant-temperature bath surrounding the reaction vessel completely
absorbs all the evolved heat so that the temperature of the gases does not change. The
volume of the system decreases by one third (3 mol gaseous reactants n 2 mol gaseous
At 25°C the change in internal energy
for the combustion of methane is
Ϫ887 kJ/mol CH4. The change in heat
content is Ϫ890 kJ/mol CH4 (see
Section 15-1). The small difference is
due to work done on the system as it is
compressed by the atmosphere.
614 CHAPTER 15: Chemical Thermodynamics
Figure 15-7 An illustration of the
one-third decrease in volume that
accompanies the reaction of H2 with
O2 at constant temperature. The
temperature is above 100°C. 2H2(g) + O2(g)
P = 1 atm P = 1 atm
2H2O(g)
2 mol H2
+
1 mol O2
2 mol
H2O(g)
Constant pressure,
T of gases > 100°C
Constant
temperature bath
(heated mineral oil)
products). The surroundings exert a constant pressure of one atmosphere and do work on
the system by compressing it. The internal energy of the system increases by an amount
equal to the amount of work done on it.
The work done on or by a system depends on the external pressure and the volume.
When the external pressure is constant during a change, the amount of work done is equal
to this pressure times the change in volume. The work done on a system equals ϪP ⌬V
or ϪP(V2 Ϫ V1).
V2 is the final volume, and V1 is the
initial volume.
15-10 Changes in Internal Energy, ⌬E 615
Compression (volume decreases) Expansion (volume increases)
Work is done by the surroundings on Work is done by the system on the
the system, so the sign of w is surroundings, so the sign of w is
positive negative
V2 is less than V1, so ⌬V ϭ (V2 Ϫ V1) is V2 is greater than V1, so ⌬V ϭ (V2 Ϫ V1)
negative is positive
w ϭ ϪP ⌬V is positive w ϭ ϪP ⌬V is negative
(Ϫ) ϫ (ϩ) ϫ (Ϫ) ϭ ϩ (Ϫ) ϫ (ϩ) ϫ (ϩ) ϭ Ϫ
Can be due to a decrease in number of Can be due to an increase in number of
moles of gas (⌬n negative) moles of gas (⌬n positive)
We substitute ϪP ⌬V for w in the equation ⌬E ϭ q ϩ w to obtain
⌬E ϭ q Ϫ P ⌬V
In constant-volume reactions, no P ⌬V work is done. Volume does not change, so
nothing “moves through a distance,” and d ϭ 0 and fd ϭ 0. The change in internal energy
of the system is just the amount of heat absorbed or released at constant volume, qv.
⌬E ϭ qv
Figure 15-8 shows the same phase change process as in Figure 15-6, but at constant volume
condition, so no work is done.
Solids and liquids do not expand or contract significantly when the pressure changes
(⌬V Ϸ 0). In reactions in which equal numbers of moles of gases are produced and
consumed at constant temperature and pressure, essentially no work is done. By the ideal
gas law, P ⌬V ϭ (⌬n)RT and ⌬n ϭ 0, where ⌬n equals the number of moles of gaseous
products minus the number of moles of gaseous reactants. Thus, the work term w has a
significant value at constant pressure only when there are different numbers of moles of
gaseous products and reactants so that the volume of the system changes.
EXAMPLE 15-13 Predicting the Sign of Work
For each of the following chemical reactions carried out at constant temperature and constant
pressure, predict the sign of w and tell whether work is done on or by the system. Consider the
reaction mixture to be the system.
(a) Ammonium nitrate, commonly used as a fertilizer, decomposes explosively.
2NH4NO3(s) 88n 2N2(g) ϩ 4H2O(g) ϩ O2(g)
This reaction was responsible for an explosion in 1947 that destroyed nearly the entire port of
Texas City, Texas, and killed 576 people.
(b) Hydrogen and chlorine combine to form hydrogen chloride gas.
H2(g) ϩ Cl2(g) 88n 2HCl(g)
Do not make the error of setting work
equal to V ⌬P.
A subscript v indicates a constantvolume
process; a subscript p indicates
a constant-pressure process.
Figure 15-8 A system that absorbs
heat at constant volume. Some dry
ice [CO2(s)] is placed into a rigid
flask, which is then sealed. As the
dry ice absorbs heat from the
surroundings, some CO2(s) sublimes
to form CO2(g). In contrast to the
case in Figure 15-6, this system
cannot expand (⌬V ϭ 0), so no work
is done, and the pressure in the flask
increases. Thus, the heat absorbed at
constant volume, qv, is equal to ⌬E
for the process.
(c) Sulfur dioxide is oxidized to sulfur trioxide, one step in the production of sulfuric acid.
2SO2(g) ϩ O2(g) 88n 2SO3(g)
Plan
For a process at constant pressure, w ϭ ϪP ⌬V ϭ Ϫ(⌬n)RT. For each reaction, we evaluate ⌬n,
the change in the number of moles of gaseous substances in the reaction.
⌬n ϭ (no. of moles of gaseous products) Ϫ (no. of moles of gaseous reactants)
Because both R and T (on the Kelvin scale) are positive quantities, the sign of w is opposite
from that of ⌬n; it tells us whether the work is done on (w ϭ ϩ) or by (w ϭ Ϫ) the system.
Solution
(a) ⌬n ϭ [2 mol N2(g) ϩ 4 mol H2O(g) ϩ 1 mol O2(g)] Ϫ 0 mol
ϭ 7 mol Ϫ 0 mol ϭ ϩ7 mol
⌬n is positive, so w is negative. This tells us that work is done by the system. The large
amount of gas formed by the reaction pushes against the surroundings (as happened with devastating
effect in the Texas City disaster).
(b) ⌬n ϭ [2 mol HCl(g)] Ϫ [1 mol H2(g) ϩ 1 mol Cl2(g)]
ϭ 2 mol Ϫ 2 mol ϭ 0 mol
Thus, w ϭ 0, and no work is done as the reaction proceeds. We can see from the balanced
equation that for every two moles (total) of gas that react, two moles of gas are formed, so the
volume neither expands nor contracts as the reaction occurs.
(c) ⌬n ϭ [2 mol SO3(g)] Ϫ [2 mol SO2(g) ϩ 1 mol O2(g)]
ϭ 2 mol Ϫ 3 mol ϭ Ϫ1 mol
⌬n is negative, so w is positive. This tells us that work is done on the system as the reaction
proceeds. The surroundings push against the diminishing volume of gas.
You should now work Exercises 75 and 76.
A bomb calorimeter is a device that measures the amount of heat evolved or absorbed
by a reaction occurring at constant volume (Figure 15-9). A strong steel vessel (the bomb)
is immersed in a large volume of water. As heat is produced or absorbed by a reaction
inside the steel vessel, the heat is transferred to or from the large volume of water. Thus,
only rather small temperature changes occur. For all practical purposes, the energy changes
associated with the reactions are measured at constant volume and constant temperature.
⌬n refers to the balanced equation.
616 CHAPTER 15: Chemical Thermodynamics
Here there are no gaseous reactants.
The decomposition of NH4NO3
produces large amounts of gas,
which expands rapidly as the very
fast reaction occurs. This explosive
reaction was the main cause of the
damage to New York’s World
Trade Center in 1993 and the
destruction of the Federal Building
in Oklahoma City (shown here) in
1995.
The “calorie content” of a food can
be determined by burning it in excess
oxygen inside a bomb calorimeter
and determining the heat released.
1 “nutritional Calorie” ϭ 1 kcal ϭ
4.184 kJ.
No work is done when a reaction is carried out in a bomb calorimeter, even if gases are
involved, because ⌬V ϭ 0. Therefore,
⌬E ϭ qv (constant volume)
EXAMPLE 15-14 Bomb Calorimeter
A 1.000-gram sample of ethanol, C2H5OH, was burned in a bomb calorimeter whose heat
capacity had been determined to be 2.71 kJ/°C. The temperature of 3000 grams of water rose
from 24.284°C to 26.225°C. Determine ⌬E for the reaction in joules per gram of ethanol, and
then in kilojoules per mole of ethanol. The specific heat of water is 4.184 J/gи°C. The combustion
reaction is
C2H5OH(ᐉ) ϩ 3O2(g) 88n 2CO2(g) ϩ 3H2O(ᐉ)
Plan
The amount of heat given off by the system (in the sealed compartment) raises the temperature
of the calorimeter and its water. The amount of heat absorbed by the water can be calculated
using the specific heat of water; similarly, we use the heat capacity of the calorimeter to find
the amount of heat absorbed by the calorimeter. The sum of these two amounts of heat is the
total amount of heat released by the combustion of 1.000 gram of ethanol. We must then scale
that result to correspond to one mole of ethanol.
Solution
The increase in temperature is
__? °C ϭ 26.225°C Ϫ 24.284°C ϭ 1.941°C rise
The amount of heat responsible for this increase in temperature of 3000 grams of water is
Heat to warm water ϭ 1.941°C ϫ ᎏ
4
g
.1
и
8
°
4
C
J
ᎏ ϫ 3000 g ϭ 2.436 ϫ 104 J ϭ 24.36 kJ
The amount of heat responsible for the warming of the calorimeter is
Heat to warm calorimeter ϭ 1.941°C ϫ ᎏ
2.7
°
1
C
kJ
ᎏ ϭ 5.26 kJ
The total amount of heat absorbed by the calorimeter and by the water is
Total amount of heat ϭ 24.36 kJ ϩ 5.26 kJ ϭ 29.62 kJ
Combustion of one gram of C2H5OH liberates 29.62 kJ of energy in the form of heat, that is
⌬E ϭ qv ϭ Ϫ29.62 kJ/g ethanol
The negative sign indicates that energy is released by the system to the surroundings. Now we
may evaluate ⌬E in kJ/mol of ethanol by converting grams of C2H5OH to moles.
ϭ ϫ ϭ Ϫ1365 kJ/mol ethanol
⌬E ϭ Ϫ1365 kJ/mol ethanol
This calculation shows that for the combustion of ethanol at constant temperature and constant
volume, the change in internal energy is Ϫ1365 kJ/mol ethanol.
You should now work Exercises 64 and 65.
46.07 g C2H5OH
ᎏᎏ
1 mol C2H5OH
Ϫ29.62 kJ
ᎏᎏ
g
__? kJ
ᎏᎏ
mol ethanol
Benzoic acid, C6H5COOH, is often
used to determine the heat capacity of
a calorimeter. It is a solid that can be
compressed into pellets. Its heat of
combustion is accurately known:
3227 kJ/mol benzoic acid, or 26.46 kJ/g
benzoic acid. Another way to measure
the heat capacity of a calorimeter is to
add a known amount of heat
electrically.
15-10 Changes in Internal Energy, ⌬E 617
Thermometer
Ignition
wires
Stirrer
Insulated
outer
container
Steel
container
Sample
dish
Steel
bomb
Water
You should look again at the
Saunders Interactive General
Chemistry CD-ROM, Screen 6.18,
Measuring Heats of Reaction:
Calorimetry.
Figure 15-9 A bomb calorimeter
measures qv, the amount of heat
given off or absorbed by a reaction
occurring at constant volume. The
amount of energy introduced via the
ignition wires is measured and taken
into account.
The balanced chemical equation involves one mole of ethanol, so we can write the unit
factor ᎏ
1 m
1
o
m
l
o
e
l
th
rx
a
n
nol
ᎏ. Then we express the result of Example 15-14 as
⌬E ϭ ᎏ
m
Ϫ
o
1
l
3
e
6
th
5
a
k
n
J
ol
ᎏ ϫ ᎏ
1 m
1
o
m
l
o
e
l
th
rx
a
n
nol
ᎏ ϭ Ϫ1365 kJ/mol rxn
RELATIONSHIP BETWEEN ⌬H AND ⌬E
The fundamental definition of enthalpy, H, is
H ϭ E ϩ PV
For a process at constant temperature and pressure,
⌬H ϭ ⌬E ϩ P ⌬V (constant T and P)
From Section 15-10, we know that ⌬E ϭ q ϩ w, so
⌬H ϭ q ϩ w ϩ P ⌬V (constant T and P)
15-11
(a) This small piece of candy is
mostly sucrose, C12H22O11, a sugar.
(b) When the piece of candy is
heated together with potassium
chlorate, KClO3 (a good oxidizing
agent), a highly product-favored
reaction occurs.
C12H22O11(s) ϩ 12O2(g) 88n
12CO2(g) ϩ 11H2O(g)
If that amount of sucrose is
completely metabolized to carbon
dioxide and water vapor in your
body, the same amount of energy
is released, though more slowly.
618 CHAPTER 15: Chemical Thermodynamics
(a)
(b)
At constant pressure, w ϭ ϪP ⌬V, so
⌬H ϭ q ϩ (ϪP ⌬V ) ϩ P ⌬V
⌬H ϭ qp (constant T and P)
The difference between ⌬E and ⌬H is the amount of expansion work (P ⌬V work) that
the system can do. Unless there is a change in the number of moles of gas present, this
difference is extremely small and can usually be neglected. For an ideal gas, PV ϭ nRT.
At constant temperature and constant pressure, P ⌬V ϭ (⌬n)RT, a work term. Substituting
gives
⌬H ϭ ⌬E ϩ (⌬n)RT or ⌬E ϭ ⌬H Ϫ (⌬n)RT (constant T and P)
As usual, ⌬n refers to the number of
moles of gaseous products minus the
number of moles of gaseous reactants.
15-11 Relationship of ⌬H and ⌬E 619
Problem-Solving Tip: Two Equations Relate ⌬H and ⌬E—Which
One Should Be Used?
The relationship ⌬H ϭ ⌬E ϩ P ⌬V is valid for any process that takes place at constant
temperature and pressure. It is very useful for physical changes that involve volume
changes, such as expansion or compression of a gas. When a chemical reaction occurs
and causes a change in the number of moles of gas, it is more convenient to use the relationship
in the form ⌬H ϭ ⌬E ϩ (⌬n)RT. You should always remember that ⌬n refers to
the change in number of moles of gas in the balanced chemical equation.
In Example 15-14 we found that the change in internal energy, ⌬E, for the combustion
of ethanol is Ϫ1365 kJ/mol ethanol at 298 K. Combustion of one mole of ethanol at
298 K and constant pressure releases 1367 kJ of heat. Therefore (see Section 15-5)
⌬H ϭ Ϫ1367 ᎏ
mol e
k
t
J
hanol
ᎏ
The difference between ⌬H and ⌬E is due to the work term, ϪP ⌬V or Ϫ(⌬n)RT. In this
balanced equation there are fewer moles of gaseous products than of gaseous reactants:
⌬n ϭ 2 Ϫ 3 ϭ Ϫ1.
C2H5OH(ᐉ) ϩ 3O2(g) 88n 2CO2(g) ϩ 3H2O(ᐉ)
Thus, the atmosphere does work on the system (compresses it). Let us find the work done
on the system per mole of reaction.
w ϭ ϪP ⌬V ϭ Ϫ(⌬n)RT
ϭ Ϫ(Ϫ1 mol) ΂ᎏ
8
m
.3
o
1
lи
4
K
J
ᎏ΃ (298 K) ϭ ϩ2.48 ϫ 103 J
w ϭ ϩ2.48 kJ or (⌬n)RT ϭ Ϫ2.48 kJ
We can now calculate ⌬E for the reaction from ⌬H and (⌬n)RT values.
⌬E ϭ ⌬H Ϫ (⌬n)RT ϭ [Ϫ1367 Ϫ (Ϫ2.48)] ϭ Ϫ1365 kJ/mol rxn
This value agrees with the result that we obtained in Example 15-14. The size of the work
term (ϩ2.48 kJ) is very small compared with ⌬H (Ϫ1367 kJ/mol rxn). This is true for
The positive sign is consistent with the
fact that work is done on the system.
The balanced equation involves one
mole of ethanol, so this is the amount
of work done when one mole of
ethanol undergoes combustion.
many reactions. Of course, if ⌬n ϭ 0, then ⌬H ϭ ⌬E, and the same amount of heat would
be absorbed or given off by the reaction whether it is carried out at constant pressure or
at constant volume.
SPONTANEITY OF PHYSICAL AND CHEMICAL CHANGES
Another major concern of thermodynamics is predicting whether a particular process can
occur under specified conditions to give predominantly products. We may summarize this
concern in the question “Which would be more stable at the given conditions—the reactants
or the products?” A change for which the collection of products is thermodynamically
more stable than the collection of reactants under the given conditions is said to be productfavored,
or spontaneous, under those conditions. A change for which the products are
thermodynamically less stable than the reactants under the given conditions is described as
reactant-favored, or nonspontaneous, under those conditions. Some changes are spontaneous
under all conditions; others are nonspontaneous under all conditions. The great
majority of changes, however, are spontaneous under some conditions but not under
others. We use thermodynamics to predict conditions for which the latter type of reactions
can occur to give predominantly products.
The concept of spontaneity has a very specific interpretation in thermodynamics. A
spontaneous chemical reaction or physical change is one that can happen without any
continuing outside influence. Any spontaneous change has a natural direction, like the
rusting of a piece of iron, the burning of a piece of paper, or the melting of ice at room
temperature. We can think of a spontaneous process as one for which products are favored
over reactants at the specified conditions. Although a spontaneous reaction might occur
rapidly, thermodynamic spontaneity is not related to speed. The fact that a process is spontaneous
does not mean that it will occur at an observable rate. It may occur rapidly, at a
moderate rate, or very slowly. The rate at which a spontaneous reaction occurs is addressed
by kinetics (Chapter 16). We now study the factors that influence spontaneity of a physical
or chemical change.
THE TWO ASPECTS OF SPONTANEITY
Many product-favored reactions are exothermic. For instance, the combustion (burning)
reactions of hydrocarbons such as methane and octane are all exothermic and highly
product-favored (spontaneous). The enthalpy contents of the products are lower than
those of the reactants. Not all exothermic changes are spontaneous, however, nor are all
spontaneous changes exothermic. As an example, consider the freezing of water, which is
an exothermic process (heat is released). This process is spontaneous at temperatures below
0°C, but it certainly is not spontaneous at temperatures above 0°C. Likewise, we can find
conditions at which the melting of ice, an endothermic process, is spontaneous. Spontaneity
is favored but not required when heat is released during a chemical reaction or a
physical change.
Another factor, related to the disorder of reactants and products, also plays a role in
determining spontaneity. The dissolution of ammonium nitrate, NH4NO3, in water is
spontaneous. Yet a beaker in which this process occurs becomes colder (see Figure 15-2).
15-12
620 CHAPTER 15: Chemical Thermodynamics
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 6.2, Product-Favored Systems,
and Screen 20.3, Directionality of
Reactions: Matter and Energy Dispersal.
The system (consisting of the water, the solid NH4NO3, and the resulting hydrated NH4
ϩ
and NO3
Ϫ ions) absorbs heat from the surroundings as the endothermic process occurs.
Nevertheless, the process is spontaneous because the system becomes more disordered as
the regularly arranged ions of crystalline ammonium nitrate become more randomly
distributed hydrated ions in solution (Figure 15-10). An increase in disorder in the system
favors the spontaneity of a reaction. In this particular case, the increase in disorder overrides
the effect of endothermicity.
Two factors affect the spontaneity of any physical or chemical change:
1. Spontaneity is favored when heat is released during the change (exothermic).
2. Spontaneity is favored when the change causes an increase in disorder.
The balance of these two effects is considered in Section 15-15.
THE SECOND LAW OF THERMODYNAMICS
We now know that two factors determine whether a reaction is spontaneous under a given
set of conditions. The effect of one factor, the enthalpy change, is that spontaneity is
favored (but not required) by exothermicity, and nonspontaneity is favored (but not
required) by endothermicity. The effect of the other factor is summarized by the Second
Law of Thermodynamics.
In spontaneous changes, the universe tends toward a state of greater disorder.
The Second Law of Thermodynamics is based on our experiences. Some examples
illustrate this law in the macroscopic world. When a mirror is dropped, it can shatter.
When a drop of food coloring is added to a glass of water, it diffuses until a homogeneously
colored solution results. When a truck is driven down the street, it consumes fuel
and oxygen, producing carbon dioxide, water vapor, and other emitted substances.
The reverse of any spontaneous change is nonspontaneous, because if it did occur, the
universe would tend toward a state of greater order. This is contrary to our experience.
We would be very surprised if we dropped some pieces of silvered glass on the floor and
a mirror spontaneously assembled. A truck cannot be driven along the street, even in
reverse gear, so that it sucks up CO2, water vapor, and other substances and produces fuel
and oxygen.
ENTROPY, S
The thermodynamic state function entropy, S, is a measure of the disorder of the system.
The greater the disorder of a system, the higher is its entropy. For any substance, the
particles are more highly ordered in the solid state than in the liquid state. These, in turn,
15-14
15-13
Figure 15-10 As particles leave a
crystal to go into solution, they
become more disordered. This
increase in disorder favors the
dissolution of the crystal.
15-14 Entropy, S 621
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 20.4, Entropy: Matter Dispersal
or Disorder.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 20.6, The Second Law of
Thermodynamics.
Suppose we shake a beaker
containing marbles. A disordered
arrangement (left) is more likely than
an ordered arrangement (right), in
which all marbles of the same color
remain together.
H2O
are more highly ordered than in the gaseous state. Thus, the entropy of any substance
increases as the substance goes from solid to liquid to gas (Figure 15-11).
If the entropy of a system increases during a process, the spontaneity of the process is
favored but not required. The Second Law of Thermodynamics says that the entropy of
the universe (not the system) increases during a spontaneous process, that is,
⌬Suniverse ϭ ⌬Ssystem ϩ ⌬Ssurroundings Ͼ 0 (spontaneous process)
Of the two ideal gas samples in Figure 15-12, the more ordered arrangement (Figure
15-12a) has lower entropy than the randomly mixed arrangement with the same volume
(Figure 15-12b). Because these ideal gas samples mix without absorbing or releasing heat
and without a change in total volume, they do not interact with the surroundings, so the
entropy of the surroundings does not change. In this case
⌬Suniverse ϭ ⌬Ssystem
If we open the stopcock between the two bulbs in Figure 15-12a, we expect the gases to
mix spontaneously, with an increase in the disorder of the system, that is, ⌬Ssystem is posi-
tive.
unmixed gases 88n mixed gases ⌬Suniverse ϭ ⌬Ssystem Ͼ 0
We do not expect the more homogeneous sample in Figure 15-12b to spontaneously
“unmix” to give the arrangement in Figure 15-12a (which would correspond to a decrease
in ⌬Ssystem).
mixed gases 88n unmixed gases ⌬Suniverse ϭ ⌬Ssystem Ͻ 0
The ideas of entropy, order, and disorder are related to probability. The more ways an
event can happen, the more probable that event is. In Figure 15-12b each individual red
molecule is equally likely to be in either container, as is each individual blue molecule. As
a result, there are many ways in which the mixed arrangement of Figure 15-12b can occur,
so the probability of its occurrence is high, and so its entropy is high. In contrast, there
is only one way the unmixed arrangement in Figure 15-12a can occur. The resulting probability
is extremely low, and the entropy of this arrangement is low.
The entropy of a system can decrease during a spontaneous process or increase during
a nonspontaneous process, depending on the accompanying ⌬Ssurr. If ⌬Ssys is negative
Figure 15-11 As a sample changes
from solid to liquid to gas, its
particles become increasingly less
ordered (more disordered), so its
entropy increases.
622 CHAPTER 15: Chemical Thermodynamics
Solid
Liquid Gas
Increasing disorder
Increasing entropy, S
(a) Stopcock closed
(b) Stopcock open
Figure 15-12 (a) A sample of gas
in which all molecules of one gas are
in one bulb and all molecules of the
other gas are in the other bulb. (b) A
sample of gas that contains the same
number of each kind of molecule
as in (a), but with the two kinds
randomly mixed in the two bulbs.
Sample (b) has greater disorder
(higher entropy), and is thus more
probable.
(decrease in disorder), then ⌬Suniv may still be positive (overall increase in disorder) if
⌬Ssurr is more positive than ⌬Ssys is negative. A refrigerator provides an illustration. It
removes heat from inside the box (the system) and ejects that heat, plus the heat generated
by the compressor, into the room (the surroundings). The entropy of the system
decreases because the air molecules inside the box move more slowly. The increase in the
entropy of the surroundings more than makes up for that, however, so the entropy of the
universe (refrigerator ϩ room) increases.
Similarly, if ⌬Ssys is positive but ⌬Ssurr is even more negative, then ⌬Suniv is still negative.
Such a process will be nonspontaneous.
Let’s consider the entropy changes that occur when a liquid solidifies at a temperature
below its freezing (melting) point (Figure 15-13a). ⌬Ssys is negative because a solid forms
from its liquid, yet we know that this is a spontaneous process. A liquid releases heat to
its surroundings (atmosphere) as it crystallizes. The released heat increases the motion
(disorder) of the molecules of the surroundings, so ⌬Ssurr is positive. As the temperature
decreases, the ⌬Ssurr contribution becomes more important. When the temperature is low
enough (below the freezing point), the positive ⌬Ssurr outweighs the negative ⌬Ssys. Then
⌬Suniv becomes positive, and the freezing process becomes spontaneous.
The situation is reversed when a liquid is boiled or a solid is melted (Figure 15-13b).
For example, at temperatures above its melting point, a solid spontaneously melts, and
⌬Ssys is positive. The heat absorbed when the solid (system) melts comes from its surroundings.
This decreases the motion of the molecules of the surroundings. Thus, ⌬Ssurr is
negative (the surroundings become less disordered). The positive ⌬Ssys is greater in magnitude
than the negative ⌬Ssurr, however, so ⌬Suniv is positive and the process is spontaneous.
Above the melting point, ⌬Suniv is positive for melting. Below the melting point, ⌬Suniv
is positive for freezing. At the melting point, ⌬Ssurr is equal in magnitude and opposite in
sign to ⌬Ssys. Then ⌬Suniv is zero for both melting and freezing; the system is at equilibrium.
Table 15-4 lists the entropy effects for these changes of physical state.
We have said that ⌬Suniv is positive for all spontaneous (product-favored) processes.
Unfortunately, it is not possible to make direct measurements of ⌬Suniv. Consequently,
entropy changes accompanying physical and chemical changes are reported in terms of
⌬Ssys. The subscript “sys” for system is usually omitted. The symbol ⌬S refers to the
change in enthalpy of the reacting system, just as ⌬H refers to the change in enthalpy of
the reacting system.
Can you develop a comparable table
for boiling (liquid n gas) and
condensation (gas n liquid)?
(Study Table 15-4 carefully.)
15-14 Entropy, S 623
We abbreviate these subscripts
as follows: system ϭ sys,
surroundings ϭ surr, and
universe ϭ univ.
Figure 15-13 A schematic representation of heat flow and entropy changes for (a) freezing
and (b) melting of a pure substance.
Heat
SURROUNDINGS
SYSTEM
Ssys ↓ so
∆Ssys < 0
Heat
Ssurr ↑ so ∆Ssurr > 0
Heat
SURROUNDINGS
SYSTEM
Ssys ↑ so
∆Ssys > 0
Heat
Ssurr ↓ so ∆Ssurr < 0
(a) Freezing below mp (b) Melting above mp
The Third Law of Thermodynamics establishes the zero of the entropy scale.
The entropy of a pure, perfect crystalline substance (perfectly ordered) is zero at
absolute zero (0 K).
As the temperature of a substance increases, the particles vibrate more vigorously, so
the entropy increases (Figure 15-14). Further heat input causes either increased temperature
(still higher entropy) or phase transitions (melting, sublimation, or boiling) that also
result in higher entropy. The entropy of a substance at any condition is its absolute
entropy, also called standard molar entropy. Consider the absolute entropies at 298 K
listed in Table 15-5. At 298 K, any substance is more disordered than if it were in a perfect
crystalline state at absolute zero, so tabulated S0
298 values for compounds and elements
are always positive. Notice especially that S0
298 of an element, unlike its ⌬H f
0
, is not equal
to zero. The reference state for absolute entropy is specified by the Third Law of TherEnthalpies
are measured only as
differences with respect to an arbitrary
standard state. Entropies, in contrast,
are defined relative to an absolute zero
level. In either case, the per mole
designation means per mole of substance
in the specified state.
624 CHAPTER 15: Chemical Thermodynamics
TABLE 15-4 Entropy Effects Associated with Melting and Freezing
Sign of (Magnitude of ⌬Ssys)
Compared with ⌬Suniv ‫؍‬
Change Temperature ⌬Ssys ⌬Ssurr (Magnitude of ⌬Ssurr) ⌬Ssys ؉ ⌬Ssurr Spontaneity
1. Melting (a) Ͼ mp ϩ Ϫ Ͼ Ͼ 0 Spontaneous
1. (solid n liquid) (b) ϭ mp ϩ Ϫ ϭ ϭ 0 Equilibrium
(c) Ͻ mp ϩ Ϫ Ͻ Ͻ 0 Nonspontaneous
2. Freezing (a) Ͼ mp Ϫ ϩ Ͼ Ͻ 0 Nonspontaneous
1. (liquid n solid) (b) ϭ mp Ϫ ϩ ϭ ϭ 0 Equilibrium
(c) Ͻ mp Ϫ ϩ Ͻ Ͼ 0 Spontaneous
Figure 15-14 (a) A simplified representation of a side view of a “perfect” crystal of a polar
substance of 0 K. Note the perfect alignment of the dipoles in all molecules in a perfect
crystal. This causes its entropy to be zero at 0 K. There are no perfect crystals, however,
because even the purest substances that scientists have prepared are contaminated by traces
of impurities that occupy a few of the positions in the crystal structure. Additionally, there
are some vacancies in the crystal structures of even very highly purified substances such as
those used in semiconductors (see Section 13-17). (b) A simplified representation of the
same “perfect” crystal at a temperature above 0 K. Vibrations of the individual molecules
within the crystal cause some dipoles to be oriented in directions other than those in a
perfect arrangement. The entropy of such a crystalline solid is greater than zero, because
there is disorder in the crystal.
+ – + – + – + – + –
+–+–+–+–+–
+ – + – + – + – + –
+–+–+–+–+–
+ – + – + – + – + –
+ – + – + –
+
– + –
+–+–+
–+–+
–
+ – + – + – + – +
–
+–+
–
+–
+
–+–
+ –
+
– + – + – + –
(b)(a)
TABLE 15-5 Absolute
Entropies at
298 K for a
Few Common
Substances
Substance S0 (J/molؒK)
C(diamond) 2.38
C(g) 158.0
H2O(ᐉ) 69.91
H2O(g) 188.7
I2(s) 116.1
I2(g) 260.6
modynamics. It is different from the reference state for ⌬Hf
0
(see Section 15-7). The
absolute entropies, S0
298, of various substances under standard conditions are tabulated in
Appendix K.
The standard entropy change, ⌬S0, of a reaction can be determined from the absolute
entropies of reactants and products. The relationship is analogous to Hess’s Law.
⌬S0
rxn ϭ ͚ n S0
products Ϫ ͚ n S0
reactants
S0 values are tabulated in units of J/molиK rather than the larger units involving kilojoules
that are used for enthalpy changes. The “mol” term in the units for a substance refers to
a mole of the substance, whereas for a reaction it refers to a mole of reaction. Each term
in the sums on the right-hand side of the equation has the units
ϫ ϭ
The result is usually abbreviated as J/molиK, or sometimes even as J/K. As before, we
will usually omit units in intermediate steps and then apply appropriate units to the result.
EXAMPLE 15-15 Calculation of ⌬S0
rxn
Use the values of standard molar entropies in Appendix K to calculate the entropy change at
25°C and one atmosphere pressure for the reaction of hydrazine with hydrogen peroxide. This
explosive reaction has been used for rocket propulsion. Do you think the reaction is spontaneous?
The balanced equation for the reaction is
N2H4(ᐉ) ϩ 2H2O2(ᐉ) 88n N2(g) ϩ 4H2O(g) ⌬H0
rxn ϭ Ϫ642.2 kJ/mol reaction
Plan
We use the equation for standard entropy change to calculate ⌬S0
rxn from the tabulated values
of standard molar entropies, S0
298, for the substances in the reaction.
Solution
We can list the S0
298 values that we obtain from Appendix K for each substance:
N2H4(ᐉ) H2O2(ᐉ) N2(g) H2O(g)
S0, J/molиK: 121.2 109.6 191.5 188.7
⌬S0
rxn ϭ ͚ n S0
products Ϫ ͚ n S0
reactants
ϭ [S0
N2(g) ϩ 4S0
H2O(g)] Ϫ [S0
N2H4(ᐉ) ϩ 2S0
H2O2(ᐉ)]
ϭ [1 (191.5) ϩ 4 (188.7)] Ϫ [1 (121.2) ϩ 2 (109.6)]
⌬S0
rxn ϭ ϩ605.9 J/molиK
The “mol” designation for ⌬S0
rxn refers to a mole of reaction, that is, one mole of N2H4(ᐉ),
two moles of H2O2(ᐉ), and so on. Although it may not appear to be, ϩ605.9 J/molиK is a relatively
large value of ⌬S0
sys. The positive entropy change favors spontaneity. This reaction is
also exothermic (⌬H0 is negative). As we shall see, this reaction must be spontaneous, because
both factors are favorable: the reaction is exothermic (⌬H0
rxn is negative) and the disorder of
the system increases (⌬S0
rxn is positive).
You should now work Exercise 92.
J
ᎏᎏ
(mol rxn)иK
J
ᎏᎏᎏ
(mol substance)иK
mol substance
ᎏᎏ
mol rxn
The ͚ n means that each S0 value must
be multiplied by the appropriate
coefficient, n, from the balanced
equation. These values are then added.
15-14 Entropy, S 625
Small booster rockets adjust the
course of a satellite in orbit. Some of
these small rockets are powered by
the N2H4–H2O2 reaction.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 20.5, Calculating ⌬S for a
Chemical Reaction.
Because changes in the thermodynamic quantity entropy may be understood in terms
of changes in molecular disorder, we can often predict the sign of ⌬Ssys. The following illustrations
emphasize several common types of processes that result in predictable entropy
changes for the system.
1. Phase changes. When melting occurs, the molecules or ions are taken from their quite
ordered crystalline arrangement to a more disordered one in which they are able to
move past one another in the liquid. Thus, a melting process is always accompanied
by an entropy increase (⌬Ssys Ͼ 0). Likewise, vaporization and sublimation both
take place with large increases in disorder, and hence with increases in entropy. For
the reverse processes of freezing, condensation, and deposition, entropy decreases
because order increases.
2. Temperature changes—for example, warming a gas from 25°C to 50°C. As any sample
is warmed, the molecules undergo more (random) motion; hence entropy increases
(⌬Ssys Ͼ 0) as temperature increases. Likewise, as we raise the temperature of a solid,
the particles vibrate more vigorously about their positions in the crystal, so that at
any instant there is a larger average displacement from their mean positions; this
results in an increase in entropy.
3. Volume changes. When the volume of a sample of gas increases, the molecules can
occupy more positions, and hence are more randomly arranged than when they are
closer together in a smaller volume. Hence, an expansion is accompanied by an
increase in entropy (⌬Ssys Ͼ 0). Conversely, as a sample is compressed, the molecules
are more restricted in their locations, and a situation of greater order (lower
entropy) results.
4. Mixing of substances, even without chemical reaction. Situations in which the molecules
are more “mixed up” are more disordered, and hence are at higher entropy.
We pointed out that the mixed gases of Figure 15-12b were more disordered than
the separated gases of Figure 15-12a, and that the former was a situation of higher
entropy. We see that mixing of gases by diffusion is a process for which ⌬Ssys Ͼ 0;
we know from experience that it is always spontaneous. We have already pointed
The vaporization of bromine,
Br2(ᐉ) n Br2(g) (left) and the
sublimation of iodine, I2(s) n I2(g)
(right) both lead to an increase in
disorder, so ⌬Ssys Ͼ 0 for each
process. Which do you think results
in the more positive ⌬S? Carry out
the calculation using values from
Appendix K to check whether your
prediction was correct.
626 CHAPTER 15: Chemical Thermodynamics
out (Section 14-2) that the increase in disorder (entropy increase) that accompanies
mixing often provides the driving force for solubility of one substance in another.
For example, when one mole of solid NaCl dissolves in water, NaCl(s) n NaCl(aq),
the entropy (Appendix K) increases from 72.4 J/molиK to 115.5 J/molиK, or
⌬S0 ϭ ϩ43.1 J/molиK. The term “mixing” can be interpreted rather liberally. For
example, the reaction H2(g) ϩ Cl2(g) n 2HCl(g) has ⌬S0 Ͼ 0; in the reactants, each
atom is bonded to an identical atom, a less “mixed-up” situation than in the products,
where unlike atoms are bonded together.
5. Increase in the number of particles, as in the dissociation of a diatomic gas such as
F2(g) n 2F(g). Any process in which the number of particles increases results in an
increase in entropy, ⌬Ssys Ͼ 0. Values of ⌬S0 calculated for several reactions of this
type are given in Table 15-6. As you can see, the ⌬S0 values for the dissociation
process X2 n 2X are all similar for X ϭ H, F, Cl, and N. Why is the value given
in Table 15-6 so much larger for X ϭ Br? This process starts with liquid Br2. The
total process Br2(ᐉ) n 2Br(g), for which ⌬S0 ϭ 197.5 J/molиK, can be treated as
the result of two processes. The first of these is vaporization, Br2(ᐉ) n Br2(g), for
which ⌬S0 ϭ 93.1 J/molиK. The second step is the dissociation of gaseous bromine,
Br2(g) n 2Br(g), for which ⌬S0 ϭ 104.4 J/molиK; this entropy increase is about the
same as for the other processes that involve only dissociation of a gaseous diatomic
species. Can you rationalize the even higher value given in the table for the process
I2(s) n 2I(g)?
When water, H2O, and propyl alcohol, CH3CH2CH2OH (left) are mixed to form a solution
(right), disorder increases. ⌬S Ͼ 0 for the mixing of any two molecular substances.
15-14 Entropy, S 627
TABLE 15-6 Entropy
Changes for
Some
Processes
X2 n 2X
Reaction ⌬S0 (J/molؒK)
H2(g) 88n 2H(g) 98.0
N2(g) 88n 2N(g) 114.9
O2(g) 88n 2O(g) 117.0
F2(g) 88n 2F(g) 114.5
Cl2(g) 88n 2Cl(g) 107.2
Br2(ᐉ) 88n 2Br(g) 197.5
I2(s) 88n 2I(g) 245.3
6. Changes in the number of moles of gaseous substances. Processes that result in an increase
in the number of moles of gaseous substances have ⌬Ssys Ͼ 0. Example 15-15 illustrates
this. There are no gaseous reactants, but the products include five moles of
gas. Conversely, we would predict that the process 2H2(g) ϩ O2(g) n 2H2O(g) has
a negative ⌬S0 value; here, three moles of gas is consumed while only two moles
is produced, for a net decrease in the number of moles in the gas phase. You
should be able to calculate the value of ⌬S0 for this reaction from the values in
Appendix K.
FREE ENERGY CHANGE, ⌬G, AND SPONTANEITY
Energy is the capacity to do work. If heat is released in a chemical reaction (⌬H is negative),
some of the heat may be converted into useful work. Some of it may be expended to
increase the order of the system (if ⌬S is negative). If a system becomes more disordered
(⌬S Ͼ 0), however, more useful energy becomes available than indicated by ⌬H alone.
J. Willard Gibbs (1839–1903), a prominent nineteenth-century American professor of
mathematics and physics, formulated the relationship between enthalpy and entropy in
terms of another state function that we now call the Gibbs free energy, G. It is defined
as
G ϭ H Ϫ TS
The Gibbs free energy change, ⌬G, at constant temperature and pressure, is
⌬G ϭ ⌬H Ϫ T ⌬S (constant T and P)
The amount by which the Gibbs free energy decreases is the maximum useful energy
obtainable in the form of work from a given process at constant temperature and pressure.
It is also the indicator of spontaneity of a reaction or physical change at constant T and
P. If there is a net decrease of useful energy, ⌬G is negative and the process is sponta-
15-15
(a) The entropy of an organism
decreases (unfavorable) when new
cells are formed. The energy to
sustain animal life is provided by the
metabolism of food. This energy is
released when the chemical bonds in
the food are broken. Exhalation
of gases and excretion of waste
materials increase the entropy of the
surroundings enough so that the
entropy of the universe increases
and the overall process can occur.
(b) Stored chemical energy can later
be transformed by the organism to
the mechanical energy for muscle
contraction, to the electrical energy
for brain function, or to another
needed form.
628 CHAPTER 15: Chemical Thermodynamics
(a) (b)
Do you think that the reaction
2H2(g) ϩ O2(g) 88n 2H2O(ᐉ)
would have a higher or lower value of
⌬S0 than when the water is in the gas
phase? Confirm by calculation.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 20.7, Gibbs Free Energy.
This is often called simply the Gibbs
energy change or the free energy change.
neous (product-favored). We see from the equation that ⌬G becomes more negative as
(1) ⌬H becomes more negative (the process gives off more heat) and (2) ⌬S becomes more
positive (the process results in greater disorder). If there is a net increase in free energy
of the system during a process, ⌬G is positive and the process is nonspontaneous
(reactant-favored). This means that the reverse process is spontaneous under the given
conditions. When ⌬G ϭ 0, there is no net transfer of free energy; both the forward and
reverse processes are equally favorable. Thus, ⌬G ϭ 0 describes a system at equilibrium.
The relationship between ⌬G and spontaneity may be summarized as follows.
In other words, are the reactants or
the products more stable at standard
conditions?
15-15 Free Energy Change, ⌬G, and Spontaneity 629
The free energy content of a system depends on temperature and pressure (and, for
mixtures, on concentrations). The value of ⌬G for a process depends on the states and
the concentrations of the various substances involved. It also depends strongly on temperature,
because the equation ⌬G ϭ ⌬H Ϫ T ⌬S includes temperature. Just as for other
thermodynamic variables, we choose some set of conditions as a standard state reference.
The standard state for ⌬G0 is the same as for ⌬H0—1 atm and the specified temperature,
usually 25°C (298 K). Values of standard molar free energy of formation, ⌬Gf
0
,
for many substances are tabulated in Appendix K. For elements in their standard states,
⌬Gf
0
ϭ 0. The values of ⌬Gf
0
may be used to calculate the standard free energy change of
a reaction at 298 K by using the following relationship.
⌬G0
rxn ϭ ͚ n ⌬G0
f products Ϫ ͚ n ⌬G0
f reactants (1 atm and 298 K only)
The value of ⌬G0
rxn allows us to predict the spontaneity of a very special hypothetical
reaction that we call the standard reaction.
In the standard reaction, the numbers of moles of reactants shown in the balanced
equation, all at standard conditions, are completely converted to the numbers of moles
of products shown in the balanced equation, all at standard conditions.
We must remember that it is ⌬G, and not ⌬G0, that is the general criterion for spontaneity.
⌬G depends on concentrations of reactants and products in the mixture. For most
reactions, there is an equilibrium mixture of reactants and products that is more stable than
either all reactants or all products. In Chapter 17 we will study the concept of equilibrium
and see how to find ⌬G for mixtures.
Spontaneity of Reaction
⌬G (constant T and P)
⌬G is positive Reaction is nonspontaneous (reactant-favored)
⌬G is zero System is at equilibrium
⌬G is negative Reaction is spontaneous (product-favored)
Ϫ 0 ϩ
⌬G Ͻ 0 ⌬G Ͼ 0
Reaction is spontaneous Reaction is not spontaneous
Product-favored reaction Reactant-favored reaction
Forward reaction is favored Reverse reaction is favored
EXAMPLE 15-16 Spontaneity of Standard Reaction
Diatomic nitrogen and oxygen molecules make up about 99% of all the molecules in reasonably
“unpolluted” dry air. Evaluate ⌬G0 for the following reaction at 298 K, using ⌬Gf
0
values
from Appendix K. Is the standard reaction spontaneous?
N2(g) ϩ O2(g) 88n 2NO(g) (nitrogen oxide)
Plan
The reaction conditions are 1 atm and 298 K, so we can use the tabulated values of ⌬G0
f for
each substance in Appendix K to evaluate ⌬G0
rxn in the preceding equation. The treatment of
units for calculation of ⌬G0 is the same as that for ⌬H0 in Example 15-9.
Solution
We obtain the following values of ⌬G0
f from Appendix K:
N2(g) O2(g) NO(g)
⌬G0
f, kJ/mol: 0 0 86.57
For the reverse reaction at 298 K,
⌬G0
rxn ϭ Ϫ173.1 kJ/mol. It is productfavored
but very slow at room
temperature. The NO formed in
automobile engines is oxidized to even
more harmful NO2 much more rapidly
than it decomposes to N2 and O2.
Thermodynamic spontaneity does not
guarantee that a process occurs at an
observable rate. The oxides of nitrogen
in the atmosphere represent a major
environmental problem.
630 CHAPTER 15: Chemical Thermodynamics
Problem-Solving Tip: Some Common Pitfalls in Calculating ⌬G0
rxn
Be careful of these points when you carry out calculations that involve ⌬G0:
1. The calculation of ⌬G0
rxn from tabulated values of ⌬G0
f is valid only if the reaction is
at 25°C (298 K) and one atmosphere.
2. Calculations with the equation ⌬G0 ϭ ⌬H0 Ϫ T ⌬S0 must be carried out with the
temperature in kelvins.
3. The energy term in ⌬S0 is usually in joules, whereas that in ⌬H0 is usually in kilojoules;
remember to convert one of these so that units are consistent before you
combine them.
⌬G0
rxn ϭ ͚ n ⌬G0
f products Ϫ ͚ n ⌬G0
f reactants
ϭ 2 ⌬G0
f NO(g) Ϫ [⌬G0
f N2(g) ϩ ⌬G0
f O2(g)]
ϭ 2(86.57) Ϫ [0 ϩ 0]
⌬G0
rxn ϭ ϩ173.1 kJ/mol rxn for the reaction as written
Because ⌬G0 is positive, the reaction is nonspontaneous at 298 K under standard state condi-
tions.
You should now work Exercise 99.
The value of ⌬G0 can also be calculated by the equation
⌬G0 ϭ ⌬H0 Ϫ T ⌬S0 (constant T and P)
Strictly, this last equation applies to standard conditions; however, ⌬H0 and ⌬S0 often do
not vary much with temperature, so the equation can often be used to estimate free energy
changes at other temperatures.
EXAMPLE 15-17 Spontaneity of Standard Reaction
Make the same determination as in Example 15-16, using heats of formation and absolute
entropies rather than free energies of formation.
Plan
First we calculate ⌬H0
rxn and ⌬S0
rxn. We use the relationship ⌬G0 ϭ ⌬H0 Ϫ T ⌬S0 to evaluate
the free energy change under standard state conditions at 298 K.
Solution
The values we obtain from Appendix K are:
N2(g) O2(g) NO(g)
⌬H0
f, kJ/mol: 0 0 90.25
S0, J/molиK: 191.5 205.0 210.7
⌬H0
rxn ϭ ͚ n ⌬H0
f products Ϫ ͚ n ⌬H0
f reactants
⌬H0
rxn ϭ 2 ⌬H0
f NO(g) Ϫ [⌬H0
f N2(g) ϩ ⌬H0
f O2(g)]
⌬H0
rxn ϭ [2(90.25) Ϫ (0 ϩ 0)] ϭ 180.5 kJ/mol
⌬S0
rxn ϭ ͚ nS0
products Ϫ ͚ nS0
reactants
⌬S0
rxn ϭ 2S0
NO(g) Ϫ [S0
N2(g) ϩ S0
O2(g)]
⌬S0
rxn ϭ [2 (210.7) Ϫ (191.5 ϩ 205.0)] ϭ 24.9 J/molиK ϭ 0.0249 kJ/molиK
Now we use the relationship ⌬G0 ϭ ⌬H0 Ϫ T ⌬S0, with T ϭ 298 K, to evaluate the free energy
change under standard state conditions at 298 K.
⌬G0
rxn ϭ ⌬H0
rxn Ϫ T ⌬S0
rxn
ϭ 180.5 kJ/mol Ϫ (298 K)(0.0249 kJ/molиK)
ϭ 180.5 kJ/mol Ϫ 7.42 kJ/mol
⌬G0
rxn ϭ ϩ173.1 kJ/mol rxn, the same value obtained in Example 15-16.
You should now work Exercise 100.
THE TEMPERATURE DEPENDENCE OF SPONTANEITY
The methods developed in Section 15-15 can also be used to estimate the temperature at
which a process is in equilibrium. When a system is at equilibrium, ⌬G ϭ 0. Thus,
⌬Grxn ϭ ⌬Hrxn Ϫ T ⌬Srxn or 0 ϭ ⌬Hrxn Ϫ T ⌬Srxn
so
⌬Hrxn ϭ T ⌬Srxn or T ϭ ᎏ
⌬
⌬
H
Sr
r
x
x
n
n
ᎏ (at equilibrium)
EXAMPLE 15-18 Estimation of Boiling Point
Use the thermodynamic data in Appendix K to estimate the normal boiling point of bromine,
Br2. Assume that ⌬H and ⌬S do not change with temperature.
15-16
15-16 The Temperature Dependence of Spontaneity 631
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 20.8, Free Energy and
Temperature.
Plan
The process we must consider is
Br2(ᐉ) 88n Br2(g)
By definition, the normal boiling point of a liquid is the temperature at which pure liquid and
pure gas coexist in equilibrium at 1 atm. Therefore, ⌬G ϭ 0. We assume that ⌬Hrxn ϭ ⌬H0
rxn
and ⌬Srxn ϭ ⌬S0
rxn. We can evaluate these two quantities, substitute them in the relationship
⌬G ϭ ⌬H Ϫ T ⌬S, and then solve for T.
Solution
The required values (Appendix K) are as follows:
Br2(ᐉ) Br2(g)
⌬H0
f, kJ/mol: 0 30.91
S0, J/molиK: 152.2 245.4
⌬Hrxn ϭ ⌬H0
f Br2(g) Ϫ ⌬H0
f Br2(ᐉ)
ϭ 30.91 Ϫ 0 ϭ 30.91 kJ/mol
⌬Srxn ϭ S0
Br2(g) Ϫ S0
Br2(ᐉ)
ϭ (245.4 Ϫ 152.2) ϭ 93.2 J/molиK ϭ 0.0932 kJ/molиK
We can now solve for the temperature at which the system is in equilibrium, that is, the boiling
point of Br2.
⌬Grxn ϭ ⌬Hrxn Ϫ T ⌬Srxn ϭ 0 so ⌬Hrxn ϭ T ⌬Srxn
T ϭ ϭ ϭ 332 K (59°C)
This is the temperature at which the system is in equilibrium, that is, the boiling point of Br2.
The value listed in a handbook of chemistry and physics is 58.78°C.
You should now work Exercise 110.
The free energy change and spontaneity of a reaction depend on both enthalpy and
entropy changes. Both ⌬H and ⌬S may be either positive or negative, so we can group
reactions in four classes with respect to spontaneity (Figure 15-15).
⌬G ϭ ⌬H Ϫ T ⌬S (constant temperature and pressure)
1. ⌬H ϭ Ϫ (favorable) ⌬S ϭ ϩ (favorable) Reactions are productfavored
at all temperatures
2. ⌬H ϭ Ϫ (favorable) ⌬S ϭ Ϫ (unfavorable) Reactions become productfavored
below a definite
temperature
3. ⌬H ϭ ϩ (unfavorable) ⌬S ϭ ϩ (favorable) Reactions become productfavored
above a definite
temperature
4. ⌬H ϭ ϩ (unfavorable) ⌬S ϭ Ϫ (unfavorable) Reactions are reactantfavored
at all temperatures
30.91 kJ/mol
ᎏᎏ
0.0932 kJ/molиK
⌬Hrxn
ᎏ
⌬Srxn
632 CHAPTER 15: Chemical Thermodynamics
Actually, both ⌬H0
rxn and ⌬S0
rxn vary
with temperature, but usually not
enough to introduce significant errors
for modest temperature changes. The
value of ⌬G0
rxn, on the other hand,
is strongly dependent on the
temperature.
When ⌬H and ⌬S have opposite signs (classes 1 and 4), they act in the same direction,
so the direction of spontaneous change does not depend on temperature. When ⌬H and
⌬S have the same signs (classes 2 and 3), their effects oppose one another, so changes in
temperature can cause one factor or the other to dominate, and spontaneity depends on
temperature. For class 2, decreasing the temperature decreases the importance of the unfavorable
T ⌬S term, so the reaction becomes product-favored at lower temperatures. For
class 3, increasing the temperature increases the importance of the favorable T ⌬S term,
so the reaction becomes product-favored at higher temperatures.
We can estimate the temperature range over which a chemical reaction in class 2 or 3
is spontaneous by evaluating ⌬H0
rxn and ⌬S0
rxn from tabulated data. The temperature at
which ⌬G0
rxn ϭ 0 is the temperature limit of spontaneity. The sign of ⌬S0
rxn tells us whether
the reaction is spontaneous below or above this limit (Table 15-7).
Figure 15-15 A graphical
representation of the dependence of
⌬G and spontaneity on temperature
for each of the four classes of
reactions listed in the text and in
Table 15-7.
TABLE 15-7 Thermodynamic Classes of Reactions
⌬H ⌬S Temperature Range
Class Examples (kJ/mol) (J/molؒK) of Spontaneity
1
2H2O2(ᐉ) 88n 2H2O(ᐉ) ϩ O2(g) Ϫ196 ϩ126 All temperatures
H2(g) ϩ Br2(ᐉ) 88n 2HBr(g) Ϫ72.8 ϩ114 All temperatures
2
NH3(g) ϩ HCl(g) 88n NH4Cl(s) Ϫ176 Ϫ285 Lower temperatures (Ͻ619 K)
2H2S(g) ϩ SO2(g) 88n 3S(s) ϩ 2H2O(ᐉ) Ϫ233 Ϫ424 Lower temperatures (Ͻ550 K)
3
NH4Cl(s) 88n NH3(g) ϩ HCl(g) ϩ176 ϩ285 Higher temperatures (Ͼ619 K)
CCl4(ᐉ) 88n C(graphite) ϩ 2Cl2(g) ϩ135 ϩ235 Higher temperatures (Ͼ517 K)
4
2H2O(ᐉ) ϩ O2(g) 88n 2H2O2(ᐉ) ϩ196 Ϫ126 Nonspontaneous, all temperatures
3O2(g) 88n 2O3(g) ϩ285 Ϫ137 Nonspontaneous, all temperatures












NOT SPONTANEOUS
SPONTANEOUS
(∆H > 0)
∆H
= +, ∆S = – (Class 4)
∆H
= +, ∆S = +
(Class 3)
∆H
= –, ∆S = –
(Class 2)
(∆H < 0)
∆H
= –, ∆S = + (Class 1)
∆G
0
+
–
T
EXAMPLE 15-19 Temperature Range of Spontaneity
Mercury(II) sulfide is a dark red mineral called cinnabar. Metallic mercury is obtained by
roasting the sulfide in a limited amount of air. Estimate the temperature range in which the
standard reaction is product-favored.
HgS(s) ϩ O2(g) 88n Hg(ᐉ) ϩ SO2(g)
Plan
We evaluate ⌬H0
rxn and ⌬S0
rxn and assume that their values are independent of temperature.
We find that both factors are favorable to spontaneity.
Solution
From Appendix K:
HgS(s) O2 Hg(ᐉ) SO2(g)
⌬H0
f, kJ/mol: Ϫ58.2 2005. 706. Ϫ296.8
S0, J/molиK: Ϫ82.4 205.0 76.0 Ϫ248.1
⌬H0
rxn ϭ ⌬H0
f Hg(ᐉ) ϩ ⌬H0
f SO2(g) Ϫ [⌬H0
f HgS(s) ϩ ⌬H0
f O2(g)]
ϭ 0 Ϫ 296.8 ϩ 58.2 Ϫ 0 ϭ Ϫ238.6 kJ/mol
⌬S0
rxn ϭ S0
Hg(ᐉ) ϩ S0
SO2(g) Ϫ [S0
HgS(s) ϩ S0
O2(g)]
ϭ 76.02 ϩ 248.1 Ϫ [82.4 ϩ 205.0] ϭ ϩ36.7 J/molиK
⌬H0
rxn is negative and ⌬S0
rxn is positive, so the reaction is product-favored at all temperatures.
The reverse reaction is, therefore, nonspontaneous at all temperatures.
The fact that a reaction is product-favored at all temperatures does not mean that the reaction
occurs rapidly enough to be useful at all temperatures. As a matter of fact, Hg(ᐉ) can be
obtained from HgS(s) by this reaction at a reasonable rate only at high temperatures.
EXAMPLE 15-20 Temperature Range of Spontaneity
Estimate the temperature range for which the following standard reaction is product-favored.
SiO2(s) ϩ 2C(graphite) ϩ 2Cl2(g) 88n SiCl4(g) ϩ 2CO(g)
Plan
When we proceed as in Example 15-19, we find that ⌬S0
rxn is favorable to spontaneity, whereas
⌬H0
rxn is unfavorable. Thus, we know that the reaction becomes product-favored above some
temperature. We can set ⌬G0 equal to zero in the equation ⌬G0 ϭ ⌬H0 Ϫ T⌬S0 and solve for
the temperature at which the system is at equilibrium. This will represent the temperature above
which the reaction would be product-favored.
Solution
From Appendix K:
SiO2(s) C(graphite) Cl2(g) SiCl4(g) CO(g)
⌬H0
f, kJ/mol: Ϫ910.9 0 990 Ϫ657.0 Ϫ110.5
S0, J/molиK: Ϫ941.84 5.740 223.0 Ϫ330.6 Ϫ197.6
Heating red HgS in air produces
liquid Hg. The gaseous SO2 escapes.
Cinnabar, an important ore of
mercury, contains HgS.
634 CHAPTER 15: Chemical Thermodynamics
⌬H0
rxn ϭ [⌬H0
f SiCl4(g) ϩ 2 ⌬H0
f CO(g)] Ϫ [⌬H0
f SiO2(s) ϩ 2 ⌬H0
f C(graphite) ϩ 2 ⌬H0
f Cl2(g)]
ϭ [(Ϫ657.0) ϩ 2(Ϫ110.5)] Ϫ [(Ϫ910.9) ϩ 2(0) ϩ 2(0)]
ϭ ϩ32.9 kJ/mol
⌬S0
rxn ϭ S0
SiCl4(g) ϩ 2S0
CO(g) Ϫ [S0
SiO2(s) ϩ 2S0
C(graphite) ϩ 2S0
Cl2(g)]
ϭ [330.6 ϩ 2(197.6)] Ϫ [41.84 ϩ 2(5.740) ϩ 2(223.0)]
ϭ 226.5 J/molиK ϭ 0.2265 kJ/molиK
When ⌬G0 ϭ 0, neither the forward nor the reverse reaction is favored. Let’s find the temperature
at which ⌬G0 ϭ 0 and the system is at equilibrium.
0 ϭ ⌬G0 ϭ ⌬H0 Ϫ T ⌬S0
⌬H0 ϭ T ⌬S0
T ϭ ϭ
ϭ 145 K
At temperatures above 145 K, the T ⌬S0 term would be greater (ϪT ⌬S0 would be more negative)
than the ⌬H0 term, which would make ⌬G0 negative; so the reaction would be
product-favored above 145 K. At temperatures below 145 K, the T ⌬S0 term would be smaller
than the ⌬H0 term, which would make ⌬G0 positive; so the reaction would be reactant-favored
below 145 K.
However, 145 K (Ϫ128°C) is a very low temperature. For all practical purposes, the reaction
is product-favored at all but very low temperatures. In practice, it is carried out at 800°C
to 1000°C because of the greater reaction rate at these higher temperatures. This gives a useful
and economical rate of production of SiCl4, an important industrial chemical.
You should now work Exercises 104 and 108.
ϩ32.9 kJ/mol
ᎏᎏᎏ
ϩ0.2265 kJ/molиK
⌬H0
ᎏ
⌬S0
Key Terms 635
Key Terms
Absolute entropy (of a substance) The entropy of a substance
relative to its entropy in a perfectly ordered crystalline form at
0 K (where its entropy is zero).
Bomb calorimeter A device used to measure the heat transfer
between system and surroundings at constant volume.
Bond energy The amount of energy necessary to break one mole
of bonds in a gaseous substance, to form gaseous products at
the same temperature and pressure.
Calorimeter A device used to measure the heat transfer that
accompanies a physical or chemical change.
Endothermic process A process that absorbs heat.
Enthalpy change, ⌬H The quantity of heat transferred into or
out of a system as it undergoes a chemical or physical change
at constant temperature and pressure.
Entropy, S A thermodynamic state property that measures the
degree of disorder or randomness of a system.
Equilibrium A state of dynamic balance in which the rates of
forward and reverse processes (reactions) are equal; the state of
a system when neither the forward nor the reverse process is
thermodynamically favored.
Exothermic process A process that gives off (releases) heat.
First Law of Thermodynamics The total amount of energy in
the universe is constant (also known as the Law of Conservation
of Energy); energy is neither created nor destroyed in
ordinary chemical reactions and physical changes.
Gibbs free energy, G The thermodynamic state function of
a system that indicates the amount of energy available for the
system to do useful work at constant T and P. It is defined as
G ϭ H Ϫ TS.
Gibbs free energy change, ⌬G The indicator of spontaneity of
a process at constant T and P. ⌬G ϭ ⌬H Ϫ T ⌬S. If ⌬G is negative,
the process is product-favored (spontaneous).
Hess’s Law of heat summation The enthalpy change for a reaction
is the same whether it occurs in one step or a series of
steps.
Internal energy, E All forms of energy associated with a specific
amount of a substance.
Mole of reaction (mol rxn) The amount of reaction that corresponds
to the number of moles of each substance shown in
the balanced equation.
636 CHAPTER 15: Chemical Thermodynamics
Exercises
General Concepts
*001. State precisely the meaning of each of the following
terms. You may need to review Chapter 1 to refresh your
memory concerning terms introduced there. (a) energy;
(b) kinetic energy; (c) potential energy; (d) joule.
*002. State precisely the meaning of each of the following
terms. You may need to review Chapter 1 to refresh your
memory about terms introduced there. (a) heat; (b) temperature;
(c) system; (d) surroundings; (e) thermodynamic
state of system; (f) work.
*003. (a) Give an example of the conversion of heat into work.
(b) Give an example of the conversion of work into heat.
*004. Distinguish between endothermic and exothermic
processes. If we know that a reaction is endothermic in
one direction, what can be said about the reaction in the
reverse direction?
*005. According to the First Law of Thermodynamics, the total
amount of energy in the universe is constant. Why, then,
do we say that we are experiencing a declining supply of
energy?
*006. Use the First Law of Thermodynamics to describe what
occurs when an incandescent light is turned on.
*007. Use the First Law of Thermodynamics to describe what
occurs when sunlight strikes a black surface.
*008. Which of the following are examples of state functions?
(a) your bank balance; (b) your mass; (c) your weight;
(d) the heat lost by perspiration during a climb up a mountain
along a fixed path.
*009. What is a state function? Would Hess’s Law be a law if
enthalpy were not a state function?
Enthalpy and Changes in Enthalpy
*010. (a) Distinguish between ⌬H and ⌬H0 for a reaction.
(b) Distinguish between ⌬H0
rxn and ⌬H0
f.
*011. A reaction is characterized by ⌬Hrxn ϭ Ϫ500 kJ/mol.
Nonspontaneous change See Reactant-favored change.
Pressure–volume work Work done by a gas when it expands
against an external pressure or work done on a system as gases
are compressed or consumed in the presence of an external pres-
sure.
Product-favored change A change for which the collection of
products is more stable than the collection of reactants under
the given conditions; also called spontaneous change.
Reactant-favored change A change for which the collection of
reactants is more stable than the collection of products under
the given conditions; also called nonspontaneous change.
Second Law of Thermodynamics The universe tends toward
a state of greater disorder in spontaneous processes.
Spontaneous change See Product-favored change.
Standard enthalpy change, ⌬H0 The enthalpy change in which
the number of moles of reactants specified in the balanced
chemical equation, all at standard states, is converted completely
to the specified number of moles of products, all at
standard states.
Standard entropy, S0 (of a substance) The absolute entropy
of a substance in its standard state at 298 K.
Standard entropy change, ⌬S0 The entropy change in which
the number of moles of reactants specified in the balanced chemical
equation, all at standard states, is converted completely to
the specified number of moles of products, all at standard states.
Standard molar enthalpy of formation, ⌬Hf
0
(of a substance)
The enthalpy change for the formation of one mole of a substance
in a specified state from its elements in their standard
states.
Standard reaction A reaction in which the numbers of moles of
reactants shown in the balanced equation, all in their standard
states, are completely converted to the numbers of moles of products
shown in the balanced equation, also all at their standard
states.
Standard state (of a substance) See Thermodynamic standard
state of a substance.
State function A variable that defines the state of a system; a
function that is independent of the pathway by which a process
occurs.
Surroundings Everything in the environment of the system.
System The substances of interest in a process; the part of the
universe under investigation.
Thermochemical equation A balanced chemical equation
together with a designation of the corresponding value of ⌬Hrxn.
Sometimes used with changes in other thermodynamic quanti-
ties.
Thermodynamics The study of the energy transfers accompanying
physical and chemical processes.
Thermodynamic state of a system A set of conditions that
completely specifies all of the properties of the system.
Thermodynamic standard state of a substance The most stable
state of the substance at one atmosphere pressure and at
some specific temperature (25°C unless otherwise specified).
Third Law of Thermodynamics The entropy of a hypothetical
pure, perfect, crystalline substance at absolute zero
temperature is zero.
Universe The system plus the surroundings.
Work The application of a force through a distance; for physical
changes or chemical reactions at constant external pressure,
the work done on the system is ϪP ⌬V.
Exercises 637
Does the reaction mixture absorb heat from the surroundings
or release heat to them?
*012. For each of the following reactions, (a) does the enthalpy
increase or decrease; (b) is Hreactant Ͼ Hproduct or is
Hproduct Ͼ Hreactant; (c) is ⌬H positive or negative?
i(i) Al2O3(s) 88n 2Al(s) ϩ ᎏ3
2
ᎏO2(g) (endothermic)
(ii) Sn(s) ϩ Cl2(g) 88n SnCl2(s) (exothermic)
*013. (a) The combustion of 0.0222 g of isooctane vapor,
C8H18(g), at constant pressure raises the temperature of
a calorimeter 0.400°C. The heat capacity of the calorimeter
and water combined is 2.48 kJ/°C. Find the molar heat
of combustion of gaseous isooctane.
C8H18(g) ϩ 12ᎏ1
2
ᎏO2(g) 88n 8CO2(g) ϩ 9H2O(ᐉ)
(b) How many grams of C8H18(g) must be burned to
obtain 362 kJ of heat energy?
*014. Methanol, CH3OH, is an efficient fuel with a high octane
rating that can be produced from coal and hydrogen.
CH3OH(g) ϩ ᎏ3
2
ᎏO2(g) 88n CO2(g) ϩ 2H2O(ᐉ)
⌬H ϭ Ϫ764 kJ/mol rxn
(a) Find the heat evolved when 90.0 g CH3OH(g) burns
in excess oxygen. (b) What mass of O2 is consumed when
945 kJ of heat is given out?
*015. How much heat is liberated when 0.0662 mole of sodium
reacts with excess water according to the following equa-
tion?
2Na(s) ϩ 2H2O(ᐉ) 88n H2(g) ϩ 2NaOH(aq)
⌬H ϭ Ϫ368 kJ/mol rxn
*016. What is ⌬H for the reaction
PbO(s) ϩ C(s) 88n Pb(s) ϩ CO(g)
if 35.7 kJ must be supplied to convert 74.6 g lead(II) oxide
to lead?
*017. The standard molar enthalpy of formation, ⌬H0
f, listed in
Appendix K is zero for almost all elements. A few entries
are not zero; explain why, and give two examples.
*018. Why is the standard molar enthalpy of formation, ⌬H0
f,
for liquid water different than is ⌬H0
f for water vapor,
both at 25°C. Which formation reaction is more exothermic?
Does your answer indicate that H2O(ᐉ) is at a higher
or lower enthalpy than H2O(g)?
*019. Methylhydrazine is burned with dinitrogen tetroxide in
the attitude-control engines of the space shuttles.
CH6N2(ᐉ) ϩ ᎏ5
4
ᎏN2O4(ᐉ) 88n
CO2(g) ϩ 3H2O(ᐉ) ϩ ᎏ9
4
ᎏN2(g)
The two substances ignite instantly on contact, producing
a flame temperature of 3000 K. The energy liberated
per 0.100 g of CH6N2 at constant atmospheric pressure
after the products are cooled back to 25°C is 750 J. (a)
Find ⌬H for the reaction as written. (b) How many kilojoules
are liberated when 44.0 g of N2 is produced?
*020. Which is more exothermic, the combustion of one mole
of methane to form CO2(g) and liquid water or the combustion
of one mole of methane to form CO2(g) and
steam? Why? (No calculations are necessary.)
*021. Which is more exothermic, the combustion of one mole
of gaseous benzene, C6H6, or the combustion of one mole
of liquid benzene? Why? (No calculations are necessary.)
Thermochemical Equations, ⌬H0
f, and Hess’s Law
*022. Explain the meaning of each word in the term “thermodynamic
standard state of a substance.”
*023. Explain the meaning of each word in the term “standard
molar enthalpy of formation.”
*024. From the data in Appendix K, determine the form that
represents the standard state for each of the following elements:
(a) chlorine; (b) chromium; (c) nitrogen; (d) iodine;
(e) sulfur.
*025. From the data in Appendix K, determine the form that
represents the standard state for each of the following
elements: (a) oxygen; (b) carbon; (c) phosphorus;
(d) rubidium; (e) calcium.
*026. Write the balanced chemical equation whose ⌬H 0
rxn value
is equal to ⌬Hf
0
for each of the following substances:
(a) calcium hydroxide, Ca(OH)2(s); (b) benzene, C6H6(ᐉ);
(c) sodium carbonate, Na2CO3(s); (d) calcium fluoride,
CaF2(s); (e) phosphine, PH3(g); (f ) propane, C3H8(g);
(g) atomic sulfur, S(g).
*027. Write the balanced chemical equation whose ⌬H 0
rxn value
is equal to ⌬Hf
0
for each of the following substances:
(a) hydrogen sulfide, H2S(g); (b) lead(II) chloride,
A space shuttle.
638 CHAPTER 15: Chemical Thermodynamics
PbCl2(s); (c) atomic oxygen, O(g); (d) benzoic acid,
C6H5COOH(s); (e) hydrogen peroxide, H2O2(ᐉ);
(f ) dinitrogen tetroxide, N2O4(g).
0*28. We burn 14.4 g of lithium in excess oxygen at constant
atmospheric pressure to form Li2O. Then we bring the
reaction mixture back to 25°C. In this process 605 kJ of
heat is given off. What is the standard molar enthalpy of
formation of Li2O?
0*29. We burn 14.4 g of magnesium in excess nitrogen at constant
atmospheric pressure to form Mg3N2. Then we
bring the reaction mixture back to 25°C. In this process
136.7 kJ of heat is given off. What is the standard molar
enthalpy of formation of Mg3N2?
*030. From the following enthalpies of reaction,
4HCl(g) ϩ O2(g) 88n 2H2O(ᐉ) ϩ 2Cl2(g)
⌬H ϭ Ϫ202.4 kJ/mol rxn
ᎏ1
2
ᎏH2(g) ϩ ᎏ1
2
ᎏF2(g) 88n HF(ᐉ)
⌬H ϭ Ϫ600.0 kJ/mol rxn
H2(g) ϩ ᎏ1
2
ᎏO2(g) 88n H2O(ᐉ)
⌬H ϭ Ϫ285.8 kJ/mol rxn
find ⌬Hrxn for 2HCl(g) ϩ F2(g) 88n 2HF(ᐉ) ϩ Cl2(g).
*031. From the following enthalpies of reaction,
CaCO3(s) 88n CaO(s) ϩ CO2(g)
⌬H ϭ ϩ178.1 kJ/mol rxn
CaO(s) ϩ H2O(ᐉ) 88n Ca(OH)2(s)
⌬H ϭ Ϫ65.3 kJ/mol rxn
Ca(OH)2(s) 88n Ca2ϩ(aq) ϩ 2OHϪ(aq)
⌬H ϭ Ϫ16.2 kJ/mol rxn
calculate ⌬Hrxn for
Ca2ϩ(aq) ϩ 2OHϪ(aq) ϩ CO2(g) 88n
CaCO3(s) ϩ H2O(ᐉ)
*032. Given that
S(s) ϩ O2(g) 88n SO2(g) ⌬H ϭ Ϫ296.8 kJ/mol
S(s) ϩ ᎏ3
2
ᎏO2(g) 88n SO3(g) ⌬H ϭ Ϫ395.6 kJ/mol
determine the enthalpy change for the decomposition
reaction 2SO3(g) n 2SO2(g) ϩ O2(g).
*033. Aluminum reacts vigorously with many oxidizing agents.
For example,
4Al(s) ϩ 3O2(g) 88n 2Al2O3(s)
⌬H ϭ Ϫ3352 kJ/mol
4Al(s) ϩ 3MnO2(s) 88n 3Mn(s) ϩ 2Al2O3(s)
⌬H ϭ Ϫ1792 kJ/mol
Use this information to determine the enthalpy of formation
of MnO2(s).
*034. Given that
2H2(g) ϩ O2(g) 88n 2H2O(ᐉ)
⌬H ϭ Ϫ571.6 kJ/mol
C3H4(g) ϩ 4O2(g) 88n 3CO2(g) ϩ 2H2O(ᐉ)
⌬H ϭ Ϫ1937 kJ/mol.
C3H8(g) ϩ 5O2(g) 88n 3CO2(g) ϩ 4H2O(ᐉ)
⌬H ϭ Ϫ2220 kJ/mol.
determine the heat of the hydrogenation reaction
C3H4(g) ϩ 2H2(g) 88n C3H8(g)
*035. Determine the heat of formation of liquid hydrogen peroxide
at 25°C from the following thermochemical
equations.
H2(g) ϩ ᎏ1
2
ᎏO2(g) 88n H2O(g)
⌬H0 ϭ 2Ϫ241.82 kJ/mol
2H(g) ϩ O(g) 88n H2O(g)
⌬H0 ϭ 2Ϫ926.92 kJ/mol
2H(g) ϩ 2O(g) 88n H2O2(g)
⌬H0 ϭ Ϫ1070.60 kJ/mol
2O(g) 88n O2(g)
⌬H0 ϭ 2Ϫ498.34 kJ/mol
H2O2(ᐉ) 88n H2O2(g)
⌬H0 ϭ Ϫ2251.46 kJ/mol
*036. Use data in Appendix K to find the enthalpy of reaction
for
(a) NH4NO3(s) 88n N2O(g) ϩ 2H2O(g)
(b) 2FeS2(s) ϩ ᎏ1
2
1ᎏO2(g) 88n Fe2O3(s) ϩ 4SO2(g)
(c) SiO2(s) ϩ 3C(s) 88n SiC(s) ϩ 2CO(g)
*037. Repeat Exercise 36 for
(a) CaCO3(s) 88n CaO(s) ϩ CO2(g)
(b) 2HI(g) ϩ F2(g) 88n 2HF(g) ϩ I2(s)
(c) SF6(g) ϩ 3H2O(ᐉ) 88n 6HF(g) ϩ SO3(g)
*038. The internal combustion engine uses heat produced during
the burning of a fuel. Propane, C3H8(g), is sometimes
used as the fuel. Gasoline is the most commonly used fuel.
Assume that the gasoline is pure octane, C8H18(ᐉ) and
the fuel and oxygen are completely converted into CO2(g)
and H2O(g). For each of these fuels, determine the heat
released per gram of fuel burned.
*039. Propane, C3H8(g), is used as the fuel for some modern
internal combustion engines. Methane, CH4(g), has been
proposed by the movie industry as the post-Apocalypse
fuel when gasoline and propane are supposedly no longer
available. Assume that the fuel and oxygen are completely
converted into CO2(g) and H2O(g). For each of these
fuels determine the heat released per gram of fuel burned.
Compare your answers to the answers for Exercise 38.
Exercises 639
*040. The thermite reaction, used for welding iron, is the reaction
of Fe3O4 with Al.
8Al(s) ϩ 3Fe3O4(s) 88n 4Al2O3(s) ϩ 9Fe(s)
⌬H0 ϭ Ϫ3350 kJ/mol rxn
Because this large amount of heat cannot be rapidly dissipated
to the surroundings, the reacting mass may reach
temperatures near 3000°C. How much heat is released
by the reaction of 23.0 g of Al with 57.6 g of Fe3O4?
*043. Natural gas is mainly methane, CH4(g). Assume that
gasoline is octane, C8H18(ᐉ), and that kerosene is
C10H22(ᐉ). (a) Write the balanced equations for the combustion
of each of these three hydrocarbons in excess O2.
The products are CO2(g) and H2O(ᐉ). (b) Calculate
⌬H0
rxn at 25°C for each combustion reaction. ⌬H0
f for
C10H22 is Ϫ300.9 kJ/mol. (c) When burned at standard
conditions, which of these three fuels would produce the
most heat per mole? (d) When burned at standard conditions,
which of the three would produce the most heat
per gram?
Bond Energies
*044. (a) How is the heat released or absorbed in a gas phase
reaction related to bond energies of products and reactants?
(b) Hess’s Law states that
⌬H0
rxn ϭ ͚ n ⌬H0
f products Ϫ ͚ n ⌬H0
f reactants
The relationship between ⌬H0
rxn and bond energies for a
gas phase reaction is
⌬H0
rxn ϭ
͚ bond energiesreactants Ϫ ͚ bond energiesproducts
It is not true, in general, that ⌬H0
f for a substance is equal
to the negative of the sum of the bond energies of the
substance. Why?
*045. (a) Suggest a reason for the fact that different amounts of
energy are required for the successive removal of the three
hydrogen atoms of an ammonia molecule, even though
all NXH bonds in ammonia are equivalent. (b) Suggest
why the NXH bonds in different compounds such as
ammonia, NH3; methylamine, CH3NH2; and ethylamine,
C2H5NH2, have slightly different bond energies.
*046. Use tabulated bond energies to estimate the enthalpy of
reaction for each of the following gas phase reactions.
(a) H2CUCH2 ϩ Br2 88n BrH2CXCH2Br
(b) H2O2 88n H2O ϩ ᎏ1
2
ᎏO2
*047. Use tabulated bond energies to estimate the enthalpy of
reaction for each of the following gas phase reactions.
(a) N2 ϩ 3H2 88n 2NH3
(b) CH4 ϩ Cl2 88n CH3Cl ϩ HCl
(c) CO ϩ H2O 88n CO2 ϩ H2
*048. Use the bond energies listed in Table 15-2 to estimate
the heat of reaction for
*049. Estimate ⌬H for the burning of one mole of butane, using
the bond energies listed in Tables 15-2 and 15-3.
F(g)C
F
Cl
Cl F(g)Fϩ F(g)C
F
F
F Cl(g)Clϩ
*041. When a welder uses an acetylene torch, the combustion
of acetylene liberates the intense heat needed for welding
metals together. The equation for this combustion
reaction is
2C2H2(g) ϩ 5O2(g) 88n 4CO2(g) ϩ 2H2O(g)
The heat of combustion of acetylene is Ϫ1255.5 kJ/mol
of C2H2. How much heat is liberated when 3.462 kg of
C2H2 is burned?
*042. Silicon carbide, or carborundum, SiC, is one of the hardest
substances known and is used as an abrasive. It has the
structure of diamond with half of the carbons replaced by
silicon. It is prepared industrially by reduction of sand
(SiO2) with carbon in an electric furnace.
SiO2(s) ϩ 3C(graphite) 88n SiC(s) ϩ 2CO(g)
⌬H0 for this reaction is 624.6 kJ, and the ⌬H0
f for SiO2(s)
and CO(g) are Ϫ910.9 kJ/mol and Ϫ110.5 kJ/mol, respectively.
Calculate ⌬H0
f for silicon carbide.
The thermite reaction.
640 CHAPTER 15: Chemical Thermodynamics
*050. (a) Use the bond energies listed in Table 15-2 to estimate
the heats of formation of HCl(g) and HF(g). (b) Compare
your answers to the standard heats of formation in
Appendix K.
*051. (a) Use the bond energies listed in Table 15-2 to estimate
the heats of formation of H2O(g) and O3(g). (b) Compare
your answers to the standard heats of formation in
Appendix K.
*052. Using data in Appendix K, calculate the average SXF
bond energy in SF6(g).
*053. Using data in Appendix K, calculate the average HXS
bond energy in H2S(g).
*054. Using data in Appendix K, calculate the average OXF
bond energy in OF2(g).
0*55. Methane undergoes several different exothermic reactions
with gaseous chlorine. One of these forms
chloroform, CHCl3(g).
CH4(g) ϩ 3Cl2(g) 88n CHCl3(g) ϩ 3HCl(g)
⌬H0
rxn ϭ Ϫ305.2 kJ/mol rxn
Average bond energies per mole of bonds are: CXH ϭ
413 kJ; ClXCl ϭ 242 kJ; HXCl ϭ 432 kJ. Use these to
calculate the average CXCl bond energy in chloroform.
Compare this with the value in Table 15-2.
0*56. Ethylamine undergoes an endothermic gas phase dissociation
to produce ethylene (or ethene) and ammonia.
The following average bond energies per mole of bonds
are given: CXH ϭ 413 kJ; CXC ϭ 346 kJ; CUC ϭ
602 kJ; NXH ϭ 391 kJ. Calculate the CXN bond energy
in ethylamine. Compare this with the value in Table
15-2.
Calorimetry
*057. What is a coffee-cup calorimeter? How do coffee-cup
calorimeters give us useful information?
*058. A calorimeter contained 75.0 g of water at 16.95°C. A
93.3-g sample of iron at 65.58°C was placed in it, giving
a final temperature of 19.68°C for the system. Calculate
the heat capacity of the calorimeter. Specific heats are
4.184 J/gи°C for H2O and 0.444 J/gи°C for Fe.
*059. A student wishes to determine the heat capacity of a
coffee-cup calorimeter. After she mixes 100.0 g of water
C
H
H
H C C C
H
H
N
H
N
H
H
H
H H
H
H H
heat
ϩ
⌬Hrxn ϭ ϩ53.6 kJ/mol rxn0
O5H H(g)ϩO(g)C4O
C
H
H
H C
H
H
C
H
H
C
H
H
H(g) O(g)ϩ O13
2
at 58.5°C with 100.0 g of water, already in the calorimeter,
at 22.8°C, the final temperature of the water is 39.7°C.
(a) Calculate the heat capacity of the calorimeter in J/°C.
Use 4.18 J/gи°C as the specific heat of water. (b) Why is
it more useful to express the value in J/°C rather than
units of J/(g calorimeterи°C)?
*060. A coffee-cup calorimeter is used to determine the specific
heat of a metallic sample. The calorimeter is filled with
50.0 mL of water at 25.0°C (density ϭ 0.997 g/mL). A
36.5-gram sample of the metallic material is taken from
water boiling at 100.0°C and placed in the calorimeter.
The equilibrium temperature of the water and sample is
32.5°C. The calorimeter constant is known to be 1.87
J/°C. Calculate the specific heat of the metallic material.
*061. A 5.1-gram piece of gold jewelry is removed from water
at 100.0°C and placed in a coffee-cup calorimeter containing
16.9 g of water at 22.5°C. The equilibrium
temperature of the water and jewelry is 23.2°C. The
calorimeter constant is known from calibration experiments
to be 1.87 J/°C. What is the specific heat of
this piece of jewelry? The specific heat of pure gold is
0.129 J/g и °C. Is the jewelry pure gold?
*062. A coffee-cup calorimeter having a heat capacity of
472 J/°C is used to measure the heat evolved when the
following aqueous solutions, both initially at 22.6°C, are
mixed: 100. g of solution containing 6.62 g of lead(II)
nitrate, Pb(NO3)2, and 100. g of solution containing
6.00 g of sodium iodide, NaI. The final temperature
is 24.2°C. Assume that the specific heat of the mixture is
the same as that for water, 4.18 J/gи°C. The reaction is
Pb(NO3)2(aq) ϩ 2NaI(aq) 88n PbI2(s) ϩ 2NaNO3(aq)
(a) Calculate the heat evolved in the reaction. (b) Calculate
the ⌬H for the reaction under the conditions of the
experiment.
*063. A coffee-cup calorimeter is used to determine the heat of
reaction for the acid–base neutralization
CH3COOH(aq) ϩ NaOH(aq) 88n
NaCH3COO(aq) ϩ H2O(ᐉ)
When we add 20.00 mL of 0.625 M NaOH at 21.400°C
to 30.00 mL of 0.500 M CH3COOH already in the
calorimeter at the same temperature, the resulting temperature
is observed to be 24.347°C. The heat capacity
of the calorimeter has previously been determined to be
27.8 J/°C. Assume that the specific heat of the mixture is
the same as that of water, 4.18 J/gи°C, and that the density
of the mixture is 1.02 g/mL. (a) Calculate the amount
of heat given off in the reaction. (b) Determine ⌬H for
the reaction under the conditions of the experiment.
*064. In a bomb calorimeter compartment surrounded by
945 g of water, the combustion of 1.048 g of benzene,
C6H6(ᐉ), raised the temperature of the water from
23.640°C to 32.692°C. The heat capacity of the calorimeter
is 891 J/°C. (a) Write the balanced equation for the
Exercises 641
combustion reaction, assuming that CO2(g) and H2O(ᐉ)
are the only products. (b) Use the calorimetric data to
calculate ⌬E for the combustion of benzene in kJ/g and
in kJ/mol.
*065. A 2.00-g sample of hydrazine, N2H4, is burned in a bomb
calorimeter that contains 6.40 ϫ 103 g of H2O, and the
temperature increases from 25.00°C to 26.17°C. The heat
capacity of the calorimeter is 3.76 kJ/°C. Calculate ⌬E
for the combustion of N2H4 in kJ/g and in kJ/mol.
0*66. A strip of magnesium metal having a mass of 1.22 g dissolves
in 100. mL of 6.02 M HCl, which has a specific
gravity of 1.10. The hydrochloric acid is initially at
23.0°C, and the resulting solution reaches a final temperature
of 45.5°C. The heat capacity of the calorimeter
in which the reaction occurs is 562 J/°C. Calculate ⌬H
for the reaction under the conditions of the experiment,
assuming the specific heat of the final solution is the same
as that for water, 4.18 J/gи°C.
Mg(s) ϩ 2HCl(aq) 88n MgCl2(aq) ϩ H2(g)
*067. When 3.16 g of salicylic acid, C7H6O3, is burned in a
bomb calorimeter containing 5.00 kg of water originally
at 23.00°C, 69.3 kJ of heat is evolved. The calorimeter
constant is 3612 J/°C. Calculate the final temperature.
*068. A 6.620-gram sample of decane, C10H22(ᐉ), was burned
in a bomb calorimeter whose heat capacity had been
determined to be 2.45 kJ/°C. The temperature of 1250.0
grams of water rose from 24.6°C to 26.4°C. Calculate ⌬E
for the reaction in joules per gram of decane and in kilojoules
per mole of decane. The specific heat of water is
4.184 J/gи°C.
*069. A nutritionist determines the caloric value of a 10.00gram
sample of beef fat by burning it in a bomb
calorimeter. The calorimeter held 2.500 kg of water, the
heat capacity of the bomb is 1.360 kJ/°C, and the temperature
of the calorimeter increased from 25.0°C to
56.9°C. (a) Calculate the number of joules released per
gram of beef fat. (b) One nutritional Calorie is 1 kcal or
4184 joules. What is the dietary, caloric value of beef fat,
in nutritional Calories per gram?
Internal Energy and Changes in Internal Energy
*070. (a) What are the sign conventions for q, the amount of
heat added to or removed from a system? (b) What are
the sign conventions for w, the amount of work done on
or by a system?
*071. What happens to ⌬E for a system during a process in
which (a) q Ͻ 0 and w Ͻ 0, (b) q ϭ 0 and w Ͼ 0, and
(c) q Ͼ 0 and w Ͻ 0?
*072. What happens to ⌬E for a system during a process in
which (a) q Ͼ 0 and w Ͼ 0, (b) q ϭ w ϭ 0, and (c) q Ͻ 0
and w Ͼ 0?
*073. A system performs 720 Lиatm of pressure–volume work
(1 Lиatm ϭ 101.325 J) on its surroundings and absorbs
5500. J of heat from its surroundings. What is the change
in internal energy of the system?
*074. A system receives 93 J of electrical work, performs 227 J
of pressure–volume work, and releases 155 J of heat.
What is the change in internal energy of the system?
*075. For each of the following chemical and physical changes
carried out at constant pressure, state whether work is
done by the system on the surroundings or by the surroundings
on the system, or whether the amount of work
is negligible.
(a) C6H6(ᐉ) 88n C6H6(g)
(b) ᎏ1
2
ᎏN2(g) ϩ ᎏ3
2
ᎏH2(g) 88n NH3(g)
(c) SiO2(s) ϩ 3C(s) 88n SiC(s) ϩ 2CO(g)
*076. Repeat Exercise 75 for
(a) 2SO2(g) ϩ O2(g) 88n 2SO3(g)
(b) CaCO3(s) 88n CaO(s) ϩ CO2(g)
(c) CO2(g) ϩ H2O(ᐉ) ϩ CaCO3(s) 88n
Ca2ϩ(aq)ϩ 2HCO3
Ϫ(aq)
*077. Assuming that the gases are ideal, calculate the amount
of work done (in joules) in each of the following reactions.
In each case, is the work done on or by the system?
(a) A reaction in the Mond process for purifying nickel
that involves formation of the gas nickel(0) tetracarbonyl
at 50–100°C. Assume one mole of nickel is used and a
constant temperature of 75°C is maintained.
Ni(s) ϩ 4CO(g) 88n Ni(CO)4(g)
(b) The conversion of one mole of brown nitrogen dioxide
into colorless dinitrogen tetroxide at 10.0°C.
2NO2(g) 88n N2O4(g)
*078. Assuming that the gases are ideal, calculate the amount
of work done (in joules) in each of the following reactions.
In each case, is the work done on or by the system?
(a) The oxidation of one mole of HCl(g) at 200°C.
4HCl(g) ϩ O2(g) 88n 2Cl2(g) ϩ 2H2O(g)
(b) The decomposition of one mole of nitric oxide (an air
pollutant) at 300°C.
2NO(g) 88n N2(g) ϩ O2(g)
0*79. When an ideal gas expands at constant temperature, there
is no change in molecular kinetic energy (kinetic energy
is proportional to temperature), and there is no change
in potential energy due to intermolecular attractions
(these are zero for an ideal gas). Thus, for the isothermal
(constant temperature) expansion of an ideal gas, ⌬E ϭ 0.
Suppose we allow an ideal gas to expand isothermally
from 2.00 L to 5.00 L in two steps: (a) against a constant
external pressure of 3.00 atm until equilibrium is reached,
then (b) against a constant external pressure of 2.00 atm
until equilibrium is reached. Calculate q and w for this
two-step expansion.
642 CHAPTER 15: Chemical Thermodynamics
Entropy and Entropy Changes
*080. A car uses gasoline as a fuel. Describe the burning of the
fuel in terms of chemical and physical changes. Relate
your answer to the Second Law of Thermodynamics.
*081. State the Second Law of Thermodynamics. We cannot
use ⌬Suniv directly as a measure of the spontaneity of a
reaction. Why?
*082. State the Third Law of Thermodynamics. What does it
mean?
*083. Explain why ⌬S may be referred to as a contributor to
spontaneity.
*084. For each of the following processes, tell whether the
entropy of the system increases, decreases, or remains
constant:
(a) Melting one mole of ice to water at 0°C
(b) Freezing one mole of water to ice at 0°C
(c) Freezing one mole of water to ice at Ϫ10°C
(d) Freezing one mole of water to ice at 0°C and then
cooling it to Ϫ10°C
*085. When solid sodium chloride is cooled from 25°C to 0°C,
the entropy change is Ϫ4.4 J/molиK. Is this an increase
or decrease in randomness? Explain this entropy change
in terms of what happens in the solid at the molecular
level.
*086. When a one-mole sample of argon gas at 0°C is compressed
to one half its original volume, the entropy
change is Ϫ5.76 J/molиK. Is this an increase or a decrease
in randomness? Explain this entropy change in terms of
what happens in the gas at the molecular level.
*087. Which of the following processes are accompanied by an
increase in entropy of the system? (No calculation is necessary.)
(a) A building is constructed from bricks, mortar,
lumber, and nails. (b) A building collapses into bricks,
mortar, lumber, and nails. (c) Iodine sublimes, I2(s) n
I2(g). (d) White silver sulfate, Ag2SO4, precipitates from
a solution containing silver ions and sulfate ions. (e) A
marching band is gathered into formation. (f ) A partition
is removed to allow two gases to mix.
*088. Which of the following processes are accompanied by
an increase in entropy of the system? (No calculation is
necessary.) (a) Thirty-five pennies are removed from a
bag and placed heads-up on a table. (b) The pennies of
part (a) are swept off the table and back into the bag.
(c) Water freezes. (d) Carbon tetrachloride, CCl4, evaporates.
(e) The reaction PCl5(g) n PCl3(g) ϩ Cl2(g)
occurs. (f ) The reaction PCl3(g) ϩ Cl2(g) n PCl5(g)
occurs.
*089. For each of the following processes, tell whether the
entropy of the universe increases, decreases, or remains
constant. (a) melting one mole of ice to water at 0°C; (b)
freezing one mole of water to ice at 0°C; (c) freezing one
mole of water to ice at Ϫ10°C; (d) freezing one mole of
water to ice at 0°C and then cooling it to Ϫ10°C.
0*90. Consider the boiling of a pure liquid at constant pressure.
Is each of the following greater than, less than, or equal
to zero? (a) ⌬Ssys; (b) ⌬Hsys; (c) ⌬Tsys.
*091. Use S0 data from Appendix K to calculate the value of
⌬S0
298 for each of the following reactions. Compare the
signs and magnitudes for these ⌬S0
298 values and explain
your observations.
(a) 2NO(g) ϩ H2(g) 88n N2O(g) ϩ H2O(g)
(b) 2N2O5(g) 88n 4NO2(g) ϩ O2(g)
(c) 2NH4NO3(s) 88n 2N2(g) ϩ 4H2O(g) ϩ O2(g)
*092. Use S0 data from Appendix K to calculate the value of
⌬S0
298 for each of the following reactions. Compare the
signs and magnitudes for these ⌬S0
298 values and explain
your observations.
(a) 4HCl(g) ϩ O2(g) 88n 2Cl2(g) ϩ 2H2O(g)
(b) PCl3(g) ϩ Cl2(g) 88n PCl5(g)
(c) 2NO(g) 88n N2(g) ϩ O2(g)
Gibbs Free Energy Changes and Spontaneity
*093. (a) What are the two factors that favor spontaneity of a
process? (b) What is Gibbs free energy? What is change
in Gibbs free energy? (c) Most spontaneous reactions are
exothermic, but some are not. Explain. (d) Explain how
the signs and magnitudes of ⌬H and ⌬S are related to the
spontaneity of a process and how they affect it.
*094. Which of the following conditions would predict a
process that is (a) always spontaneous, (b) always nonspontaneous,
or (c) spontaneous or nonspontaneous
depending on the temperature and magnitudes of ⌬H
and ⌬S? (i) ⌬H Ͼ 0, ⌬S Ͼ 0; (ii) ⌬H Ͼ 0, ⌬S Ͻ 0;
(iii) ⌬H Ͻ 0, ⌬S Ͼ 0; (iv) ⌬H Ͻ 0, ⌬S Ͻ 0.
*095. For the decomposition of O3(g) to O2(g)
2O3(g) 88n 3O2(g)
⌬H0 ϭ Ϫ285.4 kJ/mol and ⌬S0 ϭ 137.55 J/molиK at
25°C. Calculate ⌬G0 for the reaction. Is the reaction
spontaneous? Is either or both of the driving forces (⌬H0
and ⌬S0) for the reaction favorable?
*096. Calculate ⌬G0 at 25°C for the reaction
2NO2(g) 88n N2O4(g)
given ⌬H0 ϭ Ϫ57.20 kJ/mol and ⌬S0 ϭ Ϫ175.83
J/molиK. Is this reaction spontaneous? What is the driving
force for spontaneity?
*097. The standard Gibbs free energy of formation is Ϫ286.06
kJ/mol for NaI(s), Ϫ261.90 kJ/mol for Naϩ(aq), and
Ϫ51.57 kJ/mol for IϪ(aq) at 25°C. Calculate ⌬G0 for the
reaction
H2O
NaI(s) 8888n Naϩ(aq) ϩ IϪ(aq)
0*98. Use the following equations to find ⌬Gf
0
for HBr(g) at
25°C.
Br2(ᐉ) 88n Br2(g) ⌬G0 ϭ 003.14 kJ/mol
HBr(g) 88n H(g) ϩ Br(g) ⌬G0 ϭ 339.09 kJ/mol
Exercises 643
Br2(g) 88n 2Br(g) ⌬G0 ϭ 161.7 kJ/mol
H2(g) 88n 2H(g) ⌬G0 ϭ 406.494 kJ/mol
*099. Use values of standard free energy of formation, ⌬G0
f,
from Appendix K, to calculate the standard free energy
change for each of the following reactions at 25°C and
1 atm.
(a) 3NO2(g) ϩ H2O(ᐉ) 88n 2HNO3(ᐉ) ϩ NO(g)
(b) SnO2(s) ϩ 2CO(g) 88n 2CO2(g) ϩ Sn(s)
(c) 2Na(s) ϩ 2H2O(ᐉ) 88n 2NaOH(aq) ϩ H2(g)
*100. Make the same calculations as in Exercise 99, using
values of standard enthalpy of formation and absolute
entropy instead of values of ⌬G0
f.
*101. Calculate ⌬G0 for the reduction of the oxides of iron and
copper by carbon at 700 K represented by the equations
(a) 2Fe2O3(s) ϩ 3C(graphite) 88n 4Fe(s) ϩ 3CO2(g)
(b) 2CuO(s) ϩ C(graphite) 88n 2Cu(s) ϩ CO2(g)
Values of ⌬G0
f at 700K are Ϫ92 kJ/mol for CuO(s), Ϫ632
kJ/mol for Fe2O3(s), and Ϫ395 kJ/mol for CO2(g).
(c) Which oxide can be reduced using carbon in a wood
fire (which has a temperature of about 700 K), assuming
standard state conditions?
Temperature Range of Spontaneity
*102. Are the following statements true or false? Justify your
answers. (a) An exothermic reaction is spontaneous. (b) If
⌬H and ⌬S are both positive, then ⌬G will decrease when
the temperature increases. (c) A reaction for which ⌬Ssys
is positive is spontaneous.
*103. For the reaction
C(s) ϩ O2(g) 88n CO2(g)
⌬H0 ϭ Ϫ393.51 kJ/mol and ⌬S0 ϭ 2.86 J/mol и K at
25°C. (a) Does this reaction become more or less favorable
as the temperature increases? (b) For the reaction
C(s) ϩ ᎏ1
2
ᎏO2(g) 88n CO(g)
⌬H0 ϭ Ϫ110.52 kJ/mol and ⌬S0 ϭ 89.36 J/molиK at
25°C. Does this reaction become more or less favorable
as the temperature increases? (c) Compare the temperature
dependencies of these reactions.
*104. (a) Calculate ⌬H0, ⌬G0, and ⌬S0 for the reaction
2H2O2(ᐉ) 88n 2H2O(ᐉ) ϩ O2(g)
at 25°C. (b) Is there any temperature at which H2O2(ᐉ)
is stable at 1 atm?
*105. When is it true that ⌬S ϭ ?
*106. Dissociation reactions are those in which molecules break
apart. Why do high temperatures favor the spontaneity
of most dissociation reactions?
⌬H
ᎏ
T
*107. Estimate the temperature range over which each of the
following standard reactions is spontaneous.
(a) 2Al(s) ϩ 3Cl2(g) 88n 2AlCl3(s)
(b) 2NOCl(g) 88n 2NO(g) ϩ Cl2(g)
(c) 4NO(g) ϩ 6H2O(g) 88n 4NH3(g) ϩ 5O2(g)
(d) 2PH3(g) 88n 3H2(g) ϩ 2P(g)
*108. Estimate the temperature range over which each of the
following standard reactions is spontaneous.
(a) The reaction by which sulfuric acid droplets from polluted
air convert water-insoluble limestone or marble
(calcium carbonate) to slightly soluble calcium sulfate,
which is slowly washed away by rain:
CaCO3(s) ϩ H2SO4(ᐉ) 88n
CaSO4(s) ϩ H2O(ᐉ) ϩ CO2(g)
(b) The reaction by which Antoine Lavoisier achieved the
first laboratory preparation of oxygen in the late eighteenth
century: the thermal decomposition of the
red-orange powder, mercury(II) oxide, to oxygen and the
silvery liquid metal, mercury:
2HgO(s) 88n 2Hg(ᐉ) ϩ O2(g)
(c) The reaction of coke (carbon) with carbon dioxide to
form the reducing agent, carbon monoxide, which is used
to reduce some metal ores to metals:
CO2(g) ϩ C(s) 88n 2CO(g)
(d) The reverse of the reaction by which iron rusts:
2Fe2O3(s) 88n 4Fe(s) ϩ 3O2(g)
*109. Estimate the normal boiling point of tin(IV) chloride,
SnCl4, at 1 atm pressure, using Appendix K.
*110. (a) Estimate the normal boiling point of water, at 1 atm
pressure, using Appendix K. (b) Compare the temperature
obtained with the known boiling point of water. Can
you explain the discrepancy?
*111. Sublimation and subsequent deposition onto a cold surface
are a common method of purification of I2 and other
solids that sublime readily. Estimate the sublimation temperature
(solid to vapor) of the dark violet solid iodine,
I2, at 1 atm pressure, using the data of Appendix K.
Sublimation and deposition of I2.
644 CHAPTER 15: Chemical Thermodynamics
*112. (a) Is the reaction C(diamond) n C(graphite) spontaneous
at 25°C and 1 atm? (b) Now are you worried about
your diamonds turning to graphite? Why or why not?
(c) Is there a temperature at which diamond and graphite
are in equilibrium? If so, what is this temperature?
(d) How do you account for the formation of diamonds
in nature? (Hint: Diamond has a higher density than
graphite.)
Mixed Exercises
*113. An ice calorimeter, shown here, can be used to measure
the amount of heat released or absorbed by a reaction
that is carried out at a constant temperature of 0°C. If
heat is transferred from the system to the bath, some of
the ice melts. A given mass of liquid water has a smaller
volume than the same mass of ice, so the total volume of
the ice and water mixture decreases. Measuring the volume
decrease using the scale at the left indicates the
amount of heat released by the reacting system. As long
as some ice remains in the bath, the temperature remains
at 0°C. In Example 15-2 we saw that the reaction
CuSO4(aq) ϩ 2NaOH(aq) 88n
Cu(OH)2(s) ϩ Na2SO4(aq)
releases 846 J of heat at constant temperature and
pressure when 50.0 mL of 0.400 M CuSO4 solution and
50.0 mL of 0.600 M NaOH solution are allowed to react.
(Because no gases are involved in the reaction, the volume
change of the reaction mixture is negligible.)
Calculate the change in volume of the ice and water mixture
that would be observed if we carried out the same
experiment in an ice calorimeter. The density of H2O(ᐉ)
at 0°C is 0.99987 g/mL and that of ice is 0.917 g/mL.
The heat of fusion of ice at 0°C is 334 J/g.
it is easier to measure the standard enthalpy of combustion
by reaction of the compound with excess O2(g) to
form CO2(g) and H2O(ᐉ). From the following standard
enthalpies of combustion at 25°C, determine ⌬Hf
0
for the
compound. (a) cyclohexane, C6H12(ᐉ), a useful organic
solvent: ⌬H0
combustion ϭ Ϫ3920 kJ/mol; (b) phenol,
C6H5OH(s), used as a disinfectant and in the production
of thermo-setting plastics: ⌬H0
combustion ϭ Ϫ3053 kJ/mol.
*115. Standard entropy changes cannot be measured directly in
the laboratory. They are calculated from experimentally
obtained values of ⌬G0 and ⌬H0. From the data given
here, calculate ⌬S0 at 298 K for each of the following
reactions.
(a) OF2(g) ϩ H2O(g) 88n O2(g) ϩ 2HF(g)
oxygen difluoride
⌬H0 ϭ Ϫ323.2 kJ/mol ⌬G0 ϭ Ϫ358.4 kJ/mol
(b) CaC2(s) ϩ 2H2O(ᐉ) 88n Ca(OH)2(s) ϩ C2H2(g)
calcium carbide acetylene
⌬H0 ϭ Ϫ125.4 kJ/mol ⌬G0 ϭ Ϫ145.4 kJ/mol
(c) CaO(s) ϩ H2O(ᐉ) 88n Ca(OH)2(aq)
⌬H0 ϭ 81.5 kJ/mol ⌬G0 ϭ Ϫ26.20 kJ/mol
*116. Calculate q, w, and ⌬E for the vaporization of 10.0 g of
liquid ethanol (C2H5OH) at 1.00 atm at 78.0°C, to form
gaseous ethanol at 1.00 atm at 78.0°C. Make the following
simplifying assumptions: (a) the density of liquid
ethanol at 78.0°C is 0.789 g/mL, and (b) gaseous ethanol
is adequately described by the ideal gas equation. The
heat of vaporization of ethanol is 855 J/g.
*117. We add 0.100 g of CaO(s) to 125 g H2O at 23.6°C in a
coffee-cup calorimeter. The following reaction occurs.
What will be the final temperature of the solution?
CaO(s) ϩ H2O(ᐉ) 88n Ca(OH)2(aq)
⌬H0 ϭ 81.5 kJ/mol rxn
*118. (a) The accurately known molar heat of combustion of
naphthalene, C10H8(s), ⌬H ϭ Ϫ5156.8 kJ/mol C10H8, is
used to calibrate calorimeters. The complete combustion
of 0.01520 g of C10H8 at constant pressure raises the temperature
of a calorimeter by 0.212°C. Find the heat
capacity of the calorimeter. (b) The initial temperature of
the calorimeter (part a) is 22.102°C; 0.1040 g of C8H18(ᐉ),
octane (molar heat of combustion ⌬H ϭ Ϫ5451.4 kJ/mol
C8H18), is completely burned in the calorimeter. Find the
final temperature of the calorimeter.
CONCEPTUAL EXERCISES
*119. When a gas expands suddenly, it may not have time to
absorb a significant amount of heat: q ϭ 0. Assume that
1.00 mol N2 expands suddenly, doing 3000 J of work.
*114. It is difficult to prepare many compounds directly from
their elements, so ⌬Hf
0
values for these compounds cannot
be measured directly. For many organic compounds,
Ice in
water
Insulation
System
Exercises 645
(a) What is ⌬E for the process? (b) The heat capacity of
N2 is 20.9 J/molи°C. How much does its temperature fall
during this expansion? (This is the principle of most
snow-making machines, which use compressed air mixed
with water vapor.)
*120. As a rubber band is stretched, it gets warmer; when
released, it gets cooler. To obtain the more nearly linear
arrangement of the rubber band’s polymeric material
from the more random relaxed rubber band requires that
there be rotation about carbon–carbon single bonds.
Based on these data, give the sign of ⌬G, ⌬H, and ⌬S for
the stretching of a rubber band and for the relaxing of a
stretched rubber band. What drives the spontaneous
process?
*121. (a) The decomposition of mercury(II) oxide has been used
as a method for producing oxygen, but this is not a recommended
method. Why not? (b) Write the balanced
equation for the decomposition of mercury(II) oxide.
(c) Calculate the ⌬H0, ⌬S0, and ⌬G0 for the reaction.
(d) Is the reaction spontaneous at room temperature?
*122. (a) A student heated a sample of a metal weighing 32.6 g
to 99.83°C and put it into 100.0 g of water at 23.62°C in
a calorimeter. The final temperature was 24.41°C. The
student calculated the specific heat of the metal, but
neglected to use the heat capacity of the calorimeter. The
specific heat of water is 4.184 J/gи°C. What was his
answer? The metal was known to be chromium, molybdenum,
or tungsten. By comparing the value of the
specific heat to those of the metals (Cr, 0.460; Mo, 0.250;
W, 0.135 J/gи°C), the student identified the metal. What
was the metal? (b) A student at the next laboratory bench
did the same experiment, obtained the same data, and
used the heat capacity of the calorimeter in his calculations.
The heat capacity of the calorimeter was 410 J/°C.
Was his identification of the metal different?
*123. A sugar cube dissolves in a cup of coffee in an endothermic
process. (a) Is the entropy change of the system (sugar
plus coffee) greater than, less than, or equal to zero?
(b) Is the entropy change of the universe greater than,
less than, or equal to zero? (c) Is the entropy change of
the surroundings greater than, less than, or equal to zero?
BUILDING YOUR KNOWLEDGE
*124. Energy to power muscular work is produced from stored
carbohydrate (glycogen) or fat (triglycerides). Metabolic
consumption and production of energy are described with
the nutritional “Calorie,” which is equal to 1 kilocalorie.
Average energy output per minute for various activities
follows: sitting, 1.7 kcal; walking, level, 3.5 mph, 5.5 kcal;
cycling, level, 13 mph, 10 kcal; swimming, 8.4 kcal;
running, 10 mph, 19 kcal. Approximate energy values of
some common foods are also given: large apple, 100 kcal;
8-oz cola drink, 105 kcal; malted milkshake, 8 oz milk,
500 kcal; ᎏ3
4
ᎏ cup pasta with tomato sauce and cheese,
195 kcal; hamburger on bun with sauce, 350 kcal;
10-oz sirloin steak, including fat, 1000 kcal. To maintain
body weight, fuel intake should balance energy output.
Prepare a table showing (a) each given food, (b) its fuel
value, and (c) the minutes of each activity that would balance
the kcal of each food.
*125. From its heat of fusion, calculate the entropy change associated
with the melting of one mole of ice at its melting
point. From its heat of vaporization, calculate the entropy
change associated with the boiling of one mole of water
at its boiling point. Are your calculated values consistent
with the simple model that we use to describe order in
solids, liquids, and gases?
*126. The energy content of dietary fat is 39 kJ/g, and for protein
and carbohydrate it is 17 and 16 kJ/g, respectively. A
70.0-kg (155-lb) person utilizes 335 kJ/hr while resting
and 1250 kJ/h while walking 6 km/h. How many hours
would the person need to walk per day instead of resting
if he or she consumed 100 g (about ᎏ1
4
ᎏ lb) of fat instead of
100 g of protein?
*127. The enthalpy change for melting one mole of water at
273 K is ⌬H0
273 ϭ 6010 J/mol, whereas that for vaporizing
a mole of water at 373 K is ⌬H0
373 ϭ 40,660 J/mol.
Why is the second value so much larger?
*128. A 436-g chunk of lead was removed from a beaker of boiling
water, quickly dried, and dropped into a Styrofoam
cup containing 50.0 g of water at 25.0°C. As the system
reached equilibrium, the water temperature rose to
40.8°C. Calculate the heat capacity and the specific heat
of the lead.
*129. Methane, CH4(g), is the main constituent of natural gas.
In excess oxygen, methane burns to CO2(g) and H2O(ᐉ),
whereas in limited oxygen, the products are CO(g) and
H2O(ᐉ). Which would result in a higher temperature: a
gas–air flame or a gas–oxygen flame? How can you tell?
*130. A 0.483-g sample of butter was burned in a bomb
calorimeter whose heat capacity was 4572 J/°C, and the
temperature was observed to rise from 24.76 to 27.93°C.
Calculate the fuel value of butter in (a) kJ/g; (b) nutritional
Calories/g (one nutritional Calorie is equal to one
kilocalorie); (c) nutritional Calories/5-gram pat.
A methane flame.
Chemical
Kinetics16
16OUTLINE
16-1 The Rate of a Reaction
Factors That Affect Reaction Rates
16-2 Nature of the Reactants
16-3 Concentrations of Reactants:
The Rate-Law Expression
16-4 Concentration versus Time:
The Integrated Rate Equation
16-5 Collision Theory of Reaction
Rates
16-6 Transition State Theory
16-7 Reaction Mechanisms and the
Rate-Law Expression
16-8 Temperature: The Arrhenius
Equation
16-9 Catalysts
OBJECTIVES
After you have studied this chapter, you should be able to
• Express the rate of a chemical reaction in terms of changes in concentrations of
reactants and products with time
• Describe the experimental factors that affect the rates of chemical reactions
• Use the rate-law expression for a reaction—the relationship between concentration and
rate
• Use the concept of order of a reaction
• Apply the method of initial rates to find the rate-law expression for a reaction
• Use the integrated rate-law expression for a reaction—the relationship between
concentration and time
• Analyze concentration-versus-time data to determine the order of a reaction
• Describe the collision theory of reaction rates
• Describe the main aspects of transition state theory and the role of activation energy in
determining the rate of a reaction
• Explain how the mechanism of a reaction is related to its rate-law expression
• Predict the rate-law expression that would result from a proposed reaction mechanism
• Identify reactants, products, intermediates, and catalysts in a multistep reaction
mechanism
• Explain how temperature affects rates of reactions
• Use the Arrhenius equation to relate the activation energy for a reaction to changes in
its rate constant with changing temperature
• Explain how a catalyst changes the rate of a reaction
• Describe homogeneous catalysis and heterogeneous catalysis
A burning building is an example
of a rapid, highly exothermic
reaction. Firefighters use basic
principles of chemical kinetics to
battle the fire. When water is
sprayed onto a fire, its evaporation
absorbs a large amount of energy;
this lowers the temperature and
slows the reaction. Other common
methods for extinguishing fires
include covering them with CO2 (as
with most household extinguishers),
which decreases the supply of
oxygen, and backburning (for grass
and forest fires), which removes
combustible material. In both cases,
the removal of a reactant slows (or
stops) the reaction.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 6.3, Thermodynamics and
Kinetics, and Screen 20.2, Reaction
Spontaneity (Thermodynamics and
Kinetics).
648 CHAPTER 16: Chemical Kinetics
WW
e are all familiar with processes in which some quantity changes with time—
a car travels at 40 miles/hour, a faucet delivers water at 3 gallons/minute, or
a factory produces 32,000 tires/day. Each of these ratios is called a rate. The
rate of a reaction describes how fast reactants are used up and products are formed.
Chemical kinetics is the study of rates of chemical reactions, the factors that affect reaction
rates, and the mechanisms (the series of steps) by which reactions occur.
Our experience tells us that different chemical reactions occur at very different rates.
For instance, combustion reactions—such as the burning of methane, CH4, in natural gas
and the combustion of isooctane, C8H18, in gasoline—proceed very rapidly, sometimes
even explosively.
CH4(g) ϩ 2O2(g) 88n CO2(g) ϩ 2H2O(g)
2C8H18(g) ϩ 25O2(g) 88n 16CO2(g) ϩ 18H2O(g)
On the other hand, the rusting of iron occurs only very slowly.
In our study of thermodynamics, we learned to assess whether a particular reaction was
favorable. The question of whether substantial reaction would occur in a certain time
period is addressed by kinetics. If a reaction is not thermodynamically favored, it will not
occur appreciably under the given conditions. Even though a reaction is thermodynamically
favored, it might not occur at a measurable rate.
The reactions of strong acids with strong bases are thermodynamically favored and
occur at very rapid rates. Consider, for example, the reaction of hydrochloric acid solution
with solid magnesium hydroxide. It is thermodynamically spontaneous at standard
state conditions, as indicated by the negative ⌬G0
rxn value. It also occurs rapidly.
2HCl(aq) ϩ Mg(OH)2(s) 88n MgCl2(aq) ϩ 2H2O(ᐉ) ⌬G0
rxn ϭ Ϫ97 kJ/mol
The reaction of diamond with oxygen is also thermodynamically spontaneous.
C(diamond) ϩ O2(g) 88n CO2(g) ⌬G0
rxn ϭ Ϫ397 kJ/mol
However, we know from experience that diamonds exposed to air, even over long periods,
do not react to form carbon dioxide. The reaction does not occur at an observable rate
near room temperature.
The reaction of graphite with oxygen is also spontaneous, with a similar value of ⌬G0
rxn,
Ϫ394 kJ/mol. Once it is started, this reaction occurs rapidly. These observations of reaction
speeds are explained by kinetics, not thermodynamics.
THE RATE OF A REACTION
Rates of reactions are usually expressed in units of moles per liter per unit time. If we
know the chemical equation for a reaction, its rate can be determined by following the
change in concentration of any product or reactant that can be detected quantitatively.
To describe the rate of a reaction, we must determine the concentration of a reactant
or product at various times as the reaction proceeds. Devising effective methods for this
is a continuing challenge for chemists who study chemical kinetics. If a reaction is slow
enough, we can take samples from the reaction mixture after successive time intervals and
then analyze them. For instance, if one reaction product is an acid, its concentration can
be determined by titration (Section 11-2) after each time interval. The reaction of ethyl
acetate with water in the presence of a small amount of strong acid produces acetic acid.
The extent of the reaction at any time can be determined by titration of the acetic acid.
16-1
The rusting of iron is a complicated
process. It can be represented in
simplified form as
4Fe(s) ϩ 3O2(g) 88n 2Fe2O3(s)
Recall that the units kJ/mol refer to
the numbers of moles of reactants and
products in the balanced equation.
This is one of the reactions that occurs
in a human digestive system when an
antacid containing relatively insoluble
magnesium hydroxide neutralizes
excess stomach acid.
A methane flame is a rapid reaction.
Figure 16-1 (a) Blue dye is
reacting with bleach, which converts
it into a colorless product. The color
decreases and eventually disappears.
The rate of the reaction could be
determined by repeatedly measuring
both the color intensity and the
elapsed time. The concentration of
dye could be calculated from the
intensity of the blue color. (Solvent
molecules and reaction product
molecules have been omitted from
the diagram for clarity).
(b) A spectroscopic method for
determining reaction rates. Light of
a wavelength that is absorbed by
some substance whose concentration
is changing is passed through a
reaction chamber. Recording the
change in light intensity gives a
measure of the changing
concentration of a reactant or
product as the reaction progresses.
16-1 The Rate of a Reaction 649
This approach is suitable only if the reaction is sufficiently slow that the time elapsed
during withdrawal and analysis of the sample is negligible. Sometimes the sample is withdrawn
and then quickly cooled (“quenched”). This slows the reaction (Section 16-8) so
much that the desired concentration does not change significantly while the analysis is
performed.
It is more convenient, especially when a reaction is rapid, to use a technique that continually
monitors the change in some physical property of the system. If one of the reactants
or products is colored, the increase (or decrease) in intensity of its color might be used
to measure a decrease or increase in its concentration. Such an experiment is a special
case of spectroscopic methods. These methods involve passing light (visible, infrared, or
ultraviolet) through the sample. The light should have a wavelength that is absorbed by
some substance whose concentration is changing (Figure 16-1). An appropriate lightsensing
apparatus provides a signal that depends on the concentration of the absorbing
substance. Modern techniques that use computer-controlled pulsing and sensing of lasers
have enabled scientists to sample concentrations at intervals on the order of picoseconds
OCH2CH3(aq)CH3C
O
H2O(ᐉ)
Hϩ
ϩ OH(aq) CH3CH2OH(aq)CH3C
O
ϩ
ethanolacetic acidethyl acetate
(a)
(b)
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 15.2, Rates of Chemical
Reactions.
(1 picosecond ϭ 10Ϫ12 second) or even femtoseconds (1 femtosecond ϭ 10Ϫ15 second).
Such studies have yielded information about very fast reactions, such as energy transfer
resulting from absorption of light in photosynthesis.
If the progress of a reaction causes a change in the total number of moles of gas present,
the change in pressure of the reaction mixture (held at constant temperature and constant
volume) lets us measure how far the reaction has gone. For instance, the decomposition
of dinitrogen pentoxide, N2O5(g), has been studied by this method.
2N2O5(g) 88n 4NO2(g) ϩ O2(g)
For every two moles of N2O5 gas that react, a total of five moles of gas is formed (four
moles of NO2 and one mole of O2). The resulting increase in pressure can be related by
the ideal gas equation to the total number of moles of gas present. This indicates the
extent to which the reaction has occurred.
Once we have measured the changes in concentrations of reactants or products with
time, how do we describe the rate of a reaction? Consider a hypothetical reaction.
aA ϩ bB 88n cC ϩ dD
In this generalized representation, a represents the coefficient of substance A in the
balanced chemical equation, b is the coefficient of substance B, and so on. For example,
in an earlier equation given for the decomposition of N2O5, a ϭ 2, A represents N2O5,
c ϭ 4, C represents NO2, and so on.
The amount of each substance present can be given by its concentration, usually
expressed as molarity (mol/L) and designated by brackets. The rate at which the reaction
proceeds can be described in terms of the rate at which one of the reactants disappears,
Ϫ⌬[A]/⌬t or Ϫ⌬[B]/⌬t, or the rate at which one of the products appears, ⌬[C]/⌬t or
⌬[D]/⌬t. The reaction rate must be positive because it describes the forward (left-to-right)
reaction, which consumes A and B. The concentrations of reactants A and B decrease in
the time interval ⌬t. Thus, ⌬[A]/⌬t and ⌬[B]/⌬t would be negative quantities. The purpose
of a negative sign in the definition of a rate of reaction is to make the rate a positive
quantity.
If no other reaction takes place, the changes in concentration are related to one another.
For every a mol/L that [A] decreases, [B] must decrease by b mol/L, [C] must increase by
c mol/L, and so on. We wish to describe the rate of reaction on a basis that is the same
regardless of which reactant or product we choose to measure. To do this, we can describe
the number of moles of reaction that occur per liter in a given time. For instance, this is
accomplished for reactant A as
΂ ΃΂ ΃ϭ Ϫ΂ ΃΂ ΃
The units for rate of reaction are , which we usually shorten to ᎏ
L
m
иti
o
m
l
e
ᎏ or
molиLϪ1иtimeϪ1. The units ᎏ
m
L
ol
ᎏ represent molarity, M, so the units for rate of reaction
can also be written as ᎏ
ti
M
me
ᎏ or MиtimeϪ1. Similarly, we can divide each concentration
change by its coefficient in the balanced equation. Bringing signs to the beginning of each
term, we write the rate of reaction based on the rate of change of concentration of each
species.
mol rxn
ᎏ
Lиtime
⌬[A]
ᎏ
⌬t
1 mol rxn
ᎏᎏ
a mol A
rate of decrease
ᎏᎏ
in [A]
1 mol rxn
ᎏᎏ
a mol A
In reactions involving gases, rates of
reactions may be related to rates of
change of partial pressures. Pressures
of gases and concentrations of gases
are directly proportional.
PV ϭ nRT or P ϭ RT ϭ MRT
where M is molarity.
n
ᎏ
V
650 CHAPTER 16: Chemical Kinetics
The Greek “delta,” ⌬, stands for
“change in,” just as it did in
Chapter 15.
in terms of in terms of
reactants products
rate of rate of rate of rate of
rate of reaction ϭ Ϫᎏ
1
a
ᎏ
΂decrease
΃ϭ Ϫᎏ
1
b
ᎏ
΂decrease
΃ϭ ᎏ
1
c
ᎏ
΂increase
΃ϭ ᎏ
1
d
ᎏ
΂increase
΃in [A] in [B] in [C] in [D]
rate of reaction ϭ Ϫᎏ
1
a
ᎏ ΂ᎏ
⌬
⌬
[A
t
]
ᎏ΃ ϭ Ϫᎏ
1
b
ᎏ ΂ᎏ
⌬
⌬
[B
t
]
ᎏ΃ ϭ ᎏ
1
c
ᎏ ΂ᎏ
⌬
⌬
[C
t
]
ᎏ΃ ϭ ᎏ
1
d
ᎏ ΂ᎏ
⌬
⌬
[D
t
]
ᎏ΃
This representation gives several equalities, any one of which can be used to relate changes
in observed concentrations to the rate of reaction.
The expressions just given describe the average rate over a time period ⌬t. The rigorous
expressions for the rate at any instant involve the derivatives of concentrations with respect
to time.
Ϫᎏ
1
a
ᎏ ΂ᎏ
d[
d
A
t
]
ᎏ΃, ᎏ
1
c
ᎏ ΂ᎏ
d[
d
C
t
]
ᎏ΃, and so on.
The shorter the time period, the closer is to the corresponding
derivative.
⌬(concentration)
ᎏᎏ
⌬t
16-1 The Rate of a Reaction 651






























Problem-Solving Tip: Signs and Divisors in Expressions for Rate
As an analogy to these chemical reaction rate expressions, suppose we make sardine sandwiches
by the following procedure:
2 bread slices ϩ 3 sardines ϩ 1 pickle 88n 1 sandwich
As time goes by, the number of sandwiches increases, so ⌬(sandwiches) is positive; the
rate of the process is given by ⌬(sandwiches)/⌬(time). Alternatively, we would count the
decreasing number of pickles at various times. Because ⌬(pickles) is negative, we must
multiply by (Ϫ1) to make the rate positive; rate ϭ Ϫ⌬(pickles)/⌬(time). If we measure
the rate by counting slices of bread, we must also take into account that bread slices are
consumed twice as fast as sandwiches are produced, so rate ϭ Ϫᎏ1
2
ᎏ(⌬(bread)/⌬(time)). Four
different ways of describing the rate all have the same numerical value.
rate ϭ ΂ ΃ϭ Ϫ ΂ ΃ϭ Ϫ ΂ ΃ϭ Ϫ΂ ΃⌬(pickles)
ᎏᎏ
⌬t
⌬(sardines)
ᎏᎏ
⌬t
1
ᎏ
3
⌬(bread)
ᎏ
⌬t
1
ᎏ
2
⌬(sandwiches)
ᎏᎏ
⌬t
Consider as a specific chemical example the gas-phase reaction that occurs when we
mix 1.000 mole of hydrogen and 2.000 moles of iodine chloride at 230°C in a closed
1.000-liter container.
H2(g) ϩ 2ICl(g) 88n I2(g) ϩ 2HCl(g)
The coefficients tell us that one mole of H2 disappears for every two moles of ICl that
disappear and for every one mole of I2 and two moles of HCl that are formed. In other
terms, the rate of disappearance of moles of H2 is one-half the rate of disappearance of
moles of ICl, and so on. So we write the rate of reaction as
rate of rate of rate of rate of
rate of reaction ϭ
΂decrease
΃ϭ ᎏ
1
2
ᎏ
΂decrease
΃ ϭ
΂increase
΃ϭ ᎏ
1
2
ᎏ
΂increase
΃r
in [H2] in [ICl] in [I2] in [HCl]
rate of reaction ϭ Ϫ΂ᎏ
⌬[
⌬
H
t
2]
ᎏ΃ ϭ Ϫᎏ
1
2
ᎏ ΂ᎏ
⌬[
⌬
IC
t
l]
ᎏ΃ ϭ ΂ᎏ
⌬
⌬
[I
t
2]
ᎏ΃ ϭ ᎏ
1
2
ᎏ ΂ᎏ
⌬[H
⌬t
Cl]
ᎏ΃r
Table 16-1 lists the concentrations of reactants remaining at 1-second intervals, beginning
with the time of mixing (t ϭ 0 seconds). The average rate of reaction over each
1-second interval is indicated in terms of the rate of decrease in concentration of hydrogen.
Verify for yourself that the rate of loss of ICl is twice that of H2. Therefore, the rate of
reaction could also be expressed as rate ϭ Ϫᎏ1
2
ᎏ (⌬[ICl]/⌬t). Increases in concentration of
products could be used instead. Figure 16-2 shows graphically the rates of change of
concentrations of all reactants and products.
For example, the average rate over the
interval from 1 to 2 seconds can be
calculated as
Ϫᎏ
⌬[
⌬
H
t
2]
ᎏ ϭ
Ϫ
ϭ 0.148 molиLϪ1иsϪ1
ϭ 0.148 MиsϪ1
This does not mean that the reaction
proceeds at this rate during the entire
interval.
(0.526 Ϫ 0.674) molиLϪ1
ᎏᎏᎏ
(2 Ϫ 1) s
TABLE 16-1 Concentration and Rate Data for Reaction of 2.000 M ICl and
1.000 M H2 at 230°C
Average Rate During
[ICl] [H2]
One Time Interval ϭ Ϫ ᎏ
⌬[
⌬
H
t
2]
ᎏ
Time (t)
(mol/L) (mol/L) (MиsϪ1) (seconds)
2.000 1.000 0
0.326
1.348 0.674 1
0.148
1.052 0.526 2
0.090
0.872 0.436 3
0.062
0.748 0.374 4
0.046
0.656 0.328 5
0.035
0.586 0.293 6
0.028
0.530 0.265 7
0.023
0.484 0.242 8
Figure 16-2 Plot of concentrations
of all reactants and products versus
time in the reaction of 1.000 M H2
with 2.000 M ICl at 230°C, from
data in Table 16-1 (and a few more
points).
[I2]
[H2]
[HCl]
[ICl]
2.000 M
1.000 M
0
0 5 10 15
Time (s)
Concentrationsofreactantsandproducts
Figure 16-3 is a plot of the hydrogen concentration versus time, using data of Table
16-1. The initial rate, or the rate at the instant of mixing the reactants, is the negative of
the slope at t ϭ 0. The instantaneous rate of reaction at time t (2.0 seconds, for example)
is the negative of the slope of the tangent to the curve at time t. We see that the rate
decreases with time; lower concentrations of H2 and ICl result in slower reaction. Had
we plotted concentration of a product versus time, the rate would have been related to
the positive slope of the tangent at time t.
EXAMPLE 16-1 Rate of Reaction
At some time, we observe that the reaction 2N2O5(g) n 4NO2(g) ϩ O2(g) is forming NO2
at the rate of 0.0072 ᎏ
m
Lи
o
s
l
ᎏ.
(a) What is the rate of change of [O2], ᎏ
⌬[
⌬
O
t
2]
ᎏ, in ᎏ
m
Lи
o
s
l
ᎏ, at this time?
(b) What is the rate of change of [N2O5], ᎏ
⌬[N
⌬
2
t
O5]
ᎏ, in ᎏ
m
Lи
o
s
l
ᎏ, at this time?
(c) What is the rate of reaction at this time?
Plan
We can use the mole ratios from the balanced equation to determine the rates of change of
other products and reactants. The rate of reaction can then be derived from any one of these
individual rates.
Solution
(a) The balanced equation gives the reaction ratio 1 mol ᎏ
O
4
2
ᎏ mol NO2.
rate of change of [O2] ϭ ϭ ϫ ϭ 0.0018
mol O2
ᎏ
Lиs
1 mol O2
ᎏᎏ
4 mol NO2
0.0072 mol NO2
ᎏᎏ
Lиs
⌬[O2]
ᎏ
⌬t
Figure 16-3 Plot of H2
concentration versus time for the
reaction of 1.000 M H2 with
2.000 M ICl. The instantaneous rate
of reaction at any time, t, equals the
negative of the slope of the tangent
to this curve at time t. The initial
rate of the reaction is equal to the
negative of the initial slope (t ϭ 0).
The determination of the
instantaneous rate at t ϭ 2 seconds is
illustrated. (If you do not recall how
to find the slope of a straight line,
refer to Figure 16-5.)
16-1 The Rate of a Reaction 653
Suppose a driver travels 40 miles in an
hour; we describe his average speed
(rate) as 40 mi/h. This does not
necessarily mean that he drove at a
steady speed. He might have stopped
at a few traffic signals, made a fuel
stop, driven sometimes faster,
sometimes slower—his instantaneous
rate (the rate at which he was traveling
at any instant) was quite changeable.
1.000 M
0.600 M
0.300 M
0
0 2 4 6 8 10 12 14
Time (s)
3.95
⌬t
⌬t
⌬[H2]
Ϫslope ϭ
Ϫslope ϭ initial rate
ϭ Rate at (t ϭ 2.0 s) ϭ ϭ 0.107 M sϪ1
Ϫ⌬[H2]
1.15
[H2]
(3.95 Ϫ 1.15) s
Ϫ(0.300 Ϫ 0.600) M
(b) The balanced equation shows that 2 mol N2O5 is consumed for every 4 mol NO2 that is
formed. Because [N2O5] is decreasing as [NO2] increases, we should write the reaction ratio as
.
rate of change of [N2O5] ϭ ϭ ϫ
ϭ Ϫ0.0036
The rate of change of [N2O5] with time, ᎏ
⌬[N
⌬
2
t
O5]
ᎏ, is Ϫ0.0036 ᎏ
mol
L
N
иs
2O5
ᎏ, a negative number,
because N2O5, a reactant, is being used up.
(c) The rate of reaction can be calculated from the rate of decrease of any reactant concentration
or the rate of increase of any product concentration.
rate of reaction ϭ Ϫᎏ
1
2
ᎏ ΂ ΃ϭ Ϫᎏ
1
2
ᎏ ΂Ϫ0.0036 ΃ϭ 0.0018
rate of reaction ϭ ΂ ΃ϭ ΂0.0072 ΃ϭ 0.0018
rate of reaction ϭ ΂ ΃ϭ 0.0018
We see that the rate of reaction is the same, no matter which reactant or product we use to
determine it. Remember that the mol in these units is interpreted as “moles of reaction.”
You should now work Exercise 10.
FACTORS THAT AFFECT REACTION RATES
Often we want a reaction to take place rapidly enough to be practical but not so rapidly
as to be dangerous. The controlled burning of fuel in an internal combustion engine is
an example of such a process. On the other hand, we want some undesirable reactions,
such as the spoiling of food, to take place more slowly.
Four factors have marked effects on the rates of chemical reactions. They are (1) nature
of the reactants, (2) concentrations of the reactants, (3) temperature, and (4) the presence
of a catalyst. Understanding their effects can help us control the rates of reactions in desirable
ways. The study of these factors gives important insight into the details of the
processes by which a reaction occurs. This kind of study is the basis for developing theories
of chemical kinetics. Now we study these factors and the related theories—collision
theory and transition state theory.
mol
ᎏ
Lиs
⌬[O2]
ᎏ
⌬t
1
ᎏ
1
mol
ᎏ
Lиs
mol
ᎏ
Lиs
1
ᎏ
4
⌬[NO2]
ᎏ
⌬t
1
ᎏ
4
mol
ᎏ
Lиs
mol
ᎏ
Lиs
⌬[N2O5]
ᎏᎏ
⌬t
mol N2O5
ᎏᎏ
Lиs
Ϫ2 mol N2O5
ᎏᎏ
4 mol NO2
0.0072 mol NO2
ᎏᎏ
Lиs
⌬[N2O5]
ᎏᎏ
⌬t
Ϫ2 mol N2O5
ᎏᎏ
4 mol NO2
654 CHAPTER 16: Chemical Kinetics
NATURE OF THE REACTANTS
The physical states of reacting substances are important in determining their reactivities.
A puddle of liquid gasoline can burn smoothly, but gasoline vapors can burn explosively.
Two immiscible liquids may react slowly at their interface, but if they are intimately mixed
to provide better contact, the reaction speeds up. White phosphorus and red phosphorus
are different solid forms (allotropes) of elemental phosphorus. White phosphorus ignites
when exposed to oxygen in the air. By contrast, red phosphorus can be kept in open
containers for long periods of time without noticeable reaction.
Samples of dry solid potassium sulfate, K2SO4, and dry solid barium nitrate, Ba(NO3)2,
can be mixed with no appreciable reaction occurring for several years. But if aqueous solutions
of the two are mixed, a reaction occurs rapidly, forming a white precipitate of barium
sulfate.
Ba2ϩ(aq) ϩ SO4
2Ϫ(aq) 88n BaSO4(s) (net ionic equation)
Chemical identities of elements and compounds affect reaction rates. Metallic sodium,
with its low ionization energy, reacts rapidly with water at room temperature; metallic
calcium has a higher ionization energy and reacts only slowly with water at room temperature.
Solutions of a strong acid and a strong base react rapidly when they are mixed
because the interactions involve mainly electrostatic attractions between ions in solution.
Reactions that involve the breaking of covalent bonds are usually slower.
The extent of subdivision of solids or liquids can be crucial in determining reaction
rates. Large chunks of most metals do not burn. But many powdered metals, with larger
surface areas and hence more atoms exposed to the oxygen of the air, burn easily. One
pound of fine iron wire rusts much more rapidly than a solid one-pound chunk of iron.
Violent explosions sometimes occur in grain elevators, coal mines, and chemical plants in
which large amounts of powdered substances are produced. These explosions are examples
of the effect of large surface areas on rates of reaction. The rate of reaction depends
on the surface area or degree of subdivision. The ultimate degree of subdivision would
make all reactant molecules (or ions or atoms) accessible to react at any given time. This
situation can be achieved when the reactants are in the gaseous state or in solution.
16-2
Two allotropes of phosphorus.
White phosphorus (above) ignites
and burns rapidly when exposed to
oxygen in the air, so it is stored
under water. Red phosphorus (below)
reacts with air much more slowly,
and can be stored in contact with air.
16-2 Nature of the Reactants 655
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 15.3, Control of Reaction Rates
(1): Surface Area.
Powdered chalk (mostly calcium
carbonate, CaCO3) reacts rapidly with
dilute hydrochloric acid because it has
a large total surface area. A stick of
chalk has a much smaller surface area,
so it reacts much more slowly.
CONCENTRATIONS OF REACTANTS:
THE RATE-LAW EXPRESSION
As the concentrations of reactants change at constant temperature, the rate of reaction
changes. We write the rate-law expression (often called simply the rate law) for a reaction
to describe how its rate depends on concentrations; this rate law is experimentally
deduced for each reaction from a study of how its rate varies with concentration.
The rate-law expression for a reaction in which A, B, . . . are reactants has the
general form
rate ϭ k[A]x[B]y . . .
The constant k is called the specific rate constant (or just the rate constant) for
the reaction at a particular temperature. The values of the exponents, x and y, and
of the rate constant, k bear no necessary relationship to the coefficients in the balanced
chemical equation for the overall reaction and must be determined experimentally.
The powers to which the concentrations are raised, x and y, are usually integers or zero
but are occasionally fractional or even negative. A power of one means that the rate is
directly proportional to the concentration of that reactant. A power of two means that the
rate is directly proportional to the square of that concentration. A power of zero means
that the rate does not depend on the concentration of that reactant, so long as some of
the reactant is present. The value of x is said to be the order of the reaction with respect
to A, and y is the order of the reaction with respect to B. The overall order of the reaction
is x ϩ y. Examples of observed rate laws for some reactions follow.
1. 3NO(g) 88n N2O(g) ϩ NO2(g)
rate ϭ k[NO]2 second order in NO; second order overall
2. 2NO2(g) ϩ F2(g) 88n 2NO2F(g)
rate ϭ k[NO2][F2] first order in NO2 and first order in F2;
second order overall
3. 2NO2(g) 88n 2NO(g) ϩ O2(g) rate ϭ k[NO2]2 second order in NO2;
second order overall
4. H2O2(aq) ϩ 3IϪ(aq) ϩ 2Hϩ(aq) 88n 2H2O(ᐉ) ϩ I3
Ϫ(aq)
rate ϭ k[H2O2][IϪ] first order in H2O2 and first order in IϪ;
zero order in Hϩ; second order overall
We see that the orders (exponents) in the rate law expression may or may not match
the coefficients in the balanced equation. There is no way to predict reaction orders
from the balanced overall chemical equation. The orders must be determined exper-
imentally.
It is important to remember the following points about this specific rate constant, k.
16-3
The word order is used in kinetics in
its mathematical meaning. This use is
unrelated to the order–disorder
discussion of entropy (Chapter 15).
656 CHAPTER 16: Chemical Kinetics
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 15.4, Control of Reaction Rates
(2): Concentration Dependence.
More details about values and units of
k will be discussed in later sections.
Any number raised to the zero power
is one. Here [Hϩ]0 ϭ 1.
1. Once the reaction orders are known, experimental data must be used to determine
the value of k for the reaction at appropriate conditions.
2. The value we determine is for a specific reaction, represented by a balanced equa-
tion.
3. The units of k depend on the overall order of the reaction.
4. The value we determine does not change with concentrations of either reactants
or products.
5. The value we determine does not change with time (Section 16-4).
6. The value we determine refers to the reaction at a particular temperature and
changes if we change the temperature (Section 16-8).
7. The value we determine depends on whether a catalyst is present (Section 16-9).
EXAMPLE 16-2 Interpretation of the Rate Law
For a hypothetical reaction
A ϩ B ϩ C 88n products
the rate law is determined to be
rate ϭ k[A][B]2
What happens to the reaction rate when we make each of the following concentration
changes?
(a) We double the concentration of A without changing the concentration of B or C. (b) We
double the concentration of B without changing the concentration of A or C. (c) We double
the concentration of C without changing the concentration of A or B. (d) We double all three
concentrations simultaneously.
Plan
We interpret the rate law to predict the changes in reaction rate. We remember that changing
concentrations does not change the value of k.
Solution
(a) We see that rate is directly proportional to the first power of [A]. We do not change [B] or
[C]. Doubling [A] (i.e., increasing [A] by a factor of 2) causes the reaction rate to increase by
a factor of 21 ϭ 2 so the reaction rate doubles.
(b) We see that rate is directly proportional to the second power of [B]. We do not change [A]
or [C]. Doubling [B] (i.e., increasing [B] by a factor of 2) causes the reaction rate to increase
by a factor of 22 ϭ 4.
(c) The reaction rate is independent of [C], so changing [C] causes no change in reaction rate.
(d) Doubling all concentrations would cause the changes described in (a), (b), and (c) simultaneously.
The rate would increase by a factor of 2 due to the change in [A], by a factor of 4 due
to the change in [B], and be unaffected by the change in [C]. The result is that the reaction
rate increases by a factor of 21 ϫ 22 ϭ 8.
You should now work Exercises 14 and 15.
16-3 Concentrations of Reactants: The Rate-Law Expression 657
We can use the method of initial rates to deduce the rate law from experimentally
measured rate data. Usually we know the concentrations of all reactants at the start
of the reaction. We can then measure the initial rate of the reaction, corresponding
to these initial concentrations. The following tabulated data refer to the hypothetical
reaction
A ϩ 2B 88n C
at a specific temperature. The brackets indicate the concentrations of the reacting species
at the beginning of each experimental run listed in the first column—that is, the initial
concentrations for each experiment.
Initial Initial Initial Rate of
Experiment [A] [B] Formation of C
1 1.0 ϫ 10Ϫ2 M 1.0 ϫ 10Ϫ2 M 1.5 ϫ 10Ϫ6 MиsϪ1
2 1.0 ϫ 10Ϫ2 M 2.0 ϫ 10Ϫ2 M 3.0 ϫ 10Ϫ6 MиsϪ1
3 2.0 ϫ 10Ϫ2 M 1.0 ϫ 10Ϫ2 M 6.0 ϫ 10Ϫ6 MиsϪ1
Because we are describing the same reaction in each experiment, each is governed by
the same rate-law expression. This expression has the form
rate ϭ k[A]x[B]y
Let’s compare the initial rates of formation of product (reaction rates) for different experimental
runs to see how changes in concentrations of reactants affect the rate of reaction.
This lets us evaluate x and y, and then k.
We see that the initial concentration of A is the same in experiments 1 and 2; for these
trials, any change in reaction rate would be due to different initial concentrations of B.
Comparing these two experiments, we see that [B] has been changed by a factor of
ϭ 2.0 ϭ [B] ratio
The rate changes by a factor of
ϭ 2.0 ϭ rate ratio
The exponent y can be deduced from
rate ratio ϭ ([B] ratio)
y
2.0 ϭ (2.0)
y
so y ϭ 1
The reaction is first order in [B]. Thus far we know that the rate expression is
rate ϭ k[A]x[B]1
To evaluate x, we observe that the concentrations of [A] are different in experiments 1
and 3. For these two trials, the initial concentration of B is the same, so any change in
reaction rate would be due to different initial concentrations of A. Comparing these two
experiments, we see that [A] has been multiplied by a factor of
ϭ 2.0 ϭ [A] ratio
2.0 ϫ 10Ϫ2
ᎏᎏ
1.0 ϫ 10Ϫ2
3.0 ϫ 10Ϫ6
ᎏᎏ
1.5 ϫ 10Ϫ6
2.0 ϫ 10Ϫ2
ᎏᎏ
1.0 ϫ 10Ϫ2
In such an experiment we often keep
some initial concentrations the same
and vary others by simple factors, such
as 2 or 3. This makes it easier to
access the effect of each change on
the rate.
658 CHAPTER 16: Chemical Kinetics
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 15.5, Determination of Rate
Equations (1): Method of Initial Rates.
Rate increases by a factor of
ϭ 4.0 ϭ rate ratio
The exponent x can be deduced from
rate ratio ϭ ([A] ratio)x
4.0 ϭ (2.0)x so x ϭ 2
The reaction is second order in [A]. We can now write its rate-law expression.
rate ϭ k[A]2[B]
The specific rate constant, k, can be evaluated by substituting any of the three sets of data
into the rate-law expression. Using the data from experiment 1 gives
rate1 ϭ k[A]1
2[B]1 or k ϭ
k ϭ ϭ 1.5 MϪ2иsϪ1
At the temperature at which the measurements were made, the rate-law expression for
this reaction is
rate ϭ k[A]2[B] or rate ϭ 1.5 MϪ2иsϪ1 [A]2[B]
We can check our result by evaluating k from one of the other sets of data.
1.5 ϫ 10Ϫ6 MиsϪ1
ᎏᎏᎏᎏ
(1.0 ϫ 10Ϫ2 M)2(1.0 ϫ 10Ϫ2 M)
rate1
ᎏ
[A]1
2[B]1
6.0 ϫ 10Ϫ6
ᎏᎏ
1.5 ϫ 10Ϫ6
Remember that the specific rate
constant k does not change with
concentration. Only a temperature
change or the introduction of a catalyst
can change the value of k.
The units of k depend on the overall
order of the reaction, consistent
with converting the product of
concentrations on the right to
concentration/time on the left. For any
reaction that is third order overall, the
units of k are MϪ2 иtimeϪ1.
16-3 Concentrations of Reactants: The Rate-Law Expression 659
An Alternative Method
We can also use a simple algebraic approach to find the exponents in a rate-law expression.
Consider the set of rate data given earlier for the hypothetical reaction
A ϩ 2B 88n C
Problem-Solving Tip: Be Sure to Use the Rate of Reaction
The rate-law expression should always give the dependence of the rate of reaction on
concentrations. The data for the preceding calculation describe the rate of formation of
the product C; the coefficient of C in the balanced equation is one, so the rate of reaction
is equal to the rate of formation of C. If the coefficient of the measured substance
had been 2, then before we began the analysis we should have divided each value of the
“initial rate of formation” by 2 to obtain the initial rate of reaction. For instance, suppose
we measure the rate of formation of AB in the reaction
A2 ϩ B2 88n 2AB
Then
rate of increase ϭ ΂ ΃ϭ (rate of formation of AB)
Then we would analyze how this reaction rate changes as we change the concentrations
of reactants.
1
ᎏ
2
⌬[AB]
ᎏ
⌬t
1
ᎏ
2
When heated in air, steel wool glows
but does not burn rapidly, due to the
low O2 concentration in air (about
21%). When the glowing steel wool
is put into pure oxygen, it burns
vigorously because of the much
greater accessibility of O2 reactant
molecules.
Initial Rate of
Initial Initial Formation of C
Experiment [A] [B] (MиsϪ1)
1 1.0 ϫ 10Ϫ2 M 1.0 ϫ 10Ϫ2 M 1.5 ϫ 10Ϫ6
2 1.0 ϫ 10Ϫ2 M 2.0 ϫ 10Ϫ2 M 3.0 ϫ 10Ϫ6
3 2.0 ϫ 10Ϫ2 M 1.0 ϫ 10Ϫ2 M 6.0 ϫ 10Ϫ6
Because we are describing the same reaction in each experiment, all the experiments
are governed by the same rate-law expression,
rate ϭ k[A]x[B]y
The initial concentration of A is the same in experiments 1 and 2, so any change in
the initial rates for these experiments would be due to different initial concentrations of
B. To evaluate y, we solve the ratio of the rate-law expressions of these two experiments
for y. We can divide the first rate-law expression by the corresponding terms in the second
rate-law expression.
ᎏ
r
r
a
a
t
t
e
e
1
2
ᎏ ϭ ᎏ
k
k
[
[
A
A
]
]
1
2
x
x
[
[
B
B
]
]
1
2
y
y
ᎏ
The value of k always cancels from such a ratio because it is constant at a particular temperature.
The initial concentrations of A are equal, so they too cancel. Thus, the expression
simplifies to
ᎏ
r
r
a
a
t
t
e
e
1
2
ᎏ ϭ ΂ᎏ
[
[
B
B
]
]
1
2
ᎏ΃
y
The only unknown in this equation is y. We substitute data from experiments 1 and 2 into
the equation, which gives us
ϭ ΂ ΃
y
0.5 ϭ (0.5)y
so y ϭ 1
Thus far, we know that the rate-law expression is
rate ϭ k[A]x[B]1
Next we evaluate x. In experiments 1 and 3, the initial concentration of B is the same, so
any change in the initial rates for these experiments would be due to the different initial
concentrations of A. We solve the ratio of the rate-law expressions of these two experiments
for x. We divide the third rate-law expression by the corresponding terms in the
first rate-law expression.
ᎏ
r
r
a
a
t
t
e
e
3
1
ᎏ ϭ ᎏ
k
k
[
[
A
A
]
]
3
1
x
x
[
[
B
B
]
]
3
1
1
1
ᎏ
The value k cancels, and so do the concentrations of B because they are equal. Thus, the
expression simplifies to
ᎏ
r
r
a
a
t
t
e
e
3
1
ᎏ ϭ ᎏ
[
[
A
A
]
]
3
1
x
x
ᎏ ϭ ΂ᎏ
[
[
A
A
]
]
3
1
ᎏ΃
x
1.0 ϫ 10Ϫ2 M
ᎏᎏ
2.0 ϫ 10Ϫ2 M
1.5 ϫ 10Ϫ6 MиsϪ1
ᎏᎏᎏ
3.0 ϫ 10Ϫ6 MиsϪ1
It does not matter which way we take
the ratio. We would get the same value
for y if we divide the second rate-law
expression by the first—try it!
660 CHAPTER 16: Chemical Kinetics
ϭ ΂ ΃
x
4.0 ϭ (2.0)x so x ϭ 2
The power to which [A] is raised in the rate-law expression is 2, so the rate-law expression
for this reaction is the same as that obtained earlier.
rate ϭ k[A]2[B]1 or rate ϭ k[A]2[B]
EXAMPLE 16-3 Method of Initial Rates
Given the following data, determine the rate-law expression and the value of the rate constant
for the reaction
2A ϩ B ϩ C 88n D ϩ E
Initial Initial Initial Initial Rate of
Experiment [A] [B] [C] Formation of E
1 0.20 M 0.20 M 0.20 M 2.4 ϫ 10Ϫ6 MиminϪ1
2 0.40 M 0.30 M 0.20 M 9.6 ϫ 10Ϫ6 MиminϪ1
3 0.20 M 0.30 M 0.20 M 2.4 ϫ 10Ϫ6 MиminϪ1
4 0.20 M 0.40 M 0.60 M 7.2 ϫ 10Ϫ6 MиminϪ1
Plan
The rate law is of the form Rate ϭ k[A]x[B]y[C]z. We must evaluate x, y, z, and k. We use the
reasoning outlined earlier; in this presentation the first method is used.
Solution
Dependence on [B]: In experiments 1 and 3, the initial concentrations of A and C are unchanged.
Thus, any change in the rate would be due to the change in concentration of B. But we see
that the rate is the same in experiments 1 and 3, even though the concentration of B is different.
Thus, the reaction rate is independent of [B], so y ϭ 0. We can neglect changes in [B] in the
subsequent reasoning. The rate law must be
rate ϭ k[A]x[C]z
Dependence on [C]: Experiments 1 and 4 involve the same initial concentration of A; thus the
observed change in rate must be due entirely to the changed [C]. So we compare experiments
1 and 4 to find z.
[C] has been multiplied by a factor of ϭ 3.0 ϭ [C] ratio
The rate changes by a factor of
ϭ 3.0 ϭ rate ratio
The exponent z can be deduced from
rate ratio ϭ ([C] ratio)z
3.0 ϭ (3.0)z so z ϭ 1 The reaction is first order in [C].
7.2 ϫ 10Ϫ6
ᎏᎏ
2.4 ϫ 10Ϫ6
0.60
ᎏ
0.20
2.0 ϫ 10Ϫ2 M
ᎏᎏ
1.0 ϫ 10Ϫ2 M
6.0 ϫ 10Ϫ6 MиsϪ1
ᎏᎏ
1.5 ϫ 10Ϫ6 MиsϪ1
The coefficient of E in the balanced
equation is 1, so the rate of reaction is
equal to the rate of formation of E.
16-3 Concentrations of Reactants: The Rate-Law Expression 661
[B]0 ϭ 1
The alternative algebraic method
outlined above can also be used.
Now we know that the rate law is of the form
rate ϭ k[A]x[C]
Dependence on [A]: We use experiments 1 and 2 to evaluate x, because [A] is changed,
[B] does not matter, and [C] is unaltered. The observed rate change is due only to the
changed [A].
[A] has been multiplied by a factor of ϭ 2.0 ϭ [A] ratio
The rate changes by a factor of
ϭ 4.0 ϭ rate ratio
The exponent x can be deduced from
rate ratio ϭ ([A] ratio)x
4.0 ϭ (2.0)x so x ϭ 2 The reaction is second order in [A].
From these results we can write the complete rate-law expression.
rate ϭ k[A]2[B]0[C]1 or rate ϭ k[A]2[C]
We can evaluate the specific rate constant, k, by substituting any of the four sets of data into
the rate-law expression we have just derived. Data from experiment 2 give
rate2 ϭ k[A]2
2[C]2
k ϭ ϭ ϭ 3.0 ϫ 10Ϫ4 MϪ2иminϪ1
The rate-law expression can also be written with the value of k incorporated.
rate ϭ 3.0 ϫ 10Ϫ4 MϪ2иminϪ1 [A]2[C]
This expression allows us to calculate the rate at which this reaction occurs with any known
concentrations of A and C (provided some B is present). As we shall see presently, changes in
temperature change reaction rates. This value of k is valid only at the temperature at which the
data were collected.
You should now work Exercises 17 and 18.
9.6 ϫ 10Ϫ6 MиminϪ1
ᎏᎏᎏ
(0.40 M)2(0.20 M)
rate2
ᎏᎏ
[A]2
2[C]2
9.6 ϫ 10Ϫ6
ᎏᎏ
2.4 ϫ 10Ϫ6
0.40
ᎏ
0.20
662 CHAPTER 16: Chemical Kinetics
Problem-Solving Tip: Check the Rate Law You Have Derived
If the rate law that you deduce from initial rate data is correct, it will not matter which
set of data you use to calculate k. As a check, you can calculate k several times, once
from each set of experimental concentration and rate data. If the reaction orders in your
derived rate law are correct, then all sets of experimental data will give the same value
of k (within rounding error); but if the orders are wrong, then the k values will vary
considerably.
EXAMPLE 16-4 Method of Initial Rates
Use the following initial rate data to determine the form of the rate-law expression for the
reaction
3A ϩ 2B 88n 2C ϩ D
Initial Initial Initial Rate of
Experiment [A] [B] Formation of D
1 1.00 ϫ 10Ϫ2 M 1.00 ϫ 10Ϫ2 M 6.00 ϫ 10Ϫ3 MиminϪ1
2 2.00 ϫ 10Ϫ2 M 3.00 ϫ 10Ϫ2 M 1.44 ϫ 10Ϫ1 MиminϪ1
3 1.00 ϫ 10Ϫ2 M 2.00 ϫ 10Ϫ2 M 1.20 ϫ 10Ϫ2 MиminϪ1
Plan
The rate law is of the form rate ϭ k[A]x[B]y. No two experiments have the same initial [B], so
let’s use the alternative method presented earlier to evaluate x and y.
Solution
The initial concentration of A is the same in experiments 1 and 3. We divide the third ratelaw
expression by the corresponding terms in the first one
ϭ
The initial concentrations of A are equal, so they cancel, as does k. Simplifying and then substituting
known values of rates and [B],
ϭ or ϭ ΂ ΃
y
2.0 ϭ (2.0)y or y ϭ 1 The reaction is first order in [B].
No two of the experimental runs have the same concentrations of B, so we must proceed
somewhat differently. Let us compare experiments 1 and 2. The observed change in rate must
be due to the combination of the changes in [A] and [B]. We can divide the second rate-law
expression by the corresponding terms in the first one, cancel the equal k values, and collect
terms.
ϭ ϭ ΂ ΃
x
΂ ΃
y
Now let’s insert the known values for rates and concentrations and the known [B] exponent
of 1.
ϭ ΂ ΃
x
΂ ΃
1
24.0 ϭ (2.00)x
(3.00)
8.00 ϭ (2.00)x
or x ϭ 3 The reaction is third order in [A].
The rate-law expression has the form rate ϭ k[A]3[B].
You should now work Exercise 22.
3.00 ϫ 10Ϫ2 M
ᎏᎏ
1.00 ϫ 10Ϫ2 M
2.00 ϫ 10Ϫ2 M
ᎏᎏ
1.00 ϫ 10Ϫ2 M
1.44 ϫ 10Ϫ1 MиminϪ1
ᎏᎏᎏ
6.00 ϫ 10Ϫ3 MиminϪ1
[B]2
ᎏ
[B]1
[A]2
ᎏ
[A]1
k[A]2
x[B]2
y
ᎏᎏ
k[A]1
x[B]1
y
rate2
ᎏ
rate1
2.00 ϫ 10Ϫ2 M
ᎏᎏ
1.00 ϫ 10Ϫ2 M
1.20 ϫ 10Ϫ2 Mиmin
ᎏᎏᎏ
6.00 ϫ 10Ϫ3 Mиmin
[B]3
y
ᎏ
[B]1
y
rate3
ᎏ
rate1
k[A]3
x[B]3
y
ᎏᎏ
k[A]1
x[B]1
y
rate3
ᎏ
rate1
16-3 Concentrations of Reactants: The Rate-Law Expression 663
CONCENTRATION VERSUS TIME:
THE INTEGRATED RATE EQUATION
Often we want to know the concentration of a reactant that would remain after some
specified time, or how long it would take for some amount of the reactants to be used up.
The equation that relates concentration and time is the integrated rate equation.
We can also use it to calculate the half-life, t1/2, of a reactant—the time it takes
for half of that reactant to be converted into product. The integrated rate equation
and the half-life are different for reactions of different order.
We will look at relationships for some simple cases. If you know calculus, you may be
interested in the derivation of the integrated rate equations. This development is presented
in the Enrichments at the end of this section.
First-Order Reactions
For reactions involving aA n products that are first order in A and first order overall, the
integrated rate equation is
ln΂ ΃ϭ akt (first order)
[A]0 is the initial concentration of reactant A, and [A] is its concentration at some time,
t, after the reaction begins. Solving this relationship for t gives
t ϭ ln΂ ΃
By definition, [A] ϭ ᎏ1
2
ᎏ[A]0 at t ϭ t1/2. Thus
t1/2 ϭ ln ϭ ln 2
t1/2 ϭ ϭ (first order)
This relates the half-life of a reactant in a first-order reaction and its rate constant, k. In
such reactions, the half-life does not depend on the initial concentration of A. This is not
true for reactions having overall orders other than first order.
EXAMPLE 16-5 Half-Life: First-Order Reaction
Compound A decomposes to form B and C in a reaction that is first order with respect to A
and first order overall. At 25°C, the specific rate constant for the reaction is 0.0450 sϪ1. What
is the half-life of A at 25°C?
0.693
ᎏ
ak
ln 2
ᎏ
ak
1
ᎏ
ak
[A]0
ᎏ
ᎏ1
2
ᎏ[A]0
1
ᎏ
ak
[A]0
ᎏ
[A]
1
ᎏ
ak
[A]0
ᎏ
[A]
16-4
a represents the coefficient of reactant
A in the balanced overall equation.
664 CHAPTER 16: Chemical Kinetics
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 15.6, Concentration-Time
Relationships.
Nuclear decay (Chapter 26) is a very
important first-order process. Exercises
at the end of that chapter involve
calculations of nuclear decay rates.
When time t1/2 has elapsed, half of the
original [A]0 has reacted, so half of it
remains.
A 88n B ϩ C
Plan
We use the equation given earlier for t1/2 for a first-order reaction. The value of k is given in
the problem; the coefficient of reactant A is a ϭ 1.
Solution
t1/2 ϭ ϭ ϭ 15.4 s
After 15.4 seconds of reaction, half of the original reactant remains, so that [A] ϭ ᎏ1
2
ᎏ[A]0.
You should now work Exercise 39.
EXAMPLE 16-6 Concentration Versus Time: First-Order Reaction
The reaction 2N2O5(g) n 2N2O4(g) ϩ O2(g) obeys the rate law Rate ϭ k[N2O5], in which the
specific rate constant is 0.00840 sϪ1 at a certain temperature. (a) If 2.50 moles of N2O5 were
placed in a 5.00-liter container at that temperature, how many moles of N2O5 would remain
after 1.00 minute? (b) How long would it take for 90% of the original N2O5 to react?
Plan
We apply the integrated first-order rate equation.
ln ΂ᎏ
[
[
N
N
2
2
O
O
5
5
]
]
0
ᎏ΃ϭ akt
(a) First, we must determine [N2O5]0, the original molar concentration of N2O5. Then we
solve for [N2O5], the molar concentration after 1.00 minute. We must remember to express k
and t using the same time units. Finally, we convert molar concentration of N2O5 to moles
remaining. (b) We solve the integrated first-order equation for the required time.
Solution
The original concentration of N2O5 is
[N2O5]0 ϭ ϭ 0.500 M
The other quantities are
a ϭ 2 k ϭ 0.00840 sϪ1 t ϭ 1.00 min ϭ 60.0 s [N2O5] ϭ __?
The only unknown in the integrated rate equation is [N2O5] after 1.00 minute. Let us solve
for the unknown. Because ln x/y ϭ ln x Ϫ ln y,
ln ϭ ln [N2O5]0 Ϫ ln [N2O5] ϭ akt
ln [N2O5] ϭ ln [N2O5]0 Ϫ akt
ϭ ln (0.500) Ϫ (2)(0.00840 sϪ1)(60.0 s) ϭ Ϫ0.693 Ϫ 1.008
ln [N2O5] ϭ Ϫ1.701
Taking the inverse natural logarithm of both sides gives
[N2O5] ϭ 1.82 ϫ 10Ϫ1 M
[N2O5]0
ᎏ
[N2O5]
2.50 mol
ᎏᎏ
5.00 L
0.693
ᎏᎏ
1(0.0450 sϪ1)
ln 2
ᎏ
ak
inv ln x ϭ eln x
16-4 Concentration versus Time: The Integrated Rate Equation 665
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 15.8, Half-Life: First-Order
Reactions.
Thus, after 1.00 minute of reaction, the concentration of N2O5 is 0.182 M. The number of
moles of N2O5 left in the 5.00-L container is
__? mol N2O5 ϭ 5.00 L ϫ ϭ 0.910 mol N2O5
(b) Because the integrated first-order rate equation involves a ratio of concentrations, we do
not need to obtain the numerical value of the required concentration. When 90.0% of the
original N2O5 has reacted, 10.0% remains, or
[N2O5] ϭ (0.100)[N2O5]0
We make this substitution into the integrated rate equation and solve for the elapsed time, t.
ln ϭ akt
ln ϭ (2)(0.00840 sϪ1)t
ln (10.0) ϭ (0.0168 sϪ1)t
2.302 ϭ (0.0168 sϪ1)t or t ϭ ϭ 137 seconds
You should now work Exercises 36 and 40.
2.302
ᎏᎏ
0.0168 sϪ1
[N2O5]0
ᎏᎏ
(0.100)[N2O5]0
[N2O5]0
ᎏ
[N2O5]
0.182 mol
ᎏᎏ
L
666 CHAPTER 16: Chemical Kinetics
Problem-Solving Tip: Does Your Answer Make Sense?
We know that the amount of N2O5 in Example 16-6 must be decreasing. The calculated
result, 0.910 mol N2O5 after 1.00 minute, is less than the initial amount, 2.50 mol N2O5,
which is a reasonable result. If our solution had given a result that was larger than the
original, we should recognize that we must have made some error. For example, if we
had incorrectly written the equation as
lnᎏ
[
[
N
N
2
2
O
O
5
5
]
]
0
ᎏ ϭ akt
we would have obtained [N2O5] ϭ 1.37 M, corresponding to 6.85 mol N2O5. This would
be more N2O5 than was originally present, which we should immediately recognize as
an impossible answer.
Second-Order Reactions
For reactions involving aA n products that are second order with respect to A and second
order overall, the integrated rate equation is
Ϫ ϭ akt ΂ ΃
second order in A,
ᎏᎏᎏ
second order overall
1
ᎏ
[A]0
1
ᎏ
[A]
Figure 16-4 (a) Plot of concentration versus time for a first-order reaction. During the
first half-life, 1.73 seconds, the concentration of A falls from 1.00 M to 0.50 M. An
additional 1.73 seconds is required for the concentration to fall by half again, from 0.50 M
to 0.25 M, and so on. For a first-order reaction, t1/2 ϭ ϭ ; t1/2 does not depend
on the concentration at the beginning of that time period. (b) Plot of concentration versus
time for a second-order reaction. The same values are used for a, [A]0, and k as in part (a).
During the first half-life, 2.50 seconds, the concentration of A falls from 1.00 M to 0.50 M.
The concentration falls by half again from 2.50 to 7.50 seconds, so the second half-life is
5.00 seconds. The half-life beginning at 0.25 M is 10.00 seconds. For a second-order
reaction, t1/2 ϭ ; t1/2 is inversely proportional to the concentration at the beginning
of that time period.
1
ᎏ
ak[A]0
0.693
ᎏ
ak
ln 2
ᎏ
ak
For t ϭ t1/2, we have [A] ϭ ᎏ1
2
ᎏ[A]0, so
Ϫ ϭ akt1/2
Simplifying and solving for t1/2, we obtain the relationship between the rate constant and
t1/2.
t1/2 ϭ ΂ ΃
In this case t1/2 depends on the initial concentration of A. Figure 16-4 illustrates the different
behavior of half-life for first- and second-order reactions.
EXAMPLE 16-7 Half-Life: Second-Order Reaction
Compounds A and B react to form C and D in a reaction that was found to be second order
in A and second order overall. The rate constant at 30°C is 0.622 liter per mole per minute.
What is the half-life of A when 4.10 ϫ 10Ϫ2 M A is mixed with excess B?
A ϩ B 88n C ϩ D rate ϭ k[A]2
second order in A,
ᎏᎏᎏ
second order overall
1
ᎏ
ak[A]0
1
ᎏ
[A]0
1
ᎏ
ᎏ1
2
ᎏ[A]0
[A](M)
1.0
0.8
0.6
0.4
0.2
0 5 10
0
t (seconds)
(a)
[A](M)
1.0
0.8
0.6
0.4
0.2
0 5 10
0
t (seconds)
(b)
15 20
First-order reaction
a ϭ 1; [A]0 ϭ 1.0 M
k ϭ 0.40 sϪ1
Second-order reaction
a ϭ 1; [A]0 ϭ 1.0 M
k ϭ 0.40 MϪ1
иsϪ1
You should carry out the algebraic
steps to solve for t1/2.
Plan
As long as some B is present, only the concentration of A affects the rate. The reaction is
second order in [A] and second order overall, so we use the appropriate equation for the half-
life.
Solution
t1/2 ϭ ϭ ϭ 39.2 min
EXAMPLE 16-8 Concentration Versus Time: Second-Order Reaction
The gas-phase decomposition of NOBr is second order in [NOBr], with k ϭ 0.810 MϪ1иsϪ1
at 10°C. We start with 4.00 ϫ 10Ϫ3 M NOBr in a flask at 10°C. How many seconds does it
take to use up 1.50 ϫ 10Ϫ3 M of this NOBr?
2NOBr(g) 88n 2NO(g) ϩ Br2(g) rate ϭ k[NOBr]2
Plan
We first determine the concentration of NOBr that remains after 1.50 ϫ 10Ϫ3 M is used up.
Then we use the second-order integrated rate equation to determine the time required to
reach that concentration.
Solution
__? M NOBr remaining ϭ (0.00400 Ϫ 0.00150) M ϭ 0.00250 M ϭ [NOBr]
We solve the integrated rate equation ᎏ
[NO
1
Br]
ᎏ Ϫ ᎏ
[NO
1
Br]0
ᎏ ϭ akt for t.
t ϭ ΂ Ϫ ΃ϭ ΂ Ϫ ΃
ϭ (400 MϪ1 Ϫ 250 MϪ1)
ϭ 92.6 s
You should now work Exercise 34.
EXAMPLE 16-9 Concentration Versus Time: Second-Order Reaction
Consider the reaction of Example 16-8 at 10°C. If we start with 2.40 ϫ 10Ϫ3 M NOBr, what
concentration of NOBr will remain after 5.00 minutes of reaction?
Plan
We use the integrated second-order rate equation to solve for the concentration of NOBr
remaining at t ϭ 5.00 minutes.
Solution
Again, we start with the expression Ϫ ϭ akt. Then we put in the known
values and solve for [NOBr].
Ϫ ϭ (2)(0.810 MϪ1иsϪ1)(5.00 min)΂ ΃
60 s
ᎏ
1 min
1
ᎏᎏ
2.40 ϫ 10Ϫ3 M
1
ᎏ
[NOBr]
1
ᎏᎏ
[NOBr]0
1
ᎏ
[NOBr]
1
ᎏᎏ
1.62 MϪ1иsϪ1
1
ᎏᎏ
0.00400 M
1
ᎏᎏ
0.00250 M
1
ᎏᎏᎏ
(2)(0.810 MϪ1иsϪ1
1
ᎏᎏ
[NOBr]0
1
ᎏ
[NOBr]
1
ᎏ
ak
1
ᎏᎏᎏᎏᎏ
(1)(0.622 MϪ1иminϪ1)(4.10 ϫ 10Ϫ2 M)
1
ᎏ
ak[A]0
The coefficient of NOBr is a ϭ 2.
668 CHAPTER 16: Chemical Kinetics
Ϫ 4.17 ϫ 102 MϪ1 ϭ 486 MϪ1
ϭ 486 MϪ1 ϩ 417 MϪ1 ϭ 903 MϪ1
[NOBr] ϭ ϭ 1.11 ϫ 10Ϫ3 M (46.2% remains unreacted)
Thus, 53.8% of the original concentration of NOBr reacts within the first 5 minutes. This is
reasonable because, as you can easily verify, the reaction has an initial half-life of 257 seconds,
or 4.29 minutes.
You should now work Exercises 35 and 38.
Zero-Order Reaction
For a reaction aA n products that is zero order, the reaction rate is independent of
concentrations. We can write the rate-law expression as
rate ϭ Ϫ ΂ ΃ϭ k
The corresponding integrated rate equation is
[A] ϭ [A]0 Ϫ akt (zero order)
and the half-life is
t1/2 ϭ (zero order)
Table 16-2 summarizes the relationships that we have presented in Sections 16-3 and
16-4.
[A]0
ᎏ
2ak
⌬[A]
ᎏ
⌬t
1
ᎏ
a
1
ᎏᎏ
903 MϪ1
1
ᎏ
[NOBr]
1
ᎏ
[NOBr]
16-4 Concentration versus Time: The Integrated Rate Equation 669
TABLE 16-2 Summary of Relationships for Various Orders of the Reaction
aA n Products
Order
Zero First Second
Rate-law expression rate ϭ k rate ϭ k[A] rate ϭ k[A]2
Integrated rate equation [A] ϭ [A]0 Ϫ akt ln ᎏ
[
[
A
A
]
]
0
ᎏ ϭ akt or ᎏ
[A
1
]
ᎏ Ϫ ᎏ
[A
1
]0
ᎏ ϭ akt
log ᎏ
[
[
A
A
]
]
0
ᎏ ϭ ᎏ
2.
a
3
k
0
t
3
ᎏ
Half-life, t1/2 ᎏ
[
2
A
a
]
k
0
ᎏ ᎏ
ln
ak
2
ᎏ ϭ ᎏ
0.
a
6
k
93
ᎏ ᎏ
ak[
1
A]0
ᎏ
One method of assessing reaction order is based on comparing successive half-lives. As
we have seen, t1/2 for a first-order reaction does not depend on initial concentration. We
can measure the time required for different concentrations of a reactant to fall to half of
their original values. If this time remains constant, it is an indication that the reaction is
first order for that reactant and first order overall (see Figure 16-4a). By contrast, for other
orders of reaction, t1/2 would change depending on initial concentration. For a secondorder
reaction, successively measured t1/2 values would increase by a factor of 2 as [A]0
decreases by a factor of 2 (see Figure 16-4b). [A]0 is measured at the beginning of each
particular measurement period.
You should test this method using the
concentration-versus-time data of
Example 16-10, plotted in Figure
16-8b.
670 CHAPTER 16: Chemical Kinetics
Problem-Solving Tip: Which Equation Should Be Used?
How can you tell which equation to use to solve a particular problem?
1. You must decide whether to use the rate-law expression or the integrated rate equation.
Remember that
the rate-law expression relates rate and concentration
whereas
the integrated rate equation relates time and concentration.
When you need to find the rate that corresponds to particular concentrations, or
the concentrations needed to give a desired rate, you should use the rate-law
expression. When time is involved in the problem, you should use the integrated
rate equation.
2. You must choose the form of the rate-law expression or the integrated rate equation
—zero, first, or second order—that is appropriate to the order of the reaction. These
are summarized in Table 16-2. One of the following usually helps you decide.
a. The statement of the problem may state explicitly what the order of the reaction
is.
b. The rate-law expression may be given, so that you can tell the order of the reaction
from the exponents in that expression.
c. The units of the specific rate constant, k, may be given; you can interpret these
stated units to tell you the order of the reaction.
Order Units of k
0 MиtimeϪ1
1 timeϪ1
2 MϪ1иtimeϪ1
E n r i c h m e n t
Calculus Derivation of Integrated Rate Equations
The derivation of the integrated rate equation is an example of the use of calculus in chemistry.
The following derivation is for a reaction that is assumed to be first order in a reactant
A and first order overall. If you do not know calculus, you can still use the results of this
derivation, as we have already shown in this section. For the reaction
16-4 Concentration versus Time: The Integrated Rate Equation 671
Ϫ ΂ ΃ represents the average
rate over a finite time interval ⌬t.
Ϫ ΂ ΃ involves a change over an
infinitesimally short time interval dt, so
it represents the instantaneous rate.
d[A]
ᎏ
dt
1
ᎏ
a
⌬[A]
ᎏ
⌬t
1
ᎏ
a
aA 88n products
the rate is expressed as
rate ϭ Ϫ ΂ ΃
For a first-order reaction, the rate is proportional to the first power of [A].
Ϫ ΂ ΃ϭ k[A]
In calculus terms, we express the change during an infinitesimally short time dt as the derivative
of [A] with respect to time.
Ϫ ϭ k[A]
Separating variables, we obtain
Ϫ ϭ (ak)dt
We integrate this equation with limits: As the reaction progresses from time ϭ 0 (the start
of the reaction) to time ϭ t elapsed, the concentration of A goes from [A]0, its starting value,
to [A], the concentration remaining after time t:
Ϫ͵
[A]
[A]0
ϭ ak ͵
t
0
dt
The result of the integration is
Ϫ(ln [A] Ϫ ln [A]0) ϭ ak(t Ϫ 0) or ln [A]0 Ϫ ln [A] ϭ akt
Remembering that ln(x) Ϫ ln(y) ϭ ln(x/y), we obtain
lnᎏ
[
[
A
A
]
]
0
ᎏ ϭ akt (first order)
This is the integrated rate equation for a reaction that is first order in reactant A and first
order overall.
Integrated rate equations can be derived similarly from other simple rate laws. For a
reaction aA n products that is second order in reactant A and second order overall, we can
write the rate equation as
Ϫ ϭ k[A]2
Again, using the methods of calculus, we can separate variables, integrate, and rearrange to
obtain the corresponding integrated second-order rate equation.
ᎏ
[A
1
]
ᎏ Ϫ ᎏ
[A
1
]0
ᎏ ϭ akt (second order)
d[A]
ᎏ
adt
d[A]
ᎏ
[A]
d[A]
ᎏ
[A]
d[A]
ᎏ
dt
1
ᎏ
a
⌬[A]
ᎏ
⌬t
1
ᎏ
a
⌬[A]
ᎏ
⌬t
1
ᎏ
a
672 CHAPTER 16: Chemical Kinetics
For a reaction aA n products that is zero order overall, we can write the rate equation
as
Ϫ ϭ k
In this case, the calculus derivation already described leads to the integrated zero-order rate
equation
[A] ϭ [A]0 Ϫ akt (zero order)
d[A]
ᎏ
adt
E n r i c h m e n t
Using Integrated Rate Equations to Determine Reaction Order
The integrated rate equation can help us to analyze concentration-versus-time data to determine
reaction order. A graphical approach is often used. We can rearrange the integrated
first-order rate equation
ln ᎏ
[
[
A
A
]
]
0
ᎏ ϭ akt
as follows. The logarithm of a quotient, ln (x/y), is equal to the difference of the logarithms,
ln x Ϫ ln y, so we can write
ln [A]0 Ϫ ln [A] ϭ akt or ln [A] ϭ Ϫakt ϩ ln [A]0
Recall that the equation for a straight line may be written as
y ϭ mx ϩ b
where y is the variable plotted along the ordinate (vertical axis), x is the variable plotted
along the abscissa (horizontal axis), m is the slope of the line, and b is the intercept of the
line with the y axis (Figure 16-5). If we compare the last two equations, we find that ln [A]
can be interpreted as y, and t as x.
ln [A] ϭ Ϫakt ϩ ln [A]0
g g g g
y ϭ m x ϩ b






Ά
(Enrichment, continued)
⌬y ϭ y2 Ϫ y1
⌬x ϭ x2 Ϫ x1
⌬y
⌬x
point x1, y1
slope ϭ m ϭ
point x2, y2
y ϭ b
x
y
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 15.7, Determination of Rate
Equations (2): Graphical Methods.
Figure 16-5 Plot of the equation
y ϭ mx ϩ b, where m and b are
constant. The slope of the line
(positive in this case) is equal to m;
the intercept on the y axis is equal
to b.
It is not possible for all of the plots
suggested here to yield straight lines
for a given reaction. The nonlinearity
of the plots may not become obvious,
however, if the reaction times used are
too short. In practice, all three lines
might seem to be straight; we should
then suspect that we need to observe
the reaction for a longer time.
16-4 Concentration versus Time: The Integrated Rate Equation 673
The quantity Ϫak is a constant as the reaction proceeds, so it can be interpreted as m. The
initial concentration of A is fixed, so ln [A]0 is a constant for each experiment, and ln [A]0
can be interpreted as b. Thus, a plot of ln [A] versus time for a first-order reaction would
be expected to give a straight line (Figure 16-6) with the slope of the line equal to Ϫak and
the intercept equal to ln [A]0.
We can proceed in a similar fashion with the integrated rate equation for a reaction that
is second order in A and second order overall. We rearrange
Ϫ ϭ akt to read ϭ akt ϩ
Again comparing this with the equation for a straight line, we see that a plot of 1/[A] versus
time would be expected to give a straight line (Figure 16-7). The line would have a slope
equal to ak and an intercept equal to 1/[A]0.
For a zero-order reaction, we can rearrange the integrated rate equation
[A]0 Ϫ [A] ϭ akt to [A] ϭ Ϫakt ϩ [A]0
Comparing this with the equation for a straight line, we see that a straight-line plot would
be obtained by plotting concentration versus time, [A] versus t. The slope of this line is
Ϫak, and the intercept is [A]0.
This discussion suggests another way to deduce an unknown rate-law expression from
experimental concentration data. The following approach is particularly useful for any
decomposition reaction, one that involves only one reactant.
aA 88n products
We plot the data in various ways as suggested above. If the reaction followed zero-order
kinetics, then a plot of [A] versus t would give a straight line. But if the reaction followed
first-order kinetics, then a plot of ln [A] versus t would give a straight line whose slope could
be interpreted to derive a value of k. If the reaction were second order in A and second
order overall, then neither of these plots would give a straight line, but a plot of 1/[A] versus
t would. If none of these plots gave a straight line (within expected scatter due to experimental
error), we would know that none of these is the correct order (rate law) for the
reaction. Plots to test for other orders can be devised, as can graphical tests for rate-law
expressions involving more than one reactant, but those are subjects for more advanced
texts. The graphical approach that we have described is illustrated in the following example.
1
ᎏ
[A]0
1
ᎏ
[A]
1
ᎏ
[A]0
1
ᎏ
[A]
Figure 16-6 Plot of ln [A] versus time for a reaction aA n products that follows firstorder
kinetics. The observation that such a plot gives a straight line would confirm that
the reaction is first order in [A] and first order overall, that is, rate ϭ k[A]. The slope is
equal to Ϫak. Because a and k are positive numbers, the slope of the line is always
negative. Logarithms are dimensionless, so the slope has the units (time)Ϫ1. The
logarithm of a quantity less than 1 is negative, so data points for concentrations less than
1 molar would appear below the time axis.
Figure 16-7 Plot of 1/[A] versus time for a reaction aA n products that follows a
second-order kinetics. The observation that such a plot gives a straight line would
confirm that the reaction is second order in [A] and second order overall, that is, rate ϭ
k[A]2. The slope is equal to ak. Because a and k are positive numbers, the slope of the
line is always positive. Because concentrations cannot be negative, 1/[A] is always positive,
and the line is always above the time axis.
Time
ln [A]0
slope = –ak
ln[A]
First order
Time
l͞[A]0
slope = ak
l͞[A]
Second order
Figure 16-8 Data conversion and
plots for Example 16-10. (a) The
data are used to calculate the two
columns ln [A] and 1/[A]. (b) Test
for zero-order kinetics: a plot of [A]
versus time. The nonlinearity of this
plot shows that the reaction does not
follow zero-order kinetics. (c) Test
for first-order kinetics: a plot of ln
[A] versus time. The observation
that this plot gives a straight line
indicates that the reaction follows
first-order kinetics. (d) Test for
second-order kinetics: a plot of 1/[A]
versus time. If the reaction had
followed second-order kinetics, this
plot would have resulted in a
straight line and the plot in part (c)
would not.
EXAMPLE 16-10 Graphical Determination of Reaction Order
We carry out the reaction A n B ϩ C at a particular temperature. As the reaction proceeds,
we measure the molarity of the reactant, [A], at various times. The observed data are tabulated
in the margin. (a) Plot [A] versus time. (b) Plot ln [A] versus time. (c) Plot 1/[A] versus
time. (d) What is the order of the reaction? (e) Write the rate-law expression for the reaction.
(f) What is the value of k at this temperature?
Plan
For parts (a)–(c), we use the observed data to make the required plots, calculating related
values as necessary. (d) We can determine the order of the reaction by observing which of
these plots gives a straight line. (e) Knowing the order of the reaction, we can write the
rate-law expression. (f ) The value of k can be determined from the slope of the straightline
plot.
Solution
(a) The plot of [A] versus time is given in Figure 16-8b.
(b) We first use the given data to calculate the ln [A] column in Figure 16-8a. These data
are then used to plot ln [A] versus time, as shown in Figure 16-8c.
(c) The given data are used to calculate the 1/[A] column in Figure 16-8a. Then we plot
1/[A] versus time, as shown in Figure 16-8d.
(d) It is clear from the answer to part (b) that the plot of ln [A] versus time gives a straight
line. This tells us that the reaction is first order in [A].
(e) In the form of a rate-law expression, the answer to part (d) gives rate ϭ k[A].
Time (min)
2.5
[A]
0 2 4 6 8 10
(b) Example 16-10(a).
2.0
1.5
1.0
0.5
0
Time
(min) [A] ln [A] 1͞[A]
2.000
1.107
0.612
0.338
0.187
0.103
0.693
–0.491
–1.085
–1.677
–2.273
1.63
2.95
5.35
9.71
0.00
2.00
4.00
6.00
8.00
(a) Data for Example 16-10.
10.00
0.102
0.5000
0.9033
Time (min)
1
ln[A]
0 2 6 8 10 12
(c) Example 16-10(b).
Q
P
1.5 8.5
0.5
0
–0.5
–1
–1.5
–2
–2.5
–3
0.2
–1.83
Time (min)
12
1/[A]
0 2 4 6 8 10
(d) Example 16-10(c).
0
10
8
6
4
2
(Enrichment, continued)
Time [A]
(min) (mol/L)
0.00 2.000
2.00 1.107
4.00 0.612
6.00 0.338
8.00 0.187
10.00 0.103
(f ) We use the straight-line plot in Figure 16-8c to find the value of the rate constant for
this first-order reaction from the relationship
slope ϭ Ϫak or k ϭ Ϫ
To determine the slope of the straight line, we pick any two points, such as P and Q, on
the line. From their coordinates, we calculate
slope ϭ ϭ ϭ Ϫ0.300 minϪ1
k ϭ Ϫ ϭ Ϫ ϭ 0.300 minϪ1
You should now work Exercises 44 and 45.
The graphical interpretations of concentration-versus-time data for some common
reaction orders are summarized in Table 16-3.
Ϫ0.300 minϪ1
ᎏᎏ
1
slope
ᎏ
a
(Ϫ1.83) Ϫ (0.27)
ᎏᎏ
(8.50 Ϫ 1.50) min
change in ordinate
ᎏᎏᎏ
change in abscissa
slope
ᎏ
a
16-4 Concentration versus Time: The Integrated Rate Equation 675
Problem-Solving Tip: Some Warnings About the Graphical Method
for Determining Reaction Order
1. If we were dealing with real experimental data, there would be some error in each
of the data points on the plot. For this reason, we should not use experimental data
points to determine the slope. (Random experimental errors of only 10% can introduce
errors of more than 100% in slopes based on only two points.) Rather we
should draw the best straight line and then use points on that line to find its slope.
Errors are further minimized by choosing points that are widely separated.
2. Remember that the ordinate is the vertical axis and the abscissa is the horizontal
one. If you are not careful to keep the points in the same order in the numerator
and denominator, you will get the wrong sign for the slope.
TABLE 16-3 Graphical Interpretations for Various Orders of the Reaction
aA n Products
Order
Zero First Second
Plot that gives straight line [A] vs. t ln [A] vs. t vs. t
Direction of straight-line slope down with time down with time up with time
Interpretation of slope Ϫak Ϫak ak
Interpretation of intercept [A]0 ln [A]0
1
ᎏ
[A]0
1
ᎏ
[A]
COLLISION THEORY OF REACTION RATES
The fundamental notion of the collision theory of reaction rates is that for a reaction
to occur, molecules, atoms, or ions must first collide. Increased concentrations of reacting
species result in greater numbers of collisions per unit time. However, not all collisions
result in reaction; that is, not all collisions are effective collisions. For a collision to be
effective, the reacting species must (1) possess at least a certain minimum energy necessary
to rearrange outer electrons in breaking bonds and forming new ones and (2) have
the proper orientations toward one another at the time of collision.
Collisions must occur in order for a chemical reaction to proceed, but they do not
guarantee that a reaction will occur.
A collision between atoms, molecules, or ions is not like one between two hard billiard
balls. Whether or not chemical species “collide” depends on the distance at which
they can interact with one another. For instance, the gas-phase ion–molecule reaction
CH4
ϩ ϩ CH4 n CH5
ϩ ϩ CH3 can occur with a fairly long-range contact. This is because
the interactions between ions and induced dipoles are effective over a relatively long
distance. By contrast, the reacting species in the gas reaction CH3 ϩ CH3 n C2H6 are
both neutral. They interact appreciably only through very short-range forces between
induced dipoles, so they must approach one another very closely before we could say that
they “collide.”
Recall (Chapter 12) that the average kinetic energy of a collection of molecules is
proportional to the absolute temperature. At higher temperatures, more of the molecules
possess sufficient energy to react (Section 16-8).
The colliding molecules must have the proper orientation relative to one another and
have sufficient energy to react. If colliding molecules have improper orientations, they do
not react even though they may possess sufficient energy. Figure 16-9 depicts some possible
collisions between molecules of NO and N2O, which can react to form NO2 and N2.
NO ϩ N2O 88n NO2 ϩ N2
16-5
Zn(s) ϩ 2Hϩ(aq) 88n
Zn2ϩ(aq) ϩ H2(g)
Dilute sulfuric acid reacts slowly
with zinc metal (left), whereas more
concentrated acid reacts rapidly
(right). The Hϩ(aq) concentration is
higher in the more concentrated
acid, and so more Hϩ(aq) ions
collide with Zn per unit time.
676 CHAPTER 16: Chemical Kinetics
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 15.9, Microscopic View of
Reactions (1): Collision Theory.
Only the collision in Figure 16-9a is in the correct orientation to transfer an oxygen atom
from the linear N2O molecule to form the angular NO2 molecule. For some reactions,
the presence of a heterogeneous catalyst (Section 16-9) can increase the fraction of
colliding molecules that have the proper orientations.
TRANSITION STATE THEORY
Chemical reactions involve the making and breaking of chemical bonds. The energy associated
with a chemical bond is a form of potential energy. Reactions are accompanied by
changes in potential energy. Consider the following hypothetical, one-step reaction at a
certain temperature.
16-6
Figure 16-9 Some possible
collisions between N2O and NO
molecules in the gas phase. (a) A
collision that could be effective in
producing the reaction. (b, c)
Collisions that would be ineffective.
The molecules must have the proper
orientations relative to one another
and have sufficient energy to react.
16-6 Transition State Theory 677
Ineffective
orientation
of collision
Ineffective
orientation
of collision
Effective
orientation
of collision
Products
Reaction
occurs
CollisionReactants
CollisionReactants
CollisionReactants
Reactants
Reactants
No reaction
No reaction
NO + N2O → NO2 + N2
(a)
(b)
(c)
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 15.10, Microscopic View of
Reactions (2): Transition State Theory.
A ϩ B2 88n AB ϩ B
Figure 16-10 shows plots of potential energy versus the progress of the reaction. In
Figure 16-10a the ground state energy of the reactants, A and B2, is higher than the ground
state energy of the products, AB and B. The energy released in the reaction is the difference
between these two energies, ⌬E. It is related to the change in enthalpy, ⌬H0
rxn (Section
15-11).
Quite often, for reaction to occur, some covalent bonds must be broken so that others
can be formed. This can occur only if the molecules collide with enough kinetic energy to
overcome the potential energy stabilization of the bonds. According to the transition
state theory, the reactants pass through a short-lived, high-energy intermediate state,
called a transition state, before the products are formed.
A ϩ BXB 88n A,B,B 88n AXB ϩ B
reactants transition state products
A ϩ B2 AB2 AB ϩ B
The “progress of reaction” represents
how far the reaction has proceeded
along the pathway leading from
reactants to products. This coordinate
is sometimes labeled the “reaction
coordinate.”
678 CHAPTER 16: Chemical Kinetics
Progress of reaction
A - - - B - - - B
Ea
forward
reaction
Products
AB + B
∆Erxn
Reactants
A + B2
(a)
Progress of reaction
A - - - B - - - B
Ea
reverse
reaction Products
AB + B
∆ErxnReactants
A + B2
Energy
(b)
Endothermic reaction
Exothermic reaction
Energy
Ea
reverse
reaction
Ea
forward
reaction
Figure 16-10 A potential energy
diagram. (a) A reaction that releases
energy (exothermic). An example of
an exothermic gas-phase reaction is
H ϩ I2 88n HI ϩ I
(b) A reaction that absorbs energy
(endothermic). An example of an
endothermic gas-phase reaction is
I ϩ H2 88n HI ϩ H
The activation energy, Ea, is the kinetic energy that reactant molecules must have to
allow them to reach the transition state. If A and B2 molecules do not possess the necessary
amount of energy, Ea, when they collide, reaction cannot occur. If they do possess
sufficient energy to “climb the energy barrier” to reach the transition state, the reaction
can proceed. When the atoms go from the transition state arrangement to the product
molecules, energy is released. If the reaction results in a net release of energy (see Figure
16-10a), more energy than the activation energy is returned to the surroundings and the
reaction is exothermic. If the reaction results in a net absorption of energy (see Figure 16-10b),
an amount less than Ea is given off when the transition state is converted to products and
the reaction is endothermic. The net release of energy is ⌬Erxn.
For the reverse reaction to occur, some molecules on the right (AB) must have kinetic
energy equal to the reverse activation energy, Ea reverse, to allow them to reach the transition
state. As you can see from the potential energy diagrams in Figure 16-10,
Ea forward Ϫ Ea reverse ϭ ⌬Ereaction
As we shall see, increasing the temperature changes the rate by altering the fraction of
molecules that can get over a given energy barrier (Section 16-8). Introducing a catalyst
increases the rate by providing a different pathway that has a lower activation energy
(Section 16-9).
As a specific example that illustrates the ideas of collision theory and transition state
theory, consider the reaction of iodide ions with methyl chloride.
IϪ ϩ CH3Cl 88n CH3I ϩ ClϪ
Many studies have established that this reaction proceeds as shown in Figure 16-11a. The
IϪ ion must approach the CH3Cl from the “back side” of the CXCl bond, through the
middle of the three hydrogen atoms. A collision of an IϪ ion with a CH3Cl molecule from
any other direction would not lead to reaction. But a sufficiently energetic collision with
Figure 16-11 (a) A collision that
could lead to reaction of IϪ ϩ
CH3Cl to give CH3I ϩ ClϪ. The IϪ
must approach along the “back side”
of the CXCl bond. (b) Two
collisions that are not in the
“correct” orientation to cause a
reaction.
I
I–
I–
I–
Cl–
–
Before collision Transition state After reaction
Cl Cl I
ClCl
(a)
(b)
Remember that ⌬Erxn relates product
energy to reactant energy, regardless of
the pathway. ⌬Erxn is negative when
energy is given off; ⌬Erxn is positive
when energy is absorbed from the
surroundings.
The CH3Cl and CH3I molecules each
have tetrahedral molecular geometry.
the appropriate orientation could allow the new IXC bond to form at the same time that
the CXCl bond is breaking. This collection of atoms, which we represent as
is what we call the transition state of this reaction (Figure 16-11a). From this state, either
of two things could happen: (1) the IXC bond could finish forming and the CXCl bond
could finish breaking with ClϪ leaving, leading to products, or (2) the IXC bond could
fall apart with IϪ leaving and the CXCl bond re-forming, leading back to reactants.
REACTION MECHANISMS AND THE
RATE-LAW EXPRESSION
The step-by-step pathway by which a reaction occurs is called its mechanism. Some reactions
take place in a single step, but most reactions occur in a series of elementary steps.
The reaction orders for any single elementary step are equal to the coefficients for that
step.
In many mechanisms, however, one step is much slower than the others.
A reaction can never occur faster than its slowest step.
This slow step is called the rate-determining step. The speed at which the slow step
occurs limits the rate at which the overall reaction occurs.
As an analogy, suppose you often drive a distance of 120 miles at the speed limit of 60
mi/h, requiring 2 hours. But one day there is an accident along the route, causing a slowdown
for several hours. After passing the accident scene, you resume the posted speed of
60 mi/h. If the total time for this trip was 4 hours, then the average speed would be only
120 miles/4 hours, or 30 mi/h. Even though you drove for many miles at the same high
speed, 60 mi/h, the overall rate was limited by the slow step, passing the accident scene.
The balanced equation for the overall reaction is equal to the sum of all the individual
steps, including any steps that might follow the rate-determining step. We emphasize again
that the rate-law exponents do not necessarily match the coefficients of the overall balanced
equation.
For the general overall reaction
aA ϩ bB 88n cC ϩ dD
the experimentally determined rate-law expression has the form
rate ϭ k[A]x[B]y
The values of x and y are related to the coefficients of the reactants in the slowest
step, influenced in some cases by earlier steps.
16-7
CI
H H
H
Cl
We can view this transition state as
though carbon is only partially bonded
to I and only partially bonded to Cl.
680 CHAPTER 16: Chemical Kinetics
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 15.12, Reaction Mechanisms.
Using a combination of experimental data and chemical intuition, we can postulate a
mechanism by which a reaction could occur. We can never prove absolutely that a proposed
mechanism is correct. All we can do is postulate a mechanism that is consistent with experimental
data. We might later detect reaction-intermediate species that are not explained
by the proposed mechanism. We must then modify the mechanism or discard it and
propose a new one.
As an example, the reaction of nitrogen dioxide and carbon monoxide has been found
to be second order with respect to NO2 and zero order with respect to CO below 225°C.
NO2(g) ϩ CO(g) 88n NO(g) ϩ CO2(g) rate ϭ k[NO2]2
The balanced equation for the overall reaction shows the stoichiometry but does not necessarily
mean that the reaction simply occurs by one molecule of NO2 colliding with one
molecule of CO. If the reaction really took place in that one step, then the rate would be
first order in NO2 and first order in CO, or rate ϭ k[NO2][CO]. The fact that the experimentally
determined orders do not match the coefficients in the overall balanced equation
tells us that the reaction does not take place in one step.
The following proposed two-step mechanism is consistent with the observed rate-law
expression.
(1) NO2 ϩ NO2 88n N2O4 (slow)
(2) N2O4 ϩ CO 88n NO ϩ CO2 ϩ NO2 (fast)
NO2 ϩ CO 88n NO ϩ CO2 (overall)
The rate-determining step of this mechanism involves a bimolecular collision between two
NO2 molecules. This is consistent with the rate expression involving [NO2]2. Because the
CO is involved only after the slow step has occurred, the reaction rate would not depend
on [CO] (i.e., the reaction would be zero order in CO) if this were the actual mechanism.
In this proposed mechanism, N2O4 is formed in one step and is completely consumed in
a later step. Such a species is called a reaction intermediate.
In other studies of this reaction, however, nitrogen trioxide, NO3, has been detected
as a transient (short-lived) intermediate. The mechanism now thought to be correct is
(1) NO2 ϩ NO2 88n NO3 ϩ NO (slow)
(2) NO3 ϩ CO 88n NO2 ϩ CO2 (fast)
NO2 ϩ CO 88n NO ϩ CO2 (overall)
In this proposed mechanism two molecules of NO2 collide to produce one molecule each
of NO3 and NO. The reaction intermediate NO3 then collides with one molecule of CO
and reacts very rapidly to produce one molecule each of NO2 and CO2. Even though two
NO2 molecules are consumed in the first step, one is produced in the second step. The
net result is that only one NO2 molecule is consumed in the overall reaction.
Each of these proposed mechanisms meets both criteria for a plausible mechanism:
(1) The steps add to give the equation for the overall reaction, and (2) the mechanism is
consistent with the experimentally determined rate-law expression (in that two NO2 molecules
and no CO molecules are reactants in the slow step). The NO3 that has been detected
is evidence in favor of the second mechanism, but this does not unequivocally prove that
mechanism; it may be possible to think of other mechanisms that would involve NO3 as
an intermediate and would also be consistent with the observed rate law.
You should be able to distinguish among various species that can appear in a reaction
mechanism. So far, we have seen three such species:
Some reaction intermediates are so
unstable that it is very difficult to
prove experimentally that they exist.
16-7 Reaction Mechanisms and the Rate-Law Expression 681
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 15.13, Reaction Mechanisms
and Rate Equations.
1. Reactant: More is consumed than is formed.
2. Product: More is formed than is consumed.
3. Reaction intermediate: Formed in earlier steps, then consumed in an equal amount
in later steps.
The gas-phase reaction of nitrogen oxide and bromine is known to be second order
in NO and first order in Br2.
2NO(g) ϩ B2(g) 88n 2NOBr(g) rate ϭ k[NO]2[Br2]
A one-step collision involving two NO molecules and one Br2 molecule would be consistent
with the experimentally determined rate-law expression. However, the likelihood of
all three molecules colliding simultaneously is far less than the likelihood of two colliding.
Routes involving only bimolecular collisions or unimolecular decompositions are thought to be more
favorable in reaction mechanisms. The mechanism is believed to be
(1) NO ϩ Br2 34 NOBr2 (fast, equilibrium)
(2) NOBr2 ϩ NO 88n 2NOBr (slow)
2NO ϩ Br2 88n 2NOBr (overall)
The first step involves the collision of one NO molecule (reactant) and one Br2 molecule
(reactant) to produce the intermediate species NOBr2. The NOBr2 can react rapidly,
however, to re-form NO and Br2. We say that this is an equilibrium step. Eventually another
NO molecule (reactant) can collide with a short-lived NOBr2 molecule and react to
produce two NOBr molecules (product).
To analyze the rate law that would be consistent with this proposed mechanism, we
again start with the slow (rate-determining) step, step 2. Denoting the rate constant for
this step as k2, we could express the rate of this step as
rate ϭ k2[NOBr2][NO]
However, NOBr2 is a reaction intermediate, so its concentration at the beginning of the
second step may not be easy to measure directly. Because NOBr2 is formed in a fast equilibrium
step, we can relate its concentration to the concentrations of the original reactants.
When a reaction or reaction step is at equilibrium, its forward (f ) and reverse (r) rates are
equal.
rate1f ϭ rate1r
Because this is an elementary step, we can write the rate expression for both directions
from the equation for the elementary step
k1f[NO][Br2] ϭ k1r[NOBr2]
and then rearrange for [NOBr2].
[NOBr2] ϭ ᎏ
k
k
1
1
r
f
ᎏ[NO][Br2]
When we substitute the right side of this equation for [NOBr2] in the rate expression for
the rate-determining step, rate ϭ k2[NOBr2][NO], we arrive at the experimentally determined
rate-law expression.
rate ϭ k2΂ᎏ
k
k
1
1
r
f
ᎏ[NO][Br2]΃[NO] or rate ϭ k[NO]2[Br2]
Think how unlikely it is for three
moving billiard balls to collide
simultaneously.
Any fast step that precedes a slow step
reaches equilibrium.
682 CHAPTER 16: Chemical Kinetics
The product and quotient of constants
k2, k1f, and k1r is another constant, k.
The rate-law expression of step 2 (the
rate-determining step) determines the
rate law for the overall reaction.
Similar interpretations apply to most other overall third- or higher-order reactions, as
well as many lower-order reactions. When several steps are about equally slow, however,
the analysis of experimental data is more complex. Fractional or negative reaction orders
result from complex multistep mechanisms.
One of the earliest kinetic studies involved the gas-phase reaction of hydrogen and
iodine to form hydrogen iodide. The reaction was found to be first order in both hydrogen
and iodine.
H2(g) ϩ I2(g) 88n 2HI(g) rate ϭ k[H2][I2]
The mechanism that was accepted for many years involved collision of single molecules
of H2 and I2 in a simple one-step reaction. Current evidence indicates a more complex
process, however. Most kineticists now accept the following mechanism.
(1) I2 34 2I (fast, equilibrium)
(2) IH2 ϩ H2 34 H2I (fast, equilibrium)
(3) H2I ϩ I 88n 2HI (slow)
H2I ϩ I2 88n 2HI (overall)
In this case neither original reactant appears in the rate-determining step, but both appear
in the rate-law expression. Each step is a reaction in itself. Transition state theory tells us
that each step has its own activation energy. Because step 3 is the slowest, we know that
its activation energy is the highest, as shown in Figure 16-12.
In summary
The experimentally determined reaction orders of reactants indicate the number of
molecules of those reactants involved in (1) the slow step only, if it occurs first, or
(2) the slow step and any fast equilibrium steps preceding the slow step.
Figure 16-12 A graphical representation of the relative energies of activation for a
postulated mechanism for the gas-phase reaction
H2 ϩ I2 88n 2HI
16-7 Reaction Mechanisms and the Rate-Law Expression 683
H2
H2 ϩ I2
2HI
H2I
ϩ 2I ϩ I
Ea2
Ea3
overall
Ea
Ea1
Energy
Progress of reaction
Net energy released
Apply the algebraic approach described
earlier to show that this mechanism is
consistent with the observed rate-law
expression.
TEMPERATURE: THE ARRHENIUS EQUATION
The average kinetic energy of a collection of molecules is proportional to the absolute
temperature. At a particular temperature, T1, a definite fraction of the reactant molecules
have sufficient kinetic energy, KE ϾEa, to react to form product molecules on collision.
At a higher temperature, T2, a greater fraction of the molecules possess the necessary activation
energy, and the reaction proceeds at a faster rate. This is depicted in Figure 16-13a.
From experimental observations, Svante Arrhenius developed the mathematical relationship
among activation energy, absolute temperature, and the specific rate constant of
a reaction, k, at that temperature. The relationship, called the Arrhenius equation, is
k ϭ AeϪEa/RT
or, in logarithmic form,
ln k ϭ ln A Ϫ ᎏ
R
E
T
a
ᎏ
In this expression, A is a constant having the same units as the rate constant. It is equal
to the fraction of collisions with the proper orientations when all reactant concentrations
are one molar. R is the universal gas constant, expressed with the same energy units in its
numerator as are used for Ea. For instance, when Ea is known in J/mol, the value R ϭ
8.314 J/molиK is appropriate. Here the unit “mol” is interpreted as “mole of reaction,”
as described in Chapter 15. One important point is the following: The greater the value
of Ea, the smaller the value of k and the slower the reaction rate (other factors being equal).
16-8
Figure 16-13 (Left) The effect of temperature on the number of molecules that have
kinetic energies greater than Ea. At T2, a higher fraction of molecules possess at least Ea,
the activation energy. The area between the distribution curve and the horizontal axis is
proportional to the total number of molecules present. The total area is the same at T1 and
T2. The shaded areas represent the number of particles that exceed the energy of activation,
Ea. (Right) Consider two hypothetical reactions 1 and 2, where the activation energy of
reaction 1 is less than that of reaction 2—that is, Ea1 Ͻ Ea2. At any given temperature,
a larger fraction of the molecules have energies that exceed Ea1 than that exceed Ea2, so
reaction 1 would have a higher specific rate constant, k, than reaction 2 at the same reactant
concentrations.
684 CHAPTER 16: Chemical Kinetics
Ea
T1
T2
T2 Ͼ T1
Fractionofmoleculeswith
agivenkineticenergy
Kinetic energy
Fractionofmoleculeswith
agivenkineticenergy
Kinetic energy
Ea1 Ea2
e ϭ 2.718 is the base of natural
logarithms (ln).
Antimony powder reacts with bromine more rapidly at 75°C (left) than at 25°C (right).
16-8 Temperature: The Arrhenius Equation 685
This is because fewer collisions take place with sufficient energy to get over a high-energy
barrier (see Figure 16-13b).
The Arrhenius equation predicts that increasing T results in a faster reaction for the
same Ea and concentrations.
ᎏ
inc
I
r
f
e
T
ases
ᎏ :F ᎏ
de
E
c
a
r
/
e
R
a
T
ses
ᎏ :F ᎏ
i
Ϫ
nc
E
r
a
e
/
a
R
se
T
s
ᎏ :F ᎏ
in
eϪ
cr
E
e
a/
a
R
s
T
es
ᎏ :F ᎏ
incr
k
eases
ᎏ :F ᎏ
s
R
p
e
e
a
e
c
d
t
s
io
u
n
p
ᎏ
Let’s look at how the rate constant varies with temperature for a given single reaction.
Assume that the activation energy and the factor A do not vary with temperature. We can
write the Arrhenius equation for two different temperatures. Then we subtract one equation
from the other and rearrange the result to obtain
ln ϭ ΂ Ϫ ΃
Let’s substitute some typical values into this equation. The activation energy for many
reactions that occur near room temperature is about 50 kJ/mol. For such a reaction, a
temperature increase from 300 K to 310 K would result in
ln ϭ ΂ Ϫ ΃ϭ 0.647
ϭ 1.91 Ϸ 2
Chemists sometimes use the rule of thumb that near room temperature the rate of a reaction
approximately doubles with a 10°C rise in temperature. Such a “rule” must be used
with care, however, because it obviously depends on the activation energy.
k2
ᎏ
k1
1
ᎏ
310 K
1
ᎏ
300 K
50,000 J/mol
ᎏᎏ
(8.314 J/molиK)
k2
ᎏ
k1
1
ᎏ
T2
1
ᎏ
T1
Ea
ᎏ
R
k2
ᎏ
k1
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 15.11, Control of Reaction Rates
(3): Temperature Dependence.
Which egg will cook faster—the
one in ice water (left) or the one in
boiling water (right)? Why—in
terms of what you have learned in
this chapter?
EXAMPLE 16-11 Arrhenius Equation
The specific rate constant, k, for the following first-order reaction is 9.16 ϫ 10Ϫ3 sϪ1 at 0.0°C.
The activation energy of this reaction is 88.0 kJ/mol. Determine the value of k at 2.0°C.
N2O5 88n NO2 ϩ NO3
Plan
First we tabulate the values, remembering to convert temperature to the Kelvin scale.
Ea ϭ 88,000 J/mol R ϭ 8.314 J/molиK
k1 ϭ 9.16 ϫ 10Ϫ3 sϪ1 at T1 ϭ 0.0°C ϩ 273 ϭ 273 K
k2 ϭ __? at T2 ϭ 2.0°C ϩ 273 ϭ 275 K
We use these values in the “two-temperature” form of the Arrhenius equation.
Solution
ln ϭ ΂ᎏ
T
1
1
ᎏ Ϫ ᎏ
T
1
2
ᎏ΃
ln΂ ΃ϭ ΂ᎏ
273
1
K
ᎏ Ϫ ᎏ
275
1
K
ᎏ΃ϭ 0.282
Taking inverse (natural) logarithms of both sides,
ϭ 1.32
k2 ϭ 1.32(9.16 ϫ 10Ϫ3 sϪ1) ϭ 1.21 ϫ 10Ϫ2 sϪ1
We see that a very small temperature difference, only 2°C, causes an increase in the rate constant
(and hence in the reaction rate for the same concentrations) of about 32%. Such sensitivity of
rate to temperature change makes the control and measurement of temperature extremely
important in chemical reactions.
You should now work Exercise 53.
EXAMPLE 16-12 Activation Energy
The gas-phase decomposition of ethyl iodide to give ethylene and hydrogen iodide is a firstorder
reaction.
C2H5I 88n C2H4 ϩ HI
At 600. K, the value of k is 1.60 ϫ 10Ϫ5 sϪ1. When the temperature is raised to 700. K, the
value of k increases to 6.36 ϫ 10Ϫ3 sϪ1. What is the activation energy for this reaction?
Plan
We know k at two different temperatures. We solve the two-temperature forms of the Arrhenius
equation for Ea and evaluate.
Solution
k1 ϭ 1.60 ϫ 10Ϫ5 sϪ1 at T1 ϭ 600. K k2 ϭ 6.36 ϫ 10Ϫ3 sϪ1 at T2 ϭ 700. K
R ϭ 8.314 J/molиK Ea ϭ _?_
k2
ᎏᎏ
9.16 ϫ 10Ϫ3 sϪ1
88,000 J/mol
ᎏᎏ
8.314 ᎏ
mo
J
lиK
ᎏ
k2
ᎏᎏ
9.16 ϫ 10Ϫ3 sϪ1
Ea
ᎏ
R
k2
ᎏ
k1
686 CHAPTER 16: Chemical Kinetics
We arrange the Arrhenius equation for Ea.
ln ϭ ΂ᎏ
T
1
1
ᎏ Ϫ ᎏ
T
1
2
ᎏ΃ so Ea ϭ
Substituting,
R ln ᎏ
k
k
2
1
ᎏ
ᎏᎏ
΂ᎏ
T
1
1
ᎏ Ϫ ᎏ
T
1
2
ᎏ΃
Ea
ᎏ
R
k2
ᎏ
k1
Compare this approach to that
described in the earlier Enrichment
section for determining k.
16-8 Temperature: The Arrhenius Equation 687
Ea ϭ ϭ ϭ 2.09 ϫ 105 J/mol
΂8.314 ᎏ
mo
J
lиK
ᎏ΃(5.98)
ᎏᎏᎏ
2.38 ϫ 10Ϫ4 KϪ1
΂8.314 ᎏ
mo
J
lиK
ᎏ΃ ln ΂ᎏ
6
1
.
.
3
6
6
0
ϫ
ϫ
1
1
0
0
Ϫ
Ϫ
3
5
s
s
Ϫ
Ϫ
1
1
ᎏ΃
ᎏᎏᎏᎏᎏ
΂ᎏ
60
1
0.K
ᎏ Ϫ ᎏ
7
1
00
ᎏ.K΃
You should now work Exercise 54.
The determination of Ea in the manner illustrated in Example 16-12 may be subject to
considerable error because it depends on the measurement of k at only two temperatures.
Any error in either of these k values would greatly affect the resulting value of Ea. A more
reliable method that uses many measured values for the same reaction is based on a graphical
approach. Let us rearrange the single-temperature logarithmic form of the Arrhenius
equation and compare it with the equation for a straight line.
ln k ϭ Ϫ΂ ΃΂ ΃ϩ ln A
A A
g g g g
y ϭ m x ϩ b
The value of the collision frequency factor, A, is very nearly constant over moderate
temperature changes. Thus, ln A can be interpreted as the constant term in the equation
(the intercept). The slope of the straight line obtained by plotting ln k versus 1/T equals
ϪEa/R. This allows us to determine the value of the activation energy from the slope
(Figure 16-14). Exercises 57 and 58 use this method.
1
ᎏ
T
Ea
ᎏ
R
Figure 16-14 A graphical method for determining activation energy, Ea. At each of several
different temperatures, the rate constant, k, is determined by methods such as those in
Sections 16-3 and 16-4. A plot of ln k versus 1/T gives a straight line with negative slope.
The slope of this straight line is ϪEa/R. Use of this graphical method is often desirable,
because it partially compensates for experimental errors in individual k and T values.
1nk
1/T
Ea
R
slope ϭ
Ά
Ά
or 209 kJ/molΆ
Ά
CATALYSTS
Catalysts are substances that can be added to reacting systems to increase the rate
of reaction. They allow reactions to occur via alternative pathways that increase
reaction rates by lowering activation energies.
The activation energy is lowered in all catalyzed reactions, as depicted in Figures 16-15
and 16-16. A catalyst does take part in the reaction, but all of it is re-formed in later steps.
Thus, a catalyst does not appear in the balanced equation for the reaction.
For constant T and the same concentrations,
ᎏ
dec
If
re
E
a
a
ses
ᎏ :F ᎏ
de
E
c
a
r
/
e
R
a
T
ses
ᎏ :F ᎏ
i
Ϫ
nc
E
r
a
e
/
a
R
se
T
s
ᎏ :F ᎏ
in
eϪ
cr
E
e
a
/
a
R
s
T
es
ᎏ :F ᎏ
incr
k
eases
ᎏ :F ᎏ
s
R
p
e
e
a
e
c
d
t
s
io
u
n
p
ᎏ
16-9
Figure 16-15 Potential energy
diagrams showing the effect of a
catalyst. The catalyst provides a
different mechanism, corresponding
to a lower-energy pathway, for the
formation of the products. A
catalyzed reaction typically occurs
in several steps, each with its own
barrier, but the overall energy
barrier for the net reaction, EaЈ, is
lower than that for the uncatalyzed
reaction, Ea. The value of ⌬Erxn
depends only on the states of the
reactants and products, so it is the
same for either path.
688 CHAPTER 16: Chemical Kinetics
Progress of
uncatalyzed reaction
Progress of
catalyzed reaction
Energy
Energy
Ea
forward Ea
reverse
∆Erxn ∆Erxn
Ea′
forward
Ea′
reverse
Ea′ < Ea
Figure 16-16 When a catalyst is present, the energy barrier is lowered. Thus, more
molecules possess the minimum kinetic energy necessary for reaction. This is analogous
to allowing more students to pass a course by lowering the requirements.
Fractionofmoleculeswith
agivenkineticenergy
Kinetic energy
Minimum kinetic energy
for catalyzed reaction
Minimum kinetic energy
for uncatalyzed reaction
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 15.14, Catalysis and Reaction
Rate.
:F :F :F :F :F
We can describe two categories of catalysts: (1) homogeneous catalysts and (2) heterogeneous
catalysts.
Homogeneous Catalysis
A homogeneous catalyst exists in the same phase as the reactants. Ceric ion, Ce4ϩ, at
one time was an important laboratory oxidizing agent that was used in many redox
titrations (Section 11-8). For example, Ce4ϩ oxidizes thallium(I) ions in solution; this reaction
is catalyzed by the addition of a very small amount of a soluble salt containing
manganese(II) ions, Mn2ϩ. The Mn2ϩ acts as a homogeneous catalyst.
Mn2ϩ
2Ce4ϩ ϩ Tlϩ 8888n 2Ce3ϩ ϩ Tl3ϩ
This reaction is thought to proceed by the following sequence of elementary steps.
Ce4ϩ ϩ Mn2ϩ 88n Ce3ϩ ϩ Mn3ϩ step 1
Ce4ϩ ϩ Mn3ϩ 88n Ce3ϩ ϩ Mn4ϩ step 2
Mn4ϩ ϩ Tlϩ 88n Mn2ϩ ϩ Tl3ϩ step 3
2Ce4ϩ ϩ Tlϩ 88n 2Ce3ϩ ϩ Tl3ϩ overall
Some of the Mn2ϩ catalyst reacts in step 1, but an equal amount is regenerated in step 3
and is thus available to react again. The two ions shown in blue, Mn3ϩ and Mn4ϩ, are
reaction intermediates. Mn3ϩ ions are formed in step 1 and consumed in an equal amount
in step 2; similarly, Mn4ϩ ions are formed in step 2 and consumed in an equal amount in
step 3.
We can now summarize the species that can appear in a reaction mechanism.
1. Reactant: More is consumed than is formed.
2. Product: More is formed than is consumed.
3. Reaction intermediate: Formed in earlier steps, then consumed in an equal amount
in later steps.
4. Catalyst: Consumed in earlier steps, then regenerated in an equal amount in later
steps.
Strong acids function as homogeneous catalysts in the acid-catalyzed hydrolysis of esters
(a class of organic compounds—Section 27-14). Using ethyl acetate (a component of nail
polish removers) as an example of an ester, we can write the overall reaction as follows.
This is a thermodynamically favored reaction, but because of its high energy of activation,
it occurs only very, very slowly when no catalyst is present. In the presence of strong
acids, however, the reaction occurs more rapidly. In this acid-catalyzed hydrolysis, different
intermediates with lower activation energies are formed. The sequence of steps in the
postulated mechanism follows.
OCH2CH3(aq)CH3C
O
H2O
Hϩ
ϩ OH(aq) CH3CH2OH(aq)CH3C
O
ϩ
ethanolacetic acidethyl acetate
“Hydrolysis” means reaction with
water.
16-9 Catalysts 689
We see that Hϩ is a reactant in step 1, but it is completely regenerated in step 5. Hϩ
is therefore a catalyst. The species shown in brackets in steps 1 through 4 are reaction
intermediates. Ethyl acetate and water are the reactants, and acetic acid and ethanol are
the products of the overall catalyzed reaction.
Heterogeneous Catalysis
A heterogeneous catalyst (also known as a contact catalyst) is present in a different
phase from the reactants. Such catalysts are usually solids, and they lower activation energies
by providing surfaces on which reactions can occur. The first step in the catalytic
process is usually adsorption, in which one or more of the reactants become attached to
the solid surface. Some reactant molecules may be held in particular orientations, or some
bonds may be weakened; in other molecules, some bonds may be broken to form atoms
or smaller molecular fragments. This causes activation of the reactants. As a result, reac-
OCH2CH3CCH3
O
Hϩϩ OCH2CH3CCH3
OH
ϩ
ϩ
OCH2CH3CCH3 H2Oϩ C
Oϩ
HH
OCH2CH3CH3
OH OH
step 1
OH HOCH2CH3CCH3
ϩOH
ϩOH
step 4
step 5
overall
step 2
step 3C O
HO
ϩ
CH2CH3CH3
OH
H
C
Oϩ
HH
OCH2CH3CH3
OH
C O
HO
ϩ
CH2CH3CH3
OH
H
ϩ
ethanol
OH HϩCCH3
O
OHCCH3 ϩ
acetic acid
OCH2CH3CCH3
O
H2O
Hϩ
ϩ OHC CH3CH2OHCH3
O
ϩ
ethanolacetic acidethyl acetate
Groups of atoms that are involved in
the change in each step are shown in
blue. The catalyst, Hϩ, is shown in
red.
690 CHAPTER 16: Chemical Kinetics
The petroleum industry uses numerous heterogeneous catalysts. Many of them contain
highly colored compounds of transition metal ions. Several are shown here.
All intermediates in this sequence of
elementary steps are charged species,
but this is not always the case.
tion occurs more readily than would otherwise be possible. In a final step, desorption, the
product molecules leave the surface, freeing reaction sites to be used again. Most contact
catalysts are more effective as small particles, because they have relatively large surface
areas.
Transition metals and their compounds function as effective catalysts in many homogeneous
and heterogeneous reactions. Vacant d orbitals in many transition metal ions can
accept electrons from reactants to form intermediates. These subsequently decompose to
form products. Three transition metals, Pt, Pd, and Ni, are often used as finely divided
solids to provide surfaces on which heterogeneous reactions can occur.
The catalytic converters (Figure 16-17) built into automobile exhaust systems contain
two types of heterogeneous catalysts, powdered noble metals and powdered transition
metal oxides. They catalyze the oxidation of unburned hydrocarbon fuel (reaction 1) and
of partial combustion products such as carbon monoxide (reaction 2, shown in Figure
16-18).
Pt
1. 2C8H18(g) ϩ 25O2(g) 8888n 16CO2(g) ϩ 18H2O(g)
NiO
h
isooctane (a component of gasoline)
Pt
2. 2CO(g) ϩ O2(g) 8888n 2CO2(g)
NiO
It is desirable to carry out these reactions in automobile exhaust systems. Carbon
monoxide is very poisonous. The latter reaction is so slow that a mixture of CO and O2
gas at the exhaust temperature would remain unreacted for thousands of years in the
absence of a catalyst! Yet the addition of only a small amount of a solid, finely divided
transition metal catalyst promotes the production of up to a mole of CO2 per minute.
Because this reaction is a very simple but important one, it has been studied extensively
by surface chemists. It is one of the best understood heterogeneously catalyzed reactions.
The major features of the catalytic process are shown in Figure 16-18.
The same catalysts also catalyze another reaction, the decomposition of nitrogen oxide,
NO, into harmless N2 and O2.
Pt
2NO(g) 8888n N2(g) ϩ O2(g)
NiO
At the high temperatures of the
combustion of any fuel in air, nitrogen
and oxygen combine to form nitrogen
oxide.
16-9 Catalysts 691
Figure 16-17 (a) The arrangement of a catalytic converter in an automobile. (b) A cutaway
view of a catalytic converter, showing the pellets of catalyst.
Exhaust manifold
Exhaust
pipe
Catalytic
converter
Tailpipe
Air
compressor
in
O2, CO, NO,
hydrocarbons
out
O2, CO2,
N2, H2O
(a) (b)
Nitrogen oxide is a serious air pollutant because it is oxidized to nitrogen dioxide, NO2,
which reacts with water to form nitric acid and with other products of the incomplete
combustion of hydrocarbons to form nitrates. The latter are eye irritants in photochemical
smog.
These three reactions, catalyzed in catalytic converters, are all exothermic and thermodynamically
favored. Unfortunately, other energetically favored reactions are also
accelerated by the mixed catalysts. All fossil fuels contain sulfur compounds, which are
oxidized to sulfur dioxide during combustion. Sulfur dioxide, itself an air pollutant, undergoes
further oxidation to form sulfur trioxide as it passes through the catalytic bed.
See the Chemistry in Use essay
“Nitrogen Oxides and Photochemical
Smog” in Chapter 24.
692 CHAPTER 16: Chemical Kinetics
Catalyst surface
Adsorption: CO and O2 reactant molecules become
bound to the surface:
CO(g) CO(surface) and O2(g) O2(surface)
The CO molecules are linked through their C atoms
to one or more metal atoms on the surface. The O2
molecules are more weakly bound.
(a) Activation: The O2 molecules dissociate into O atoms,
which are held in place more tightly:
O2(surface) 2O(surface)
The CO molecules stick to the surface, but they migrate
easily across the surface.
(b)
Desorption: The weakly bound CO2 product molecules
leave the surface:
CO2(surface) CO2(g)
Fresh reactant molecules can then replace them to
start the cycle again [back to step (a)].
(d) Reaction: O atoms react with bound CO molecules,
to form CO2 molecules:
CO(surface) + O(surface) CO2(surface)
The resulting CO2 molecules bind to the surface
very poorly.
(c)
Figure 16-18 A schematic representation of the catalysis of the reaction
2CO(g) ϩ O2(g) 88n 2CO2(g)
on a metallic surface.
Pt
2SO2(g) ϩ O2(g) 8888n 2SO3(g)
NiO
Sulfur trioxide is probably a worse pollutant than sulfur dioxide, because SO3 is the acid
anhydride of strong, corrosive sulfuric acid. Sulfur trioxide reacts with water vapor in the
air, as well as in auto exhausts, to form sulfuric acid droplets. This problem must be overcome
if the current type of catalytic converter is to see continued use. These same catalysts
also suffer from the problem of being “poisoned”—that is, made inactive—by lead.
Leaded fuels contain tetraethyl lead, Pb(C2H5)4, and tetramethyl lead, Pb(CH3)4. Such
fuels are not suitable for automobiles equipped with catalytic converters and are excluded
by U. S. law from use in such cars.
Reactions that occur in the presence of a solid catalyst, as on a metal surface (heterogeneous
catalysis) often follow zero-order kinetics. For instance, the rate of decomposition
of NO2(g) at high pressures on a platinum metal surface does not change if we add more
NO2. This is because only the NO2 molecules on the surface can react. If the metal surface
is completely covered with NO2 molecules, no additional molecules can be adsorbed until
the ones already there have reacted and the products have desorbed. Thus, the rate of the
Maintaining the continued efficiency
of all three reactions in a “three-way”
catalytic converter is a delicate matter.
It requires control of such factors as
the O2 supply pressure and the order
in which the reactants reach the
catalyst. Modern automobile engines
use microcomputer chips, based on an
O2 sensor in the exhaust stream, to
control air valves.
16-9 Catalysts 693
A 30% hydrogen peroxide, H2O2, solution at room temperature decomposes very slowly
to form O2 and H2O. (a) This reaction takes place more rapidly if the solution is heated.
(b) A very small amount of a transition metal oxide is added. (c) This oxide catalyzes the
decomposition reaction. The catalyzed reaction is rapid, so the exothermic reaction
quickly heats the solution to the boiling point of water, forming steam. The temperature
increase further accelerates the decomposition. Never use a syringe with a metal tip to
withdraw a sample from a 30% hydrogen peroxide solution. (d) The bombardier beetle
uses a catalyzed decomposition of hydrogen peroxide as a means of defense. An enzyme
produced by the beetle catalyzes the rapid exothermic reaction. The resulting steam,
along with other irritating chemicals, is ejected.
(a) (b) (c)
(d)
(text continues on page 696)
The EnvironmentCCHEMISTRY IN USE
Ozone
Ozone, O3, is such a powerful oxidizing agent that in significant
concentrations it destroys many plastics, metals, and
rubber, as well as both plant and animal tissues. We therefore
try to minimize exposure to ozone in our immediate
environment. In the upper atmosphere, however, ozone plays
a very important role in the absorption of harmful radiation
from the sun. Maintaining appropriate concentrations of
ozone—minimizing its production where ozone is harmful
and preventing its destruction where ozone is helpful—is an
important challenge in environmental chemistry.
Ozone is formed in the upper atmosphere as some O2
molecules absorb high-energy electromagnetic radiation
from the sun and dissociate into oxygen atoms; these then
combine with other O2 molecules to form ozone.
O2(g) ϩ UV radiation 88n 2O(g) (step 1—occurs once)
O2(g) ϩ O(g) 88n O3(g) (step 2—occurs twice)
3O2(g) ϩ UV radiation 88n 2O3(g) (net reaction)
Although it decomposes in the upper atmosphere, the ozone
supply is continuously replenished by this process. Its concentration
in the stratosphere (Ϸ 7–31 miles above the earth’s
surface) is about 10 ppm (parts per million), whereas it is only
about 0.04 ppm near the earth’s surface.
The high-altitude ozone layer is responsible for absorbing
much of the dangerous ultraviolet light from the sun in
the 20–30 Å wavelength range.
O3(g) ϩ UV radiation 88n O2(g) ϩ O(g)
(step 1—occurs once)
O2(g) ϩ O(g) 88n O3(g) (step 2—occurs once)
No net reaction
We see that each time this sequence takes place, it absorbs
one photon of ultraviolet light; however, the process regenerates
as much ozone as it uses up. Each stratospheric ozone
molecule can thus absorb a significant amount of ultraviolet
light. If this high-energy radiation reached the surface of the
earth in higher intensity, it would be very harmful to plants
and animals (including humans). It has been estimated that
the incidence of skin cancer would increase by 2% for every
1% decrease in the concentration of ozone in the stratos-
phere.
Chlorofluorocarbons (CFCs) are chemically inert, nonflammable,
nontoxic compounds that are superb solvents and
have been used in many industrial processes; they are excellent
coolants for air conditioners and refrigerators. Two
CFCs that have been widely used are Freon-11 and Freon-
12 (Freon is a DuPont trade name).
Cl
Cl
C FCl
Cl
F
C FCl
Freon-11 is CCl3F Freon-12 is CCl2F2
1.0
0.5
0
63˚S
0.5
1.0
1.5
2.0
2.5
Ozone,ppm
Chlorineoxide,ppb
Approximate latitude 72˚S
O3
ClO
A plot that shows the decrease in [O3] as [ClO]
increases over Antarctica.
The compound known as HCFC-134, a
fluorocarbon currently used in home and
automobile air conditioners.
The CFCs are so unreactive that they do not readily decompose,
that is, break down into simpler compounds, when they
are released into the atmosphere. Over time the CFCs are
carried into the stratosphere by air currents, where they are
exposed to large amounts of ultraviolet radiation.
In 1974, Mario Molina and Sherwood Rowland of the
University of California–Irvine demonstrated in their laboratory
that when CFCs are exposed to ultraviolet radiation
they break down to form chlorine radicals (SOClT).
h
Molina and Rowland predicted that these very reactive radicals
could cause problems by catalyzing the destruction of
ozone in the stratosphere.
Each spring since 1979, researchers have observed a thinning
of the ozone layer over Antarctica. Each spring (autumn
in the Northern Hemisphere) beginning in 1983, satellite
images have shown a “hole” in the ozone layer over the South
Pole. During August and September 1987, a NASA research
team flew a plane equipped with sophisticated analytical
instruments into the ozone hole 25 times. Their measurements
demonstrated that as the concentration of the chlorine
oxide radicals, ClXO increased, the concentration of ozone
decreased.
By September 1992, this ozone hole was nearly three
times the area of the United States. In December 1994, three
years of data from NASA’s Upper Atmosphere Research
Satellite (UARS) provided conclusive evidence that CFCs are
primarily responsible for this destruction of the ozone layer.
Considerable thinning of the ozone layer in the Northern
Hemisphere has also been observed.
The following is a simplified representation of the chain
reaction that is now believed to account for most of the
ozone destruction in the stratosphere.
A sufficient supply of oxygen atoms, O, is available in the
upper atmosphere for the second step to occur. The net reaction
results in the destruction of a molecule of ozone. The
chlorine radical that initiates the first step of this reaction
sequence is regenerated in the second step, however, and so
a single chlorine radical can act as a catalyst to destroy many
(step 1)ϩ ϩCl Cl
Cl (step 2)ϩ ϩCl
(net reaction)ϩ
O3 O2O
OO O2
OO3(g) 2O2(g)
Cl
C
F
F Cl
Cl
F
C ClF
UV
radiation
(a chlorine radical)ϩ
thousands of O3 molecules. Other well-known reactions also
destroy ozone in the stratosphere, but the evidence shows
conclusively that the CFCs are the principal culprits.
Since January 1978, the use of CFCs in aerosol cans in
the United States has been banned; increasingly strict laws
prohibit the release into the atmosphere of CFCs from
sources such as automobile air conditioners and discarded
refrigerators. The Montreal Protocol, signed by 24 countries
in 1989, called for reductions in production and use of many
CFCs. International agreements have since called for a complete
ban on CFC production. Efforts to develop suitable
replacement substances and controls for existing CFCs continue.
The good news is that scientists expect the ozone hole
to decrease and possibly disappear during the twenty-first
century if current international treaties remain in effect and
if they are implemented throughout the world. These are two
very large ifs.
Additional information on stratospheric ozone can be
found at the EPA Web site, http://www.epa.gov/docs/ozone.
The Web site for this textbook will direct you to additional
information about ozone.
A computer-generated image of part of the Southern Hemisphere on
October 17, 1994, reveals the ozone “hole” (black and purple areas)
over Antarctica and the tip of South America. Relatively low ozone levels
(blue and green areas) extend into much of South America as well as
Central America. Normal ozone levels are shown in yellow, orange, and
red. The ozone hole is not stationary but moves about as a result of air
currents. (Courtesy NASA)
reaction is controlled only by the availability of reaction sites on the Pt surface, and not
by the total number of NO2 molecules available.
Some other important reactions that are catalyzed by transition metals and their oxides
follow.
1. The Haber process for the production of ammonia (Section 17-7).
Fe, Fe oxides
N2 ϩ 3H2 8888888888n 2NH3
high T, P
2. The contact process for the production of sulfur trioxide in the manufacture of
sulfuric acid (Section 24-11).
V2O5
2SO2 ϩ O2 88888n 2SO3
400°C
3. The chlorination of benzene (Section 27-16).
FeBr3
C6H6 ϩ Cl2 88888n C6H5Cl ϩ HCl
benzene chlorobenzene
4. The hydrogenation of unsaturated hydrocarbons (Section 27-17).
Pt
RCHUCH2 ϩ H2 888n RCH2CH3 R ϭ organic groups
Enzymes as Biological Catalysts
Enzymes are proteins that act as catalysts for specific biochemical reactions in living
systems. The reactants in enzyme-catalyzed reactions are called substrates. Thousands
of vital processes in our bodies are catalyzed by many distinct enzymes. For instance, the
enzyme carbonic anhydrase catalyzes the combination of CO2 and water (the substrates),
facilitating most of the transport of carbon dioxide in the blood. This combination reaction,
ordinarily uselessly slow, proceeds rapidly in the presence of carbonic anhydrase; a
single molecule of this enzyme can promote the conversion of more than 1 million molecules
of carbon dioxide each second. Each enzyme is extremely specific, catalyzing only a
few closely related reactions—or, in many cases, only one particular reaction—for only
certain substrates. Modern theories of enzyme action attribute this to the requirement of
very specific matching of shapes (molecular geometries) for a particular substrate to bind
to a particular enzyme (Figure 16-19).
When heated, a sugar cube (sucrose,
melting point 185°C) melts but does
not burn. A sugar cube rubbed in
cigarette ash burns before it melts.
The cigarette ash contains trace
amounts of metal compounds that
catalyze the combustion of sugar.
696 CHAPTER 16: Chemical Kinetics
A space-filling model of the enzyme lysozyme. This enzyme catalyzes the hydrolysis of
polysaccharides (complex carbohydrates) found in bacterial cell walls, that is, it breaks the
link between two adjacent sugar units in the polysaccharide. The intact polysaccharide,
shown here in green, fits into the active site, which is a cleft in the surface of the lysozyme
molecule. The arrangement of hydrogen-bonding groups in the active site of the enzyme
surface matches that on the polysaccharide, but other molecules with different hydrogenbonding
groups do not fit as well, so the enzyme exhibits very specific bonding for
polysaccharides. After the hydrolysis reaction has taken place, the product molecules do
not fit the active site as well, so they are released, making way for another polysaccharide to
be bound to the enzyme active site.
Enzyme-catalyzed reactions are important examples of zero-order reactions; that is, the
rate of such a reaction is independent of the concentration of the substrate (provided some
substrate is present).
rate ϭ k
The active site on an enzyme can bind to only one substrate molecule at a time (or one
pair, if the reaction links two reactant molecules), no matter how many other substrate
molecules are available in the vicinity.
Ammonia is a very important industrial chemical that is used as a fertilizer and in the
manufacture of many other chemicals. The reaction of nitrogen with hydrogen is a thermodynamically
spontaneous reaction (product-favored), but without a catalyst it is very
slow, even at high temperatures. The Haber process for its preparation involves the use
of iron as a catalyst at 450°C to 500°C and high pressures.
Fe
N2(g) ϩ 3H2(g) 888n 2NH3(g) ⌬G0 ϭ Ϫ194.7 kJ/mol (at 500°C)
Even so, iron is not a very effective catalyst.
In contrast, the reaction between N2 and H2 to form NH3 is catalyzed at room temperature
and atmospheric pressure by a class of enzymes, called nitrogenases, that are present
in some bacteria. Legumes are plants that support these bacteria; they are able to obtain
nitrogen as N2 from the atmosphere and convert it to ammonia.
In comparison with manufactured catalysts, most enzymes are tremendously efficient
under very mild conditions. If chemists and biochemists could develop catalysts with a
small fraction of the efficiency of enzymes, such catalysts could be a great boon to the
world’s health and economy. One of the most active areas of current chemical research
involves attempts to discover or synthesize catalysts that can mimic the efficiency of naturally
occurring enzymes such as nitrogenases. Such a development would be important in
industry. It would eliminate the costs of the high temperature and high pressure that are
necessary in the Haber process. This could decrease the cost of food grown with the aid
of ammonia-based fertilizers. Ultimately this would help greatly to feed the world’s
growing population.
Transition metal ions are present in
the active sites of some enzymes.
Enzyme bound
to substrate
Substrate
cleaved
Enzyme
unbound
Substrate
Enzyme
ϩ
Figure 16-19 A schematic representation of a simplified mechanism (lock-and-key) for
enzyme reaction. The substrates (reactants) fit the active sites of the enzyme molecule much
as keys fit locks. When the reaction is complete, the products do not fit the active sites as
well as the reactants did. They separate from the enzyme, leaving it free to catalyze the
reaction of additional reactant molecules. The enzyme is not permanently changed by the
process. The illustration here is for a process in which a complex reactant molecule is split
to form two simpler product molecules. The formation of simple sugars from complex
carbohydrates is a similar reaction. Some enzymes catalyze the combination of simple
molecules to form more complex ones.
The process is called nitrogen fixation.
The ammonia can be used in the
synthesis of many nitrogen-containing
biological compounds such as proteins
and nucleic acids.
698 CHAPTER 16: Chemical Kinetics
Key Terms
Activation energy The kinetic energy that reactant molecules
must have to allow them to reach the transition state so that a
reaction can occur.
Arrhenius equation An equation that relates the specific rate
constant to activation energy and temperature.
Catalyst A substance that increases the rate at which a reaction
occurs. It remains unchanged when the reaction is complete.
Chemical kinetics The study of rates and mechanisms of chemical
reactions and of the factors on which they depend.
Collision theory A theory of reaction rates that states that effective
collisions between reactant molecules must take place for
reaction to occur.
Contact catalyst See Heterogeneous catalyst.
Effective collision A collision between molecules that results in
reaction; one in which molecules collide with proper orientations
and with sufficient energy to react.
Elementary step An individual step in the mechanism by which
a reaction occurs. For each elementary step, the reaction orders
do match the reactant coefficients in that step.
Enzyme A protein that acts as a catalyst in a biological system.
Half-life of a reactant The time required for half of that reactant
to be converted into product(s).
Heterogeneous catalyst A catalyst that exists in a different
phase (solid, liquid, or gas) from the reactants; a contact cata-
lyst.
Homogeneous catalyst A catalyst that exists in the same phase
(liquid or gas) as the reactants.
Integrated rate equation An equation that relates the concentration
of a reactant remaining to the time elapsed; has different
mathematical forms for different orders of reaction.
Method of initial rates A method of determining the rate-law
expression by carrying out a reaction with different initial concentrations
and analyzing the resulting changes in initial rates.
Order of a reactant The power to which the reactant’s concentration
is raised to the rate-law expression.
Order of a reaction The sum of the powers to which all concentrations
are raised in the rate-law expression; also called
overall order of a reaction.
Rate constant (also called specific rate constant) An experimentally
determined proportionality constant that is different
for different reactions and that, for a given reaction, changes
only with temperature or the presence of a catalyst; k in the
rate-law expression, Rate ϭ k[A]x[B]y.
Rate-determining step The slowest step in a reaction mechanism;
the step that limits the overall rate of reaction.
Rate-law expression (also called rate law) An equation that
relates the rate of a reaction to the concentrations of the reactants
and the specific rate constant; rate ϭ k[A]x[B]y. The
exponents of reactant concentrations do not necessarily match the
coefficients in the overall balanced chemical equation. The ratelaw
expression must be determined from experimental data.
Rate of reaction The change in concentration of a reactant or
product per unit time.
Reaction coordinate The progress along the pathway from
reactants to products; sometimes called “progress of reaction”.
Reaction intermediate A species that is produced and then
entirely consumed during a multistep reaction; usually short-
lived.
Reaction mechanism The sequence of steps by which reactants
are converted into products.
Substrate A reactant in an enzyme-catalyzed reaction.
Thermodynamically favorable (spontaneous) reaction A
reaction that occurs with a net release of free energy, G; a reaction
for which ⌬G is negative (see Section 15-15).
Transition state A relatively high-energy state in which bonds
in reactant molecules are partially broken and new ones are partially
formed.
Transition state theory A theory of reaction rates that states
that reactants pass through high-energy transition states before
forming products.
Exercises
General Concepts
*01. Briefly summarize the effects of each of the four factors
that affect rates of reactions.
*02. Describe the basic features of collision theory and transition
state theory.
*03. What is a rate-law expression? Describe how it is determined
for a particular reaction.
*04. Distinguish between reactions that are thermodynamically
favorable and reactions that are kinetically favorable. What
can be said about relationships between the two?
*05. What is meant by the order of a reaction?
*06. What, if anything, can be said about the relationship
between the coefficients of the balanced overall equation
for a reaction and the powers to which concentrations are
raised in the rate-law expression? To what are these powers
related?
*07. Express the rate of reaction in terms of the rate of change
of each reactant and product in the following reactions.
(a) H2O2(aq) ϩ 2Hϩ(aq) ϩ 2IϪ(aq) 88n
I2(aq) ϩ 2H2O(ᐉ)
(b) 2NO(g) ϩ Br2(g) 88n 2NOBr(g)
(c) CH3COOH(aq) ϩ OHϪ(aq) 88n
CH3COOϪ(aq) ϩ H2O(ᐉ)
Exercises 699
*08. Express the rate of reaction in terms of the rate of change
of each reactant and each product in the following.
(a) 3ClOϪ(aq) 88n ClO3
Ϫ(aq) ϩ 2ClϪ(aq)
(b) 2SO2(g) ϩ O2(g) 88n 2SO3(g)
(c) C2H4(g) ϩ Br2(g) 88n C2H4Br2(g)
*09. At a given time, N2 is reacting with H2 at a rate of 0.25
M/min to produce NH3. At that same time, what is the
rate at which the other reactant is changing and the rate
at which the product is changing?
N2 ϩ 3H2 88n 2NH3
*10. The following equation shows the production of NO and
H2O by oxidation of ammonia. At a given time, NH3 is
reacting at a rate of 1.10 M/min. At that same time, which
is the rate at which the other reactant is changing and the
rate at which each product is changing?
4NH3 ϩ 5O2 88n 4NO ϩ 6H2O
*11. Why do large crystals of sugar burn more slowly than finely
ground sugar?
*12. Some fireworks are bright because of the burning of magnesium.
Speculate on how fireworks might be constructed
using magnesium. How might the sizes of the pieces of
magnesium be important? What would you expect to occur
if pieces that were too large were used? Too small?
tion of O2 is doubled. The initial rate in the second experiment
will be ______ times that of the first.
2NO ϩ O2 88n 2NO2
*15. The rate-law expression for the following reaction is found
to be rate ϭ k [N2O5]. What is the overall reaction order?
2N2O5(g) 88n 4NO2(g) ϩ O2(g)
*16. Use times expressed in seconds to give the units of the rate
constant for reactions that are overall (a) first order; (b)
second order; (c) third order; (d) of order 1ᎏ1
2
ᎏ.
*17. Rate data were obtained at 25°C for the following reaction.
What is the rate-law expression for this reaction?
A ϩ 2B 88n C ϩ 2D
[A] [B] Initial Rate of
Expt. [mol/L] [mol/L] Formation of C
1 0.10 0.10 3.0 ϫ 10Ϫ4 MиminϪ1
2 0.30 0.30 9.0 ϫ 10Ϫ4 MиminϪ1
3 0.30 0.10 3.0 ϫ 10Ϫ4 MиminϪ1
4 0.40 0.20 6.0 ϫ 10Ϫ4 MиminϪ1
*18. Rate data were obtained for the following reaction at 25°C.
What is the rate-law expression for the reaction?
2A ϩ B ϩ 2C 88n D ϩ 2E
Initial Initial Initial Initial Rate of
Expt. [A] [B] [C] Formation of D
1 0.10 M 0.20 M 0.10 M 5.0 ϫ 10Ϫ4 MиminϪ1
2 0.20 M 0.20 M 0.30 M 1.5 ϫ 10Ϫ3 MиminϪ1
3 0.30 M 0.20 M 0.10 M 5.0 ϫ 10Ϫ4 MиminϪ1
4 0.40 M 0.60 M 0.30 M 4.5 ϫ 10Ϫ3 MиminϪ1
*19. (a) A certain reaction is zero order in reactant A and second
order in reactant B. If the concentrations of both
reactants are doubled, what happens to the reaction rate?
(b) What would happen to the reaction rate if the reaction
in part (a) were first order in A and first order in B?
*20. The rate expression for the following reaction is shown to
be rate ϭ k[A]2[B2]. If, during a reaction, the concentrations
of both A and B2 were suddenly halved, the rate of
the reaction would ______ by a factor of ______.
A ϩ B2 88n products
*21. Rate data were collected for the following reaction at a
particular temperature.
A ϩ B 88n products
Rate-Law Expression
*13. If doubling the initial concentration of a reactant doubles
the initial rate of reaction, what is the order of the reaction
with respect to the reactant? If this concentration
change causes the rate to increase by a factor of 8, what is
the order? If the concentration changes and the rate
remains the same, what is the order?
*14. The rate expression for the following reaction at a certain
temperature is rate ϭ k[NO]2[O2]. Two experiments
involving this reaction are carried out at the same temperature.
In the second experiment the initial
concentration of NO is halved while the initial concentra-
700 CHAPTER 16: Chemical Kinetics
Initial Rate
Expt. [A]0 (mol/L) [B]0 (mol/L) of Reaction
1 0.10 0.10 0.0090 MиsϪ1
2 0.20 0.10 0.0360 MиsϪ1
3 0.10 0.20 0.0180 MиsϪ1
4 0.10 0.30 0.0270 MиsϪ1
(a) What is the rate-law expression for this reaction?
(b) Describe the order of the reaction with respect to each
reactant and to the overall order.
*22. Rate data were collected for the following reaction at a particular
temperature.
2ClO2(aq) ϩ 2OHϪ(aq) 88n
ClO3
Ϫ(aq) ϩ ClO2
Ϫ(aq) ϩ H2O(ᐉ)
[ClO2]0 [OHϪ]0 Initial Rate
Expt. (mol/L) (mol/L) of Reaction
1 0.012 0.012 2.07 ϫ 10Ϫ4 MиsϪ1
2 0.024 0.012 8.28 ϫ 10Ϫ4 MиsϪ1
3 0.012 0.024 4.14 ϫ 10Ϫ4 MиsϪ1
4 0.024 0.024 1.66 ϫ 10Ϫ3 MиsϪ1
(a) What is the rate-law expression for this reaction?
(b) Describe the order of the reaction with respect to each
reactant and to the overall order.
*23. The reaction (C2H5)2(NH)2 ϩ I2 n (C2H5)2N2 ϩ 2HI
gives the following initial rates.
Initial Rate
[(C2H5)2(NH)2]0 [I2]0 of Formation
Expt. (mol/L) (mol/L) of (C2H5)2N2
1 0.015 0.015 03.15 MиsϪ1
2 0.015 0.045 09.45 MиsϪ1
3 0.030 0.045 18.90 MиsϪ1
Write the rate-law expression.
*24. Given the following data for the reaction A ϩ B n C, write
the rate-law expression.
Initial Initial Initial Rate of
Expt. [A] [B] Formation of C
1 0.20 M 0.10 M 5.0 ϫ 10Ϫ6 MиsϪ1
2 0.30 M 0.10 M 7.5 ϫ 10Ϫ6 MиsϪ1
3 0.40 M 0.20 M 4.0 ϫ 10Ϫ5 MиsϪ1
*25. Given the following data for the reaction A ϩ B n C, write
the rate-law expression.
Initial Initial Initial Rate of
Expt. [A] [B] Formation of C
1 0.25 0.15 8.00 ϫ 10Ϫ5 M/s
2 0.25 0.30 3.20 ϫ 10Ϫ4 M/s
3 0.50 0.60 5.12 ϫ 10Ϫ3 M/s
*26. Given the following data for the reaction A ϩ B n C, write
the rate-law expression.
Initial Initial Initial Rate of
Expt. [A] [B] Formation of C
1 0.10 M 0.10 M 2.00 ϫ 10Ϫ4 M/s
2 0.20 M 0.10 M 8.00 ϫ 10Ϫ4 M/s
3 0.40 M 0.20 M 2.56 ϫ 10Ϫ2 M/s
*27. Consider a chemical reaction between compounds A and
B that is first order in A and first order in B. From the
information shown here, fill in the blanks.
Expt. Rate (MиsϪ1) [A] [B]
1 0.24 0.20 M 0.050 M
2 0.20 ______ M 0.030 M
3 0.80 0.40 M ______ M
*28. Consider a chemical reaction of compounds A and B that
was found to be first order in A and second order in B.
From the following information, fill in the blanks.
Expt. Rate (MиsϪ1) [A] [B]
1 0.150 1.00 M 0.200 M
2 ______ 2.00 M 0.200 M
3 ______ 2.00 M 0.400 M
*29. The following rearrangement reaction is first order:
CH3NC 88n CH3CN
In a table of kinetics data we find the following values listed
for this reaction: A ϭ 3.98 ϫ 1013 sϪ1, Ea ϭ 160 kJ/mol.
(a) Calculate the value of the specific rate constant at room
temperature, 25°C. (b) Calculate the value of the specific
rate constant at 115°C.
Exercises 701
*30. The following gas-phase decomposition reaction is first
order:
C2H5Cl 88n C2H4 ϩ HCl
In a table of kinetics data we find the following values listed
for this reaction: A ϭ 1.58 ϫ 1013 sϪ1, Ea ϭ 237 kJ/mol.
(a) Calculate the value of the specific rate constant at room
temperature, 25°C. (b) Calculate the value of the specific
rate constant at 275°C.
*31. The rate of decomposition of NO2 by the following reaction
at some temperature is 5.4 ϫ 10Ϫ5 mol NO2/Lиs when
[NO2] ϭ 0.0110 mol/L.
2NO2(g) 88n 2NO(g) ϩ O2(g)
(a) Assume that the rate law is rate ϭ k[NO2]. What rate
of disappearance of NO2 would be predicted when
[NO2] ϭ 0.00550 mol/L? (b) Now assume that the rate law
is rate ϭ k[NO2]2. What rate of disappearance of NO2
would be predicted when [NO2] ϭ 0.00550 mol/L? (c) The
rate when [NO2] ϭ 0.00550 mol/L is observed to be
1.4 ϫ 10Ϫ5 mol NO2/Lиs. Which rate law is correct? (d)
Calculate the rate constant. (Reminder: Express the rate of
reaction in terms of rate of disappearance of NO2.)
Integrated Rate Equations and Half-Life
*32. What is meant by the half-life of a reactant?
*33. The rate law for the reaction of sucrose in water,
C12H22O11 ϩ H2O 88n 2C6H12O6
is rate ϭ k[C12H22O11]. After 2.57 hours at 25°C, 6.00 g/L
of C12H22O11 has decreased to 5.40 g/L. Evaluate k for
this reaction at 25°C.
*34. The rate constant for the decomposition of nitrogen
dioxide
2NO2 88n 2NO ϩ O2
with a laser beam is 1.70 MϪ1иminϪ1. Find the time,
in seconds, needed to decrease 2.00 mol/L of NO2 to
1.25 mol/L.
*35. The second-order rate constant for the following gas-phase
reaction is 0.0442 MϪ1иsϪ1. We start with 0.135 mol C2F4
in a 2.00-liter container, with no C4F8 initially present.
2C2F4 88n C4F8
(a) What will be the concentration of C2F4 after 1.00 hour?
(b) What will be the concentration of C4F8 after 1.00 hour?
(c) What is the half-life of the reaction for the initial C2F4
concentration given in part (a)?
(d) How long will it take for half of the C2F4 that remains
after 1.00 hour to disappear?
*36. The decomposition reaction of carbon disulfide, CS2, to
carbon monosulfide, CS, and sulfur is first order with
k ϭ 2.8 ϫ 10Ϫ7 sϪ1 at 1000°C.
CS2 88n CS ϩ S
(a) What is the half-life of this reaction at 1000°C?
(b) How many days would pass before a 2.00-gram sample
of CS2 had decomposed to the extent that 0.75 gram of
CS2 remained?
(c) Refer to part (b). How many grams of CS would be
present after this length of time?
(d) How much of a 2.00-gram sample of CS2 would remain
after 45.0 days?
*37. The first-order rate constant for the conversion of cyclobutane
to ethylene at 1000°C is 87 sϪ1.
(a) What is the half-life of this reaction at 1000°C? (b) If
you started with 2.00 g of cyclobutane, how long would it
take to consume 1.50 g of it? (Hint: Write the ratio of
concentrations, [A]0/[A], in terms of mass, molecular
weight, and volume.) (c) How much of an initial 1.00-g
sample of cyclobutane would remain after 1.00 s?
*38. For the reaction
2NO2 88n 2NO ϩ O2
the rate equation is
rate ϭ 1.4 ϫ 10Ϫ10 MϪ1иsϪ1[NO2]2 at 25°C
(a) If 3.00 mol of NO2 is initially present in a sealed 2.00-L
vessel at 25°C, what is the half-life of the reaction?
(b) Refer to part (a). What concentration and how many
grams of NO2 remain after 115 years?
(c) Refer to part (b). What concentration of NO would
have been produced during the same period of time?
*39. The first-order rate constant for the radioactive decay of
radium-223 is 0.0606 dayϪ1. What is the half-life of
radium-223?
*40. Cyclopropane rearranges to form propene
CH H
HH
C C
H H
H
C
C
cyclopropane propene
C
H
HH H H
C
H
H
H H
CH H
C
H
H
C C
H
H
H
H
C
cyclobutane ethylene
2
702 CHAPTER 16: Chemical Kinetics
in a reaction that follows first-order kinetics. At 800 K, the
specific rate constant for this reaction is 2.74 ϫ 10Ϫ3 sϪ1.
Suppose we start with a cyclopropane concentration of
0.290 M. How long will it take for 99.0% of the cyclopropane
to disappear according to this reaction?
*41. The rate constant for the first-order reaction
N2O5 88n 2NO2 ϩ ᎏ1
2
ᎏO2
is 1.20 ϫ 10Ϫ2 sϪ1 at 45°C, and the initial concentration
of N2O5 is 0.00500 M. (a) How long will it take for the
concentration to decrease to 0.00110 M? (b) How much
longer will it take for a further decrease to 0.000900 M?
*42. It is found that 47.0 minutes is required for the concentration
of substance A to decrease from 0.75 M to 0.20 M.
What is the rate constant for this first-order decomposi-
tion?
A 88n B ϩ C
*43. The thermal decomposition of ammonia at high temperatures
was studied in the presence of inert gases. Data at
2000 K are given for a single experiment.
NH3 88n NH2 ϩ H
t (hours) [NH3] (mol/L)
00 8.000 ϫ 10Ϫ7
25 6.750 ϫ 10Ϫ7
50 5.840 ϫ 10Ϫ7
75 5.150 ϫ 10Ϫ7
Plot the appropriate concentration expressions against
time to find the order of the reaction. Find the rate
constant of the reaction from the slope of the line. Use the
given data and the appropriate integrated rate equation to
check your answer.
*44. The following data were obtained from a study of the
decomposition of a sample of HI on the surface of a gold
wire. (a) Plot the data to find the order of the reaction, the
rate constant, and the rate equation. (b) Calculate the HI
concentration in mmol/L at 600. s.
t (seconds) [HI] (mmol/L)
000. 5.46
250. 4.10
500. 2.73
750. 1.37
*45. The decomposition of SO2Cl2 in the gas phase,
SO2Cl2 88n SO2 ϩ Cl2
can be studied by measuring the concentration of Cl2 as
the reaction proceeds. We begin with [SO2Cl2]0 ϭ 0.250
M. Holding the temperature constant at 320°C, we
monitor the Cl2 concentration, with the following results.
t (hours) [Cl2] (mol/L)
0.00 0.000
2.00 0.037
4.00 0.068
6.00 0.095
8.00 0.117
10.00 0.137
12.00 0.153
14.00 0.168
16.00 0.180
18.00 0.190
20.00 0.199
(a) Plot [Cl2] versus t. (b) Plot [SO2Cl2] versus t. (c) Determine
the rate law for this reaction. (d) What is the value,
with units, for the specific rate constant at 320°C? (e) How
long would it take for 95% of the original SO2Cl2 to react?
*46. At some temperature, the rate constant for the decomposition
of HI on a gold surface is 0.080 MиsϪ1.
2HI(g) 88n H2(g) ϩ I2(g)
(a) What is the order of the reaction? (b) How long will it
take for the concentration of HI to drop from 1.50 M to
0.30 M?
Activation Energy, Temperature, and Catalysts
*47. Draw typical reaction energy diagrams for one-step reactions
that release energy and that absorb energy.
Distinguish between the net energy change, ⌬E, for each
kind of reaction and the activation energy. Indicate potential
energies of products and reactants for both kinds of
reactions.
*48. Use graphs to illustrate how the presence of a catalyst can
affect the rate of a reaction.
*49. How do homogeneous catalysts and heterogeneous catalysts
differ?
*50. (a) Why should one expect an increase in temperature to
increase the initial rate of reaction? (b) Why should one
expect a reaction in the gaseous state to be faster than the
same reaction in the solid state?
*51. What is the activation energy for a reaction if its rate constant
is found to triple when the temperature is raised from
300 K to 310 K?
*52. What is the activation energy for a reaction if its rate constant
is found to triple when the temperature is raised from
600 K to 610 K?
Exercises 703
*53. For a gas-phase reaction, Ea ϭ 103 kJ/mol, and the rate
constant is 0.0850 minϪ1 at 273 K. Find the rate constant
at 323 K.
*54. The rate constant of a reaction is tripled when the temperature
is increased from 298 K to 308 K. Find Ea.
*55. The rate constant for the decomposition of N2O
2N2O(g) 88n 2N2(g) ϩ O2(g)
is 2.6 ϫ 10Ϫ11 sϪ1 at 300°C and 2.1 ϫ 10Ϫ10 sϪ1 at 330°C.
Calculate the activation energy for this reaction. Prepare
a reaction coordinate diagram like Figure 16-10 using
Ϫ164.1 kJ/mol as the ⌬Erxn.
*56. For a particular reaction, ⌬E0 ϭ 51.51 kJ/mol, k ϭ 8.0 ϫ
10Ϫ7 sϪ1 at 0.0°C, and k ϭ 8.9 ϫ 10Ϫ4 sϪ1 at 50.0°C. Prepare
a reaction coordinate diagram like Figure 16-10 for
this reaction.
*57. You are given the rate constant as a function of temperature
for the exchange reaction
Mn(CO)5(CH3CN)ϩ ϩ NC5H5 88n
Mn(CO)5(NC5H5)ϩ ϩ CH3CN
T (K) k (minϪ1)
298 0.0409
308 0.0818
318 0.157
(a) Calculate Ea from a plot of log k versus 1/T. (b) Use
the graph to predict the value of k at 311 K. (c) What is
the numerical value of the collision frequency factor, A, in
the Arrhenius equation?
*58. The rearrangement of cyclopropane to propene described
in Exercise 40 has been studied at various temperatures.
The following values for the specific rate constant have
been determined experimentally.
T (K) k (sϪ1)
600. 3.30 ϫ 10Ϫ9
650. 2.19 ϫ 10Ϫ7
700. 7.96 ϫ 10Ϫ6
750. 1.80 ϫ 10Ϫ4
800. 2.74 ϫ 10Ϫ3
850. 3.04 ϫ 10Ϫ2
900. 2.58 ϫ 10Ϫ1
(a) From the appropriate plot of these data, determine the
value of the activation energy for this reaction. (b) Use the
graph to estimate the value of k at 500 K. (c) Use the graph
to estimate the temperature at which the value of k would
be equal to 5.00 ϫ 10Ϫ5 sϪ1.
*59. Biological reactions nearly always occur in the presence of
enzymes as catalysts. The enzyme catalase, which acts on
peroxides, reduces the Ea for the reaction from 72 kJ/mol
(uncatalyzed) to 28 kJ/mol (catalyzed). By what factor does
the reaction rate increase at normal body temperature,
37.0°C, for the same reactant (peroxide) concentration?
Assume that the collision factor, A, remains constant.
*60. The enzyme carbonic anhydrase catalyzes the hydration of
carbon dioxide.
CO2 ϩ H2O 88n H2CO3
This reaction is involved in the transfer of CO2 from tissues
to the lungs via the bloodstream. One enzyme molecule
hydrates 106 molecules of CO2 per second. How many
kilograms of CO2 are hydrated in one hour in 1 L by
1.0 ϫ 10Ϫ6 M enzyme?
*61. The following gas-phase reaction follows first-order
kinetics.
ClO2F 88n ClOF ϩ O
The activation energy of this reaction is 186 kJ/mol. The
value of k at 322°C is 6.76 ϫ 10Ϫ4 sϪ1. (a) What would
be the value of k for this reaction at 25°C? (b) At what
temperature would this reaction have a k value of
3.00 ϫ 10Ϫ2 sϪ1?
*62. The following gas-phase reaction is first order.
N2O5 88n NO2 ϩ NO3
The activation energy of this reaction is 88 kJ/mol. The
value of k at 0°C is 9.16 ϫ 10Ϫ3 sϪ1. (a) What would be
the value of k for this reaction at room temperature, 25°C?
(b) At what temperature would this reaction have a k value
of 3.00 ϫ 10Ϫ2 sϪ1?
Reaction Mechanisms
*63. Define reaction mechanism. Why do we believe that only
bimolecular collisions and unimolecular decompositions
are important in most reaction mechanisms?
*64. The rate equation for the reaction
Cl2(aq) ϩ H2S(aq) 88n S(s) ϩ 2HCl(aq)
is found to be rate ϭ k[Cl2][H2S]. Which of the following
mechanisms are consistent with the rate law?
(a) Cl2 88n Clϩ ϩ ClϪ (slow)
ClϪ ϩ H2S 88n HCl ϩ HSϪ (fast)
Clϩ ϩ HSϪ 88n HCl ϩ S (fast)
Cl2 ϩ H2S 88n S ϩ 2HCl (overall)
(b) Cl2 ϩ H2S 88n HCl ϩ Clϩ ϩ HSϪ (slow)
Clϩ ϩ HSϪ 88n HCl ϩ S (fast)
Cl2 ϩ H2S 88n S ϩ 2HCl (overall)
704 CHAPTER 16: Chemical Kinetics
(c) Cl2 34 Cl ϩ Cl (fast, equilibrium)
Cl ϩ H2S 34 HCl ϩ HS (fast, equilibrium)
HS ϩ Cl 88n HCl ϩ S (slow)
Cl2 ϩ H2S 88n S ϩ 2HCl (overall)
*65. Write the overall reaction and the rate expressions that correspond
to the following reaction mechanisms. Be sure to
eliminate intermediates from the answers:
(a) A ϩ B 34 C ϩ D (fast, equilibrium)
(a) C ϩ E 88n F (slow)
(b) A 34 B ϩ C (fast, equilibrium)
(a) C ϩ D 34 E (fast, equilibrium)
(a) E 88n F (slow)
*66. Write the overall reaction and the rate expressions that correspond
to the following mechanisms. Be sure to eliminate
intermediates from the answers:
(a) 2 A ϩ B 34 D (fast, equilibrium)
(a) D ϩ B 88n E ϩ F (slow)
(a) F 88n G (fast)
(b) A ϩ B 34 C (fast, equilibrium)
(a) C ϩ D 34 F (fast, equilibrium)
(a) F 88n G (slow)
*67. The ozone, O3, of the stratosphere can be decomposed by
reaction with nitrogen oxide (commonly called nitric
oxide), NO, from high-flying jet aircraft.
O3(g) ϩ NO(g) 88n NO2(g) ϩ O2(g)
The rate expression is rate ϭ k[O3][NO]. Which of the
following mechanisms are consistent with the observed rate
expression?
(a) NO ϩ O3 88n NO3 ϩ O (slow)
NO3 ϩ O 88n NO2 ϩ O2 (fast)
O3 ϩ NO 88n NO2 ϩ O2 (overall)
(b) NO ϩ O3 88n NO2 ϩ O2 (slow)
(one step)
(c) O3 88n O2 ϩ O (slow)
O ϩ NO 88n NO2 (fast)
O3 ϩ NO 88n NO2 ϩ O2 (overall)
(d) NO 88n N ϩ O (slow)
O ϩ O3 88n 2O2 (fast)
O2 ϩ N 88n NO2 (fast)
O3 ϩ NO 88n NO2 ϩ O2 (overall)
(e) NO 34 N ϩ O (fast, equilibrium)
O ϩ O3 88n 2O2 (slow)
O2 ϩ N 88n NO2 (fast)
O3 ϩ NO 88n NO2 ϩ O2 (overall)
*68. A proposed mechanism for the decomposition of ozone,
2O3 n 3O2, is
O3 34 O2 ϩ O (fast, equilibrium)
O ϩ O3 88n 2O2 (slow)
Derive the rate equation for the net reaction.
*69. A mechanism for the gas-phase reaction
H2 ϩ I2 88n 2HI
was discussed in the chapter. (a) Show that this mechanism
predicts the correct rate law, rate ϭ k[H2][I2].
I2 34 2I (fast, equilibrium)
I ϩ H2 34 H2I (fast, equilibrium)
H2I ϩ I 88n 2HI (slow)
(b) Identify any reaction intermediates in this proposed
mechanism.
*70. The combination of Cl atoms is catalyzed by N2(g). The
following mechanism is suggested.
N2 ϩ Cl 34 N2Cl (fast, equilibrium)
N2Cl ϩ Cl 88n Cl2 ϩ N2 (slow)
(a) Identify any reaction intermediates in this proposed
mechanism. (b) Is this mechanism consistent with the
experimental rate law, rate ϭ k[N2][Cl]2?
*71. The reaction between NO and Br2 was discussed in Section
16-7. The following mechanism has also been
proposed.
2NO 34 N2O2 (fast, equilibrium)
N2O2 ϩ Br2 88n 2NOBr (slow)
Is this mechanism consistent with the observation that the
reaction is second order in NO and first order in Br2?
*72. The following mechanism for the reaction between H2 and
CO to form formaldehyde, H2CO, has been proposed.
H2 34 2H (fast, equilibrium)
H ϩ CO 88n HCO (slow)
H ϩ HCO 88n H2CO (fast)
(a) Write the balanced equation for the overall reaction.
(b) The observed rate dependence is found to be one-half
order in H2 and first order in CO. Is this proposed reaction
mechanism consistent with the observed rate
dependence?
*73. The reaction between nitrogen dioxide and ozone,
2NO2 ϩ O3 88n N2O5 ϩ O2
has been studied at 231 K. The experimental rate equation
is rate ϭ k[NO2][O3]. (a) What is the order of the reaction?
(b) Is either of the following proposed mechanisms
Exercises 705
consistent with the given kinetic data? Show how you
arrived at your answer.
(a) NO2 ϩ NO2 34 N2O4 (fast, equilibrium)
N2O4 ϩ O3 88n N2O5 ϩ O2 (slow)
(b) NO2 ϩ O3 88n NO3 ϩ O2 (slow)
NO3 ϩ NO2 88n N2O5 (fast)
Mixed Exercises
*74. Some reactions occur faster than others due to differences
in the shapes of the reactants. Use the collision theory to
explain these observations.
*75. How it is possible for two reactant molecules to collide
with the correct orientation and still not react?
*76. (a) What is the transition state in a reactant mechanism?
(b) Are the energy of activation and the transition state
related concepts? Explain. (c) How does the activation
energy affect the rate of reaction?
*77. Refer to the reaction and data in Exercise 61. We begin
with 3.60 mol of ClO2F in a 3.00-L container. (a) How
many moles of ClO2F would remain after 1.00 min at
25°C? (b) How much time would be required for 99.0%
of the ClO2F to decompose at 25°C?
*78. Refer to the reaction and data in Exercise 62. We begin
with 3.60 mol of N2O5 in a 3.00-L container. (a) How
many moles of N2O5 would remain after 1.00 min at 25°C?
(b) How much time would be required for 99.0% of the
N2O5 to decompose at 25°C?
*79. The decomposition of gaseous dimethyl ether
CH3OCH3 88n CH4 ϩ CO ϩ H2
follows first-order kinetics. Its half-life is 25.0 min at
500°C. (a) Starting with 10.00 g of dimethyl ether at
500°C, how many grams would remain after 150 min? (b)
In part (a), how many grams would remain after 180 min?
(c) In part (b), what fraction remains and what fraction
reacts? (d) Calculate the time, in minutes, required to
decrease 9.20 mg of dimethyl ether to 2.40 mg.
*80. The rate of the hemoglobin (Hb)–carbon monoxide
reaction,
4Hb ϩ 3CO 88n Hb4(CO)3
has been studied at 20°C. Concentrations are expressed in
micromoles per liter (␮mol/L).
Concentration
(␮mol/L) Rate of
Disappearance
[Hb] [CO] of Hb (␮molиLϪ1 иsϪ1)
3.36 1.00 0.941
6.72 1.00 1.88
6.72 3.00 5.64
(a) Write the rate equation for the reaction. (b) Calculate
the rate constant for the reaction. (c) Calculate the rate, at
the instant when [Hb] ϭ 1.50 and [CO] ϭ 0.600 ␮mol/L.
*81. How does an enzyme change the speed with which a reaction
reaches equilibrium? Can an enzyme change the final
equilibrium concentrations? Explain.
CONCEPTUAL EXERCISES
*82. Write the net ionic equation for the following reaction.
Construct a potential energy diagram, like Figure 16-10,
for this reaction.
HCl(aq) ϩ NaOH(aq) 88n NaCl(aq) ϩ H2O(ᐉ)
*83. Starting with only two molecules of each reactant in a reaction
that is first order in each reactant, show how the
collision theory predicts that the rate of reaction will double
if the amount of either reactant is doubled.
*84. A sentence in an introductory chemistry textbook reads,
“Dioxygen reacts with itself to form trioxygen, ozone,
according to the following equation, 3O2 n 2O3.” As a
student of chemistry, what would you write to criticize this
sentence?
*85. A stream of gaseous H2 is directed onto finely divided
platinum powder in the open air. The metal immediately
glows white-hot and continues to do so as long as the
stream continues. Explain.
*86. Is the activation energy of a reaction expected to be higher
or lower when the same reactants are in the gaseous state
rather than the liquid or solid state? Explain.
BUILDING YOUR KNOWLEDGE
*87. The following explanation of the operation of a pressure
cooker appears in a cookbook: “Boiling water in the presence
of air can never produce a temperature higher than
212°F, no matter how high the heat source. But in a pressure
cooker, the air is withdrawn first, so the boiling water
can be maintained at higher temperatures.” Support or criticize
this explanation.
*88. A cookbook gives the following general guideline for use
of a pressure cooker. “For steaming vegetables, cooking
time at a gauge pressure of 15 pounds per square inch (psi)
is ᎏ1
3
ᎏ that at atmospheric pressure.” Remember that gauge
pressure is measured relative to the external atmospheric
pressure, which is 15 psi at sea level. From this information,
estimate the activation energy for the process of
steaming vegetables. (Hint: Clausius and Clapeyron may
be able to help you.)
706 CHAPTER 16: Chemical Kinetics
*89. For most reactions that involve an enzyme, the rate of
product formation versus reactant concentration increases
as reactant concentration increases until a maximum value
is obtained, after which further increases do not yield increased
rates. Using a description like that in Figure 16-19,
describe how the reaction may be first order with respect
to substrate but the amount of enzyme can also be a determining
factor.
*90. Using the mechanism and energy values shown in Figure
16-12, prepare Lewis formulas that illustrate the species
that are likely to be present at each of the peaks and troughs
in the graphical representation given in Figure 16-12.
(Hint: You may need to label some bonds as being weaker,
stretched, in the process of being formed, and so on.)
*91. The activation energy for the reaction
2HI(g) 88n H2(g) ϩ I2(g)
is 179 kJ/mol. Construct a diagram similar to Figure 16-10
for this reaction. (Hint: Calculate ⌬H0 from values in
Appendix K. How does ⌬H0 compare with ⌬E0 for this
reaction?)
*92. The activation energy for the reaction between O3 and NO
is 9.6 kJ/mol.
O3(g) ϩ NO(g) 88n NO2(g) ϩ O2(g)
(a) Use the thermodynamic quantities in Appendix K to
calculate ⌬H 0 for this reaction. (b) Prepare an activation
energy plot similar to Figure 16-10 for this reaction. (Hint:
How does ⌬H0 compare with ⌬E0 for this reaction?)
17Chemical
Equilibrium
OUTLINE
17-1 Basic Concepts
17-2 The Equilibrium Constant
17-3 Variation of Kc with the Form
of the Balanced Equation
17-4 The Reaction Quotient
17-5 Uses of the Equilibrium
Constant, Kc
17-6 Factors That Affect Equilibria
17-7 The Haber Process: A Practical
Application of Equilibrium
17-8 Application of Stress to a
System at Equilibrium
17-9 Partial Pressures and the
Equilibrium Constant
17-10 Relationship between KP
and Kc
17-11 Heterogeneous Equilibria
17-12 Relationship between ⌬G0
rxn
and the Equilibrium Constant
17-13 Evaluation of Equilibrium
Constants at Different
Temperatures
OBJECTIVES
After you have studied this chapter, you should be able to
• Explain the basic ideas of chemical equilibrium
• Explain what an equilibrium constant is and what it tells us
• Explain what a reaction quotient is and what it tells us
• Use equilibrium constants to describe systems at equilibrium
• Recognize the factors that affect equilibria and predict the resulting effects
• Use the equilibrium constant expressed in terms of partial pressures (KP) and
relate it to Kc
• Describe heterogeneous equilibria and write their equilibrium constants
• Use the relationships between thermodynamics and equilibrium
• Estimate equilibrium constants at different temperatures
BASIC CONCEPTS
Most chemical reactions do not go to completion. That is, when reactants are mixed in
stoichiometric quantities, they are not completely converted to products. Reactions that
do not go to completion and that can occur in either direction are called reversible reac-
tions.
Reversible reactions can be represented in general terms as follows, where the capital
letters represent formulas and the lowercase letters represent the stoichiometric coefficients
in the balanced equation.
aA ϩ bB 34 cC ϩ dD
17-1
Gaseous ammonia, NH3, and
gaseous hydrogen chloride, HCl,
react to form solid NH4Cl, the
white smoke. In the reverse
reaction, solid NH4Cl decomposes
when heated, to form gaseous NH3
and HCl.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 16.2, The Principle of
Microscopic Reversibility.
The double arrow (34) indicates that the reaction is reversible—that is, both the forward
and reverse reactions occur simultaneously. In discussions of chemical equilibrium, the
substances that appear on the left side of the balanced chemical equation are called the
“reactants,” and those on the right side are called the “products.” In fact, the reaction can
proceed in either direction. When A and B react to form C and D at the same rate at which
C and D react to form A and B, the system is at equilibrium.
Chemical equilibrium exists when two opposing reactions occur simultaneously at
the same rate.
Chemical equilibria are dynamic equilibria; that is, individual molecules are continually
reacting, even though the overall composition of the reaction mixture does not change.
In a system at equilibrium, the equilibrium is said to lie toward the right if more C and
D are present than A and B, and to lie toward the left if more A and B are present.
Consider a case in which the coefficients in the equation for a reaction are all 1. When
substances A and B react, the rate of the forward reaction decreases as time passes because
the concentration of A and B decrease.
A ϩ B 88n C ϩ D (1)
As the concentrations of C and D build up, they start to re-form A and B.
C ϩ D 88n A ϩ B (2)
As more C and D molecules are formed, more can react, and so the rate of reaction
between C and D increases with time. Eventually, the two reactions occur at the same
rate, and the system is at equilibrium (Figure 17-1).
A ϩ B 34 C ϩ D
If a reaction begins with only C and D present, the rate of reaction (2) decreases with
time, and the rate of reaction (1) increases with time until the two rates are equal.
The dynamic nature of chemical
equilibrium can be proved
experimentally by inserting radioactive
atoms into a small percentage of
molecules and following them through
the reaction. Even when the initial
mixture is at equilibrium, radioactive
atoms eventually appear in both
reactant and product molecules.
17-1 Basic Concepts 709
[C] and [D]
[A] and [B]
te
Concentrations
Time
Equilibrium is
established
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 16.3, The Equilibrium State.
Figure 17-1 Variation in the
concentrations of species present in
the A ϩ B 34 C ϩ D system as
equilibrium is approached, beginning
with equal concentrations of A and
B only. For this reaction, production
of products is favored. As we will
see, this corresponds to a value of
the equilibrium constant greater
than 1. Brackets, [ ], represent the
concentration, in moles per liter, of
the species enclosed within them.
The time axis may be in any
convenient units—seconds, minutes,
hours, and so on.
The SO2–O2–SO3 System
Consider the reversible reaction of sulfur dioxide with oxygen to form sulfur trioxide at
1500 K.
2SO2(g) ϩ O2(g) 34 2SO3(g)
Suppose 0.400 mole of SO2 and 0.200 mole of O2 are injected into a closed 1.00-liter
container. When equilibrium is established (at time te, Figure 17-2a), we find that 0.056
mole of SO3 has formed and that 0.344 mole of SO2 and 0.172 mole of O2 remain unreacted.
The reaction does not go to completion. These changes are summarized in the
following reaction summary, using molarity units rather than moles. (They are numerically
identical here because the volume of the reaction vessel is 1.00 liter.) The net reaction
is represented by the changes in concentrations.
2SO2(g) ϩ O2(g) 34 2SO3(g)
initial conc’n 0.400 M Ϫ0.200 M Ϫ0
change due to rxn Ϫ0.056 M Ϫ0.028 M ϩ0.056 M
equilibrium conc’n 0.344 M Ϫ0.172 M Ϫ0.056 M
In another experiment, 0.500 mole of SO3 is introduced alone into a closed 1.00-liter
container. When equilibrium is established (at time te, Figure 17-2b), 0.076 mole of SO3,
0.212 mole of O2, and 0.424 mole of SO2 are present. These equilibrium amounts differ
from those in the previous case, but they are related in an important way, as we will see
in the next section. This time the reaction proceeds from right to left as the equation is
written. The changes in concentration are in the same 2Ϻ1Ϻ2 ratio as in the previous case,
as required by the coefficients of the balanced equation. The time required to reach equilibrium
may be longer or shorter.
2SO2(g) ϩ O2(g) 34 2SO3(g)
initial conc’n 0 0 0.500 M
change due to rxn ϩ0.424 M ϩ0.212 M Ϫ0.424 M
equilibrium conc’n 0.424 M 0.212 M 0.076 M
The numbers in this discussion were
determined experimentally.
A setup such as this is called a
“reaction summary.” The ratio in the
“change due to rxn” line is determined
by the coefficients in the balanced
equation.
710 CHAPTER 17: Chemical Equilibrium
te
Concentrations
Time
(a) Starting with SO2 and O2.
0.400 M
0.100 M
te
0.056 M
0.172 M
0.344 M
[SO2]
[O2]
[SO3]
Time
Concentrations
0.400 M
[SO2]
0.424 M
0.212 M
0.076 M
0.100 M
[O2]
[SO3]
te
(b) Starting with SO3.
0.500 M
Figure 17-2 Establishment of equilibrium in the 2SO2 ϩ O2 34 2SO3 system.
(a) Beginning with stoichiometric amounts of SO2 and O2 and no SO3. (b) Beginning
with only SO3 and no SO2 or O2. Greater changes in concentrations occur to establish
equilibrium when starting with SO3 than when starting with SO2 and O2. The equilibrium
favors SO2 and O2.
The results of these experiments are summarized in the following table and in Figure
17-2.
Initial Concentrations Equilibrium Concentrations
[SO2] [O2] [SO3] [SO2] [O2] [SO3]
Experiment 1 0.400 M 0.200 M 0 M 0.344 M 0.172 M 0.056 M
Experiment 2 0 M 0 M 0.500 M 0.424 M 0.212 M 0.076 M
THE EQUILIBRIUM CONSTANT
Suppose a reversible reaction occurs by a one-step mechanism.
2A ϩ B 34 A2B
The rate of the forward reaction is Ratef ϭ kf[A]2[B]; the rate of the reverse reaction is
Rater ϭ kr[A2B]. In these expressions, kf and kr are the specific rate constants of the forward
and reverse reactions, respectively. By definition, the two rates are equal at equilibrium
(Ratef ϭ Rater). So we write
kf[A]2[B] ϭ kr[A2B] (at equilibrium)
Dividing both sides of this equation by kr and by [A]2[B] gives
ϭ
At any specific temperature, both kf and kr are constants, so kf/kr is also a constant.
This ratio is given a special name and symbol, the equilibrium constant, Kc or simply K.
Kc ϭ (at equilibrium)
Even if the overall reaction occurs by a multistep mechanism, we can show that the
equilibrium constant is the product and ratio of the rate constants for each step of the
mechanism. Regardless of the mechanism by which this reaction occurs, the concentrations
of reaction intermediates cancel out and the equilibrium constant expression has the
same form. For a reaction in general terms, the equilibrium constant can always be written
as follows:
The equilibrium constant Kc, is defined as the product of the equilibrium concentrations
(moles per liter) of the products, each raised to the power that corresponds to
its coefficient in the balanced equation, divided by the product of the equilibrium
concentrations of reactants, each raised to the power that corresponds to its coefficient
in the balanced equation.
For aA ϩ bB cC ϩ dD,
reactants
reactants
products
products
Kc ϭ
[C]eq
c[D]eq
d
[A]eq
a[B]eq
b
[A2B]
ᎏ
[A]2[B]
[A2B]
ᎏ
[A]2[B]
kf
ᎏ
kr
17-2
The subscript c refers to
concentrations. The brackets, [ ], in
this expression indicate equilibrium
concentrations in moles per liter.
17-2 The Equilibrium Constant 711
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 16.4, The Equilibrium Constant.
In general, numerical values for Kc can come only from experiments. Some equilibrium
constant expressions and their numerical values at 25°C are
N2(g) ϩ O2(g) 34 2NO(g) Kc ϭ ϭ 4.5 ϫ 10Ϫ31
CH4(g) ϩ Cl2(g) 34 CH3Cl(g) ϩ HCl(g) Kc ϭ ϭ 1.2 ϫ 1018
N2(g) ϩ 3H2(g) 34 2NH3(g) Kc ϭ ϭ 3.6 ϫ 108
We have used the subscript “eq” to emphasize that the concentrations in the equilibrium
constant expression are those at equilibrium. For the remainder of this text, we shall omit
these subscripts, remembering that calculations with Kc values always involve equilibrium
values of concentrations.
The thermodynamic definition of the equilibrium constant involves activities rather
than concentrations. The activity of a component of an ideal mixture is the ratio of its
concentration or partial pressure to a standard concentration (1 M) or pressure (1 atm).
For now, we can consider the activity of each species to be a dimensionless quantity whose
numerical value can be determined as follows.
1. For any pure liquid or pure solid, the activity is taken as 1.
2. For components of ideal solutions, the activity of each component is taken to be
the ratio of its molar concentration to a standard concentration of 1 M, so the units
cancel.
3. For gases in an ideal mixture, the activity of each component is taken to be the ratio
of its partial pressure to a standard pressure of 1 atm, so again the units cancel.
Because of the use of activities, the equilibrium constant has no units; the values we put into
Kc are numerically equal to molar concentrations, but are dimensionless, that is, they have no
units. In this text, calculations have usually included units along with numbers. Calculations
involving equilibrium are frequently carried out without units; we will follow that
practice in this text.
The magnitude of Kc is a measure of the extent to which reaction occurs. For any
balanced chemical equation, the value of Kc
1. is constant at a given temperature;
2. changes if the temperature changes;
3. does not depend on the initial concentrations.
A value of Kc much greater than 1 indicates that the “numerator concentrations” (products)
would be much greater than the “denominator concentrations” (reactants); this means
that at equilibrium most of the reactants would be converted into products. For example,
the reaction CH4(g) ϩ Cl2(g) 34 CH3Cl(g) ϩ HCl(g) shown earlier goes nearly to completion;
in Chapter 15, we called such a reaction “product-favored.” On the other hand, if
Kc is quite small, equilibrium is established when most of the reactants remain unreacted
and only small amounts of products are formed. The reaction N2(g) ϩ O2(g) 34 2NO(g)
shown earlier reaches equilibrium with only a tiny amount of NO present; in Chapter 15,
we called such a reaction “reactant-favored.”
[NH3]eq
2
ᎏᎏ
[N2]eq[H2]eq
3
[CH3Cl]eq[HCl]eq
ᎏᎏ
[CH4]eq[Cl2]eq
[NO]eq
2
ᎏᎏ
[N2]eq[O2]eq
These are three very important ideas.
712 CHAPTER 17: Chemical Equilibrium
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 16.6, Writing Equilibrium
Expressions.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 16.5, The Meaning of the
Equilibrium Constant.
For a given chemical reaction at a specific temperature, the product of the concentrations
of the products formed by the reaction, each raised to the appropriate power, divided
by the product of the concentrations of the reactants, each raised to the appropriate power,
always has the same value, that is, Kc. This does not mean that the individual equilibrium
concentrations for a given reaction are always the same, but it does mean that this particular
numerical combination of their values (Kc) is constant.
Consider again the SO2–O2–SO3 equilibrium described in Section 17-1. We can use
the equilibrium concentrations from either experiment to calculate the value of the equilibrium
constant for this reaction at 1500 K.
From experiment 1: 2SO2(g) ϩ O2(g) 34 2SO3(g)
equilibrium conc’n 0.344 M 0.172 M 0.056 M
Substituting the numerical values (without units) into the equilibrium expression gives the
value of the equilibrium constant.
Alternatively, Kc ϭ ϭ ϭ 0.15
From experiment 2: 2SO2(g) ϩ O2(g) 34 2SO3(g)
equilibrium conc’n 0.424 M 0.212 M 0.076 M
Kc ϭ ϭ 0.15
No matter what combinations of reactant and product concentrations we start with, the
resulting equilibrium concentrations at 1500 K for this reversible reaction would always
give the same value of Kc, 0.15.
For the reversible reaction written as it is with SO2 and O2 as reactants and SO3 as the
product, Kc is 0.15 at 1500 K.
EXAMPLE 17-1 Calculation of Kc
Some nitrogen and hydrogen are placed in an empty 5.00-liter container at 500°C. When equilibrium
is established, 3.01 mol of N2, 2.10 mol of H2, and 0.565 mol of NH3 are present.
Evaluate Kc for the following reaction at 500°C.
N2(g) ϩ 3H2(g) 34 2NH3(g)
Plan
The equilibrium concentrations are obtained by dividing the number of moles of each reactant
and product by the volume, 5.00 liters. Then we substitute these equilibrium concentrations
into the equilibrium constant expression.
Solution
The equilibrium concentrations are
[N2] ϭ 3.01 mol/5.00 L ϭ 0.602 M
[H2] ϭ 2.10 mol/5.00 L ϭ 0.420 M
[NH3] ϭ 0.565 mol/5.00 L ϭ 0.113 M
We substitute these numerical values into the expression for Kc.
Kc ϭ ϭ ϭ 0.286
(0.113)2
ᎏᎏ
(0.602)(0.420)3
[NH3]2
ᎏᎏ
[N2][H2]3
(0.076)2
ᎏᎏ
(0.424)2(0.212)
(0.056)2
ᎏᎏ
(0.344)2(0.172)
[SO3]2
ᎏᎏ
[SO2]2[O2]
Remember that the concentrations in
Kc calculations are equilibrium values of
molar concentration.
17-2 The Equilibrium Constant 713
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 16.8, Determining an
Equilibrium Constant.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 16.9, Systems at Equilibrium.
Thus, for the reaction of N2 and H2 to form NH3 at 500°C, we can write
Kc ϭ ϭ 0.286
The small value of Kc indicates that the equilibrium lies to the left.
The value of Kc from Example 17-1 is much different from the value given earlier for
the same reaction at 25°C. For this reaction, products are favored at the lower temperature
(Kc ϭ 3.6 ϫ 108 at 25°C), whereas reactants are favored at the higher temperature
(Kc ϭ 0.286 at 500°C). The dependence of Kc on temperature will be discussed later in
this chapter; for now, we note that it can depend strongly on temperature.
EXAMPLE 17-2 Calculation of Kc
We put 10.0 moles of N2O into a 2.00-L container at some temperature, where it decomposes
according to
2N2O(g) 34 2N2(g) ϩ O2(g)
At equilibrium, 2.20 moles of N2O remain. Calculate the value of Kc for the reaction.
Plan
We express all concentrations in moles per liter. The mole ratio from the balanced chemical
equation allows us to find the changes in concentrations of the other substances in the reaction.
We use the reaction summary to find the equilibrium concentrations to use in the Kc
expression.
Solution
At equilibrium, 2.20 mol N2O remain, so
__? mol N2O reacting ϭ 10.00 mol N2O initial Ϫ 2.20 mol N2O remaining
ϭ 7.80 mol N2O reacting
The initial [N2O] ϭ (10.00 mol)/(2.00 L) ϭ 5.00 M; the concentration of N2O that reacts is
(7.80 mol)/(2.00 L) ϭ 3.90 M. From the balanced chemical equation, each 2 mol N2O that
react produces 2 mol N2 and 1 mol O2, or a reaction ratio of
1 mol N2O reactingS1 mol N2 formedSᎏ1
2
ᎏ mol O2 formed
We can now write the reaction summary.
2N2O(g) 34 2N2(g) ϩ O2(g)
initial 5.00 M 0 0
change due to rxn Ϫ3.90 M ϩ3.90 M ϩᎏ1
2
ᎏ(3.90) ϭ 1.95 M
equilibrium 1.10 M 3.90 M 1.95 M
We put these equilibrium concentrations into the equilibrium constant expression and evaluate
Kc.
Kc ϭ ϭ ϭ 24.5
You should now work Exercises 24 and 28.
(3.90)2(1.95)
ᎏᎏ
(1.10)2
[N2]2[O2]
ᎏᎏ
[N2O]2
[NH3]2
ᎏᎏ
[N2][H2]3
714 CHAPTER 17: Chemical Equilibrium
Reversing an equation is the same as
multiplying all coefficients by Ϫ1. This
reverses the roles of “reactants” and
“products.”
We see that KЉc is the square root of Kc.
Kc
1/2 means the square root of Kc.
17-3 Variation of Kc with the Form of the Balanced Equation 715
VARIATION OF Kc WITH THE FORM OF THE BALANCED
EQUATION
The value of Kc depends on the form of the balanced equation for the reaction. We wrote
the equation for the reaction of SO2 and O2 to form SO3, and its equilibrium constant
expression as
2SO2(g) ϩ O2(g) 34 2SO3(g) and Kc ϭ ϭ 0.15
Suppose we write the equation for the same reaction in reverse. The equation and its equilibrium
constant, written this way, are
2SO3(g) 34 2SO2(g) ϩ O2(g) and KЈc ϭ ϭ ᎏ
K
1
c
ᎏ ϭ ᎏ ϭ 6.7
We see that KЈc, the equilibrium constant for the reaction written in reverse, is the reciprocal
of Kc, the equilibrium constant for the original reaction.
If the equation for the reaction were written as
SO2(g) ϩ ᎏ1
2
ᎏO2(g) 34 SO3(g) KЉc ϭ ϭ Kc
1/2 ϭ 0.39
If an equation for a reaction is multiplied by any factor, n, then the original value of Kc
is raised to the nth power. Thus, we must always write the balanced chemical equation to
accompany the value of Kc for a chemical reaction.
EXAMPLE 17-3 Variation of the Form of Kc
You are given the following reaction and its equilibrium constant at a given temperature.
2HBr(g) ϩ Cl2(g) 34 2HCl(g) ϩ Br2(g) Kc ϭ 4.0 ϫ 104
Write the expression for, and calculate the numerical value of, the equilibrium constant for
each of the following at the same temperature.
(a) 4HBr(g) ϩ 2Cl2(g) 34 4HCl(g) ϩ 2Br2(g)
(b) HBr(g) ϩ ᎏ1
2
ᎏCl2(g) 34 HCl(g) ϩ ᎏ1
2
ᎏBr2(g)
Plan
We recall the definition of the equilibrium constant. For the original equation,
Kc ϭ ϭ 4.0 ϫ 104
Solution
(a) The original equation has been multiplied by 2, so Kc must be squared.
KЈc ϭ KЈc ϭ (Kc)2 ϭ (4.0 ϫ 104)2 ϭ 1.6 ϫ 109
(b) The original equation has been multiplied by ᎏ1
2
ᎏ (divided by 2), so Kc must be raised to the
ᎏ1
2
ᎏ power. The value of KЉc is the square root of the original Kc value.
[HCl]4[Br2]2
ᎏᎏ
[HBr]4[Cl2]2
[HCl]2[Br2]
ᎏᎏ
[HBr]2[Cl2]
[SO3]
ᎏᎏ
[SO2][O2]1/2
1
ᎏ
0.15
[SO2]2[O2]
ᎏᎏ
[SO3]2
[SO3]2
ᎏᎏ
[SO2]2[O2]
17-3
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 16.7, Manipulating Equilibrium
Expressions.
A coefficient of ᎏ1
2
ᎏ refers to ᎏ1
2
ᎏ of a mole,
not ᎏ1
2
ᎏ of a molecule.
KЉc ϭ ϭ ͙Kෆcෆ ϭ ͙4ෆ.0ෆ ϫෆ 1ෆ0ෆ4ෆ ϭ 2.0 ϫ 102
You should now work Exercise 30.
THE REACTION QUOTIENT
The reaction quotient, Q, for the general reaction is given as follows.
not necessarily
For aA ϩ bB 34 cC ϩ dD, Q ϭ ᎏ
[
[
C
A
]
]
c
a
[
[
D
B]
]
b
d
ᎏ equilibrium
concentrations
The reaction quotient has the same form as the equilibrium constant, but it involves specific
values that are not necessarily equilibrium concentrations. If they are equilibrium concentrations,
then Q ϭ Kc. The concept of the reaction quotient is very useful. We can compare
the magnitude of Q with that of Kc for a reaction under given conditions to decide whether
the forward or the reverse reaction must occur to a greater extent to establish equilib-
rium.
We can think of the reaction quotient, Q, as a measure of the progress of the reaction.
When the mixture contains only reactants, the concentrations in the numerator are zero,
so Q ϭ 0. As the reaction proceeds to the right, the product concentrations (numerator)
increase and the reactant concentrations (denominator) decrease, so Q increases to an infinitely
large value when all reactants have been consumed and only products remain.
The value of Kc is a particular value of Q that represents equilibrium mixtures for the
reaction.
If at any time Q Ͻ Kc, the forward reaction must occur to a greater extent than the
reverse reaction for equilibrium to be established. This is because when Q Ͻ Kc, the
numerator of Q is too small and the denominator is too large. To increase the numerator
and to reduce the denominator, A and B must react to produce C and D. Conversely, if
Q Ͼ Kc, the reverse reaction must occur to a greater extent than the forward reaction for
equilibrium to be reached. When the value of Q reaches the value of Kc, the system is at
equilibrium, so no further net reaction occurs.
17-4
[HCl][Br2]1/2
ᎏᎏ
[HBr][Cl2]1/2
The reaction quotient is sometimes
called the mass action expression.
When the forward reaction occurs
to a greater extent than the reverse
reaction, we say that a net forward
reaction has occurred.
716 CHAPTER 17: Chemical Equilibrium
m886
66m88
Reaction proceeds toward right,
forming more products
Reaction proceeds toward left,
forming more reactants
Q = 0
(reactants only)
Increasing Q
Kc
Q = ∞
(products only)
Q Ͻ Kc Forward reaction predominates until equilibrium is established.
Q ϭ Kc System is at equilibrium.
Q Ͼ Kc Reverse reaction predominates until equilibrium is established.
In Example 17-4 we calculate the value for Q and compare it with the known value of
Kc to predict the direction of the reaction that leads to equilibrium.
EXAMPLE 17-4 The Reaction Quotient
At a very high temperature, Kc ϭ 65.0 for the following reaction.
2HI(g) 34 H2(g) ϩ I2(g)
The following concentrations were detected in a mixture. Is the system at equilibrium? If not,
in which direction must the reaction proceed for equilibrium to be established?
[HI] ϭ 0.500 M, [H2] ϭ 2.80 M, and [I2] ϭ 3.40 M
Plan
We substitute these concentrations into the expression for the reaction quotient to calculate
Q. Then we compare Q with Kc to see whether the system is at equilibrium.
Solution
Q ϭ ϭ ϭ 38.1
But Kc ϭ 65.0, so Q Ͻ Kc. The system is not at equilibrium. For equilibrium to be established,
the value of Q must increase until it equals Kc. This can occur only if the numerator increases
and the denominator decreases. Thus, the forward (left-to-right) reaction must occur to a greater
extent than the reverse reaction; that is, some HI must react to form more H2 and I2 to reach
equilibrium.
You should now work Exercises 38, 40, and 42.
USES OF THE EQUILIBRIUM CONSTANT, Kc
We have seen (Section 17-2) how to calculate the value of Kc from one set of equilibrium
concentrations. Once that value has been obtained, the process can be turned around to
calculate equilibrium concentrations from the equilibrium constant.
EXAMPLE 17-5 Finding Equilibrium Concentrations
The equation for the following reaction and the value of Kc at a given temperature are given.
An equilibrium mixture in a 1.00-liter container contains 0.25 mol of PCl5 and 0.16 mol of
PCl3. What equilibrium concentration of Cl2 must be present?
PCl3(g) ϩ Cl2(g) 34 PCl5(g) Kc ϭ 1.9
Plan
We write the equilibrium constant expression and its value. Only one term, [Cl2], is unknown.
We solve for it.
Solution
Because the volume of the container is 1.00 liter, the molar concentration (mol/L) of each
substance is numerically equal to the number of moles. The equilibrium constant expression
and its numeric value are
17-5
(2.80)(3.40)
ᎏᎏ
(0.500)2
[H2][I2]
ᎏ
[HI]2
These concentrations could be present
if we started with a mixture of HI, H2,
and I2.
17-5 Uses of the Equilibrium Constant, Kc 717
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 16.10, Estimating Equilibrium
Concentrations.
The equilibrium constant is a
“constant” only if the temperature
does not change.
Reaction proceeds to the right
Increasing Q
Q = 38.1 K = 65.0
Kc ϭ ϭ 1.9
[Cl2] ϭ ϭ ϭ 0.82 M
Often we know the starting concentrations and want to know how much of each reactant
and each product would be present at equilibrium. The next two examples illustrate
this important kind of calculation.
EXAMPLE 17-6 Finding Equilibrium Concentrations
For the following reaction, the equilibrium constant is 49.0 at a certain temperature. If 0.400
mol each of A and B are placed in a 2.00-liter container at that temperature, what concentrations
of all species are present at equilibrium?
A ϩ B 34 C ϩ D
Plan
First we find the initial concentrations. Then we write the reaction summary and represent
the equilibrium concentrations algebraically. Finally we substitute the algebraic representations
of equilibrium concentrations into the Kc expression and find the equilibrium
concentrations.
Solution
The initial concentrations are
[A] ϭ ᎏ
0.
2
4
.
0
0
0
0
m
L
ol
ᎏ ϭ 0.200 M [C] ϭ 0 M
[B] ϭ ᎏ
0.
2
4
.
0
0
0
0
m
L
ol
ᎏ ϭ 0.200 M [D] ϭ 0 M
We know that the reaction can only proceed to the right because only “reactants” are present.
The reaction summary includes the values, or symbols for the values, of (1) initial concentrations,
(2) changes in concentrations, and (3) concentrations at equilibrium.
Let x ϭ moles per liter of A that react; then x ϭ moles per liter of B that react and x ϭ
moles per liter of C and D that are formed.
A ϩ B 34 C ϩ D
initial 0.200 M 0.200 M 0 M 0 M
change due to rxn Ϫx M Ϫx M ϩx M ϩx M
at equilibrium (0.200 Ϫ x)M (0.200 Ϫ x)M x M x M
Now Kc is known, but concentrations are not. But the equilibrium concentrations have all been
expressed in terms of the single variable x. We substitute the equilibrium concentrations (not
the initial ones) into the Kc expression and solve for x.
Kc ϭ ϭ 49.0
ϭ ϭ 49.0
This quadratic equation has a perfect square on both sides. We solve it by taking the square
roots of both sides of the equation and then rearranging for x.
x2
ᎏᎏ
(0.200 Ϫ x)2
(x)(x)
ᎏᎏᎏ
(0.200 Ϫ x)(0.200 Ϫ x)
[C][D]
ᎏ
[A][B]
(0.25)
ᎏᎏ
(1.9)(0.16)
[PCl5]
ᎏ
Kc[PCl3]
[PCl5]
ᎏᎏ
[PCl3][Cl2]
We have represented chemical
formulas by single letters to simplify
the notation in these calculations.
The coefficients in the equation are
all 1’s, so the reaction ratio must be
1Ϻ1Ϻ1Ϻ1.
718 CHAPTER 17: Chemical Equilibrium
ᎏ
0.200
x
Ϫ x
ᎏ ϭ 7.00
x ϭ 1.40 Ϫ 7.00x 8.00x ϭ 1.40 x ϭ ᎏ
1
8
.
.
4
0
0
0
ᎏ ϭ 0.175
Now we know the value of x, so the equilibrium concentrations are
[A] ϭ (0.200 Ϫ x) M ϭ 0.025 M ; [C] ϭ x M ϭ 0.175 M
[B] ϭ (0.200 Ϫ x) M ϭ 0.025 M ; [D] ϭ x M ϭ 0.175 M
To check our answers we use the equilibrium concentrations to calculate Q and verify that its
value is equal to Kc.
Q ϭ ϭ ϭ 49 Recall that Kc ϭ 49.0
The ideas developed in Example 17-6 may be applied to cases in which the reactants
are mixed in nonstoichiometric amounts. This is shown in Example 17-7.
EXAMPLE 17-7 Finding Equilibrium Concentrations
Consider the same system as in Example 17-6 at the same temperature. If 0.600 mol of A and
0.200 mol of B are mixed in a 2.00-liter container and allowed to reach equilibrium, what are
the equilibrium concentrations of all species?
Plan
We proceed as we did in Example 17-6. The only difference is that now we have nonstoichiometric
amounts of reactants.
Solution
As in Example 17-6, we let x ϭ mol/L of A that react; then x ϭ mol/L of B that react, and x ϭ
mol/L of C and D formed.
A ϩ B 34 C ϩ D
initial 0.300 M 0.100 M 0 M 0 M
change due to rxn Ϫx M Ϫx M ϩx M ϩx M
equilibrium (0.300 Ϫ x) M (0.100 Ϫ x) M x M x M
The initial concentrations are governed by the amounts of reactants mixed together. But changes
in concentrations due to reaction must occur in the 1Ϻ1Ϻ1Ϻ1 ratio required by the coefficients
in the balanced equation.
Kc ϭ ϭ 49.0 so ϭ 49.0
We can arrange this quadratic equation into the standard form.
ϭ 49.0
x2 ϭ 1.47 Ϫ 19.6x ϩ 49.0x2
48.0x2 Ϫ 19.6x ϩ 1.47 ϭ 0
Quadratic equations can be solved by use of the quadratic formula.
x2
ᎏᎏᎏ
0.0300 Ϫ 0.400x ϩ x2
(x)(x)
ᎏᎏᎏ
(0.300 Ϫ x)(0.100 Ϫ x)
[C][D]
ᎏ
[A][B]
(0.175)(0.175)
ᎏᎏ
(0.025)(0.025)
[C][D]
ᎏ
[A][B]
We see that the equilibrium
concentrations of products are
much greater than those of reactants
because Kc is much greater than 1.
17-5 Uses of the Equilibrium Constant, Kc 719
The left side of this equation is not a
perfect square.
x ϭ
In this case a ϭ 48.0, b ϭ Ϫ19.6, and c ϭ 1.47. Substituting these values gives
x ϭ
ϭ ᎏ
19.6
9
Ϯ
6.0
10.1
ᎏ ϭ 0.309 or 0.099
Solving a quadratic equation always yields two roots. One root (the answer) has physical
meaning. The other root, while mathematically correct, is extraneous; that is, it has no physical
meaning. The value of x is defined as the number of moles of A per liter that react and
the number of moles of B per liter that react. No more B can be consumed than was initially
present (0.100 M), so x ϭ 0.309 is the extraneous root. Thus, x ϭ 0.099 is the root that has
physical meaning, and the extraneous root is 0.309. The equilibrium concentrations are
[A] ϭ (0.300 Ϫ x) M ϭ 0.201 M ; [B] ϭ (0.100 Ϫ x) M ϭ 0.001 M ;
[C] ϭ [D] ϭ x M ϭ 0.099 M
You should now work Exercises 34, 44, and 46.
Ϫ(Ϫ19.6) Ϯ ͙(Ϫෆ1ෆ9ෆ.6ෆ)2ෆ Ϫෆ 4ෆ(4ෆ8ෆ.0ෆ)(ෆ1ෆ.4ෆ7ෆ)ෆ
ᎏᎏᎏᎏᎏ
2(48.0)
Ϫb Ϯ ͙b2ෆ Ϫෆ 4ෆacෆ
ᎏᎏ
2a
Check Example 17-6:
Q ϭ ϭ
Q ϭ 49 ϭ Kc
Check Example 17-7
Q ϭ
Q ϭ 49 ϭ Kc
(0.099)(0.099)
ᎏᎏ
(0.201)(0.001)
(0.175)(0.175)
ᎏᎏ
(0.025)(0.025)
[C][D]
ᎏ
[A][B]
720 CHAPTER 17: Chemical Equilibrium
The following table summarizes Examples 17-6 and 17-7.
Initial Concentrations (M) Equilibrium Concentrations (M)
[A] [B] [C] [D] [A] [B] [C] [D]
Example 17-6 0.200 0.200 0 0 0.025 0.025 0.175 0.175
Example 17-7 0.300 0.100 0 0 0.201 0.001 0.099 0.099
Problem-Solving Tip: Solving Quadratic Equations—Which Root
Shall We Use?
Quadratic equations can be rearranged into standard form.
ax2 ϩ bx ϩ c ϭ 0
All can be solved by the quadratic formula, which is
x ϭ (Appendix A)
This formula gives two roots, both of which are mathematically correct. A foolproof way
to determine which root of the equation has physical meaning is to substitute the value
of the variable into the expressions for the equilibrium concentrations. For the extraneous
root, one or more of these substitutions will lead to a negative concentration,
which is physically impossible (there cannot be less than none of a substance present!).
The correct root will give all positive concentrations. In Example 17-7, substitution of
the extraneous root x ϭ 0.309 would give [A] ϭ (0.300 Ϫ 0.309) M ϭ Ϫ0.009 M and
[B] ϭ (0.100 Ϫ 0.309) M ϭ Ϫ0.209 M. Either of these concentration values is impossible,
so we would know that 0.309 is an extraneous root. You should apply this check
to subsequent calculations that involve solving a quadratic equation.
Ϫb Ϯ ͙b2ෆ Ϫෆ 4ෆacෆ
ᎏᎏ
2a
The data from the table can be substituted into the reaction quotient expression, Q, as
a check. Even though the reaction is initiated by different relative amounts of reactants
in the two cases, the ratios of equilibrium concentrations of products to reactants (each
raised to the first power) agree within roundoff error.
FACTORS THAT AFFECT EQUILIBRIA
Once a reacting system has reached equilibrium, it remains at equilibrium until it is
disturbed by some change of conditions. The guiding principle is known as LeChatelier’s
Principle (Section 13-6).
If a change of conditions (stress) is applied to a system at equilibrium, the system
shifts in the direction that reduces the stress to move toward a new state of equi-
librium.
The reaction quotient, Q, helps us predict the direction of this shift. Three types of changes
can disturb the equilibrium of a reaction.
1. Changes in concentration
2. Changes in pressure or volume (for reactions that involve gases)
3. Changes in temperature
We now study the effects of these types of stresses from a qualitative, or descriptive,
point of view. In Section 17-8 we expand our discussion with quantitative examples.
Changes in Concentration
Consider the following system starting at equilibrium.
A ϩ B 34 C ϩ D Kc ϭ ᎏ
[
[
C
A
]
]
[
[
D
B]
]
ᎏ
When more of any reactant or product is added to the system, the value of Q changes, so
it no longer matches Kc, and the reaction is no longer at equilibrium. The stress due to
the added substance is relieved by shifting the equilibrium in the direction that consumes
some of the added substance, moving the value of Q back toward Kc. Let us compare the
mass action expressions for Q and Kc. If more A or B is added, then Q Ͻ Kc, and the
forward reaction occurs more rapidly and to a greater extent than the reverse reaction
until equilibrium is reestablished. If more C or D is added, Q Ͼ Kc, and the reverse reaction
occurs more rapidly and to a greater extent until equilibrium is reestablished.
Adding or removing reactants or products changes the value of Q. It does not change
the value of Kc.
We can understand LeChatelier’s Principle in the kinetic terms we used to introduce
equilibrium. The rate of the forward reaction is proportional to the reactant concentrations
raised to some powers,
Ratef ϭ kf[A]x[B]y
17-6
LaChatelier is pronounced “le-SHOT-
lee-ay.”
Remember that the value of an
equilibrium constant changes only
with temperature.
For reactions involving gases at
constant temperature, changes in
volume cause changes in pressure,
and vice versa.
17-6 Factors that Affect Equilibria 721
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 16.11, Disturbing a Chemical
Equilibrium (1): LeChatelier’s Principle.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 16.13, Disturbing a Chemical
Equilibrium (3): Addition or Removal of
a Reagent.
When we add more A to an equilibrium mixture, this rate increases so that it no longer
matches the rate of the reverse reaction. As the reaction proceeds to the right, consuming
some A and B and forming more C and D, the forward rate diminishes and the reverse
rate increases until they are again equal. At that point, a new equilibrium condition has
been reached, with more C and D than were present in the original equilibrium mixture.
Not all of the added A has been consumed when the new equilibrium is reached, however.
If a reactant or product is removed from a system at equilibrium, the reaction that
produces that substance occurs more rapidly and to a greater extent than its reverse. If
some C or D is removed, then Q Ͻ Kc, and the forward reaction is favored until equilibrium
is reestablished. If some A or B is removed, the reverse reaction is favored.
Direction of Shift of
Stress Q A ϩ B 34 C ϩ D
Increase concentration of A or B Q Ͻ K left 88n right
Increase concentration of C or D Q Ͼ K left right
Decrease concentration of A or B Q Ͼ K left right
Decrease concentration of C or D Q Ͻ K left 88n right
88n
88n
The terminology used here is not
as precise as we might like, but it is
widely used. When we say that the
equilibrium is “shifted to the left,”
we mean that the reaction to the left
occurs to a greater extent than the
reaction to the right.
This tabulation summarizes a lot of
useful information. Study it carefully.
722 CHAPTER 17: Chemical Equilibrium
Effects of changes in concentration on the equilibrium
[Co(OH2)6]2ϩ ϩ 4ClϪ 34 [CoCl4]2Ϫ ϩ 6H2O
A solution of CoCl2T6H2O in isopropyl alcohol and water is purple (not shown) due to
the mixture of [Co(OH2)6]2ϩ (pink) and [CoCl4]2Ϫ (blue). When we add concentrated HCl,
the excess ClϪ shifts the reaction to the right (blue, right). Adding AgNO3(aq) removes
some ClϪ by precipitation of AgCl(s) and favors the reaction to the left (produces more
[Co(OH2)6]2ϩ; the resulting solution is pink (left). Each insert shows the structure of the
cobalt complex species present in higher concentration; other ions and solvent molecules
are not shown.
When a new equilibrium condition is established, (1) the rates of the forward and
reverse reactions are again equal; and (2) Kc is again satisfied by the new concentrations
of reactants and products.
Practical applications of changes of this type are of great economic importance. Removing
a product of a reversible reaction forces the reaction to produce more product than could
be obtained if the reaction were simply allowed to reach equilibrium.
EXAMPLE 17-8 Applying a Stress to a System at Equilibrium
Given the following reaction at equilibrium in a closed container at 500°C, predict the effect
of each of the following changes on the amount of NH3 present at equilibrium: (a) forcing
more H2 into the system; (b) removing some NH3 from the system.
N2(g) ϩ 3H2(g) 34 2NH3(g)
Plan
We apply LeChatelier’s Principle to each part of the question individually.
Solution
(a) Adding a substance favors the reaction that uses up that substance (forward in this case).
More NH3 is formed.
(b) Removing a substance favors the reaction that produces that substance (forward in this case).
More NH3 is formed.
Note that at the new equilibrium, the concentration [NH3] would still be less than at the original
equilibrium.
Changes in Volume and Pressure
Changes in pressure have little effect on the concentrations of solids or liquids because
they are only slightly compressible. Changes in pressure do cause significant changes in
concentrations of gases, however. Such changes therefore affect the value of Q for reactions
in which the number of moles of gaseous reactants differs from the number of moles
of gaseous products. For an ideal gas,
PV ϭ nRT or P ϭ ΂ᎏ
V
n
ᎏ΃(RT )
The term (n/V) represents concentration, that is, mol/L. At constant temperature, n, R,
and T are constants. Thus, if the volume occupied by a gas decreases, its partial pressure
increases and its concentration (n/V) increases. If the volume of a gas increases, both its
partial pressure and its concentration decrease.
Consider the following gaseous system at equilibrium.
A(g) 34 2D(g) Kc ϭ ᎏ
[
[
D
A
]
]
2
ᎏ
17-6 Factors that Affect Equilibria 723
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 16.14, Disturbing a Chemical
Equilibrium (4): Volume Changes.
Reaction proceeds to the right
Increasing Q
Q K
(a)
Reaction proceeds to the right
Increasing Q
Q K
(b)
At constant temperature, a decrease in volume (increase in pressure) increases the concentrations
of both A and D. In the expression for Q, the concentration of D is squared and
the concentration of A is raised to the first power. As a result, the numerator of Q increases
more than the denominator as pressure increases. Thus, Q Ͼ Kc, and this equilibrium
shifts to the left. Conversely, an increase in volume (decrease in pressure) shifts this reaction
to the right until equilibrium is reestablished, because Q Ͻ Kc. We can summarize
the effect of pressure (volume) changes on this gas-phase system at equilibrium.
Stress Q* Direction of Shift of A(g) 34 2D(g)
Volume decrease, Q Ͼ Kc Toward smaller number of moles of gas (left for
pressure increase this reaction)
Volume increase, Q Ͻ Kc Toward larger number of moles of gas (right for
pressure decrease this reaction)
*In Q for this reaction, there are more moles of gaseous product than gaseous reactant.
In general, for reactions that involve gaseous reactants or products, LeChatelier’s Principle
allows us to predict the following results.
1. If there is no change in the total number of moles of gases in a reaction, a volume
(pressure) change does not affect the position of equilibrium; Q is unchanged by
the volume (or pressure) change.
2. If a reaction involves a change in the total number of moles of gases, changing the
volume (or pressure) of a reaction mixture changes the value of Q; it does not
change the value of Kc. For such a reaction:
(a) A decrease in volume (increase in pressure) shifts a reaction in the direction
that produces the smaller total number of moles of gas, until Q again equals Kc.
(b) An increase in volume (decrease in pressure) shifts a reaction in the direction
that produces the larger total number of moles of gas, until Q again equals Kc.
The foregoing argument applies only when pressure changes are due to volume
changes. It does not apply if the total pressure of a gaseous system is raised by merely
pumping in an inert gas, for example, He. If the gas that is introduced is not involved in
the reaction, the partial pressure of each reacting gas remains constant, so the system
remains at equilibrium.
EXAMPLE 17-9 Applying a Stress to a System at Equilibrium
(a) Given the following reaction at equilibrium in a closed container at 500°C, predict the effect
of increasing the pressure by decreasing the volume.
N2(g) ϩ 3H2(g) 34 2NH3(g)
(b) Make the same prediction for the following reaction at high temperature.
H2(g) ϩ I2(g) 34 2HI(g)
Plan
We apply LeChatelier’s Principle to predict the effect on each reaction.
Study this tabulation carefully. How
would these conclusions change for
a reaction in which there are more
moles of gaseous reactants than moles
of gaseous products?
One practical application of these ideas
is illustrated in the next section by the
Haber process.
724 CHAPTER 17: Chemical Equilibrium
Solution
(a) Increasing the pressure favors the reaction that produces the smaller number of moles of
gas (forward in this case).
More NH3 is formed.
(b) This reaction involves the same number of moles of gas on both sides, so a pressure (volume)
change does not disturb the equilibrium. There is no effect on the position of equilibrium.
Changes in Temperature
Consider the following exothermic reaction at equilibrium:
A ϩ B 34 C ϩ D ϩ heat (⌬H is negative)
Heat is produced by the forward (exothermic) reaction. Suppose we increase the temperature
at constant pressure by adding heat to the system. This favors the reaction to the
left, removing some of the extra heat. Lowering the temperature favors the reaction to
the right as the system replaces some of the heat that was removed.
By contrast, for an endothermic reaction at equilibrium,
W ϩ X ϩ heat 34 Y ϩ Z (⌬H is positive)
17-6 Factors that Affect Equilibria 725
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 16.12, Disturbing a Chemical
Equilibrium (2): Temperature Changes.
The gas-phase equilibrium for the exothermic reaction
2NO2(g) 34 N2O4(g) ⌬H0
rxn ϭ Ϫ57.2 kJ/mol rxn
The two flasks contain the same total amounts of gas. NO2 is brown, whereas N2O4 is
colorless. The higher temperature (50°C) of the flask on the right favors the reverse
reaction; this mixture is more highly colored because it contains more NO2. The flask
on the left, at the temperature of ice water, contains less brown NO2 gas.
an increase in temperature at constant pressure favors the reaction to the right. A decrease
in temperature favors the reaction to the left.
The value of any equilibrium constant changes as the temperature changes. Changing the
temperature of a reaction at equilibrium thus causes Q to differ from Kc, but this is now
because Kc has changed. The reaction then proceeds in the direction that moves Q toward
the new value of Kc. As we will see in Section 17-13,
the Kc values of exothermic reactions decrease with increasing T, and the Kc values
of endothermic reactions increase with increasing T. No other stresses affect the
value of Kc.
Effect of temperature changes on
the equilibrium of the endothermic
reaction
[Co(OH2)6]2ϩ ϩ 4ClϪ ϩ heat 34
[CoCl4]2Ϫ ϩ 6H2O
We begin with a purple equilibrium
mixture of the pink and blue
complexes at room temperature (not
shown). In hot water the forward
reaction (endothermic) is favored
and Kc is higher, so the solution is
blue (right). At 0°C, the reverse
reaction (exothermic) is favored and
Kc is lower, so the solution is pink
(left). Each insert shows the structure
of the cobalt complex species present
in highest concentration; other ions
and solvent molecules are not
shown.
Reaction proceeds more to right at higher temperature
Kc
at lower temperature
Kc
at higher temperature
For an endothermic reaction
Reaction proceeds more to left at higher temperature
Kc
at higher temperature
Kc
at lower temperature
For an exothermic reaction
Addition of a Catalyst
Adding a catalyst to a system changes the rate of the reaction (Section 16-9), but this
cannot shift the equilibrium in favor of either products or reactants. Because a catalyst
affects the activation energy of both forward and reverse reactions equally, it changes both
rate constants by the same factor, so their ratio, Kc, does not change.
Adding a catalyst to a reaction at equilibrium has no effect; it changes neither Q nor
Kc.
The same equilibrium mixture is achieved with or without the catalyst, but the equilibrium
is established more quickly in the presence of a catalyst.
Not all reactions attain equilibrium; they may occur too slowly, or else products or
reactants may be continually added or removed. Such is the case with most reactions in
biological systems. On the other hand, some reactions, such as typical acid–base neutralizations,
achieve equilibrium very rapidly.
EXAMPLE 17-10 Applying a Stress to a System at Equilibrium
Given the following reaction at equilibrium in a closed container at 500°C, predict the effect
of each of the following changes on the amount of NH3 present at equilibrium: (a) increasing
the temperature; (b) lowering the temperature; (c) introducing some platinum catalyst.
N2(g) ϩ 3H2(g) 34 2NH3(g) ⌬H0 ϭ Ϫ92 kJ/mol rxn
Plan
We apply LeChatelier’s Principle to each part of the question individually.
Solution
(a) The negative value for ⌬H tells us that the forward reaction is exothermic. Increasing the
temperature favors the endothermic reaction (reverse in this case).
Some NH3 is used up.
(b) Lowering the temperature favors the exothermic reaction (forward in this case).
More NH3 is formed.
(c) A catalyst does not favor either reaction.
It would have no effect on the amount of NH3.
You should now work Exercises 54, 57, and 58.
Now we shall illustrate the commercial importance of these changes.
Can you use the Arrhenius equation
(Section 16-8) to show that lowering
the activation energy barrier increases
forward and reverse rates by the same
factor?
17-6 Factors that Affect Equilibria 727
THE HABER PROCESS: A PRACTICAL APPLICATION
OF EQUILIBRIUM
Nitrogen, N2, is very unreactive. The Haber process is the economically important industrial
process by which atmospheric N2 is converted to ammonia, NH3, a soluble, reactive
compound. Innumerable dyes, plastics, explosives, fertilizers, and synthetic fibers are made
from ammonia. The Haber process provides insight into kinetic and thermodynamic
factors that influence reaction rates and the positions of equilibria. In this process the
reaction between N2 and H2 to produce NH3 is never allowed to reach equilibrium, but
moves toward it.
N2(g) ϩ 3H2(g) 34 2NH3(g) ⌬H0 ϭ Ϫ92 kJ/mol
Kc ϭ ϭ 3.6 ϫ 108 (at 25°C)
The process is diagrammed in Figure 17-3. The reaction is carried out at about 450°C
under pressures ranging from 200 to 1000 atmospheres. Hydrogen is obtained from coal
gas or petroleum refining and nitrogen from liquefied air.
[NH3]2
ᎏᎏ
[N2][H2]3
17-7
Fritz Haber (1868–1934) developed
the process to provide a cheaper and
more reliable source of explosives as
Germany prepared for World War I.
(Britain controlled the seas and thus
the access to the natural nitrates in
India and Chile that were needed to
prepare explosives.) The current use of
the process is more humanitarian: most
NH3 is used to produce fertilizers. In
the United States, approximately 135
pounds of NH3 is required per person
per year.
728 CHAPTER 17: Chemical Equilibrium
Heat
exchanger
Catalyst
Heating
coil
N2
and H2
Cooling coil
Liquid ammonia
Recirculating
pump
Uncombined
N2
and H2
H2
, N2
, and ammonia
Figure 17-3 A simplified
representation of the Haber process
for synthesizing ammonia.
The value of Kc is 3.6 ϫ 108 at 25°C. This very large value of Kc indicates that at equilibrium
virtually all of the N2 and H2 (mixed in a 1Ϻ3 mole ratio) would be converted into
NH3. At 25°C, the reaction occurs so slowly, however, that no measurable amount of NH3
is produced within a reasonable time. Thus, the large equilibrium constant (a thermodynamic
factor) indicates that the reaction proceeds toward the right almost completely. It
tells us nothing, however, about how fast the reaction occurs (a kinetic factor).
There are four moles of gas on the left side of the equation and only two moles of gas
on the right, so increasing the pressure favors the production of NH3. The Haber process
is therefore carried out at very high pressures, as high as the equipment will safely stand.
The reaction is exothermic (⌬H0
rxn is negative), so increasing the temperature favors
the decomposition of NH3 (the reverse reaction). But, the rates of both forward and reverse
reactions increase as the temperature increases.
The addition of a catalyst of finely divided iron and small amounts of selected oxides
also speeds up both the forward and reverse reactions. This allows NH3 to be produced
not only faster but at a lower temperature, which increases the yield of NH3 and extends
the life of the equipment.
Table 17-1 shows the effects of increases in temperature and pressure on the equilibrium
yield of NH3, starting with 1Ϻ3 mole ratios of N2ϺH2. Kc decreases by more than
ten orders of magnitude, from 3.6 ϫ 108 at 25°C to only 1.4 ϫ 10Ϫ2 at 758°C. This tells
us that the reaction proceeds very far to the left at high temperatures. Casual examination
of the data might suggest that the reaction should be run at lower temperatures, because
a higher percentage of the N2 and H2 is converted into NH3. The reaction occurs so
slowly, however, even in the presence of a catalyst, that it cannot be run economically at
temperatures below about 450°C.
The emerging reaction mixture is cooled down, and NH3 (bp ϭ Ϫ33.43°C) is removed
as a liquid. This prevents the reaction from reaching equilibrium and favors the forward
reaction. The unreacted N2 and H2 are recycled. Excess N2 is used to favor the reaction
to the right.
APPLICATION OF STRESS TO A SYSTEM
AT EQUILIBRIUM
We can use equilibrium constants to determine new equilibrium concentrations that result
from adding one or more species to, or removing one or more species from, a system at
equilibrium.
17-8
In practice, the mixed reactants are
compressed by special pumps and
injected into the heated reaction vessel.
Ten orders of magnitude is 1010, that
is, 10 billion.
1 ϫ 1010 ϭ 10,000,000,000
17-8 Application of Stress to a System at Equilibrium 729
TABLE 17-1 Effect of T and P on Yield of Ammonia
Mole % NH3 in Equilibrium Mixture
°C Kc 10 atm 100 atm 1000 atm
209 650 51 82 98
467 0.5 4 25 80
758 0.014 0.5 5 13
A nighttime photo of a large plant
for the commercial production of
ammonia, NH3. Such an installation
can produce up to 7000 metric tons
of ammonia per day. There are
nearly 100 such plants in the world.
EXAMPLE 17-11 Applying a Stress to a System at Equilibrium
Some hydrogen and iodine are mixed at 229°C in a 1.00-liter container. When equilibrium is
established, the following concentrations are present: [HI] ϭ 0.490 M, [H2] ϭ 0.080 M, and
[I2] ϭ 0.060 M. If an additional 0.300 mol of HI is then added, what concentrations will be
present when the new equilibrium is established?
H2(g) ϩ I2(g) 34 2HI(g)
Plan
We use the initial equilibrium concentrations to calculate the value of Kc. Then we determine
the new concentrations after some HI has been added and calculate Q. The value of Q tells us
which reaction is favored. Then we can represent the new equilibrium concentrations. We
substitute these representations into the Kc expression and solve for the new equilibrium
concentrations.
Solution
Calculate the value of Kc from the first set of equilibrium concentrations.
Kc ϭ ϭ ϭ 50
When we add 0.300 mol of HI to the 1.00-liter container, the [HI] instantaneously increases
by 0.300 M.
H2(g) ϩ I2(g) 34 2HI(g)
equilibrium 0.080 M 0.060 M 0.490 M
mol/L added 0 M 0 M ϩ0.300 M
new initial conc’n 0.080 M 0.060 M 0.790 M
Substitution of these new initial concentrations into the reaction quotient gives
Q ϭ ϭ ϭ 130
Because Q Ͼ Kc, the reaction proceeds to the left to establish a new equilibrium. The new equilibrium
concentrations can be determined as follows. Let x ϭ mol/L of H2 formed; so
x ϭ mol/L of I2 formed, and 2x ϭ mol/L of HI consumed.
H2(g) ϩ I2(g) 34 2HI(g)
new initial conc’n 0.080 M 0.060 M 0.790 M
change due to rxn ϩx M ϩx M Ϫ2x M
new equilibrium (0.080 ϩ x) M (0.060 ϩ x) M (0.790 Ϫ 2x) M
Substitution of these values into Kc allows us to evaluate x.
Kc ϭ 50 ϭ ϭ
0.24 ϩ 7.0x ϩ 50x2 ϭ 0.624 Ϫ 3.16x ϩ 4x2
46x2 ϩ 10.2x Ϫ 0.38 ϭ 0
Solution by the quadratic formula gives x ϭ 0.032 and Ϫ0.25.
Clearly, x ϭ Ϫ0.25 is the extraneous root, because x cannot be less than zero in this case.
This reaction does not consume a negative quantity of HI, because the reaction is proceeding
toward the left. Thus, x ϭ 0.032 is the root with physical meaning, so the new equilibrium
concentrations are
0.624 Ϫ 3.16x ϩ 4x2
ᎏᎏᎏ
0.0048 ϩ 0.14x ϩ x2
(0.790 Ϫ 2x)2
ᎏᎏᎏ
(0.080 ϩ x)(0.060 ϩ x)
(0.790)2
ᎏᎏ
(0.080)(0.060)
[HI]2
ᎏ
[H2][I2]
(0.490)2
ᎏᎏ
(0.080)(0.060)
[HI]2
ᎏ
[H2][I2]
Equal concentrations of H2 and I2
must be formed by the new progress of
the reaction.
To “consume a negative quantity of
HI” would be to form HI. The value
x ϭ Ϫ0.25 would lead to [H2] ϭ
(0.080 ϩ x) M ϭ (0.080 Ϫ 0.25) M ϭ
Ϫ0.17 M. A negative concentration
is impossible, so x ϭ Ϫ0.25 is the
extraneous root.
730 CHAPTER 17: Chemical Equilibrium
It is obvious that adding some HI
favors the reaction to the left. If more
than one substance is added to the
reaction mixture, it might not be
obvious which reaction will be favored.
Calculating Q always lets us make the
decision.
[H2] ϭ (0.080 ϩ x) M ϭ (0.080 ϩ 0.032) M ϭ 0.112 M
[I2] ϭ (0.060 ϩ x) M ϭ (0.060 ϩ 0.032) M ϭ 0.092 M
[HI] ϭ (0.790 Ϫ 2x) M ϭ (0.790 Ϫ 0.064) M ϭ 0.726 M
In summary,
Original Equilibrium Stress Applied New Equilibrium
[H2] ϭ 0.080 M [H2] ϭ 0.112 M
[I2] ϭ 0.060 M Add 0.300 M HI [I2] ϭ 0.092 M
[HI] ϭ 0.490 M
· Ά[HI] ϭ 0.726 M
We see that some of the additional HI is consumed, but not all of it. More HI remains after
the new equilibrium is established than was present before the stress was imposed. The new
equilibrium [H2] and [I2] are substantially greater than the original equilibrium concentrations,
however.
You should now work Exercise 64.
We can also use the equilibrium constant to calculate new equilibrium concentrations
that result from decreasing the volume (increasing the pressure) of a gaseous system that
was initially at equilibrium.
EXAMPLE 17-12 Applying a Stress to a System at Equilibrium
At 22°C the equilibrium constant, Kc, for the following reaction is 4.66 ϫ 10Ϫ3. (a) If 0.800 mol
of N2O4 were injected into a closed 1.00-liter container at 22°C, how many moles of each gas
would be present at equilibrium? (b) If the volume were halved (to 0.500 L) at constant temperature,
how many moles of each gas would be present after the new equilibrium has been
established?
N2O4(g) 34 2NO2(g) Kc ϭ ᎏ
[
[
N
N
O
2O
2]
4
2
]
ᎏ ϭ 4.66 ϫ 10Ϫ3
Plan
(a) We are given the value of Kc and the initial concentration of N2O4. We write the reaction
summary, which gives the representation of the equilibrium concentrations. Then we substitute
these into the Kc expression and solve for the new equilibrium concentrations.
(b) We obtain the new initial concentrations by adjusting the equilibrium concentrations from
part (a) for the volume change. Then we solve for the new equilibrium concentrations as we
did in part (a).
Solution
(a) Let x ϭ mol/L of N2O4 consumed and 2x ϭ mol/L of NO2 formed.
N2O4(g) 34 2NO2(g)
initial 0.800 M 0 M
change due to rxn Ϫx M ϩ2x M
equilibrium (0.800 Ϫ x) M 2x M
17-8 Application of Stress to a System at Equilibrium 731
Kc ϭ ᎏ
[
[
N
N
O
2O
2]
4
2
]
ᎏ ϭ 4.66 ϫ 10Ϫ3 ϭ ᎏ
0.8
(
0
2
0
x)
Ϫ
2
x
ᎏ ϭ ᎏ
0.80
4
0
x2
Ϫ x
ᎏ
3.73 ϫ 10Ϫ3 Ϫ 4.66 ϫ 10Ϫ3x ϭ 4x2
4x2 ϩ 4.66 ϫ 10Ϫ3x Ϫ 3.73 ϫ 10Ϫ3 ϭ 0
Solving by the quadratic formula gives x ϭ 3.00 ϫ 10Ϫ2 and x ϭ Ϫ3.11 ϫ 10Ϫ2. We use
x ϭ 3.00 ϫ 10Ϫ2.
The original equilibrium concentrations are
[NO2] ϭ 2x M ϭ 6.00 ϫ 10Ϫ2 M
[N2O4] ϭ (0.800 Ϫ x) M ϭ (0.800 Ϫ 3.00 ϫ 10Ϫ2) M ϭ 0.770 M
__? mol NO2 ϭ 1.00 L ϫ ϭ 6.00 ϫ 10Ϫ2 mol NO2
__? mol N2O4 ϭ 1.00 L ϫ ϭ 0.770 mol N2O4
(b) When the volume of the reaction vessel is halved, the concentrations are doubled, so the
new initial concentrations of N2O4 and NO2 are 2(0.770 M) ϭ 1.54 M and 2(6.00 ϫ 10Ϫ2 M) ϭ
0.120 M, respectively.
N2O4(g) 34 2NO2(g)
new initial 1.54 M 0.120 M
change due to rxn ϩx M Ϫ2x M
new equilibrium (1.54 ϩ x) M (0.120 Ϫ 2x) M
Kc ϭ ϭ 4.66 ϫ 10Ϫ3 ϭ
Rearranging into the standard form of a quadratic equation gives
x2 Ϫ 0.121x ϩ 1.81 ϫ 10Ϫ3 ϭ 0
Solving as before gives x ϭ 0.104 and x ϭ 0.017.
The maximum value of x is 0.060 M, because 2x may not exceed the concentration of NO2
that was present after the volume was halved. Thus, x ϭ 0.017 M is the root with physical
significance. The new equilibrium concentrations in the 0.500-liter container are
[NO2] ϭ (0.120 Ϫ 2x) M ϭ (0.120 Ϫ 0.034) M ϭ 0.086 M
[N2O4] ϭ (1.54 ϩ x) M ϭ (1.54 ϩ 0.017) M ϭ 1.56 M
__? mol NO2 ϭ 0.500 L ϫ ϭ 0.043 mol NO2
__? mol N2O4 ϭ 0.500 L ϫ ϭ 0.780 mol N2O4
In summary,
First Equilibrium Stress New Equilibrium
0.770 mol of N2O4 Decrease volume from 0.780 mol of N2O4
0.0600 mol of NO2 1.00 L to 0.500 L 0.043 mol of NO2
1.56 mol N2O4
ᎏᎏ
L
0.086 mol NO2
ᎏᎏ
L
(0.120 Ϫ 2x)2
ᎏᎏ
1.54 ϩ x
[NO2]2
ᎏ
[N2O4]
0.770 mol N2O4
ᎏᎏ
L
6.00 ϫ 10Ϫ2 mol NO2
ᎏᎏᎏ
L
The value of x is the number of moles
per liter of N2O4 that react. So x must
be positive and cannot be greater than
0.800 M.
0 Ͻ x Ͻ 0.800 M
LeChatelier’s Principle tells us that
a decrease in volume (increase in
pressure) favors the production of
N2O4.
Q ϭ ϭ 9.35 ϫ 10Ϫ3
Q Ͼ K
shift left
І m8888888
The root x ϭ 0.104 would give a
negative concentration for NO2, which
is impossible.
(0.120)2
ᎏ
1.54
732 CHAPTER 17: Chemical Equilibrium
Both [N2O4] and [NO2] increase because of the large decrease in volume. However, the
number of moles of N2O4 increases, while the number of moles of NO2 decreases. We predict this
from LeChatelier’s Principle.
You should now work Exercise 66.
΂ᎏ
V
n
ᎏ΃ is ΂ᎏ
no.
L
mol
ᎏ΃
KP has no units for the same reasons
that Kc has no units.
17-9 Partial Pressures and the Equilibrium Constant 733
Problem-Solving Tip: There Are Several Ways to Solve
Equilibrium Problems
When a stress is applied to a system originally at equilibrium, it is no longer at equilibrium.
As we did in Example 17-12(b), we can apply the stress to the old equilibrium values
and then treat these as the new “initial values.” Alternatively, we could adjust the original
concentrations values to reflect the stress, and then treat these as the new “initial values.”
We could consider [N2O4]initial in Example 17-12(b) as the original 0.800 mol of N2O4
from part (a) in the new volume, 0.500 L. That is, [N2O4]initial ϭ 0.800 mol/0.500 L ϭ
1.60 M, with no NO2 having yet been formed. [NO2]initial ϭ 0 M. From that starting
point, the reaction would proceed to the right.
PARTIAL PRESSURES AND THE EQUILIBRIUM
CONSTANT
It is often more convenient to measure pressures rather than concentrations of gases.
Solving the ideal gas equation, PV ϭ nRT, for pressure gives
P ϭ (RT) or P ϭ M(RT)
The pressure of a gas is directly proportional to its concentration (n/V). For reactions in
which all substances that appear in the equilibrium constant expression are gases, we sometimes
prefer to express the equilibrium constant in terms of partial pressures in atmospheres
(KP) rather than in terms of concentrations (Kc).
In general for a reaction involving gases,
aA(g) ϩ bB(g) 34 cC(g) ϩ dD(g) KP ϭ
For instance, for the following reversible reaction,
N2(g) ϩ 3H2(g) 34 2NH3(g) KP ϭ
EXAMPLE 17-13 Calculation of KP
In an equilibrium mixture at 500°C, we find PNH3
ϭ 0.147 atm, PN2
ϭ 6.00 atm, and PH2
ϭ
3.70 atm. Evaluate KP at 500°C for the following reaction.
N2(g) ϩ 3H2(g) 34 2NH3(g)
(PNH3
)2
ᎏᎏ
(PN2
)(PH2
)3
(PC)c(PD)d
ᎏᎏ
(PA)a(PB)b
n
ᎏ
V
17-9
Plan
We are given equilibrium partial pressures of all reactants and products. So we write the expression
for KP and substitute partial pressures in atmospheres into it.
Solution
KP ϭ ϭ ϭ 7.11 ϫ 10Ϫ5
You should now work Exercises 70 and 73.
(0.147)2
ᎏᎏ
(6.00)(3.70)3
(PNH3
)2
ᎏᎏ
(PN2
)(PH2
)3
⌬n refers to the numbers of moles of
gaseous substances in the balanced
equation, not in the reaction vessel.
734 CHAPTER 17: Chemical Equilibrium
Problem-Solving Tip: In KP Calculations, Gas Pressures Must Be
Expressed in Atmospheres
One error that students sometimes make when solving KP problems is to express pressures
in torr. Remember that these pressures must be expressed in atmospheres.
RELATIONSHIP BETWEEN KP AND Kc
If the ideal gas equation is rearranged, the molar concentration of a gas is
΂ ΃ϭ or M ϭ
Substituting P/RT for n/V in the Kc expression for the N2–H2–NH3 equilibrium gives
the relationship between Kc and KP for this reaction.
Kc ϭ ᎏ
[N
[N
2]
H
[H
3]
2
2
]3
ᎏ ϭ ϭ ϫ
ϭ KP (RT)2 and KP ϭ Kc (RT)Ϫ2
In general the relationship between Kc and KP is
KP ϭ Kc(RT)⌬n or Kc ϭ KP(RT)Ϫ⌬n ⌬n ϭ (ngas prod) Ϫ (ngas react)
For reactions in which equal numbers of moles of gases appear on both sides of the equation,
⌬n ϭ 0 and KP ϭ Kc.
In Example 17-1, we saw that for the ammonia reaction at 500°C (or 773 K),
Kc ϭ 0.286. We can describe this equilibrium in terms of partial pressures using KP.
NH3(g) ϩ 3H2(g) 34 2NH3(g) ⌬n ϭ 2 Ϫ 4 ϭ Ϫ2
KP ϭ Kc(RT)⌬n ϭ (0.286)[(0.0821)(773)]Ϫ2 ϭ 7.10 ϫ 10Ϫ5
This is essentially the same value we obtained in Example 17-13.
΂ᎏ
R
1
T
ᎏ΃
2
ᎏ
΂ᎏ
R
1
T
ᎏ΃
4
(PNH3
)2
ᎏᎏ
(PN2
)(PH2
)3
΂ᎏ
P
R
N
T
H3
ᎏ΃
2
ᎏᎏ
΂ᎏ
P
R
N
T
2
ᎏ΃΂ᎏ
P
R
H
T
2
ᎏ΃
3
P
ᎏ
RT
P
ᎏ
RT
n
ᎏ
V
17-10
For gas-phase reactions, we can calculate the amounts of substances present at equilibrium
using either KP or Kc. The results are the same by either method (when they are
both expressed in the same terms). To illustrate, let’s solve the following problem by both
methods.
EXAMPLE 17-14 Using Kc and KP
We place 10.0 grams of SbCl5 in a 5.00-liter container at 448°C and allow the reaction to
attain equilibrium. How many grams of SbCl5 are present at equilibrium? Solve this problem
(a) using Kc and molar concentrations and (b) using KP and partial pressures.
SbCl5(g) 34 SbCl3(g) ϩ Cl2(g) at 448°C, Kc ϭ 2.51 ϫ 10Ϫ2 and KP ϭ 1.48
(a) Plan (using Kc)
We calculate the initial concentration of SbCl5, write the reaction summary and represent the
equilibrium concentrations; then we substitute into the Kc expression to obtain the equilibrium
concentrations.
(a) Solution (using Kc)
Because we are given Kc, we use concentrations. The initial concentration of SbCl5 is
[SbCl5] ϭ ᎏ
10.0
5.0
g
0
S
L
bCl5
ᎏ ϫ ᎏ
1
29
m
9
o
g
l
ᎏ ϭ 0.00669 M SbCl5
Let x ϭ mol/L of SbCl5 that react. In terms of molar concentrations, the reaction summary is
SbCl5 34 SbCl3 ϩ Cl2
initial 0.00669 M 0 0
change due to rxn Ϫx M ϩx M ϩx M
equilibrium (0.00669 Ϫ x) M x M x M
Kc ϭ
2.51 ϫ 10Ϫ2 ϭ
x2 ϭ 1.68 ϫ 10Ϫ4 Ϫ 2.51 ϫ 10Ϫ2x
x2 ϩ 2.51 ϫ 10Ϫ2x Ϫ 1.68 ϫ 10Ϫ4 ϭ 0
(x)(x)
ᎏᎏ
0.00669 Ϫ x
[SbCl3][Cl2]
ᎏᎏ
[SbCl5]
17-10 Relationship Between Kp and Kc 735
Problem-Solving Tip: Be Careful About the Value of R
To decide which value of R to use when you convert between Kc and KP, you can reason
as follows. Kc involves molar concentrations, for which the units are mol/L; KP involves
pressures expressed in atm. Thus the appropriate value of R to use for these conversions
must include these units. We use 0.08206 , rounded to the number of places
appropriate to the problem.
LTatm
ᎏ
molTK
Solving by the quadratic formula gives
x ϭ 5.49 ϫ 10Ϫ3 and Ϫ3.06 ϫ 10Ϫ2 (extraneous root)
[SbCl5] ϭ (0.00669 Ϫ x) M ϭ (0.00669 Ϫ 0.00549) M ϭ 1.20 ϫ 10Ϫ3 M
__? g SbCl5 ϭ 5.00 L ϫ ᎏ
1.20 ϫ 1
L
0Ϫ3 mol
ᎏ ϫ ᎏ
2
m
99
ol
g
ᎏ ϭ 1.79 g SbCl5
Let us now solve the same problem using KP and partial pressures.
(b) Plan (using KP)
Calculate the initial partial pressure of SbCl5, and write the reaction summary. Substitution of
the representations of the equilibrium partial pressures into KP gives their values.
(b) Solution (using KP)
We calculate the initial pressure of SbCl5 in atmospheres, using PV ϭ nRT.
PSbCl5
ϭ ϭ ϭ 0.396 atm
Clearly, PSbCl3
ϭ 0 and PCl2
ϭ 0 because only SbCl5 is present initially. We write the reaction
summary in terms of partial pressures in atmospheres, because KP refers to pressures in atmo-
spheres.
Let y ϭ decrease in pressure (atm) of SbCl5 due to reaction. In terms of partial pressures,
the reaction summary is
SbCl5 34 SbCl3 ϩ Cl2
initial 0.396 atm 0 0
change due to rxn Ϫy atm ϩy atm ϩy atm
equilibrium (0.396 Ϫ y) atm y atm y atm
KP ϭ ϭ 1.48 ϭ
0.586 Ϫ 1.48y ϭ y2 y2 ϩ 1.48y Ϫ 0.586 ϭ 0
Solving by the quadratic formula gives
y ϭ 0.325 and Ϫ1.80 (extraneous root)
PSbCl5
ϭ (0.396 Ϫ y) ϭ (0.396 Ϫ 0.325) ϭ 0.071 atm
We use the ideal gas law, PV ϭ nRT, to calculate the number of moles of SbCl5.
n ϭ ϭ ϭ 0.0060 mol SbCl5
__? g SbCl5 ϭ 0.0060 mol ϫ ᎏ
2
m
99
ol
g
ᎏ ϭ 1.8 g SbCl5
You should now work Exercises 75 and 78.
(0.071 atm)(5.00 L)
ᎏᎏᎏ
΂0.0821 ᎏ
m
Lи
o
a
l
t
и
m
K
ᎏ΃(721 K)
PV
ᎏ
RT
( y)( y)
ᎏᎏ
0.396 Ϫ y
(PSbCl3
)(PCl2
)
ᎏᎏ
PSbCl5
΄(10.0 g)΂ᎏ
1
29
m
9
o
g
l
ᎏ΃΅΂0.0821 ᎏ
m
Lи
o
a
l
t
и
m
K
ᎏ΃(721 K)
ᎏᎏᎏᎏᎏ
5.00 L
nRT
ᎏ
V
The partial pressure of each substance is
proportional to the number of moles of
that substance.
If we did not know the value of KP,
we could calculate it from the known
value of Kc, using the relationship
KP ϭ Kc(RT)⌬n.
We see that within roundoff range
the same result is obtained by both
methods.
736 CHAPTER 17: Chemical Equilibrium
HETEROGENEOUS EQUILIBRIA
Thus far, we have considered only equilibria involving species in a single phase, that is,
homogeneous equilibria. Heterogeneous equilibria involve species in more than one
phase. Consider the following reversible reaction at 25°C.
2HgO(s) 34 2Hg(ᐉ) ϩ O2(g)
When equilibrium is established for this system, a solid, a liquid, and a gas are present.
Neither solids nor liquids are significantly affected by changes in pressure. The fundamental
definition of the equilibrium constant in thermodynamics is in terms of the
activities of the substances involved.
For any pure solid or pure liquid, the activity is taken as 1 (Section 17-2), so terms
for pure liquids and pure solids do not appear in the K expressions for heterogeneous
equilibria.
Thus, for the reaction
2HgO(s) 34 2Hg(ᐉ) ϩ O2(g) Kc ϭ [O2] and KP ϭ PO2
These equilibrium constant expressions indicate that equilibrium exists at a given temperature
for one and only one concentration and one partial pressure of oxygen in contact with
liquid mercury and solid mercury(II) oxide.
17-11
17-11 Heterogeneous Equilibria 737
A photograph of the reaction
2HgO(s) 34 2Hg(ᐉ) ϩ O2(g)
The reaction is not at equilibrium here, because O2 gas has been allowed to escape.
EXAMPLE 17-15 Kc and KP for Heterogeneous Equilibrium
Write both Kc and KP for the following reversible reactions.
(a) 2SO2(g) ϩ O2(g) 34 2SO3(g)
(b) 2NH3(g) ϩ H2SO4(ᐉ) 34 (NH4)2SO4(s)
(c) S(s) ϩ H2SO3(aq) 34 H2S2O3(aq)
Plan
We apply the definitions of Kc and KP to each reaction.
Solution
(a) Kc ϭ KP ϭ
(b) Kc ϭ ϭ [NH3]Ϫ2 KP ϭ ϭ (PNH3
)Ϫ2
(c) Kc ϭ KP undefined; no gases involved
EXAMPLE 17-16 Heterogeneous Equilibria
The value of KP is 27 for the thermal decomposition of potassium chlorate at a given high
temperature. What is the partial pressure of oxygen in a closed container in which the following
system is at equilibrium at the given temperature? (This can be a dangerous reaction.)
heat
2KClO3(s) 888n 2KCl(s) ϩ 3O2(g)
Plan
Because two solids, KClO3 and KCl, and only one gas, O2, are involved, we see that KP involves
only the partial pressure of O2, that is, KP ϭ (PO2
)3.
Solution
We are given
KP ϭ (PO2
)3 ϭ 27
Let x atm ϭ PO2
at equilibrium. Then we have
(PO2
)3 ϭ 27 ϭ x3 x ϭ 3.0 atm
This tells us that the partial pressure of oxygen at equilibrium is 3.0 atm.
You should now work Exercises 80 and 91.
RELATIONSHIP BETWEEN ⌬G0
rxn AND THE
EQUILIBRIUM CONSTANT
Let us consider in thermodynamic terms what may happen when two substances are mixed
together at constant temperature and pressure. First, as a result of mixing, there is usually
an increase in entropy due to the increase in disorder. If the two substances can react with
17-12
[H2S2O3]
ᎏᎏ
[H2SO3]
1
ᎏ
(PNH3
)2
1
ᎏ
[NH3]2
(PSO3
)2
ᎏᎏ
(PSO2
)2(PO2
)
[SO3]2
ᎏᎏ
[SO2]2[O2]
738 CHAPTER 17: Chemical Equilibrium
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 20.9, Thermodynamics and the
Equilibrium Constant.
each other, the chemical reaction begins, heat is released or absorbed, and the concentrations
of the substances in the mixture change. An additional change in entropy, which
depends on changes in the nature of the reactants and products, also begins to occur. The
evolution or absorption of heat energy, the changes in entropy, and the changes in concentrations
all continue until equilibrium is established. Equilibrium may be reached with
large amounts of products formed, with virtually of all of the reactants remaining
unchanged, or at any intermediate combination of concentrations.
The standard free energy change for a reaction is ⌬G0
rxn. This is the free energy change
that would accompany complete conversion of all reactants initially present at standard
conditions to all products at standard conditions—the standard reaction (Section 15-15).
The free energy change for any other concentrations or pressures is ⌬Grxn (no superscript
zero). The two quantities are related by the equation
⌬Grxn ϭ ⌬G0
rxn ϩ RT ln Q
R is the universal gas constant, T is the absolute temperature, and Q is the reaction quotient
(Section 17-4). When a system is at equilibrium, ⌬Grxn ϭ 0 (Section 15-15) and Q ϭ K
(Section 17-4). Recall that the reaction quotient may represent nonequilibrium concentrations
(or partial pressures) of products and reactants. As reaction occurs, the free energy
of the mixture and the concentrations change until at equilibrium ⌬Grxn ϭ 0, and the
concentrations of reactants and products satisfy the equilibrium constant. At that point,
Q becomes equal to K (Section 17-4). Then
0 ϭ ⌬G0
rxn ϩ RT ln K (at equilibrium)
Rearranging gives
⌬G0
rxn ϭ ϪRT ln K
This equation shows the relationship between the standard free energy change and the
thermodynamic equilibrium constant.
For the following generalized reaction, the thermodynamic equilibrium constant is
defined in terms of the activities of the species involved.
aA ϩ bB 34 cC ϩ dD K ϭ
where aA is the activity of substance A, and so on. The mass action expression to which
it is related involves concentration terms for species in solution and partial pressures for
gases.
When the relationship ⌬G0
rxn ϭ ϪRT ln K is used with
1. all gaseous reactants and products, K represents KP;
2. all solution reactants and products, K represents Kc;
3. a mixture of solution and gaseous reactants, K represents the thermodynamic
equilibrium constant, and we do not make the distinction between KP and Kc.
(aC)c(aD)d
ᎏᎏ
(aA)a(aB)b
Thermodynamic standard states are
(1) pure solids or pure liquids at 1 atm,
(2) solutions of one-molar
concentrations, and (3) gases at partial
pressures of 1 atm.
The energy units of R must match
those of ⌬G0. We usually use
R ϭ 8.314 J/molTK
We will encounter equilibrium
calculations of solution and gaseous
species in Chapter 21,
“Electrochemistry.”
17-12 Relationship Between ⌬G0
rxn and the Equilibrium Constant 739
Figure 17-4 displays the relationships between free energy and equilibrium. The left
end of each curve represents the total free energy of the reactants and the right end of
each curve represents the total free energy of the products at standard state conditions.
The difference between them is ⌬G0
rxn; like K, ⌬G0
rxn depends only on temperature and is
a constant for any given reaction.
From the preceding equation that relates ⌬G0
rxn and K, we see that when ⌬G0
rxn is negative,
ln K must be positive, and K is greater than 1. This tells us that products are favored
over reactants at equilibrium. This case is illustrated in Figure 17-4a. When ⌬G0
rxn is positive,
ln K must be negative, and K is less than 1. This tells us that reactants are favored
over products at equilibrium (Figure 17-4b). In the rare case of a chemical reaction for
which ⌬G0
rxn ϭ 0, then K ϭ 1 and the numerator and the denominator must be equal in
the equilibrium constant expression, (i.e., [C]c[D]d . . . ϭ [A]a[B]b . . .). These relationships
are summarized as follows.
⌬G0
rxn K Product Formation
⌬G0
rxn Ͻ 0 K Ͼ 1 Products favored over reactants at equilibrium
⌬G0
rxn ϭ 0 K ϭ 1 At equilibrium when [C]c[D]d . . . ϭ [A]a[B]b . . . (very rare)
⌬G0
rxn Ͼ 0 K Ͻ 1 Reactants favored over products at equilibrium
The direction of approach to equilibrium and the actual free energy change (⌬Grxn)
are not constants. They vary with the conditions and the initial concentrations. If the initial
concentrations correspond to Q Ͻ K, equilibrium is approached from left to right on the
curves in Figure 17-4, and the forward reaction predominates. If Q Ͼ K, equilibrium is
approached from right to left, and the reverse reaction predominates.
The magnitude of ⌬G0
rxn indicates the extent to which a chemical reaction occurs under
standard state conditions, that is, how far the reaction goes toward the formation of products
before equilibrium is reached. The more negative the ⌬G0
rxn value, the larger is the
Figure 17-4 Variation in total free
energy for a reversible reaction
carried out at constant T. The
standard free energy change, ⌬G0,
represents the free energy change
for the standard reaction—the
complete conversion of reactants
into products. In (a) this change is
negative, indicating that the
standard reaction is product-favored
(spontaneous); the collection of just
products would be more stable than
the collection of just reactants. The
mixture of reactants and products
corresponding to the minimum
of the curve is even more stable,
however, and represents the
equilibrium mixture. Because ⌬G0 is
negative, K Ͼ 1, and the equilibrium
mixture contains more products than
reactants. At any point on the curve,
comparing Q and K indicates the
direction in which the reaction must
proceed to approach equilibrium,
that is, which way is “downhill” in
free energy. The plot in (b) is for
positive ⌬G0 (the standard reaction
is reactant-favored). In this case
K Ͻ 1, and the equilibrium mixture
contains more reactants than
products.
Increasingfreeenergy,G
Reaction coordinate
Reactants
only
(a)
Products
only
⌬G0
Ͻ 0
Q Ͻ K
Q Ͼ K
Q ϭ K
Equilibrium
mixture
Increasingfreeenergy,G
Reaction coordinate
Reactants
only
Standard reaction is
spontaneous in reverse direction
(reactant-favored)
⌬G0
Ͼ 0, K Ͻ 1
Standard reaction is
spontaneous in forward direction
(product-favored)
⌬G0
Ͻ 0, K Ͼ 1
(b)
Products
only
⌬G0
Ͼ 0
Q Ͼ K
Q Ͻ K
Q ϭ K
Equilibrium
mixture
value of K and the more favorable is the formation of products. We think of some reactions
as going to “completion.” These generally have very negative ⌬G0
rxn values. The
more positive the ⌬G0
rxn value, the smaller is the value of K and the less favorable is the
formation of products.
EXAMPLE 17-17 K versus ⌬G0
rxn
Use the data in Appendix K to calculate KP for the following reaction at 25°C.
2N2O(g) 34 2N2(g) ϩ O2(g)
Plan
The temperature is 25°C, so we evaluate ⌬G0
rxn for the reaction from ⌬Gf
0
values in Appendix
K. The reaction involves only gases, so K is KP. This means that ⌬G0
rxn ϭ ϪRT ln KP. We solve
for KP.
Solution
⌬G0
rxn ϭ [2 ⌬Gf
0
N2(g) ϩ ⌬Gf
0
O2(g)] Ϫ [2 ⌬Gf
0
N2O(g)]
ϭ [2(0) ϩ 0] Ϫ [2(104.2)] ϭ Ϫ208.4 kJ/mol, or Ϫ2.084 ϫ 105 J/mol
This is a gas-phase reaction, so ⌬G0
rxn is related to KP by
⌬G0
rxn ϭ ϪRT ln KP
ln KP ϭ ϭ ϭ 84.1
KP ϭ e84.1 ϭ 3.3 ϫ 1036
The very large value of KP tells us that the equilibrium lies very far to the right. This reaction
is so slow at 25°C, however, that very little N2O decomposes to N2 and O2 at that
temperature.
You should now work Exercise 84.
EXAMPLE 17-18 K versus ⌬G0
rxn
In Examples 15-16 and 15-17 we evaluated ⌬G0
rxn for the following reaction at 25°C and found
it to be ϩ173.1 kJ/mol. Calculate KP at 25°C for this reaction.
N2(g) ϩ O2(g) 34 2NO(g)
Plan
In this example we use ⌬G0
rxn ϭ ϪRT ln KP.
Solution
⌬G0
rxn ϭ ϪRT ln KP
ln KP ϭ ϭ
ϭ Ϫ69.9
KP ϭ eϪ69.9 ϭ 4.4 ϫ 10Ϫ31
1.731 ϫ 105 J/mol
ᎏᎏᎏ
Ϫ(8.314 J/molTK)(298 K)
⌬G0
rxn
ᎏ
ϪRT
Ϫ2.084 ϫ 105 J/mol
ᎏᎏᎏ
Ϫ(8.314 J/molTK)(298K)
⌬G0
rxn
ᎏ
ϪRT
Units cancel when we express ⌬G0 in
joules per mole. We interpret this as
meaning “per mole of reaction”—that
is, for the number of moles of each
substance shown in the balanced
equation.
On some calculators, we evaluate ex as
follows: Enter the value of x, then
press INV followed by ln x.
17-12 Relationship Between ⌬G0
rxn and the Equilibrium Constant 741
This very small number indicates that at equilibrium almost no N2 and O2 are converted to
NO at 25°C. For all practical purposes, the reaction does not occur at 25°C.
You should now work Exercise 86.
A very important application of the relationships in this section is the use of measured
K values to calculate ⌬G0
rxn.
EXAMPLE 17-19 K versus ⌬G0
rxn
The equilibrium constant, KP, for the following reaction is 5.04 ϫ 1017 at 25°C. Calculate
⌬G0
298 for the hydrogenation of ethylene to form ethane.
C2H4(g) ϩ H2(g) 34 C2H6(g)
Plan
We use the relationship between ⌬G0 and KP.
Solution
⌬G0
298 ϭ ϪRT ln KP
ϭ Ϫ(8.314 J/molTK)(298 K) ln (5.04 ϫ 1017)
ϭ Ϫ1.01 ϫ 105 J/mol
ϭ Ϫ101 kJ/mol
You should now work Exercise 88(a).
EVALUATION OF EQUILIBRIUM CONSTANTS AT
DIFFERENT TEMPERATURES
Chemists have determined equilibrium constants for thousands of reactions. It would be
an impossibly huge task to catalog such constants at every temperature of interest for each
reaction. Fortunately, there is no need to do this. If we determine the equilibrium constant,
KT1
, for a reaction at one temperature, T1, and also its ⌬H0, we can then estimate the equilibrium
constant at a second temperature, T2, using the van’t Hoff equation.
ln ΂ ΃ϭ ΂ Ϫ ΃
Thus, if we know ⌬H0 for a reaction and K at a given temperature (say 298 K), we can
use the van’t Hoff equation to calculate the value of K at any other temperature.
EXAMPLE 17-20 Evaluation of KP at Different Temperatures
We found in Example 17-18 that KP ϭ 4.4 ϫ 10Ϫ31 at 25°C (298 K) for the following reaction.
⌬H0 ϭ 180.5 kJ/mol for this reaction. Evaluate KP at 2400. K.
1
ᎏ
T2
1
ᎏ
T1
⌬H0
ᎏ
R
KT2
ᎏ
KT1
17-13
742 CHAPTER 17: Chemical Equilibrium
Compare the form of this equation to
the Arrhenius equation (Section 16-8)
and to the Clausius–Clapeyron
equation (Section 13-9).
17-13 Evaluation of Equilibrium Constants at Different Temperatures 743
Problem-Solving Tip: Use the Correct K
The K values that appear in the van’t Hoff equation represent the thermodynamic equilibrium
constant (see Section 17-12). For a gas-phase reaction (such as that in Example
17-20), K represents KP; if the value of Kc were given, we would have to convert it to
KP (see Section 17-10) before using the van’t Hoff equation.
N2(g) ϩ O2(g) 34 2NO(g)
Plan
We are given KP at one temperature, 25°C, and the value of ⌬H0. We are given the second
temperature, 2400. K. These data allow us to evaluate the right side of the van’t Hoff
equation, which gives us ln (KT2
/KT1
). Because we know KT1
, we can find the value for
KT2
.
Solution
Let T1 ϭ 298 K and T2 ϭ 2400. K. Then
ln ΂ ΃ϭ ΂ Ϫ ΃
Let us first evaluate the right side of the equation.
ln ΂ ΃ϭ ΂ Ϫ ΃ϭ 63.8
Now take the inverse logarithm of both sides.
ϭ e63.8 ϭ 5.1 ϫ 1027
Solving for KT2
and substituting the known value of KT1
, we obtain
KT2
ϭ (5.1 ϫ 1027)(KT1
) ϭ (5.1 ϫ 1027)(4.4 ϫ 10Ϫ31) ϭ 2.2 ϫ 10Ϫ3 at 2400. K
You should now work Exercise 88(b–f).
KT2
ᎏ
KT1
1
ᎏ
2400. K
1
ᎏ
298 K
1.805 ϫ 105 J/mol
ᎏᎏᎏ
8.314 J/molTK
KT2
ᎏ
KT1
1
ᎏ
T2
1
ᎏ
T1
⌬H0
ᎏ
R
KT2
ᎏ
KT1
In Example 17-20 we see that KT2
(KP at 2400. K) is quite small, which tells us that the
equilibrium favors N2 and O2 rather than NO. Nevertheless, KT2
is very much larger than
KT1
, which is 4.4 ϫ 10Ϫ31. At 2400. K, significantly more NO is present at equilibrium,
relative to N2 and O2, than at 298 K. So automobiles emit small amounts of NO into
the atmosphere, sufficient to cause severe air pollution problems. Catalytic converters
(Section 16-9) are designed to catalyze the breakdown of NO into N2 and O2.
2NO(g) 34 N2(g) ϩ O2(g)
This reaction is spontaneous. Catalysts do not shift the position of equilibrium. They
favor neither consumption nor production of NO. They merely allow the system to reach
equilibrium more rapidly. The time factor is very important because the NO stays in the
automobile exhaust system for only a very short time.
The KP value could be converted to Kc
using the relationship Kc ϭ KP(RT)Ϫ⌬n
(Section 17-10).
2400 K is a typical temperature
inside the combustion chambers of
automobile engines. Large quantities
of N2 and O2 are present during
gasoline combustion, because the
gasoline is mixed with air.
744 CHAPTER 17: Chemical Equilibrium
Key Terms
Activity (of a component of an ideal mixture) A dimensionless
quantity whose magnitude is equal to molar concentration
in an ideal solution, equal to partial pressure (in atmospheres)
in an ideal gas mixture, and defined as 1 for pure solids or
liquids.
Chemical equilibrium A state of dynamic balance in which the
rates of forward and reverse reactions are equal; there is no net
change in concentrations of reactants or products while a system
is at equilibrium.
Dynamic equilibrium An equilibrium in which processes occur
continuously, with no net change.
Equilibrium constant, (Kc) A quantity that indicates the extent
to which a reversible reaction occurs. Kc varies with tempera-
ture.
Heterogeneous equilibria Equilibria involving species in more
than one phase.
Homogeneous equilibria Equilibria involving only species in a
single phase, that is, all gases, all liquids, or all solids.
LeChatelier’s Principle If a stress (change of conditions) is
applied to a system at equilibrium, the system shifts in the direction
that reduces the stress, to move toward a new state of
equilibrium.
Mass action expression For a reversible reaction,
aA ϩ bB 34 cC ϩ dD
the product of the molar concentrations of the products (species
on the right), each raised to the power that corresponds to its coefficient
in the balanced chemical equation, divided by the product
of the concentrations of the reactants (species on the left), each
raised to the power that corresponds to its coefficient in the
balanced chemical equation. At equilibrium the mass action expression
is equal to Kc; at other conditions, it is Q.
ᎏ
[
[
C
A
]
]
c
a
[
[
D
B]
]
b
d
ᎏ ϭ Q or, at equilibrium, Kc
Reaction quotient, Q The mass action expression under any set
of conditions (not necessarily equilibrium); its magnitude relative
to Kc determines the direction in which reaction must occur
to establish equilibrium.
Reversible reactions Reactions that do not go to completion
and occur in both the forward and reverse directions.
van’t Hoff equation The relationship between ⌬H0 for a reaction
and its equilibrium constants at two different temperatures.
Exercises
Basic Concepts
001. Define and illustrate the following terms: (a) reversible
reaction, (b) chemical equilibrium, (c) equilibrium con-
stant.
002. Equilibrium constants do not have units. Explain.
003. Distinguish between the terms “static equilibrium” and
“dynamic equilibrium.” Which kind does chemical equilibrium
represent?
004. (a) Describe three examples of static equilibrium. (b) Describe
three examples of dynamic equilibrium (besides
chemical equilibrium).
005. Explain the significance of (a) a very large value of K,
(b) a very small value of K, and (c) a value of K of
about 1.
006. What can be said about the magnitude of the equilibrium
constant in a reaction whose equilibrium lies far to the
right? To the left?
007. What is the relationship between equilibrium and the rates
of opposing processes?
008. What does the value of an equilibrium constant tell us
about the time required for the reaction to reach equilib-
rium?
009. When giving the value of an equilibrium constant, it is
necessary also to write the balanced chemical equation.
Why? Give examples to illustrate your explanation.
010. (a) How is the equilibrium constant related to the forward
and reverse rate constants? (b) Can the rate expressions for
forward and reverse reactions be written from the balanced
chemical equation? Explain. (c) Can the equilibrium constant
expression be written from the balanced chemical
equation? Explain.
011. (a) Sketch a set of curves similar to Figure 17-2 for concentration
changes with time for a reaction
2A(g) ϩ B(g) 34 C(g) ϩ 2D(g)
assuming that K is much greater than 1. In each case,
assume that A and B start at the same concentration and
that no C or D are present.
(b) Repeat part (a) for the case that K is much less than 1.
012. (a) Sketch a set of curves similar to Figure 17-2 for concentration
changes with time for a reaction
3A(g) ϩ B 34 2C(g)
assuming that K is much greater than 1. In each case,
assume that A and B start at the same concentration and
that no C is present.
(b) Repeat part (a) for the case that K is much less than 1.
013. At some temperature, the reaction
N2(g) ϩ 3H2(g) 34 2NH3(g)
Exercises 745
has an equilibrium constant Kc numerically equal to 1.
State whether each of the following is true or false, and
explain why.
(a) An equilibrium mixture must have the H2 concentration
three times that of N2 and the NH3 concentration
twice that of H2.
(b) An equilibrium mixture must have the H2 concentration
three times that of N2.
(c) A mixture in which the H2 concentration is three times
that of N2 and the NH3 concentration is twice that of N2
could be an equilibrium mixture.
(d) A mixture in which the concentration of each reactant
and each product is 1 M is an equilibrium mixture.
(e) Any mixture in which the concentrations of all reactants
and products are equal is an equilibrium mixture.
(f ) An equilibrium mixture must have equal concentrations
of all reactants and products.
014. Why do we omit concentrations of pure solids and pure
liquids from equilibrium constant expressions?
015. Consider the following compounds, in the states indicated,
as possible reactants or products in a chemical reaction.
Which of these compounds would be omitted from the
equilibrium constant expression? H2O(s), H2O(ᐉ),
H2O(g), HCl(g), HCl(aq), NaHCO3(s), N2(g), NH3(ᐉ),
CO(g), and Fe2O3(s).
016. Consider the following compounds, in the states indicated,
as possible reactants or products in a chemical reaction.
Which of these compounds would be omitted from the
equilibrium constant expression? Explain. CaCO3(s),
H2SO4(ᐉ), NaOH(s), NaOH(aq), O2(g), CH3COOH(ᐉ),
HI(g), I2(s), C(graphite), and SO3(g).
Equilibrium Constant Expression and Value of K
017. Write the expression for Kc for each of the following
reactions:
(a) CO2(g) ϩ H2(g) 34 CO(g) ϩ H2O(g)
(b) 2NO2(g) 34 2NO(g) ϩ O2(g)
(c) 2CHCl3(g) ϩ 3H2(g) 34 2CH4(g) ϩ 3Cl2(g)
(d) H2(g) ϩ I2(g) 34 2HI(g)
(e) 2NOCl(g) 34 2NO(g) ϩ Cl2(g)
018. Write the expression for Kc for each of the following
reactions:
(a) 2H2O(g) ϩ 2SO2(g) 34 2H2S(g) ϩ 3O2(g)
(b) 4NH3(g) ϩ 5O2(g) 34 4NO(g) ϩ 6H2O(g)
(c) PCl3(g) ϩ Cl2(g) 34 PCl5(g)
(d) N2O4(g) 34 2NO2(g)
(e) 2SO2(g) ϩ O2(g) 34 2SO3(g)
019. Write the expression for Kc for each of the following
reactions:
(a) 2CO(g) ϩ O2(g) 34 2CO2(g)
(b) SrCO3(s) 34 SrO(s) ϩ CO2(g)
(c) 2HBr(g) 34 H2(g) ϩ Br2(ᐉ)
(d) P4(g) ϩ 3O2(g) 34 P4O6(s)
(e) N2(g) ϩ O2(g) 34 2NO(g)
020. Write the expression for Kc for each of the following
reactions:
(a) 2H2O2(g) 34 2H2O(g) ϩ O2(g)
(b) 2ZnS(s) ϩ 3O2(g) 34 2ZnO(s) ϩ 2SO2(g)
(c) NH3(g) ϩ HCl(g) 34 NH4Cl(s)
(d) NaF(s) ϩ H2SO4(ᐉ) 34 NaHSO4(s) ϩ HF(g)
(e) 2Cl2(g) ϩ 2H2O(g) 34 4HCl(g) ϩ O2(g)
021. Write the expression for Kc for each of the following
reactions:
(a) TlCl3(s) 34 TlCl(s) ϩ Cl2(g)
(b) CuCl4
2Ϫ(aq) 34 Cu2ϩ(aq) ϩ 4ClϪ(aq)
(c) 3O2(g) 34 2O3(g)
(d) 4H3Oϩ(aq) ϩ 2ClϪ(aq) ϩ MnO2(s) 34
Mn2ϩ(aq) ϩ 6H2O(ᐉ) ϩ Cl2(aq)
022. On the basis of the equilibrium constant values, choose the
reactions in which the products are favored.
(a) NH3(aq) ϩ H2O(ᐉ) 34 NH4
ϩ(aq) ϩ OHϪ(aq)
K ϭ 1.8 ϫ 10Ϫ5
(b) Auϩ(aq) ϩ 2CNϪ(aq) 34 [Au(CN)2]Ϫ(aq)
K ϭ 2 ϫ 1038
(c) PbC2O4(s) 34 Pb2ϩ(aq) ϩ C2O4
2Ϫ(aq) K ϭ 4.8
(d) HSϪ(aq) ϩ Hϩ(aq) 34 H2S(aq) K ϭ 1.0 ϫ 107
023. On the basis of the equilibrium constant values, choose the
reactions in which the reactants are favored.
(a) H2O(ᐉ) 34 Hϩ(aq) ϩ OHϪ(aq)
K ϭ 1.0 ϫ 10Ϫ14
(b) [AlF6]3Ϫ(aq) 34 Al3ϩ(aq) ϩ 6FϪ(aq)
K ϭ 2 ϫ 10Ϫ24
(c) Ca3(PO4)2(s) 34 3Ca2ϩ(aq) ϩ 2PO4
3Ϫ(aq)
K ϭ 10Ϫ25
(d) 2Fe3ϩ(aq) ϩ 3S2Ϫ(aq) 34 Fe2S3(s)
K ϭ 1 ϫ 1088
Calculation of K
024. The reaction between nitrogen and oxygen to form NO(g)
is represented by the chemical equation
N2(g) ϩ O2(g) 34 2NO(g)
Equilibrium concentrations of the gases at 1500 K are
1.7 ϫ 10Ϫ3 mol/L for O2, 6.4 ϫ 10Ϫ3 mol/L for N2, and
1.1 ϫ 10Ϫ5 mol/L for NO. Calculate the value of Kc at
1500 K from these data.
025. At elevated temperatures, BrF5 establishes the following
equilibrium.
2BrF5(g) 34 Br2(g) ϩ 5F2(g)
The equilibrium concentrations of the gases at 1500 K
are 0.0064 mol/L for BrF5, 0.0018 mol/L for Br2, and
0.0090 mol/L for F2. Calculate the value of Kc.
026. At some temperature the reaction
PCl3(g) ϩ Cl2(g) 34 PCl5(g)
is at equilibrium when the concentrations of PCl3, Cl2, and
PCl5 are 10, 9, and 12 mol/L, respectively. Calculate the
value of Kc for this reaction at that temperature.
746 CHAPTER 17: Chemical Equilibrium
027. For the reaction
CO(g) ϩ H2O(g) 34 CO2(g) ϩ H2(g)
the value of the equilibrium constant, Kc, is 1.845 at a given
temperature. We place 0.500 mole CO and 0.500 mole
H2O in a 1.00-L container at this temperature, and allow
the reaction to reach equilibrium. What will be the equilibrium
concentrations of all substances present?
028. Given: A(g) ϩ B(g) 34 C(g) ϩ 2D(g)
One mole of A and one mole of B are placed in a 0.400-liter
container. After equilibrium has been established,
0.20 mole of C is present in the container. Calculate the
equilibrium constant, Kc, for the reaction.
029. Nitrogen reacts with hydrogen to form ammonia.
N2(g) ϩ 3H2(g) 34 2NH3(g)
An equilibrium mixture at a given temperature is found to
contain 0.31 mol/L N2, 0.50 mol/L H2, and 0.14 mol/L
NH3. Calculate the value of Kc at the given temperature.
030. For the following equation, Kc ϭ 7.9 ϫ 1011 at 500 K.
H2(g) ϩ Br2(g) 34 2HBr(g)
(a) ᎏ1
2
ᎏH2(g) ϩ ᎏ1
2
ᎏBr2(g) 34 HBr(g) Kc ϭ ?
(b) 2HBr(g) 34 H2(g) ϩ Br2(g) Kc ϭ ?
(c) 4HBr(g) 34 2H2(g) ϩ 2Br2(g) Kc ϭ ?
031. The equilibrium constant for the reaction
2SO2 ϩ O2 34 2SO3
is Kc ϭ 279 at a given high temperature. What is the value
of the equilibrium constant for each of the following
reactions at this temperature?
(a) 2SO3 34 2SO2 ϩ O2
(b) SO2 ϩ ᎏ1
2
ᎏO2 34 SO3
032. A sealed tube initially contains 9.84 ϫ 10Ϫ4 mol H2 and
1.38 ϫ 10Ϫ3 mol I2. It is kept at 350°C until the reaction
H2(g) ϩ I2(g) 34 2HI(g)
comes to equilibrium. At equilibrium, 4.73 ϫ 10Ϫ4 mol I2
is present. Calculate (a) the numbers of moles of H2 and
HI present at equilibrium; (b) the equilibrium constant,
Kc, for the reaction.
033. NO and O2 are mixed in a container of fixed volume kept
at 1000 K. Their initial concentrations are 0.0200 mol/L
and 0.0300 mol/L, respectively. When the reaction
2NO(g) ϩ O2(g) 34 2NO2(g)
has come to equilibrium, the concentration of NO2 is
2.2 ϫ 10Ϫ3 mol/L. Calculate (a) the concentration of NO
at equilibrium, (b) the concentration of O2 at equilibrium,
and (c) the equilibrium constant, Kc, for the reaction.
034. Antimony pentachloride decomposes in a gas-phase reaction
at high temperatures.
SbCl5(g) 34 SbCl3(g) ϩ Cl2(g)
(a) At some temperature, an equilibrium mixture in a
5.00-L container is found to contain 6.91 g of SbCl5,
16.45 g of SbCl3, and 5.11 g of Cl2. Evaluate Kc.
(b) If 10.0 grams of SbCl5 is placed in the 5.00-liter
container and allowed to establish equilibrium at the
temperature in part (a), what will be the equilibrium
concentrations of all species?
035. At standard temperature and pressure, the reaction indicated
by the following equation has an equilibrium constant,
Kc, equal to 0.021
2HI(g) 34 H2(g) ϩ I2(g)
Calculate the equilibrium constant, Kc, for the reverse
equation.
036. The following reaction has an equilibrium constant, Kc,
equal to 1538 at 1800°C
2NO(g) ϩ O2(g) 34 2NO2(g)
Calculate the equilibrium constant, Kc, for the reverse
equation.
The Reaction Quotient, Q
037. Define the reaction quotient, Q. Distinguish between Q
and K.
038. Why is it useful to compare Q with K? What is the situation
when (a) Q ϭ K? (b) Q Ͻ K? (c) Q Ͼ K?
039. How does the form of the reaction quotient compare with
that of the equilibrium constant? What is the difference
between these two expressions?
040. If the reaction quotient is larger than the equilibrium constant,
what will happen to the reaction? What will happen
if Q Ͻ K?
041. Kc ϭ 19.9 for the reaction
Cl2(g) ϩ F2(g) 34 2ClF(g)
What will happen in a reaction mixture originally
containing [Cl2] ϭ 0.4 mol/L, [F2] ϭ 0.2 mol/L, and
[ClF] ϭ 7.3 mol/L?
42. The concentration equilibrium constant for the gas-phase
reaction
H2CO 34 H2 ϩ CO
has the numerical value 0.50 at a given temperature. A
mixture of H2CO, H2, and CO is introduced into a flask
at this temperature. After a short time, analysis of a small
sample of the reaction mixture shows the concentrations
to be [H2CO] ϭ 0.50 M, [H2] ϭ 1.50 M, and [CO] ϭ
Exercises 747
0.25 M. Classify each of the following statements about
this reaction mixture as true or false.
(a) The reaction mixture is at equilibrium.
(b) The reaction mixture is not at equilibrium, but no
further reaction will occur.
(c) The reaction mixture is not at equilibrium, but will
move toward equilibrium by using up more H2CO.
(d) The forward rate of this reaction is the same as the
reverse rate.
043. The value of Kc at 25°C for
C(graphite) ϩ CO2(g) 34 2CO(g)
is 3.7 ϫ 10Ϫ23. Describe what will happen if 2.0 mol of
CO and 2.0 mol of CO2 are mixed in a 1.0-L container
with a suitable catalyst to make the reaction “go” at this
temperature.
Uses of the Equilibrium Constant, Kc
044. For the reaction described by the equation
N2(g) ϩ C2H2(g) 34 2HCN(g)
Kc ϭ 2.3 ϫ 10Ϫ4 at 300°C. What is the equilibrium
concentration of hydrogen cyanide if the initial concentrations
of N2 and acetylene (C2H2) were 2.2 mol/L and
1.0 mol/L, respectively?
045. The equilibrium constant, Kc, for the reaction
Br2(g) ϩ F2(g) 34 2BrF(g)
is 55.3. What are the equilibrium concentrations of all
these gases if the initial concentrations of bromine and
fluorine were both 0.250 mol/L?
046. Kc ϭ 96.2 at 400 K for the reaction
PCl3(g) ϩ Cl2(g) 34 PCl5(g)
What is the concentration of Cl2 at equilibrium if the initial
concentrations were 0.20 mol/L for PCl3 and 7.0 mol/L
for Cl2?
047. Kc ϭ 5.85 ϫ 10Ϫ3 at 25°C for the reaction
N2O4(g) 34 2NO2(g)
Fifteen (15.0) grams of N2O4 is confined in a 5.00-L flask
at 25°C. Calculate (a) the number of moles of NO2 present
at equilibrium and (b) the percentage of the original N2O4
that is dissociated.
048. The reaction of iron and water vapor results in an equi-
librium
3Fe(s) ϩ 4H2O(g) 34 Fe3O4(s) ϩ 4H2(g)
and an equilibrium constant, Kc, of 4.6 at 850°C. What is
the concentration of hydrogen present at equilibrium if
the reaction is initiated with 33 g of H2O and excess Fe
in a 16.0-liter container?
049. The reaction of iron and water vapor results in an equi-
librium
3Fe(s) ϩ 4H2O(g) 34 Fe3O4(s) ϩ 4H2(g)
and an equilibrium constant, Kc, of 4.6 at 850°C. What is
the concentration of water present at equilibrium if the
reaction is initiated with 10. g of H2 and excess iron oxide,
Fe3O4, in a 16.0-liter container?
050. Carbon dioxide reacts with hot carbon in the form of
graphite. The equilibrium constant, Kc, for the reaction is
10.0 at 850°C.
CO2(g) ϩ C(graphite) 34 2CO(g)
If 22 g of carbon monoxide is placed in a 2.5-L reaction
vessel and heated to 850°C, what is the mass of carbon
dioxide at equilibrium?
051. Carbon dioxide reacts with hot carbon in the form of
graphite. The equilibrium constant, Kc, for the reaction is
10.0 at 850°C.
CO2(g) ϩ C(graphite) 34 2CO(g)
If 22 g of carbon dioxide and 50. grams of graphite are
placed in a 2.5-L reaction vessel and heated to 850°C, what
is the mass of carbon monoxide at equilibrium?
052. A 62.5-gram sample of HI was placed in a 1.50-L reaction
vessel and allowed to come to equilibrium as illustrated in
the following equation
2HI(g) 34 H2(g) ϩ I2(g)
The equilibrium constant, Kc, is 0.830. Calculate the
concentration of each species present at equilibrium.
Factors That Influence Equilibrium
53. State LeChatelier’s Principle. Which factors have an effect
on a system at equilibrium? How does the presence of a
catalyst affect a system at chemical equilibrium? Explain
your answer.
54. What will be the effect of increasing the total pressure on
the equilibrium conditions for (a) a chemical equation that
has more moles of gaseous products than gaseous reactants,
(b) a chemical equation that has more moles of
gaseous reactants than gaseous products, (c) a chemical
equation that has the same number of moles of gaseous
reactants and gaseous products, and (d) a chemical equation
in which all reactants and products are pure solids,
pure liquids, or in aqueous solution?
055. Suppose the following exothermic reaction is allowed to
reach equilibrium.
A(g) ϩ 3B(g) 34 2C(g) ϩ 3D(g)
Then we make each of the following changes, and allow
the reaction to reestablish equilibrium. Tell whether the
748 CHAPTER 17: Chemical Equilibrium
amount of B present at the new equilibrium will be
(i) greater than, (ii) less than, or (iii) the same as the amount
of B before the change was imposed.
(a) The temperature is increased while the volume is kept
constant; (b) more A is added; (c) more C is added; (d) a
small amount of D is removed; (e) the pressure is increased
by decreasing the volume.
056. Suppose the following exothermic reaction is allowed to
reach equilibrium.
A(g) ϩ 3B(g) 34 2C(g) ϩ 3D(g)
Then we make each of the following changes and allow
the reaction to reestablish equilibrium. Tell whether the
value of the equilibrium constant will be (i) greater than,
(ii) less than, or (iii) the same as before the change was
imposed.
(a) The temperature is increased while the volume is kept
constant; (b) more A is added; (c) more C is added; (d) a
small amount of D is removed; (e) the pressure is increased
by decreasing the volume.
057. What would be the effect on the equilibrium position of
an equilibrium mixture of Br2, F2, and BrF5 if the total
pressure of the system were increased?
2BrF5(g) 34 Br2(g) ϩ 5F2(g)
058. What would be the effect on the equilibrium position of
an equilibrium mixture of carbon, oxygen, and carbon
monoxide if the total pressure of the system were
increased?
2C(s) ϩ O2(g) 34 2CO(g)
059. A weather indicator can be made with a hydrate of
cobalt(II) chloride, which changes color as a result of the
following reaction.
[Co(OH2)6]Cl2(s) 34 [Co(OH2)4]Cl2(s) ϩ 2H2O(g)
pink blue
Does a blue color indicate “moist” or “dry” air? Explain.
060. Predict whether the equilibrium for the photosynthesis
reaction described by the equation
6CO2(g) ϩ 6H2O(ᐉ) 34 C6H12O6(s) ϩ 6O2(g)
⌬H0 ϭ 2801.69 kJ/mol
would (i) shift to the right, (ii) shift to the left, or (iii)
remain unchanged if (a) [CO2] were increased; (b) PO2
were
increased; (c) one half of the C6H12O6 were removed; (d)
the total pressure were decreased; (e) the temperature were
increased; (f ) a catalyst were added.
061. What would be the effect of decreasing the temperature
on each of the following systems at equilibrium?
(a) H2(g) ϩ I2(g) 34 2HI(g); ⌬H0 ϭ Ϫ9.45 kJ/mol
(b) PCl5(g) 34 PCl3(g) ϩ Cl2(g); ⌬H0 ϭ 92.5 kJ/mol
(c) 2SO2(g) ϩ O2(g) 34 2SO3(g); ⌬H0 ϭ Ϫ198 kJ/mol
(d) 2NOCl(g) 34 2NO(g) ϩ Cl2(g); ⌬H0 ϭ 75 kJ/mol
(e) C(s) ϩ H2O(g) 34 CO(g) ϩ H2(g);
⌬H0 ϭ 131 kJ/mol
062. What would be the effect of decreasing the pressure by
increasing the volume on each of the following systems at
equilibrium?
(a) 2CO(g) ϩ O2(g) 34 2CO2(g)
(b) 2NO(g) 34 N2(g) ϩ O2(g)
(c) N2O4(g) 34 2NO2(g)
(d) Ni(s) ϩ 4CO(g) 34 Ni(CO)4(g)
(e) N2(g) ϩ 3H2(g) 34 2NH3(g)
063. The value of Kc is 0.020 at 2870°C for the reaction
N2(g) ϩ O2(g) 34 2NO(g)
There are 0.800 mole of N2, 0.500 mole of O2, and 0.400
mole of NO in a 1.00-liter container at 2870°C. Is the
system at equilibrium or must the forward or reverse reaction
occur to a greater extent to bring the system to
equilibrium?
064. Given: A(g) ϩ B(g) 34 C(g) ϩ D(g)
(a) At equilibrium a 1.00-liter container was found to
contain 1.60 moles of C, 1.60 moles of D, 0.40 mole of A,
and 0.40 mole of B. Calculate the equilibrium constant for
this reaction.
(b) If 0.20 mole of B and 0.20 mole of C are added to this
system, what will the new equilibrium concentration of A
be?
065. Given: A(g) ϩ B(g) 34 C(g) ϩ D(g)
When one mole of A and one mole of B are mixed and
allowed to reach equilibrium at room temperature, the
mixture is found to contain ᎏ2
3
ᎏ mole of C.
(a) Calculate the equilibrium constant.
(b) If two moles of A were mixed with two moles of B and
allowed to reach equilibrium, how many moles of C would
be present at equilibrium?
066. Given: A(g) 34 B(g) ϩ C(g)
(a) When the system is at equilibrium at 200°C, the
concentrations are found to be: [A] ϭ 0.30 M, [B] ϭ [C] ϭ
0.25 M. Calculate Kc.
(b) If the volume of the container in which the system is
at equilibrium is suddenly doubled at 200°C, what will the
new equilibrium concentrations be?
(c) Refer back to part (a). If the volume of the container
is suddenly halved at 200°C, what will the new equilibrium
concentrations be?
*67. The equilibrium constant, Kc for the dissociation of phosphorus
pentachloride is 9.3 ϫ 10Ϫ2 at 252°C. How many
moles and grams of PCl5 must be added to a 2.0-liter flask
to obtain a Cl2 concentration of 0.15 M?
PCl5(g) 34 PCl3(g) ϩ Cl2(g)
068. At 25°C, Kc is 5.84 ϫ 10Ϫ3 for the dissociation of dinitrogen
tetroxide to nitrogen dioxide.
Exercises 749
N2O4(g) 34 2NO2(g)
(a) Calculate the equilibrium concentrations of both gases
when 3.50 grams of N2O4 is placed in a 2.00-liter flask at
25°C.
(b) What will be the new equilibrium concentrations if the
volume of the system is suddenly increased to 4.00 liters
at 25°C?
(c) What will be the new equilibrium concentrations if the
volume is decreased to 1.00 liter at 25°C?
K in Terms of Partial Pressures
069. Write the KP expression for each reaction in Exercise 17.
070. Under what conditions are Kc and KP for a reaction numerically
equal? Are Kc and KP numerically equal for any of
the reactions in Exercises 17 and 18? Which ones?
071. 0.0100 mol of NH4Cl and 0.0100 mol of NH3 are placed
in a closed 2.00-L container and heated to 603 K. At this
temperature, all the NH4Cl vaporizes. When the reaction
NH4Cl(g) 34 NH3(g) ϩ HCl(g)
has come to equilibrium, 5.8 ϫ 10Ϫ3 mol of HCl is
present. Calculate (a) Kc and (b) KP for this reaction at
603 K.
072. CO2 is passed over graphite at 500 K. The emerging gas
stream contains 4.0 ϫ 10Ϫ3 mol percent CO. The total
pressure is 1.00 atm. Assume that equilibrium is attained.
Find KP for the reaction
C(graphite) ϩ CO2(g) 34 2CO(g)
073. At 425°C, the equilibrium partial pressures of H2, I2, and
HI are 0.06443 atm, 0.06540 atm, and 0.4821 atm, respectively.
Calculate KP for the following reaction at this
temperature.
2HI(g) 34 H2(g) ϩ I2(g)
074. The equilibrium constant, KP, for the reaction indicated
by the following equation is 0.715 at 47°C.
N2O4(g) 34 2NO2(g)
Calculate the partial pressures of N2O4 and NO2 in an
experiment in which 3.0 moles of N2O4 is placed in a
5.0-L flask and allowed to establish equilibrium at 47°C.
075. The equilibrium constant, KP, is 1.92 at 252°C for the
decomposition reaction of phosphorus pentachloride indicated
in the following equation.
PCl5(g) 34 PCl3(g) ϩ Cl2(g)
Calculate the partial pressures of all species present after
6.0 moles of PCl5 is placed in an evacuated 3.0-liter
container and equilibrium is reached at 252°C.
076. The following equilibrium partial pressures were measured
at 750°C: PH2
ϭ 0.387 atm, PCO2
ϭ 0.152 atm, PCO ϭ
0.180 atm, and PH2O ϭ 0.252 atm. What is the value of
the equilibrium constant, KP, for the reaction?
H2 ϩ CO2 34 CO ϩ H2O
077. For the reaction
H2(g) ϩ Cl2(g) 34 2HCl(g)
Kc ϭ 193 at 2500 K. What is the value of KP?
078. For the reaction
Br2(g) 34 2Br(g)
KP ϭ 2550 at 4000 K. What is the value of Kc?
079. A stream of gas containing H2 at an initial partial pressure
of 0.200 atm is passed through a tube in which CuO is
kept at 500 K. The reaction
CuO(s) ϩ H2(g) 34 Cu(s) ϩ H2O(g)
comes to equilibrium. For this reaction, KP ϭ 1.6 ϫ 109.
What is the partial pressure of H2 in the gas leaving the
tube? Assume that the total pressure of the stream is
unchanged.
Relationships Among K, ⌬G0, ⌬H0, and T
080. In the distant future, when hydrogen may be cheaper than
coal, steel mills may make iron by the reaction
Fe2O3(s) ϩ 3H2(g) 34 2Fe(s) ϩ 3H2O(g)
For this reaction, ⌬H ϭ 96 kJ/mol and Kc ϭ 8.11 at
1000 K. (a) What percentage of the H2 remains unreacted
after the reaction has come to equilibrium at 1000 K?
(b) Is this percentage greater or less if the temperature is
decreased to below 1000 K?
081. What kind of equilibrium constant can be calculated from
a ⌬G0 value for a reaction involving only gases?
082. What must be true of the value of ⌬G0 for a reaction if
(a) K ϾϾ 1; (b) K ϭ 1; (c) K ϽϽ 1?
083. A mixture of 3.00 mol of Cl2 and 3.00 mol of CO is
enclosed in a 5.00-L flask at 600°C. At equilibrium, 3.3%
of the Cl2 has been consumed.
CO(g) ϩ Cl2(g) 34 COCl2(g)
(a) Calculate Kc for the reaction at 600°C.
(b) Calculate ⌬G0 for the reaction at this temperature.
084. (a) Use the tabulated thermodynamic values of ⌬Hf
0
and
S0 to calculate the value of KP at 25°C for the gas-phase
reaction
CO ϩ H2O 34 CO2 ϩ H2
(b) Calculate the value of KP for this reaction at 200°C, by
the same method as in part (a).
(c) Repeat the calculation of part (a), using tabulated values
of ⌬Gf
0
.
750 CHAPTER 17: Chemical Equilibrium
085. The equilibrium constant Kc of the reaction
H2(g) ϩ Br2(g) 34 2HBr(g)
is 1.6 ϫ 105 at 1297 K and 3.5 ϫ 104 at 1495 K. (a) Is ⌬H0
for this reaction positive or negative? (b) Find Kc for the
reaction
ᎏ1
2
ᎏH2(g) ϩ ᎏ1
2
ᎏBr2(g) 34 HBr(g)
at 1297 K. (c) Pure HBr is placed in a container of constant
volume and heated to 1297 K. What percentage of the
HBr is decomposed to H2 and Br2 at equilibrium?
086. The air pollutant sulfur dioxide can be partially removed
from stack gases in industrial processes and converted to
sulfur trioxide, the acid anhydride of commercially important
sulfuric acid. Write the equation for the reaction,
using the smallest whole-number coefficients. Calculate
the value of the equilibrium constant for this reaction at
25°C, from values of ⌬Gf
0
in Appendix K.
087. The value of ⌬H0 for the reaction in Exercise 86 is Ϫ197.6
kJ/mol. (a) Predict qualitatively (i.e., without calculation)
whether the value of KP for this reaction at 500°C would
be greater than, the same as, or less than the value at room
temperature (25°C). (b) Now calculate the value of KP at
500°C.
*88. The following is an example of an alkylation reaction that
is important in the production of isooctane (2,2,4trimethylpentane)
from two components of crude oil:
isobutane and isobutene. Isooctane is an antiknock additive
for gasoline.
The thermodynamic equilibrium constant, K, for this reaction
at 25°C is 4.3 ϫ 106, and ⌬H0 is Ϫ78.58 kJ/mol.
(a) Calculate ⌬G0 at 25°C.
(b) Calculate K at 800°C.
(c) Calculate ⌬G0 at 800°C.
(d) How does the spontaneity of the forward reaction at
800°C compare with that at 25°C?
(e) Why do you think the reaction mixture is heated in the
industrial preparation of isooctane?
(f ) What is the purpose of the catalyst? Does it affect the
forward reaction more than the reverse reaction?
CH3 C
CH3
CH3 H
CH2 CH3C
CH3
isooctane
CH3 HC
CH3
CH3
CH3 CH2C
CH3
isobutane isobutene
ϩ
catalyst
heat
089. At sufficiently high temperatures, chlorine gas dissociates,
according to
Cl2(g) 34 2Cl(g)
At 800°C, KP for this reaction is 5.63 ϫ 10Ϫ7.
(a) A sample originally contained Cl2 at 1 atm and 800°C.
Calculate the percentage dissociation of Cl2 when this
reaction has reached equilibrium.
(b) At what temperature would Cl2 (originally at 1 atm
pressure) be 1% dissociated into Cl atoms?
Mixed Exercises
090. At 700°C, KP is 1.50 for the reaction
C(s) ϩ CO2(g) 34 2CO(g)
Suppose the total gas pressure at equilibrium is 1.00 atm.
What are the partial pressures of CO and CO2?
091. At Ϫ10°C, the solid compound Cl2(H2O)8 is in equilibrium
with gaseous chlorine, water vapor, and ice. The
partial pressures of the two gases in equilibrium with a
mixture of Cl2(H2O)8 and ice are 0.20 atm for Cl2 and
0.00262 atm for water vapor. Find the equilibrium constant
KP for each of these reactions.
(a) Cl2(H2O)8(s) 34 Cl2(g) ϩ 8H2O(g)
(b) Cl2(H2O)8(s) 34 Cl2(g) ϩ 8H2O(s)
Why are your two answers so different?
092. A flask contains NH4Cl(s) in equilibrium with its decomposition
products.
NH4Cl(s) 34 NH3(g) ϩ HCl(g)
For this reaction, ⌬H ϭ 176 kJ/mol. How is the mass of
NH3 in the flask affected by each of the following disturbances?
(a) The temperature is decreased. (b) NH3 is
added. (c) HCl is added. (d) NH4Cl is added, with no
appreciable change in the gas volume. (e) A large amount
of NH4Cl is added, decreasing the volume available to the
gases.
093. The equilibrium constant for the reaction
H2(g) ϩ Br2(ᐉ) 34 2HBr(g)
is KP ϭ 4.5 ϫ 1018 at 25°C. The vapor pressure of liquid
Br2 at this temperature is 0.28 atm. (a) Find KP at 25°C
for the reaction
H2(g) ϩ Br2(g) 34 2HBr(g)
(b) How will the equilibrium in part (a) be shifted by an
increase in the volume of the container if (1) liquid Br2 is
absent; (2) liquid Br2 is present? Explain why the effect is
different in these two cases.
094. Given that KP is 4.6 ϫ 10Ϫ14 at 25°C for the reaction
2Cl2(g) ϩ 2H2O(g) 34 4HCl(g) ϩ O2(g)
⌬H0 ϭ ϩ115 kJ/mol
Exercises 751
Calculate KP and Kc for the reaction at 400°C and at
800°C.
095. Kc ϭ 19.9 for the reaction
Cl2(g) ϩ F2(g) 34 2ClF(g)
What will happen in a reaction mixture originally
containing [Cl2] ϭ 0.200 mol/L, [F2] ϭ 0.300 mol/L, and
[ClF] ϭ 0.950 mol/L?
096. A mixture of CO, H2, CH4, and H2O is kept at 1133 K
until the reaction
CO(g) ϩ 3H2(g) 34 CH4(g) ϩ H2O(g)
has come to equilibrium. The volume of the container is
0.100 L. The equilibrium mixture contains 1.21 ϫ 10Ϫ4 mol
CO, 2.47 ϫ 10Ϫ4 mol H2, 1.21 ϫ 10Ϫ4 mol CH4, and 5.63
ϫ 10Ϫ8 mol H2O. Calculate KP for this reaction at 1133 K.
097. What would the pressure of hydrogen be at equilibrium
when PWCl6
ϭ 0.012 atm and PHCl ϭ 0.10 atm? KP ϭ
1.37 ϫ 1021 at 900 K.
WCl6(g) ϩ 3H2(g) 34 W(s) ϩ 6HCl(g)
CONCEPTUAL EXERCISES
098. What is the relationship between the forward and reverse
reactions in an equilibrium?
099. The term “equilibrium” brings to mind the word “equal.”
What is the relationship between the two terms?
100. The masses of participants in a chemical equilibrium are
not the same on both sides of the reaction. Does the equilibrium
concept violate the Law of Conservation of
Matter? Explain.
101. A sample of benzoic acid, a solid carbon-containing acid,
is in equilibrium with an aqueous solution of benzoic acid.
A tiny quantity of D2O, water containing the isotope 2H,
deuterium, is added to the solution. The solution is allowed
to stand at constant temperature for several hours, after
which some of the solid benzoic acid is removed and analyzed.
The benzoic acid is found to contain a tiny quantity
of deuterium, D, and the formula of the deuteriumcontaining
molecules is C6H5COOD. Explain how this
can happen.
102. Imagine yourself the size of atoms and molecules inside a
beaker containing the following equilibrium mixture with
a K greater than 1.
[Co(OH2)6]2ϩ(aq) ϩ 4ClϪ(aq) 34
pink CoCl4
2Ϫ(aq) ϩ 6H2O(ᐉ)
blue
Write a brief description of what you observe around you
before and after additional water is added to the mixture.
BUILDING YOUR KNOWLEDGE
103. Hemoglobin, Hb, has four Fe atoms per molecule that, on
the average, pick up roughly three molecules of O2.
Hb(aq) ϩ 3O2(g) 34 Hb(O2)3(aq)
Discuss mountain or space sickness in terms of this equi-
librium.
Astronauts incur the risk of space sickness.
104. At room temperature, the equilibrium constant for the
reaction
2SO2 ϩ O2 34 2SO3
is 6.98 ϫ 1024. Calculate ⌬G0
rxn and ⌬Gf
0
(SO3), given
the additional information that ⌬Gf
0
(SO2) ϭ
Ϫ300.194 kJ/mol. Check your answer by looking up
⌬Gf
0
(SO3) in Appendix K.
105. At 25°C, 550.0 g of deuterium oxide, D2O (20.0 g/mol;
density 1.10 g/mL), and 498.5 g of H2O (18.0 g/mol; density
0.997 g/mL) are mixed. The volumes are additive.
47.0% of the H2O reacts to form HDO. Calculate Kc at
25°C for the reaction
H2O ϩ D2O 34 2HDO
106. At its normal boiling point of 100°C, the heat of vaporization
of water is 40.66 kJ/mol. What is the equilibrium
vapor pressure of water at 50°C? (You may wish to review
Example 15-20.)
107. Use the data in the preceding exercise to calculate the temperature
at which the vapor pressure of water is 1.50 atm.
18 Ionic Equilibria I:
Acids and Bases
OUTLINE
18-1 A Review of Strong Electrolytes
18-2 The Autoionization of Water
18-3 The pH and pOH Scales
18-4 Ionization Constants for Weak
Monoprotic Acids and Bases
18-5 Polyprotic Acids
18-6 Solvolysis
18-7 Salts of Strong Bases and
Strong Acids
18-8 Salts of Strong Bases and
Weak Acids
18-9 Salts of Weak Bases and
Strong Acids
18-10 Salts of Weak Bases and Weak
Acids
18-11 Salts that Contain Small,
Highly Charged Cations
OBJECTIVES
After you have studied this chapter, you should be able to
• Recognize strong electrolytes and calculate concentrations of their ions
• Understand the autoionization of water
• Understand the pH and pOH scales and how they are used
• Use ionization constants for weak monoprotic acids and bases
• Discuss the concepts of solvolysis and hydrolysis
• Describe how polyprotic acids ionize in steps and how to calculate concentrations
of all species in solutions of polyprotic acids
• Apply acid–base equilibrium concepts to salts of strong bases and weak acids
• Apply acid–base equilibrium concepts to salts of weak bases and strong acids
• Apply acid–base equilibrium concepts to salts of weak bases and weak acids
• Apply acid–base equilibrium concepts to salts of small, highly charged cations
AA
queous solutions are very important. Nearly three fourths of the earth’s surface
is covered with water. Enormous numbers of chemical reactions occur in the
oceans and smaller bodies of water. Body fluids of plants and animals are mostly
water. Life processes (chemical reactions) of all plants and animals occur in aqueous solutions
or in contact with water. Before we were born, all of us developed in sacs filled with
aqueous solutions, which protected and nurtured us until we had developed to the point
that we could live in the atmosphere.
Many foods contain weak acids.
Citrus fruits contain citric acid and
ascorbic acid (vitamin C).
A REVIEW OF STRONG ELECTROLYTES
In previous discussions we have seen that water-soluble compounds may be classified as
either electrolytes or nonelectrolytes. Electrolytes are compounds that ionize (or dissociate
into their constituent ions) to produce aqueous solutions that conduct an electric
current. Nonelectrolytes exist as molecules in aqueous solution, and such solutions do
not conduct an electric current.
Strong electrolytes are ionized or dissociated completely, or very nearly completely,
in dilute aqueous solutions. Strong electrolytes include strong acids, strong bases, and
most soluble salts. You should review the discussions of these substances in Sections 4-2
and 10-8. The common strong acids and strong bases are listed again in Table 18-1. See
Section 4-2, part 5, for the solubility guidelines for ionic compounds.
Concentrations of ions in aqueous solutions of strong electrolytes can be calculated
directly from the molarity of the strong electrolyte, as the following example illustrates.
EXAMPLE 18-1 Calculation of Concentrations of Ions
Calculate the molar concentrations of Ba2ϩ and OH Ϫ ions in 0.030 M barium hydroxide.
Plan
Write the equation for the dissociation of Ba(OH)2, and construct the reaction summary.
Ba(OH)2 is a strong base that is completely dissociated.
Solution
From the equation for the dissociation of barium hydroxide, we see that one mole of Ba(OH)2
produces one mole of Ba2ϩ ions and two moles of OHϪ ions.
(strong base) Ba(OH)2(s) 88n Ba2ϩ (aq) ϩ 2OHϪ (aq)
initial 0.030 M
change due to rxn Ϫ0.030 M ϩ0.030 M ϩ2(0.030) M
final 0 M 0.030 M 0.060 M
[Ba2ϩ] ϭ 0.030 M and [OHϪ] ϭ 0.060 M
You should now work Exercises 4 and 6.
THE AUTOIONIZATION OF WATER
Careful experiments on its electrical conductivity have shown that pure water ionizes to
a very slight extent.
H2O(ᐉ) ϩ H2O(ᐉ) 34 H3Oϩ(aq) ϩ OHϪ(aq)
Because the H2O is pure, its activity is 1, so we do not include its concentration in the
equilibrium constant expression. This equilibrium constant is known as the ion product
for water and is usually represented as Kw.
Kw ϭ [H3Oϩ][OHϪ]
18-2
18-1
Recall that we use a single arrow (n)
to indicate that a reaction goes to
completion, or nearly to completion,
in the indicated direction.
Recall that we use a double arrow (34)
to indicate that the reaction can go in
either direction to reach equilibrium.
754 CHAPTER 18: Ionic Equilibria I: Acids and Bases
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 14.1, Solutions and Their
Behavior.
TABLE 18-1 Common
Strong Acids
and Strong
Bases
Strong Acids
HCl HNO3
HBr HClO4
HI HClO3
H2SO4
Strong Bases
LiOH
NaOH
KOH Ca(OH)2
RbOH Sr(OH)2
CsOH Ba(OH)2
The formation of an H3Oϩ ion by the ionization of water is always accompanied by the
formation of an OHϪ ion. Thus, in pure water the concentration of H3Oϩ is always equal
to the concentration of OHϪ. Careful measurements show that, in pure water at 25°C,
[H3Oϩ] ϭ [OHϪ] ϭ 1.0 ϫ 10Ϫ7 mol/L
Substituting these concentrations into the Kw expression gives
Kw ϭ [H3Oϩ][OHϪ] ϭ (1.0 ϫ 10Ϫ7)(1.0 ϫ 10Ϫ7)
ϭ 1.0 ϫ 10Ϫ14 (at 25°C)
Although the expression Kw ϭ [H3Oϩ][OHϪ] ϭ 1.0 ϫ 10Ϫ14 was obtained for pure
water, it is also valid for dilute aqueous solutions at 25°C. This is one of the most useful relationships
chemists have discovered. It gives a simple relationship between H3Oϩ and OHϪ
concentrations in all dilute aqueous solutions.
The value of Kw is different at different temperatures (Table 18-2), but the relationship
Kw ϭ [H3Oϩ][OHϪ] is still valid.
In this text, we shall assume a temperature of 25°C for all calculations involving
aqueous solutions unless we specify another temperature.
EXAMPLE 18-2 Calculation of Ion Concentrations
Calculate the concentrations of H3Oϩ and OHϪ ions in a 0.050 M HNO3 solution.
Plan
Write the equation for the ionization of HNO3, a strong acid, and construct the reaction
summary, which gives the concentrations of H3Oϩ (and NO3
Ϫ) ions directly. Then use the
relationship Kw ϭ [H3Oϩ][OHϪ] ϭ 1.0 ϫ 10Ϫ14 to find the concentration of OHϪ ions.
Solution
The reaction summary for the ionization of HNO3, a strong acid, is
(strong acid) HNO3 ϩ H2O 88n H3Oϩ ϩ NO3
Ϫ
initial 0.050 M Ϸ0 M 0 M
change due to rxn Ϫ0.050 M ϩ0.050 M ϩ0.050 M
at equil 0 M 0.050 M 0.050 M
[H3Oϩ] ϭ [NO3
Ϫ] ϭ 0.050 M
The [OHϪ] is determined from the equation for the autoionization of water and its Kw.
2H2O 34 H3Oϩ(aq) ϩ OHϪ
initial 0.050 M
change due to rxn Ϫ2x M ϩx M ϩx M
at equil (0.050 ϩ x) M x M
Kw ϭ [H3Oϩ][OHϪ]
1.0 ϫ 10Ϫ14 ϭ (0.050 ϩ x)(x)
Solutions in which the concentration
of solute is less than about 1 mol/L are
usually called dilute solutions.
18-2 The Autoionization of Water 755
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 17.3, The Acid–Base Properties
of Water.
TABLE 18-2 Kw at Some
Temperatures
Temperature
(°C) Kw
0 1.1 ϫ 10Ϫ15
10 2.9 ϫ 10Ϫ15
25 1.0 ϫ 10Ϫ14
37* 2.4 ϫ 10Ϫ14
45 4.0 ϫ 10Ϫ14
60 9.6 ϫ 10Ϫ14
*Normal human body temperature.
Because the product (0.050 ϩ x)(x) is a very small number, we know that x must be very small.
Thus, it will not matter whether we add x to 0.050; we can assume that (0.050 ϩ x) Ϸ 0.050.
We substitute this approximation into the equation and solve.
1.0 ϫ 10Ϫ14 ϭ (0.050)(x) or x ϭ ϭ 2.0 ϫ 10Ϫ13 M ϭ [OHϪ]
We see that the assumption that x is much smaller than 0.050 was a good one.
You should now work Exercise 14.
In solving Example 18-2 we assumed that all of the H3Oϩ (0.050 M) came from the
ionization of HNO3 and neglected the H3Oϩ formed by the ionization of H2O. The
ionization of H2O produces only 2.0 ϫ 10Ϫ13 M H3Oϩ and 2.0 ϫ 10Ϫ13 M OHϪ in this
solution. Thus, we were justified in assuming that the [H3Oϩ] is derived solely from the
strong acid. A more concise way to carry out the calculation to find the [OHϪ] concentration
is to write directly
Kw ϭ [H3Oϩ][OHϪ] ϭ 1.0 ϫ 10Ϫ14 or [OHϪ] ϭ
Then we substitute to obtain
[OHϪ] ϭ ϭ 2.0 ϫ 10Ϫ13 M
From now on, we shall use this more direct approach for such calculations.
When nitric acid is added to water, large numbers of H3Oϩ ions are produced. The
large increase in [H3Oϩ] shifts the water equilibrium far to the left (LeChatelier’s Principle),
and the [OHϪ] decreases.
H2O(ᐉ) ϩ H2O(ᐉ) 34 H3Oϩ(aq) ϩ OHϪ(aq)
In acidic solutions the H3Oϩ concentration is always greater than the OHϪ concentration.
We should not conclude that acidic solutions contain no OHϪ ions. Rather, the
[OHϪ] is always less than 1.0 ϫ 10Ϫ7 M in such solutions. The reverse is true for basic
solutions, in which the [OHϪ] is always greater than 1.0 ϫ 10Ϫ7 M. By definition, “neutral”
aqueous solutions at 25°C are solutions in which [H3Oϩ] ϭ [OHϪ] ϭ 1.0 ϫ 10Ϫ7 M.
1.0 ϫ 10Ϫ14
ᎏᎏ
0.050
1.0 ϫ 10Ϫ14
ᎏᎏ
[H3Oϩ]
1.0 ϫ 10Ϫ14
ᎏᎏ
0.050
Recall that
[OHϪ]from H2O ϭ [H3Oϩ]from H2O
in all aqueous solutions. So we know
that [H3Oϩ]from H2O must also be
2.0 ϫ 10Ϫ13 M.
756 CHAPTER 18: Ionic Equilibria I: Acids and Bases
Solution General Condition At 25°C
acidic [H3Oϩ] Ͼ [OHϪ] [H3Oϩ] Ͼ 1.0 ϫ 10Ϫ7 [OHϪ] Ͻ 1.0 ϫ 10Ϫ7
neutral [H3Oϩ] ϭ [OHϪ] [H3Oϩ] ϭ 1.0 ϫ 10Ϫ7 [OHϪ] ϭ 1.0 ϫ 10Ϫ7
basic [H3Oϩ] Ͻ [OHϪ] [H3Oϩ] Ͻ 1.0 ϫ 10Ϫ7 [OHϪ] Ͼ 1.0 ϫ 10Ϫ7
THE pH AND pOH SCALES
The pH and pOH scales provide a convenient way to express the acidity and basicity of
dilute aqueous solutions. The pH and pOH of a solution are defined as
pH ϭ Ϫlog [H3Oϩ] or [H3Oϩ] ϭ 10ϪpH
pOH ϭ Ϫlog [OHϪ] or [OHϪ] ϭ 10ϪpOH
18-3
Note that we use pH rather than pH3O. At the time the pH concept was developed, H3Oϩ
was represented as Hϩ. Various “p” terms are used. In general a lowercase “p” before a
symbol means “negative logarithm of the symbol.” Thus, pH is the negative logarithm of
the H3Oϩ concentration, pOH is the negative logarithm of the OHϪ concentration, and
pK refers to the negative logarithm of an equilibrium constant. It is convenient to describe
the autoionization of water in terms of pKw.
pKw ϭ Ϫlog Kw
EXAMPLE 18-3 Calculation of pH
Calculate the pH of a solution in which the H3Oϩ concentration is 0.050 mol/L.
Plan
We are given the value for [H3Oϩ]; so we take the negative logarithm of this value.
Solution
[H3Oϩ] ϭ 0.050 M ϭ 5.0 ϫ 10Ϫ2 M
pH ϭ Ϫlog [H3Oϩ] ϭ Ϫlog [5.0 ϫ 10Ϫ2] ϭ 1.30
This answer contains only two significant figures. The “1” in 1.30 is not a significant figure; it
comes from the power of ten.
You should now work Exercise 22.
EXAMPLE 18-4 Calculation of H3Oϩ Concentration from pH
The pH of a solution is 3.301. What is the concentration of H3Oϩ in this solution?
Plan
By definition, pH ϭ Ϫlog [H3Oϩ]. We are given the pH, so we solve for [H3Oϩ].
When dealing with pH, we always use
the base-10 (common) logarithm, not
the base-e (natural) logarithm. This is
because pH is defined using base-10
logarithms.
In a pH value, only the digits after the
decimal point are significant figures.
18-3 The pH and pOH Scales 757
pH Range for a Few Common Substances
Substance pH Range
Gastric contents (human) 1.6–3.0
Soft drinks 2.0–4.0
Lemons 2.2–2.4
Vinegar 2.4–3.4
Tomatoes 4.0–4.4
Beer 4.0–5.0
Urine (human) 4.8–8.4
Milk (cow’s) 6.3–6.6
Saliva (human) 6.5–7.5
Blood plasma (human) 7.3–7.5
Egg white 7.6–8.0
Milk of magnesia 10.5
Household ammonia 11–12
Moreacidic
Morebasic
The pH of some common
substances is shown by a universal
indicator. Refer to Figure 18-2 to
interpret the indicator colors.
Solution
From the definition of pH, we write
Ϫlog [H3Oϩ] ϭ 3.301
Multiplying through by Ϫ1 gives
log [H3Oϩ] ϭ Ϫ3.301
Taking the inverse logarithm (antilog) of both sides of the equation gives
[H3Oϩ] ϭ 10Ϫ3.301 so [H3Oϩ] ϭ 5.00 ϫ 10Ϫ4 M
You should now work Exercise 26.
A convenient relationship between pH and pOH in all dilute solutions at 25°C can be
easily derived. We start with the Kw expression.
[H3Oϩ][OHϪ] ϭ 1.0 ϫ 10Ϫ14
Taking the logarithm of both sides of this equation gives
log [H3Oϩ] ϩ log [OHϪ] ϭ log (1.0 ϫ 10Ϫ14)
Multiplying both sides of this equation by Ϫ1 gives
(Ϫlog [H3Oϩ]) ϩ (Ϫlog [OHϪ]) ϭ Ϫlog (1.0 ϫ 10Ϫ14)
or
pH ϩ pOH ϭ 14.00
We can now relate [H3Oϩ] and [OHϪ] as well as pH and pOH.
[H3Oϩ][OHϪ] ϭ 1.0 ϫ 10Ϫ14 and pH ϩ pOH ϭ 14.00 (at 25°C)
From this relationship, we see that pH and pOH can both be positive only if both are less
than 14. If either pH or pOH is greater than 14, the other is obviously negative.
Please study carefully the following summary. It will be helpful.
At any temperature, pH ϩ pOH ϭ
pKw.
Remember these relationships!
758 CHAPTER 18: Ionic Equilibria I: Acids and Bases
Solution General Condition At 25°C
acidic [H3Oϩ] Ͼ [OHϪ] [H3Oϩ] Ͼ 1.0 ϫ 10Ϫ7 M Ͼ [OHϪ]
pH Ͻ pOH pH Ͻ 7.00 Ͻ pOH
neutral [H3Oϩ] ϭ [OHϪ] [H3Oϩ] ϭ 1.0 ϫ 10Ϫ7 M ϭ [OHϪ]
pH ϭ pOH pH ϭ 7.00 ϭ pOH
basic [H3Oϩ] Ͻ [OHϪ] [H3Oϩ] Ͻ 1.0 ϫ 10Ϫ7 M Ͻ [OHϪ]
pH Ͼ pOH pH Ͼ 7.00 Ͼ pOH
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 17.4, The pH Scale.
To develop familiarity with the pH and pOH scales, consider a series of solutions
in which [H3Oϩ] varies from 10 M to 1.0 ϫ 10Ϫ15 M. Obviously, [OHϪ] will vary from
1.0 ϫ 10Ϫ15 M to 10 M in these solutions. Table 18-3 summarizes these scales.
EXAMPLE 18-5 Calculations Involving pH and pOH
Calculate [H3Oϩ], pH, [OHϪ], and pOH for a 0.015 M HNO3 solution.
Plan
We write the equation for the ionization of the strong acid HNO3, which gives us [H3Oϩ].
Then we calculate pH. We use the relationships [H3Oϩ][OHϪ] ϭ 1.0 ϫ 10Ϫ14 and
pH ϩ pOH ϭ 14.00 to find pOH and [OHϪ].
Solution
HNO3 ϩ H2O 88n H3Oϩ ϩ NO3
Ϫ
Because nitric acid is a strong acid (it ionizes completely), we know that
[H3Oϩ] ϭ 0.015 M
pH ϭ Ϫlog [H3Oϩ] ϭ Ϫlog (0.015) ϭ Ϫ(Ϫ1.82) ϭ 1.82
We also know that pH ϩ pOH ϭ 14.00. Therefore,
pOH ϭ 14.00 Ϫ pH ϭ 14.00 Ϫ 1.82 ϭ 12.18
Because [H3Oϩ][OHϪ] ϭ 1.0 ϫ 10Ϫ14, [OHϪ] is easily calculated.
[OHϪ] ϭ ϭ ϭ 6.7 ϫ 10Ϫ13 M
You should now work Exercise 27.
1.0 ϫ 10Ϫ14
ᎏᎏ
0.015
1.0 ϫ 10Ϫ14
ᎏᎏ
[H3Oϩ]
All ions are hydrated in aqueous
solution. We often omit the
designations (ᐉ), (g), (s), (aq), and
so on.
18-3 The pH and pOH Scales 759
TABLE 18-3 Relationships Among [H3O؉], pH, pOH, and [OH؊]
[H3O؉] pH pOH [OH؊]
10Ϫ15 15 Ϫ1 101
10Ϫ14 14 0 1
10Ϫ13 13 1 10Ϫ1
10Ϫ12 12 2 10Ϫ2
10Ϫ11 11 3 10Ϫ3
10Ϫ10 10 4 10Ϫ4
10Ϫ9 9 5 10Ϫ5
10Ϫ8 8 6 10Ϫ6
10Ϫ7 7 7 10Ϫ7
10Ϫ6 6 8 10Ϫ8
10Ϫ5 5 9 10Ϫ9
10Ϫ4 4 10 10Ϫ10
10Ϫ3 3 11 10Ϫ11
10Ϫ2 2 12 10Ϫ12
10Ϫ1 1 13 10Ϫ13
1 0 14 10Ϫ14
101 Ϫ1 15 10Ϫ15
H3Oϩ
concentration
OHϪ
concentration
Neutral
IncreasingacidityIncreasingbasicity
EXAMPLE 18-6 Calculations Involving pH and pOH
Calculate [H3Oϩ], pH, [OHϪ], and pOH for a 0.015 M Ca(OH)2 solution.
Plan
We write the equation for the ionization of the strong base Ca(OH)2, which gives us [OHϪ].
Then we calculate pOH. We use the relationships [H3Oϩ][OHϪ] ϭ 1.0 ϫ 10Ϫ14 and
pH ϩ pOH ϭ 14.00 to find pH and [H3Oϩ].
Solution
H2O
Ca(OH)2 8888n Ca2ϩ ϩ 2OH Ϫ
Because calcium hydroxide is a strong base (it dissociates completely), we know that
[OHϪ] ϭ 2 ϫ 0.015 M ϭ 0.030 M
pOH ϭ Ϫlog [OHϪ] ϭ Ϫlog (0.030) ϭ Ϫ(Ϫ1.52) ϭ 1.52
We also know that pH ϩ pOH ϭ 14.00. Therefore,
pH ϭ 14.00 Ϫ pOH ϭ 14.00 Ϫ 1.52 ϭ 12.48
Because [H3Oϩ][OHϪ] ϭ 1.0 ϫ 10Ϫ14, [H3Oϩ] is easily calculated.
[H3Oϩ] ϭ ϭ ϭ 3.3 ϫ 10Ϫ13 M
You should now work Exercises 28 and 29.
The pH of a solution can be determined using a pH meter (Figure 18-1) or by the
indicator method. Acid–base indicators are intensely colored complex organic compounds
that have different colors in solutions of different pH (Section 19-4). Many are weak acids
or weak bases that are useful over rather narrow ranges of pH values. Universal indicators
are mixtures of several indicators; they show several color changes over a wide range of
pH values.
1.0 ϫ 10Ϫ14
ᎏᎏ
0.030
1.0 ϫ 10Ϫ14
ᎏᎏ
[OHϪ]
A pH meter is much more accurate
than an indicator for obtaining pH
values.
760 CHAPTER 18: Ionic Equilibria I: Acids and Bases
Figure 18-1 A pH meter gives the pH of the
solution directly. When the electrode is dipped
into a solution, the meter displays the pH. The
pH meter is based on the glass electrode. This
sensing device generates a voltage that is
proportional to the pH of the solution in which
the electrode is placed. The instrument has an
electrical circuit to amplify the voltage from the
electrode and a meter that relates the voltage to
the pH of the solution. Before being used, a pH
meter must be calibrated with a series of
solutions of known pH.
In the indicator method we prepare a series of solutions of known pH (standard solutions).
We add a universal indicator to each; solutions with different pH have different
colors (Figure 18-2). We then add the same universal indicator to the unknown solution
and compare its color to those of the standard solutions. Solutions with the same pH have
the same color.
Universal indicator papers can also be used to determine pH. A drop of solution is
placed on a piece of paper or a piece of the paper is dipped into a solution. The color of
the paper is then compared with a color chart on the container to establish the pH of the
solution.
IONIZATION CONSTANTS FOR WEAK MONOPROTIC
ACIDS AND BASES
We have discussed strong acids and strong bases. There are relatively few of these. Weak
acids are much more numerous than strong acids. For this reason you were asked to learn
the list of common strong acids (see Table 18-1). You may assume that nearly all other
acids you encounter in this text will be weak acids. Table 18-4 contains names, formulas,
ionization constants, and pKa values for a few common weak acids; Appendix F contains
a longer list of Ka values. Weak acids ionize only slightly in dilute aqueous solution. Our
classification of acids as strong or weak is based on the extent to which they ionize in dilute
aqueous solution.
Several weak acids are familiar to us. Vinegar is a 5% solution of acetic acid,
CH3COOH. Carbonated beverages are saturated solutions of carbon dioxide in water,
which produces carbonic acid.
CO2 ϩ H2O 34 H2CO3
Citrus fruits contain citric acid, C3H5O(COOH)3. Some ointments and powders used for
medicinal purposes contain boric acid, H3BO3. These everyday uses of weak acids suggest
that there is a significant difference between strong and weak acids. The difference is that
strong acids ionize completely in dilute aqueous solution, whereas weak acids ionize only slightly.
18-4
Figure 18-2 Solutions containing
a universal indicator. A universal
indicator shows a wide range of
colors as pH varies. The pH values
are given by the black numbers.
These solutions range from quite
acidic (upper left) to quite basic
(lower right).
18-4 Ionization Constants for Weak Monoprotic Acids and Bases 761
1 2 3 4
5 6 7 8
1211109
How can you tell that carbonated
beverages are saturated CO2 solutions?
Would you think of using sulfuric or
nitric acid for any of these purposes?
Let us consider the reaction that occurs when a weak acid, such as acetic acid, is dissolved
in water. The equation for the ionization of acetic acid is
CH3COOH(aq) ϩ H2O(ᐉ) 34 H3Oϩ(aq) ϩ CH3COOϪ(aq)
The equilibrium constant for this reaction could be represented as
Kc ϭ
We should recall that the thermodynamic definition of K is in terms of activities. In dilute
solutions, the activity of the (nearly) pure H2O is essentially 1. The activity of each
dissolved species is numerically equal to its molar concentration. Thus the ionization
constant of a weak acid, Ka, does not include a term for the concentration of water.
We often use HA as a general representation for a monoprotic acid and AϪ for its
conjugate base. The equation for the ionization of a weak acid can be written as
HA 34 Hϩ ϩ AϪ
For example, for acetic acid, we can write either
Ka ϭ
or
Ka ϭ ϭ 1.8 ϫ 10Ϫ5
This expression tells us that in dilute aqueous solutions of acetic acid, the concentration
of H3Oϩ multiplied by the concentration of CH3COOϪ and then divided by the concentration
of nonionized acetic acid is equal to 1.8 ϫ 10Ϫ5.
Ionization constants for weak acids (and bases) must be calculated from experimentally
determined data. Measurements of pH, conductivity, or depression of freezing point provide
data from which these constants can be calculated.
EXAMPLE 18-7 Calculation of Ka and pKa from Equilibrium
Concentrations
Nicotinic acid is a weak monoprotic organic acid that we can represent as HA.
HA ϩ H2O 34 H3Oϩ ϩ AϪ
[H3Oϩ][CH3COOϪ]
ᎏᎏᎏ
[CH3COOH]
[Hϩ][AϪ]
ᎏᎏ
[HA]
[H3Oϩ][CH3COOϪ]
ᎏᎏᎏ
[CH3COOH][H2O]
Recall that pKa means Ϫlog Ka.
762 CHAPTER 18: Ionic Equilibria I: Acids and Bases
TABLE 18-4 Ionization Constants and pKa Values for Some Weak Monoprotic
Acids
Acid Ionization Reaction Ka at 25°C pKa
hydrofluoric acid HF ϩ H2O 34 H3Oϩ ϩ FϪ 7.2 ϫ 10Ϫ4 3.14
nitrous acid HNO2 ϩ H2O 34 H3Oϩ ϩ NO2
Ϫ 4.5 ϫ 10Ϫ4 3.35
acetic acid CH3COOH ϩ H2O 34 H3Oϩ ϩ CH3COOϪ 1.8 ϫ 10Ϫ5 4.74
hypochlorous acid HOCl ϩ H2O 34 H3Oϩ ϩ OClϪ 3.5 ϫ 10Ϫ8 7.45
hydrocyanic acid HCN ϩ H2O 34 H3Oϩ ϩ CNϪ 4.0 ϫ 10Ϫ10 9.40
The thermodynamic approach is that
the activity of the (nearly) pure H2O is
essentially 1. The activity of each
dissolved species is numerically equal
to its molar concentration.
A dilute solution of nicotinic acid was found to contain the following concentrations at equilibrium
at 25°C. What is the value of Ka? [HA] ϭ 0.049 M; [H3Oϩ] ϭ [AϪ] ϭ 8.4 ϫ 10Ϫ4 M.
Plan
We are given equilibrium concentrations, and so we substitute these into the expression for Ka.
Solution
HA ϩ H2O 34 H3Oϩ ϩ AϪ Ka ϭ
Ka ϭ ϭ 1.4 ϫ 10Ϫ5
The equilibrium constant expression is
Ka ϭ ϭ 1.4 ϫ 10Ϫ5 pKa ϭ 4.85
You should now work Exercise 32.
EXAMPLE 18-8 Calculation of Ka from Percent Ionization
In 0.0100 M solution, acetic acid is 4.2% ionized. Calculate its ionization constant.
Plan
We write the equation for the ionization of acetic acid and its equilibrium constant expression.
Next we use the percent ionization to complete the reaction summary and then substitute into
the Ka expression.
Solution
The equations for the ionization of CH3COOH and its ionization constant are
CH3COOH ϩ H2O 34 H3Oϩ ϩ CH3COOϪ and Ka ϭ
Because 4.2% of the CH3COOH ionizes,
MCH3COOH that ionizes ϭ 0.042 ϫ 0.0100 M ϭ 4.2 ϫ 10Ϫ4 M
Each mole of CH3COOH that ionizes forms one mole of H3Oϩ and one mole of CH3COOϪ.
We represent this in the reaction summary.
CH3COOH ϩ H2O 34 H3Oϩ ϩ CH3COOϪ
initial 0.0100 M Ϸ0 M 0 M
change Ϫ4.2 ϫ 10Ϫ4 M ϩ4.2 ϫ 10Ϫ4 M ϩ4.2 ϫ 10Ϫ4 M
at equil 9.58 ϫ 10Ϫ3 M 4.2 ϫ 10Ϫ4 M 4.2 ϫ 10Ϫ4 M
Substitution of these values into the Ka expression gives the value for Ka.
Ka ϭ ϭ ϭ 1.8 ϫ 10Ϫ5
You should now work Exercise 38.
(4.2 ϫ 10Ϫ4)(4.2 ϫ 10Ϫ4)
ᎏᎏᎏ
9.58 ϫ 10Ϫ3
[H3Oϩ][CH3COOϪ]
ᎏᎏᎏ
[CH3COOH]
[H3Oϩ][CH3COOϪ]
ᎏᎏᎏ
[CH3COOH]
[H3Oϩ][AϪ]
ᎏᎏ
[HA]
(8.4 ϫ 10Ϫ4)(8.4 ϫ 10Ϫ4)
ᎏᎏᎏ
(0.049)
[H3Oϩ][AϪ]
ᎏᎏ
[HA]
The structure of nicotinic acid is
Nicotinic acid, also called niacin, is
a necessary vitamin in our diets. It
is not physiologically related to
nicotine.
HC
N
CH
C
C
H
O
B
O
OO HHC CO
18-4 Ionization Constants for Weak Monoprotic Acids and Bases 763
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 17.7, Determining Ka and Kb
Values.
Some common household weak
acids. A strip of paper impregnated
with a universal indicator is
convenient for estimating the
pH of a solution.
EXAMPLE 18-9 Calculation of Ka from pH
The pH of a 0.115 M solution of chloroacetic acid, ClCH2COOH, is measured to be 1.92.
Calculate Ka for this weak monoprotic acid.
Plan
For simplicity, we represent ClCH2COOH as HA. We write the ionization equation and the
expression for Ka. Next we calculate [H3Oϩ] for the given pH and complete the reaction
summary. Finally, we substitute into the Ka expression.
Solution
The ionization of this weak monoprotic acid and its ionization constant expression may be
represented as
HA ϩ H2O 34 H3Oϩ ϩ AϪ and Ka ϭ
We calculate [H3Oϩ] from the definition of pH.
pH ϭ Ϫlog [H3Oϩ]
[H3Oϩ] ϭ 10ϪpH ϭ 10Ϫ1.92 ϭ 0.012 M
We use the usual reaction summary as follows. At this point, we know the original [HA] and
the equilibrium [H3Oϩ]. From this information, we fill out the “change” line and then deduce
the other equilibrium values.
HA ϩ H2O 34 H3Oϩ ϩ AϪ
initial 0.115 M Ϸ0 M 0 M
change due to rxn Ϫ0.012 M ϩ0.012 M 0.012 M
at equil 0.103 M 0.012 M 0.012 M
Now that all concentrations are known, Ka can be calculated.
Ka ϭ ϭ ϭ 1.4 ϫ 10Ϫ3
You should now work Exercise 40.
(0.012)(0.012)
ᎏᎏ
0.103
[H3Oϩ][AϪ]
ᎏᎏ
[HA]
[H3Oϩ][AϪ]
ᎏᎏ
[HA]
764 CHAPTER 18: Ionic Equilibria I: Acids and Bases
Problem-Solving Tip: Filling in Reaction Summaries
In Examples 18-8 and 18-9 the value of an equilibrium concentration was used to determine
the change in concentration. You should become proficient at using a variety of
data to determine values that are related via a chemical equation. Let’s review what we
did in Example 18-9. Only the equilibrium expression, initial concentrations, and the
equilibrium concentration of H3Oϩ were known when we started the reaction summary.
The following steps show how we filled in the remaining values in the order indicated
by the numbered red arrows.
1. [H3Oϩ]equil ϭ 0.012 M, so we record this value
2. [H3Oϩ]initial Ϸ 0, so change in [H3Oϩ] due to rxn must be ϩ0.012 M
3. Formation of 0.012 M H3Oϩ consumes 0.012 M HA, so the change in [HA] ϭ
Ϫ0.012 M
Because ionization constants are equilibrium constants for ionization reactions, their
values indicate the extents to which weak electrolytes ionize. At the same concentrations,
acids with larger ionization constants ionize to greater extents (and are stronger acids)
than acids with smaller ionization constants. From Table 18-4, we see that the order of
decreasing acid strength for these five weak acids is
HF Ͼ HNO2 Ͼ CH3COOH Ͼ HOCl Ͼ HCN
Conversely, in Brønsted–Lowry terminology (Section 10-4), the order of increasing base
strength of the anions of these acids is
FϪ Ͻ NO2
Ϫ Ͻ CH3COOϪ Ͻ OClϪ Ͻ CNϪ
If we know the value of the ionization constant for a weak acid, we can calculate the
concentrations of the species present in solutions of known initial concentrations.
EXAMPLE 18-10 Calculation of Concentrations from Ka
Calculate the concentrations of the various species in 0.10 M hypochlorous acid, HOCl. For
HOCl, Ka ϭ 3.5 ϫ 10Ϫ8.
Plan
We write the equation for the ionization of the weak acid and its Ka expression. Then we represent
the equilibrium concentrations algebraically and substitute into the Ka expression.
Solution
The equation for the ionization of HOCl and its Ka expression are
HOCl ϩ H2O 34 H3Oϩ ϩ OClϪ and Ka ϭ ϭ 3.5 ϫ 10Ϫ8
We would like to know the concentrations of H3Oϩ, OClϪ, and nonionized HOCl in solution.
An algebraic representation of concentrations is required, because there is no other obvious
way to obtain the concentrations.
Let x ϭ mol/L of HOCl that ionizes. Then, write the “change” line and complete the reaction
summary.
HOCl ϩ H2O 34 H3Oϩ ϩOClϪ
initial 0.10 M Ϸ0 M 0 M
change due to rxn Ϫx M ϩx M ϩx M
at equil (0.10 Ϫ x) M x M x M
[H3Oϩ][OClϪ]
ᎏᎏ
[HOCl]
Recall that in Brønsted–Lowry
terminology, an acid forms its
conjugate base by losing Hϩ.
We have written the formula for
hypochlorous acid as HOCl rather
than HClO to emphasize that its
structure is HXOXCl.
We neglect the 1.0 ϫ 10Ϫ7 mol/L of
H3Oϩ produced by the ionization of
pure water. Recall (see Section 18-2)
that the addition of an acid to water
suppresses the ionization of H2O, so
[H3Oϩ] from H2O is even less than
1.0 ϫ 10Ϫ7 M.
18-4 Ionization Constants for Weak Monoprotic Acids and Bases 765
HA H3Oϩ AϪH2O
initial
change due to rxn
at equil 0.103 M 0.012 M
Ϫ0.012 M
0.115 M
ϩ 0.012 M
0.012 M
0.012 M
At equilibrium, [H3Oϩ] ϭ 0.012 M so
≈ 0 M 0 M
ϩ ϩ
1
6
53
24
4. [HA]equil ϭ [HA]orig ϩ [HA]chg ϭ 0.115 M ϩ (Ϫ0.012 M) ϭ 0.103 M
5. Formation of 0.012 M H3Oϩ also gives 0.012 M AϪ
6. [AϪ]equil ϭ [AϪ]orig ϩ [AϪ]chg ϭ 0 M ϩ 0.012 M ϭ 0.012 M
Substituting these algebraic representations into the Ka expression gives
Ka ϭ ϭ ϭ 3.5 ϫ 10Ϫ8
This is a quadratic equation, but it is not necessary to solve it by the quadratic formula. The
small value of the equilibrium constant, Ka, tells us that not very much of the original acid
ionizes. Thus we can assume that x ϽϽ 0.10. If x is small enough compared with 0.10, it will
not matter (much) whether we subtract it, and we can assume that (0.10 Ϫ x) is very nearly
equal to 0.10. The equation then becomes
ᎏ
0
x
.1
2
0
ᎏ Ϸ 3.5 ϫ 10Ϫ8 x2 Ϸ 3.5 ϫ 10Ϫ9 so x Ϸ 5.9 ϫ 10Ϫ5
In our algebraic representation we let
[H3Oϩ] ϭ x M ϭ 5.9 ϫ 10Ϫ5 M [OClϪ] ϭ x M ϭ 5.9 ϫ 10Ϫ5 M
[HOCl] ϭ (0.10 Ϫ x) M ϭ (0.10 Ϫ 0.000059) M ϭ 0.10 M
[OHϪ] ϭ ϭ ϭ 1.7 ϫ 10Ϫ10 M
You should now work Exercise 42.
1.0 ϫ 10Ϫ14
ᎏᎏ
5.9 ϫ 10Ϫ5
Kw
ᎏ
[H3Oϩ]
(x)(x)
ᎏᎏ
(0.10 Ϫ x)
[H3Oϩ][OClϪ]
ᎏᎏ
[HOCl]
766 CHAPTER 18: Ionic Equilibria I: Acids and Bases
Problem-Solving Tip: Simplifying Quadratic Equations
We often encounter quadratic or higher-order equations in equilibrium calculations.
With modern programmable calculators, solving such problems by iterative methods is
often feasible. But frequently a problem can be made much simpler by using some mathematical
common sense.
When the linear variable (x) in a quadratic equation is added to or subtracted from a
much larger number, it can often be disregarded if it is sufficiently small. A reasonable
rule of thumb for determining whether the variable can be disregarded in equilibrium
calculations is this: If the exponent of 10 in the K value is Ϫ3 or less (Ϫ4, Ϫ5, Ϫ6, etc.),
then the variable may be small enough to disregard when it is added to or subtracted
from a number greater than 0.05. Solve the problem neglecting x; then compare the
value of x with the number it would have been added to (or subtracted from). If x is
more than 5% of that number, the assumption was not justified, and you should solve
the equation using the quadratic formula.
Let’s examine the assumption as it applies to Example 18-10. Our quadratic equation
is
ᎏ
(0.
(
1
x
0
)(
Ϫ
x)
x)
ᎏ ϭ 3.5 ϫ 10Ϫ8
Because 3.5 ϫ 10Ϫ8 is a very small Ka value, we know that the acid ionizes only slightly.
Thus, x must be very small compared with 0.10, so we can write (0.10 Ϫ x) Ϸ 0.10.
The equation then becomes ᎏ
0
x
.1
2
0
ᎏ Ϸ 3.5 ϫ 10Ϫ8. To solve this, we rearrange and take the
From our calculations we can draw some conclusions. In a solution containing only a
weak monoprotic acid, the concentration of H3Oϩ is equal to the concentration of the anion
of the acid. Unless the solution is very dilute, such as less than 0.050 M, the concentration
of nonionized acid is approximately equal to the molarity of the solution. When the value
of Ka for the weak acid is greater than Ϸ10Ϫ3, then the extent of ionization will be large
enough to make a significant difference between the concentration of nonionized acid and
the molarity of the solution. In such cases we cannot make the simplifying assumption.
EXAMPLE 18-11 Percent Ionization
Calculate the percent ionization of a 0.10 M solution of acetic acid.
Plan
Write the ionization equation and the expression for Ka. Next, follow the procedure used in
Example 18-10 to find the concentration of acid that ionized. Then, substitute the concentration
of acid that ionized into the expression for percent ionization. Percentage is defined as
(part/whole) ϫ 100%, so the percent ionization is
% ionization ϭ ϫ 100%
Solution
The equations for the ionization of CH3COOH and its Ka are
CH3COOH ϩ H2O 34 H3Oϩ ϩ CH3COOϪ Ka ϭ ϭ 1.8 ϫ 10Ϫ5
We proceed as we did in Example 18-10. Let x ϭ [CH3COOH]ionized.
CH3COOH ϩ H2O 34 H3Oϩ ϩ CH3COOϪ
initial 0.10 M Ϸ0 M 0 M
change due to rxn Ϫx M ϩx M ϩx M
at equil (0.10 Ϫ x) M x M x M
Substituting into the ionization constant expression gives
Ka ϭ ϭ ϭ 1.8 ϫ 10Ϫ5
If we make the simplifying assumption that (0.10 Ϫ x) Ϸ 0.10, we have
ᎏ
0
x
.1
2
0
ᎏ ϭ 1.8 ϫ 10Ϫ5 x2 ϭ 1.8 ϫ 10Ϫ6 x ϭ 1.3 ϫ 10Ϫ3
(x)(x)
ᎏᎏ
(0.10 Ϫ x)
[H3Oϩ][AϪ]
ᎏᎏ
[HA]
[H3Oϩ][AϪ]
ᎏᎏ
[HA]
[CH3COOH]ionized
ᎏᎏᎏ
[CH3COOH]initial
We could write the original [H3Oϩ] as
1.0 ϫ 10Ϫ7 M. In very dilute solutions
of weak acids, we might have to take
this into account. In this acid solution,
(1.0 ϫ 10Ϫ7 ϩ x) Ϸ x.
18-4 Ionization Constants for Weak Monoprotic Acids and Bases 767
square roots of both sides. To check, we see that the result, x ϭ 5.9 ϫ 10Ϫ5, is only
0.059% of 0.10. This error is much less than 5%, so our assumption is justified. You
may also wish to use the quadratic formula to verify that the answer obtained this way
is correct to within roundoff error.
The preceding argument is purely algebraic. We could use our chemical intuition to
reach the same conclusion. A small Ka value (10Ϫ3 or less) tells us that the extent of
ionization is very small; therefore, nearly all of the weak acid exists as nonionized molecules.
The amount that ionizes is insignificant compared with the concentration of
nonionized weak acid.
The pH of a soft drink is measured
with a modern pH meter. Many soft
drinks are quite acidic due to the
dissolved CO2 and other ingredients.
This gives [CH3COOH]ionized ϭ x ϭ 1.3 ϫ 10Ϫ3 M. Now we can calculate the percent ionization
for 0.10 M CH3COOH solution.
% ionization ϭ ϫ 100% ϭ ϫ 100% ϭ 1.3%
Our assumption that (0.10 Ϫ x) Ϸ 0.10 is reasonable because (0.10 Ϫ x) ϭ (0.10 Ϫ 0.0013). This
is only about 1% different than 0.10. When Ka for a weak acid is significantly greater than
10Ϫ3, however, this assumption would introduce considerable error.
You should now work Exercise 48.
In dilute solutions, acetic acid exists primarily as nonionized molecules, as do all weak
acids; there are relatively few hydronium and acetate ions. In 0.10 M solution, CH3COOH
is 1.3% ionized; for each 1000 molecules of CH3COOH originally placed in the solution,
there are 13 H3Oϩ ions, 13 CH3COOϪ ions, and 987 nonionized CH3COOH molecules.
For weaker acids of the same concentration, the number of molecules of nonionized acid
would be even larger.
By now we should have gained some “feel” for the strength of an acid by looking
at its Ka value. Consider 0.10 M solutions of HCl (a strong acid), CH3COOH (see
Example 18-11), and HOCl (see Example 18-10). If we calculate the percent ionization
for 0.10 M HOCl (as we did for 0.10 M CH3COOH in Example 18-11), we find
that it is 0.059% ionized. In 0.10 M solution, HCl is very nearly completely ionized.
The data in Table 18-5 show that the [H3Oϩ] in 0.10 M HCl is approximately 77 times
greater than that in 0.10 M CH3COOH and approximately 1700 times greater than
that in 0.10 M HOCl.
Many scientists prefer to use pKa values rather than Ka values for weak acids. Recall
that in general, “p” terms refer to negative logarithms. The pKa value for a weak acid is
just the negative logarithm of its Ka value.
EXAMPLE 18-12 pKa Values
The Ka values for acetic acid and hydrofluoric acid are 1.8 ϫ 10Ϫ5 and 7.2 ϫ 10Ϫ4, respectively.
What are their pKa values?
Plan
pKa is defined as the negative logarithm of Ka (i.e., pKa ϭ Ϫlog Ka) so we take the negative
logarithm of each Ka.
Solution
For CH3COOH,
pKa ϭ Ϫlog Ka ϭ Ϫlog (1.8 ϫ 10Ϫ5) ϭ Ϫ(Ϫ4.74) ϭ 4.74
For HF,
pKa ϭ Ϫlog Ka ϭ Ϫlog (7.2 ϫ 10Ϫ4) ϭ Ϫ(Ϫ3.14) ϭ 3.14
You should now work Exercise 50.
1.3 ϫ 10Ϫ3 M
ᎏᎏ
0.10 M
[CH3COOH]ionized
ᎏᎏᎏ
[CH3COOH]initial
Note that we need not solve explicitly
for the equilibrium concentrations
[H3Oϩ] and [CH3COOϪ] to answer
the question. From the setup, we see
that these are both 1.3 ϫ 10Ϫ3 M.
768 CHAPTER 18: Ionic Equilibria I: Acids and Bases
From Example 18-12, we see that the stronger acid (HF in this case) has the larger Ka
value and the smaller pKa value. Conversely, the weaker acid (CH3COOH in this case)
has the smaller Ka value and the larger pKa value. The generalization is
The larger the value of Ka, the smaller is the value of pKa, and the stronger is the
acid.
EXAMPLE 18-13 Acid Strengths and Ka Values
Given the following list of weak acids and their Ka values, arrange the acids in order of (a)
increasing acid strength and (b) increasing pKa values.
Acid Ka
HOCl 3.5 ϫ 10Ϫ8
HCN 4.0 ϫ 10Ϫ10
HNO2 4.5 ϫ 10Ϫ4
Plan
(a) We see that HNO2 is the strongest acid in this group because it has the largest Ka value.
HCN is the weakest because it has the smallest Ka value.
(b) We do not need to calculate pKa values to answer the question. We recall that the weakest
acid has the largest pKa value and the strongest acid has the smallest pKa value, so the order
of increasing pKa values is just the reverse of the order in part (a).
Solution
(a) Increasing acid strength: HCN Ͻ HOCl Ͻ HNO2
(b) Increasing pKa values: HNO2 Ͻ HOCl Ͻ HCN
You should now work Exercise 111.
You may know that hydrofluoric acid dissolves glass. But HF is not a strong acid. The
reaction of glass with hydrofluoric acid occurs because silicates react with HF to produce
silicon tetrafluoride, SiF4, a very volatile compound. This reaction tells us nothing about
the acid strength of hydrofluoric acid.
A similar statement is true for weak
bases; that is, a stronger base has the
greater Kb value and the smaller pKb
value.
TABLE 18-5 Comparison of Extents of Ionization of Some Acids
Ionization Percent
Acid Solution Constant [H3Oϩ] pH Ionization
0.10 M HCl very large 0.10 M 1.00 Ϸ100
0.10 M CH3COOH 1.8 ϫ 10Ϫ5 0.0013 M 2.89 1.3
0.10 M HOCl 3.5 ϫ 10Ϫ8 0.000059 M 4.23 0.059
Do not confuse the strength of an acid
with its reactivity. Acid strength refers
to the extent of ionization of the acid,
and not the reactions that it undergoes.
An inert solid has been suspended in
the liquid to improve the quality of
this photograph of a pH meter.
Thus far we have focused our attention on acids. Very few common weak bases are
soluble in water. Aqueous ammonia is the most frequently encountered example. From
our earlier discussion of bonding in covalent compounds (Section 8-8), we recall that there
is one unshared pair of electrons on the nitrogen atom in NH3. When ammonia dissolves
in water, it accepts Hϩ from a water molecule in a reversible reaction (Section 10-4). We
say that NH3 ionizes slightly when it undergoes this reaction. Aqueous solutions of NH3
are basic because OHϪ ions are produced.
Amines are derivatives of NH3 in which one or more H atoms have been replaced by
organic groups, as the following structures indicate.
Thousands of amines are known, and many are very important in biochemical processes.
Low-molecular-weight amines are soluble weak bases. The ionization of trimethylamine,
for example, forms trimethylammonium ions and OHϪ ions.
Now let us consider the behavior of ammonia in aqueous solutions. The reaction of
ammonia with water and its ionization constant expression are
NH3 ϩ H2O 34 NH4
ϩ ϩ OHϪ
and
Kb ϭ ϭ 1.8 ϫ 10Ϫ5
The fact that Kb for aqueous NH3 has the same value as Ka for CH3COOH is pure coincidence.
It does tell us that in aqueous solutions of the same concentration, CH3COOH
and NH3 are ionized to the same extent. Table 18-6 lists Kb and pKb values for a few
common weak bases. Appendix G includes a longer list of Kb values.
We use Kb’s for weak bases in the same way we used Ka’s for weak acids and pKb values
for weak bases in the same way we used pKa values for weak acids.
EXAMPLE 18-14 pH of a Weak Base Solution
Calculate the [OHϪ], pH, and percent ionization for a 0.20 M aqueous NH3 solution.
Plan
Write the equation for the ionization of aqueous NH3 and represent the equilibrium concentrations
algebraically. Then, substitute into the Kb expression and solve for [OHϪ] and
[NH3]ionized.
[NH4
ϩ][OHϪ]
ᎏᎏ
[NH3]
H OH
H
ϩ OHϪϩ
H
ϩ transfer
ϩ
CH3CH3
CH3
N
trimethylamine
(CH3)3N
CH3 CH3
CH3
N
trimethylammonium ion
(CH3)3NHϩ
H H
H
N
ammonia
NH3
H3C H
H
N
methylamine
CH3NH2
H3C H
CH3
N
dimethylamine
(CH3)2NH
H3C CH3
CH3
N
trimethylamine
(CH3)3N
NH3 ϩ H2O NH4
ϩ ϩ OHϪ
The subscript “b” indicates that the
substance ionizes as a base. We do not
include [H2O] in the Kb expression for
the same reasons described for Ka.
770 CHAPTER 18: Ionic Equilibria I: Acids and Bases
Trimethylamine, (CH3)3N
Ammonia, NH3
Solution
The equation for the ionization of aqueous ammonia and the algebraic representations of equilibrium
concentrations follow. Let x ϭ [NH3]ionized.
NH3 ϩ H2O 34 NH4
ϩ ϩ OHϪ
initial 0.20 M 0 M Ϸ0 M
change due to rxn Ϫx M ϩx M ϩx M
at equil (0.20 Ϫ x) M x M x M
Substitution into the ionization constant expression gives
Kb ϭ ϭ 1.8 ϫ 10 Ϫ 5 ϭ
Again, we can simplify this equation. The small value of Kb tells us that the base is only slightly
ionized, so we can assume that x ϽϽ 0.20, or (0.20 Ϫ x) Ϸ 0.20, and we have
ᎏ
0
x
.2
2
0
ᎏ ϭ 1.8 ϫ 10Ϫ5 x2 ϭ 3.6 ϫ 10Ϫ6 x ϭ 1.9 ϫ 10Ϫ3 M
Then [OHϪ] ϭ x ϭ 1.9 ϫ 10Ϫ3 M, pOH ϭ 2.72, and pH ϭ 11.28.
[NH3]ionized ϭ x, so the percent ionization may be calculated.
% ionization ϭ ϫ 100% ϭ ϫ 100% ϭ 0.95% ionized
You should now work Exercises 54 and 110.
EXAMPLE 18-15 Household Ammonia
The pH of a household ammonia solution is 11.50. What is its molarity?
Plan
We are given the pH of an aqueous NH3 solution. Use pH ϩ pOH ϭ 14.00 to find pOH,
which we can convert to [OHϪ]. Then, complete the ionization reaction summary, and substitute
the representations of equilibrium concentrations into the Kb expression.
Solution
At equilibrium pH ϭ 11.50; we know that pOH ϭ 2.50, so [OHϪ] ϭ 10Ϫ2.50 ϭ 3.2 ϫ 10Ϫ3 M.
This [OHϪ] results from the reaction, so we can write the change line. Then, letting x represent
the initial concentration of NH3, we can complete the reaction summary.
1.9 ϫ 10Ϫ3
ᎏᎏ
0.20
[NH3]ionized
ᎏᎏ
[NH3]initial
(x)(x)
ᎏᎏ
(0.20 Ϫ x)
[NH4
ϩ][OHϪ]
ᎏᎏ
[NH3]
The value of x is only about 1% of
the original concentration, so the
assumption is justified.
18-4 Ionization Constants for Weak Monoprotic Acids and Bases 771
TABLE 18-6 Ionization Constants and pKb Values for Some Weak Bases
Base Ionization Reaction Kb at 25°C pKb
ammonia NH3 ϩ H2O 34 NH4
ϩ ϩ OHϪ 1.8 ϫ 10Ϫ5 4.74
methylamine (CH3)NH2 ϩ H2O 34 (CH3)NH3
ϩ ϩ OHϪ 5.0 ϫ 10Ϫ4 3.30
dimethylamine (CH3)2NH ϩ H2O 34 (CH3)2NH2
ϩ ϩ OHϪ 7.4 ϫ 10Ϫ4 3.13
trimethylamine (CH3)3N ϩ H2O 34 (CH3)3NHϩ ϩ OHϪ 7.4 ϫ 10Ϫ5 4.13
pyridine C5H5N ϩ H2O 34 C5H5NHϩ ϩ OHϪ 1.5 ϫ 10Ϫ9 8.82
aniline C6H5NH2 ϩ H2O 34 C6H5NH3
ϩ ϩ OHϪ 4.2 ϫ 10Ϫ10 9.38
Measurement of the pH of a
solution of household ammonia.
NH2C
C
C C
C C
H
H H
H
H
aniline
C
C C
C C
H
H H
HN
H
pyridine
Substituting these values into the Kb expression for aqueous NH3 gives
Kb ϭ ϭ ϭ 1.8 ϫ 10Ϫ5
This suggests that (x Ϫ 3.2 ϫ 10Ϫ3) Ϸ x. So we can approximate.
ϭ 1.8 ϫ 10Ϫ5 and x ϭ 0.57 M NH3
The solution is 0.57 M NH3. Our assumption that (x Ϫ 3.2 ϫ 10Ϫ3) Ϸ x is justified.
You should now work Exercises 54, 56, and 58.
(3.2 ϫ 10Ϫ3)(3.2 ϫ 10Ϫ3)
ᎏᎏᎏ
x
(3.2 ϫ 10Ϫ3)(3.2 ϫ 10Ϫ3)
ᎏᎏᎏ
(x Ϫ 3.2 ϫ 10Ϫ3)
[NH4
ϩ][OHϪ]
ᎏᎏ
[NH3]
NH3 NH4
ϩ OHϪH2O
initial
change
at equil (x Ϫ 3.2 ϫ 10Ϫ3) M 3.2 ϫ 10Ϫ3 M
Ϫ3.2 ϫ 10Ϫ3 M ϩ3.2 ϫ 10Ϫ3 M
3.2 ϫ 10Ϫ3 M
ϩ3.2 ϫ 10Ϫ3 M
At equilibrium [OHϪ] ϭ 3.2 ϫ 10Ϫ3 M, so
x M 0 M ≈0 M
ϩ ϩ
1
2
34
56
Each Ka expression includes [H3Oϩ],
so each Ka expression must be satisfied
by the total concentration of H3Oϩ in
the solution.
772 CHAPTER 18: Ionic Equilibria I: Acids and Bases
Problem-Solving Tip: It Is Not Always x that Can Be Neglected
Students sometimes wonder about the approximation in Example 18-15, thinking that
only x can be neglected. We can consider neglecting one term in an expression only
when the expression involves addition or subtraction. The judgment we must make is
whether either of the terms is sufficiently smaller than the other that ignoring it would
not significantly affect the result. In Example 18-15, x represents the initial concentration
of NH3; 3.2 ϫ 10Ϫ3 represents the concentration that ionizes, which cannot be
greater than the original x. We know that NH3 is a weak base (Kb ϭ 1.8 ϫ 10Ϫ5), so only
a small amount of the original ionizes. We can safely assume that 3.2 ϫ 10Ϫ3 ϽϽ x, so
(x Ϫ 3.2 ϫ 10Ϫ3) Ϸ x, the approximation that we used in solving the example.
Remember that we can never neglect a term in multiplication or division.
POLYPROTIC ACIDS
Thus far we have considered only monoprotic weak acids. Acids that can furnish two or
more hydronium ions per molecule are called polyprotic acids. The ionizations of
polyprotic acids occur stepwise, that is, one proton at a time. An ionization constant expression
can be written for each step, as the following example illustrates. Consider phosphoric
acid as a typical polyprotic acid. It contains three acidic hydrogen atoms and ionizes in
three steps.
H3PO4 ϩ H2O 34 H3Oϩ ϩ H2PO4
Ϫ Ka1 ϭ ϭ 7.5 ϫ 10Ϫ3
H2PO4
Ϫ ϩ H2O 34 H3Oϩ ϩ HPO4
2Ϫ Ka2 ϭ ϭ 6.2 ϫ 10Ϫ8
HPO4
2Ϫ ϩ H2O 34 H3Oϩ ϩ PO4
3Ϫ Ka3 ϭ ϭ 3.6 ϫ 10Ϫ13
[H3Oϩ][PO4
3Ϫ]
ᎏᎏ
[HPO4
2Ϫ]
[H3Oϩ][HPO4
2Ϫ]
ᎏᎏ
[H2PO4
Ϫ]
[H3Oϩ][H2PO4
Ϫ]
ᎏᎏ
[H3PO4]
18-5
We see that Ka1 is much greater than Ka2 and that Ka2 is much greater than Ka3. This is
generally true for polyprotic inorganic acids (Appendix F). Successive ionization constants
often decrease by a factor of approximately 104 to 106, although some differences are
outside this range. Large decreases in the values of successive ionization constants mean
that each step in the ionization of a polyprotic acid occurs to a much lesser extent than
the previous step. Thus, the [H3Oϩ] produced in the first step is very large compared
with the [H3Oϩ] produced in the second and third steps. As we shall see, except in
extremely dilute solutions of H3PO4, the concentration of H3Oϩ may be assumed to be
that furnished by the first step in the ionization alone.
EXAMPLE 18-16 Solutions of Weak Polyprotic Acid
Calculate the concentrations of all species present in 0.100 M H3PO4.
Plan
Because H3PO4 contains three acidic hydrogens per formula unit, we show its ionization in
three steps. For each step, write the appropriate ionization equation, with its Ka expression and
value. Then, represent the equilibrium concentrations from the first ionization step, and substitute
into the Ka1 expression. Repeat the procedure for the second and third steps in order.
Solution
First we calculate the concentrations of all species that are formed in the first ionization step.
Let x ϭ mol/L of H3PO4 that ionize; then x ϭ [H3Oϩ]1st ϭ [H2PO4
Ϫ].
H3PO4 ϩ H2O 34 H3Oϩ ϩ H2PO4
Ϫ
(0.100 Ϫ x) M x M x M
Substitution into the expression for Ka1 gives
Ka1 ϭ ϭ 7.5 ϫ 10Ϫ3 ϭ
This equation must be solved by the quadratic formula because Ka1 is too large to neglect x
relative to 0.100 M. Solving gives the positive root x ϭ 2.4 ϫ 10Ϫ2. Thus, from the first step in
the ionization of H3PO4,
x M ϭ [H3Oϩ]1st ϭ [H2PO4
Ϫ] ϭ 2.4 ϫ 10Ϫ2 M
(0.100 Ϫ x) M ϭ [H3PO4] ϭ 7.6 ϫ 10Ϫ2 M
For the second step, we use the [H3Oϩ] and [H2PO4
Ϫ] from the first step. Let y ϭ mol/L
of H2PO4
2Ϫ that ionize; then y ϭ [H3Oϩ]2nd ϭ [HPO4
2Ϫ].
H2PO4
Ϫ ϩ H2O 34 H3Oϩ ϩ HPO4
2Ϫ
(2.4 ϫ 10Ϫ2 Ϫ y) M (2.4 ϫ 10Ϫ2 ϩ y)M y M
from 1st step from 2nd step
Substitution into the expression for Ka2 gives
Ka2 ϭ ϭ 6.2 ϫ 10Ϫ8 ϭ
Examination of this equation suggests that y ϽϽ 2.4 ϫ 10Ϫ2, so
ϭ 6.2 ϫ 10Ϫ8 y ϭ 6.2 ϫ 10Ϫ8 M ϭ [HPO4
2Ϫ] ϭ [H3Oϩ]2nd
(2.4 ϫ 10Ϫ2)(y)
ᎏᎏ
(2.4 ϫ 10Ϫ2)
(2.4 ϫ 10 Ϫ 2 ϩ y)(y)
ᎏᎏᎏ
(2.4 ϫ 10 Ϫ 2 Ϫ y)
[H3Oϩ][HPO4
2Ϫ]
ᎏᎏ
[H2PO4
Ϫ]
(x)(x)
ᎏᎏ
(0.100 Ϫ x)
[H3Oϩ][H2PO4
Ϫ]
ᎏᎏ
[H3PO4]
x ϭ Ϫ3.1 ϫ 10Ϫ2 is the extraneous
root of the quadratic equation.
18-5 Polyprotic Acids 773
We see that [HPO4
2Ϫ] ϭ Ka2 and [H3Oϩ]2nd ϽϽ [H3Oϩ]1st. In general, in solutions of
reasonable concentration of weak polyprotic acids for which Ka1 ϾϾ Ka2 and that contain no
other electrolytes, the concentration of the anion produced in the second ionization step is always equal
to Ka2.
For the third step, we use [H3Oϩ] from the first step and [HPO4
2Ϫ] from the second step.
Let z ϭ mol/L of HPO4
2Ϫ that ionize; then z ϭ [H3Oϩ]3rd ϭ [PO4
3Ϫ].
HPO4
2Ϫ ϩ H2O 34 H3Oϩ ϩ PO4
3Ϫ
(6.2 ϫ 10Ϫ8 Ϫ z) M (2.4 ϫ 10Ϫ2 ϩ z) M z M
from 2nd step from 3rd step from 1st step from 3rd step
Ka3 ϭ ϭ 3.6 ϫ 10Ϫ13 ϭ
We make the usual simplifying assumption, and find that
z M ϭ 9.3 ϫ 10Ϫ19 M ϭ [PO4
3Ϫ] ϭ [H3Oϩ]3rd
The [H3Oϩ] found in the three steps can be summarized:
[H3Oϩ]from 1st step ϭ 2.4 ϫ 10Ϫ2 M ϭ 0.024 M
[H3Oϩ]from 2nd step ϭ 6.2 ϫ 10Ϫ8 M ϭ 0.000000062 M
[H3Oϩ]from 3rd step ϭ 9.3 ϫ 10Ϫ19 M ϭ 0.00000000000000000093 M
[H3Oϩ]total ϭ 2.4 ϫ 10Ϫ2 M ϭ 0.024 M
We see that the H3Oϩ furnished by the second and third steps of ionization is negligible
compared with that from the first step.
You should now work Exercise 60.
We have calculated the concentrations of the species formed by the ionization of
0.100 M H3PO4. These concentrations are compared in Table 18-7. The concentration
of [OHϪ] in 0.100 M H3PO4 is included. It was calculated from the known [H3Oϩ] using
the ion product for water, [H3Oϩ][OHϪ] ϭ 1.0 ϫ 10Ϫ14.
Nonionized H3PO4 is present in greater concentration than any other species in
0.100 M H3PO4 solution. The only other species present in significant concentrations are
H3Oϩ and H2PO4
Ϫ. Similar statements can be made for other weak polyprotic acids for
which the last K is very small.
(2.4 ϫ 10Ϫ2 ϩ z)(z)
ᎏᎏᎏ
(6.2 ϫ 10Ϫ8 Ϫ z)
[H3Oϩ][PO4
3Ϫ]
ᎏᎏ
[HPO4
2Ϫ]
The pH of solutions of most
polyprotic acids is governed by
the first ionization step.
[H3Oϩ]2nd ϭ y ϭ 6.2 ϫ 10Ϫ8 was
disregarded in the second step and is
also disregarded here.
774 CHAPTER 18: Ionic Equilibria I: Acids and Bases
TABLE 18-7 Concentrations of the Species
in 0.10 M H3PO4
(Example 18-16)
Species Concentration (mol/L)
H3PO4 7.6 ϫ 10Ϫ22 ϭ 0.076
H3Oϩ 2.4 ϫ 10Ϫ22 ϭ 0.024
H2PO4
Ϫ 2.4 ϫ 10Ϫ22 ϭ 0.024
HPO4
2Ϫ 6.2 ϫ 10Ϫ82 ϭ 0.000000062
OHϪ 4.2 ϫ 10Ϫ13 ϭ 0.00000000000042
PO4
3Ϫ 9.3 ϫ 10Ϫ19 ϭ 0.00000000000000000093
Phosphoric acid is a typical weak polyprotic acid. Let us now describe solutions of
sulfuric acid, a very strong polyprotic acid.
EXAMPLE 18-17 Solutions of Strong Polyprotic Acid
Calculate concentrations of all species present in 0.10 M H2SO4. Ka2 ϭ 1.2 ϫ 10Ϫ2.
Plan
Because the first ionization step of H2SO4 is complete, we read the concentrations for the first
step from the balanced equation. The second ionization step is not complete, and so we write
the ionization equation, the Ka2 expression, and the algebraic representations of equilibrium
concentrations. Then we substitute into Ka2 for H2SO4.
Solution
As we pointed out, the first ionization step of H2SO4 is complete.
100%
H2SO4 ϩ H2O 8888n H3Oϩ ϩ HSO4
Ϫ
0.10 M ::::::F 0.10 M 0.10 M
The second ionization step is not complete, however.
HSO4
Ϫ ϩ H2O 34 H3Oϩ ϩ SO4
2Ϫ and Ka2 ϭ ϭ 1.2 ϫ 10Ϫ2
Let x ϭ [HSO4
Ϫ] that ionizes. [H3Oϩ] is the sum of the concentrations produced in the first
and second steps. So we represent the equilibrium concentrations as
HSO4
Ϫ ϩ H2O 34 H3Oϩ ϩ SO4
2Ϫ
(0.10 Ϫ x) M (0.10 ϩ x) M x M
from 1st step from 2nd step
Substitution into the ionization constant expression for Ka2 gives
Ka2 ϭ ϭ 1.2 ϫ 10Ϫ2 ϭ
Clearly, x cannot be disregarded because Ka2 is too large. This equation must be solved by the
quadratic formula, which gives x ϭ 0.010 and x ϭ Ϫ0.12 (extraneous). So [H3Oϩ]2nd ϭ
[SO4
2Ϫ] ϭ 0.010 M. The concentrations of species in 0.10 M H2SO4 are
[H2SO4] Ϸ 0 M [HSO4
Ϫ] ϭ (0.10 Ϫ x) M ϭ 0.09 M [SO4
2Ϫ] ϭ 0.010 M
[H3Oϩ] ϭ (0.10 ϩ x) M ϭ 0.11 M
[OHϪ] ϭ ϭ ϭ 9.1 ϫ 10Ϫ14 M
In 0.10 M H2SO4 solution, the extent of the second ionization step is 10%.
You should now work Exercise 62.
In Table 18-8 we compare 0.10 M solutions of these two polyprotic acids. Their acidities
are very different.
1.0 ϫ 10Ϫ14
ᎏᎏ
0.11
Kw
ᎏ
[H3Oϩ]
(0.10 ϩ x)(x)
ᎏᎏ
0.10 Ϫ x
[H3Oϩ][SO4
2Ϫ]
ᎏᎏ
[HSO4
Ϫ]
[H3Oϩ][SO4
2Ϫ]
ᎏᎏ
[HSO4
Ϫ]
18-5 Polyprotic Acids 775
SOLVOLYSIS
Solvolysis is the reaction of a substance with the solvent in which it is dissolved. The
solvolysis reactions that we will consider in this chapter occur in aqueous solutions so they
are called hydrolysis reactions. Hydrolysis is the reaction of a substance with water. Some
hydrolysis reactions involve reaction with H3Oϩ or OHϪ ions. One common kind of
hydrolysis involves reaction of the anion of a weak acid with water to form nonionized acid
molecules and OHϪ ions. This upsets the H3Oϩ/OHϪ balance in water and produces
basic solutions. This reaction is usually represented as
AϪ ϩ H2O 34 HA ϩ OHϪ (excess OHϪ, so solution is basic)
anion of weak acid
weak acid
Recall that in
neutral solutions [H3Oϩ] ϭ [OHϪ] ϭ 1.0 ϫ 10Ϫ7 M
basic solutions [H3Oϩ] Ͻ [OHϪ] or [OHϪ] Ͼ 1.0 ϫ 10Ϫ7 M
acidic solutions [H3Oϩ] Ͼ [OHϪ] or [H3Oϩ] Ͼ 1.0 ϫ 10Ϫ7 M
In Brønsted–Lowry terminology anions of strong acids are extremely weak bases,
whereas anions of weak acids are stronger bases (Section 10-4). To refresh your memory,
consider the following examples.
Nitric acid, a common strong acid, is essentially completely ionized in dilute aqueous
solution. Dilute aqueous solutions of HNO3 contain equal concentrations of H3Oϩ and
NO3
Ϫ ions. In dilute aqueous solution nitrate ions show almost no tendency to react with
H3Oϩ ions to form nonionized HNO3; thus, NO3
Ϫ is a very weak base.
100%
HNO3 ϩ H2O 8888n H3Oϩ ϩ NO3
Ϫ
On the other hand, acetic acid (a weak acid) is only slightly ionized in dilute aqueous
solution. Acetate ions have a strong tendency to react with H3Oϩ to form CH3COOH
molecules. Acetic acid ionizes only slightly.
18-6
We often omit (aq) from molecules
and ions in aqueous solutions to save
space and time.
TABLE 18-8 Comparison of 0.10 M
Solutions of Two Polyprotic
Acids (Examples 18-16, 18-17)
0.10 M 0.10 M
H3PO4 H2SO4
Ka1 7.5 ϫ 10Ϫ3 Ϸvery large
Ka2 6.2 ϫ 10Ϫ8 Ϸ1.2 ϫ 10Ϫ2
Ka3 3.6 ϫ 10Ϫ13 Ϸ—
[H3Oϩ] 2.4 ϫ 10Ϫ2 M Ϸ0.11 M
pH 1.62 Ϸ0.96
[nonionized molecules] 7.6 ϫ 10Ϫ2 M Ϸ0 M
Examples of conjugate acid–base pairs.
Acid Conjugate Base
strong (HCl) 88n very weak (ClϪ)
weak (HCN) 88n stronger, but still
weak (CNϪ)
Base Conjugate Acid
strong (OHϪ) 88n very weak (H2O)
weak (NH3) 88n stronger, but still
weak (NH4
ϩ)
CH3COOH ϩ H2O 34 H3Oϩ ϩ CH3COOϪ
Hence, the CH3COOϪ ion is a stronger base than the NO3
Ϫ ion, but it is still weak.
In dilute solutions, strong acids and strong bases are completely ionized or dissociated.
In the following sections we consider dilute aqueous solutions of salts. Based on our classification
of acids and bases, we can identify four different kinds of salts.
1. Salts of strong bases and strong acids
2. Salts of strong bases and weak acids
3. Salts of weak bases and strong acids
4. Salts of weak bases and weak acids
SALTS OF STRONG BASES AND STRONG ACIDS
We could also describe these as salts that contain the cation of a strong base and the anion
of a strong acid. Salts derived from strong bases and strong acids give neutral solutions
because neither the cation nor the anion reacts appreciably with H2O. Consider an aqueous
solution of NaCl, which is the salt of the strong base NaOH and the strong acid HCl.
Sodium chloride is ionic even in the solid state. It dissociates into hydrated ions in H2O.
H2O ionizes slightly to produce equal concentrations of H3Oϩ and OHϪ ions.
H2O
NaCl (solid) 8888n Naϩ ϩ ClϪ
100%
H2O ϩ H2O 34 OHϪ ϩ H3Oϩ
We see that aqueous solutions of NaCl contain four ions, Naϩ, ClϪ, H3Oϩ and OHϪ.
The cation of the salt, Naϩ, is such a weak acid that it does not react appreciably with
water. The anion of the salt, ClϪ, is such a weak base that it does not react appreciably
with water. Solutions of salts of strong bases and strong acids are therefore neutral because
neither ion of such a salt reacts to upset the H3Oϩ/OHϪ balance in water.
SALTS OF STRONG BASES AND WEAK ACIDS
When salts derived from strong bases and weak acids are dissolved in water, the resulting
solutions are always basic. This is because anions of weak acids react with water to form
hydroxide ions. Consider a solution of sodium acetate, NaCH3COO, which is the salt of
the strong base NaOH and the weak acid CH3COOH. It is soluble and dissociates
completely in water.
Acetate ion is the conjugate base of a weak acid, CH3COOH. Thus, it combines to some
extent with H3Oϩ to form CH3COOH. As H3Oϩ is removed from the solution, causing
NaCH3COO(solid) Naϩ ϩ CH3COOϪ
CH3COOH ϩ H2O
H3OϩOHϪ ϩ
(result is excess OHϪ)
H2O
100%
H2O ϩH2O
Equilibrium is shifted
18-8
18-7
18-8 Salts of Strong Bases and Weak Acids 777
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 17.10, Acid–Base Properties of
Salts.
Salt of
strong base/strong acid
Neutral
aqueous solution
Salt of
strong base/weak acid
Basic
aqueous solution
m8888m8888
more H2O to ionize, an excess of OHϪ builds up. So the solution becomes basic. The
net result of the preceding equations can be written as a single equation. This equation
describes the hydrolysis of acetate ions.
CH3COOϪ ϩ H2O 34 CH3COOH ϩ OHϪ
The equilibrium constant for this reaction is called a (base) hydrolysis constant, or Kb for
CH3COOϪ.
Kb ϭ (Kb for CH3COOϪ)
We can evaluate this equilibrium constant from other known expressions. We multiply
the preceding expression by [H3Oϩ]/[H3Oϩ] to give
Kb ϭ ϫ ϭ ϫ
We recognize that
Kb ϭ ϫ ϭ ϭ
which gives
Kb ϭ ϭ 5.6 ϫ 10Ϫ10
We have calculated Kb, the hydrolysis constant for the acetate ion, CH3COOϪ.
We can do the same kind of calculations for the anion of any weak monoprotic acid
and find that Kb ϭ Kw/Ka, where Ka refers to the ionization constant for the weak monoprotic
acid from which the anion is derived.
This equation can be rearranged to
Kw ϭ KaKb (valid for any conjugate acid–base pair in aqueous solution)
If either Ka or Kb is known, the other can be calculated.
EXAMPLE 18-18 Kb for the Anion of a Weak Acid
(a) Write the equation for the reaction of the base CNϪ with water. (b) The value of the ionization
constant for hydrocyanic acid, HCN, is 4.0 ϫ 10Ϫ10. What is the value of Kb for the
cyanide ion, CNϪ?
Plan
(a) The base CNϪ accepts Hϩ from H2O to form the weak acid HCN and OHϪ ions. (b) We
know that KaKb ϭ Kw. So we solve for Kb and substitute into the equation.
Solution
(a) CNϪ ϩ H2O 34 HCN ϩ OHϪ
(b) We are given Ka ϭ 4.0 ϫ 10Ϫ10 for HCN, and we know that Kw ϭ 1.0 ϫ 10Ϫ14.
[CH3COOH][OHϪ]
ᎏᎏᎏ
[CH3COOϪ]
1.0 ϫ 10Ϫ14
ᎏᎏ
1.8 ϫ 10Ϫ5
Kw
ᎏᎏ
Ka (CH3COOH)
Kw
ᎏ
1
1
ᎏᎏ
Ka (CH3COOH)
[H3Oϩ][OHϪ]
ᎏᎏ
1
[CH3COOH]
ᎏᎏᎏ
[H3Oϩ][CH3COOϪ]
[H3Oϩ]
ᎏ
[H3Oϩ]
[CH3COOH][OHϪ]
ᎏᎏᎏ
[CH3COOϪ]
[CH3COOH][OHϪ]
ᎏᎏᎏ
[CH3COOϪ]
This is like the reaction of a molecular
weak base
NH3 ϩ H2O 34 NH4
ϩ ϩ OHϪ
(excess OHϪ is produced; the solution
becomes basic)
Just as we can write a Kb expression
to describe the extent of ionization of
NH3, we can do the same for the
acetate ion.
778 CHAPTER 18: Ionic Equilibria I: Acids and Bases
Base hydrolysis constants, Kb’s,
for anions of weak acids can be
determined experimentally. The values
obtained from experiments agree with
the calculated values. Please note that
this Kb refers to a reaction in which
the anion of a weak acid acts as a base.
Small Ka
Large Ka
Large Kb
Small Kb
Stronger
conjugateacid
Stronger
conjugatebase
For conjugate acid–base pairs
Kb(CNϪ) ϭ ϭ ᎏ
1
4
.
.
0
0
ϫ
ϫ
1
1
0
0
Ϫ
Ϫ
1
1
4
0
ᎏ ϭ 2.5 ϫ 10Ϫ5
You should now work Exercises 78 and 80.
EXAMPLE 18-19 Calculations Based on Hydrolysis
Calculate [OHϪ], pH, and the percent hydrolysis for 0.10 M solutions of (a) sodium acetate,
NaCH3COO, and (b) sodium cyanide, NaCN. Both NaCH3COO and NaCN are soluble ionic
salts that are completely dissociated in H2O. From the text, Kb for CH3COOϪ ϭ 5.6 ϫ 10Ϫ10;
from Example 18-18, Kb for CNϪ ϭ 2.5 ϫ 10Ϫ5.
Plan
We recognize that both NaCH3COO and NaCN are salts of strong bases and weak acids. The
anions in such salts hydrolyze to give basic solutions. As we have done before, we first write
the appropriate chemical equation and equilibrium constant expression. Then we complete the
reaction summary, substitute the algebraic representations of equilibrium concentrations into
the equilibrium constant expression, and solve for the unknown concentration(s).
Solution
(a) The overall equation for the reaction of CH3COOϪ with H2O and its equilibrium constant
expression are
CH3COOϪ ϩ H2O 34 CH3COOH ϩ OHϪ Kb ϭ
ϭ 5.6 ϫ 10Ϫ10
Let x ϭ mol/L of CH3COOϪ that hydrolyzes. Then x ϭ [CH3COOH] ϭ [OHϪ].
CH3COOϪ ϩ H2O 88n CH3COOH ϩ OHϪ
initial 0.10 M 0 M Ϸ0 M
change due to rxn Ϫx M ϩx M ϩx M
at equil (0.10 Ϫ x) M x M x M
Because the value of Kb (5.6 ϫ 10Ϫ10) is quite small, we know that the reaction does not go
very far. We can assume x ϽϽ 0.10, so (0.10 Ϫ x) Ϸ 0.10; this lets us simplify the equation to
5.6 ϫ 10Ϫ10 ϭ ᎏ
(
0
x
.
)
1
(x
0
)
ᎏ so x ϭ 7.5 ϫ 10Ϫ6
x ϭ 7.5 ϫ 10Ϫ6 M ϭ [OHϪ] pOH ϭ 5.12 and pH ϭ 8.88
The 0.10 M NaCH3COO solution is distinctly basic.
% hydrolysis ϭ ϫ 100% ϭ ϫ 100%
ϭ 0.0075% hydrolysis
(b) Perform the same kind of calculation for 0.10 M NaCN. Let y ϭ mol/L of CNϪ that
hydrolyzes. Then y ϭ [HCN] ϭ [OHϪ].
7.5 ϫ 10Ϫ6 M
ᎏᎏ
0.10 M
[CH3COOϪ]hydrolyzed
ᎏᎏᎏ
[CH3COOϪ]initial
[CH3COOH][OHϪ]
ᎏᎏᎏ
[CH3COOϪ]
Kw
ᎏ
Ka
If we did not know Kb, we could use
Ka(HCN) to find Kb(CNϪ).
18-8 Salts of Strong Bases and Weak Acids 779
CNϪ ϩ H2O 34 HCN ϩ OHϪ
initial 0.10 M
change due to rxn Ϫy M ϩy M ϩy M
at equil (0.10 Ϫ y) M y M y M
Kb ϭ ᎏ
[HC
[C
N
N
][O
Ϫ]
HϪ]
ᎏ ϭ 2.5 ϫ 10Ϫ5
Substitution into this expression gives
ᎏ
(0.
(
1
y
0
)(
Ϫ
y)
y)
ᎏ ϭ 2.5 ϫ 10Ϫ5 so y ϭ 1.6 ϫ 10Ϫ3 M
y ϭ [OHϪ] ϭ 1.6 ϫ 10Ϫ3 M pOH ϭ 2.80 and pH ϭ 11.20
The 0.10 M NaCN solution is even more basic than the 0.10 M NaCH3COO solution in
part (a).
% hydrolysis ϭ ϫ 100% ϭ ϫ 100%
ϭ 1.6% hydrolysis
You should now work Exercises 83 and 84.
The 0.10 M solution of NaCN is much more basic than the 0.10 M solution of
NaCH3COO because CNϪ is a much stronger base than CH3COOϪ. This is expected
because HCN is a much weaker acid than CH3COOH so that the Kb for CNϪ is much
larger than the Kb for CH3COOϪ.
The percent hydrolysis for 0.10 M CNϪ (1.6%) is about 210 times greater than the
percent hydrolysis for 0.10 M CH3COOϪ (0.0075%). In Table 18-9 we compare 0.10 M
solutions of CH3COOϪ, CNϪ, and NH3 (the familiar molecular weak base). We see that
CH3COOϪ is a much weaker base than NH3, whereas CNϪ is a slightly stronger base
than NH3.
1.6 ϫ 10Ϫ3 M
ᎏᎏ
0.10 M
[CNϪ]hydrolyzed
ᎏᎏ
[CNϪ]initial
Ammonium chloride is an ionic salt
that is soluble in water.
TABLE 18-9 Data for 0.10 M Solutions of NaCH3COO, NaCN, and NH3
0.10 M NaCH3COO 0.10 M NaCN 0.10 M aq NH3
Ka for parent acid 1.8 ϫ 10Ϫ5 4.0 ϫ 10Ϫ10
Kb for anion 5.6 ϫ 10Ϫ10 2.5 ϫ 10Ϫ5 Kb for NH3 ϭ 1.8 ϫ 10Ϫ5
[OHϪ] 7.5 ϫ 10Ϫ6 M 1.6 ϫ 10Ϫ3 M 1.3 ϫ 10Ϫ3 M
% hydrolysis 0.0075% 1.6% 1.3% ionized
pH 8.88 11.20 11.11
The pH of 0.10 M NaCH3COO is
8.88. The pH of 0.10 M NaCN is
11.20. An inert solid has been
suspended in the liquids to improve
the quality of photographs of pH
meters.
SALTS OF WEAK BASES AND STRONG ACIDS
The second common kind of hydrolysis reaction involves the reaction of the cation of a
weak base with water to form nonionized molecules of the weak base and H3Oϩ ions.
This upsets the H3Oϩ/OHϪ balance in water, giving an excess of H3Oϩ, and making such
solutions acidic. Consider a solution of ammonium chloride, NH4Cl, the salt of aqueous
NH3 and HCl.
18-9
Ammonium ions from NH4Cl react to some extent with OHϪ to form nonionized NH3
and H2O molecules. This reaction removes OHϪ from the system, so it causes more H2O
to ionize to produce an excess of H3Oϩ. The net result of the preceding equations can
be written as a single equation with its equilibrium constant expression.
NH4
ϩ ϩ H2O 34 NH3 ϩ H3Oϩ Ka ϭ
The expression Kw ϭ KaKb is valid for any conjugate acid–base pair in aqueous solution.
We use it for the NH4
ϩ/NH3 pair.
Ka (NH4
ϩ) ϭ ϭ ϭ 5.6 ϫ 10Ϫ10 ϭ
The fact that Ka for the ammonium ion, NH4
ϩ, is the same as Kb for the acetate ion
should not be surprising. Recall that the ionization constants for CH3COOH and aqueous
NH3 are equal (by coincidence). Thus, we expect CH3COOϪ to hydrolyze to the same
extent as NH4
ϩ does.
EXAMPLE 18-20 pH of a Soluble Salt
Calculate the pH of a 0.20 M solution of ammonium nitrate, NH4NO3. Ka for NH4
ϩ ϭ
5.6 ϫ 10Ϫ10.
Plan
We recognize that NH4NO3 is the salt of a weak base, NH3, and a strong acid, HNO3, and
that the cations of such salts hydrolyze to give acidic solutions. We proceed as we did in Example
18-19.
Solution
The cation of the weak base reacts with H2O. Let x ϭ mol/L of NH4
ϩ that hydrolyzes. Then
x ϭ [NH3] ϭ [H3Oϩ].
NH4
ϩ ϩ H2O 34 NH3 ϩ H3Oϩ
initial 0.20 M 0 M Ϸ0 M
change due to rxn Ϫx M ϩx M ϩx M
at equil (0.20 Ϫ x) M x M x M
Substituting into the Ka expression gives
Ka ϭ ϭ ϭ 5.6 ϫ 10Ϫ10
Making the usual simplifying assumption gives x ϭ 1.1 ϫ 10Ϫ5 M ϭ [H3Oϩ] and pH ϭ 4.96.
The 0.20 M NH4NO3 solution is distinctly acidic.
You should now work Exercise 90.
(x)(x)
ᎏᎏ
(0.20 Ϫ x)
[NH3][H3Oϩ]
ᎏᎏ
[NH4
ϩ]
[NH3][H3Oϩ]
ᎏᎏ
[NH4
ϩ]
1.0 ϫ 10Ϫ14
ᎏᎏ
1.8 ϫ 10Ϫ5
Kw
ᎏ
Kb (NH3)
[NH3][H3Oϩ]
ᎏᎏ
[NH4
ϩ]
NH4Cl(solid) NH4
ϩ ϩ ClϪ
NH3 ϩ H2O
ϩ H3OϩOHϪ
(result is excess H3Oϩ)
H2O
100%
H2O ϩ H2O
Equilibrium is shifted
Similar equations can be written for
cations derived from other weak bases
such as CH3NH3
ϩ and (CH3)2NH2
ϩ.
If you wish to derive Ka ϭ Kw/Kb for
this case, multiply the Ka expression by
[OHϪ]/[OHϪ] and simplify.
Ammonium nitrate is widely used as a
fertilizer because of its high nitrogen
content. It contributes to soil acidity.
18-9 Salts of Weak Bases and Strong Acids 781
The pH of 0.20 M NH4NO3
solution is 4.96.
Salt of
weak base/strong acid
Acidic
aqueous solution
m8888
Our Daily LivesCCHEMISTRY IN USE
Taming Acids with Harmless Salts
From yellowing paper in old books and newsprint to heartburn
and environmental spills, many Americans encounter
unwanted or excessive amounts of acid. Neutralizing unwanted
acids with hydroxide bases might appear to be a good
way to combat these acids. But even more effective chemicals
exist that can neutralize acids without the risks posed by
hydroxide bases. These acid-neutralizing chemicals are salts
of weak acids and strong bases. Such salts can neutralize acids
because hydrolysis makes their aqueous solutions basic. More
importantly, there is a significant advantage in using relatively
harmless salts such as sodium hydrogen carbonate (baking
soda) rather than stronger bases such as sodium hydroxide
(lye). For example, if we used too much sodium hydroxide to
neutralize sulfuric acid spilled from a car battery, any excess
lye left behind would pose an environmental and human
health threat about equal to that of the spilled sulfuric acid.
(Lye is the major ingredient in such commercial products as
oven cleaners and Draෆno.) We would not be concerned, however,
if a little baking soda were left on the ground after the
sulfuric acid from the car battery had been neutralized.
The same principle applies to acid indigestion. Rather
than swallow lye (ugh!) or some other strong base to neutralize
excess stomach acid, most people take antacids.
Antacids typically contain salts such as calcium carbonate,
sodium hydrogen carbonate (sodium bicarbonate), and magnesium
carbonate, all of which are salts of weak acids. These
salts hydrolyze to form hydroxide ions that reduce the degree
of acidity in the stomach. Physicians also prescribe these
and similar salts to treat peptic ulcers. The repeated use
of antacids should always be under the supervision of a
physician.
Salts of weak acids and strong bases can be used effectively
against a major acid spill in much the same way they
are used against sulfuric acid from a car battery or excess
stomach acid. In a recent major acid spill, a tank car filled
with nitric acid was punctured by the coupling of another rail
car, spilling 22,000 gallons of concentrated nitric acid onto
the ground. Many thousand residents living near the spill
were evacuated. There were no fatalities or serious injuries,
and there was no major environmental damage; resident fire
fighters neutralized the concentrated nitric acid by using airport
snow blowers to spread relatively harmless sodium
carbonate (washing soda) over the contaminated area.
Salts of weak acids and strong bases are also being used
to combat the destructive aging process of paper. Think how
serious this problem is for the Library of Congress, which
loses 70,000 books each year to the decomposition of aging
paper. Many of the twenty million books in the Library of
Congress have a life expectancy of only 25 to 40 years. Paper
ages because of the hydrolysis of aluminum sulfate. Aluminum
sulfate has been used in the paper manufacturing
process since the 1850s because it is an inexpensive sizing
compound (it keeps ink from spreading out on paper). Aluminum
sulfate is the salt of an insoluble weak base and a
strong acid; it hydrolyzes in the water in paper (typically 4–
7% H2O) to give an acidic environment. The acid eats away
at cellulose fibers, which causes the paper to turn yellow and
eventually disintegrate. To combat this aging, the Library of
Congress individually treats its collections with solutions of
salts of weak acids and strong bases at great cost. Meanwhile,
the paper industry is fighting this aging process by increasing
its output of alkaline paper. Some alkaline paper contains
calcium carbonate, the same salt found in several brands of
antacids. Calcium carbonate increases the pH of paper to
between 7.5 and 8.5. Special manufacturing techniques produce
calcium carbonate that is very fine and that has uniform
particle size. Alkaline papers are expected to last about 300
years, in contrast to the average 25- to 40-year life expectancy
of standard acidic paper.
Salts that hydrolyze to produce basic solutions can settle
upset stomachs, prevent yellowing pages, and neutralize
major and minor acid spills. A knowledge of hydrolysis is very
useful and has many applications.
Ronald DeLorenzo
Middle Georgia College
SALTS OF WEAK BASES AND WEAK ACIDS
Salts of weak bases and weak acids are the fourth class of salts. Most are soluble. Salts of
weak bases and weak acids contain cations that would give acidic solutions and anions that
would give basic solutions. Will solutions of such salts be neutral, basic, or acidic? They
may be any one of the three depending on the relative strengths of the weak molecular
acid and weak molecular base from which each salt is derived. Thus, salts of this class may
be divided into three types that depend on the relative strengths of their parent weak bases
and weak acids.
Salts of Weak Bases and Weak Acids for Which Kb ϭ Ka
The common example of a salt of this type is ammonium acetate, NH4CH3COO, the salt
of aqueous NH3 and CH3COOH. The ionization constants for both aqueous NH3 and
CH3COOH are 1.8 ϫ 10Ϫ5. We know that ammonium ions react with water to produce
H3Oϩ.
NH4
ϩ ϩ H2O 34 NH3 ϩ H3Oϩ Ka ϭ ϭ 5.6 ϫ 10Ϫ10
We also recall that acetate ions react with water to produce OHϪ.
CH3COOϪ ϩ H2O 34 CH3COOH ϩ OHϪ Kb ϭ
ϭ 5.6 ϫ 10Ϫ10
Because these K values are equal, the NH4
ϩ produces just as many H3Oϩ ions as the
CH3COOϪ produces OHϪ ions. Thus, we predict that ammonium acetate solutions are
neutral, and they are. There are very few salts that have cations and anions with equal K
values.
Salts of Weak Bases and Weak Acids for Which Kb Ͼ Ka
Salts of weak bases and weak acids for which Kb is greater than Ka are always basic because
the anion of the weaker acid hydrolyzes to a greater extent than the cation of the stronger
base.
Consider NH4CN, ammonium cyanide. Ka for HCN (4.0 ϫ 10Ϫ10) is much smaller
than Kb for NH3 (1.8 ϫ 10Ϫ5), so Kb for CNϪ (2.5 ϫ 10Ϫ5) is much larger than Ka for
NH4
ϩ (5.6 ϫ 10Ϫ10). This tells us that the CNϪ ions hydrolyze to a much greater extent
than do NH4
ϩ ions, and so ammonium cyanide solutions are distinctly basic. Stated differently,
CNϪ is much stronger as a base than NH4
ϩ is as an acid.
NH4
ϩ ϩ H2O 34 NH3 ϩ H3Oϩ
The second reaction occurs to greater
2H2O
extent І the solution is basic.
CNϪ ϩ H2O 34 HCN ϩ OHϪ
Salts of Weak Bases and Weak Acids for Which Kb Ͻ Ka
Salts of weak bases and weak acids for which Kb is less than Ka are acidic because the
cation of the weaker base hydrolyzes to a greater extent than the anion of the stronger
[CH3COOH][OHϪ]
ᎏᎏᎏ
[CH3COOϪ]
[NH3][H3Oϩ]
ᎏᎏ
[NH4
ϩ]
18-10
18-10 Salts of Weak Bases and Weak Acids 783
Salt of
weak base/weak acid
Neutral
aqueous solution
IF
parent Kb ϭ parent Ka
Salt of
weak base/weak acid
Basic
aqueous solution
IF
parent Kb Ͼ parent Ka
Salt of
weak base/weak acid
Acidic
aqueous solution
IF
parent Kb Ͻ parent Ka
m8888m8888m8888
XAAnAA
X
acid. Consider ammonium fluoride, NH4F, the salt of aqueous ammonia and hydrofluoric
acid.
Kb for aqueous NH3 is 1.8 ϫ 10Ϫ5 and Ka for HF is 7.2 ϫ 10Ϫ4. So the Ka value for
NH4
ϩ (5.6 ϫ 10Ϫ10) is slightly larger than the Kb value for FϪ (1.4 ϫ 10Ϫ11). This tells
us that NH4
ϩ ions hydrolyze to a slightly greater extent than FϪ ions. In other words,
NH4
ϩ is slightly stronger as an acid than FϪ is as a base. Ammonium fluoride solutions
are slightly acidic.
NH4
ϩ ϩ H2O 34 NH3 ϩ H3Oϩ
The first reaction occurs to greater
2H2O
extent І the solution is acidic.
FϪ ϩ H2O 34 HF ϩ OHϪ
SALTS THAT CONTAIN SMALL, HIGHLY
CHARGED CATIONS
Solutions of certain common salts of strong acids are acidic. For this reason, many homeowners
apply iron(II) sulfate, FeSO4T7H2O, or aluminum sulfate, Al2(SO4)3T18H2O, to
the soil around “acid-loving” plants such as azaleas, camelias, and hollies. You have probably
tasted the sour, “acid” taste of alum, KAl(SO4)2T12H2O, a substance that is frequently
added to pickles.
Each of these salts contains a small, highly charged cation and the anion of a strong
acid. Solutions of such salts are acidic because these cations hydrolyze to produce excess
hydronium ions. Consider aluminum chloride, AlCl3, as a typical example. When solid
anhydrous AlCl3 is added to water, the water becomes very warm as the Al3ϩ ions become
hydrated in solution. In many cases, the interaction between positively charged ions and
the negative ends of polar water molecules is so strong that salts crystallized from aqueous
solution contain definite numbers of water molecules. Salts containing Al3ϩ, Fe2ϩ, Fe3ϩ,
and Cr3ϩ ions usually crystallize from aqueous solutions with six water molecules bonded
to each metal ion. These salts contain the hydrated cations [Al(OH2)6]3ϩ, [Fe(OH2)6]2ϩ,
[Fe(OH2)6]3ϩ, and [Cr(OH2)6]3ϩ, respectively, in the solid state. Such species also exist
in aqueous solutions. Each of these species is octahedral, meaning that the metal ion (Mnϩ)
is located at the center of a regular octahedron, and the O atoms in six H2O molecules
are located at the corners (Figure 18-3). In the metal–oxygen bonds of the hydrated cation,
electron density is decreased around the O end of each H2O molecule by the positively
charged metal ion. This weakens the HXO bonds in coordinated H2O molecules relative
to the HXO bonds in noncoordinated H2O molecules. Consequently, the coordinated
H2O molecules can donate Hϩ to solvent H2O molecules to form H3Oϩ ions. This
produces acidic solutions (Figure 18-4).
The equation for the hydrolysis of hydrated Al3ϩ is written as follows.
[Al(OH2)6]3ϩ ϩ H2O 34 [Al(OH)(OH2)5]2ϩ ϩ H3Oϩ
Ka ϭ ϭ 1.2 ϫ 10Ϫ5
Removing an Hϩ converts a coordinated water molecule to a coordinated hydroxide ion
and decreases the positive charge on the hydrated species.
Hydrolysis of hydrated small, highly charged cations may occur beyond the first step.
In many cases these reactions are quite complex. They may involve two or more cations
reacting with each other to form large polymeric species. For most common hydrated
cations, consideration of the first hydrolysis constant is adequate for our calculations.
[[Al(OH)(OH2)5]2ϩ][H3Oϩ]
ᎏᎏᎏᎏ
[[Al(OH2)6]3ϩ]
18-11
Figure 18-3 (a) Lewis structures
of hydrated aluminum ions,
[Al(OH2)6]3ϩ, and hydrated iron(II)
ions, [Fe(OH2)6]2ϩ. (b) Ball-andstick
models of these ions.
784 CHAPTER 18: Ionic Equilibria I: Acids and Bases
XAAnAA
X
OH2
OH2
OH2
OH2
H2O
H2O
3ϩ
Al
OH2
OH2
OH2
OH2
H2O
H2O
2ϩ
Fe
(a)
(b)
EXAMPLE 18-21 Percent Hydrolysis
Calculate the pH and percent hydrolysis in 0.10 M AlCl3 solution. Ka ϭ 1.2 ϫ 10Ϫ5 for
[Al(OH2)6]3ϩ (often abbreviated Al3ϩ).
Plan
We recognize that AlCl3 produces a hydrated, small, highly charged cation that hydrolyzes to
give an acidic solution. We represent the equilibrium concentrations and proceed as we did in
earlier examples.
Solution
The equation for the reaction and its hydrolysis constant can be represented as
[Al(OH2)6]3ϩ ϩ H2O 34 [Al(OH)(OH2)5]2ϩ ϩ H3Oϩ
Ka ϭ ϭ 1.2 ϫ 10Ϫ5
Let x ϭ mol/L of [Al(OH2)6]3ϩ that hydrolyzes. Then x ϭ [Al(OH)(OH2)5]2ϩ ϭ [H3Oϩ].
[Al(OH2)6]3ϩ ϩ H2O 34 [Al(OH)(OH2)5]2ϩ ϩ H3Oϩ
initial 0.10 M
change due to rxn Ϫx M ϩx M ϩx M
at equil (0.10 Ϫ x) M x M x M
ᎏ
(0.
(
1
x
0
)(
Ϫ
x)
x)
ᎏ ϭ 1.2 ϫ 10Ϫ5 so x ϭ 1.1 ϫ 10Ϫ3
[H3Oϩ] ϭ 1.1 ϫ 10Ϫ3 M, pH ϭ 2.96, and the solution is quite acidic.
% hydrolysis ϭ ϫ 100% ϭ ϫ 100% ϭ 1.1% hydrolyzed
As a reference point, CH3COOH is 1.3% ionized in 0.10 M solution (see Example 18-11). In
0.10 M solution AlCl3 is 1.1% hydrolyzed. The acidities of the two solutions are very similar.
You should now work Exercise 96.
1.1 ϫ 10Ϫ3 M
ᎏᎏ
0.10 M
[Al3ϩ]hydrolyzed
ᎏᎏ
[Al3ϩ]initial
[[Al(OH)(OH2)5]2ϩ][H3Oϩ]
ᎏᎏᎏᎏ
[[Al(OH2)6]3ϩ]
[Al(OH2)6]3ϩ is often abbreviated as
Al3ϩ. Recall that x represents the
concentration of Al3ϩ that hydrolyzes.
The pH of 0.10 M AlCl3 is 2.96. The
pH of 0.10 M CH3COOH is 2.89.
18-11 Salts that Contain Small, Highly Charged Cations 785
The blue color and the acidity of a
0.10 M CuSO4 solution are both due
to the hydrated Cu2ϩ ion.
Figure 18-4 Hydrolysis of
hydrated aluminum ions to produce
H3Oϩ —that is, the removal of a
proton from a coordinated H2O
molecule by a noncoordinated one.
H
O
H2O
OH2
3+
+
H2O OH2
OH2 H2O
OH
H2O
H2O
OH2
OH2
OH2
2+
AlAl
H
H
+ transfer
+ H3O+
786 CHAPTER 18: Ionic Equilibria I: Acids and Bases
TABLE 18-10 Ionic Radii and Hydrolysis Constants for Some Cations
Cation Ionic Radius (Å) Hydrated Cation Ka
Liϩ 0.90 [Li(OH2)4]ϩ 1 ϫ 10Ϫ14
Be2ϩ 0.59 [Be(OH2)4]2ϩ 1.0 ϫ 10Ϫ5
Naϩ 1.16 [Na(OH2)6]ϩ (?) 10Ϫ14
Mg2ϩ 0.85 [Mg(OH2)6]2ϩ 3.0 ϫ 10Ϫ12
Al3ϩ 0.68 [Al(OH2)6]3ϩ 1.2 ϫ 10Ϫ5
Fe2ϩ 0.76 [Fe(OH2)6]2ϩ 3.0 ϫ 10Ϫ10
Fe3ϩ 0.64 [Fe(OH2)6]3ϩ 4.0 ϫ 10Ϫ3
Co2ϩ 0.74 [Co(OH2)6]2ϩ 5.0 ϫ 10Ϫ10
Co3ϩ 0.63 [Co(OH2)6]3ϩ 1.7 ϫ 10Ϫ2
Cu2ϩ 0.96 [Cu(OH2)6]2ϩ 1.0 ϫ 10Ϫ8
Zn2ϩ 0.74 [Zn(OH2)6]2ϩ 2.5 ϫ 10Ϫ10
Hg2ϩ 1.10 [Hg(OH2)6]2ϩ 8.3 ϫ 10Ϫ7
Bi3ϩ 0.74 [Bi(OH2)6]3ϩ 1.0 ϫ 10Ϫ2
Key Terms
Amines Derivatives of ammonia in which one or more hydrogen
atoms has been replaced by organic groups.
Hydrolysis The reaction of a substance with water.
Hydrolysis constant An equilibrium constant for a hydrolysis
reaction.
Indicator An organic compound that exhibits different colors in
solutions of different acidities.
Ion product for water An equilibrium constant for the ionization
of water,
Kw ϭ [H3Oϩ][OHϪ] ϭ 1.0 ϫ 10Ϫ14 at 25°C
Ionization constant An equilibrium constant for the ionization
of a weak electrolyte.
Monoprotic acid An acid that can form only one hydronium ion
per molecule; may be strong or weak.
pH The negative logarithm of the concentration (mol/L) of the
H3Oϩ (or Hϩ) ion; the commonly used scale ranges from 0 to
14.
pKa The negative logarithm of Ka, the ionization constant for a
weak acid.
pKb The negative logarithm of Kb, the ionization constant for a
weak base.
pKw The negative logarithm of the ion product for water.
pOH The negative logarithm of the concentration (mol/L)
of the OHϪ ion; the commonly used scale ranges from 14
to 0.
Polyprotic acid An acid that can form two or more hydronium
ions per molecule.
Solvolysis The reaction of a substance with the solvent in which
it is dissolved.
Pepto-Bismol contains
BiO(HOC4H6COO), bismuth
subsalicylate, a hydrolyzed bismuth
salt. Such salts “coat” polar surfaces
such as glass and the lining of the
stomach.
Smaller, more highly charged cations are stronger acids than larger, less highly charged
cations (Table 18-10). This is because the smaller, more highly charged cations interact
with coordinated water molecules more strongly.
For isoelectronic cations in the same period in the periodic table, the smaller, more
highly charged cation is the stronger acid. (Compare Ka values for hydrated Liϩ and Be2ϩ
and for hydrated Naϩ, Mg2ϩ, and Al3ϩ.) For cations with the same charge from the same
group in the periodic table, the smaller cation hydrolyzes to a greater extent. (Compare
Ka values for hydrated Be2ϩ and Mg2ϩ.) If we compare cations of the same element in
different oxidation states, the smaller, more highly charged cation is the stronger acid.
(Compare Ka values for hydrated Fe2ϩ and Fe3ϩ and for hydrated Co2ϩ and Co3ϩ.)
Exercises 787
Exercises
NOTE All exercises in this chapter assume a temperature of 25°C
unless they specify otherwise. All logarithms are common (base 10).
Review of Strong Electrolytes
*001. List names and formulas for (a) the common strong acids;
(b) six weak bases; (c) the common strong bases; (d) ten
soluble ionic salts.
*002. (a) How are a strong acid and a weak acid similar? How
are they different? (b) How are a strong base and a weak
base similar? How are they different?
*003. Which of the following are strong electrolytes: Group IA
hydroxide, Group IIIA hydroxide, Zn(OH)2, Be(OH)2,
H3AsO3, HBr, H3PO4, salt of a Group IA metal?
*004. Calculate the molarity of each of the following solutions.
(a) 17.52 g of NaCl in 125 mL of solution; (b) 50.5 g of
H2SO4 in 575 mL of solution; (c) 0.135 g of phenol,
C6H5OH, in 1.5 L of solution.
*005. Square brackets, [ ], are often used in some mathematical
statements in chemistry. What is the meaning
associated with square brackets?
*006. Calculate the concentrations of the constituent ions in
solutions of the following compounds in the indicated
concentrations. (a) 0.25 M HBr; (b) 0.055 M KOH;
(c) 0.0155 M CaCl2.
*007. Calculate the concentrations of the constituent ions in
solutions of the following compounds in the indicated
concentrations. (a) 0.050 M Sr(OH)2; (b) 0.050 M
HClO3; (c) 0.0040 M K2SO4.
*008. Calculate the concentrations of the constituent ions in
the following solutions. (a) 2.55 g of KOH in 1.50 L of
solution; (b) 0.720 g of Ba(OH)2 in 250. mL of solution;
(c) 1.64 g of Ca(NO3)2 in 100. mL of solution.
*009. Calculate the concentrations of the constituent ions in the
following solutions. (a) 1.77 g of Al2(SO4)3 in 400. mL of
solution; (b) 75.8 g of CaCl2T6H2O in 8.00 L of solution;
(c) 18.4 g of HBr in 675 mL of solution.
The Autoionization of Water
*010. (a) Write a chemical equation showing the ionization of
water. (b) Write the equilibrium constant expression for
this equation. (c) What is the special symbol used for this
equilibrium constant? (d) What is the relationship
between [Hϩ] and [OHϪ] in pure water? (e) How can this
relationship be used to define the terms “acidic” and
“basic”?
*011. Use Kw to explain the relationship between the hydronium
ion concentration and the hydroxide ion concentration
in aqueous solutions.
*012. (a) Why is the concentration of OHϪ produced by
the ionization of water neglected in calculating the concentration
of OHϪ in a 0.10 M solution of NaOH?
(b) Demonstrate that it (OHϪ from H2O) may be
neglected.
*013. Calculate the concentrations of OHϪ in the solutions
described in Exercises 6(a), 7(b), and 9(c), and compare
them with the OHϪ concentration in pure water.
*014. Calculate the concentrations of H3Oϩ in the solutions
described in Exercises 6(b), 7(a), and 8(b), and compare
them with the H3Oϩ concentration in pure water.
*015. Calculate [OHϪ] that is in equilibrium with
(a) [H3Oϩ] ϭ 2.9 ϫ 10Ϫ4 mol/L
(b) [H3Oϩ] ϭ 8.5 ϫ 10Ϫ9 mol/L
*016. Calculate [H3Oϩ] that is in equilibrium with
[OHϪ] ϭ 5.62 ϫ 10Ϫ6 mol/L
The pH and pOH Scales
*017. Write mathematical definitions for pH and pOH. What
is the relationship between pH and pOH? How can pH
be used to define the terms “acidic” and “basic”?
*018. What are the logarithms of the following numbers?
(a) 0.000052; (b) 15.2; (c) 5.8 ϫ 10Ϫ12; (d) 4.9 ϫ 10Ϫ7.
*019. A sample of milk is found to have a pH of 6.50. What
are the concentrations of H3Oϩ and OHϪ ions in this
sample?
*020. The normal pH of human blood ranges from 7.35 to 7.45.
Calculate the concentrations of H3Oϩ and OHϪ ions in
human blood that has a pH of 7.45.
*021. Calculate the pH of a 1.25 ϫ 10Ϫ4 M solution of HClO4,
a strong acid, at 25°C.
*022. Calculate the pH of the following solutions. (a) 6.00 ϫ
10Ϫ1 M HCl; (b) 0.030 M HNO3; (c) 0.75 gTLϪ1 HClO4.
*023. Calculate the pH of the following solutions. (a) 7.5 ϫ
10Ϫ2 M HBr; (b) 0.0062 M HI; (c) 2.84 g HNO3 in
250. mL solution.
*024. Calculate the pH of a 1.8 ϫ 10Ϫ4 M solution of NaOH.
*025. Calculate the pH of a 1.5 ϫ 10Ϫ11 M solution of HCl.
*026. A solution of HNO3 has a pH of 3.32. What is the molarity
of the solution?
*027. Complete the following table. Is there an obvious relationship
between pH and pOH? What is it?
Solution [H3Oϩ] [OHϪ] pH pOH
0.15 M HI ______ ______ ______ ______
0.060 M RbOH ______ ______ ______ ______
0.020 M Ba(OH)2 ______ ______ ______ ______
0.00030 M HClO4 ______ ______ ______ ______
788 CHAPTER 18: Ionic Equilibria I: Acids and Bases
*028. Calculate the following values for each solution.
Solution [H3Oϩ] [OHϪ] pH pOH
(a) 0.055 M NaOH ______ ______ ______ ______
(b) 0.055 M HCl ______ ______ ______ ______
(c) 0.055 M Ca(OH)2 ______ ______ ______ ______
*029. Complete the following table by appropriate calculations.
[H3Oϩ] pH [OHϪ] pOH
(a) ______ 03.84 ______ ______
(b) ______ 12.61 ______ ______
(c) ______ ______ ______ 2.90
(d) ______ ______ ______ 9.47
*030. Predict which acid of each pair is the stronger acid. Briefly
explain how you arrived at your answer. (a) H3PO4 or
H3AsO4; (b) HClO3 or HIO3; (c) HCl or HI. (Hint:
Review Chapter 10.)
*031. Predict which acid of each pair is the stronger acid. Briefly
explain how you arrived at your answer. (a) H2O or H2S;
(b) H2SO3 or H2SO2; (c) H2CO3 or HCO3
Ϫ. (Hint:
Review Chapter 10.)
*032. Write a chemical equation that represents the ionization
of a weak acid, HA. Write the equilibrium constant
expression for this reaction. What is the special symbol
used for this equilibrium constant?
*033. What is the relationship between the strength of an acid
and the numerical value of Ka? What is the relationship
between the acid strength and the value of pKa?
*034. (a) What is the pH of pure water at body temperature,
37°C? Refer to Table 18-2. (b) Is this acidic, basic, or neutral?
Why?
*035. Fill in the blanks in this table for the given solutions.
Temp.
Concentration (mol/L)
Sol’n (°C) [H3O؉] [OH؊] pH
(a) 25 1.0 ϫ 10Ϫ6 __________ ____
(b) 00 __________ __________ 4.75
(c) 60 __________ __________ 7.00
(d) 25 __________ 4.5 ϫ 10Ϫ9 ____
*036. Write a chemical equation that represents the equilibrium
between water and a weak base, B. Write the equilibrium
constant expression for this reaction. What is the special
symbol used for this equilibrium constant?
*037. What is the relationship between base strength and the
value of Kb? What is the relationship between base
strength and the value of pKb?
*038. A 0.0830 M solution of a monoprotic acid is known to be
1.07% ionized. What is the pH of the solution? Calculate
the value of Ka for this acid.
*039. A 0.055 M aqueous solution of a weak, monoprotic acid
is 0.85% ionized. Calculate the value of the ionization
constant, Ka, for this acid.
*040. The pH of a 0.025 M solution of butanoic acid,
C3H7COOH, is 3.21. What is the value of the ionization
constant, Ka, for butanoic acid?
*041. The pH of a 0.35 M solution of uric acid is 2.17. What
is the value of Ka for uric acid, a monoprotic acid?
*042. Calculate the concentrations of all the species present in
a 0.35 M benzoic acid solution. (See Appendix F.)
*043. Find the concentrations of the various species present in
a 0.25 M solution of HOBr. What is the pH of the solution?
(See Appendix F.)
*044. Hydrofluoric acid can be used to etch glass. Calculate the
pH of a 0.33 M HF solution.
*045. The ionization constant, Ka, for propanoic acid,
C2H5COOH, is 1.3 ϫ 10Ϫ5. What is the percent ionization
in a 0.65 M solution of propanoic acid? Propanoic
acid is a monoprotic acid.
*046. Calculate the pH and pOH of a household ammonia solution
that contains 2.5 moles of NH3 per liter of solution.
*047. Calculate the percent ionization in a 0.50 M NH3
solution.
*048. What is the percent ionization in a 0.0500 M solution of
formic acid, HCOOH?
*049. What is the percent ionization in (a) a 0.100 M
CH3COOH solution and (b) a 0.0100 M CH3COOH
solution?
*050. The Ka values for two weak acids are 8.3 ϫ 10Ϫ5 and
3.5 ϫ 10Ϫ10, respectively. What are their pKa values?
*051. What is the concentration of OIϪ in equilibrium with
[H3Oϩ] ϭ 0.035 mol/L and [HOI] ϭ 0.427 mol/L?
*052. Pyridine is 0.053% ionized in 0.00500 M solution. What
is the pKb of this monobasic compound?
*053. A 0.068 M solution of benzamide has a pOH of 2.91.
What is the value of pKb for this monobasic compound?
*054. In a 0.0100 M aqueous solution of methylamine,
CH3NH2, the equilibrium concentrations of the species
are [CH3NH2] ϭ 0.0080 mol/L and [CH3NH3
ϩ] ϭ
[OHϪ] ϭ 2.0 ϫ 10Ϫ3 mol/L. Calculate Kb for this weak
base.
CH3NH2(aq) ϩ H2O(ᐉ) 34 CH3NH3
ϩ ϩ OHϪ
*055. What is the concentration of NH3 in equilibrium with
[NH4
ϩ] ϭ 0.010 mol/L and [OHϪ] ϭ 1.2 ϫ 10Ϫ5 mol/L?
*056. Calculate [OHϪ], percent ionization, and pH for (a) 0.10
M aqueous ammonia, and (b) 0.15 M methylamine
solution.
*057. Calculate [H3Oϩ], [OHϪ], pH, pOH, and percent ionization
for 0.21 M aqueous ammonia solution.
Exercises 789
*058. Because Kb is larger for triethylamine
(C2H5)3N(aq) ϩ H2O(ᐉ) 34 (C2H5)3NHϩ ϩ OHϪ
Kb ϭ 5.2 ϫ 10Ϫ4
than for trimethylamine
(CH3)3N(aq) ϩ H2O(ᐉ) 34 (CH3)3NHϩ ϩ OHϪ
Kb ϭ 7.4 ϫ 10Ϫ5
an aqueous solution of triethylamine should have a larger
concentration of OHϪ ion than an aqueous solution of
trimethylamine of the same concentration. Confirm this
statement by calculating the [OHϪ] for 0.012 M solutions
of both weak bases.
*059. The equilibrium constant of the following reaction is
1.35 ϫ 10Ϫ15.
2D2O 34 D3Oϩ ϩ ODϪ
D is deuterium, 2H. Calculate the pD of pure deuterium
oxide (heavy water). What is the relationship between
[D3Oϩ] and [ODϪ] in pure D2O? Is pure D2O acidic,
basic, or neutral?
Polyprotic Acids
*060. Calculate the concentrations of the various species in a
0.100 M H3AsO4 solution. Compare the concentrations
with those of the analogous species in 0.100 M H3PO4
solution (Example 18-16 and Table 18-7).
*061. Citric acid, the acid in lemons and other citrus fruits, has
the structure
which we may abbreviate as C3H5O(COOH)3 or H3A. It
is a triprotic acid. Write the chemical equations for the
three stages in the ionization of citric acid with the appropriate
Ka expressions.
HO COOH
CH2COOH
CH2COOH
C
*062. Calculate the concentrations of H3Oϩ, OHϪ, HSeO4
Ϫ,
and SeO4
2Ϫ in 0.12 M H2SeO4, selenic acid, solution.
*063. Some kidney stones are crystalline deposits of calcium
oxalate, a salt of oxalic acid, (COOH)2. Calculate the
concentrations of H3Oϩ, OHϪ, COOCOOHϪ, and
(COOϪ)2 in 0.12 M (COOH)2. Compare the concentrations
with those obtained in Exercise 62. How can you
explain the difference between the concentrations of
HSeO4
Ϫ and COOCOOHϪ? between SeO4
2Ϫ and
(COOϪ)2?
*064. Rust stains can be removed from painted surfaces with a
solution of oxalic acid, (COOH)2. Calculate the pH of a
0.050 M oxalic acid solution.
*065. Calculate the pH and pOH of a carbonated soft drink
that is 0.0032 M carbonic acid solution. Assume that there
are no other acidic or basic components.
Hydrolysis
*066. Define and illustrate the following terms clearly and concisely:
(a) solvolysis; (b) hydrolysis.
*067. Predict which base of each pair is the stronger base.
Briefly explain how you arrived at your answer. (a) NO2
Ϫ
or NO3
Ϫ; (b) BrO3
Ϫ or IO3
Ϫ; (c) HSO3
Ϫ or HSO4
Ϫ.
(Hint: You may wish to review Chapter 10.)
*068. Predict which base of each pair is the stronger base.
Briefly explain how you arrived at your answer. (a) PH3
or NH3; (b) BrϪ or FϪ; (c) ClO3
Ϫ or ClO2
Ϫ. (Hint: You
may wish to review Chapter 10.)
*069. Some anions, when dissolved, undergo no significant
reaction with water molecules. What is the relative base
strength of such an anion compared with water? What
effect will dissolution of such anions have on the pH of
the solution?
*070. Some cations in aqueous solution undergo no significant
reactions with water molecules. What is the relative acid
strength of such a cation compared with water? What
effect will dissolution of such cations have on the pH of
the solution?
*071. How can salts be classified conveniently into four classes?
For each class, write the name and formula of a salt that
fits into that category. Use examples other than those used
in illustrations in this chapter.
790 CHAPTER 18: Ionic Equilibria I: Acids and Bases
Salts of Strong Bases and Strong Acids
*072. What determines whether the aqueous solution of a salt
is acidic, basic, or neutral?
*073. Why do salts of strong bases and strong acids give
neutral aqueous solutions? Use KNO3 to illustrate. Write
names and formulas for three other salts of strong bases
and strong acids.
*074. Which of the following salts is the salt of a strong base
and a strong acid? (a) Na3PO4; (b) K2CO3; (c) LiNO3;
(d) BaSO4; (e) NaClO3.
*075. Which of the following salts is the salt of a strong base
and a strong acid? (a) Ba3(PO4)2; (b) KNO3; (c) NaCl;
(d) CaCO3; (e) LiClO4.
Salts of Strong Bases and Weak Acids
*076. Why do salts of strong bases and weak acids give basic
aqueous solutions? Use sodium hypochlorite, NaOCl, to
illustrate. (Clorox, Purex, and other “chlorine bleaches”
are 5% NaOCl.)
Salts of Weak Bases and Strong Acids
*085. Why do salts of weak bases and strong acids give acidic
aqueous solutions? Illustrate with NH4NO3, a common
fertilizer.
*086. Write names and formulas for four salts of weak bases
and strong acids.
*087. Use values found in Table 18-6 and in Appendix G to
calculate hydrolysis constants for the following cations of
weak bases: (a) NH4
ϩ; (b) CH3NH3
ϩ, methylammonium
ion; (c) C6H5NH3
ϩ, anilinium ion.
*088. Use the values found in Table 18-6 and Appendix G to
calculate hydrolysis constants for the following cations
of weak bases: (a) (CH3)2NH2
ϩ, dimethylammonium ion;
(b) C5H5NHϩ, pyridinium ion; (c) (CH3)3NHϩ, trimethylammonium
ion.
*089. Can you make a general statement relating parent base
strength and extent of hydrolysis of the cations of Exercise
87 by using hydrolysis constants calculated in that
exercise?
*090. Calculate the pH of 0.12 M solutions of (a) NH4NO3;
(b) (CH3)NH3NO3; (c) C6H5NH3NO3.
Salts of Weak Bases and Weak Acids
*091. Why are some aqueous solutions of salts of weak acids
and weak bases neutral, whereas others are acidic and still
others are basic?
*092. Write the names and formulas for three salts of a weak
acid and a weak base that give (a) neutral, (b) acidic, and
(c) basic aqueous solutions.
*093. If both the cation and anion of a salt react with water
when dissolved, what determines whether the solution
will be acidic, basic, or neutral? Classify aqueous solutions
of the following salts as acidic, basic, or neutral.
(a) NH4F(aq) and (b) CH3NH3OI(aq).
Salts That Contain Small, Highly Charged Cations
*094. Choose the hydrated cations that react with water to give
acidic solutions. (a) [Be(H2O)4]2ϩ; (b) [Al(H2O)6]3ϩ;
(c) [Fe(H2O)6]3ϩ; (d) [Cu(H2O)6]2ϩ. Write chemical
equations for the reactions.
*095. Why do some salts that contain cations related to insoluble
bases (metal hydroxides) and anions related to strong
acids give acidic aqueous solutions? Use Fe(NO3)3 to
illustrate.
*096. Calculate pH and percent hydrolysis for the following
(Table 18-10). (a) 0.15 M Al(NO3)3, aluminum nitrate;
(b) 0.075 M Co(ClO4)2; cobalt (II) perchlorate; (c) 0.15
M MgCl2, magnesium chloride.
0*97. Given pH values for solutions of the following concentrations,
calculate hydrolysis constants for the
hydrated cations: (a) 0.00050 M CeCl3, cerium(III)
chloride, pH ϭ 5.99; (b) 0.10 M Cu(NO3)2, copper(II)
nitrate, pH ϭ 4.50; (c) 0.10 M Sc(ClO4)3, scandium
perchlorate, pH ϭ 3.44.
*077. Some anions react with water to upset the H3Oϩ/OHϪ
balance. What is the relative base strength of such an
anion compared with water? What effect will dissolution
of such anions have on the pH of the solution?
*078. Calculate the equilibrium constant for the reaction of
azide ions, N3
Ϫ, with water.
*079. Write names and formulas for three salts of strong bases
and weak acids other than those that appear in Section
18-8.
*080. Calculate hydrolysis constants for the following anions of
weak acids: (a) NO2
Ϫ; (b) OClϪ; (c) HCOOϪ. What is
the relationship between Ka, the ionization constant for
a weak acid, and Kb, the hydrolysis constant for the anion
of the weak acid? (See Appendix F.)
*081. Calculate the equilibrium constant for the reaction of
hypobromite ions (OBrϪ) with water.
*082. Calculate the pH of 1.50 M solutions of the following
salts: (a) NaCH3COO; (b) KOBr; (c) LiCN.
*083. Calculate the pH of 0.12 M solutions of the following
salts: (a) NaNO2; (b) NaOCl; (c) NaHCOO.
*084. (a) What is the pH of a 0.12 M solution of KOI? (b) What
is the pH of a 0.12 M solution of KF?
Mixed Exercises
*098. A solution of nitrous acid, HNO2, is 0.20 M. (a) Calculate
the pH. (b) What is the concentration of the nitrite
ion, NO2
Ϫ?
*099. Calculate the pH of the following solutions. (a) 0.0050 M
Ca(OH)2; (b) 0.20 M chloroacetic acid, ClCH2COOH,
Ka ϭ 1.4 ϫ 10Ϫ3; (c) 0.040 M pyridine, C5H5N.
*100. Classify aqueous solutions of the following salts as acidic,
basic, or essentially neutral. Justify your choice.
(a) (NH4)HSO4; (b) (NH4)2SO4; (c) KCl; (d) LiBrO;
(e) AlCl3.
*101. Repeat Exercise 100 for (a) NaClO4; (b) NH4Cl; (c) KCl;
(d) NH4CN. (See Appendix F.)
*102. In aqueous solution some cations react with water to upset
the H3Oϩ/OHϪ balance. What is the relative acid
strength of such a cation compared with water? What
effect will dissolution of these cations have on the pH of
the solution?
*103. Some plants require acidic soils for healthy growth.
Which of the following could be added to the soil around
such plants to increase the acidity of the soil? Write
equations to justify your answers. (a) FeSO4; (b) Na2SO4;
(c) Al2(SO4)3; (d) Fe2(SO4)3; (e) BaSO4. Arrange the salts
that give acidic solutions in order of increasing acidity.
(Assume equal molarities of the salt solutions.)
*104. Some of the following salts are used in detergents and
other cleaning materials because they produce basic solutions.
Which of the following could not be used for
this purpose? Write equations to justify your answers.
(a) Na2CO3; (b) Na2SO4; (c) (NH4)2SO4; (d) Na3PO4.
*105. Calculate the pH of each of the following solutions.
(a) 0.038 g of barium hydroxide in 250. mL of solution;
(b) 0.050 g of hydrogen iodide in 750. mL of solution;
(c) 0.00055 g of HCl in 1.00 L of solution.
*106. For each of the following pairs, tell which solution would
have the lower pH. Tell how you arrive at each answer.
(a) 0.010 M NH4Br, ammonium bromide, and 0.010 M
NH4NO3, ammonium nitrate; (b) 0.010 M ammonium
perchlorate, NH4ClO4, and 0.010 M ammonium fluoride,
NH4Cl; (c) 0.010 M NH4Cl and 0.050 M NH4Cl.
(Hint: Think before you calculate.)
CONCEPTUAL EXERCISES
*107. How could we demonstrate that 0.010 M solutions of HCl
and HNO3 contain essentially no molecules of nonionized
acid?
*108. How could we demonstrate that 0.010 M solutions of HF
and HNO2 contain relatively few ions?
*109. Carbonic acid, H2CO3, is diprotic and therefore has two
ionization constants, Ka1 ϭ 4.2 ϫ 10Ϫ7 and Ka2 ϭ 4.8 ϫ
10Ϫ11. The pH of a carbonic acid solution can be calculated
without using Ka2. Explain using a 0.100 M solution
of carbonic acid.
*110. Answer the following questions for 0.15 M solutions of
the weak bases listed in Table 18-6. (a) In which solution
is (i) the pH highest; (ii) the pH lowest; (iii) the pOH
highest; (iv) the pOH lowest? (b) Which solution contains
(i) the highest concentration of the cation of the
weak base; (ii) the lowest concentration of the cation of
the weak base?
*111. Answer the following questions for 0.15 M solutions of
the weak acids listed in Table 18-4. Which solution contains
(a) the highest concentration of H3Oϩ; (b) the
highest concentration of OHϪ; (c) the lowest concentration
of H3Oϩ; (d) the lowest concentration of OHϪ;
(e) the highest concentration of nonionized acid molecules;
(f) the lowest concentration of nonionized acid
molecules?
BUILDING YOUR KNOWLEDGE
*112. Ascorbic acid, C5H7O4COOH, also known as vitamin C,
is an essential vitamin for all mammals. Among mammals,
only humans, monkeys, and guinea pigs cannot synthesize
it in their bodies. Ka for ascorbic acid is 7.9 ϫ 10Ϫ5.
Calculate [H3Oϩ] and pH in a 0.110 M solution of ascorbic
acid.
Exercises 791
*113. Arrange the following common kitchen samples from
most acidic to most basic.
carrot juice, pH 5.1 blackberry juice, pH 3.4
soap, pH 11.0 red wine, pH 3.7
egg white, pH 7.8 milk of magnesia, pH 10.5
sauerkraut, pH 3.5 lime juice, pH 2.0
*114. The buildup of lactic acid in muscles causes pain during
extreme physical exertion. The Ka for lactic acid,
C2H5OCOOH, is 8.4 ϫ 10Ϫ4. Calculate the pH of a
0.110 M solution of lactic acid. Can you make a simplifying
assumption in this case?
*115. A 0.0100 molal solution of acetic acid freezes at
Ϫ0.01938°C. Use this information to calculate the ionization
constant for acetic acid. A 0.0100 molal solution
is sufficiently dilute that it may be assumed to be 0.0100
molar without introducing a significant error.
19 Ionic Equilibria II:
Buffers and Titration
Curves
OUTLINE
19-1 The Common Ion Effect and
Buffer Solutions
19-2 Buffering Action
19-3 Preparation of Buffer Solutions
19-4 Acid–Base Indicators
Titration Curves
19-5 Strong Acid/Strong Base
Titration Curves
19-6 Weak Acid/Strong Base
Titration Curves
19-7 Weak Acid/Weak Base Titration
Curves
19-8 Summary of Acid–Base
Calculations
I
OBJECTIVES
After you have studied this chapter, you should be able to
• Explain the common ion effect and give illustrations of its operation
• Recognize buffer solutions and describe their chemistry
• Describe how to prepare a buffer solution of a specified pH
• Carry out calculations related to buffer solutions and their action
• Explain what acid–base indicators are and how they function
• Describe what species are present at various stages of titration curves for (a) strong
acids and strong bases, (b) weak acids and strong bases, and (c) weak acids and weak
bases
• Carry out calculations based on titration curves for (a) strong acids and strong bases
and (b) weak acids and strong bases
I
n the previous chapter we calculated the acidity or basicity of aqueous solutions of
strong acids, strong bases, weak acids, weak bases, and their salts. In this chapter we
will study (1) solutions that have both weak acids and weak bases present, (2) indicators,
and (3) titration curves.
It can be a challenge to recognize the type of solution present. Recognition can be
even more difficult if the solution is formed by a partial or total neutralization reaction.
To help you to recognize the various solutions, a summary table (Table 19-7) is included
in Section 19.8. We encourage you to look at the table often as you progress through this
chapter and as you review this and the previous chapter.
THE COMMON ION EFFECT AND BUFFER SOLUTIONS
In laboratory reactions, in industrial processes, and in the bodies of plants and animals, it
is often necessary to keep the pH nearly constant despite the addition of acids or bases.
19-1
The red anthocyanin pigment in the
common geranium is a naturally
occurring acid–base indicator.
The oxygen-carrying capacity of the hemoglobin in your blood and the activity of the
enzymes in your cells are very sensitive to the pH of your body fluids. Our bodies use a
combination of compounds known as a buffer system to keep the pH within a narrow range.
A buffer solution contains a conjugate acid–base pair with both the acid and base
in reasonable concentrations. The acidic component reacts with added strong bases.
The basic component reacts with added strong acids.
The operation of a buffer solution depends on the common ion effect, a special case of
LeChatelier’s Principle.
When a solution of a weak electrolyte is altered by adding one of its ions from
another source, the ionization of the weak electrolyte is suppressed. This behavior
is termed the common ion effect.
Many types of solutions exhibit this behavior. Two of the most frequently encountered
kinds are
1. A solution of a weak acid plus a soluble ionic salt of the weak acid (e.g., CH3COOH
plus NaCH3COO)
2. A solution of a weak base plus a soluble ionic salt of the weak base (e.g., NH3 plus
NH4Cl)
Weak Acids Plus Salts of Weak Acids
Consider a solution that contains acetic acid and sodium acetate, a soluble ionic salt of
CH3COOH. The NaCH3COO is completely dissociated into its constituent ions, but
CH3COOH is only slightly ionized.
Both CH3COOH and NaCH3COO are sources of CH3COOϪ ions. The completely
dissociated NaCH3COO provides a high [CH3COOϪ]. This shifts the ionization equilibrium
of CH3COOH far to the left as CH3COOϪ combines with H3Oϩ to form
nonionized CH3COOH and H2O. The result is a drastic decrease in [H3Oϩ] in the
solution.
Solutions that contain a weak acid plus a salt of the weak acid are always less acidic
than solutions that contain the same concentration of the weak acid alone.
EXAMPLE 19-1 Weak Acid/Salt of Weak Acid Buffer Solution
Calculate the concentration of H3Oϩ and the pH of a buffer solution that is 0.10 M in
CH3COOH and 0.20 M in NaCH3COO.
NaCH3COO Naϩ CH3COOϪ
(to completion)
CH3COOH H2Oϩ ϩ
ϩ
H3Oϩ CH3COOϪ (reversible)
H2O
794 CHAPTER 19: Ionic Equilibria II: Buffers and Titration Curves
Buffer systems resist changes in pH.
LeChatelier’s Principle (Section 17-6)
is applicable to equilibria in aqueous
solution.
Plan
Write the appropriate equations for both NaCH3COO and CH3COOH and the ionization
constant expression for CH3COOH. Then, represent the equilibrium concentrations algebraically
and substitute into the Ka expression.
Solution
The appropriate equations and ionization constant expression are
rxn 1 NaCH3COO 88n Naϩ ϩ CH3COOϪ (to completion)
rxn 2 CH3COOH ϩ H2O 34 H3Oϩ ϩ CH3COOϪ (reversible)
Ka ϭ ϭ 1.8 ϫ 10Ϫ5
This Ka expression is valid for all solutions that contain CH3COOH. In solutions that contain
both CH3COOH and NaCH3COO, CH3COOϪ ions come from two sources. The ionization
constant is satisfied by the total CH3COOϪ concentration.
Because NaCH3COO is completely dissociated, the [CH3COOϪ] from NaCH3COO will be
0.20 mol/L. Let x ϭ [CH3COOH] that ionizes; then x is also equal to [H3Oϩ] and equal to
[CH3COOϪ] from CH3COOH. The total concentration of CH3COOϪ is (0.20 ϩ x) M. The
concentration of nonionized CH3COOH is (0.10 Ϫ x) M.
Substitution into the ionization constant expression for CH3COOH gives
Ka ϭ ϭ ϭ 1.8 ϫ 10Ϫ5
The small value of Ka suggests that x is very small. This leads to two assumptions.
(x)(0.20 ϩ x)
ᎏᎏ
(0.10 Ϫ x)
[H3Oϩ][CH3COOϪ]
ᎏᎏᎏ
[CH3COOH]
[H3Oϩ][CH3COOϪ]
ᎏᎏᎏ
[CH3COOH]
19-1 The Common Ion Effect and Buffer Solutions 795
The general equations
NaA 88n Naϩ ϩ AϪ
and
HA 34 Hϩ ϩ AϪ
may be used in place of these
equations.
You can verify the validity of the
assumption by substituting the value
for x, 9.0 ϫ 10Ϫ6, into the original
equation.
rxn 1 NaCH3COO Naϩ
0.20 M 0.20 M 0.20 M
(0.10 Ϫ x) M x M x M
rxn 2 CH3COOH H2O H3Oϩ
CH3COOϪ
CH3COOϪϩ ϩ
ϩ
Total [CH3COOϪ] ϭ (0.20 ϩ x) M
Assumption Implication
x ϽϽ 0.20, so Most of the CH3COOϪ comes from NaCH3COO
(0.20 ϩ x) Ϸ 0.20 (rxn 1) and very little CH3COOϪ comes from
ionization of CH3COOH (rxn 2)
x ϽϽ 0.10, so Very little of the CH3COOH ionizes (rxn 2)
(0.10 Ϫ x) Ϸ 0.10
It is reasonable to assume that x (from the ionization of CH3COOH) is small, because
CH3COOH is a weak acid (rxn 2), and its ionization is further suppressed by the high concentration
of CH3COOϪ formed by the soluble salt, NaCH3COO (rxn 1).
Introducing these assumptions gives
ᎏ
0
0
.2
.1
0
0
x
ᎏ ϭ 1.8 ϫ 10Ϫ5 and x ϭ 9.0 ϫ 10Ϫ6
x M ϭ [H3Oϩ] ϭ 9.0 ϫ 10Ϫ6 M so pH ϭ 5.05
You should now work Exercise 8.
To see how much the acidity of the 0.10 M CH3COOH solution is reduced by making
it also 0.20 M in NaCH3COO, refer back to Example 18-11. There we found that in
0.10 M CH3COOH the H3Oϩ concentration is 1.3 ϫ 10Ϫ3 mol/L (pH ϭ 2.89).
Let us calculate the percent ionization in the solution of Example 19-1.
% ionization ϭ ϫ 100%
ϭ ϫ 100% ϭ 0.0090% ionized
This compares with 1.3% ionization in 0.10 M CH3COOH (Example 18-11). Table
19-1 compares these solutions. The third column shows that [H3Oϩ] is 140 times greater
in 0.10 M CH3COOH than in the solution to which 0.20 mol/L NaCH3COO has been
added (common ion effect).
The calculation of the pH of a solution containing significant amounts of both a weak
acid and the salt of the weak acid may be carried out as we have done in Example 19-1.
Alternatively, one may proceed as follows. We can start by writing the equation for the
ionization of the weak monoprotic acid and its Ka as we did previously.
HA ϩ H2O 34 H3Oϩ ϩ AϪ and ϭ Ka
Solving this expression for [H3Oϩ] gives
[H3Oϩ] ϭ Ka ϫ
Consider a solution in which the concentrations of both the weak acid and its anion (from
an added salt) are some reasonable values, such as greater than 0.050 M. Under these
conditions the concentration of the anion, [AϪ], in the solution can be assumed to be
entirely due to the dissolved salt. With these restrictions, the preceding expression becomes
[H3Oϩ] ϭ Ka ϫ
[HA] is the concentration of nonionized weak acid (in most cases this is the total acid
concentration) and [conjugate base] is the concentration of the anion from the dissolved
salt.
If we take the logarithm of both sides of the preceding equation, we obtain
log [H3Oϩ] ϭ log Ka ϩ log
[acid]
ᎏ
[salt]
[HA]
ᎏᎏ
[conjugate base]
[HA]
ᎏ
[AϪ]
[H3Oϩ][AϪ]
ᎏᎏ
[HA]
9.0 ϫ 10Ϫ6 M
ᎏᎏ
0.10 M
[CH3COOH]ionized
ᎏᎏᎏ
[CH3COOH]initial
796 CHAPTER 19: Ionic Equilibria II: Buffers and Titration Curves
HA and AϪ represent the weak acid
and its conjugate base, respectively.
These are the kinds of assumptions we
made in Example 19-1.
The two solutions of Table 19-1, in
the presence of universal indicator.
The CH3COOH solution is on the
left.
TABLE 19-1 Comparison of [H3O؉] and pH in Acetic Acid and Sodium
Acetate–Acetic Acid Solutions
Solution % CH3COOH Ionized [H3Oϩ] pH
0.10 M CH3COOH 1.3% 1.3 ϫ 10Ϫ3 M 2.89
0.10 M CH3COOH ⌬pH ϭ 2.16
and 0.0090% 9.0 ϫ 10Ϫ6 M 5.05
0.20 M NaCH3COO










Multiplying by Ϫ1 gives
Ϫlog [H3Oϩ] ϭ Ϫlog Ka Ϫ log
Recalling that in Chapter 18 we defined Ϫlog [H3Oϩ] (or for the general expression Ϫlog
[Hϩ]) as pH and Ϫlog Ka as pKa, the preceding equation becomes
pH ϭ pKa ϩ log where pKa ϭ Ϫlog Ka (acid/salt buffer)
This equation is known as the Henderson–Hasselbalch equation. Workers in the biological
sciences use it frequently. In general terms, we write
pH ϭ pKa ϩ log (acid/salt buffer)
EXAMPLE 19-2 Weak Acid/Salt of Weak Acid Buffer Solution
(via the Henderson–Hasselbalch Equation)
Use the Henderson–Hasselbalch equation to calculate the pH of the buffer solution in
Example 19-1.
Plan
The Henderson–Hasselbalch equation is pH ϭ pKa ϩ log ᎏ
[conju
[a
g
c
a
i
t
d
e
]
base]
ᎏ. The value for pKa
for acetic acid can be calculated from the value of Ka found in Example 19-1 and many other
places. The solution in Example 19-1 is 0.10 M in CH3COOH and 0.20 M in NaCH3COO.
The values used for [conjugate base] and [acid] are their initial concentrations after mixing but
before reaction.
Solution
The appropriate values needed for the Henderson–Hasselbalch equation are
pKa ϭ Ϫlog Ka ϭ Ϫlog 1.8 ϫ 10Ϫ5 ϭ 4.74
[conjugate base] ϭ [CH3COOϪ] ϭ [NaCH3COO]initial ϭ 0.20 M
[acid] ϭ [CH3COOH]initial ϭ 0.10 M
pH ϭ pKa ϩ log ϭ 4.74 ϩ log
ϭ 4.74 ϩ log 2.0 ϭ 4.74 ϩ 0.30 ϭ 5.04
You should now work Exercise 9.
(0.20)
ᎏ
(0.10)
[conjugate base]
ᎏᎏ
[acid]
[conjugate base]
ᎏᎏ
[acid]
[salt]
ᎏ
[acid]
[acid]
ᎏ
[salt]
19-1 The Common Ion Effect and Buffer Solutions 797
The relationship is valid only for
solutions that contain a weak
monoprotic acid and a soluble, ionic
salt of the weak acid with a univalent
cation, both in reasonable
concentrations.
The values pKa ϭ 4.74, the log of 2.0,
and pH ϭ 5.04 all have two significant
figures.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 18.9, pH of Buffer Solutions.
Weak Bases Plus Salts of Weak Bases
Let us consider the second common kind of buffer solution, containing a weak base and
its salt. A solution that contains aqueous NH3 and ammonium chloride, NH4Cl, a soluble
ionic salt of NH3, is typical. The NH4Cl is completely dissociated, but aqueous NH3 is
only slightly ionized.
Both NH4Cl and aqueous NH3 are sources of NH4
ϩ ions. The completely dissociated
NH4Cl provides a high [NH4
ϩ]. This shifts the ionization equilibrium of aqueous NH3
far to the left, as NH4
ϩ ions combine with OHϪ ions to form nonionized NH3 and H2O.
The result is that [OHϪ] is decreased significantly.
Solutions that contain a weak base plus a salt of the weak base are always less basic
than solutions that contain the same concentration of the weak base alone.
EXAMPLE 19-3 Weak Base/Salt of Weak Base Buffer Solution
Calculate the concentration of OHϪ and the pH of a solution that is 0.20 M in aqueous NH3
and 0.10 M in NH4Cl.
Plan
Write the appropriate equations for both NH4Cl and NH3 and the ionization constant expression
for NH3. Then, represent the equilibrium concentrations algebraically, and substitute into
the Kb expression.
Solution
The appropriate equations and algebraic representations of concentrations are
Substitution into the Kb expression for aqueous NH3 gives
Kb ϭ ϭ 1.8 ϫ 10Ϫ5 ϭ
The small value of Kb suggests that x is very small. This leads to two assumptions.
(0.10 ϩ x)(x)
ᎏᎏ
(0.20 Ϫ x)
[NH4
ϩ][OHϪ]
ᎏᎏ
[NH3]
rxn 1 NH4Cl NH4
ϩ
0.10 M 0.10 M 0.10 M
(0.20 Ϫ x) M x M x M
rxn 2 NH3 H2O NH4
ϩ
ClϪ
OHϪϩ ϩ
ϩ
Total [NH4
ϩ] ϭ (0.10 ϩ x) M
rxn 1
rxn 2
NH4Cl NH4
ϩ
NH4
ϩ OHϪ (reversible)
(to completion)
ϩ
ϩ
ϩ
ClϪ
NH3 H2O
H2O
798 CHAPTER 19: Ionic Equilibria II: Buffers and Titration Curves
Assumption Implication
x ϽϽ 0.10, so (0.10 ϩ x) Ϸ 0.10 Most of the NH4
ϩ comes from NH4Cl (rxn 1) and
very little NH4
ϩ comes from ionization of NH3 (rxn 2)
x ϽϽ 0.20, so (0.20 Ϫ x) Ϸ 0.20 Very little of the NH3 ionizes (rxn 2)
It is reasonable to assume that x (from the ionization of NH3) is small, because NH3 is a weak
base (rxn 2), and its ionization is further suppressed by the high concentration of NH4
ϩ formed
by the soluble salt, NH4Cl (rxn 1).
Introducing these assumptions gives
ᎏ
0
0
.
.
1
2
0
0
x
ᎏ ϭ 1.8 ϫ 10Ϫ5 M and x ϭ 3.6 ϫ 10Ϫ5 M
x M ϭ [OHϪ] ϭ 3.6 ϫ 10Ϫ5 M so pOH ϭ 4.44 and pH ϭ 9.56
You should now work Exercise 10.
In Example 18-14 we calculated [OHϪ] and pH for 0.20 M aqueous NH3. Compare those
results with the values obtained in Example 19-3 (Table 19-2). The concentration of OHϪ
is 53 times greater in the solution containing only 0.20 M aqueous NH3 than in the solution
to which 0.10 mol/L NH4Cl has been added. This is another demonstration of the
common ion effect.
We can derive a relationship for [OHϪ] in a solution containing a weak base, B, plus
a salt that contains the cation, BHϩ, of the weak base, just as we did for weak acids. In
general terms the equation for the ionization of a monoprotic weak base and its Kb expression
are
B ϩ H2O 34 BHϩ ϩ OHϪ and ϭ Kb
Solving the Kb expression for [OHϪ] gives
[OHϪ] ϭ Kb ϫ
Taking the logarithm of both sides of the equation gives
log [OHϪ] ϭ log Kb ϩ log
Multiplication by Ϫ1 and rearrangement gives another form of the Henderson–Hasselbalch
equation for solutions containing a weak base plus a salt of the weak base.
pOH ϭ pKb ϩ log where pKb ϭ Ϫlog Kb (base/salt buffer)
[BHϩ]
ᎏ
[B]
[B]
ᎏ
[BHϩ]
[B]
ᎏ
[BHϩ]
[BHϩ][OHϪ]
ᎏᎏ
[B]
19-1 The Common Ion Effect and Buffer Solutions 799
TABLE 19-2 Comparison of [OHϪ] and pH in Ammonia and Ammonium
Chloride—Ammonia Solutions
Solution % NH3 Ionized [OHϪ] pH
0.20 M aq NH3 0.95% 1.9 ϫ 10Ϫ3 M 11.28
0.20 M aq NH3 ⌬pH ϭ Ϫ1.72
and 0.018% 3.6 ϫ 10Ϫ5 M 9.56
0.10 M aq NH4Cl










The percent ionization of NH3 in this
solution is
ϫ 100% ϭ 0.018%.
3.6 ϫ 10Ϫ5 Mionized
ᎏᎏᎏ
0.20 Moriginal
B and BHϩ represent the weak base
and its conjugate acid, respectively—
for example, NH3 and NH4
Ϫ.
The two solutions in Table 19-2, in
the presence of universal indicator.
The NH3 solution is on the left.
Can you calculate the percentage of
NH3 that is ionized in these two
solutions?
The Henderson–Hasselbalch equation is valid for solutions of weak bases plus salts of
weak bases with univalent anions in reasonable concentrations. In general terms we can
also write this equation as
pOH ϭ pKb ϩ log (base/salt buffer)
BUFFERING ACTION
A buffer solution is able to react with either H3Oϩ or OHϪ ions, whichever is added.
Thus, a buffer solution resists changes in pH. When we add a modest amount of a strong
base or a strong acid to a buffer solution, the pH changes very little.
The two common kinds of buffer solutions are the ones we have just discussed—namely,
solutions containing (1) a weak acid plus a soluble ionic salt of the weak acid and (2) a
weak base plus a soluble ionic salt of the weak base.
Solutions of a Weak Acid and a Salt of the Weak Acid
A solution containing acetic acid, CH3COOH, and sodium acetate, NaCH3COO, is an
example of this kind of buffer solution. The acidic component is CH3COOH. The basic
component is NaCH3COO because the CH3COOϪ ion is the conjugate base of
CH3COOH. The operation of this buffer depends on the equilibrium
CH3COOH ϩ H2O 34 H3Oϩ ϩ CH3COOϪ
high conc high conc (from salt)
If we add a strong acid such as HCl to this solution, it produces H3Oϩ. As a result of the
added H3Oϩ, the reaction occurs to the left, to use up most of the added H3Oϩ and
reestablish equilibrium. Because the [CH3COOϪ] in the buffer solution is high, this can
occur to a great extent. The net reaction is
H3Oϩ ϩ CH3COOϪ 88n CH3COOH ϩ H2O (Ϸ100%)
or, as a formula unit equation,
HCl ϩ NaCH3COO 88n CH3COOHϩNaCl (Ϸ100%)
added acid base weak acid salt
This reaction goes nearly to completion because CH3COOH is a weak acid; even when
mixed from separate sources, its ions have a strong tendency to form nonionized
CH3COOH molecules rather than remain separate.
When a strong base, such as NaOH, is added to the CH3COOH–NaCH3COO buffer
solution, it is consumed by the acidic component, CH3COOH. This occurs in the
following way. The additional OHϪ causes the water autoionization reaction to proceed
to the left.
2H2O 34 H3Oϩ ϩ OHϪ (shifts left)
19-2
[conjugate acid]
ᎏᎏ
[base]
800 CHAPTER 19: Ionic Equilibria II: Buffers and Titration Curves
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 18.8, Buffer Solutions.
Three common examples of buffers.
Many medications are buffered to
minimize digestive upset. Most body
fluids, including blood plasma,
contain very efficient natural buffer
systems. Buffer capsules are used in
laboratories to prepare solutions of
specified pH.
The net effect is to neutralize most
of the H3Oϩ from HCl by forming
nonionized CH3COOH molecules.
This slightly decreases the ratio
[CH3COOϪ]/[CH3COOH], which
governs the pH of the solution.
This uses up some H3Oϩ, causing more CH3COOH to ionize.
CH3COOH ϩ H2O 34 CH3COOϪ ϩ H3Oϩ (shifts right)
Because the [CH3COOH] is high, this can occur to a great extent. The net result is the
neutralization of OHϪ by CH3COOH.
OHϪ ϩ CH3COOH 88n CH3COOϪϩH2O (Ϸ100%)
added base acid weaker base water
EXAMPLE 19-4 Buffered Solutions
If we add 0.010 mol of solid NaOH to 1.00 liter of a buffer solution that is 0.100 M in
CH3COOH and 0.100 M in NaCH3COO, how much will [H3Oϩ] and pH change? Assume
that there is no volume change due to the addition of solid NaOH.
Plan
Calculate [H3Oϩ] and pH for the original buffer solution. Then, write the reaction summary
that shows how much of the CH3COOH is neutralized by NaOH. Calculate [H3Oϩ] and pH
for the resulting buffer solution. Finally, calculate the change in pH.
Solution
For the 0.100 M CH3COOH and 0.100 M NaCH3COO solution, we can write
[H3Oϩ] ϭ Ka ϫ ϭ 1.8 ϫ 10Ϫ5 ϫ ᎏ
0
0
.
.
1
1
0
0
0
0
ᎏ ϭ 1.8 ϫ 10Ϫ5 M; pH ϭ 4.74
When solid NaOH is added, it reacts with CH3COOH to form more NaCH3COO.
NaOH ϩ CH3COOH 88n NaCH3COO ϩ H2O
start 0.010 mol 0.100 mol 0.100 mol
change due to rxn Ϫ0.010 mol Ϫ0.010 mol ϩ0.010 mol
after rxn 0 mol 0.090 mol 0.110 mol
The volume of the solution is 1.00 liter, so we now have a solution that is 0.09 M in CH3COOH
and 0.110 M in NaCH3COO. In this solution,
[H3Oϩ] ϭ Ka ϫ ϭ 1.8 ϫ 10Ϫ5 ϫ ᎏ
0
0
.
.
0
1
9
1
0
0
ᎏ ϭ 1.5 ϫ 10Ϫ5 M; pH ϭ 4.82
The addition of 0.010 mol of solid NaOH to 1.00 liter of this buffer solution decreases
[H3Oϩ] from 1.8 ϫ 10Ϫ5 M to 1.5 ϫ 10Ϫ5 M and increases pH from 4.74 to 4.82, a change
of 0.08 pH unit, which is a very slight change.
You should now work Exercise 22.
Addition of 0.010 mole of solid NaOH to one liter of 0.100 M CH3COOH (pH ϭ
2.89 from Table 19-1) would give a solution that is 0.090 M in CH3COOH and 0.010 M
in NaCH3COO. The pH of this solution is 3.78, which is 0.89 pH unit higher than that
of the 0.100 M CH3COOH solution.
By contrast, adding 0.010 mole of NaOH to enough pure H2O to give one liter
produces a 0.010 M solution of NaOH: [OHϪ] ϭ 1.0 ϫ 10Ϫ2 M and pOH ϭ 2.00. The
pH of this solution is 12.00, an increase of 5.00 pH units above that of pure H2O.
[acid]
ᎏ
[salt]
[acid]
ᎏ
[salt]
19-2 Buffering Action 801
The net effect is to neutralize most
of the OHϪ from NaOH. This
slightly increases the ratio
[CH3COOϪ]/[CH3COOH], which
governs the pH of the solution.
This is enough NaOH to neutralize
10% of the acid.
pH ϩ pOH ϭ 14
In summary, 0.010 mole of NaOH
added to 1.00 L of the CH3COOH/NaCH3COO buffer, pH 4.74 88n 4.82
added to 1.00 L of 0.100 M CH3COOH, pH 2.89 88n 3.78
added to 1.00 L of pure H2O, pH 7.00 88n 12.00
In similar fashion we could calculate the effects of adding 0.010 mole of pure HCl(g)
instead of pure NaOH to 1.00 liter of each of these three solutions. This would result in
the following changes in pH.
added to 1.00 L of the CH3COOH/NaCH3COO buffer, pH 4.74 88n 4.66
added to 1.00 L of 0.100 M CH3COOH, pH 2.89 88n 2.00
added to 1.00 L of pure H2O, pH 7.00 88n 2.00
The results of adding NaOH or HCl to these solutions (Table 19-3) demonstrate the
efficiency of the buffer solution. We recall that each change of 1 pH unit means that the
[H3Oϩ] and [OHϪ] change by a factor of 10. In these terms, the effectiveness of the buffer
solution in controlling pH is even more dramatic.
Solutions of a Weak Base and a Salt of the Weak Base
An example of this type of buffer solution is one that contains the weak base ammonia,
NH3, and its soluble ionic salt ammonium chloride, NH4Cl. The reactions responsible
for the operation of this buffer are
H2O
NH4Cl 8888n NH4
ϩ ϩ ClϪ (to completion)
NH3 ϩ H2O 3::4 NH4
ϩ ϩ OHϪ (reversible)
high conc high conc
from salt
If a strong acid such as HCl is added to this buffer solution, the resulting H3Oϩ shifts
the equilibrium reaction
2H2O 34 H3Oϩ ϩ OHϪ (shifts left)
802 CHAPTER 19: Ionic Equilibria II: Buffers and Titration Curves
TABLE 19-3 Changes in pH Caused by Addition of Pure Acid or Base to One
Liter of Solution
When We Add When We Add
0.010 mol NaOH(s) 0.010 mol HCl(g)
pH [H3Oϩ] pH [H3Oϩ]
We Have 1.00 L of Changes Decreases Changes Increases
Original Solution by by a factor of by by a factor of
buffer solution
(0.10 M NaCH3COO
ϩ0.08 pH unit 1.2 Ϫ0.08 pH unit 1.2
and
0.10 M CH3COOH)
0.10 M CH3COOH ϩ0.91 8.1 Ϫ0.89 7.8
pure H2O ϩ5.00 100,000 Ϫ5.00 100,000





strongly to the left. As a result of the diminished OHϪ concentration, the reaction
NH3 ϩ H2O 34 NH4
ϩ ϩ OHϪ (shifts right)
shifts markedly to the right. Because the [NH3] in the buffer solution is high, this can
occur to a great extent. The net reaction is
H3Oϩ ϩ NH3 88n NH4
ϩ ϩ H2O (Ϸ100%)
added acid base weak acid water
When a strong base such as NaOH is added to the original buffer solution, it is neutralized
by the more acidic component, NH4Cl, or NH4
ϩ, the conjugate acid of ammonia.
NH3 ϩ H2O 34 NH4
ϩ ϩ OHϪ (shifts left)
Because the [NH4
ϩ] is high, this can occur to a great extent. The result is the neutralization
of OHϪ by NH4
ϩ.
OHϪ ϩ NH4
ϩ 88n NH3 ϩ H2O (Ϸ100%)
or, as a formula unit equation,
NaOH ϩ NH4Cl 88n NH3 ϩ H2O ϩ NaCl (Ϸ100%)
added base acid weak base water
Summary Changes in pH are minimized in buffer solutions because the basic
component can react with H3Oϩ ions and the acidic component can react with OHϪ
ions.
PREPARATION OF BUFFER SOLUTIONS
Buffer solutions can be prepared by mixing other solutions. When solutions are mixed,
the volume in which each solute is contained increases, so solute concentrations change.
These changes in concentration must be considered. If the solutions are dilute, we may
assume that their volumes are additive.
EXAMPLE 19-5 Preparation of Buffer Solutions
Calculate the concentration of H3Oϩ and the pH of a buffer solution prepared by mixing 200. mL
of 0.10 M NaF and 100. mL of 0.050 M HF. Ka ϭ 7.2 ϫ 10Ϫ4 for HF.
Plan
Calculate the number of millimoles (or moles) of NaF and HF and then the molarity of each
solute in the solution after mixing. Write the appropriate equations for both NaF and HF,
represent the equilibrium concentrations algebraically, and substitute into the Ka expression
for HF.
Solution
When two dilute solutions are mixed, we assume that their volumes are additive. The volume
of the new solution will be 300. mL. Mixing a solution of a weak acid with a solution of its
salt does not form any new species. So we have a straightforward buffer calculation. We calculate
the number of millimoles (or moles) of each compound and the molarities in the new
solution.
19-3
19-3 Preparation of Buffer Solutions 803
The net effect is to neutralize most of
the H3Oϩ from HCl. This slightly
increases the ratio [NH4
ϩ]/[NH3],
which governs the pH of the solution.
The net effect is to neutralize most of
the OHϪ from NaOH. This slightly
decreases the ratio [NH4
ϩ]/[NH3],
which governs the pH of the solution.
Alternatively, one could substitute into
the Henderson–Hasselbalch equation
and solve for pH, then [H3Oϩ].
_?_ mmol NaF ϭ 200. mL ϫ ϭ 20. mmol HF
in 300. mL
_?_ mmol HF ϭ 100. mL ϫ ϭ 5.0 mmol HF
The molarities of NaF and HF in the solution are
ϭ 0.067 M NaF and ϭ 0.017 M HF
The appropriate equations and algebraic representations of concentrations are
Substituting into the Ka expression for hydrofluoric acid gives
Ka ϭ ϭ ϭ 7.2 ϫ 10Ϫ4
Can we assume that x is negligible compared with 0.067 and 0.017 in this expression? When
in doubt, solve the equation using the simplifying assumption. Then decide whether the assumption
was valid. Assume that (0.067 ϩ x) Ϸ 0.067 and (0.017 Ϫ x) Ϸ 0.017.
ᎏ
0
0
.
.
0
0
6
1
7
7
x
ᎏ ϭ 7.2 ϫ 10Ϫ4 x ϭ 1.8 ϫ 10Ϫ4 M ϭ [H3Oϩ]
pH ϭ 3.74
You should now work Exercise 34.
We often need a buffer solution of a given pH. One method by which such solutions
can be prepared involves adding a salt of a weak base (or weak acid) to a solution of the
weak base (or weak acid).
EXAMPLE 19-6 Preparation of Buffer Solutions
Calculate the numbers of moles and grams of NH4Cl that must be used to prepare 500. mL
of a buffer solution that is 0.10 M in aqueous NH3 and has a pH of 9.15.
Plan
Convert the given pH to the desired [OHϪ] by the usual procedure. Write the appropriate
equations for the reactions of NH4Cl and NH3 and represent the equilibrium concentrations.
Then, substitute into the Kb expression, and solve for the concentration of NH4Cl required.
Solution
Because the desired pH ϭ 9.15, pOH ϭ 14.00 Ϫ 9.15 ϭ 4.85. So [OHϪ] ϭ 10ϪpOH ϭ
10Ϫ4.85 ϭ 1.4 ϫ 10Ϫ5 M OHϪ desired. Let x mol/L be the necessary molarity of NH4Cl.
Because [OHϪ] ϭ 1.4 ϫ 10Ϫ5 M, this must be the [OHϪ] produced by ionization of NH3. The
equations and representations of equilibrium concentrations follow.
(x)(0.067 ϩ x)
ᎏᎏ
(0.017 Ϫ x)
[H3Oϩ][FϪ]
ᎏᎏ
[HF]
NaF Naϩ
0.067 M 0.067 M 0.067 M
(0.017 Ϫ x) M x M x M
HF H2O H3Oϩ
FϪ
FϪϩ ϩ
ϩ
Total [FϪ] ϭ (0.067 ϩ x) M
5.0 mmol HF
ᎏᎏ
300. mL
20. mmol NaF
ᎏᎏ
300. mL
0.050 mmol HF
ᎏᎏ
mL
0.10 mmol NaF
ᎏᎏ
mL
804 CHAPTER 19: Ionic Equilibria II: Buffers and Titration Curves







Our assumption is valid.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 18.10, Preparing Buffer
Solutions.
19-3 Preparation of Buffer Solutions 805
Substitution into the Kb expression for aqueous ammonia gives
Kb ϭ ϭ 1.8 ϫ 10Ϫ5 ϭ
NH4Cl is 100% dissociated, so x ϾϾ 1.4 ϫ 10Ϫ5. Then (x ϩ 1.4 ϫ 10Ϫ5) Ϸ x.
ϭ 1.8 ϫ 10Ϫ5 x ϭ 0.13 M ϭ [NH4
ϩ] ϭ MNH4Cl
Now we calculate the number of moles of NH4Cl that must be added to prepare 500. mL
(0.500 L) of buffer solution.
_?_ mol NH4Cl ϭ 0.500 L ϫ ϭ 0.065 mol NH4Cl
_?_ g NH4Cl ϭ 0.065 mol ϫ ϭ 3.5 g NH4Cl
You should now work Exercise 37.
We have used Ka or Kb expressions or the Henderson–Hasselbalch equation in its acid–
salt or base–salt form to find the pH of buffered solutions. Each of these approaches
involves a ratio of concentrations, for instance . Although the ratios are
[conjugate base]
ᎏᎏ
[acid]
53.5 g NH4Cl
ᎏᎏ
mol NH4Cl
0.13 mol NH4Cl
ᎏᎏ
L
(x)(1.4 ϫ 10Ϫ5)
ᎏᎏ
0.10
(x ϩ 1.4 ϫ 10Ϫ5)(1.4 ϫ 10Ϫ5)
ᎏᎏᎏᎏ
0.10 Ϫ 1.4 ϫ 10Ϫ5
[NH4
ϩ][OHϪ]
ᎏᎏ
[NH3]
NH4
ϩNH4Cl
(0.10 Ϫ 1.4 ϫ 10Ϫ5) M 1.4 ϫ 10Ϫ5 M 1.4 ϫ 10Ϫ5 M
x M x M x M
NH3 H2O NH4
ϩ
ClϪ
OHϪϩ ϩ
ϩ
Total [NH4
ϩ] ϭ (x ϩ 1.4 ϫ 10Ϫ5) M
100%
Here x does not represent a change in
concentration, but rather the initial
concentration of NH4Cl. We do not
assume that x ϽϽ 1.4 ϫ 10Ϫ5, but
rather the reverse.
Preparation of the buffer solution in Example 19-6. We add 3.5 grams of NH4Cl to a
500-mL volumetric flask and dissolve it in a little of the 0.10 M NH3 solution. We then
dilute to 500 mL with the 0.10 M NH3 solution.
ϭ ᎏᎏᎏ ϭ
mol conjugate base
ᎏᎏᎏ
mol acid
[conjugate base]
ᎏᎏ
[acid]
written in terms of molarities, it is not always necessary to use concentrations in the calculation.
Both reagents are present in a single buffer solution, so the solution volume cancels
from the molarity ratio. For example:
mol acid
ᎏ
L soln
mol conjugate base
ᎏᎏᎏ
L soln
Our Daily LivesCCHEMISTRY IN USE
Fun with Carbonates
Carbonates react with acids to produce carbon dioxide. This
property of carbonates has been exploited in many ways, both
serious and silly.
One of the giddiest applications of this behavior of carbonates
is in Mad Dawg, a foaming bubble gum developed
in the early 1990s. If you chew a piece of this gum, large
quantities of foam are produced so that it is difficult to keep
the colorful lather from oozing out of your mouth. The froth
begins to form as your teeth mix saliva with the gum’s ingredients
(sodium hydrogen carbonate, citric acid, malic acid,
food coloring, and flavoring).
How is this foam produced? When citric acid and malic
acid dissolve in saliva, they produce hydrogen ions which
decompose the sodium hydrogen carbonate (baking soda) to
produce carbon dioxide, a gas. These bubbles of carbon dioxide
produce the foam. Large quantities of foam are produced
because citric and malic acids taste sour, which stimulates
salivation.
A common medical recipe for a similar combination of
ingredients is found in Alka Seltzer tablets; these contain
sodium hydrogen carbonate, citric acid, and aspirin. The acid
and carbonate react in water to produce carbon dioxide,
which gives the familiar fizz of Alka Seltzer.
Makeup artists add baking soda to cosmetics to produce
monster-flesh makeup. When the hero throws acid (which
is actually vinegar, a dilute solution of acetic acid) into the
monster’s face, the acetic acid reacts with sodium hydrogen
carbonate to produce the disgustingly familiar scenes of
“dissolving flesh” that we see in horror movies. The ability
of baking soda to produce carbon dioxide delights children
of all ages as it creates monsters in the movies.
Many early fire extinguishers utilized the reaction of
sodium hydrogen carbonate with acids. A metal cylinder was
filled with a solution of sodium hydrogen carbonate and
water; a bottle filled with sulfuric acid was placed above the
water layer. Inverting the extinguisher activated it by causing
the acid to spill into the carbonate solution. The pressure
produced by gaseous carbon dioxide gas pushed the liquid
contents out through a small hose.
Kitchen oven fires can usually be extinguished by throwing
baking soda onto the flame. When heated, carbonates
decompose to produce carbon dioxide, which smothers fires
by depriving them of oxygen.
Chefs frequently use the heat-sensitive nature of carbonates
to test the freshness of a box of baking soda. Pouring
some boiling water over a little fresh baking soda results in
active bubbling. Less active bubbling means the baking soda
is unlikely to work well in a baking recipe.
Ronald DeLorenzo
Middle Georgia College
Alka Seltzer™.
Thus we see that a molarity ratio in the acid–salt Henderson–Hasselbalch equation can
be treated as a mole (or millimole) ratio. A similar conclusion can be reached for the base–
salt version of the Henderson–Hasselbalch equation or for the Ka or Kb expressions that
were used in previous buffer calculations.
In practice, a common method of preparing a buffer solution is by partial neutralization
of a weak acid solution by adding a strong base solution. For example,
HA ϩ NaOH 88n NaA ϩ H2O (partial)
If an appreciable amount of the weak acid remains unneutralized, then this solution
contains significant concentrations of a weak acid and its conjugate base, just as though we
had added the salt from a separate solution; thus, it is a buffer solution. Example 19-7
illustrates the preparation of a buffer by this method.
EXAMPLE 19-7 Buffer Preparation by Partial Neutralization
Calculate the pH of a solution obtained by mixing 400. mL of a 0.200 M acetic acid solution
and 100. mL of a 0.300 M sodium hydroxide solution.
Plan
Sodium hydroxide, NaOH, is a strong base, so it reacts with acetic acid, CH3COOH, to form
sodium acetate, NaCH3COO. If an appreciable amount of excess acetic acid is still present
after the sodium hydroxide has reacted, the excess acetic acid and the newly formed sodium
acetate solution form a buffered solution.
Solution
We first calculate how much of the weak acid has been neutralized. The numbers of millimoles
of CH3COOH and NaOH mixed are calculated as
mmol CH3COOH ϭ (0.200 mmol/mL) ϫ 400. mL ϭ 80.0 mmol
mmol NaOH ϭ (0.300 mmol/mL) ϫ 100. mL ϭ 30.0 mmol
Not enough NaOH is present to neutralize all of the CH3COOH, so NaOH is the limiting
reactant.
NaOH ϩ CH3COOH 88n NaCH3COO ϩ H2O
start 30.0 mol 80.0 mmol 0 —
change Ϫ30.0 mmol Ϫ30.0 mmol ϩ30.0 mmol —
after rxn 0.0 mmol 50.0 mmol 30.0 mmol
Because NaCH3COO is a soluble salt, it provides 30.0 mmol CH3COOϪ to the solution. This
solution contains a significant amount of CH3COOH not yet neutralized and a significant
amount of its conjugate base, CH3COOϪ. We recognize this as a buffer solution and can use
the Henderson–Hasselbalch equation to find the pH.
pH ϭ pKa ϩ log ϭ pKa ϩ log
ϭ 4.74 ϩ log ᎏ
3
5
0
0
.
.
0
0
ᎏ ϭ 4.74 ϩ log(0.600) ϭ 4.74 ϩ (Ϫ0.222)
ϭ 4.52
You should now work Exercises 52b to 52f.
mmol CH3COOϪ
ᎏᎏᎏ
mmol CH3COOH
mmol conjugate base
ᎏᎏᎏ
mmol acid
19-3 Preparation of Buffer Solutions 807
We could solve this problem using the
Ka expression as we did in Example
19-5.
ACID–BASE INDICATORS
In Section 11-2 we described acid–base titrations and the use of indicators to tell us when
to stop a titration. Detection of the end point in an acid–base titration is only one of the
important uses of indicators.
An indicator is an organic dye; its color depends on the concentration of H3Oϩ ions,
or pH, in the solution. By the color an indicator displays, it “indicates” the acidity or
basicity of a solution. Figure 19-1 displays solutions that contain three common indicators
in solutions over the pH range 3 to 11. Carefully study Figure 19-1 and its legend.
The first indicators used were vegetable dyes. Litmus is a familiar example. Most of
the indicators that we use in the laboratory today are synthetic compounds; that is, they
have been made in laboratories by chemists. Phenolphthalein is the most common acid–
base indicator. It is colorless in solutions of pH less than 8 ([H3Oϩ] Ͼ 10Ϫ8 M) and turns
bright pink as pH approaches 10.
Many acid–base indicators are weak organic acids, HIn, where “In” represents various
complex organic groups. Bromthymol blue is such an indicator. Its ionization constant is
19-4
808 CHAPTER 19: Ionic Equilibria II: Buffers and Titration Curves
Phenolphthalein was the active
component of the laxative Ex-Lax. It is
sometimes added to laboratory ethyl
alcohol to discourage consumption.
Figure 19-1 Three common
indicators in solutions that cover
the pH range 3 to 11 (the black
numbers). (a) Methyl red is red
at pH 4 and below; it is yellow at
pH 7 and above. Between pH 4 and
pH 7 it changes from red to redorange,
to orange, to yellow.
(b) Bromthymol blue is yellow at
pH 6 and below; it is blue at pH 8
and above. Between pH 6 and 8 it
changes from yellow to yellowgreen,
to green, to blue-green, to
blue. (c) Phenolphthalein is colorless
below pH 8 and bright pink above
pH 10. It changes from colorless to
pale pink, to pink, to bright pink in
the pH range 8 to 10.
3 4 5 6 7 8 9 10 11
(a)
3 4 5 6 7 8 9 10 11
(b)
3 4 5 6 7 8 9 10 11
(c)
7.9 ϫ 10Ϫ8. We can represent its ionization in dilute aqueous solution and its ionization
constant expression as
HIn ϩ H2O 34 H3Oϩ ϩ InϪ Ka ϭ ϭ 7.9 ϫ 10Ϫ8
ocolor 1o ocolor 2o
m for bromthymol blue n
oyellowo oblueo
HIn represents nonionized acid molecules, and InϪ represents the anion (conjugate base)
of HIn. The essential characteristic of an acid–base indicator is that HIn and InϪ must
have quite different colors. The relative amounts of the two species determine the color
of the solution. Adding an acid favors the reaction to the left and gives more HIn molecules
(color 1). Adding a base favors the reaction to the right and gives more InϪ ions
(color 2). The ionization constant expression can be rearranged.
ϭ Ka so ϭ
This shows clearly how the [InϪ]/[HIn] ratio depends on [H3Oϩ] (or on pH) and the Ka
value for the indicator. As a rule of thumb, when [InϪ]/[HIn] Ն 10, color 2 is observed;
conversely, when [InϪ]/[HIn] Յ ᎏ
1
1
0
ᎏ, color 1 is observed.
Universal indicators are mixtures of several acid–base indicators that display a continuous
range of colors over a wide range of pH values. Figure 18-2 shows concentrated
solutions of a universal indicator in flat dishes so that the colors are very intense. The
juice of red (purple) cabbage is a universal indicator. Figure 19-2 shows the color of red
cabbage juice in solutions within the pH range 1 to 13.
One important use of universal indicators is in commercial indicator papers, which are
small strips of paper impregnated with solutions of universal indicators. A strip of the
paper is dipped into the solution of interest, and the color of the indicator on the paper
indicates the pH of the solution. The photographs on page 757 (solutions of universal
indicators) and page 763 (an indicator paper) illustrate the use of universal indicators to
estimate pH. We shall describe the use of indicators in titrations more fully in Sections
19-5 and 19-6.
Ka
ᎏ
[H3Oϩ]
[InϪ]
ᎏ
[HIn]
[H3Oϩ][InϪ]
ᎏᎏ
[HIn]
[H3Oϩ][InϪ]
ᎏᎏ
[HIn]
19-4 Acid–Base Indicators 809
Bromthymol blue indicator is yellow
in acidic solutions and blue in basic
solutions.
Figure 19-2 The juice of the red
(purple) cabbage is a naturally
occurring universal indicator. From
left to right are solutions of pH 1, 4,
7, 10, and 13.
TITRATION CURVES
STRONG ACID/STRONG BASE TITRATION CURVES
A titration curve is a plot of pH versus the amount (usually volume) of acid or base added.
It displays graphically the change in pH as acid or base is added to a solution and shows
how pH changes near the equivalence point.
The point at which the color of an indicator changes in a titration is known as the end
point. It is determined by the Ka value for the indicator (Section 19-4). Table 19-4 shows
a few acid–base indicators and the pH ranges over which their colors change. Typically,
color changes occur over a range of 1.5 to 2.0 pH units.
The equivalence point is the point at which chemically equivalent amounts of acid
and base have reacted.
Ideally, the end point and the equivalence point in a titration should coincide.
In practice, we try to select an indicator whose range of color change includes the equivalence
point. We use the same procedures in both standardization and analysis to minimize
any error arising from a difference between end point and equivalence point.
Review Section 11-2 before you consider the titration of 100.0 mL of a 0.100 M solution
of HCl with a 0.100 M solution of NaOH. As we know, NaOH and HCl react in a
1:1 ratio. We calculate the pH of the solution at several stages as NaOH is added.
1. Before any NaOH is added to the 0.100 M HCl solution:
100%
HCl ϩ H2O 8888n H3Oϩ ϩ ClϪ
[H3Oϩ] ϭ 0.100 M so pH ϭ 1.00
2. After 20.0 mL of 0.100 M NaOH has been added:
HCl ϩ NaOH 88n NaCl ϩ H2O
start 10.0 mmol 2.0 mmol 0 mmol
change Ϫ2.0 mmol Ϫ2.0 mmol ϩ2.0 mmol
after rxn 8.0 mmol 0 mmol 2.0 mmol
19-5
810 CHAPTER 19: Ionic Equilibria II: Buffers and Titration Curves
Titrations are usually done with
50-mL or smaller burets. We have
used 100. mL of solution in this
example to simplify the arithmetic.
If one prefers, these calculations can be
done using moles and liters in the
place of millimoles and milliliters.
TABLE 19-4 Range and Color Changes of Some Common Acid–Base Indicators
pH Scale
Indicators 1 2 3 4 5 6 7 8 9 10 11 12 13
methyl orange
methyl red
bromthymol blue
neutral red
phenolphthalein
m8888 red 8888n3.1 88888 4.4 m8888888888888888888888 yellow 8888888888888888888888n
m8888888 red 8888888n 4.4 888888888 6.2 m8888888888888888 yellow 88888888888888888n
m888888888888 yellow 888888888888n 6.2 8888 7.6 m8888888888888 blue 88888888888888n
m88888888888888 red 88888888888888n 6.8 88888 8.0 m88888888888 yellow 88888888888n
m8888888888888888 colorless 8888888888888888n 8.0 88888888 10.0 m88 bright pink 88n colorless beyond 13.0
The concentration of unreacted HCl in the total volume of 120. mL is
MHCl ϭ ϭ 0.067 M HCl
[H3Oϩ] ϭ 6.7 ϫ 10Ϫ2 M so pH ϭ 1.17
3. After 50.0 mL of 0.100 M NaOH has been added (midpoint of the titration):
HCl ϩ NaOH 88n NaCl ϩ H2O
start 10.0 mmol 5.0 mmol 0 mmol
change Ϫ5.0 mmol Ϫ5.0 mmol ϩ5.0 mmol
after rxn 5.0 mmol 0 mmol 5.0 mmol
MHCl ϭ ϭ 0.033 M HCl
[H3Oϩ] ϭ 3.3 ϫ 10Ϫ2 M so pH ϭ 1.48
4. After 100. mL of 0.100 M NaOH has been added:
HCl ϩ NaOH 88n NaCl ϩ H2O
start 10.0 mmol 10.0 mmol 0 mmol
change Ϫ10.0 mmol Ϫ10.0 mmol ϩ10.0 mmol
after rxn 0 mmol 0 mmol 10.0 mmol
We have added enough NaOH to neutralize the HCl exactly so this is the equivalence
point. A strong acid and a strong base react to give a neutral salt solution so
pH ϭ 7.00.
5. After 110.0 mL of 0.100 M NaOH has been added:
HCl ϩ NaOH 88n NaCl ϩ H2O
start 10.0 mmol 11.0 mmol 0 mmol
change Ϫ10.0 mmol Ϫ10.0 mmol ϩ10.0 mmol
after rxn 0 mmol 1.0 mmol 10.0 mmol
The pH is determined by the excess NaOH.
5.00 mmol HCl
ᎏᎏ
150. mL
8.00 mmol HCl
ᎏᎏ
120. mL
19-5 Strong Acid/Strong Base Titration Curves 811
Some household products. Each
solution contains a few drops of a
universal indicator. A color of yellow
or red indicates a pH less than 7. A
green to purple color indicates a pH
greater than 7.
The end point of the titration of 0.100 M HCl with 0.100 M NaOH using another
indicator, bromthymol blue.
MNaOH ϭ ϭ 0.0048 M NaOH
[OHϪ] ϭ 4.8 ϫ 10Ϫ3 M so pOH ϭ 2.32 and pH ϭ 11.68
Table 19-5 displays the data for the titration of 100.0 mL of 0.100 M HCl by 0.100
M NaOH solution. A few additional points have been included to show the shape of the
curve better. These data are plotted in Figure 19-3a. This titration curve has a long “vertical
section” over which the pH changes very rapidly with the addition of very small amounts
of base. The pH changes from 3.60 (99.5 mL NaOH added) to 10.40 (100.5 mL of NaOH
added) in the vicinity of the equivalence point (100.0 mL NaOH added). The midpoint
of the vertical section (pH ϭ 7.00) is the equivalence point. We can separate the calculations
on this kind of titration into four distinct types that correspond to four regions of
the titration curves.
1. Before any strong base is added, the pH depends on the strong acid alone.
2. After some strong base has been added, but before the equivalence point, the
remaining (excess) strong acid determines the pH.
3. At the equivalence point, the solution is neutral.
4. Beyond the equivalence point, excess strong base determines the pH.
Ideally, the indicator color change should occur at pH ϭ 7.00. For practical purposes,
indicators with color changes in the pH range 4 to 10 can be used in the titration of strong
acids and strong bases because the vertical portion of the titration curve is so long. Figure
19-3 shows the ranges of color changes for methyl red and phenolphthalein, two widely
used indicators. Both fall within the vertical section of the NaOH/HCl titration curve.
When a strong acid is added to a solution of a strong base, the titration curve is inverted,
but its essential characteristics are the same (Figure 19-3b).
In Figure 19-3a we see that the curve rises very slowly before the equivalence point.
It then rises very rapidly near the equivalence point because there is no hydrolysis. The
curve becomes almost flat beyond the equivalence point.
1.0 mmol NaOH
ᎏᎏ
210. mL
812 CHAPTER 19: Ionic Equilibria II: Buffers and Titration Curves
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 18.12, Titration Curves.
TABLE 19-5 Titration Data for 100.0 mL of 0.100 M HCl versus NaOH
mL of 0.100 M mmol NaOH mmol Excess
NaOH Added Added Acid or Base pH
0.0 0.00 10.0 H3Oϩ 1.00
20.0 2.00 8.0 1.17
50.0 5.00 5.0 1.48
90.0 9.00 1.0 2.28
99.0 9.90 0.10 3.30
99.5 9.95 0.05 3.60
100.0 10.00 0.00 (eq. pt.) 7.00
100.5 10.05 0.05 OHϪ 10.40
110.0 11.00 1.00 11.68
120.0 12.00 2.00 11.96
19-6 Weak Acid/Strong Base Titration Curves 813
WEAK ACID/STRONG BASE TITRATION CURVES
When a weak acid is titrated with a strong base, the curve is quite different. The solution
is buffered before the equivalence point. It is basic at the equivalence point because salts
of weak acids and strong bases hydrolyze to give basic solutions. So, we can separate the
calculations on this kind of titration into four distinct types, which correspond to four
regions of the titration curves.
1. Before any base is added, the pH depends on the weak acid alone.
2. After some base has been added, but before the equivalence point, a series of weak
acid/salt buffer solutions determines the pH.
3. At the equivalence point, hydrolysis of the anion of the weak acid determines
the pH.
4. Beyond the equivalence point, excess strong base determines the pH.
Consider the titration of 100.0 mL of 0.100 M CH3COOH with 0.100 M NaOH solution.
(The strong electrolyte is added to the weak electrolyte.)
1. Before any base is added, the pH is 2.89 (Example 18-11 and Table 18-5).
2. As soon as some NaOH is added, but before the equivalence point, the solution is
buffered because it contains both CH3COOH and NaCH3COO.
19-6
pH
mL NaOH added
(a)
1.0
0
0 10 20 30 40 50 60 70 80 90 100 110 120
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
11.0
12.0
13.0
Range of phenolphthalein indicator
Equivalence point
Range of methyl red indicator
pH
mL HCl added
(b)
1.0
0
0 10 20 30 40 50 60 70 80 90 100 110 120
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
11.0
12.0
13.0
Range of phenolphthalein indicator
Equivalence point
Range of methyl red indicator
Figure 19-3 (a) The titration curve for 100. mL of 0.100 M HCl with 0.100 M NaOH.
Note that the “vertical” section of the curve is quite long. The titration curves for other
strong acids and bases are identical with this one if the same concentrations of acid and
bases are used and if both are monoprotic. (b) The titration curve for 100. mL of 0.100 M
NaOH with 0.100 M HCl. This curve is similar to that in part (a), but inverted.
NaOH ϩ CH3COOH 88n NaCH3COO ϩ H2O
lim amt excess
For instance, after 20.0 mL of 0.100 M NaOH solution has been added, we have
NaOH ϩ CH3COOH 88n NaCH3COO ϩ H2O
start 2.00 mmol 10.00 mmol 0 mmol
change Ϫ2.00 mmol Ϫ2.00 mmol ϩ2.00 mmol
after rxn 0 mmol 8.00 mmol 2.00 mmol
We recognize that this is a buffer solution, so we can use the ratio (mmol conjugate
base)/(mmol acid) in the Henderson–Hasselbalch equation as we did in
Example 19-7.
pH ϭ pKa ϩ log ϭ pKa ϩ log
ϭ 4.74 ϩ log ᎏ
2
8
.
.
0
0
0
0
ᎏ ϭ 4.74 ϩ log(0.250) ϭ 4.74 ϩ (Ϫ0.602) ϭ 4.14
After some NaOH has been added, the solution contains both NaCH3COO and
CH3COOH, and so it is buffered until the equivalence point is reached. All points
before the equivalence point are calculated in the same way.
3. At the equivalence point, the solution is 0.0500 M in NaCH3COO.
NaOH ϩ CH3COOH 88n NaCH3COO ϩ H2O
start 10.0 mmol 10.0 mmol 0 mmol
change Ϫ10.0 mmol Ϫ10.0 mmol ϩ10.0 mmol
after rxn 0 mmol 0 mmol 10.0 mmol
MNaCH3COO ϭ ϭ 0.0500 M NaCH3COO
The pH of a 0.0500 M solution of NaCH3COO is 8.72 (Example 18-19 shows a
similar calculation). The solution is distinctly basic at the equivalence point because
of the hydrolysis of the acetate ion.
4. Beyond the equivalence point, the concentration of the excess NaOH determines
the pH of the solution just as it did in the titration of a strong acid.
Table 19-6 lists several points on the titration curve, and Figure 19-4 shows the titration
curve for 100.0 mL of 0.100 M CH3COOH titrated with a 0.100 M solution of
NaOH. This titration curve has a short vertical section (pH Ϸ 7 to 10), and the indicator
range is limited. Phenolphthalein is the indicator commonly used to titrate weak acids
with strong bases (see Table 19-4).
The titration curves for weak bases and strong acids are similar to those for weak acids
and strong bases except that they are inverted (recall that strong is added to weak). Figure
19-5 displays the titration curve for 100.0 mL of 0.100 M aqueous ammonia titrated with
0.100 M HCl solution.
10.0 mmol NaCH3COO
ᎏᎏᎏ
200. mL
mmol CH3COOϪ
ᎏᎏᎏ
mmol CH3COOH
mmol conjugate base
ᎏᎏᎏ
mmol acid
814 CHAPTER 19: Ionic Equilibria II: Buffers and Titration Curves
We could use the Ka expression for
these calculations as we did in Example
19-5.
Just before the equivalence point, the
solution contains relatively high
concentrations of NaCH3COO and
relatively low concentrations of
CH3COOH. Just after the equivalence
point, the solution contains relatively
high concentrations of NaCH3COO
and relatively low concentrations of
NaOH, both basic components. In
both regions our calculations are only
approximations. Exact calculations of
pH in these regions are beyond the
scope of this text.
19-6 Weak Acid/Strong Base Titration Curves 815
Figure 19-4 The titration curve for 100. mL of 0.100 M
CH3COOH with 0.100 M NaOH. The “vertical” section of
this curve is much shorter than those in Figure 19-3 because
the solution is buffered before the equivalence point.
Figure 19-5 The titration curve for 100. mL of 0.100 M
aqueous ammonia with 0.100 M HCl. The vertical section of
the curve is relatively short because the solution is buffered
before the equivalence point. The curve is very similar to that
in Figure 19-4, but inverted.
pH
mL NaOH added
1
0
0 10 20 30 40 50 60 70 80 90 100 110 120
2
3
4
5
6
7
8
9
10
11
12
Indicator
range Equivalence point
Buffered solution
(CH3COOH/NaCH3COO)
pH
mL HCl added
1
0
0 10 20 30 40 50 60 70 80 90 100 110 120
2
3
4
5
6
7
8
9
10
11
12
Indicator
range
Equivalence point
(NH3/NH4Cl)
Buffered solution
TABLE 19-6 Titration Data for 100.0 mL of 0.100 M CH3COOH with
0.100 M NaOH
mL 0.100 M mmol Base mmol Excess
NaOH Added Added Acid or Base pH
0.0 mL 0 10.0 CH3COOH 2.89
20.0 mL 2.00 8.00 4.14
50.0 mL 5.00 5.00 4.74
75.0 mL buffered 7.50 2.50 5.22
90.0 mL region 9.00 1.00 5.70
95.0 mL 9.50 0.50 6.02
99.0 mL 9.90 0.10 6.74
100.0 mL 10.0 0 (equivalence point) 8.72
101.0 mL 10.1 0.10 OHϪ 10.70
110.0 mL 11.0 1.0 11.68
120.0 mL 12.0 2.0 11.96











WEAK ACID/WEAK BASE TITRATION CURVES
In titration curves for weak acids and weak bases, pH changes near the equivalence point
are too small for color indicators to be used. The solution is buffered both before and
after the equivalence point. Figure 19-6 shows the titration curve for 100.0 mL of 0.100 M
CH3COOH solution titrated with 0.100 M aqueous NH3. The calculation of values on
the curve in Figure 19-6 other than the initial pH and the pH at the equivalence point is
beyond the scope of this text.
19-7
Problem-Solving Tip: Titration Curves
You can consider a titration curve in four parts.
1. Initial solution (before any titrant is added).
2. Region before the equivalence point. This may or may not be buffered. The solution
is buffered in this region if the substance being titrated is a weak acid or weak
base.
3. Equivalence point. Its location depends on the concentrations of the acid and the
base solutions; its pH depends on the strengths of the acid and base.
4. Region beyond the equivalence point. This becomes nearly flat as more and more
excess reactant is added. We often calculate only one or two points in this region.
Recognizing the four regions of a titration curve allows you to decide which kind of
calculation is required.
SUMMARY OF ACID–BASE CALCULATIONS
In this and the previous chapter we have discussed several different types of acidic or alkaline
solutions. Table 19-7 summarizes those many different types of solutions.
19-8
Figure 19-6 The titration curve
for 100. mL of 0.100 M CH3COOH
with 0.100 M aqueous NH3. Because
the solution is buffered before and
after the equivalence point, the
vertical section of the curve is too
short to be noticed. Color indicators
cannot be used in such titrations.
Instead, physical methods such as
conductivity measurements can be
used to detect the end point.
12
11
10
9
8
7
6pH
5
4
3
2
1
0
0 10 20 30 40 50 60
mL NH3 added
(NH3/NH4CH3COO)
Buffered solution
Buffered solution
(CH3COOH/NH4CH3COO)
Equivalence point
70 80 90 100 110 120
19-8 Summary of Acid–Base Calculations 817
TABLE 19-7 A Review of Our Understanding of Nonneutral Aqueous Solutions
Type of Aq. Solution Example(s) Resulting Chemistry or Type of Solution; Calculation Section
strong acid HNO3 or HCl Complete ionization; reaction goes to completion to form H3Oϩ 18-1
(represented as HX) HX ϩ H2O 88n H3Oϩ ϩ XϪ
strong base NaOH or Complete ionization; reaction goes to completion to form OHϪ 18-1
[represented as Ba(OH)2 MOH 88n Mϩ ϩ OHϪ
MOH or M(OH)2] M(OH)2 88n M2ϩ ϩ 2OHϪ
weak acid CH3COOH or Partial ionization; HA ϩ H2O 34 H3Oϩ ϩ AϪ 18-4
(represented as HA) HCN Must solve equilibrium expression involving Ka for the acid
using equilibrium concentrations: Ka ϭ ᎏ
[H3
[
O
H
ϩ
A
][
]
AϪ]
ᎏ
weak base NH3 Partial ionization; B ϩ H2O 34 BHϩ ϩ OHϪ 18-4
(represented as B) Must solve equilibrium expression involving Kb for the base,
using equilibrium concentrations: Kb ϭ ᎏ
[BHϩ
[
]
B
[O
]
HϪ]
ᎏ
Salt of strong NaNO3 No hydrolysis—neutral solution 18-7
base & strong acid (salt of NaOH
(represented as MX) and HNO3)
Salt of strong base NaCN Hydrolysis of conjugate base (AϪ) of weak acid :::F basic solution 18-8
& weak acid (salt of NaOH AϪ ϩ H2O 34 HA ϩ OHϪ
(represented as MA) and HCN)
Solve weak base equilibrium for AϪ using Kb for AϪ ϭ ᎏ
Ka
K
fo
w
r HA
ᎏ
Salt of weak base NH4NO3 Hydrolysis of conjugate acid (BHϩ) of weak base :::F acidic solution 18-9
& strong acid (salt of NH3 BHϩ ϩ H2O 34 B ϩ H3Oϩ
(represented as and HNO3)
Solve weak acid equilibrium for BHϩ using Ka for BHϩ ϭ ᎏ
Kb
K
fo
w
r B
ᎏBHX)
Salt of weak base & NH4CN Hydrolysis of conjugate base (AϪ) of weak acid :::F basic solution and 18-10
weak acid (salt of NH3 Hydrolysis of conjugate acid (BHϩ) of weak base :::F acidic solution.
(represented as BHA) and HCN) This solution can be basic, neutral, or acidic, depending on
which hydrolysis occurs to a greater extent; use Kb and Ka
values for the hydrolysis reactions to tell which ion will be the
dominant factor. (We did not do calculations for this type of salt.)
Weak acid & salt of its CH3COOH ϩ Salt ionizes completely; MA 88n Mϩ ϩ AϪ 19-1,
conjugate base NaCH3COO Mixture of weak acid (HA) and its conjugate base (AϪ) in 19-2,
(represented as (or the product significant concentrations gives a buffer. 19-3
HA & MA) of the partial Use Henderson–Hasselbalch equation for acid–salt buffer
neutralization in (with initial concentrations or mole ratio):
a titration of a
weak acid with a pH ϭ pKa ϩ log ᎏ
[conju
[a
g
c
a
i
t
d
e
]
base]
ᎏ
strong base
Weak base & salt of NH3 ϩ NH4Cl Salt ionizes completely; NH4Cl 88n NH4
ϩ ϩ ClϪ 19-1,
its conjugate (or the product Mixture of weak base (NH3) and its conjugate acid (NH4
ϩ) in 19-2,
acid of the partial significant concentrations gives a buffer. 19-3
(represented as neutralization in Use Henderson–Hasselbalch equation for base–salt buffer
B & BHX) a titration of a (with initial concentrations or mole ratio):
weak base with a
strong acid) pOH ϭ pKb ϩ log ᎏ
[conju
[b
g
a
a
s
t
e
e
]
acid]
ᎏ
818 CHAPTER 19: Ionic Equilibria II: Buffers and Titration Curves
Key Terms
Buffer solution A solution that resists changes in pH when
strong acids or strong bases are added. A buffer solution contains
an acid and its conjugate base, so it can react with added
base or acid. Common buffer solutions contain either (1) a weak
acid and a soluble ionic salt of the weak acid or (2) a weak base
and a soluble ionic salt of the weak base.
Common ion effect Suppression of ionization of a weak electrolyte
by the presence in the same solution of a strong
electrolyte containing one of the same ions as the weak elec-
trolyte.
End point The point at which an indicator changes color and a
titration should be stopped.
Equivalence point The point at which chemically equivalent
amounts of reactants have reacted.
Henderson–Hasselbalch equation An equation that enables us
to calculate the pH or pOH of a buffer solution directly.
For acid–salt buffer pH ϭ pKa ϩ log ᎏ
[co
[
n
a
j
c
.
i
b
d
a
]
se]
ᎏ
For base–salt buffer pOH ϭ pKb ϩ log ᎏ
[co
[
n
b
j
a
.
s
a
e
c
]
id]
ᎏ
Indicator (for acid–base titrations) An organic compound that
exhibits different colors in solutions of different acidities; used
to indicate the point at which reaction between an acid and a
base is complete.
Titration A procedure in which one solution is added to another
solution until the chemical reaction between the two solutes is
complete; usually the concentration of one solution is known
and that of the other is unknown.
Titration curve (for acid–base titration) A plot of pH versus
volume of acid or base solution added.
Exercises
Note All exercises in this chapter assume a temperature of 25°C unless
they specify otherwise. Values of Ka and Kb can be found in Appendix F
or will be specified in the exercise.
Basic Ideas
*01. (a) What is the relationship between pH and pOH?
(b) What is the relationship between Ka and pH? (c) What
is the relationship between Ka and pOH?
*02. Write the balanced equation for an acid–base reaction
that would produce each of the following salts; predict
whether an aqueous solution of each salt is acidic, basic,
or neutral. (a) NaNO3; (b) KCN; (c) Al2(SO4)3;
(d) Mg(CH3COO)2; (e) (NH4)2SO4.
*03. Write the balanced equation for an acid–base reaction that
would produce each of the following salts; predict whether
an aqueous solution of each salt is acidic, basic, or neutral.
(a) CaF2; (b) ZnS; (c) AsCl3; (d) Sr3(PO4)2; (e) NH4NO3.
*04. Under what circumstances can it be predicted that a neutral
solution is produced by an acid–base reaction? (Hint:
This question has more than one answer.)
The Common Ion Effect and Buffer Solutions
*05. Which of the following solutions are buffered? Each
solution was prepared by mixing and diluting appropriate
quantities of the two solutes to yield the concentrations
indicated. Explain your decision for each solution.
(a) 0.10 M HCN and 0.10 M NaCN; (b) 0.10 M NaOH
and 0.10 M NaCl; (c) 0.10 M NH3 and 0.10 M NH4Br;
(d) 0.10 M NaOH and 0.90 M KOH.
*06. Which of the following solutions are buffered? Each
solution was prepared by mixing and diluting appropriate
quantities of the two solutes to yield the concentrations
indicated. Explain your decision for each solution.
(a) 1.0 M HCN and 0.20 M NaCN; (b) 0.10 M NaF and
0.10 M HF; (c) 0.10 M NH4Cl and 0.90 M NH4Br;
(d) 0.10 M NaCl and 0.20 M HF.
*07. Suppose that you have a solution that is 0.50 M in
methylamine, CH3NH2, and 0.00050 M in the salt methylammonium
chloride, CH3NH3Cl. Would you expect this
to be an effective buffer solution? Why or why not?
*08. Calculate pH for each of the following buffer solutions.
(a) 0.10 M HF and 0.25 M KF; (b) 0.050 M CH3COOH
and 0.025 M Ba(CH3COO)2.
*09. The pKa of HOCl is 7.45. Calculate the pH of a solution
that is 0.0444 M HOCl and 0.0888 M NaOCl.
*10. Calculate the concentration of OHϪ and the pH for the
following buffer solutions. (a) 0.30 M NH3(aq) and 0.20
M NH4NO3; (b) 0.15 M NH3(aq) and 0.20 M (NH4)2SO4.
*11. Calculate the concentration of OHϪ and the pH for the
following solutions. (a) 0.45 M NH3(aq) and 0.25 M
NH4NO3; (b) 0.10 M aniline, C6H5NH2, and 0.20 M
anilinium chloride, C6H5NH3Cl.
*12. Buffer solutions are especially important in our body
fluids and metabolism. Write net ionic equations to illustrate
the buffering action of (a) the H2CO3/NaHCO3
buffer system in blood and (b) the NaH2PO4/Na2HPO4
buffer system inside cells.
*13. Calculate the ratio of [NH3]/[NH4
ϩ] concentrations that
gives (a) solutions of pH ϭ 9.75 and (b) solutions of pH ϭ
9.10.
*14. We prepare two solutions as follows. In solution A 0.50
mole of potassium acetate is added to 0.25 mole of acetic
acid and diluted to a final volume of 1.00 liter. In solution
Exercises 819
B 0.25 mole of potassium acetate is added to 0.50 mole of
acetic acid and diluted to 1.00 liter. (a) Which solution is
expected to have the lower pH? (b) Explain how you can
reach your conclusion without calculating the pH of each
solution.
*15. Compare the pH of a 0.15 M acetic acid solution to the
pH of a solution composed of 0.15 M acetic acid to which
0.15 mole of sodium acetate per liter has been added.
*16. Calculate the mass of sodium acetate that must be added
to one liter of a 0.125 M acetic acid solution to produce a
buffer with the pH of 4.75. Assume that there is no volume
change.
*17. A solution is produced by dissolving 0.075 mole of formic
acid and 0.075 mole of sodium formate in sufficient water
to produce 1.00 L of solution. (a) Calculate the pH of this
buffer solution. (b) Calculate the pH of the solution after
an additional 0.010 mole of sodium formate is dissolved.
Buffering Action
*18. Consider the ionization of formic acid, HCOOH.
HCOOH ϩ H2O 34 HCOOϪ ϩ H3Oϩ
What effect does the addition of sodium formate,
NaHCOO, have on the fraction of formic acid molecules
that undergo ionization in aqueous solution?
*19. Briefly describe why the pH of a buffer solution remains
nearly constant when small amounts of acid or base are
added. Over what pH range do we observe the best buffering
action (nearly constant pH)?
*20. What is the pH of a solution that is 0.10 M in HClO4 and
0.10 M KClO4? Is this a buffer solution?
*21. (a) Find the pH of a solution that is 0.50 M in formic acid
and 0.40 M in sodium formate. (b) Find the pH after
0.050 mol HCl has been added to 1.0 liter of the solution.
*22. One liter of 0.400 M NH3 solution also contains 12.78 g
of NH4Cl. How much will the pH of this solution change
if 0.142 mole of gaseous HCl is bubbled into it?
*23. (a) Find the pH of a solution that is 0.12 M in nitrous
acid and 0.15 M in sodium nitrite. (b) Find the pH after
0.15 mol of NaOH has been added to 1.0 liter of the
solution.
*24. (a) Find the pH of a solution that is 1.00 M in NH3 and
0.80 M in NH4Cl. (b) Find the pH of the solution after
0.10 mol of HCl has been added to 1.0 liter of the solution.
(c) A solution was prepared by adding NaOH to pure
water to give 1.00 liter of solution whose pH ϭ 9.34. Find
the pH of this solution after 0.10 mol of HCl has been
added to it.
*25. (a) Calculate the concentrations of CH3COOH and
CH3COOϪ in a solution in which their total concentration
is 0.200 mol/L and the pH is 4.50. (b) If 0.0100 mol
of solid NaOH is added to 1.00 L of this solution, how
much does the pH change?
*26. A solution contains bromoacetic acid, BrCH2COOH, and
sodium bromoacetate, NaBrCH2COO, with a total concentration
of 0.30 mol/L. If the pH is 3.10, what are the
concentrations of the acid and the salt? Ka ϭ 2.0 ϫ 10Ϫ3
for BrCH2COOH.
*27. Calculate the concentration of propionate ion,
CH3CH2COOϪ, in equilibrium with 0.020 M
CH3CH2COOH (propionic acid) and 0.10 M Hϩ from
hydrochloric acid. Ka ϭ 1.3 ϫ 10Ϫ5 for CH3CH2COOH.
*28. Calculate the concentration of C2H5NH3
ϩ in equilibrium
with 0.015 M C2H5NH2 (ethylamine) and 0.0010 M OHϪ
ion from sodium hydroxide. Kb ϭ 4.7 ϫ 10Ϫ4 for ethylamine.
*29. When chlorine gas is dissolved in water to make “chlorine
water,” HCl (a strong acid) and HOCl (a weak acid) are
produced in equal amounts.
Cl2(g) ϩ H2O(ᐉ) 88n HCl(aq) ϩ HOCl(aq)
What is the concentration of OClϪ ion in a solution containing
0.010 mol of each acid in 1.00 L of solution?
Preparation of Buffer Solutions
*30. A buffer solution of pH 5.30 is to be prepared from propionic
acid and sodium propionate. The concentration of
sodium propionate must be 0.50 mol/L. What should be
the concentration of the acid? Ka ϭ 1.3 ϫ 10Ϫ5 for
CH3CH2COOH.
*31. We need a buffer with pH 9.00. It can be prepared from
NH3 and NH4Cl. What must be the [NH4
ϩ]/[NH3] ratio?
*32. What volumes of 0.150 M acetic acid and 0.100 M NaOH
solutions must be mixed to prepare 1.00 L of a buffer solution
of pH 4.50 at 25°C?
*33. One liter of a buffer solution is prepared by dissolving
0.150 mol of NaNO2 and 0.070 mol of HCl in water. What
is the pH of this solution? If the solution is diluted twofold
with water, what is the pH?
*34. One liter of a buffer solution is made by mixing 500. mL
of 1.25 M acetic acid and 500. mL of 0.500 M calcium
acetate. What is the concentration of each of the following
in the buffer solution? (a) CH3COOH; (b) Ca2ϩ;
(c) CH3COOϪ; (d) Hϩ. (e) What is the pH?
*35. What must be the concentration of benzoate ion,
C6H5COOϪ, in a 0.045 M benzoic acid, C6H5COOH,
solution so that the pH is 5.00?
*36. What must be the concentration of chloroacetic acid,
ClCH2COOH, in a 0.015 M NaCH2ClCOO solution so
that the pH is 3.00? Ka ϭ 1.4 ϫ 10Ϫ3 for ClCH2COOH.
*37. What must be the concentration of NH4
ϩ in a 0.075 M
NH3 solution so that the pH is 8.80?
Acid–Base Indicators
*38. (a) What are acid–base indicators? (b) What are the essential
characteristics of acid–base indicators? (c) What
820 CHAPTER 19: Ionic Equilibria II: Buffers and Titration Curves
determines the color of an acid–base indicator in an aqueous
solution?
*39. Ka is 7.9 ϫ 10Ϫ8 for bromthymol blue, an indicator that
can be represented as HIn. HIn molecules are yellow, and
InϪ ions are blue. What color will bromthymol blue be
in a solution in which (a) [H3Oϩ] ϭ 1.0 ϫ 10Ϫ4 M and
(b) pH ϭ 10.30?
*40. The indicator metacresol purple changes from yellow to
purple at pH 8.2. At this point it exists in equal concentrations
as the conjugate acid and the conjugate base. What
are Ka and pKa for metacresol purple, a weak acid represented
as HIn?
*41. A series of acid–base indicators can be used to estimate the
pH of an unknown solution. Use the values given in Table
19-4 to determine the possible range of pH values of the
following solution. The solution was colorless with phenolphthalein,
yellow with methyl orange, and yellow with
methyl red.
*42. A series of acid–base indicators can be used to estimate the
pH of an unknown solution. Use the values given in Table
19-4 to determine the possible range of pH values of the
following solution. The solution was colorless in phenolphthalein,
blue in bromthymol blue, and yellow in
methyl orange.
*43. Use Table 19-4 to choose one or more indicators that could
be used to “signal” reaching a pH of (a) 3.5; (b) 7.0;
(c) 10.3; (d) 8.0.
*44. A solution of 0.020 M acetic acid is to be titrated with a
0.015 M NaOH solution. What is the pH at the equivalence
point? Choose an appropriate indicator for the
titration.
*45. Demonstrate mathematically that neutral red is red in solutions
of pH 3.00, whereas it is yellow in solutions of pH
10.00. HIn is red, and InϪ is yellow. Ka is 2.0 ϫ 10Ϫ7.
Strong Acid/Strong Base Titration Curves
*46. Make a rough sketch of the titration curve expected for
the titration of a strong acid with a strong base. What
determines the pH of the solution at the following points?
(a) No base added; (b) half-equivalence point; (c) equivalence
point; (d) excess base added. Compare your curve
with Figure 19-3.
For Exercises 47, 52, 57, and 58, calculate and tabulate [H3Oϩ],
[OHϪ], pH, and pOH at the indicated points as we did in Table
19-4. In each case assume that pure acid (or base) is added to
exactly 1 L of a 0.0100 molar solution of the indicated base (or
acid). This simplifies the arithmetic because we may assume that
the volume of each solution is constant throughout the titration.
Plot each titration curve with pH on the vertical axis and moles
of base (or acid) added on the horizontal axis.
*47. Solid NaOH is added to 1 L of 0.0500 M HCl solution.
Number of moles of NaOH added: (a) none; (b) 0.00500;
(c) 0.01500; (d) 0.02500 (50% titrated); (e) 0.03500;
(f ) 0.04500; (g) 0.04750; (h) 0.0500 (100% titrated);
(i) 0.0525; (j) 0.0600; (k) 0.0750 (50% excess NaOH).
Consult Table 19-4, and list the indicators that could be
used in this titration.
*48. A 25.0-mL sample of 0.125 M HNO3 is titrated with 0.100 M
NaOH. Calculate the pH of the solution (a) before the
addition of NaOH and after the addition of (b) 5.0 mL;
(c) 12.5 mL; (d) 25.0 mL; (e) 31.2 mL; (f) 37.5 mL of
NaOH.
*49. A 33.0-mL sample of 0.245 M HNO3 solution is titrated
with 0.213 M KOH. Calculate the pH of the solution
(a) before the addition of KOH and after the addition of
(b) 5.55 mL, (c) 12.0 mL, (d) 24.5 mL, (e) 35.2 mL,
(f ) 38.8 mL of KOH solution.
*50. A 44.0-mL sample of 0.145 M HCl solution is titrated with
0.213 M NaOH. Calculate the pH of the solution (a) before
the addition of NaOH and after the addition of (b) 5.55 mL,
(c) 12.0 mL, (d) 20.5 mL, (e) 27.2 mL, (f) 31.8 mL of
NaOH solution.
Weak Acid/Strong Base Titration Curves
*51. Make a rough sketch of the titration curve expected for the
titration of a weak monoprotic acid with a strong base.
What determines the pH of the solution at the following
points? (a) No base added; (b) half-equivalence point;
(c) equivalence point; (d) excess base added. Compare your
curve to Figure 19-4.
*52. Solid NaOH is added to exactly 1 L of 0.0200 M
CH3COOH solution. Number of moles NaOH added:
(a) none; (b) 0.00400; (c) 0.00800; (d) 0.01000 (50%
titrated); (e) 0.01400; (f ) 0.01800; (g) 0.01900; (h) 0.0200
(100% titrated); (i) 0.0210; (j) 0.0240; (k) 0.0300 (50%
excess NaOH). Consult Table 19-4, and list the indicators
that could be used in this titration.
*53. A 44.0-mL sample of 0.202 M CH3COOH solution is
titrated with 0.185 M NaOH. Calculate the pH of the
solution (a) before the addition of any NaOH solution and
after the addition of (b) 15.5 mL, (c) 20.0 mL, (d) 24.0 mL,
(e) 27.2 mL, (f) 48.0 mL, (g) 50.2 mL of NaOH solution.
*54. A 32.44-mL sample of 0.182 M CH3COOH solution is
titrated with 0.185 M NaOH. Calculate the pH of the solution
(a) before the addition of any NaOH solution and after
the addition of (b) 15.55 mL, (c) 20.0 mL, (d) 24.02 mL,
(e) 27.2 mL, (f ) 31.91 mL, (g) 33.12 mL of NaOH solu-
tion.
*55. A solution contains an unknown weak monoprotic acid,
HA. It takes 46.24 mL of NaOH solution to titrate
50.00 mL of the HA solution to the equivalence point.
To another 50.00-mL sample of the same HA solution,
23.12 mL of the same NaOH solution is added. The pH
of the resulting solution in the second experiment is 5.14.
What are Ka and pKa of HA?
*56. Calculate the pH at the equivalence point of the titration
of 100.0 mL of each of the following with 0.150 M KOH:
Exercises 821
(a) 1.000 M acetic acid; (b) 0.100 M acetic acid; (c) 0.0100
M acetic acid.
Mixed Exercises
*57. Gaseous HCl is added to 1 L of 0.0100 M aqueous ammonia
solution. Number of moles HCl added: (a) none;
(b) 0.00100; (c) 0.00300; (d) 0.00500 (50% titrated);
(e) 0.00700; (f ) 0.00900; (g) 0.00950; (h) 0.0100 (100%
titrated); (i) 0.0105; (j) 0.0120; (k) 0.0150 (50% excess
HCl). Consult Table 19-4, and list the indicators that could
be used in this titration.
*58. Gaseous NH3 is added to exactly 1 L of 0.0100 M HNO3
solution. Number of moles NH3 added: (a) none;
(b) 0.00100; (c) 0.00400; (d) 0.00500 (50% titrated);
(e) 0.00900; (f) 0.00950; (g) 0.0100 (100% titrated);
(h) 0.0105; (i) 0.0130. What is the major difference between
the titration curve for the reaction of HNO3 and NH3 and
the other curves you have plotted? Consult Table 19-4.
Can you suggest a satisfactory indicator for this titration?
*59. Compare the pH of 0.33 M NaCl with the pH of 0.33 M
NaCN.
*60. Compare the pH of 0.33 M NaCl with the pH of 0.33 M
NH4Cl.
CONCEPTUAL EXERCISES
*61. The pH of an equal molar acetic acid/sodium acetate buffer
is 4.74. Draw a molecular representation of a small portion
of this buffer solution. (You may omit the water
molecules.) Draw another molecular representation of the
solution after a very small amount of NaOH has been
added.
*62. Suppose you were asked on a laboratory test to outline a
procedure to prepare a buffered solution of pH 8.0 using
hydrocyanic acid, HCN. You realize that a pH of 8.0 is
basic, and you find that the Ka of hydrocyanic acid is
4.0 ϫ 10Ϫ10. What is your response?
*63. The odor of cooked fish is due to the presence of amines.
This odor is lessened by adding lemon juice, which contains
citric acid. Why does this work?
*64. The end point of a titration is not the same as the equivalence
point of a titration. Differentiate between these two
concepts.
BUILDING YOUR KNOWLEDGE
*65. Acetylsalicylic acid, the active ingredient in aspirin, has a
Ka value of 3.0 ϫ 10Ϫ4. We dissolve 0.0100 mole of acetylsalicylic
acid in sufficient water to make 1.00 L of solution
and then titrate it with 0.500 M NaOH solution. What is
the pH at each of these points in the titration? (a) before
any of the NaOH solution is added; (b) at the equivalence
point; (c) when a volume of NaOH solution has been added
that is equal to half the amount required to reach the equivalence
point.
*66. What is the pH of a solution that is a mixture of HOCl
and HOI, each at 0.15 M concentration?
*67. An unknown amount of water is mixed with 350. mL of a
6.0 M NaOH solution. A 75.0-mL sample of the resulting
solution is titrated to neutrality with 52.5 mL of 6.00 M
HCl. (a) Calculate the concentration of the diluted NaOH
solution. (b) What was the concentration of the NaOH
solution before it was diluted? (c) What volume of water
was added? Assume that the volumes were additive.
*68. A 3.5-L container of HCl had no concentration noted on
its label. A 20.00-mL sample of this unknown HCl solution
is titrated to a pH of 7.0 by 34.0 mL of 3.00 M NaOH
solution. Determine the volume of this HCl solution
required to prepare 1.5 L of 0.75 M HCl solution.
*69. Calculate the pH at the equivalence point for the titration
of a solution containing 150.0 mg of ethylamine,
C2H5NH2, with 0.1000 M HCl solution. The volume of
the solution at the equivalence point is 250. mL. Select a
suitable indicator. Kb for ethylamine appears in Exercise 28.
20 Ionic Equilibria III:
The Solubility
Product Principle
OUTLINE
20-1 Solubility Product Constants
20-2 Determination of Solubility
Product Constants
20-3 Uses of Solubility Product
Constants
20-4 Fractional Precipitation
20-5 Simultaneous Equilibria
Involving Slightly Soluble
Compounds
20-6 Dissolving Precipitates
OBJECTIVES
After you have studied this chapter, you should be able to
• Write solubility product constant expressions
• Explain how Ksp’s are determined
• Use Ksp’s in chemical calculations
• Recognize some common, slightly soluble compounds
• Describe fractional precipitation and how it can be used to separate ions
• Explain how simultaneous equilibria can be used to control solubility
• Describe some methods for dissolving precipitates
SS
o far we have discussed mainly compounds that are quite soluble in water. Although
most compounds dissolve in water to some extent, many are so slightly soluble that
they are called “insoluble compounds.” We shall now consider those that are only
very slightly soluble. As a rough rule of thumb, compounds that dissolve in water to the
extent of 0.020 mole/liter or more are classified as soluble. Refer to the solubility guidelines
(Table 4-8) as necessary.
Slightly soluble compounds are important in many natural phenomena. Our bones and
teeth are mostly calcium phosphate, Ca3(PO4)2, a slightly soluble compound. Also, many
natural deposits of Ca3(PO4)2 rock are mined and converted into agricultural fertilizer.
Limestone caves have been formed by acidic water slowly dissolving away calcium
carbonate, CaCO3. Sinkholes are created when acidic water dissolves away most of the
underlying CaCO3. The remaining limestone can no longer support the weight above it,
so it collapses, and a sinkhole is formed.
SOLUBILITY PRODUCT CONSTANTS
Suppose we add one gram of solid barium sulfate, BaSO4, to 1.0 liter of water at 25°C
and stir until the solution is saturated. Very little BaSO4 dissolves. Careful measurements
of conductivity show that one liter of a saturated solution of barium sulfate contains only
0.0025 gram of BaSO4, no matter how much more BaSO4 is added. The BaSO4 that does
dissolve is completely dissociated into its constituent ions.
20-1
In a process known as biomineralization,
some organisms produce
solids that also occur naturally as
minerals. The calcium carbonate,
CaCO3, in an eggshell has the same
crystal structure as in the mineral
calcite.
Pouring ammonium sulfide solution
into a solution of cadmium nitrate
gives a precipitate of cadmium sulfide.
(NH4)2S ϩ Cd(NO3)2 88n CdS(s)
ϩ 2NH4NO3
Cadmium sulfide is used as a pigment
in artists’ oil-based paints.
H2O
BaSO4(s) 3::4 Ba2ϩ(aq) ϩ SO4
2Ϫ(aq)
(We usually omit H2O over the arrows in equations like the preceding one.)
In equilibria that involve slightly soluble compounds in water, the equilibrium constant
is called a solubility product constant, Ksp. The activity of the solid BaSO4 is one (Section
17-11, Heterogeneous Equilibrium). Hence, the concentration of the solid is not included
in the equilibrium constant expression. For a saturated solution of BaSO4 in contact with
solid BaSO4, we write
BaSO4(s) 34 Ba2ϩ(aq) ϩ SO4
2Ϫ(aq) and Ksp ϭ [Ba2ϩ][SO4
2Ϫ]
The solubility product constant for BaSO4 is the product of the concentrations of its
constituent ions in a saturated solution.
In general, the solubility product expression for a compound is the product of
the concentrations of its constituent ions, each raised to the power that corresponds
to the number of ions in one formula unit of the compound. The quantity is constant
at constant temperature for a saturated solution of the compound. This statement
is the solubility product principle.
The existence of a substance in the solid state is indicated several ways. For example,
BaSO4(s), BaSO4, and BaSO4g are sometimes used to represent solid BaSO4. In this text
we use the (s) notation for formulas of solid substances in equilibrium with their saturated
aqueous solutions.
Barium sulfate is the “insoluble”
substance taken orally before stomach
X-rays are made because the barium
atoms absorb X-rays well. Even though
barium ions are quite toxic, barium
sulfate can still be taken orally without
danger. The compound is so insoluble
that it passes through the digestive
system essentially unchanged.
824 CHAPTER 20: Ionic Equilibria III: The Solubility Product Principle
The dissolving process of ions going
into solution. During precipitation,
ions are producing more solid as
the ions leave the solution. For a
saturated solution, the dissolving
and precipitation rates are equal. For
the very slightly soluble salts
discussed in this chapter, the ion
concentrations are very low
compared with the concentration
in this illustration.
An X-ray photo of the gastrointestinal
tract. The barium ions in
BaSO4 absorb X-radiation well.
Consider dissolving slightly soluble calcium fluoride, CaF2, in H2O.
CaF2(s) 34 Ca2ϩ(aq) ϩ 2FϪ(aq) Ksp ϭ [Ca2ϩ][FϪ]2 ϭ 3.9 ϫ 10Ϫ11
The very small amount of solid zinc phosphate, Zn3(PO4)2, that dissolves in water gives
three zinc ions and two phosphate ions per formula unit.
Zn3(PO4)2(s) 34 3Zn2ϩ(aq) ϩ 2PO4
3Ϫ(aq) Ksp ϭ [Zn2ϩ]3[PO4
3Ϫ]2 ϭ 9.1 ϫ 10Ϫ33
Generally, we may represent the dissolution of a slightly soluble compound and its Ksp
expression as
MyXz(s) 34 yMzϩ(aq) ϩ zXyϪ(aq) and Ksp ϭ [Mzϩ]y[XyϪ]z
In some cases a compound contains more than two kinds of ions. Dissolution of the
slightly soluble compound magnesium ammonium phosphate, MgNH4PO4, in water and
its solubility product expression are represented as
MgNH4PO4(s) 34 Mg2ϩ(aq) ϩ NH4
ϩ(aq) ϩ PO4
3Ϫ(aq)
Ksp ϭ [Mg2ϩ][NH4
ϩ][PO4
3Ϫ] ϭ 2.5 ϫ 10Ϫ12
We often shorten the term “solubility product constant” to “solubility product.” Thus,
the solubility products for barium sulfate, BaSO4, and for calcium fluoride, CaF2, are
written as
Ksp ϭ [Ba2ϩ][SO4
2Ϫ] ϭ 1.1 ϫ 10Ϫ10 Ksp ϭ [Ca2ϩ][FϪ]2 ϭ 3.9 ϫ 10Ϫ11
The molar solubility of a compound is the number of moles that dissolve to give one
liter of saturated solution.
20-1 Solubility Product Constants 825
Problem-Solving Tip: Names and Formulas of Several Ions Must Be
Recognized
To write the required equations for the dissolution and dissociation of salts, you must
recognize the names and formulas of common, stable ions that are often present. Table
2-3 contains the names and formulas of several ions; since then you have learned others.
Review these names and formulas carefully (formulas of ions must include their charges).
Many of the ions you need to know are present in the salts listed in Appendix H.
(a) (b)
Barium sulfate, BaSO4, occurs in the mineral barite (a). Calcium fluoride, CaF2, occurs in
the mineral fluorite (b). Both are clear, colorless crystals. Minerals are often discolored by
impurities.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 19.4, Solubility Product
Constant, and Screen 19.5,
Determining Ksp.
DETERMINATION OF SOLUBILITY PRODUCT
CONSTANTS
If the solubility of a compound is known, the value of its solubility product can be
calculated.
EXAMPLE 20-1 Solubility Product Constants
One (1.0) liter of saturated barium sulfate solution contains 0.0025 gram of dissolved BaSO4.
Calculate the solubility product constant for BaSO4.
Plan
We write the equation for the dissolution of BaSO4 and the expression for its solubility product
constant, Ksp. From the solubility of BaSO4 in H2O, we calculate its molar solubility and the
concentrations of the ions. This lets us calculate Ksp.
Solution
In saturated solutions, equilibrium exists between solid and dissolved solute. The equation for
the dissolution of barium sulfate in water and its solubility product expression are
BaSO4(s) 34 Ba2ϩ(aq) ϩ SO4
2Ϫ(aq) Ksp ϭ [Ba2ϩ][SO4
2Ϫ]
From the given solubility of BaSO4 in H2O we can calculate its molar solubility.
ϭ ϫ ϭ 1.1 ϫ 10Ϫ5 mol BaSO4/L
(dissolved)
We know the molar solubility of BaSO4. The dissolution equation shows that each formula
unit of BaSO4 that dissolves produces one Ba2ϩ ion and one SO4
2Ϫ ion.
BaSO4(s) 34 Ba2ϩ(aq) ϩ SO4
2Ϫ(aq)
1.1 ϫ 10Ϫ5 mol/L :::F 1.1 ϫ 10Ϫ5 M 1.1 ϫ 10Ϫ5 M
(dissolved)
In a saturated solution [Ba2ϩ] ϭ [SO4
2Ϫ] ϭ 1.1 ϫ 10Ϫ5 M. Substituting these values into the
Ksp expression for BaSO4 gives the calculated value of Ksp.
Ksp ϭ [Ba2ϩ][SO4
2Ϫ] ϭ (1.1 ϫ 10Ϫ5)(1.1 ϫ 10Ϫ5) ϭ 1.2 ϫ 10Ϫ10
You should now work Exercise 6.
EXAMPLE 20-2 Solubility Product Constant
One (1.00) liter of a saturated solution of silver chromate at 25°C contains 0.0435 gram of
Ag2CrO4. Calculate its solubility product constant.
Plan
We proceed as in Example 20-1.
Solution
The equation for the dissolution of silver chromate in water and its solubility product expression
are
Ag2CrO4(s) 34 2Agϩ(aq) ϩ CrO4
2Ϫ(aq) and Ksp ϭ [Agϩ]2[CrO4
2Ϫ]
1 mol BaSO4
ᎏᎏ
233 g BaSO4
2.5 ϫ 10Ϫ3 g BaSO4
ᎏᎏᎏ
1.0 L
__? mol BaSO4
ᎏᎏ
L
20-2
Unless otherwise indicated, solubility
product constants and solubility data
are given for 25°C (Appendix H).
We frequently use statements such as
“The solution contains 0.0025 gram of
BaSO4.” What we mean is that 0.0025
gram of solid BaSO4 dissolves to give
a solution that contains equal
concentrations of Ba2ϩ and SO4
2Ϫ
ions.
826 CHAPTER 20: Ionic Equilibria III: The Solubility Product Principle
1.1 ϫ 10Ϫ5 mole of solid BaSO4
dissolves to give a liter of saturated
solution.
Calculated values may differ slightly
from those found in tables due to data
or rounding differences.
The molar solubility of silver chromate is calculated first.
ϭ ϫ ϭ 1.31 ϫ 10Ϫ4 mol/L
(dissolved)
The equation for dissolution of Ag2CrO4 and its molar solubility give the concentrations of
Agϩ and CrO4
2Ϫ ions in the saturated solution.
Ag2CrO4(s) 34 2Agϩ(aq) ϩ CrO4
2Ϫ(aq)
1.31 ϫ 10Ϫ4 mol/L :::F 2.62 ϫ 10Ϫ4 M 1.31 ϫ 10Ϫ4 M
(dissolved)
Substitution into the Ksp expression for Ag2CrO4 gives the value of Ksp.
Ksp ϭ [Agϩ]2[CrO4
2Ϫ] ϭ (2.62 ϫ 10Ϫ4)2(1.31 ϫ 10Ϫ4) ϭ 8.99 ϫ 10Ϫ12
The calculated Ksp of Ag2CrO4 is 8.99 ϫ 10Ϫ12.
You should now work Exercise 8.
The molar solubility and Ksp values for BaSO4 and Ag2CrO4 are compared in Table
20-1. These data show that the molar solubility of Ag2CrO4 is greater than that of BaSO4.
The Ksp for Ag2CrO4, however, is less than the Ksp for BaSO4 because the expression for
Ag2CrO4 contains a squared term, [Agϩ]2.
If we compare Ksp values for two 1Ϻ1 compounds, for example, AgCl and BaSO4, the
compound with the larger Ksp value has the higher molar solubility. The same is true for
any two compounds that have the same ion ratio, for example, the 1Ϻ2 compounds CaF2
and Mg(OH)2 and the 2Ϻ1 compound Ag2CO3.
If two compounds have the same ion ratio, the one with the larger Ksp will have the
higher molar solubility.
Appendix H lists some Ksp values. Refer to it as needed.
1 mol Ag2CrO4
ᎏᎏ
332 g Ag2CrO4
0.0435 g Ag2CrO4
ᎏᎏᎏ
1.00 L
__? mol Ag2CrO4
ᎏᎏ
L
1.31 ϫ 10Ϫ4 mole of solid Ag2CrO4
dissolves to give a liter of saturated
solution.
20-2 Determination of Solubility Product Constants 827
Problem-Solving Tip: It’s Important to Know the Difference Between
Solubility, Molar Solubility, and Solubility
Product Expression
Often it is necessary that you understand which of these is given or which is asked for.
Recognition depends on a clear understanding of their definitions.
• The solubility of a compound is the amount of the compound that dissolves in a specified
volume of solution. Solubility is usually expressed as either grams per liter or
grams per 100. mL.
• The molar solubility of a compound is the number of moles that dissolve to give one
liter of saturated solution.
• The solubility product expression for a compound is the product of the concentrations
of its constituent ions, each raised to the power that corresponds to the number
of ions in one formula unit of the compound.
Values of solubility product constants
are usually tabulated to only two
significant figures.
Ksp(AgCl) ϭ 1.8 ϫ 10Ϫ10
Ksp(BaSO4) ϭ 1.1 ϫ 10Ϫ10
The molar solubility of AgCl is only
slightly higher than that of BaSO4.
USES OF SOLUBILITY PRODUCT CONSTANTS
When the solubility product for a compound is known, the solubility of the compound
in H2O at 25°C can be calculated as Example 20-3 illustrates.
EXAMPLE 20-3 Molar Solubilities from Ksp Values
Calculate the molar solubilities, concentrations of the constituent ions, and solubilities in grams
per liter for (a) silver chloride, AgCl (Ksp ϭ 1.8 ϫ 10Ϫ10), and (b) zinc hydroxide, Zn(OH)2
(Ksp ϭ 4.5 ϫ 10Ϫ17).
Plan
We are given the value for each solubility product constant. In each case we write the appropriate
equation, represent the equilibrium concentrations, and then substitute into the Ksp
expression.
Solution
(a) The equation for the dissolution of silver chloride and its solubility product expression are
AgCl(s) 34 Agϩ(aq) ϩ ClϪ(aq) Ksp ϭ [Agϩ][ClϪ] ϭ 1.8 ϫ 10Ϫ10
Each formula unit of AgCl that dissolves produces one Agϩ and one ClϪ. We let x ϭ mol/L
of AgCl that dissolves, that is, the molar solubility.
AgCl(s) 34 Agϩ(aq) ϩ ClϪ(aq)
x mol/L :::F x M x M
Substitution into the solubility product expression gives
Ksp ϭ [Agϩ][ClϪ] ϭ (x)(x) ϭ 1.8 ϫ 10Ϫ10 x2 ϭ 1.8 ϫ 10Ϫ10 x ϭ 1.3 ϫ 10Ϫ5
x ϭ molar solubility of AgCl ϭ 1.3 ϫ 10Ϫ5 mol/L
One liter of saturated AgCl contains 1.3 ϫ 10Ϫ5 mole of dissolved AgCl at 25°C. From the
balanced equation we know the concentrations of the constituent ions.
x ϭ molar solubility ϭ [Agϩ] ϭ [ClϪ] ϭ 1.3 ϫ 10Ϫ5 mol/L ϭ 1.3 ϫ 10Ϫ5 M
Now we can calculate the mass of dissolved AgCl in one liter of saturated solution.
ϭ ϫ ϭ 1.9 ϫ 10Ϫ3 g AgCl/L
A liter of saturated AgCl solution contains only 0.0019 g of dissolved AgCl.
(b) The equation for the dissolution of zinc hydroxide, Zn(OH)2, in water and its solubility
product expression are
143 g AgCl
ᎏᎏ
1 mol AgCl
1.3 ϫ 10Ϫ5 mol AgCl
ᎏᎏᎏ
L
__? g AgCl
ᎏᎏ
L
20-3
The values of Ksp are obtained from
Appendix H.
828 CHAPTER 20: Ionic Equilibria III: The Solubility Product Principle
TABLE 20-1 Comparison of Solubilities of BaSO4 and Ag2CrO4
Compound Molar Solubility Ksp
BaSO4 1.1 ϫ 10Ϫ5 mol/L [Ba2ϩ][SO4
2Ϫ] ϭ 1.1 ϫ 10Ϫ10
Ag2CrO4 1.3 ϫ 10Ϫ4 mol/L [Agϩ]2[CrO4
2Ϫ] ϭ 9.0 ϫ 10Ϫ12
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 19.6, Estimating Salt Solubility.
Zn(OH)2(s) 34 Zn2ϩ(aq) ϩ 2OHϪ(aq) Ksp ϭ [Zn2ϩ][OHϪ]2 ϭ 4.5 ϫ 10Ϫ17
We let x ϭ molar solubility, so [Zn2ϩ] ϭ x and [OHϪ] ϭ 2x, and we have
Zn(OH)2(s) 34 Zn2ϩ(aq) ϩ 2OHϪ(aq)
x mol/L :::F x M 2x M
Substitution into the solubility product expression gives
[Zn2ϩ][OHϪ]2 ϭ (x)(2x)2 ϭ 4.5 ϫ 10Ϫ17
4x3 ϭ 4.5 ϫ 10Ϫ17 x3 ϭ 11 ϫ 10Ϫ18 x ϭ 2.2 ϫ 10Ϫ6
x ϭ molar solubility of Zn(OH)2 ϭ 2.2 ϫ 10Ϫ6 mol Zn(OH)2/L
x ϭ [Zn2ϩ] ϭ 2.2 ϫ 10Ϫ6 M and 2x ϭ [OHϪ] ϭ 4.4 ϫ 10Ϫ6 M
We can now calculate the mass of dissolved Zn(OH)2 in one liter of saturated solution.
ϭ ϫ ϭ 2.2 ϫ 10Ϫ4 g Zn(OH)2/L
A liter of saturated Zn(OH)2 solution contains only 0.00022 g of dissolved Zn(OH)2.
You should now work Exercise 16.
99 g Zn(OH)2
ᎏᎏ
1 mol Zn(OH)2
2.2 ϫ 10Ϫ6 mol Zn(OH)2
ᎏᎏᎏ
L
__? g Zn(OH)2
ᎏᎏ
L
The [OHϪ] is twice the molar
solubility of Zn(OH)2 because each
formula unit of Zn(OH)2 produces two
OHϪ.
20-3 Uses of Solubility Product Constants 829
Problem-Solving Tip: The Dissolution of a Slightly Soluble Base Is
not a Kb Problem
The Ksp expression describes the equilibrium between a slightly soluble compound and
its ions; in Example 20-3(b) one of those ions is OH؊. A Kb expression describes the
equilibrium between a soluble basic species, for example, the ammonia molecule or the
acetate ion, and the products it forms in solution, including OH؊. Do you see why the
dissolution of Zn(OH)2 is not a Kb problem? We found that [OH؊] ϭ 4.4 ϫ 10؊6 M in
a saturated Zn(OH)2 solution. From this we find pOH ϭ 5.36 and pH ϭ 8.64. A saturated
Zn(OH)2 solution is not very basic because Zn(OH)2 is not very soluble in H2O.
The [OH؊] is 44 times greater than it is in pure water.
The Common Ion Effect in Solubility Calculations
The common ion effect applies to solubility equilibria just as it does to other ionic equilibria.
The solubility of a compound is less in a solution that contains an ion common to
the compound than it is in pure water (as long as no other reaction is caused by the presence
of the common ion).
EXAMPLE 20-4 Molar Solubilities and the Common Ion Effect
For magnesium fluoride, MgF2, Ksp ϭ 6.4 ϫ 10Ϫ9. (a) Calculate the molar solubility of magnesium
fluoride in pure water. (b) Calculate the molar solubility of MgF2 in 0.10 M sodium
fluoride, NaF, solution. (c) Compare these molar solubilities.
The tube at the left contains a
saturated solution of silver acetate,
AgCH3COO. When 1 M AgNO3 is
added to the tube, the equilibrium
AgCH3COO(s) 34
Agϩ(aq) ϩ CH3COOϪ(aq)
shifts to the left, demonstrating the
common ion effect.
Plan
For part (a), we write the appropriate chemical equations and solubility product expression,
designate the equilibrium concentrations, and then substitute into the solubility product expression.
For part (b), we recognize that NaF is a soluble ionic compound that is completely
dissociated into its ions. MgF2 is a slightly soluble compound. Both compounds produce FϪ
ions so this is a common ion effect problem. We write the appropriate chemical equations and
solubility product expression, represent the equilibrium concentrations, and substitute into the
solubility product expression. For part (c), we compare the molar solubilities by calculating
their ratio.
Solution
(a) We let x ϭ molar solubility for MgF2, a slightly soluble salt.
MgF2(s) 34 Mg2ϩ(aq) ϩ 2FϪ(aq) (reversible)
x mol/L :::F x M 2x M
Ksp ϭ [Mg2ϩ][FϪ]2 ϭ 6.4 ϫ 10Ϫ9
(x)(2x)2 ϭ 6.4 ϫ 10Ϫ9
x ϭ 1.2 ϫ 10Ϫ3
1.2 ϫ 10Ϫ3 M ϭ molar solubility of MgF2 in pure water
(b) NaF is a soluble ionic salt and, therefore, 0.10 M FϪ is produced by
H2O
NaF(s) 8888n Naϩ(aq) ϩ FϪ(aq) (complete)
0.10 M :::F 0.10 M 0.10 M
We let y ϭ molar solubility for MgF2, a slightly soluble salt.
MgF2(s) 34 Mg2ϩ(aq) ϩ 2FϪ(aq) (reversible)
y mol/L :::F y M 2y M
The total [FϪ] is 0.10 M from NaF plus 2y M from MgF2, or (0.10 ϩ 2y) M.
Ksp ϭ [Mg2ϩ][FϪ]2 ϭ 6.4 ϫ 10Ϫ9
(y)(0.10 ϩ 2y)2 ϭ 6.4 ϫ 10Ϫ9
Very little MgF2 dissolves, so y is small. This suggests that 2y ϽϽ 0.10, so 0.10 ϩ 2y Ϸ 0.10.
Then
(y)(0.10)2 ϭ 6.4 ϫ 10Ϫ9 and y ϭ 6.4 ϫ 10Ϫ7
6.4 ϫ 10Ϫ7 M ϭ molar solubility of MgF2 in 0.10 M NaF
(c) The ratio of molar solubility in water to molar solubility in 0.10 M NaF solution is:
ϭ ϭ
The molar solubility of MgF2 in 0.10 M NaF (6.4 ϫ 10Ϫ7 M) is nearly 1900 times less
than it is in pure water (1.2 ϫ 10Ϫ3 M).
You should now work Exercises 20 and 26.
1900
ᎏ
1
1.2 ϫ 10Ϫ3 M
ᎏᎏ
6.4 ϫ 10Ϫ7 M
molar solubility (in H2O)
ᎏᎏᎏᎏ
molar solubility (in NaF solution)
The assumption is found to be valid.
830 CHAPTER 20: Ionic Equilibria III: The Solubility Product Principle
The Reaction Quotient in Precipitation Reactions
Another application of the solubility product principle is the calculation of the maximum
concentrations of ions that can coexist in solution. From these calculations we can determine
whether a precipitate will form in a given solution. We compare Qsp (Section 17-4)
with Ksp.
If Qsp Ͻ Ksp Forward process is favored
No precipitation occurs; if solid is present, more solid can
dissolve
Qsp ϭ Ksp Solution is just saturated
Solid and solution are in equilibrium; neither forward nor reverse
process is favored
Qsp Ͼ Ksp Reverse process is favored; precipitation occurs to form more
solid
EXAMPLE 20-5 Predicting Precipitate Formation
If 100. mL of 0.00075 M sodium sulfate, Na2SO4, is mixed with 50. mL of 0.015 M barium
chloride, BaCl2, will a precipitate form?
Plan
We are mixing solutions of two soluble ionic salts. First we find the amount of each solute at
the instant of mixing. Next we find the molarity of each solute at the instant of mixing. Then
20-3 Uses of Solubility Product Constants 831
In Section 18-8 we discussed the hydrolysis of anions of weak acids. For example, we
found that for CH3COOϪ and CNϪ ions,
CH3COOϪ ϩ H2O 34 CH3COOH ϩ OHϪ
Kb ϭ ϭ 5.6 ϫ 10Ϫ10
CNϪ ϩ H2O 34 HCN ϩ OHϪ Kb ϭ ϭ 2.5 ϫ 10Ϫ5
We see that Kb for CNϪ, the anion of a very weak acid, is much larger than Kb for
CH3COOϪ, the anion of a much stronger acid. This tells us that in solutions of the
same concentration, CNϪ ions hydrolyze to a much greater extent than do CH3COOϪ
ions. So we might expect that hydrolysis would have a much greater effect on the solubilities
of cyanides such as AgCN than on the solubilities of acetates such as AgCH3COO.
It does.
Hydrolysis reduces the concentrations of anions of weak acids, such as FϪ, CO3
2Ϫ,
CH3COOϪ, and CNϪ, so its effect must be taken into account when we do very precise
solubility calculations. Taking into account the effect of hydrolysis on solubilities of
slightly soluble compounds is beyond the scope of this chapter, however.
[HCN][OHϪ]
ᎏᎏ
[CNϪ]
[CH3COOH][OHϪ]
ᎏᎏᎏ
[CH3COOϪ]
THE EFFECTS OF HYDROLYSIS ON SOLUBILITY
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 19.7, Can a Precipitation
Reaction Occur?
we find the concentration of each ion in the new solution. Now we ask the question “Could any
combination of the ions in this solution form a slightly soluble compound?” The answer is
“Yes, Ba2ϩ and SO4
2Ϫ could form BaSO4,” so we calculate Qsp and compare it with Ksp.
Solution
We find the amount of each solute at the instant of mixing.
__? mmol Na2SO4ϭ 100. mL ϫ ϭ 0.075 mmol Na2SO4
__? mmol BaCl2ϭ 50. mL ϫ ϭ 0.75 mmol BaCl2
When dilute aqueous solutions are mixed, their volumes can be added to give the volume of
the resulting solution.
volume of mixed solution ϭ 100. mL ϩ 50. mL ϭ 150. mL
Then we find the molarity of each solute at the instant of mixing.
MNa2SO4
ϭ ϭ 0.00050 M Na2SO4
MBaCl2
ϭ ϭ 0.0050 M BaCl2
Now we find the concentration of each ion in the new solution.
100%
Na2SO4(s) 8888n 2Naϩ(aq) ϩ SO4
2Ϫ(aq) (to completion)
0.00050 M ::::::F 00.0010 M0 00.00050 M0
100%
BaCl2(s) 8888n Ba2ϩ(aq) ϩ 2ClϪ(aq) (to completion)
0.0050 M ::::::F 00.0050 M0 00.010 M0
We consider the kinds of compounds mixed and determine whether a reaction could occur.
Both Na2SO4 and BaCl2 are soluble ionic salts. At the moment of mixing, the new solution
contains a mixture of Naϩ, SO4
2Ϫ, Ba2ϩ, and ClϪ ions. We must consider the possibility of
forming two new compounds, NaCl and BaSO4. Sodium chloride is a soluble ionic compound
so Naϩ and ClϪ do not combine in dilute aqueous solutions. BaSO4, however, is only very
slightly soluble, and solid BaSO4 will precipitate from the solution if Qsp Ͼ Ksp for BaSO4. Ksp
for BaSO4 is 1.1 ϫ 10Ϫ10. Substituting [Ba2ϩ] ϭ 0.0050 M and [SO4
2Ϫ] ϭ 0.00050 M into the
Qsp expression for BaSO4, we get
Qsp ϭ [Ba2ϩ][SO4
2Ϫ] ϭ (5.0 ϫ 10Ϫ3)(5.0 ϫ 10Ϫ4) ϭ 2.5 ϫ 10Ϫ6 (Qsp Ͼ Ksp)
Because Qsp Ͼ Ksp solid BaSO4 will precipitate until [Ba2ϩ][SO4
2Ϫ] just equals Ksp for BaSO4.
You should now work Exercises 28 and 30.
0.75 mmol BaCl2
ᎏᎏ
150. mL
0.075 mmol Na2SO4
ᎏᎏᎏ
150. mL
0.015 mmol BaCl2
ᎏᎏᎏ
mL
0.00075 mmol Na2SO4
ᎏᎏᎏ
mL
832 CHAPTER 20: Ionic Equilibria III: The Solubility Product Principle
Recall that Q has the same form as
the equilibrium constant, in this case
Ksp, but the concentrations are not
necessarily equilibrium concentrations.
When white solid potassium iodide, KI, and white solid lead(II) nitrate, Pb(NO3)2, are
stirred together, some yellow lead(II) iodide, PbI2, forms. This reaction occurs in the small
amount of water present in these solids.
Sinkholes are formed when the
underlying limestone, mostly
CaCO3, is dissolved away by acidic
water.
20-3 Uses of Solubility Product Constants 833
Problem-Solving Tip: Detection of Precipitates
The human eye is not a very sensitive detection device. As a rule of thumb, a precipitate
can be seen with the naked eye if Qsp Ͼ Ksp by a factor of 1000. In Example 20-5,
Qsp exceeds Ksp by a factor of ᎏ
1
2
.
.
1
5
ϫ
ϫ
1
1
0
0
Ϫ
Ϫ
1
6
0
ᎏ ϭ 2.3 ϫ 104 ϭ 23,000. We expect to be able
to see the BaSO4 precipitate that is formed. Modern techniques enable us to detect
smaller amounts of precipitates.
EXAMPLE 20-6 Initiation of Precipitation
What [Ba2ϩ] is necessary to start the precipitation of BaSO4 in a solution that is 0.0015 M in
Na2SO4? Assume that the Ba2ϩ comes from addition of a solid soluble ionic compound such
as BaCl2. For BaSO4, Ksp ϭ 1.1 ϫ 10Ϫ10.
Plan
These are the compounds in Example 20-5. We recognize that Na2SO4 is a soluble ionic
compound and that the molarity of SO4
2Ϫ is equal to the molarity of the Na2SO4 solution.
We are given Ksp for BaSO4, so we solve for [Ba2ϩ].
Solution
Because Na2SO4 is a soluble ionic compound, we know that [SO4
2Ϫ] ϭ 0.0015 M. We can use
Ksp for BaSO4 to calculate the [Ba2ϩ] required for Qsp to just equal Ksp.
[Ba2ϩ][SO4
2Ϫ] ϭ 1.1 ϫ 10Ϫ10
[Ba2ϩ] ϭ ᎏ
1.1
[S
ϫ
O4
1
2
0
Ϫ
Ϫ
]
10
ᎏ ϭ ᎏ
1
1
.
.
1
5
ϫ
ϫ
1
1
0
0
Ϫ
Ϫ
1
3
0
ᎏ ϭ 7.3 ϫ 10Ϫ8 M
Addition of enough BaCl2 to give a barium ion concentration of 7.3 ϫ 10Ϫ8 M just satisfies Ksp
for BaSO4; that is, Qsp ϭ Ksp. Ever so slightly more BaCl2 would be required for Qsp to exceed
Ksp and for precipitation of BaSO4 to occur. Therefore
[Ba2ϩ] Ͼ 7.3 ϫ 10Ϫ8 M (to initiate precipitation of BaSO4)
You should now work Exercise 29.
Often we wish to remove an ion from solution by forming an insoluble compound (as
in water purification). We use Ksp values to calculate the concentrations of ions remaining
in solution after precipitation has occurred.
EXAMPLE 20-7 Concentration of Common Ion
Suppose we wish to recover silver from an aqueous solution that contains a soluble silver
compound such as AgNO3 by precipitating insoluble silver chloride, AgCl. A soluble ionic
compound such as NaCl can be used as a source of ClϪ. What is the minimum concentration
of chloride ion needed to reduce the dissolved silver ion concentration to a maximum of
1.0 ϫ 10Ϫ9 M? For AgCl, Ksp ϭ 1.8 ϫ 10Ϫ10.
Plan
We are given Ksp for AgCl and the required equilibrium [Agϩ], so we solve for [ClϪ].
The recovery of silver from the
solutions used in developing and fixing
photographic film and prints presents
just such a problem. Silver is an
expensive metal, and the recovery is
profitable. Moreover, if not recovered,
the silver ions would constitute an
undesirable pollutant in water supplies.
Solution
The equation for the reaction of interest and the Ksp for AgCl are
AgCl(s) 34 Agϩ(aq) ϩ ClϪ(aq) and [Agϩ][ClϪ] ϭ 1.8 ϫ 10Ϫ10
To determine the [ClϪ] required to reduce the [Agϩ] to 1.0 ϫ 10Ϫ9 M, we solve the Ksp expression
for [ClϪ].
[ClϪ] ϭ ᎏ
1.8
[
ϫ
Ag
1
ϩ
0
]
Ϫ10
ᎏ ϭ ᎏ
1
1
.
.
8
0
ϫ
ϫ
1
1
0
0
Ϫ
Ϫ
1
9
0
ᎏ ϭ 0.18 M ClϪ
To reduce the [Agϩ] to 1.0 ϫ 10Ϫ9 M (0.00000011 g Agϩ/L), NaCl would be added until
[ClϪ] ϭ 0.18 M in the solution.
You should now work Exercise 32.
FRACTIONAL PRECIPITATION
We sometimes wish to remove some ions from solution while leaving others with similar
properties in solution. This separation process is called fractional precipitation. Consider
a solution that contains ClϪ, BrϪ, and IϪ ions. These halide ions are anions of elements
in the same family in the periodic table. We expect them to have similar properties. But
we also expect some differences in properties, and that is what we find. Consider the solubility
products for these silver halides.
Compound Solubility Product
AgCl 1.8 ϫ 10Ϫ10
AgBr 3.3 ϫ 10Ϫ13
AgI 1.5 ϫ 10Ϫ16
These Ksp values show that AgI is less soluble than AgBr and that AgBr is less soluble
than AgCl. Silver fluoride is quite soluble in water.
EXAMPLE 20-8 Concentration Required to Initiate Precipitation
Solid silver nitrate is slowly added to a solution that is 0.0010 M each in NaCl, NaBr, and NaI.
Calculate the [Agϩ] required to initiate the precipitation of each of the silver halides. For AgI,
Ksp ϭ 1.5 ϫ 10Ϫ16; for AgBr, Ksp ϭ 3.3 ϫ 10Ϫ13; and for AgCl, Ksp ϭ 1.8 ϫ 10Ϫ10.
Plan
We are given a solution that contains equal concentrations of ClϪ, BrϪ, and IϪ ions; all of
which form insoluble silver salts. Then we slowly add Agϩ ions. We use each Ksp to determine
the [Agϩ] that must be exceeded to initiate precipitation of each salt as we did in Example
20-6.
Solution
We calculate the [Agϩ] necessary to begin to precipitate each of the silver halides. The solubility
product for AgI is
[Agϩ][IϪ] ϭ 1.5 ϫ 10Ϫ16
[IϪ] ϭ 1.0 ϫ 10Ϫ3 M, so the [Agϩ] that must be exceeded to start precipitation of AgI is
20-4
NaCl, NaBr, NaI, AgNO3, and
NaNO3 are soluble compounds that
are completely dissociated in dilute
aqueous solution.
We ignore the extremely small change
in volume caused by addition of solid
AgNO3.
834 CHAPTER 20: Ionic Equilibria III: The Solubility Product Principle
Silver chloride precipitates when
chloride ions are added to a solution
containing silver ions.
Clearly cations such as Naϩ or Kϩ
must also be present in this solution.
[Agϩ] ϭ ᎏ
1.5 ϫ
[IϪ
1
]
0Ϫ16
ᎏ ϭ ᎏ
1
1
.
.
5
0
ϫ
ϫ
1
1
0
0
Ϫ
Ϫ
1
3
6
ᎏ ϭ 1.5 ϫ 10Ϫ13 M
Therefore, AgI will begin to precipitate when [Agϩ] Ͼ 1.5 ϫ 10Ϫ13 M.
Repeating this kind of calculation for silver bromide gives
[Agϩ][BrϪ] ϭ 3.3 ϫ 10Ϫ13
[Agϩ] ϭ ᎏ
3.3
[
ϫ
Br
1
Ϫ
0
]
Ϫ13
ᎏ ϭ ᎏ
3
1
.
.
3
0
ϫ
ϫ
1
1
0
0
Ϫ
Ϫ
1
3
3
ᎏ ϭ 3.3 ϫ 10Ϫ10 M
Thus, [Agϩ] Ͼ 3.3 ϫ 10Ϫ10 M is needed to start precipitation of AgBr.
For the precipitation of silver chloride to begin,
[Agϩ][ClϪ] ϭ 1.8 ϫ 10Ϫ10
[Agϩ] ϭ ᎏ
1.8
[
ϫ
Cl
1
Ϫ
0
]
Ϫ10
ᎏ ϭ ᎏ
1
1
.
.
8
0
ϫ
ϫ
1
1
0
0
Ϫ
Ϫ
1
3
0
ᎏ ϭ 1.8 ϫ 10Ϫ7 M
To precipitate AgCl, we must have [Agϩ] Ͼ 1.8 ϫ 10Ϫ7 M.
We have shown that
to precipitate AgI, [Agϩ] Ͼ 1.5 ϫ 10Ϫ13 M
to precipitate AgBr, [Agϩ] Ͼ 3.3 ϫ 10Ϫ10 M
to precipitate AgCl, [Agϩ] Ͼ 1.8 ϫ 10Ϫ7 M
This calculation tells us that when AgNO3 is added slowly to a solution that is 0.0010 M
in each of NaI, NaBr, and NaCl, AgI precipitates first, AgBr precipitates second, and AgCl
precipitates last. We can also calculate the amount of IϪ precipitated before BrϪ begins
to precipitate and the amounts of IϪ and BrϪ precipitated before ClϪ begins to precipitate
(Example 20-9).
EXAMPLE 20-9 Fractional Precipitation
Refer to Example 20-8. (a) Calculate the percentage of IϪ precipitated before AgBr precipitates.
(b) Calculate the percentages of IϪ and BrϪ precipitated before ClϪ precipitates.
Plan
From Example 20-8 we know the [Agϩ] that must be exceeded to initiate precipitation of each
of three silver halides, AgI, AgBr, and AgCl. We use each of these values of [Agϩ] with the
appropriate Ksp expression, in turn, to find the concentration of each halide ion that remains
in solution (unprecipitated). We express these halide ion concentrations as percent unprecipitated.
Then we subtract each from exactly 100% to find the percentage of each halide that
precipitates.
We are not suggesting that a 1.5 ϫ
10Ϫ13 M solution of AgNO3 be added.
We are pointing out the fact that when
sufficient AgNO3 has been added to
the solution to make [Agϩ] Ͼ 1.5 ϫ
10Ϫ13 M, AgI begins to precipitate.
20-4 Fractional Precipitation 835
As AgNO3 is added to the solution
containing ClϪ, BrϪ, and IϪ ions,
some AgBr and AgCl may precipitate
locally. As the solution is stirred, AgBr
and AgCl redissolve as long as [Agϩ] is
not large enough to exceed their Ksp
values in the bulk of the solution.
Freshly precipitated AgCl is white (left), AgBr is very pale yellow (center), and AgI is yellow
(right). Polarizabilities of these halide ions increase in the order ClϪ Ͻ BrϪ Ͻ IϪ. Colors of
the silver halides become more intense in the same direction. Solubilities of the silver
halides increase in the opposite direction.
Solution
(a) In Example 20-8 we found that AgBr begins to precipitate when [Agϩ] Ͼ 3.3 ϫ 10Ϫ10 M.
This value for [Agϩ] can be substituted into the Ksp expression for AgI to determine [IϪ]
remaining unprecipitated when AgBr begins to precipitate.
[Agϩ][IϪ] ϭ 1.5 ϫ 10Ϫ16
[IϪ]unppt’d ϭ ᎏ
1.5
[
ϫ
Ag
1
ϩ
0
]
Ϫ16
ᎏ ϭ ᎏ
1
3
.
.
5
3
ϫ
ϫ
1
1
0
0
Ϫ
Ϫ
1
1
6
0
ᎏ ϭ 4.5 ϫ 10Ϫ7 M
The percentage of IϪ unprecipitated is
% IϪ
unppt’d ϭ ϫ 100% ϭ ᎏ
4
1
.
.
5
0
ϫ
ϫ
1
1
0
0
Ϫ
Ϫ
7
3
M
M
ᎏ ϫ 100%
ϭ 0.045% IϪ unprecipitated
Therefore, 99.955% of the IϪ precipitates before AgBr begins to precipitate.
(b) Similar calculations show that just before AgCl begins to precipitate, [Agϩ] ϭ 1.8 ϫ 10Ϫ7 M,
and the [IϪ] unprecipitated is calculated as in part (a).
[Agϩ][IϪ] ϭ 1.5 ϫ 10Ϫ16
[IϪ]unppt’d ϭ ᎏ
1.5
[
ϫ
Ag
1
ϩ
0
]
Ϫ16
ᎏ ϭ ᎏ
1
1
.
.
5
8
ϫ
ϫ
1
1
0
0
Ϫ
Ϫ
1
7
6
ᎏ ϭ 8.3 ϫ 10Ϫ10 M
The percentage of IϪ unprecipitated just before AgCl precipitates is
% IϪ
unppt’d ϭ ϫ 100% ϭ ᎏ
8
1
.
.
3
0
ϫ
ϫ
1
1
0
0
Ϫ
Ϫ
1
3
0
M
M
ᎏ ϫ 100%
ϭ 0.000083% IϪ unprecipitated
Therefore, 99.999917% of the IϪ precipitates before AgCl begins to precipitate.
A similar calculation for the amount of BrϪ precipitated just before AgCl begins to precipitate
gives
[Agϩ][BrϪ] ϭ 3.3 ϫ 10Ϫ13
[BrϪ]unppt’d ϭ ᎏ
3.3
[
ϫ
Ag
1
ϩ
0
]
Ϫ13
ᎏ ϭ ᎏ
3
1
.
.
3
8
ϫ
ϫ
1
1
0
0
Ϫ
Ϫ
1
7
3
ᎏ ϭ 1.8 ϫ 10Ϫ6 M
% BrϪ
unppt’d ϭ ϫ 100% ϭ ᎏ
1
1
.
.
8
0
ϫ
ϫ
1
1
0
0
Ϫ
Ϫ
6
3
M
M
ᎏ ϫ 100%
ϭ 0.18% BrϪ unprecipitated
Thus, 99.82% of the BrϪ precipitates before AgCl begins to precipitate.
You should now work Exercises 36 and 38.
We have described the series of reactions that occurs when solid AgNO3 is added slowly
to a solution that is 0.0010 M in ClϪ, BrϪ, and IϪ. Silver iodide begins to precipitate first;
99.955% of the IϪ precipitates before any solid AgBr is formed. Silver bromide begins to
[BrϪ]unppt’d
ᎏᎏ
[BrϪ]orig
[IϪ]unppt’d
ᎏᎏ
[IϪ]orig
[IϪ]unppt’d
ᎏᎏ
[IϪ]orig
In Example 20-7 we did a similar
calculation, but we did not express the
result in terms of the percentage of an
ion precipitated.
836 CHAPTER 20: Ionic Equilibria III: The Solubility Product Principle
We have subtracted 0.045% from
exactly 100%. We have therefore not
violated the rules for significant
figures.
precipitate next; 99.82% of the BrϪ and 99.999917% of the IϪ precipitate before any solid
AgCl forms. This shows that we can separate these ions very effectively by fractional
precipitation.
SIMULTANEOUS EQUILIBRIA INVOLVING SLIGHTLY
SOLUBLE COMPOUNDS
Many weak acids and bases react with many metal ions to form insoluble compounds. In
such cases, we must take into account the weak acid or weak base equilibrium as well as
the solubility equilibrium. The most common examples involve the reaction of metal ions
with aqueous ammonia to form insoluble metal hydroxides.
EXAMPLE 20-10 Simultaneous Equilibria
If a solution is made 0.10 M in magnesium nitrate, Mg(NO3)2, and 0.10 M in aqueous ammonia,
a weak base, will magnesium hydroxide, Mg(OH)2, precipitate? Ksp for Mg(OH)2 is
1.5 ϫ 10Ϫ11, and Kb for aqueous NH3 is 1.8 ϫ 10Ϫ5.
Plan
We first write equations for the two reversible reactions and their equilibrium constant expressions.
We note that [OHϪ] appears in both equilibrium constant expressions. From the
statement of the problem we know the concentration of Mg2ϩ. We use the Kb expression for
aqueous NH3 to find [OHϪ]. Then we calculate Qsp for Mg(OH)2 and compare it with its Ksp.
Solution
Two equilibria and their equilibrium constant expressions must be considered.
Mg(OH)2(s) 34 Mg2ϩ(aq) ϩ 2OHϪ(aq) Ksp ϭ [Mg2ϩ][OHϪ]2 ϭ 1.5 ϫ 10Ϫ11
NH3(aq) ϩ H2O(ᐉ) 34 NH4
ϩ(aq) ϩ OHϪ(aq) Kb ϭ ϭ 1.8 ϫ 10Ϫ5
The [OHϪ] in 0.10 M aqueous NH3 is calculated as in Example 18.14.
NH3(aq) ϩ H2O(ᐉ) 34 NH4
ϩ(aq) ϩ OHϪ(aq)
(0.10 Ϫ x)M x M x M
ϭ 1.8 ϫ 10Ϫ5 ϭ x ϭ 1.3 ϫ 10Ϫ3 M ϭ [OHϪ]
Magnesium nitrate is a soluble ionic compound, so [Mg2ϩ] ϭ 0.10 M. Now that both [Mg2ϩ]
and [OHϪ] are known, we calculate Qsp for Mg(OH)2.
[Mg2ϩ][OHϪ]2 ϭ Qsp
(0.10)(1.3 ϫ 10Ϫ3)2 ϭ 1.7 ϫ 10Ϫ7 ϭ Qsp
Ksp ϭ 1.5 ϫ 10Ϫ11, so we see that Qsp Ͼ Ksp.
Therefore, Mg(OH)2 would precipitate until Qsp ϭ Ksp.
You should now work Exercise 46.
(x)(x)
ᎏᎏ
(0.10 Ϫ x)
[NH4
ϩ][OHϪ]
ᎏᎏ
[NH3]
[NH4
ϩ][OHϪ]
ᎏᎏ
[NH3]
20-5
We have calculated the [OHϪ]
produced by the ionization of 0.10 M
aqueous NH3. This is the equilibrium
concentration of OHϪ in this solution.
There is no reason to double this value
when we put it into the Ksp expression!
20-5 Simultaneous Equilibria Involving Slightly Soluble Compounds 837
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 19.10, Simultaneous Equilibria.
Example 20-11 shows how we can calculate the concentration of weak base that is
required to initiate precipitation of an insoluble metal hydroxide.
EXAMPLE 20-11 Simultaneous Equilibria
What concentration of aqueous ammonia is necessary to just start precipitation of Mg(OH)2
from a 0.10 M solution of Mg(NO3)2? Refer to Example 20-10.
Plan
We have the same reactions and equilibrium constant expressions as in Example 20-10. We are
given [Mg2ϩ], and so we use the Ksp expression of Mg(OH)2 to calculate the [OHϪ] necessary
to initiate precipitation. Then we find the molarity of aqueous NH3 solution that would furnish
the desired [OHϪ].
Solution
Two equilibria and their equilibrium constant expressions must be considered.
Mg(OH)2(s) 34 Mg2ϩ(aq) ϩ 2OHϪ(aq) Ksp ϭ 1.5 ϫ 10Ϫ11
NH3(aq) ϩ H2O(ᐉ) 34 NH4
ϩ(aq) ϩ OHϪ(aq) Kb ϭ 1.8 ϫ 10Ϫ5
We find the [OHϪ] necessary to initiate precipitation of Mg(OH)2 when [Mg2ϩ] ϭ 0.10 M.
Ksp ϭ [Mg2ϩ][OHϪ]2 ϭ 1.5 ϫ 10Ϫ11
[OHϪ]2 ϭ ᎏ
1.5
[M
ϫ
g
1
2
0
ϩ
Ϫ
]
11
ᎏ ϭ ᎏ
1.5 ϫ
0.1
1
0
0Ϫ11
ᎏ ϭ 1.5 ϫ 10Ϫ10 [OHϪ] ϭ 1.2 ϫ 10Ϫ5 M
І [OHϪ] Ͼ 1.2 ϫ 10Ϫ5 M to initiate precipitation of Mg(OH)2.
Now we use the equilibrium of NH3 as a weak base to find the [NH3] that will produce
1.2 ϫ 10Ϫ5 M OHϪ. Let x be the original [NH3].
NH3(aq) ϩH2O (ᐉ) 34 NH4
ϩ(aq) ϩ OHϪ(aq)
(x Ϫ 1.2 ϫ 10Ϫ5) M 1.2 ϫ 10Ϫ5 M 1.2 ϫ 10Ϫ5 M
Kb ϭ ϭ 1.8 ϫ 10Ϫ5 ϭ
1.8 ϫ 10Ϫ5 x Ϫ 2.16 ϫ 10Ϫ10 ϭ 1.44 ϫ 10Ϫ10
1.8 ϫ 10Ϫ5 x ϭ 3.6 ϫ 10Ϫ10 so x ϭ 2.0 ϫ 10Ϫ5 M ϭ [NH3]orig
The solution must be ever so slightly greater than 2.0 ϫ 10Ϫ5 M in NH3 to initiate precipitation
of Mg(OH)2 in a 0.10 M solution of Mg(NO3)2.
A solution that contains a weak base can be buffered (by addition of a salt of the weak
base) to decrease its basicity. Significant concentrations of some metal ions that form insoluble
hydroxides can be kept in such solutions.
EXAMPLE 20-12 Simultaneous Equilibria
What minimum number of moles of NH4Cl must be added to 1.0 liter of solution that is 0.10 M
in Mg(NO3)2 and 0.10 M in NH3 to prevent precipitation of Mg(OH)2?
(1.2 ϫ 10Ϫ5)(1.2 ϫ 10Ϫ5)
ᎏᎏᎏ
(x Ϫ 1.2 ϫ 10Ϫ5)
[NH4
ϩ][OHϪ]
ᎏᎏ
[NH3]
Because 1.2 ϫ 10Ϫ5 and x are of
comparable magnitude, neither can be
disregarded in the term (x Ϫ 1.2 ϫ 10Ϫ5).
838 CHAPTER 20: Ionic Equilibria III: The Solubility Product Principle
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 19.11, Solubility and pH.
Plan
These are the same compounds, in the same concentrations, that we used in Example 20-10.
Because we know [Mg2ϩ], we must find the maximum [OHϪ] that can exist in the solution
without exceeding Ksp for Mg(OH)2. Then we find the minimum concentration of NH4Cl that
is necessary to buffer the NH3 solution to keep the [OHϪ] below the calculated value.
Solution
The buffering action of NH4Cl in the presence of NH3 decreases the concentration of OHϪ.
Again we have two equilibria.
Mg(OH)2(s) 34 Mg2ϩ(aq) ϩ 2OHϪ(aq) Ksp ϭ 1.5 ϫ 10Ϫ11
NH3(aq) ϩ H2O(ᐉ) 34 NH4
ϩ(aq) ϩ OHϪ(aq) Kb ϭ 1.8 ϫ 10Ϫ5
To find the maximum [OHϪ] that can exist in solution without causing precipitation, we substitute
[Mg2ϩ] into the Ksp for Mg(OH)2.
[Mg2ϩ][OHϪ]2 ϭ 1.5 ϫ 10Ϫ11
[OHϪ]2 ϭ ᎏ
1.5
[M
ϫ
g
1
2
0
ϩ
Ϫ
]
11
ᎏ ϭ ᎏ
1.5 ϫ
0.1
1
0
0Ϫ11
ᎏ ϭ 1.5 ϫ 10Ϫ10
[OHϪ] ϭ 1.2 ϫ 10Ϫ5 M (maximum [OHϪ] possible)
To prevent precipitation of Mg(OH)2 in this solution, [OHϪ] must be equal to or less than
1.2 ϫ 10Ϫ5 M. Kb for aqueous NH3 is used to calculate the number of moles of NH4Cl necessary
to buffer 1.0 L of 0.10 M aqueous NH3 so that [OHϪ] ϭ 1.2 ϫ 10Ϫ5 M. Let x ϭ number
of mol/L of NH4Cl required.
We can assume that (x ϩ 1.2 ϫ 10Ϫ5) Ϸ x and (0.10 Ϫ 1.2 ϫ 10Ϫ5) Ϸ 0.10.
ᎏ
(x)(1.2
0.
ϫ
10
10Ϫ5)
ᎏ ϭ 1.8 ϫ 10Ϫ5
x ϭ 0.15 mol of NH4
ϩ per liter of solution
Addition of 0.15 mol of NH4Cl to 1.0 L of 0.10 M aqueous NH3 decreases [OHϪ] to
1.2 ϫ 10Ϫ5 M. Then Ksp for Mg(OH)2 is not exceeded in this solution, and so no precipitate
would form.
You should now work Exercises 42 and 44.
Examples 20-10, 20-11, and 20-12 illustrate a very important point.
All relevant equilibria must be satisifed when more than one equilibrium is required
to describe a solution.
NH4Cl(aq) NH4
ϩ(aq)
x M
H2O(ᐉ)NH3(aq) ϩ
(0.10 Ϫ 1.2 ϫ 10Ϫ5) M
NH4
ϩ(aq)
1.2 ϫ 10Ϫ5 M
x M
ClϪ(aq) (to completion)ϩ
x M
OHϪ(aq)ϩ
1.2 ϫ 10Ϫ5 M
In Example 20-10 we found that
Mg(OH)2 will precipitate from a
solution that is 0.10 M in Mg(NO3)2
and 0.10 M in NH3.
20-5 Simultaneous Equilibria Involving Slightly Soluble Compounds 839
You may wish to refer to Example 19-6
to refresh your understanding of buffer
solutions.
DISSOLVING PRECIPITATES
A precipitate dissolves when the concentrations of its ions are reduced so that Ksp is no
longer exceeded, that is, when Qsp Ͻ Ksp. The precipitate then dissolves until Qsp ϭ Ksp.
The precipitates can be dissolved by several types of reactions. All involve removing ions
from solution.
Converting an Ion to a Weak Electrolyte
Three specific illustrations follow.
1. Converting OHϪ to H2O. Insoluble Al(OH)3 dissolves in acids. Hϩ ions react with
OHϪ ions [from the saturated Al(OH)3 solution] to form the weak electrolyte H2O.
This makes [Al3ϩ][OHϪ]3 Ͻ Ksp, so that the dissolution equilibrium shifts to the
right and Al(OH)3 dissolves.
Al(OH)3(s) 34 Al3ϩ(aq) ϩ 3OHϪ(aq)
3Hϩ(aq) ϩ 3OHϪ(aq) 88n 3H2O(ᐉ)
overall rxn: Al(OH)3(s) ϩ 3Hϩ(aq) 88n Al3ϩ(aq) ϩ 3H2O(ᐉ)
2. Converting NH4
ϩ to NH3. Ammonium ions, from a salt such as NH4Cl, dissolve
insoluble Mg(OH)2. The NH4
ϩ ions combine with OHϪ ions in the saturated
Mg(OH)2 solution. This forms the weak electrolytes NH3 and H2O. The result is
[Mg2ϩ][OHϪ]2 Ͻ Ksp, and so the Mg(OH)2 dissolves.
Mg(OH)2(s) 34 Mg2ϩ(aq) ϩ 2OHϪ(aq)
2NH4
ϩ(aq) ϩ 2OHϪ(aq) 88n 2NH3(aq) ϩ 2H2O(ᐉ)
overall rxn: Mg(OH)2(s) ϩ 2NH4
ϩ(aq) 88n Mg2ϩ(aq) ϩ 2NH3(aq) ϩ 2H2O(ᐉ)
This process, dissolution of Mg(OH)2 in an NH4Cl solution, is the reverse of the
reaction we considered in Example 20-10. There, Mg(OH)2 precipitated from a
solution of aqueous NH3.
3. Converting S2Ϫ to H2S. Nonoxidizing acids dissolve most insoluble metal sulfides.
For example, 6 M HCl dissolves MnS. The Hϩ ions combine with S2Ϫ ions to form
H2S, a gas that bubbles out of the solution. The result is [Mn2ϩ][S2Ϫ] Ͻ Ksp, and
so the MnS dissolves.
MnS(s) 34 Mn2ϩ(aq) ϩ S2Ϫ(aq)
S2Ϫ(aq) ϩ H2O(ᐉ) 88n HSϪ(aq) ϩ OHϪ(aq)
2Hϩ(aq) ϩ HSϪ(aq) ϩ OHϪ(aq) 88n H2S(g) ϩ H2O(ᐉ)
overall rxn: MnS(s) ϩ 2Hϩ(aq) 88n Mn2ϩ(aq) ϩ H2S(g)
Converting an Ion to Another Species by a Redox Reaction
Most insoluble metal sulfides dissolve in hot dilute HNO3 because NO3
Ϫ ions oxidize
S2Ϫ ions, or actually their hydrolysis product (HSϪ), to elemental sulfur. This removes
HSϪ (and thus S2Ϫ) ions from the solution and promotes the dissolving of more of the
metal sulfide.
3HSϪ(aq) ϩ 2NO3
Ϫ(aq) ϩ 5Hϩ(aq) 88n 3S(s) ϩ 2NO(g) ϩ 4H2O(ᐉ)
20-6
Solubility products, like other
equilibrium constants, are
thermodynamic quantities. They tell
us nothing about how fast a given
reaction occurs, only whether it can
occur under specified conditions.
840 CHAPTER 20: Ionic Equilibria III: The Solubility Product Principle
S2Ϫ like O2Ϫ does not exist in
appreciable amounts in aqueous
solutions.
Consider copper(II) sulfide, CuS, in equilibrium with its ions. This equilibrium lies far
to the left. Removal of the S2Ϫ or HSϪ ions by oxidation to elemental sulfur favors the
reaction to the right, and so CuS(s) dissolves in hot dilute HNO3.
CuS(s) 34 Cu2ϩ(aq) ϩ S2Ϫ(aq)
S2Ϫ(aq) ϩ H2O(ᐉ) 88n HSϪ(aq) ϩ OHϪ(aq)
3HSϪ(aq) ϩ 3OHϪ(aq) ϩ 8Hϩ(aq) ϩ 2NO3
Ϫ(aq) 88n 3S(s) ϩ 2NO(g) ϩ 7H2O(ᐉ)
We multiply the first two equations by 3, add the three equations, and cancel like terms.
This gives the net ionic equation for dissolving CuS(s) in hot dilute HNO3.
3CuS(s) ϩ 2NO3
Ϫ(aq) ϩ 8Hϩ(aq) 88n 3Cu2ϩ(aq) ϩ 3S(s) ϩ 2NO(g) ϩ 4H2O(ᐉ)
Complex Ion Formation
The cations in many slightly soluble compounds can form complex ions. This often results
in dissolution of the slightly soluble compound. Some metal ions share electron pairs
donated by molecules and ions such as NH3, CNϪ, OHϪ, FϪ, ClϪ, BrϪ, and IϪ form
coordinate covalent bonds to metal ions. Coordinate covalent bonds are formed as these
electron-donating groups (ligands) replace H2O molecules from hydrated metal ions. The
It is sometimes convenient to ignore
hydrolysis of S2Ϫ ions in aqueous
solutions. Leaving out the hydrolysis
step may give the false impression that
S2Ϫ ions exist in solution. The overall
net ionic equation for this redox
reaction will be the same, however,
with or without consideration of
hydrolysis of the S2Ϫ ions.
20-6 Dissolving Precipitates 841
Copper(II) sulfide, CuS, is black. As CuS dissolves in 6 M HNO3, some NO is oxidized to
brown NO2 by O2 in the air. The resulting solution of Cu(NO3)2 is blue.
Manganese(II) sulfide, MnS, is salmon-colored. MnS dissolves in 6 M HCl. The resulting
solution of MnCl2 is pale pink.
“Ligand” is the name given to an atom
or a group of atoms bonded to the
central element in complex ions.
Ligands are Lewis bases.
decrease in the concentration of the hydrated metal ion shifts the solubility equilibrium
to the right.
Many copper(II) compounds react with excess aqueous NH3 to form the deep-blue
complex ion [Cu(NH3)4]2ϩ.
Cu2ϩ(aq) ϩ 4NH3(aq) 34 [Cu(NH3)4]2ϩ(aq)
The dissociation of this complex ion is represented as
[Cu(NH3)4]2ϩ(aq) 34 Cu2ϩ(aq) ϩ 4NH3(aq)
Kd ϭ ϭ 8.5 ϫ 10Ϫ13
Recall that Cu2ϩ(aq) is really a hydrated ion, [Cu(H2O)6]2ϩ. The preceding reaction and
its Kd expression are represented more accurately as
[Cu(NH3)4]2ϩ ϩ 6H2O 34 [Cu(H2O)6]2ϩ ϩ 4NH3
Kd ϭ ϭ 8.5 ϫ 10Ϫ13
The more effectively a ligand competes with H2O for a coordination site on the metal
ions, the smaller Kd is. This tells us that in a comparison of complexes with the same
number of ligands, the smaller the Kd value, the more stable the complex ion. Some
complex ions and their dissociation constants, Kd, are listed in Appendix I.
Copper(II) hydroxide dissolves in an excess of aqueous NH3 to form the deep-blue
complex ion [Cu(NH3)4]2ϩ. This decreases the [Cu2ϩ] so that [Cu2ϩ][OHϪ]2 Ͻ Ksp, and
so the Cu(OH)2 dissolves.
Cu(OH)2(s) 34 Cu2ϩ(aq) ϩ 2OHϪ(aq)
Cu2ϩ(aq) ϩ 4NH3(aq) 34 [Cu(NH3)4]2ϩ(aq)
overall rxn: Cu(OH)2(s) ϩ 4NH3(aq) 34 [Cu(NH3)4]2ϩ(aq) ϩ 2OHϪ(aq)
Similarly, Zn(OH)2 dissolves in excess NH3 to form [Zn(NH3)4]2ϩ ions.
Zn(OH)2(s) ϩ 4NH3(aq) 34 [Zn(NH3)4]2ϩ(aq) ϩ 2OHϪ(aq)
Amphoteric hydroxides such as Zn(OH)2 also dissolve in excess strong base by forming
complex ions (Section 10-6).
Zn(OH)2(s) ϩ 2OHϪ(aq) 34 [Zn(OH)4]2Ϫ(aq)
We see that we are able to shift equilibria (in this case, dissolve Zn(OH)2) by taking
advantage of complex ion formation.
[[Cu(H2O)6]2ϩ][NH3]4
ᎏᎏᎏ
[[Cu(NH3)4]2ϩ]
[Cu2ϩ][NH3]4
ᎏᎏ
[[Cu(NH3)4]2ϩ]As before, the outer brackets mean
molar concentrations. The inner
brackets are part of the formula of
the complex ion.
842 CHAPTER 20: Ionic Equilibria III: The Solubility Product Principle
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 19.12, Complex Ion Formation
and Solubility.
For brevity we shall omit H2O from
formulas of hydrated ions. For
example, we write [Cu(OH2)6]2ϩ as
Cu2ϩ.
Concentrated aqueous NH3 was added slowly to a solution of copper(II) sulfate, CuSO4.
Unreacted blue copper(II) sulfate solution remains in the bottom part of the test tube.
The light-blue precipitate in the middle is copper(II) hydroxide, Cu(OH)2. The top layer
contains deep-blue [Cu(NH3)4]2ϩ ions that were formed as some Cu(OH)2 dissolved in
excess aqueous NH3.
Exercises 843
Key Terms
Complex ions Ions resulting from the formation of coordinate
covalent bonds between simple cations and other ions or mol-
ecules.
Dissociation constant The equilibrium constant that applies to
the dissociation of a complex ion into a simple ion and coordinating
species (ligands).
Fractional precipitation Removal of some ions from solution
by precipitation while leaving other ions, with similar properties,
in solution.
Molar solubility The number of moles of a solute that dissolve
to produce a liter of saturated solution.
Precipitate A solid formed by mixing in solution the constituent
ions of a slightly soluble compound.
Solubility product constant, Ksp The equilibrium constant that
applies to the dissolution of a slightly soluble compound.
Solubility product principle The solubility product constant
expression for a slightly soluble compound is the product of the
concentrations of the constituent ions, each raised to the power
that corresponds to the number of ions in one formula unit.
Exercises
Consult Appendix H for solubility product constant values and
Appendix I for complex ion dissociation constants, as needed.
Solubility Product
*01. The solubility product constant values listed in Appendix
H were determined at 25°C. How would those Ksp values
change, if at all, with a change in temperature?
*02. State the solubility product principle. What is its signifi-
cance?
*03. Why do we not include a term for the solid in a solubility
product expression?
*04. What do we mean when we refer to the molar solubility
of a compound?
*05. Write the solubility product expression for each of the following
salts. (a) SnI2; (b) Bi2(SO4)3; (c) CuBr; (d) Ag3PO4.
*06. Write the solubility product expression for each of the
following salts. (a) Co3(AsO4)2; (b) Hg2I2 [contains
mercury(I) ions, Hg2
2ϩ]; (c) MgF2; (d) (Ag)2CO3.
*07. The Ksp value for BaSO4 is calculated from the expression
Ksp ϭ [Baϩ2][SO4
Ϫ2], whereas the Ksp value for Mg(OH)2
is calculated from the expression, Ksp ϭ [Mgϩ2][OHϪ]2.
Explain why the hydroxide ion concentration is squared,
but none of the other concentrations is squared.
Experimental Determination of Ksp
Values of Ksp calculated from the solubility data in these exercises
may not agree exactly with the solubility products given in
Appendix H because of rounding differences.
*08. From the solubility data given for the following compounds,
calculate their solubility product constants.
(a) SrCrO4, strontium chromate, 1.2 mg/mL
(b) BiI3, bismuth iodide, 7.7 ϫ 10Ϫ3 g/L
(c) Fe(OH)2, iron(II) hydroxide, 1.1 ϫ 10Ϫ3 g/L
(d) SnI2, tin(II) iodide, 10.9 g/L
*09. From the solubility data given for the following compounds,
calculate their solubility product constants.
(a) CuBr, copper(I) bromide, 1.0 ϫ 10Ϫ3 g/L
(b) AgI, silver iodide, 2.8 ϫ 10Ϫ8 g/10 mL
(c) Pb3(PO4)2, lead(II) phosphate, 6.2 ϫ 10Ϫ7 g/L
(d) Ag2SO4, silver sulfate, 5.0 mg/mL
*10. Construct a table like Table 20-1 for the compounds listed
in Exercise 8. Which compound has (a) the highest molar
solubility; (b) the lowest molar solubility; (c) the largest
Ksp; (d) the smallest Ksp?
*11. Construct a table like Table 20-1 for the compounds listed
in Exercise 9. Which compound has (a) the highest molar
solubility; (b) the lowest molar solubility; (c) the largest
Ksp; (d) the smallest Ksp?
*12. A solution is produced by stirring 1 gram of calcium fluoride
in one liter of water at 25°C. Careful analysis shows
that 0.0163 grams of calcium fluoride has dissolved. Calculate
the Ksp for calcium fluoride based on these data.
*13. Calculate the Ksp for zinc phosphate if 1.18 ϫ 10Ϫ4 grams
of zinc phosphate dissolved to make 2.0 liters of a saturated
solution.
Uses of Solubility Product Constants
*14. The solubility product for calcium carbonate is 4.8 ϫ 10Ϫ9.
What is the solubility of calcium carbonate in grams per
liter of solution?
*15. Calculate molar solubilities, concentrations of constituent
ions, and solubilities in grams per liter for the following
compounds at 25°C: (a) Zn(CN)2, zinc cyanide; (b) PbI2,
lead iodide; (c) Pb3(AsO4)2, lead(II) arsenate; (d) Hg2CO3,
mercury(I) carbonate [the formula for the mercury(I) ion
is Hg2
2ϩ].
*16. Calculate molar solubilities, concentrations of constituent
ions, and solubilities in grams per liter for the following
compounds at 25°C: (a) CuI, copper(I) iodide;
(b) Ba3(PO4)2, barium phosphate; (c) MgF2, magnesium
fluoride; (d) Pb3(PO4)2, lead(II) phosphate.
*17. What is the concentration of lead ions in one liter of
saturated PbCrO4 solution?
844 CHAPTER 20: Ionic Equilibria III: The Solubility Product Principle
*18. Barium sulfate is used to produce distinct X-rays of the
gastrointestinal tract. What is the maximum mass of
barium sulfate that can dissolve in 10.0 liters of water, a
volume much greater than the volume of the average
gastrointestinal tract?
*19. Calculate the molar solubility of AgBr in 0.015 M KBr
solution.
*20. Calculate the molar solubility of Ag2SO4 in 0.15 M K2SO4
solution.
*21. Construct a table similar to Table 20-1 for the compounds
listed in Exercise 15. Which compound has (a) the highest
molar solubility; (b) the lowest molar solubility; (c) the
highest solubility, expressed in grams per liter; (d) the lowest
solubility, expressed in grams per liter?
*22. Construct a table similar to Table 20-1 for the compounds
listed in Exercise 16. Which compound has (a) the highest
molar solubility; (b) the lowest molar solubility; (c) the
highest solubility, expressed in grams per liter; (d) the
lowest solubility, expressed in grams per liter?
*23. Of the three compounds BaCO3, Fe(OH)2, and Ag2CrO4,
which has (a) the highest molar solubility; (b) the lowest
molar solubility; (c) the highest solubility, expressed in
grams per liter; and (d) the lowest solubility, expressed in
grams per liter?
*24. Of the three compounds, Ag3PO4, AgI, and PbCO3, which
has (a) the highest molar solubility; (b) the lowest molar
solubility; (c) the highest solubility, expressed in grams per
liter; and (d) the lowest solubility, expressed in grams per
liter?
*25. What volume of water is required to dissolve 10. grams of
copper(II) carbonate, CuCO3?
*26. Which has the greater molar solubility in 0.20 M K2CrO4
solution: BaCrO4 or Ag2CrO4?
*27. Will a precipitate form when 1.00 g of AgNO3 is added to
50.0 mL of 0.050 M NaCl? If so, would you expect the
precipitate to be visible?
*28. Will a precipitate of PbCl2 form when 5.0 g of solid
Pb(NO3)2 is added to 1.00 L of 0.010 M NaCl? Assume
that volume change is negligible.
*29. Sodium bromide and lead nitrate are soluble in water.
Will lead bromide precipitate when 1.03 g of NaBr and
0.332 g of Pb(NO3)2 are dissolved in sufficient water to
make 1.00 L of solution?
*30. Will a precipitate of Cu(OH)2 form when 10.0 mL of
0.010 M NaOH is added to 1.00 L of 0.010 M CuCl2?
*31. A solution is 0.0100 M in Pb2ϩ ions. If 0.103 mol of solid
Na2SO4 is added to 1.00 L of this solution (with negligible
volume change), what percentage of the Pb2ϩ ions
remain in solution?
*32. A solution is 0.0100 M in Pb2ϩ ions. If 0.103 mol of solid
NaI is added to 1.00 L of this solution (with negligible volume
change), what percentage of the Pb2ϩ ions remain in
solution?
*33. A solution is 0.0100 M in Ba(NO3)2. If 0.103 mol of solid
Na3PO4 is added to 1.00 L of this solution (with negligible
volume change), what percentage of the Ba2ϩ ions
remain in solution?
Fractional Precipitation
*34. What is fractional precipitation?
*35. Solid Na2SO4 is added slowly to a solution that is 0.10 M
in Pb(NO3)2 and 0.10 M in Ba(NO3)2. In what order will
solid PbSO4 and BaSO4 form? Calculate the percentage of
Ba2ϩ that precipitates just before PbSO4 begins to precip-
itate.
*36. To a solution that is 0.10 M in Cuϩ, 0.10 M in Agϩ, and
0.10 M in Auϩ, solid NaCl is added slowly. Assume that
there is no volume change due to the addition of solid
NaCl. (a) Which compound will begin to precipitate first?
(b) Calculate [Auϩ] when AgCl just begins to precipitate.
What percentage of the Auϩ has precipitated at this point?
(c) Calculate [Auϩ] and [Agϩ] when CuCl just begins to
precipitate.
*37. A solution is 0.015 M in Pb2ϩ and 0.015 M in Agϩ. As ClϪ
is introduced to the solution by the addition of solid NaCl,
determine (a) which substance will precipitate first, AgCl
or PbCl2, and (b) the fraction of the metal ion in the first
precipitate that remains in solution at the moment the precipitation
of the second compound begins.
*38. A solution is 0.050 M in K2SO4 and 0.050 M in K2CrO4.
A solution of Pb(NO3)2 is added slowly without changing
the volume appreciably. (a) Which salt, PbSO4 or PbCrO4,
will precipitate first? (b) What is [Pb2ϩ] when the salt in
part (a) begins to precipitate? (c) What is [Pb2ϩ] when the
other lead salt begins to precipitate? (d) What are [SO4
2Ϫ]
and [CrO4
2Ϫ] when the lead salt in part (c) begins to
precipitate?
*39. Solid Pb(NO3)2 is added slowly to a solution that is 0.015 M
each in NaOH, K2CO3, and Na2SO4. (a) In what order
will solid Pb(OH)2, PbCO3, and PbSO4 begin to precipitate?
(b) Calculate the percentages of OHϪ and CO3
2Ϫ
that have precipitated when PbSO4 begins to precipitate.
*40. Suppose you have three beakers that contain, respectively,
100 mL of each of the following solutions: (i) 0.0015 M
KOH; (ii) 0.0015 M K2CO3; (iii) 0.0015 M KCN. (a) If
solid zinc nitrate, Zn(NO3)2, were added slowly to each
beaker, what concentration of Zn2ϩ would be required to
initiate precipitation? (b) If solid zinc nitrate were added
to each beaker until [Zn2ϩ] ϭ 0.0015 M, what concentrations
of OHϪ, CO3
2Ϫ, and CNϪ would remain in solution,
that is, unprecipitated? Neglect any volume change when
solid is added.
*41. Suppose you have three beakers that contain, respectively,
100 mL each of the following solutions: (i) 0.0015 M KOH;
(ii) 0.0015 M K2CO3; (iii) 0.0015 M KI. (a) If solid lead
nitrate, Pb(NO3)2, were added slowly to each beaker, what
Exercises 845
concentration of Pb2ϩ would be required to initiate precipitation?
(b) If solid lead nitrate were added to each
beaker until [Pb2ϩ] ϭ 0.0015 M, what concentrations of
OHϪ, CO3
2Ϫ, and IϪ would remain in solution, that is,
unprecipitated? Neglect any volume change when solute is
added.
Simultaneous Equilibria
*42. If a solution is made 0.080 M in Mg(NO3)2, 0.075 M in
aqueous ammonia, and 3.5 M in NH4NO3, will Mg(OH)2
precipitate? What is the pH of this solution?
*43. If a solution is made 0.090 M in Mg(NO3)2, 0.090 M in
aqueous ammonia, and 0.080 M in NH4NO3, will
Mg(OH)2 precipitate? What is the pH of this solution?
*44. Calculate the solubility of CaF2 in a solution that is
buffered at [Hϩ] ϭ 0.0050 M with [HF] ϭ 0.10 M.
*45. Calculate the solubility of AgCN in a solution that is
buffered at [Hϩ] ϭ 0.000200 M, with [HCN] ϭ 0.01 M.
*46. If a solution is 2.0 ϫ 10Ϫ5 M in Mn(NO3)2 and 1.0 ϫ
10Ϫ3 M in aqueous ammonia, will Mn(OH)2 precipitate?
*47. If a solution is 0.040 M in manganese(II) nitrate,
Mn(NO3)2, and 0.080 M in aqueous ammonia, will manganese(II)
hydroxide, Mn(OH)2, precipitate?
*48. Milk of magnesia is a suspension of the slightly soluble
compound Mg(OH)2 in water. (a) What is the molar
solubility of Mg(OH)2 in a 0.015 M NaOH solution?
(b) What is the molar solubility of Mg(OH)2 in a 0.015 M
MgCl2 solution?
*49. How many moles of Cr(OH)3 will dissolve in 1.00 L of a
solution with a pH of 5.00?
*50. Determine whether a precipitate forms when a 0.00050 M
solution of magnesium nitrate is brought to a pH of 8.75.
*51. What concentration of NH4NO3 is necessary to prevent
precipitation of Mn(OH)2 in the solution of Exercise 47?
*52. (a) What is the pH of a saturated solution of Mn(OH)2?
(b) What is the solubility in g Mn(OH)2/100 mL of
solution?
*53. (a) What is the pH of a saturated solution of Mg(OH)2?
(b) What is the solubility in g Mg(OH)2/100 mL of
solution?
Dissolution of Precipitates and Complex Ion
Formation
*54. Explain, by writing appropriate equations, how the following
insoluble compounds can be dissolved by the
addition of a solution of nitric acid. (Carbonates dissolve
in strong acids to form carbon dioxide, which is evolved
as a gas, and water.) What is the “driving force” for each
reaction? (a) Cu(OH)2; (b) Al(OH)3; (c) MnCO3;
(d) (PbOH)2CO3.
*55. Explain, by writing equations, how the following insoluble
compounds can be dissolved by the addition of a
solution of ammonium nitrate or ammonium chloride.
(a) Mg(OH)2; (b) Mn(OH)2; (c) Ni(OH)2.
*56. The following insoluble sulfides can be dissolved in 3 M
hydrochloric acid. Explain how this is possible, and write
the appropriate equations. (a) MnS; (b) CuS.
*57. The following sulfides are less soluble than those listed in
Exercise 56 and can be dissolved in hot 6 M nitric acid, an
oxidizing acid. Explain how, and write the appropriate balanced
equations. (a) PbS; (b) CuS; (c) Bi2S3.
*58. Why would MnS be expected to be more soluble in
0.10 M HCl solution than in water? Would the same be
true for Mn(NO3)2?
*59. For each pair, choose the salt that would be expected to
be more soluble in acidic solution than in pure water, and
justify your choice: (a) Hg2(CH3COO)2 or Hg2Br2;
(b) Pb(OH)2 or PbI2; (c) AgI or AgNO2.
Mixed Exercises
*60. We mix 25.0 mL of a 0.0030 M solution of BaCl2 and 50.0 mL
of a 0.050 M solution of NaF. (a) Find [Ba2ϩ] and [FϪ] in
the mixed solution at the instant of mixing (before any possible
reaction occurs). (b) Would BaF2 precipitate?
*61. A concentrated, strong acid is added to a solid mixture of
0.015-mol samples of Fe(OH)2 and Cu(OH)2 placed in
1.0 L of water. At what values of pH will the dissolution
of each hydroxide be complete? (Assume negligible volume
change.)
*62. A solution is 0.015 M in IϪ ions and 0.015 M in BrϪ ions.
Agϩ ions are introduced to the solution by the addition of
solid AgNO3. Determine (a) which compound will precipitate
first, AgI or AgBr, and (b) the percentage of the
halide ion in the first precipitate that is removed from
solution before the precipitation of the second compound
begins.
*63. Calculate the molar solubility of Ag2SO4 (a) in pure water,
(b) in 0.015 M AgNO3, and (c) in 0.015 M K2SO4.
CONCEPTUAL EXERCISES
*64. Suppose there was a deposit of gold(I) chloride in a river
bed. What minimum volume of water, at 25°C, would be
required to dissolve 1.0 gram of gold ions?
*65. (a) Are “insoluble” substances really insoluble? (b) What
do we mean when we refer to insoluble substances?
*66. Solubility product calculations are actually based on heterogeneous
equilibria. Why are pure solids and liquids
exempted from these calculations?
*67. Draw a picture of a portion of a saturated silver chloride
solution at the molecular level. Show a small amount of
solid plus some dissociated ions. You need not show water
or waters of hydration. Prepare a second drawing that
846 CHAPTER 20: Ionic Equilibria III: The Solubility Product Principle
includes the same volume of solution but twice as much
solid. Should your drawing include more, less, or the same
number of silver ions?
*68. The solubility product constants of silver chloride, AgCl,
and silver chromate, Ag2CrO4, are 1.8 ϫ 10Ϫ10 and 9.0 ϫ
10Ϫ12, respectively. Suppose that the chloride, ClϪ(aq), and
chromate, CrO4
2Ϫ(aq), ions are both present in the same
solution at concentrations of 0.010 M each. A standard
solution of silver ions, Agϩ(aq), is dispensed slowly from a
buret into this solution while it is stirred vigorously. Solid
silver chloride is white, and silver chromate is red. What
will be the concentration of ClϪ(aq) ions in the mixture
when the first tint of red color appears in the mixture?
BUILDING YOUR KNOWLEDGE
*69. How many moles of CO3
2Ϫ must be added to 0.50 liter
of a 0.10 M Sr2ϩ solution to produce a solution that is
1.0 ϫ 10Ϫ6 M Sr2ϩ? How many moles of CO3
2Ϫ are in
the final solution and how many moles of CO3
2Ϫ are in
the precipitate formed? Assume no volume change for the
solution.
*70. A fluoridated water supply contains 1 mg/L of FϪ. What
is the maximum amount of Ca2ϩ, expressed in grams per
liter, that can exist in this water supply?
*71. Many industrial operations require very large amounts of
water as a coolant in heat exchange processes. Muddy or
cloudy water is usually unsatisfactory because the dispersed
solids may clog filters or deposit sediment in pipes and
pumps. Murky water can be clarified on a large scale by
adding agents to coagulate colloidal material, and then
allowing the precipitate to settle out in holding tanks or
ponds before the clarified water is sent to plant intakes.
Recent methods employ the addition of both calcium
hydroxide and magnesium carbonate. If 56 g of Ca(OH)2
and 45 g of MgCO3 were added to 520 liters of water,
would these compounds form a precipitate?
*72. Magnesium carbonate is used in the manufacture of a highdensity
magnesite brick. This material is not well suited to
general exterior use because the magnesium carbonate easily
erodes. What percentage of 28 grams of surface-exposed
MgCO3 would be lost through the solvent action of 15
liters of water? Assume sufficient contact time for the water
to become saturated with MgCO3.
21Electrochemistry
OUTLINE
21-1 Electrical Conduction
21-2 Electrodes
Electrolytic Cells
21-3 The Electrolysis of Molten
Sodium Chloride (the Downs
Cell)
21-4 The Electrolysis of Aqueous
Sodium Chloride
21-5 The Electrolysis of Aqueous
Sodium Sulfate
21-6 Faraday’s Law of Electrolysis
21-7 Commercial Applications of
Electrolytic Cells
Voltaic or Galvanic Cells
21-8 The Construction of Simple
Voltaic Cells
21-9 The Zinc–Copper Cell
21-10 The Copper–Silver Cell
Standard Electrode Potentials
21-11 The Standard Hydrogen
Electrode
21-12 The Zinc–SHE Cell
21-13 The Copper–SHE Cell
21-14 Standard Electrode Potentials
21-15 Uses of Standard Electrode
Potentials
21-16 Standard Electrode Potentials
for Other Half-Reactions
21-17 Corrosion
21-18 Corrosion Protection
Effect of Concentrations (or Partial
Pressures) on Electrode Potentials
21-19 The Nernst Equation
21-20 Using Electrochemical Cells
to Determine Concentrations
21-21 The Relationship of E0
cell to
⌬G0 and K
Primary Voltaic Cells
21-22 Dry Cells
Secondary Voltaic Cells
21-23 The Lead Storage Battery
21-24 The Nickel–Cadmium
(Nicad) Cell
21-25 The Hydrogen–Oxygen Fuel
Cell
OBJECTIVES
After you have studied this chapter, you should be able to
• Use the terminology of electrochemistry (terms such as “cell,” “electrode,” “cathode,”
“anode”)
• Describe the differences between electrolytic cells and voltaic (galvanic) cells
• Recognize oxidation and reduction half-reactions, and know at which electrode each
occurs
• Write half-reactions and overall cell reactions for electrolysis processes
• Use Faraday’s Law of Electrolysis to calculate amounts of products formed, amounts of
current passed, time elapsed, and oxidation state
• Describe the refining and plating of metals by electrolytic methods
• Describe the construction of simple voltaic cells from half-cells and a salt bridge, and
understand the function of each component
• Write half-reactions and overall cell reactions for voltaic cells
Corrosion is an undesirable
electrochemical reaction with very
serious economic consequences. Rust
is formed when iron is oxidized in
the presence of water.
• Compare various voltaic cells to determine the relative strengths of oxidizing and
reducing agents
• Interpret standard reduction potentials
• Use standard reduction potentials, E0, to calculate the potential of a standard voltaic
cell, E0
cell
• Use standard reduction potentials to identify the cathode and the anode in a standard
cell
• Use standard reduction potentials to predict the spontaneity of a redox reaction
• Use standard reduction potentials to identify oxidizing and reducing agents in a cell or
in a redox reaction
• Describe some corrosion processes and some methods for preventing corrosion
• Use the Nernst equation to relate electrode potentials and cell potentials to different
concentrations and partial pressures
• Relate the standard cell potential (E0
cell ) to the standard Gibbs free energy change
(⌬G0) and the equilibrium constant (K)
• Distinguish between primary and secondary voltaic cells
• Describe the compositions and reactions of some useful primary and secondary cells
(batteries)
• Describe the electrochemical processes involved in discharging and recharging a lead
storage (automobile) battery
EE
lectrochemistry deals with the chemical changes produced by electric current and
with the production of electricity by chemical reactions. Many metals are purified
or are plated onto jewelry by electrochemical methods. Digital watches, automobile
starters, calculators, and pacemakers are just a few devices that depend on
electrochemically produced power. Corrosion of metals is an electrochemical process.
We learn much about chemical reactions from the study of electrochemistry. The
amount of electrical energy consumed or produced can be measured quite accurately. All
electrochemical reactions involve the transfer of electrons and are therefore oxidation–
reduction reactions. The sites of oxidation and reduction are separated physically so that
oxidation occurs at one location, and reduction occurs at the other. Electrochemical
processes require some method of introducing a stream of electrons into a reacting chemical
system and some means of withdrawing electrons. In most applications the reacting
system is contained in a cell, and an electric current enters or exits by electrodes.
We classify electrochemical cells into two types.
1. Electrolytic cells are those in which electrical energy from an external source causes
nonspontaneous chemical reactions to occur.
2. Voltaic cells are those in which spontaneous chemical reactions produce electricity
and supply it to an external circuit.
You should review oxidation–
reduction (redox) reactions in
text Section 4-5. See the Saunders
Interactive General Chemistry CD-ROM,
Screen 21.2, Redox Reactions.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 21.4, Electrochemical Cells.
We will discuss several electrochemical cells. From experimental observations we
deduce the electrode reactions and the overall reactions. We then construct simplified
diagrams of the cells.
ELECTRICAL CONDUCTION
Electric current represents transfer of charge. Charge can be conducted through metals
and through pure liquid electrolytes (that is, molten salts) or solutions containing electrolytes.
The former type of conduction is called metallic conduction. It involves the
flow of electrons with no similar movement of the atoms of the metal and no obvious
changes in the metal (Section 13-17). Ionic, or electrolytic, conduction is the conduction
of electric current by the motion of ions through a solution or a pure liquid. Positively
charged ions migrate toward the negative electrode while negatively charged ions move
toward the positive electrode. Both kinds of conduction, ionic and metallic, occur in electrochemical
cells (Figure 21-1).
ELECTRODES
Electrodes are surfaces on which oxidation or reduction half-reactions occur. They may
or may not participate in the reactions. Those that do not react are called inert electrodes.
Regardless of the kind of cell, electrolytic or voltaic, the electrodes are identified
as follows.
The cathode is defined as the electrode at which reduction occurs as electrons are
gained by some species. The anode is the electrode at which oxidation occurs as
electrons are lost by some species.
Each of these can be either the positive or the negative electrode.
21-2
21-1
850 CHAPTER 21: Electrochemistry
Many reactions that take place in
living cells are redox reactions.
These reactions can be studied with
miniature electrodes.
+
+
+
+
+
–
–
–
–
–
+–
Electrons
Battery
–+
Figure 21-1 The motion of ions through a solution is an electric current. This accounts
for ionic (electrolytic) conduction. Positively charged ions migrate toward the negative
electrode, and negatively charged ions migrate toward the positive electrode. Here the rate
of migration is greatly exaggerated for clarity. The ionic velocities are actually only slightly
greater than random molecular speeds.
ELECTROLYTIC CELLS
In some electrochemical cells nonspontaneous chemical reactions are forced to occur by the
input of electrical energy. This process is called electrolysis. An electrolytic cell consists
of a container for the reaction material with electrodes immersed in the reaction material
and connected to a source of direct current. Inert electrodes are often used so that
they do not react.
THE ELECTROLYSIS OF MOLTEN SODIUM CHLORIDE
(THE DOWNS CELL)
Solid sodium chloride does not conduct electricity. Its ions vibrate about fixed positions,
but they are not free to move throughout the crystal. Molten (melted) NaCl, however, is
an excellent conductor because its ions are freely mobile. Consider a cell in which a source
of direct current is connected by wires to two inert graphite electrodes (Figure 21-2a).
They are immersed in a container of molten sodium chloride. When the current flows,
we observe the following.
1. A pale green gas, which is chlorine, Cl2, is liberated at one electrode.
2. Molten, silvery white metallic sodium, Na, forms at the other electrode and floats
on top of the molten sodium chloride.
21-3
21-3 The Electrolysis of Molten Sodium Chloride (The Downs Cell) 851
+–
Battery
+–Electrons
Ammeter
Voltmeter
Molten NaCl
consists of Na+
and Cl–
Graphite
electrode
(inert)
Graphite electrode
(inert)
Na+
is reduced
(Na+
+ le–
→ Na);
therefore, cathode
Cl–
is oxidized
(2Cl–
→ Cl2 + 2e–
);
therefore, anode
Na(ᐉ)
Molten
NaCl Cl2(g)
Ring-shaped
iron cathode
Graphite
anode
Iron screen
– +
(a)
(b)
Figure 21-2 (a) Apparatus for electrolysis of molten sodium chloride. (b) The Downs cell,
the apparatus in which molten sodium chloride is commercially electrolyzed to produce
sodium metal and chlorine gas. The liquid Na floats on the denser molten NaCl.
Lysis means “splitting apart.” In many
electrolytic cells compounds are split
into their constituent elements.
Molten NaCl, melting point 801°C, is a
clear, colorless liquid that looks like
water.
The metal remains liquid because its
melting point is only 97.8°C. It floats
because it is less dense than the molten
NaCl.
From these observations we can deduce the processes of the cell. Chlorine must be
produced by oxidation of ClϪ ions, and the electrode at which this happens must be the
anode. Metallic sodium is produced by reduction of Naϩ ions at the cathode, where electrons
are being forced into the cell.
2ClϪ 88n Cl2(g) ϩ 2eϪ (oxidation, anode half-reaction)
2[Naϩ ϩ eϪ 88n Na(ᐉ)] (reduction, cathode half-reaction)
2Naϩ ϩ 2ClϪ 88n 2Na(ᐉ) ϩ Cl2(g) (overall cell reaction)
2NaCl(ᐉ)
The formation of metallic Na and gaseous Cl2 from NaCl is nonspontaneous except at
temperatures very much higher than 801°C. The direct current (dc) source must supply
electrical energy to force this reaction to occur. Electrons are used in the cathode halfreaction
(reduction) and produced in the anode half-reaction (oxidation). They therefore
travel through the wire from anode to cathode. The dc source forces electrons to flow
nonspontaneously from the positive electrode to the negative electrode. The anode is the
positive electrode and the cathode the negative electrode in all electrolytic cells. Figure 21-2a
is a simplified diagram of the cell.
Sodium and chlorine must not be allowed to come in contact with each other because
they react spontaneously, rapidly, and explosively to form sodium chloride. Figure 21-2b
shows the Downs cell that is used for the industrial electrolysis of sodium chloride. The
Downs cell is expensive to run, mainly because of the cost of construction, the cost of the
electricity, and the cost of heating the NaCl to melt it. Nevertheless, electrolysis of a
molten sodium salt is the most practical means by which metallic Na can be obtained,
owing to its extremely high reactivity. Once liberated by the electrolysis, the liquid Na
metal is drained off, cooled, and cast into blocks. These must be stored in an inert environment
(e.g., in mineral oil) to prevent reaction with O2 or other components of the
atmosphere.
Electrolysis of NaCl in the Downs cell is the main commercial method of producing
metallic sodium. The Cl2 gas produced in the Downs cell is cooled, compressed, and
marketed. This partially offsets the expense of producing metallic sodium, but most chlorine
is produced by the cheaper electrolysis of aqueous NaCl.
THE ELECTROLYSIS OF AQUEOUS SODIUM CHLORIDE
Consider the electrolysis of a moderately concentrated solution of NaCl in water, using
inert electrodes. The following experimental observations are made when a sufficiently
high voltage is applied across the electrodes of a suitable cell.
1. H2 gas is liberated at one electrode. The solution becomes basic in that vicinity.
2. Cl2 gas is liberated at the other electrode.
Chloride ions are obviously being oxidized to Cl2 in this cell, as they were in the electrolysis
of molten NaCl. But Naϩ ions are not reduced to metallic Na. Instead, gaseous
H2 and aqueous OHϪ ions are produced by reduction of H2O molecules at the cathode.
Water is more easily reduced than Naϩ ions. This is primarily because the reduction of
Naϩ would produce the very active metal Na, whereas the reduction of H2O produces
the more stable products H2(g) and OHϪ(aq). The active metals Li, K, Ca, and Na (Table
4-12) displace H2 from aqueous solutions, so we do not expect these metals to be produced
in aqueous solution. Later in this chapter (Section 21-14) we learn the quantitative basis
21-4
852 CHAPTER 21: Electrochemistry
In this chapter, as in Chapters 4 and
11, we often use red type to emphasize
reduction and blue type to emphasize
oxidation.
The direction of spontaneous flow for
negatively charged particles is from
negative to positive.









for predicting which of several possible oxidations or reductions is favored. The halfreactions
and overall cell reaction for this electrolysis are
2ClϪ 88n Cl2 ϩ 2eϪ (oxidation, anode)
2H2O ϩ 2eϪ 88n 2OHϪ ϩ H2 (reduction, cathode)
2H2O ϩ 2ClϪ 88n 2OHϪ ϩ H2 ϩ Cl2 (overall cell reaction as
net ionic equation)
ϩ 2Naϩ 88n ϩ 2Naϩ (spectator ions)
2H2O ϩ 2NaCl 88n 2NaOH ϩ H2 ϩ Cl2 (overall cell reaction as
formula unit equation)
The cell is illustrated in Figure 21-3. As before, the electrons flow from the anode (ϩ)
through the wire to the cathode (Ϫ).
We will omit the notation that
indicates states of substances—(s), (ᐉ),
(g), and (aq)—except where states are
not obvious. This abbreviates writing
equations.
The electrolysis of the aqueous
solution of KI, another Group IA–
Group VIIA salt. At the cathode
(left), water is reduced to H2(g)
and OHϪ ions, turning the
phenolphthalein indicator pink.
The characteristic brownish color
of aqueous I2 appears at the anode
(right).
Battery
+–
Pt anodePt cathode
H2(g) Na+
Na+
OH–
Cl–
Cl2(g)
2H2O + 2e–
→ H2(g) + 2OH–
Reduction
e–
e–
2Cl–
→ Cl2(g) + 2e–
Oxidation
+–
Figure 21-3 Electrolysis of
aqueous NaCl solution. Although
several reactions occur at both the
anode and the cathode, the net
result is the production of H2(g) and
NaOH at the cathode and Cl2(g)
at the anode. A few drops of
phenolphthalein indicator were
added to the solution. The solution
turns pink at the cathode, where
OHϪ ions are formed.










The overall cell reaction produces gaseous H2 and Cl2 and an aqueous solution of
NaOH, called caustic soda. Solid NaOH is then obtained by evaporation of the residual
solution. This is the most important commercial preparation of each of these substances.
It is much less expensive than the electrolysis of molten NaCl, because it is not necessary
to heat the solution.
THE ELECTROLYSIS OF AQUEOUS SODIUM SULFATE
In the electrolysis of aqueous sodium sulfate using inert electrodes, we observe the
following.
1. Gaseous H2 is produced at one electrode. The solution becomes basic around that
electrode.
2. Gaseous O2 is produced at the other electrode. The solution becomes acidic around
that electrode.
As in the previous example, water is reduced in preference to Naϩ at the cathode.
Observation 2 suggests that water is also preferentially oxidized relative to the sulfate ion,
SO4
2Ϫ, at the anode (Figure 21-4).
2(2H2O ϩ 2eϪ 88n H2 ϩ 2OHϪ) (reduction, cathode)
2H2O 88n O2 ϩ 4Hϩ ϩ 4eϪ (oxidation, anode)
6H2O 88n 2H2 ϩ O2 ϩ 4Hϩ ϩ 4OHϪ (overall cell reaction)
4H2O
2H2O 88n 2H2 ϩ O2 (net reaction)
The net result is the electrolysis of water. This occurs because H2O is more readily reduced
than Naϩ and more readily oxidized than SO4
2Ϫ. The ions of Na2SO4 conduct the current
through the solution, but they take no part in the reaction.
21-5
854 CHAPTER 21: Electrochemistry









Not surprisingly, the fluctuations in
commercial prices of these widely used
industrial products—H2, Cl2, and
NaOH—have often paralleled one
another.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 21.10, Electrolysis.
Gaseous H2 is produced from H2O
at an illuminated photoelectrode.
Light from the sun may soon be
used to produce hydrogen, the
ultimate clean-burning fuel.
Figure 21-4 The electrolysis of
aqueous Na2SO4 produces H2(g) at
the cathode and O2 at the anode.
Bromthymol blue indicator has been
added to the solution. This indicator
turns blue in the basic solution
near the cathode (where OHϪ is
produced) and yellow in the acidic
solution near the anode (where Hϩ
is formed).
Battery
+–
Pt anodePt cathode
H2(g) Na+
SO4
2–
OH–
H+
O2(g)
2(H2O + 2e–
→ H2(g) + 2OH–
)
Reduction
e–
e–
2H2O → O2(g) + 4H+
+ 4e–
Oxidation
+–
FARADAY’S LAW OF ELECTROLYSIS
In 1832–1833, Michael Faraday’s studies of electrolysis led to this conclusion.
The amount of substance that undergoes oxidation or reduction at each electrode
during electrolysis is directly proportional to the amount of electricity that passes
through the cell.
This is Faraday’s Law of Electrolysis. A quantitative unit of electricity is now called the
faraday.
One faraday is the amount of electricity that corresponds to the gain or loss, and
therefore the passage, of 6.022 ϫ 1023 electrons, or one mole of electrons.
A smaller electrical unit commonly used in chemistry, physics, and electronics is the
coulomb (C). One coulomb is defined as the amount of charge that passes a given point
when 1 ampere (A) of electric current flows for 1 second. One ampere of current equals
1 coulomb per second. One faraday is equal to 96,485 coulombs of charge.
1 ampere ϭ 1 ᎏ
c
s
o
e
u
c
l
o
o
n
m
d
b
ᎏ or 1 A ϭ 1 C/s
1 faraday ϭ 6.022 ϫ 1023 eϪ ϭ 96,485 C
Table 21-1 shows the amounts of several elements produced during electrolysis by the
passage of 1 faraday of electricity.
21-6
TABLE 21-1 Amounts of Elements Produced at One Electrode in Electrolysis by 1 Faraday of Electricity
Number of eϪ Product
Half-Reaction in Half-Reaction (electrode) Amount Produced
Agϩ(aq) ϩ eϪ 88n Ag(s) 1 Ag (cathode) 1 mol ϭ 107.868 g
2Hϩ(aq) ϩ 2eϪ 88n H2(g) 2 H2 (cathode) ᎏ1
2
ᎏ mol ϭ 1.008 g
Cu2ϩ(aq) ϩ 2eϪ 88n Cu(s) 2 Cu (cathode) ᎏ1
2
ᎏ mol ϭ 31.773 g
Au3ϩ(aq) ϩ 3eϪ 88n Au(s) 3 Au (cathode) ᎏ1
3
ᎏ mol ϭ 65.656 g
2ClϪ 88n Cl2(g) ϩ 2eϪ 2 Cl2 (anode) ᎏ1
2
ᎏ mol ϭ 35.453 g ϭ 11.2 LSTP
2H2O(ᐉ) 88n O2(g) ϩ 4Hϩ(aq) ϩ 4eϪ 4 O2 (anode) ᎏ1
4
ᎏ mol ϭ 8.000 g ϭ 5.60 LSTP
Michael Faraday (1791–1867) is considered the greatest experimental scientist of the
nineteenth century. As a bookbinder’s apprentice, he educated himself by extensive reading.
Intrigued by his self-study of chemistry and by a lecture given by Sir Humphry Davy, the
leading chemist of the day, Faraday applied for a position with Davy at the Royal Institution.
He subsequently became director of that laboratory. His public lectures on science were very
popular.
For comparison, a 100-watt household
light bulb uses a current of about 0.8
ampere.
EXAMPLE 21-1 Electrolysis
Calculate the mass of copper metal produced during the passage of 2.50 amperes of current
through a solution of copper(II) sulfate for 50.0 minutes.
Plan
The half-reaction that describes the reduction of copper(II) ions tells us the number of moles
of electrons required to produce one mole of copper metal. Each mole of electrons corresponds
to 1 faraday, or 9.65 ϫ 104 coulombs, of charge. The product of current and time gives the
number of coulombs.
ᎏ
c
ϫ
ur
t
r
im
en
e
t
ᎏ 88n ᎏ
co
n
u
o
lo
.
m
of
bs
ᎏ 88n ᎏ
m
p
o
a
l
s
o
se
f
d
eϪ
ᎏ 88n ᎏ
o
m
f
a
C
ss
u
ᎏ
Solution
The equation for the reduction of copper(II) ions to copper metal is
Cu2ϩϩ 2eϪ 88n Cu (reduction, cathode)
1 mol 2(6.02 ϫ 1023)eϪ 1 mol
63.5 g 2(9.65 ϫ 104 C) 63.5 g
We see that 63.5 grams of copper “plate out” for every 2 moles of electrons, or for every
2(9.65 ϫ 104 coulombs) of charge. We first calculate the number of coulombs passing through
the cell.
_?_ C ϭ 50.0 min ϫ ᎏ
1
6
m
0
i
s
n
ᎏ ϫ ᎏ
2.5
s
0 C
ᎏ ϭ 7.50 ϫ 103 C
We calculate the mass of copper produced by the passage of 7.50 ϫ 103 coulombs.
856 CHAPTER 21: Electrochemistry
_?_ g Cu ϭ 7.50 ϫ 103 C ϫ ᎏ
9.6
1
5
m
ϫ
ol
10
eϪ
4 C
ᎏ ϫ ᎏ
6
2
3.
m
5
o
g
l
C
eϪ
u
ᎏ ϭ 2.47 g Cu (about the mass of
a copper penny)
Notice how little copper is deposited by this considerable current in 50 minutes.
You should now work Exercises 26 and 32.
EXAMPLE 21-2 Electrolysis
What volume of oxygen gas (measured at STP) is produced by the oxidation of water in the
electrolysis of copper(II) sulfate in Example 21-1?
Plan
We use the same approach as in Example 21-1. Here we relate the amount of charge passed
to the number of moles, and hence the volume of O2 gas produced at STP.
ᎏ
c
ϫ
ur
t
r
im
en
e
t
ᎏ 88n ᎏ
co
n
u
o
lo
.
m
of
bs
ᎏ 88n ᎏ
m
p
o
a
l
s
o
se
f
d
eϪ
ᎏ 88n ᎏ
o
L
f
S
O
TP
2
ᎏ
Solution
The equation for the oxidation of water and the equivalence between the number of coulombs
and the volume of oxygen produced at STP are
2H2O 88n O2 ϩ 4Hϩ ϩ 4eϪ (oxidation, anode)
1 mol 4(6.02 ϫ 1023)eϪ
22.4 LSTP 4(9.65 ϫ 104 C)
The amount of electricity in Examples
21-1 and 21-2 would be sufficient to
light a 100-watt household light bulb
for about 150 minutes, or 2.5 hours.
When the number of significant
figures in the calculation warrants, the
value 96,485 coulombs is usually
rounded to 96,500 coulombs
(9.65 ϫ 104 C).
2.50 A ϭ 2.50 C/s
The number of coulombs passing through the cell is 7.50 ϫ 103 C. For every 4(9.65 ϫ 104
coulombs) passing through the cell, 22.4 L of O2 at STP is produced.
_?_ LSTP O2 ϭ 7.50 ϫ 103 C ϫ ᎏ
9.6
1
5
m
ϫ
ol
1
e
0
Ϫ
4 C
ᎏ ϫ ᎏ
22
4
.4
m
L
o
S
l
TP
eϪ
O2
ᎏ ϭ 0.435 LSTP O2
You should now work Exercise 28.
Notice how little product is formed by what seems to be a lot of electricity. This suggests
why electrolytic production of gases and metals is so costly.
COMMERCIAL APPLICATIONS OF
ELECTROLYTIC CELLS
Several elements are produced commercially by electrolysis. In Sections 21-3 to 21-5, we
described some electrolytic cells that produce sodium (the Downs cell), chlorine, hydrogen,
and oxygen. Electrolysis of molten compounds is also the common method of obtaining
other Group IA metals, IIA metals (except barium), and aluminum (Section 22-3). Impure
metals can also be refined electrolytically, as we will describe for copper in Section 22-8.
Metal-plated articles are common in our society. Jewelry and tableware are often plated
with silver. Gold is plated onto jewelry and electrical contacts. Copper is plated onto many
objects for decorative purposes (Figure 21-5). Some automobiles have steel bumpers plated
21-7
21-7 Commercial Applications of Electrolytic Cells 857
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 21.11, Coulometry.
A family memento that has been
electroplated with copper. To aid in
electroplating onto nonconductors,
such as shoes, the material is first
soaked in a concentrated electrolyte
solution to make it conductive.
– +
Cu anode
Cathode
Object to be plated
CuSO4
solution
Cu2+
e–
Copper
atom, Cu
Copper
ion, Cu2+ e–
Copper
atom, Cu
Copper
ion, Cu2+
Figure 21-5 Electroplating with
copper. The anode is made of pure
copper, which dissolves during
the electroplating process. This
replenishes the Cu2ϩ ions that are
removed from the solution as Cu
plates out on the cathode.
with thin films of chromium. A chrome bumper requires approximately 3 seconds of electroplating
to produce a smooth, shiny surface only 0.0002 mm thick. When the metal
atoms are deposited too rapidly, they are not able to form extended lattices. Rapid plating
of metal results in rough, grainy, black surfaces. Slower plating produces smooth surfaces.
“Tin cans” are steel cans plated electrolytically with tin; these are sometimes replaced by
cans plated in ᎏ1
3
ᎏ second with an extremely thin chromium film.
VOLTAIC OR GALVANIC CELLS
Voltaic, or galvanic, cells are electrochemical cells in which spontaneous oxidation–
reduction reactions produce electrical energy. The two halves of the redox reaction are
separated, requiring electron transfer to occur through an external circuit. In this way,
useful electrical energy is obtained. Everyone is familiar with some voltaic cells. The
batteries commonly used in flashlights, portable radios, photographic equipment, and
many toys and appliances are voltaic cells. Automobile batteries consist of voltaic cells
connected in series so that their voltages add. We will first consider some simple laboratory
cells used to measure the potential difference, or voltage, of a reaction under study.
We will then look at some common voltaic cells.
THE CONSTRUCTION OF SIMPLE VOLTAIC CELLS
A half-cell contains the oxidized and reduced forms of an element, or other more complex
species, in contact with one another. A common kind of half-cell consists of a piece of
metal (the electrode) immersed in a solution of its ions. Consider two such half-cells in
separate beakers (Figure 21-6). The electrodes are connected by a wire. A voltmeter can
be inserted into the circuit to measure the potential difference between the two electrodes,
or an ammeter can be inserted to measure the current flow. The electric current is the
result of the spontaneous redox reaction that occurs. We measure the potential of the cell.
The circuit between the two solutions is completed by a salt bridge. This can be any
medium through which ions can slowly pass. A salt bridge can be made by bending a piece
of glass tubing into the shape of a “U,” filling it with a hot saturated salt/5% agar solution,
and allowing it to cool. The cooled mixture “sets” to the consistency of firm gelatin.
As a result, the solution does not run out when the tube is inverted (see Figure 21-6), but
the ions in the gel are still able to move. A salt bridge serves three functions.
1. It allows electrical contact between the two solutions.
2. It prevents mixing of the electrode solutions.
3. It maintains the electrical neutrality in each half-cell as ions flow into and out of
the salt bridge.
A cell in which all reactants and products are in their thermodynamic standard states
(1 M for dissolved species and 1 atm partial pressure for gases) is called a standard
cell.
21-8
858 CHAPTER 21: Electrochemistry
These are named for Allesandro Volta
(1745–1827) and Luigi Galvani (1737–
1798), two Italian physicists of the
eighteenth century.
Neither of these meters generates
electrical energy.
Agar is a gelatinous material obtained
from algae.
When we show a concentration in a
standard cell as 1 M, it is assumed to be
exactly 1 M.
THE ZINC–COPPER CELL
Consider a standard cell made up of two half-cells, one a strip of metallic Cu immersed
in 1 M copper(II) sulfate solution and the other a strip of Zn immersed in 1 M zinc sulfate
solution (see Figure 21-6). This cell is called the Daniell cell. The following experimental
observations have been made about this cell.
1. The initial voltage is 1.100 volts.
2. The mass of the zinc electrode decreases. The concentration of Zn2ϩ increases in
the solution around the zinc electrode as the cell operates.
3. The mass of the copper electrode increases. The concentration of Cu2ϩ decreases
in the solution around this electrode as the cell operates.
The Zn electrode loses mass because some Zn metal is oxidized to Zn2ϩ ions, which
go into solution. Thus the Zn electrode is the anode. At the cathode, Cu2ϩ ions are reduced
to Cu metal. This plates out on the electrode, so its mass increases.
21-9
21-9 The Zinc–Copper Cell 859
Metallic
zinc
electrode
Cl– K+
Salt bridge
1 M CuSO41 M ZnSO4
Zn2+
Cu2+
Cu2+
+ 2e–
→ Cu
Reduction, cathode
Zn → Zn2+
+ 2e–
Oxidation, anode
– +
Voltmeter
K+ Cl–
(5% agar)
Electrons Electrons
Metallic
copper
electrode
+–
Zinc atom,
Zn
Zinc ion,
Zn2+
e–
e–
Copper
atom, Cu
Copper
ion, Cu2+
Sulfate
ion, SO4
2Ϫ
Figure 21-6 The zinc–copper voltaic cell utilizes the reaction
Zn(s) ϩ Cu2ϩ(aq) 88n Zn2ϩ(aq) ϩ Cu(s)
The standard potential of this cell is 1.10 volts. The standard cell can be represented as
Zn͉Zn2ϩ(1 M)͉͉Cu2ϩ(1 M)͉Cu.
Zn 88n Zn2ϩ ϩ 2eϪ (oxidation, anode)
Cu2ϩ ϩ 2eϪ 88n Cu (reduction, cathode)
Cu2ϩ ϩ Zn 88n Cu ϩ Zn2ϩ (overall cell reaction)
Electrons are released at the anode and consumed at the cathode. They therefore flow
through the wire from anode to cathode, as in all electrochemical cells. In all voltaic cells
the electrons flow spontaneously from the negative electrode to the positive electrode. So,
in contrast with electrolytic cells, the anode is negative and the cathode is positive. To
maintain electroneutrality and complete the circuit, two ClϪ ions from the salt bridge
migrate into the anode solution for every Zn2ϩ ion formed. Two Kϩ ions migrate into
860 CHAPTER 21: Electrochemistry
Our Daily LivesCCHEMISTRY IN USE
A Spectacular View of One Mole of
Electrons
Early in our study of chemistry, we saw that atoms are made
up of protons, neutrons, and electrons. We also discussed the
incredibly large size of Avogadro’s number, 6.022 ϫ 1023.
Although individual atoms and molecules are invisible to the
naked eye, one mole of atoms or molecules is easily detected.
Because subatomic particles are even smaller than atoms and
also invisible, you might never expect to see individual electrons.
Let’s consider the possibility, however, of seeing a
faraday of charge. A faraday of charge contains Avogadro’s
number of electrons. Would this collection of 6.022 ϫ 1023
electrons be visible? If so, what might it look like? It would
look quite spectacular!
Throughout the 1980s, scientists carefully studied data
collected during 5 million lightning flashes along the eastern
United States. The data were collected by 36 instruments
that were collectively known as the National Lightning
Detection Network. The investigating scientists found that
the electrical currents in lightning flashes over northern
Florida measured about 45,000 amps, about double the
25,000-amp currents in lightning flashes over the New England
states. This study showed that the amount of current
flowing during lightning flashes was inversely proportional
to the latitude (distance from the equator) of the storm.
One coulomb is the amount of charge that passes a point
when a one-ampere current flows for one second (1 coulomb
ϭ 1 ampereиsecond). Thus, a current of 96,500 amps
flowing for one second contains Avogadro’s number of electrons,
or one faraday of charge.
Measurements taken in northern Florida show that a typical
two-second lightning strike over that section of the
country would transfer approximately Avogadro’s number of
electrons between the clouds and the earth. So, for those living
in northern Florida, a spectacular mental view of one
mole of electrons can be obtained by visualizing a twosecond
lightning strike. Keep in mind that the average lightning
strike lasts only a small fraction of a second, and that
we can only have a mental view of a two-second lightning
strike by extrapolation of what is seen in nature. Because New
England lightning strikes produce only about half the current
of lightning strikes over northern Florida, people in New
England must try to imagine a four-second lightning strike.
Ronald DeLorenzo
Middle Georgia College
Compare the Ϫ/ϩ, anode–cathode,
and oxidation–reduction labels and the
directions of electron flow in Figures
21-2a and 21-6.
the cathode solution to replace every Cu2ϩ ion reduced. Some Zn2ϩ ions from the anode
vessel and some SO4
2Ϫ ions from the cathode vessel also migrate into the salt bridge.
Neither ClϪ nor Kϩ ions are oxidized or reduced in preference to the zinc metal or Cu2ϩ
ions.
As the reaction proceeds, the cell voltage decreases. When the cell voltage reaches zero,
the reaction has reached equilibrium, and no further net reaction occurs. At this point,
however, the metal ion concentrations in the cell are not zero. This description applies to
any voltaic cell.
Voltaic cells can be represented as follows for the zinc–copper cell.
salt bridge
g
Zn͉Zn2ϩ(1 M)͉͉Cu2ϩ (1 M)͉Cu
r p
species (and concentrations)
in contact with electrode surfaces
In this representation, a single line (͉) represents an interface at which a potential develops,
that is, an electrode. It is conventional to write the anode half-cell on the left in this
notation.
The same reaction occurs when a piece of Zn is dropped into a solution of CuSO4.
The Zn dissolves and the blue color of Cu2ϩ ions disappears. Copper forms on the Zn
and then settles to the bottom of the container. But no electricity flows in an external
circuit, because the two half-reactions are not physically separated.
Problem-Solving Tip: How to Tell the Anode from the Cathode
The correspondence between the names anode and cathode and the charge on the electrode
is different for electrolytic cells than for voltaic (galvanic) cells. Students sometimes
get confused by trying to remember which is which. Check the definitions of these two
terms in Section 21-2. The surest way to name these electrodes is to determine what
process takes place at each one.
anode I::F oxidation and cathode I::F reduction
As a memory aid, both anode and oxidation begin with a vowel, whereas both cathode
and reduction begin with a consonant.
(Left) A strip of zinc was placed in a
blue solution of copper(II) sulfate,
CuSO4. The copper has been
displaced from solution and has
fallen to the bottom of the beaker.
The resulting zinc sulfate solution is
colorless. This is the same overall
reaction as the one that occurs when
the two half-reactions are separated
in the zinc–copper cell (see Figure
21-6). (Right) No reaction occurs
when copper wire is placed in a
colorless zinc sulfate solution. The
reaction
Zn2ϩ ϩ Cu(s) 88n Zn(s) ϩ Cu2ϩ
is the reverse of the spontaneous
reaction in Figure 21-6; it has a
negative E0
cell and is nonspontaneous.
THE COPPER–SILVER CELL
Now consider a similar standard voltaic cell consisting of a strip of Cu immersed in 1 M
CuSO4 solution and a strip of Ag immersed in 1 M AgNO3 solution. A wire and a salt
bridge complete the circuit. The following observations have been made.
1. The initial voltage of the cell is 0.462 volt.
2. The mass of the copper electrode decreases. The Cu2ϩ ion concentration increases
in the solution around the copper electrode.
3. The mass of the silver electrode increases. The Agϩ ion concentration decreases in
the solution around the silver electrode.
In this cell the Cu electrode is the anode because Cu metal is oxidized to Cu2ϩ ions.
The Ag electrode is the cathode because Agϩ ions are reduced to metallic Ag (Figure
21-7).
Cu 88n Cu2ϩ ϩ 2eϪ (oxidation, anode)
2(Agϩ ϩ eϪ 88n Ag) (reduction, cathode)
Cu ϩ 2Agϩ 88n Cu2ϩ ϩ 2Ag (overall cell reaction)
As before, ions from the salt bridge migrate to maintain electroneutrality. Some NO3
Ϫ
ions (from the cathode vessel) and some Cu2ϩ ions (from the anode vessel) also migrate
into the salt bridge.
Recall that in the zinc–copper cell the copper electrode is the cathode; now in the
copper–silver cell the copper electrode is the anode.
21-10
862 CHAPTER 21: Electrochemistry
e–e–
Salt bridge
Cu(s) → Cu2+ +2e–
Oxidation, anode
– +
Voltmeter
1 M CuSO4
Ag+ + e– → Ag(s)
Reduction, cathode
1 M AgNO3
Cu2+
NO3
–
Ag(s)
K+
K+ NO3
–
– +
(5% agar)
Ag+
Cu(s)
Figure 21-7 The copper–silver voltaic cell utilizes the reaction
Cu(s) ϩ 2Agϩ(aq) 88n Cu2ϩ(aq) ϩ 2Ag(s)
The standard potential of this cell is 0.462 volt. This standard cell can be represented as
Cu͉Cu2ϩ(1 M)͉͉Agϩ(1 M)͉Ag.
Whether a particular electrode acts as an anode or a cathode in a voltaic cell depends
on what the other electrode of the cell is.
The two cells we have described show that the Cu2ϩ ion is more easily reduced (is a
stronger oxidizing agent) than Zn2ϩ, so Cu2ϩ oxidizes metallic zinc to Zn2ϩ. By contrast,
Agϩ ion is more easily reduced (is a stronger oxidizing agent) than Cu2ϩ ion, so Agϩ
oxidizes Cu atoms to Cu2ϩ. Conversely, metallic Zn is a stronger reducing agent than
metallic Cu, and metallic Cu is a stronger reducing agent than metallic Ag. We can now
arrange the species we have studied in order of increasing strength as oxidizing agents
and as reducing agents.
Zn2ϩ Ͻ Cu2ϩ Ͻ Agϩ Ag Ͻ Cu Ͻ Zn
8888888888888888888888n 88888888888888888n
Increasing strength Increasing strength
as oxidizing agents as reducing agents
21-10 The Copper–Silver Cell 863
The standard
Cu͉Cu2ϩ(1 M)͉͉Agϩ (1 M)͉Ag cell.
(Left) A spiral of copper wire was
placed in a colorless solution of
silver nitrate, AgNO3. The silver has
been displaced from solution and
adheres to the wire. The resulting
copper nitrate solution is blue. The
same reaction occurs when the two
half-reactions are separated in the
copper–silver cell (see Figure 21-7).
(Right) No reaction occurs when
silver wire is placed in a blue copper
sulfate solution. The reaction
Cu2ϩ ϩ 2Ag(s) 88n Cu(s) ϩ 2Agϩ
is the reverse of the spontaneous
reaction in Figure 21-7; it has a
negative E0
cell and is nonspontaneous.
e–
Silver atom,
Ag
Silver
ion, Ag+
e–
Copper
atom, Cu
Copper
ion, Cu2+
STANDARD ELECTRODE POTENTIALS
The potentials of the standard zinc–copper and copper–silver voltaic cells are 1.100 volts
and 0.462 volts, respectively. The magnitude of a cell’s potential measures the spontaneity
of its redox reaction. Higher (more positive) cell potentials indicate greater driving force for the
reaction as written. Under standard conditions, the oxidation of metallic Zn by Cu2ϩ ions
has a greater tendency to go toward completion than does the oxidation of metallic Cu
by Agϩ ions. It is convenient to separate the total cell potential into the individual contributions
of the two half-reactions. This lets us determine the relative tendencies of
particular oxidation or reduction half-reactions to occur. Such information gives us a quantitative
basis for specifying strengths of oxidizing and reducing agents. In the next several
sections we shall see how this is done for standard half-cells.
THE STANDARD HYDROGEN ELECTRODE
Every oxidation must be accompanied by a reduction (i.e., the electrons must have somewhere
to go). So it is impossible to determine experimentally the potential of any single
electrode. We therefore establish an arbitrary standard. The conventional reference electrode
is the standard hydrogen electrode (SHE). This electrode contains a piece of
metal electrolytically coated with a grainy black surface of inert platinum metal, immersed
in a 1 M Hϩ solution. Hydrogen, H2, is bubbled at 1 atm pressure through a glass envelope
over the platinized electrode (Figure 21-8).
By international convention, the standard hydrogen electrode is arbitrarily assigned
a potential of exactly 0.0000 . . . volt.
SHE Half-Reaction E0 (standard electrode potential)
H2 88n 2Hϩ ϩ 2eϪ exactly 0.0000 . . . V (SHE as anode)
2Hϩ ϩ 2eϪ 88n H2 exactly 0.0000 . . . V (SHE as cathode)
We then construct a standard cell consisting of a standard hydrogen electrode and some
other standard electrode (half-cell). Because the defined electrode potential of the SHE
contributes exactly 0 volt to the sum, the voltage of the overall cell then lets us determine
the standard electrode potential of the other half-cell. This is its potential with respect
to the standard hydrogen electrode, measured at 25°C when the concentration of each
ion in the solution is 1 M and the pressure of any gas involved is 1 atm.
By agreement, we always present the standard cell potential for each half-cell as a
reduction process.
THE ZINC–SHE CELL
This cell consists of an SHE in one beaker and a strip of zinc immersed in 1 M zinc chloride
solution in another beaker (Figure 21-9). A wire and a salt bridge complete the circuit.
When the circuit is closed, the following observations can be made.
21-12
21-11
864 CHAPTER 21: Electrochemistry
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 21.5, Electrochemical Cells and
Potentials.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 21.6, Standard Potentials.
H2 gas
1 atm
1 M HCl
Pt black
Figure 21-8 The standard
hydrogen electrode (SHE). A
molecular-level view of the operation
of the SHE as a cathode is shown in
Figure 21-9 and as an anode in
Figure 21-10.
The superscript in E0 indicates
thermodynamic standard-state
conditions.
21-12 The Zinc–SHE Cell 865
1. The initial potential of the cell is 0.763 volt.
2. As the cell operates, the mass of the zinc electrode decreases. The concentration of
Zn2ϩ ions increases in the solution around the zinc electrode.
3. The Hϩ concentration decreases in the SHE. Gaseous H2 is produced.
We can conclude from these observations that the following half-reactions and cell reaction
occur.
E0
(oxidation, anode) Zn 88n Zn2ϩ ϩ 2eϪ 0.763 V
(reduction, cathode) 2Hϩ ϩ 2eϪ 88n H2 0.000 V (by definition)
(cell reaction) Zn ϩ 2Hϩ 88n Zn2ϩ ϩ H2 E0
cell ϭ 0.763 V (measured)
The standard potential at the anode plus the standard potential at the cathode gives the
standard cell potential. The potential of the SHE is 0.000 volt, and the standard cell potential
is found to be 0.763 volt. So the standard potential of the zinc anode must be 0.763
volt. The Zn͉Zn2ϩ(1.0 M)͉͉Hϩ(1.0 M), H2(1 atm)͉Pt cell is depicted in Figure 21-9.
Note that in this cell the SHE is the cathode, and metallic zinc reduces Hϩ to H2. The
zinc electrode is the anode in this cell.
e–e–
H2(g)
Zn
Cl– K+
Salt bridge
1 M HCl1 M ZnCl2
Zn2+
H+
Pt
black
2H+
+ 2e–
→ H2
Reduction, cathode
Zn → Zn2+
+ 2e–
Oxidation, anode
– +
Voltmeter
+–
e–
H2 gas
bubble
H+
(aq)
solution
SHE as
cathode
Chloride
ion, Cl–
Zinc atom,
Zn
Zinc ion,
Zn2+
e–
Figure 21-9 The Zn͉Zn2ϩ(1 M)͉͉Hϩ(1 M); H2(1 atm)͉Pt cell, in which the following net
reaction occurs.
Zn(s) ϩ 2Hϩ(aq) 88n Zn2ϩ(aq) ϩ H2(g)
In this cell the standard hydrogen electrode functions as the cathode.
The idea of “electron pressure” helps to explain this process. The negative reduction
potential for the half-reaction
Zn2ϩ ϩ 2eϪ 88n Zn E0 ϭ Ϫ0.763 V
says that this reaction is less favorable than the corresponding reduction to H2,
2Hϩ ϩ 2eϪ 88n H2 E0 ϭ 0.000 V
Before they are connected, each half-cell builds up a supply of electrons waiting to be
released, thus generating an electron pressure. Let us compare these electron pressures
by reversing the two half-reactions to show production of electrons (and changing the
signs of their E0 values).
Zn 88n Zn2ϩ ϩ 2eϪ E0
oxidation ϭ ϩ0.763 V
H2 88n 2Hϩ ϩ 2eϪ E0
oxidation ϭ ϩ0.000 V
The process with the more positive E0 value is favored, so we reason that the electron
pressure generated at the Zn electrode is greater than that at the H2 electrode. As a result,
when the cell is connected, the electrons released by the oxidation of Zn flow through
the wire from the Zn electrode to the H2 electrode, where they are consumed by the reduction
of Hϩ ions. Oxidation occurs at the zinc electrode (anode), and reduction occurs at
the hydrogen electrode (cathode).
THE COPPER–SHE CELL
Another voltaic cell consists of an SHE in one beaker and a strip of Cu metal immersed
in 1 M copper(II) sulfate solution in another beaker. A wire and a salt bridge complete
the circuit. For this cell, we observe the following (Figure 21-10).
1. The initial cell potential is 0.337 volt.
2. Gaseous hydrogen is consumed. The Hϩ concentration increases in the solution of
the SHE.
3. The mass of the copper electrode increases. The concentration of Cu2ϩ ions
decreases in the solution around the copper electrode.
Thus, the following half-reactions and cell reaction occur.
E0
(oxidation, anode) H2 88n 2Hϩ ϩ 2eϪ 0.000 V (by definition)
(reduction, cathode) Cu2ϩ ϩ 2eϪ 88n Cu 0.337 V
(cell reaction) H2 ϩ Cu2ϩ 88n 2Hϩ ϩ Cu E0
cell ϭ 0.337 V (measured)
The SHE functions as the anode in this cell, and Cu2ϩ ions oxidize H2 to Hϩ ions. The
standard electrode potential of the copper half-cell is 0.337 volt as a cathode in the
Cu–SHE cell.
Again, we can think of E0
oxidation in the two half-cells as “electron pressures.”
Cu 88n Cu2ϩ ϩ 2eϪ E0
oxidation ϭ Ϫ0.337 V
H2 88n 2Hϩ ϩ 2eϪ E0
oxidation ϭ Ϫ0.000 V
Now the hydrogen electrode has the higher electron pressure. When the cell is connected,
electrons flow through the wire from the hydrogen electrode to the copper electrode. H2
is oxidized to 2Hϩ (anode), and Cu2ϩ is reduced to Cu (cathode).
21-13
866 CHAPTER 21: Electrochemistry
What we have informally called
“electron pressure” is the tendency to
undergo oxidation. This is formally
expressed as an oxidation potential;
however, we usually tabulate reduction
potentials (Table 21-2).
Recall that in the Zn–SHE cell the
SHE was the cathode.
STANDARD ELECTRODE POTENTIALS
We can develop a series of standard electrode potentials by measuring the potentials of
other standard electrodes versus the SHE in the way we described for the standard
Zn–SHE and standard Cu–SHE voltaic cells. Many electrodes involve metals or
nonmetals in contact with their ions. We saw (Section 21-12) that the standard Zn electrode
behaves as the anode versus the SHE and that the standard oxidation potential for
the Zn half-cell is 0.763 volt.
E0
oxidation
(as anode) Zn 88n Zn2ϩ ϩ 2eϪ ϩ0.763 V
reduced form 88n oxidized form ϩ neϪ (standard oxidation potential)
The reduction potential for the standard zinc electrode (to act as a cathode relative to the
SHE) is therefore the negative of this, or Ϫ0.763 volt.
E0
oxidation
(as cathode) Zn2ϩ ϩ 2eϪ 88n Zn Ϫ0.763 V
oxidized form ϩ neϪ 88n reduced form (standard reduction potential)
21-14
21-14 Standard Electrode Potentials 867
e–e–
H2(g)
(1 atm)
Salt bridge
H2 → 2H+
+ 2e–
Oxidation, anode
– +
Voltmeter
1 M HCl(aq)
Pt black
Cu2+
+ 2e–
→ Cu(s)
Reduction, cathode
1 M CuSO4(aq)
Cu2+
H+
Cu(s)
K+
K+
Cl–
Cl–
+–
e–
SHE as
anode
H2 gas
bubble
e–
Copper
atom, Cu
Copper
ion, Cu2+
Figure 21-10 The standard copper–SHE cell,
Pt͉Hϩ(1 M); H2(1 atm)͉͉Cu2ϩ(1 M)͉Cu
In this cell, the standard hydrogen electrode functions as the anode. The net reaction is
H2(g) ϩ Cu2ϩ(aq) 88n 2Hϩ(aq) ϩ Cu(s)
The activity series (Table 4-12) is
based on standard electrode potentials.
By international convention, the standard potentials of electrodes are tabulated for reduction
half-reactions. These indicate the tendencies of the electrodes to behave as cathodes
toward the SHE. Electrodes with positive E0 values for reduction half-reactions act as
cathodes versus the SHE. Those with negative E0 values for reduction half-reactions act
as anodes versus the SHE.
Electrodes with Positive E0
reduction Electrodes with Negative E0
reduction
Reduction occurs more readily than the Reduction is more difficult than the
reduction of 2Hϩ to H2. reduction of 2Hϩ to H2.
Electrode acts as a cathode versus the Electrode acts as an anode versus
SHE. the SHE.
The more positive the E0 value for a half-reaction, the greater the tendency for the
half-reaction to occur in the forward direction as written. Conversely, the more
negative the E0 value for a half-reaction, the greater the tendency for the halfreaction
to occur in the reverse direction as written.
Table 21-2 lists standard reduction potentials for a few elements.
1. The species on the left side are all cations of metals, hydrogen ions, or elemental
nonmetals. These are all oxidizing agents (oxidized forms of the elements). Their
strengths as oxidizing agents increase from top to bottom, that is, as the E0
reduction
values become more positive. Fluorine, F2, is the strongest oxidizing agent, and Liϩ
is a very weak oxidizing agent.
2. The species on the right side are free metals, hydrogen, or anions of nonmetals.
These are all reducing agents (reduced forms of the elements). Their strengths as
reducing agents increase from bottom to top, that is, as the E0
reduction values become
more negative. Metallic Li is a very strong reducing agent, and FϪ is a very weak
reducing agent.
The more positive the reduction potential, the stronger the species on the left is as
an oxidizing agent and the weaker the species on the right is as a reducing agent.
USES OF STANDARD ELECTRODE POTENTIALS
The most important application of electrode potentials is the prediction of the spontaneity
of redox reactions. Standard electrode potentials can be used to determine the spontaneity
of redox reactions in general, whether or not the reactions can take place in electrochemical
cells.
Suppose we ask the question: At standard conditions, will Cu2ϩ ions oxidize metallic
Zn to Zn2ϩ ions, or will Zn2ϩ ions oxidize metallic copper to Cu2ϩ? One of the two
possible reactions is spontaneous, and the reverse reaction is nonspontaneous. We must
determine which one is spontaneous. We already know the answer to this question from
experimental results (see Section 21-9), but let us demonstrate the procedure for predicting
the spontaneous reaction.
21-15
868 CHAPTER 21: Electrochemistry
The oxidizing agent is reduced.
The reducing agent is oxidized.
1. Choose the appropriate half-reactions from a table of standard reduction potentials.
2. Write the equation for the half-reaction with the more positive (or less negative)
E0 value for reduction first, along with its potential.
3. Then write the equation for the other half-reaction as an oxidation and write its
oxidation potential; to do this, reverse the tabulated reduction half-reaction, and
change the sign of E0. (Reversing a half-reaction or a complete reaction also changes
the sign of its potential.)
4. Balance the electron transfer. We do not multiply the potentials by the numbers used to
balance the electron transfer! The reason is that each potential represents a tendency
for a reaction process to occur relative to the SHE; this does not depend on how
many times it occurs. An electric potential is an intensive property.
5. Add the reduction and oxidation half-reactions, and add the reduction and oxidation
potentials. E0
cell will be positive for the resulting overall cell reaction. This
indicates that the reaction as written is product-favored (spontaneous). A negative E0
cell
value would indicate that the reaction is reactant-favored (nonspontaneous).
For the cell described here, the Cu2ϩ/Cu couple has the more positive reduction potential,
so we keep it as the reduction half-reaction and reverse the other half-reaction.
Following the steps outlined, we obtain the equation for the spontaneous reaction.
21-15 Uses of Standard Electrode Potentials 869
TABLE 21-2 Standard Aqueous Reduction Potentials in Aqueous Solution at 25°C
Standard Reduction
Element Reduction Half-Reaction Potential E0, volts
Li Liϩ ϩ eϪ 888n Li Ϫ3.045
K Kϩ ϩ eϪ 888n K Ϫ2.925
Ca Ca2ϩ ϩ 2eϪ 888n Ca Ϫ2.87
Na Naϩ ϩ eϪ 888n Na Ϫ2.714
Mg Mg2ϩ ϩ 2eϪ 888n Mg Ϫ2.37
Al Al3ϩ ϩ 3eϪ 888n Al Ϫ1.66
Zn Zn2ϩ ϩ 2eϪ 888n Zn Ϫ0.763
Cr Cr3ϩ ϩ 3eϪ 888n Cr Ϫ0.74
Fe Fe2ϩ ϩ 2eϪ 888n Fe Ϫ0.44
Cd Cd2ϩ ϩ 2eϪ 888n Cd Ϫ0.403
Ni Ni2ϩ ϩ 2eϪ 888n Ni Ϫ0.25
Sn Sn2ϩ ϩ 2eϪ 888n Sn Ϫ0.14
Pb Pb2 ϩ 2eϪ 888n Pb Ϫ0.126
H2 2Hϩ ϩ 2eϪ 888n H2 0.000 (reference electrode)
Cu Cu2ϩ ϩ 2eϪ 888n Cu ϩ0.337
I2 I2 ϩ 2eϪ 888n 2IϪ ϩ0.535
Hg Hg2ϩ ϩ 2eϪ 888n Hg ϩ0.789
Ag Agϩ ϩ eϪ 888n Ag ϩ0.799
Br2 Br2 ϩ 2eϪ 888n 2BrϪ ϩ1.08
Cl2 Cl2 ϩ 2eϪ 888n 2ClϪ ϩ1.360
Au Au3ϩ ϩ 3eϪ 888n Au ϩ1.50
F2 F2 ϩ 2eϪ 888n 2FϪ ϩ2.87
Increasingstrengthasoxidizingagent;
increasingeaseofreduction
Increasingstrengthasreducingagent;
increasingeaseofoxidation
Cu2ϩ ϩ 2eϪ 88n Cu ϩ0.337 V m88 reduction potential
Zn 88n Zn2ϩ ϩ 2eϪ ϩ0.763 V m88 oxidation potential
Cu2ϩ ϩ Zn 88n Cu ϩ Zn2ϩ E0
cell ϭ ϩ1.100 V
The positive E0
cell value tells us that the forward reaction is spontaneous at standard
conditions. So we conclude that copper(II) ions oxidize metallic zinc to Zn2ϩ ions as they
are reduced to metallic copper. (Section 21-9 shows that the potential of the standard
zinc–copper voltaic cell is 1.100 volts. This is the spontaneous reaction that occurs.)
The reverse reaction has a negative E0 and is nonspontaneous.
nonspontaneous
reaction: Cu ϩ Zn2ϩ 88n Cu2ϩ ϩ Zn E0
cell ϭ Ϫ1.100 volts
To make it occur, we would have to supply electrical energy with a potential difference
greater than 1.100 volts. That is, this nonspontaneous reaction would have to be carried
out in an electrolytic cell.
EXAMPLE 21-3 Predicting the Direction of Reactions
At standard conditions, will chromium(III) ions, Cr3ϩ, oxidize metallic copper to copper(II)
ions, Cu2ϩ, or will Cu2ϩ oxidize metallic chromium to Cr3ϩ ions? Write the cell reaction, and
calculate E0
cell for the spontaneous reaction.
Plan
We refer to the table of standard reduction potentials and choose the two appropriate half-
reactions.
Solution
The copper half-reaction has the more positive reduction potential, so we write it first. Then
we write the chromium half-reaction as an oxidation, balance the electron transfer, and add
the two half-reactions and their potentials.
E0
3(Cu2ϩ ϩ 2eϪ 88n Cu) (reduction) ϩ0.337 Vo
2(Cr 88n Cr3ϩ ϩ 3eϪ) (oxidation) ϩ0.74 V0o
o2Cr ϩ 3Cu2ϩ 88n 2Cr3ϩ ϩ 3Cu E0
cell ϭ ϩ1.08 V0o
Because E0
cell is positive, we know that the reaction is product-favored (spontaneous).
Cu2ϩ ions spontaneously oxidize metallic Cr to Cr3ϩ ions and are reduced to metallic Cu.
You should now work Exercise 54a.
870 CHAPTER 21: Electrochemistry
Problem-Solving Tip: The Sign of E0 Indicates Spontaneity
For a reaction that is spontaneous at standard conditions, E0 must be positive. A negative
value of E0
cell indicates that the reverse of the reaction written would be spontaneous at
standard conditions.
STANDARD ELECTRODE POTENTIALS FOR OTHER
HALF-REACTIONS
In some half-cells, the oxidized and reduced species are both in solution as ions in contact
with inert electrodes. For example, the standard iron(III) ion/iron(II) ion half-cell contains
1 M concentrations of the two ions. It involves the following half-reaction.
Fe3ϩ ϩ eϪ 88n Fe2ϩ E0 ϭ ϩ0.771 V
The standard dichromate (Cr2O7
2Ϫ) ion/chromium(III) ion half-cell consists of a 1 M
concentration of each of the two ions in contact with an inert electrode. The balanced
half-reaction in acidic solution (1.0 M Hϩ) is
Cr2O7
2Ϫ ϩ 14Hϩ ϩ 6eϪ 88n 2Cr3ϩ ϩ 7H2O E0 ϭ ϩ1.33 V
Standard reduction potentials for some other reactions are given in Table 21-3 and in
Appendix J. These potentials can be used like those of Table 21-2.
21-16
21-16 Standard Electrode Potentials for Other Half-Reactions 871
Platinum metal is often used as the
inert electrode material. These two
standard half-cells could be shown in
shorthand notation as
Pt/Fe3ϩ(1 M), Fe2ϩ(1 M) and
Pt/Cr2O7
2Ϫ(1 M), Cr3ϩ(1 M)
TABLE 21-3 Standard Reduction Potential for Selected Half-Cells
Standard Reduction
Reduction Half-Reaction Potential E0 (volts)
Zn(OH)4
2Ϫ ϩ 2eϪ 888n Zn ϩ 4OHϪ Ϫ1.22
Fe(OH)2 ϩ 2eϪ 888n Fe ϩ 2OHϪ Ϫ0.877
2H2O ϩ 2eϪ 888n H2 ϩ 2OHϪ Ϫ0.828
PbSO4 ϩ 2eϪ 888n Pb ϩ SO4
2Ϫ Ϫ0.356
NO3
Ϫ ϩ H2O ϩ 2eϪ 888n NO2
Ϫ ϩ 2OHϪ ϩ0.01
Sn4ϩ ϩ 2eϪ 888n Sn2ϩ ϩ0.15
AgCl ϩ eϪ 888n Ag ϩ ClϪ ϩ0.222
Hg2Cl2 ϩ 2eϪ 888n 2Hg ϩ 2ClϪ ϩ0.27
O2 ϩ 2H2O ϩ 4eϪ 888n 4OHϪ ϩ0.40
NiO2 ϩ 2H2O ϩ 2eϪ 888n Ni(OH)2 ϩ 2OHϪ ϩ0.49
H3AsO4 ϩ 2Hϩ ϩ 2eϪ 888n H3AsO3 ϩ H2O ϩ0.58
Fe3ϩ ϩ eϪ 888n Fe2ϩ ϩ0.771
ClOϪ ϩ H2O ϩ 2eϪ 888n ClϪ ϩ 2OHϪ ϩ0.89
NO3
Ϫ ϩ 4Hϩ ϩ 3eϪ 888n NO ϩ 2H2O ϩ0.96
O2 ϩ 4Hϩ ϩ 4eϪ 888n 2H2O ϩ1.229
Cr2O7
2Ϫ ϩ 14Hϩ ϩ 6eϪ 888n 2Cr3ϩ ϩ 7H2O ϩ1.33
MnO4
Ϫ ϩ 8Hϩ ϩ 5eϪ 888n Mn2ϩ ϩ 4H2O ϩ1.507
Cl2 ϩ 2eϪ 888n 2ClϪ ϩ1.360
PbO2 ϩ HSO4
2Ϫ ϩ 3Hϩ ϩ 2eϪ 888n PbSO4 ϩ 2H2O ϩ1.685
Increasingstrengthasoxidizingagent;
increasingeaseofreduction
Increasingstrengthasreducingagent;
increasingeaseofoxidation
EXAMPLE 21-4 Predicting the Direction of Reactions
In an acidic solution at standard conditions, will tin(IV) ions, Sn4ϩ, oxidize gaseous nitrogen
oxide, NO, to nitrate ions, NO3
Ϫ, or will NO3
Ϫ oxidize Sn2ϩ to Sn4ϩ ions? Write the cell
reaction and calculate E0
cell for the spontaneous reaction.
Plan
We refer to the table of standard reduction potentials (see Table 21-3) and choose the appropriate
half-reactions.
Solution
The NO3
Ϫ/NO reduction half-reaction has the more positive E0 value, so we write it first and
write the Sn4ϩ/Sn2ϩ half-reaction as an oxidation. We balance the electron transfer and add
the two half-reactions to obtain the equation for the spontaneous reaction. Then we add the
half-reaction potentials to obtain the overall cell potential.
E0
2(NO3
Ϫ ϩ 4Hϩ ϩ 3eϪ 88n NO ϩ 2H2O) (reduction) ϩ0.96 Vo
3(Sn2ϩ 88n Sn4ϩ ϩ 2eϪ) (oxidation) Ϫ0.15 Vo
o2NO3
Ϫ ϩ 8Hϩ ϩ 3Sn2ϩ 88n 2NO ϩ 4H2O ϩ 3Sn4ϩ E0
cell ϭ ϩ0.81 Vo
Because E0
cell is positive for this reaction,
nitrate ions spontaneously oxidize tin(II) ions to tin(IV) ions and are reduced to nitrogen
oxide in acidic solution.
You should now work Exercises 55 and 56.
872 CHAPTER 21: Electrochemistry
Problem-Solving Tip: A Common Error in E0
cell Calculations
Remember the italicized warning in step 4 of the procedure set out in Section 21-15.
We do not multiply the potentials by the numbers used to balance the electron transfer! This is
a very common error.
Problem-Solving Tip: Remember What We Mean by Standard
Conditions
When we say that a reaction takes place at standard conditions, we mean the following:
1. The temperature is the standard thermodynamic temperature, 25°C, unless stated
otherwise.
2. All reactants and products are at unit activity. This means that:
a. Any solution species that takes part in the reaction is at a concentration of exactly
1 M;
b. Any gas that takes part in the reaction is at a pressure of exactly 1 atm;
c. Any other substance that takes part in the reaction is pure.
(When we say “takes part in the reaction” we mean either as a reactant or a product.)
These are the same conditions that were described as standard conditions for thermodynamic
purposes (Section 15-6). When one or more of these conditions is not satisfied,
we must adjust our calculations for nonstandard conditions. We shall learn how to do
this in Section 21-19.
Now that we know how to use standard reduction potentials, let us use them to explain
the reaction that occurs in the electrolysis of aqueous NaCl. The first two electrolytic
cells we considered involved molten NaCl and aqueous NaCl (see Sections 21-3 and 21-4).
There was no doubt that in molten NaCl, metallic Na would be produced by reduction
of Naϩ, and gaseous Cl2 would be produced by oxidation of ClϪ. But we found that in
aqueous NaCl, H2O, rather than Naϩ, was reduced. This is consistent with the less negative
reduction potential of H2O, compared with Naϩ.
E0
2H2O ϩ 2eϪ 88n H2 ϩ 2OHϪ Ϫ0.828 V
Naϩ ϩ eϪ 88n Na Ϫ2.714 V
The more easily reduced species, H2O, is reduced.
Electrode potentials measure only the relative thermodynamic likelihood for various halfreactions.
In practice kinetic factors can complicate matters. For instance, sometimes the
electrode process is limited by the rate of diffusion of dissolved species to or from the
electrode surface. At some cathodes, the rate of electron transfer from the electrode to a
reactant is the rate-limiting step, and a higher voltage (called overvoltage) must be applied
to accomplish the reduction. As a result of these factors, a half-reaction that is thermodynamically
more favorable than some other process still might not occur at a significant
rate. In the electrolysis of NaCl(aq), ClϪ is oxidized to Cl2 gas (Ϫ1.360 V), instead of
H2O being oxidized to form O2 gas (Ϫ1.229 V), because of the overvoltage of O2 on Pt,
the inert electrode.
CORROSION
Ordinary corrosion is the redox process by which metals are oxidized by oxygen, O2, in
the presence of moisture. There are other kinds, but this is the most common. The
problem of corrosion and its prevention are of both theoretical and practical interest.
Corrosion is responsible for the loss of billions of dollars annually in metal products. The
mechanism of corrosion has been studied extensively. It is now known that the oxidation
of metals occurs most readily at points of strain (where the metals are most “active”).
Thus, a steel nail, which is mostly iron (Section 22-7), first corrodes at the tip and head
(Figure 21-11). A bent nail corrodes most readily at the bend.
21-17
21-17 Corrosion 873
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 21.9, Corrosion.
Figure 21-11 (a) A bent nail
corrodes at points of strain and
“active” metal atoms. (b) Two nails
were placed in an agar gel that
contained phenolphthalein and
potassium ferricyanide, K3[Fe(CN)6].
As the nails corroded they produced
Fe2ϩ ions at each end and at the
bend. Fe2ϩ ions react with
[Fe(CN)6]3Ϫ ions to form
Fe3[Fe(CN)6]2, an intensely bluecolored
compound. The rest of each
nail is the cathode, at which water is
reduced to H2 and OHϪ ions. The
OHϪ ions turn phenolphthalein
pink.
Rust
Pits
Rust
(a) (b)
A point of strain in a steel object acts as an anode where the iron is oxidized to iron(II)
ions, and pits are formed (Figure 21-12).
Fe 88n Fe2ϩ ϩ 2eϪ (oxidation, anode)
The electrons produced then flow through the nail to areas exposed to O2. These act as
cathodes where oxygen is reduced to hydroxide ions, OHϪ.
O2 ϩ 2H2O ϩ 4eϪ 88n 4OHϪ (reduction, cathode)
At the same time, the Fe2ϩ ions migrate through the moisture on the surface. The overall
reaction is obtained by balancing the electron transfer and adding the two half-reactions.
2(Fe 88n Fe2ϩ ϩ 2eϪ) (oxidation, anode)
O2 ϩ 2H2O ϩ 4eϪ 88n 4OHϪ (reduction, cathode)
2Fe ϩ O2 ϩ 2H2O 88n 2Fe2ϩ ϩ 4OHϪ (net reaction)
The Fe2ϩ ions can migrate from the anode through the solution toward the cathode
region, where they combine with OHϪ ions to form iron(II) hydroxide. Iron is further
oxidized by O2 to the ϩ3 oxidation state. The material we call rust is a complex hydrated
form of iron(III) oxides and hydroxides with variable water composition; it can be represented
as Fe2O3иxH2O. The overall reaction for the rusting of iron is
2Fe(s) ϩ ᎏ3
2
ᎏO2(aq) ϩ xH2O(ᐉ) 88n Fe2O3иxH2O(s)
CORROSION PROTECTION
There are several methods for protecting metals against corrosion. The most widely used
are
1. Plating the metal with a thin layer of a less easily oxidized metal
2. Connecting the metal directly to a “sacrificial anode,” a piece of another metal that
is more active and therefore preferentially oxidized
3. Allowing a protective film, such as a metal oxide, to form naturally on the surface
of the metal
4. Galvanizing, or coating steel with zinc, a more active metal
5. Applying a protective coating, such as paint
21-18
874 CHAPTER 21: Electrochemistry
Figure 21-12 The corrosion of
iron. Pitting appears at the anodic
region, where iron metal is oxidized
to Fe2ϩ. Rust appears at the cathodic
region.
Compare the potentials for the
reduction half-reactions to see which
metal is more easily oxidized. The
more positive the reduction potential
for a metal, the more stable the metal
is as the free element and the harder it
is to oxidize.
E0
reduction
Mg2ϩ ϩ 2eϪ 88n Mg Ϫ2.37 V
Zn2ϩ ϩ 2eϪ 88n Zn Ϫ0.763 V
Fe2ϩ ϩ 2eϪ 88n Fe Ϫ0.44 V
Sn2ϩ ϩ 2eϪ 88n Sn Ϫ0.14 V
Cu2ϩ ϩ 2eϪ 88n Cu ϩ0.337 V
H2O layer
Iron Anodic
region
Fe2+
O2
Rust OH– H2O
O2
Cathodic
region
O2 + 2H2O + 4e– → 4OH–
Reduction
Fe → Fe2+ + 2e–
Oxidation
e–
Overall process: 2Fe(s) + –O2(aq) + xH2O(ᐉ) → Fe2O3 • xH2O(s)3
2
The thin layer of tin on tin-plated steel cans is less easily oxidized than iron, and it
protects the steel underneath from corrosion. It is deposited either by dipping the can
into molten tin or by electroplating. Copper is also less active than iron (see Table 21-2).
It is sometimes deposited by electroplating to protect metals when food is not involved.
Whenever the layer of tin or copper is breached, the iron beneath it corrodes even more
rapidly than it would without the coating, because of the adverse electrochemical cell that
is set up.
21-18 Corrosion Protection 875
Rust is a serious economic problem.
Protection against corrosion. (Left) Galvanized objects are steel coated with zinc. (Right)
Steel is plated with chromium for appearance as well as protection against corrosion.
Figure 21-13a shows an iron pipe connected to a strip of magnesium, a more active
metal, to protect the iron from oxidation. The magnesium is preferentially oxidized. It is
called a “sacrificial anode.” Similar methods are used to protect bridges and the hulls of
ships from corrosion. Other active metals, such as zinc, are also used as sacrificial anodes.
Galvanizing (coating the iron with zinc), combines these two approaches. Even if the zinc
coating is broken so the iron is exposed, the iron is not oxidized as long as it is in contact
with the more reactive zinc (Figure 21-13c).
Aluminum, a very active metal, reacts rapidly with O2 from the air to form a surface
layer of aluminum oxide, Al2O3, that is so thin that it is transparent. This very tough,
hard substance is inert to oxygen, water, and most other corrosive agents in the environment.
In this way, objects made of aluminum form their own protective layers and need
not be treated further to inhibit corrosion.
876 CHAPTER 21: Electrochemistry
Magnesium bar (anode)
Iron pipe (cathode)
(a) (b)
O2
OH–
(aq)
O2 + 2H2O + 4e–
→ 4OH–
(reduction)
Fe (cathode)
Zn coating
(anode)
Zn(s) → Zn2+
+ 2e–
(oxidation)
Zn2+
(aq)
Water drop
Figure 21-13 (a) Cathodic protection of buried iron pipe. A magnesium or zinc bar
is oxidized instead of the iron. The “sacrificial” anode eventually must be replaced.
(b) Cathodic protection of a ship’s hull. The small yellow horizontal strips are blocks of
titanium (coated with platinum) that are attached to the ship’s hull. The hull is steel (mostly
iron). When the ship is in salt water, the titanium blocks become the anode, and the hull the
cathode, in a voltaic cell. Because oxidation always occurs at the anode, the ship’s hull (the
cathode) is protected from oxidation (corrosion). (c) Galvanizing is another method of
corrosion protection. Even if the zinc coating is broken, it is still oxidized in preference to
the less reactive iron as long as the two metals remain in contact.
Acid rain endangers structural
aluminum by dissolving this Al2O3
coating.
(c)
EFFECT OF CONCENTRATIONS (OR PARTIAL PRESSURES)
ON ELECTRODE POTENTIALS
THE NERNST EQUATION
Standard electrode potentials, designated E0, refer to standard-state conditions. These
standard-state conditions are one molar solutions for ions, one atmosphere pressure for
gases, and all solids and liquids in their standard states at 25°C. (Remember that we refer
to thermodynamic standard-state conditions, and not standard temperature and pressure as
in gas law calculations.) As any of the standard cells described earlier operates, and concentrations
or pressures of reactants change, the observed cell voltage drops. Similarly, cells
constructed with solution concentrations different from one molar, or gas pressures
different from one atmosphere, cause the corresponding potentials to deviate from standard
electrode potentials.
The Nernst equation is used to calculate electrode potentials and cell potentials for
concentrations and partial pressures other than standard-state values.
E ϭ E0 Ϫ ᎏ
2.30
n
3
F
RT
ᎏ log Q
where
E ϭ potential under the nonstandard conditions
E0 ϭ standard potential
R ϭ gas constant, 8.314 J/molиK
T ϭ absolute temperature in K
n ϭ number of moles of electrons transferred in the reaction or half-reaction
F ϭ faraday, 96,485 C/mol eϪ ϫ 1 J/(VиC) ϭ 96,485 J/Vиmol eϪ
Q ϭ reaction quotient
The reaction quotient, Q, was introduced in Section 17-4. It involves a ratio of concentrations
or pressures of products to those of reactants, each raised to the power indicated
by the coefficient in the balanced equation. The Q expression that is used in the Nernst
equation is the thermodynamic reaction quotient; it can include both concentrations and
pressures. Substituting these values into the Nernst equation at 25°C gives
E ϭ E0 Ϫ ᎏ
0.0
n
592
ᎏ log Q (Note: in terms of base-10 log)
In general, half-reactions for standard reduction potentials are written
x Ox ϩ neϪ 88n y Red
“Ox” refers to the oxidized species and “Red” to the reduced species; x and y are their
coefficients, respectively, in the balanced equation. The Nernst equation for any cathode
half-cell (reduction half-reaction) is
21-19
21-19 The Nernst Equation 877
In this equation the expression
following the minus sign represents
how much the nonstandard conditions
cause the electrode potential to deviate
from its standard value, E0. The
Nernst equation is normally presented
in terms of base-10 logarithms, as we
will do in this text.
At 25°C, the value of ᎏ
2.30
F
3 RT
ᎏ is
0.0592; at any other temperature, this
term must be recalculated. Can you
show that this term has the units
Vиmol?
E ϭ E0 Ϫ ᎏ
0.0
n
592
ᎏ log ᎏ
[
[
R
O
e
x
d
]
]
x
y
ᎏ (reduction half-reaction)
For the familiar half-reaction involving metallic zinc and zinc ions,
Zn2ϩ ϩ 2eϪ 88n Zn E0 ϭ Ϫ0.763 V
the corresponding Nernst equation is
E ϭ E0 Ϫ ᎏ
0.0
2
592
ᎏ log ᎏ
[Zn
1
2ϩ]
ᎏ (for reduction)
We substitute the E0 value into the equation to obtain
E ϭ Ϫ0.763 V Ϫ ᎏ
0.0
2
592
ᎏ log ᎏ
[Zn
1
2ϩ]
ᎏ
EXAMPLE 21-5 The Nernst Equation
Calculate the potential, E, for the Fe3ϩ/Fe2ϩ electrode when the concentration of Fe2ϩ is
exactly five times that of Fe3ϩ.
Plan
The Nernst equation lets us calculate potentials for concentrations other than one molar. The
tabulation of standard reduction potentials gives us the value of E0 for the reduction halfreaction.
We use the balanced half-reaction and the given concentration ratio to calculate the
value of Q. Then we substitute this into the Nernst equation with n equal to the number of
moles of electrons involved in the half-reaction.
Solution
The reduction half-reaction is
Fe3ϩ ϩ eϪ 88n Fe2ϩ E0 ϭ ϩ0.771 V
We are told that the concentration of Fe2ϩ is five times that of Fe3ϩ, or [Fe2ϩ] = 5[Fe3ϩ].
Calculating the value of Q,
Q ϭ ᎏ
[
[
R
O
e
x
d
]
]
x
y
ᎏ ϭ ᎏ
[
[
F
F
e
e
2
3
ϩ
ϩ
]
]
ᎏ ϭ ᎏ
5
[
[
F
F
e
e
3
3
ϩ
ϩ
]
]
ᎏ ϭ 5
The balanced half-reaction shows one mole of electrons, or n ϭ 1. Putting values into the
Nernst equation,
E ϭ E0 Ϫ ᎏ
0.0
n
592
ᎏ log Q ϭ ϩ0.771 Ϫ ᎏ
0.0
1
592
ᎏ log 5 ϭ (ϩ0.771 Ϫ 0.041) V
ϭ ϩ0.730 V
You should now work Exercise 78.
The Nernst equation can be applied to balanced equations for redox reactions. One
approach is to correct the reduction potential for each half-reaction to take into account
the nonstandard concentrations or pressures.
878 CHAPTER 21: Electrochemistry
Metallic Zn is a pure solid, so its
concentration does not appear in Q.
EXAMPLE 21-6 The Nernst Equation
A cell is constructed at 25°C as follows. One half-cell consists of the Fe3ϩ/Fe2ϩ couple in which
[Fe3ϩ] ϭ 1.00 M and [Fe2ϩ] ϭ 0.100 M; the other involves the MnO4
Ϫ/Mn2ϩ couple in acidic
solution in which [MnO4
Ϫ] ϭ 1.00 ϫ 10Ϫ2 M, [Mn2ϩ] ϭ 1.00 ϫ 10Ϫ4 M, and [Hϩ] ϭ 1.00 ϫ
10Ϫ3 M. (a) Find the electrode potential for each half-cell with these concentrations, and
(b) calculate the overall cell potential.
Plan
(a) We can apply the Nernst equation to find the reduction potential of each half-cell with the
stated concentrations. (b) As in Section 21-15, we write the half-reaction with the more positive
potential (after correction) along with its potential. We reverse the other half-reaction and
change the sign of its E value. We balance the electron transfer and then add the half-reactions
and their potentials to find the overall cell potential.
Solution
(a) For the MnO4
Ϫ/Mn2ϩ half-cell as a reduction,
MnO4
Ϫ ϩ 8Hϩ ϩ 5eϪ 88n Mn2ϩ ϩ H2O E0 ϭ ϩ1.507 V
E ϭ E0 Ϫ ᎏ
0.0
n
592
ᎏ log ᎏ
[Mn
[
O
M
4
n
Ϫ
2
]
ϩ
[H
]
ϩ]8
ᎏ
ϭ ϩ1.507 V Ϫ ᎏ
0.0
5
592
ᎏ log
ϭ ϩ1.507 V Ϫ ᎏ
0.0
5
592
ᎏ log (1.00 ϫ 1022) ϭ ϩ1.507 V Ϫ ᎏ
0.0
5
592
ᎏ (22.0)
ϭ ϩ1.246 V
(b) For the Fe3ϩ/Fe2ϩ half-cell as a reduction,
Fe3ϩ ϩ eϪ 88n Fe2ϩ E0 ϭ ϩ0.771 V
E ϭ E0 Ϫ ᎏ
0.0
n
592
ᎏ log ᎏ
[
[
F
F
e
e
2
3
ϩ
ϩ
]
]
ᎏ ϭ ϩ0.771 V Ϫ ᎏ
0.0
1
592
ᎏ log ᎏ
0
1
.1
.0
0
0
0
ᎏ
ϭ ϩ0.771 V Ϫ ᎏ
0.0
1
592
ᎏ log (0.100) ϭ ϩ0.771 V Ϫ ᎏ
0.0
1
592
ᎏ (Ϫ1.00)
ϭ ϩ0.830 V
The corrected potential for the MnO4
Ϫ/Mn2ϩ half-cell is greater than that for the Fe3ϩ/Fe2ϩ
half-cell, so we reverse the latter, balance the electron transfer, and add.
E (corrected)
MnO4
Ϫ ϩ 8Hϩ ϩ 5eϪ 88n Mn2ϩ ϩ 4H2O ϩ1.246 Vrr
5(Fe2ϩ 88n Fe3ϩ ϩ eϪ) Ϫ0.830 Vrr
MnO4
Ϫ ϩ 8Hϩ ϩ 5Fe2ϩ 88n Mn2ϩ 4H2O ϩ 5Fe3ϩ Ecell ϭ 0.416 V
The reaction in Example 21-6 is product-favored (spontaneous) under the stated conditions,
with a potential of ϩ0.416 volt when the cell starts operation. As the cell discharges
and current flows, the product concentrations, [Mn2ϩ] and [Fe3ϩ], increase. At the same
time, reactant concentrations, [MnO4
Ϫ], [Hϩ], and [Fe2ϩ], decrease. This increases
1.00 ϫ 10Ϫ4
ᎏᎏᎏᎏ
(1.00 ϫ 10Ϫ2)(1.00 ϫ 10Ϫ3)8
21-19 The Nernst Equation 879
log Qcell, so the correction factor becomes more negative. Thus, the overall Ecell decreases
(the reaction becomes less favorable). Eventually the cell potential approaches zero (equilibrium),
and the cell “runs down.” The cell is completely run down (Ecell ϭ 0) when the
term ᎏ
0.0
n
592
ᎏ log Qcell is equal in magnitude to E0
cell.
We can also find the cell potential for a nonstandard cell by first finding E0
cell or the
overall standard cell reaction, and then using the Nernst equation to correct for nonstandard
concentrations. The next example illustrates this approach.
EXAMPLE 21-7 The Nernst Equation
A cell is constructed at 25°C as follows. One half-cell consists of a chlorine/chloride, Cl2/ClϪ,
electrode with the partial pressure of Cl2 ϭ 0.100 atm and [ClϪ] ϭ 0.100 M. The other halfcell
involves the MnO4
Ϫ/Mn2ϩ couple in acidic solution with [MnO4
Ϫ] ϭ 0.100 M, [Mn2ϩ] ϭ
0.100 M, and [Hϩ] ϭ 0.100 M. Apply the Nernst equation to the overall cell reaction to determine
the cell potential for this cell.
Plan
First we determine the overall cell reaction and its standard cell potential, E0
cell, as in Examples
21-3 and 21-4. Then we apply the Nernst equation to the overall cell.
Solution
The MnO4
Ϫ/Mn2ϩ half-reaction has the more positive reduction potential, so we write it first.
Then we write the Cl2/ClϪ half-reaction as an oxidation, balance the electron transfer, and add
the two half-reactions and their potentials to obtain the overall cell reaction and its E0
cell.
E0
2(MnO4
Ϫ ϩ 8Hϩ ϩ 5eϪ 88n Mn2ϩ ϩ 4H2O) ϩ1.507 V
5(2ClϪ 88n Cl2 ϩ 2eϪ) Ϫ1.360 V
2MnO4
Ϫ ϩ 16Hϩ ϩ 10ClϪ 88n 2Mn2ϩ ϩ 8H2O ϩ 5Cl2 E0
cell ϭ ϩ0.147 V
In the overall reaction, n ϭ 10. We then apply the Nernst equation to this overall reaction by
substituting appropriate concentration and partial pressure values. Because Cl2 is a gaseous
component, its term in the Nernst equation involves its partial pressure, PCl2
.
Ecell ϭ E0
cell Ϫ ᎏ
0.0
n
592
ᎏ log
ϭ 0.147 V Ϫ ᎏ
0.0
1
5
0
92
ᎏ log
ϭ 0.147 V Ϫ ᎏ
0.0
1
5
0
92
ᎏ log(1.00 ϫ 1021)
ϭ 0.147 V Ϫ ᎏ
0.0
1
5
0
92
ᎏ (21.00) ϭ 0.017 V
You should now work Exercises 80 and 84.
The method illustrated in Example 21-7, applying the Nernst equation to the overall
cell reaction, usually involves less calculation than correcting the separate half-reactions
as in Example 21-6. We interpret our results as follows: The positive cell potentials in
Examples 21-6 and 21-7 tell us that each of these cell reactions is spontaneous in the direc-
(0.100)2(0.100)5
ᎏᎏᎏ
(0.100)2(0.100)16(0.100)10
[Mn2ϩ]2(PCl2
)5
ᎏᎏᎏ
[MnO4
Ϫ]2[Hϩ]16[ClϪ]10
880 CHAPTER 21: Electrochemistry
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 21.7, Electrochemical Cells at
Nonstandard Conditions.
Remember that, in evaluating Q
in the Nernst equation, (1) molar
concentrations are used for dissolved
species, and (2) partial pressures of
gases are expressed in atmospheres.
Now solve Example 21-6 by applying
the Nernst equation to the overall cell
reaction to determine the cell
potential.
tion written, for the concentrations given. If the resulting cell potential were negative, the
reverse reaction would be favored at those concentrations. We could then reverse the equation
for the overall cell reaction and change the sign of its potential to describe the
spontaneous operation of the cell.
21-20 Using Electrochemical Cells to Determine Concentrations 881
USING ELECTROCHEMICAL CELLS TO DETERMINE
CONCENTRATIONS
We can apply the ideas of the preceding section to measure the voltage of a cell and then
use the Nernst equation to solve for an unknown concentration. The following example
illustrates such an application.
EXAMPLE 21-8 The Nernst Equation
We construct an electrochemical cell at 25°C as follows. One half-cell is a standard Zn2ϩ/Zn
cell, that is, a strip of zinc immersed in a 1.00 M Zn2ϩ solution; the other is a nonstandard
hydrogen electrode in which a platinum electrode is immersed in a solution of unknown
hydrogen ion concentration with gaseous hydrogen bubbling through it at a pressure of 1.000 atm.
The observed cell voltage is 0.522 V. (a) Calculate the value of the reaction quotient Q.
(b) Calculate [Hϩ] in the second half-cell. (c) Determine the pH of the solution in the second
half-cell.
Plan
We saw in Section 21-12 that the zinc–hydrogen cell operated with oxidation at the zinc electrode
and reduction at the hydrogen electrode, with a standard cell potential of 0.763 V.
overall: Zn ϩ 2Hϩ 88n Zn2ϩ ϩ H2 E0
cell ϭ 0.763 V
(a) We rearrange the Nernst equation to solve for the reaction quotient, Q, from the measured
cell voltage and n ϭ 2. (b) We substitute concentrations and partial pressures in the expression
for Q. Then we can solve for the only unknown, [Hϩ]. (c) The pH can be determined from
the [Hϩ] determined in part (b).
21-20
Problem-Solving Tip: Be Careful of the Value of n
How do you know what value of n to use? Remember that n must be the number of
moles of electrons transferred in the balanced equation for the process to which you apply
the Nernst equation.
1. For a half-reaction, n represents the number of moles of electrons in that half-reaction.
In Example 21-6 we applied the Nernst equation to each half-reaction separately, so
we used n ϭ 5 for the half-reaction
MnO4
Ϫ ϩ 8Hϩ ϩ 5eϪ 88n Mn2ϩ ϩ H2O
and we used n ϭ 1 for the half-reaction
Fe3ϩ ϩ eϪ 88n Fe2ϩ
2. For an overall reaction, n represents the total number of moles of electrons transferred.
In Example 21-7 we applied the Nernst equation to an overall reaction in which 10
moles of electrons was transferred from 10 moles of ClϪ to 2 moles of MnO4
Ϫ, so
we used the value n ϭ 10.
A pH meter uses the voltage of a
cell to measure the Hϩ
concentration in a solution. Each
change of one pH unit causes a
voltage change of 0.0592 volts.
This cell is similar to the zinc–
hydrogen cell that we discussed in
Section 21-12, except that the
hydrogen concentration is not
(necessarily) 1.00 M.
Solution
(a) Ecell ϭ E0
cell Ϫ ᎏ
0.0
n
592
ᎏ log Q
Substituting and solving for Q,
0.522 V ϭ 0.763 V Ϫ ᎏ
0.0
2
592
ᎏ log Q
ᎏ
0.05
2
92 V
ᎏ log Q ϭ (0.763 Ϫ 0.522) V ϭ 0.241 V
log Q ϭ ᎏ
(2
0
)(
.0
0
5
.2
9
4
2
1
V
V)
ᎏ ϭ 8.14
Q ϭ 108.14 ϭ 1.4 ϫ 108
(b) We write the expression for Q from the balanced overall equation, and solve for [Hϩ].
Q ϭ ᎏ
[Z
[
n
H
2ϩ
ϩ
]
]
P
2
H2
ᎏ
[Hϩ]2 ϭ ᎏ
[Zn2
Q
ϩ]PH2
ᎏ ϭ ᎏ
(1
1
.
.
0
4
0
ϫ
)(1
1
.0
0
0
8
)
ᎏ ϭ 7.1 ϫ 10Ϫ9
[Hϩ] ϭ 8.4 ϫ 10Ϫ5 M
(c) pH ϭ Ϫlog [Hϩ] ϭ Ϫlog (8.4 ϫ 10Ϫ5) ϭ 4.08
You should now work Exercises 81 and 82.
882 CHAPTER 21: Electrochemistry
E n r i c h m e n t
Concentration Cells
As we have seen, different concentrations of ions in a half-cell result in different half-cell
potentials. We can use this idea to construct a concentration cell, in which both half-cells
are composed of the same species, but in different ion concentrations. Suppose we set up
such a cell using the Cu2ϩ/Cu half-cell that we introduced in Section 21-9. We put copper
electrodes into two aqueous solutions, one that is 0.10 M CuSO4 and another that is
1.00 M CuSO4. To complete the cell construction, we connect the two electrodes with a
wire and join the two solutions with a salt bridge as usual (Figure 21-14). Now the relevant
reduction half-reaction in either half-cell is
Cu2ϩ ϩ 2eϪ 88n Cu E0 ϭ ϩ0.337 V
Thus the Cu2ϩ ions in the more concentrated half-cell can be considered as the reactant,
and those in the more dilute cell as the product.
Cu2ϩ(1.00 M) 88n Cu2ϩ(0.10 M)
The overall cell potential can be calculated by applying the Nernst equation to the overall
cell reaction. We must first find E0, the standard cell potential at standard concentrations;
because the same electrode and the same type of ions are involved in both half-cells, this
E0 is always zero. Thus,
Microelectrodes have been
developed to measure concentrations
in very small volumes of solution.
Ecell ϭ E0 Ϫ ᎏ
0.0
n
592
ᎏ log
ϭ 0 Ϫ ᎏ
0.0
2
592
ᎏ log ᎏ
0
1
.
.
1
0
0
0
ᎏ ϭ ϩ0.030 V
As the reaction proceeds, [Cu2ϩ] decreases in the more concentrated half-cell and
increases in the more dilute half-cell until the two concentrations are equal; at that point
Ecell ϭ 0, and equilibrium has been reached. This equilibrium [Cu2ϩ] is the same concentration
that would have been formed if we had just mixed the two solutions directly to obtain
a solution of intermediate concentration.
In any concentration cell, the spontaneous reaction is always from the more concentrated
solution to the more dilute solution.
[dilute solution]
ᎏᎏᎏ
[concentrated solution]
21-20 Using Electrochemical Cells to Determine Concentrations 883
Electrochemical procedures that use the principles illustrated here provide a convenient
method for making many concentration measurements.
Figure 21-14 The concentration cell Cu͉Cu2ϩ(0.10 M)͉͉Cu2ϩ(1.00 M)͉Cu. The overall
reaction lowers the [Cu2ϩ] concentration in the more concentrated solution and increases
it in the more dilute solution.
e–e–
Salt bridge
Cu(s) → Cu2+
(0.10 M) + 2e–
Oxidation, anode
– +
Voltmeter
0.10 M CuSO4
Cu2+
(1.00 M) + 2e–
→ Cu(s)
Reduction, cathode
1.00 M CuSO4
Cu2+
Cl–
Cu(s)
K+
K+
Cl–
Cu(s)
Cu2+
+–
e–
Copper
atom, Cu
Copper
ion, Cu2+
e–
Copper
atom, Cu
Copper
ion, Cu2+
The overall cell potential is positive;
the reaction is spontaneous as written.
THE RELATIONSHIP OF E0
cell TO ⌬G0 AND K
In Section 17-12 we studied the relationship between the standard Gibbs free energy
change, ⌬G0, and the thermodynamic equilibrium constant, K.
⌬G0 ϭ ϪRT ln K
There is also a simple relationship between ⌬G0 and the standard cell potential, E0
cell, for
a redox reaction (reactants and products in standard states).
⌬G0 ϭ ϪnFE0
cell
⌬G0 can be thought of as the negative of the maximum electrical work that can be obtained
from a redox reaction. In this equation, n is the number of moles of electrons transferred
in the overall process (mol eϪ/mol rxn), and F is the faraday, 96,485 J/Vиmol eϪ.
Combining these relationships for ⌬G0 gives the relationship between E0
cell values and
equilibrium constants.
ϪnFE0
cell ϭ ϪRT ln K
⌬G0 ⌬G0
After multiplying by Ϫ1, we can rearrange.
nFE0
cell ϭ RT ln K or E0
cell ϭ ᎏ
RT
n
l
F
n K
ᎏ or ln K ϭ ᎏ
nF
R
E
T
0
cell
ᎏ
If any one of the three quantities ⌬G0, K, and E0
cell is known, the other two can be calculated
using these equations. It is usually much easier to determine K for a redox reaction
from electrochemical measurements than by measuring equilibrium concentrations
directly, as described in Chapter 17. Keep in mind the following for all redox reactions
at standard conditions.
Forward Reaction ⌬G0 K E0
cell
product-favored (spontaneous) Ϫ Ͼ1 ϩ
at equilibrium 0 Ͼ1 0 (all substances at standard conditions)
reactant-favored (nonspontaneous) ϩ Ͻ1 Ϫ
EXAMPLE 21-9 Calculation of ⌬G0 from Cell Potentials
Calculate the standard Gibbs free energy change, ⌬G0, in J/mol at 25°C for the following reaction
from standard electrode potentials.
3Sn4ϩ ϩ 2Cr 88n 3Sn2ϩ ϩ 2Cr3ϩ
Plan
We evaluate the standard cell potential as we have done before. Then we apply the relationship
⌬G0 ϭ ϪnFE0
cell.
21-21
884 CHAPTER 21: Electrochemistry















Recall from Chapter 15 that ⌬G0 can
be expressed in joules per mole of
reaction. Here we ask for the number
of joules of free energy change that
corresponds to the reaction of 2 moles
of chromium with 3 moles of tin(IV)
to give 3 moles of tin(II) ions and 2
moles of chromium(III) ions.
Solution
The standard reduction potential for the Sn4ϩ/Sn2ϩ couple is ϩ0.15 volt; that for the Cr3ϩ/Cr
couple is Ϫ0.74 volt. The equation for the reaction shows Cr being oxidized to Cr3ϩ, so the
sign of the E0 value for the Cr3ϩ/Cr couple is reversed. The overall reaction, the sum of the
two half-reactions, has a cell potential equal to the sum of the two half-reaction potentials.
E0
3(Sn4ϩ ϩ 2eϪ 88n Sn2ϩ) ϩ0.15 V)
2(Cr 88n Cr3ϩ ϩ 3eϪ) Ϫ(Ϫ0.74 V)
3Sn4ϩ ϩ 2Cr 88n 3Sn2ϩ ϩ 2Cr3ϩ E0
cell ϭ ϩ0.89 V)
The positive value of E0
cell indicates that the forward reaction is spontaneous.
⌬G0 ϭ ϪnFE0
cell ϭ Ϫ΂ᎏ
6
m
m
ol
o
r
l
x
e
n
Ϫ
ᎏ΃΂ᎏ
9
V
.65
иm
ϫ
ol
10
eϪ
4 J
ᎏ΃(ϩ0.89 V)
ϭ Ϫ5.2 ϫ 105 J/mol rxn or Ϫ5.2 ϫ 102 kJ/mol rxn
You should now work Exercise 95.
EXAMPLE 21-10 Calculation of K from Cell Potentials
Use the standard cell potential to calculate the value of the equilibrium constant, K, at 25°C
for the following reaction.
2Cu ϩ PtCl6
2Ϫ 88n 2Cuϩ ϩ PtCl4
2Ϫ ϩ 2ClϪ
Plan
We calculate E0
cell for the reaction as written. Then we use it to calculate K.
Solution
First we find the appropriate half-reactions. Cu is oxidized to Cuϩ, so we write the Cuϩ/Cu
couple as an oxidation and reverse the sign of its tabulated E0 value. We balance the electron
transfer and then add the half-reactions. The resulting E0
cell value can be used to calculate the
equilibrium constant, K, for the reaction as written.
2(Cu 88n Cuϩ ϩ eϪ) Ϫ(ϩ0.521 V)
PtCl6
2Ϫ ϩ 2eϪ 88n PtCl4
2Ϫ ϩ 2ClϪ ϩ0.68 V0)
2Cu ϩ PtCl6
2Ϫ 88n 2Cuϩ ϩ PtCl4
2Ϫ ϩ 2ClϪ E0
cell ϭ ϩ0.16 V0)
Then we calculate K.
ln K ϭ ᎏ
nF
R
E
T
0
cell
ᎏ ϭ ϭ 12.5
K ϭ e12.5 ϭ 2.7 ϫ 105
At equilibrium, K ϭ ϭ 2.7 ϫ 105.
The forward reaction is product-favored (spontaneous), and the equilibrium lies far to the right.
You should now work Exercises 96 and 98.
[Cuϩ]2[PtCl4
2Ϫ][ClϪ]2
ᎏᎏᎏ
[PtCl6
2Ϫ]
(2)(9.65 ϫ 104 J/Vиmol)(ϩ0.16 V)
ᎏᎏᎏᎏ
(8.314 J/molиK)(298 K)
21-21 The Relationship of E0
cell to ⌬G0 and K 885
The quite negative value of ⌬G0 tells
us that the reaction is product-favored.
This tells us nothing about the speed
with which the reaction would occur.
As the problem is stated, we must keep
the equation as written. We must
therefore accept either a positive or a
negative value of E0
cell. A negative
value of E0
cell would lead to K Ͻ 1.
PRIMARY VOLTAIC CELLS
As any voltaic cell produces current (discharges), chemicals are consumed. Primary
voltaic cells cannot be “recharged.” Once the chemicals have been consumed, further
chemical action is not possible. The electrolytes or electrodes (or both) cannot be regenerated
by reversing the current flow through the cell using an external direct current
source. The most familiar examples of primary voltaic cells are the ordinary “dry” cells
that are used as energy sources in flashlights and other small appliances.
DRY CELLS
The first dry cell was patented by Georges Leclanché (1839–1882) in 1866. The container
of this dry cell, made of zinc, also serves as one of the electrodes (Figure 21-15). The
other electrode is a carbon rod in the center of the cell. The zinc container is lined with
porous paper to separate it from the other materials of the cell. The rest of the cell is
filled with a moist mixture (the cell is not really dry) of ammonium chloride (NH4Cl),
manganese(IV) oxide (MnO2), zinc chloride (ZnCl2), and a porous, inert filler. Dry cells
are sealed to keep the moisture from evaporating. As the cell operates (the electrodes must
be connected externally), the metallic Zn is oxidized to Zn2ϩ, and the liberated electrons
flow along the container to the external circuit. Thus, the zinc electrode is the anode
(negative electrode).
Zn 88n Zn2ϩ ϩ 2eϪ (oxidation, anode)
The carbon rod is the cathode, at which ammonium ions are reduced.
2NH4
ϩ ϩ 2eϪ 88n 2NH3 ϩ H2 (reduction, cathode)
21-22
886 CHAPTER 21: Electrochemistry
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 21.8, Batteries.
Anode
Ϫ ϩ
Cathode
Insulating
washer
Steel cover
Wax seal
Sand cushion
Carbon rod
(cathode)
Porous
separator
Zinc can
(anode)
Wrapper
NH4Cl
ZnCl2,
MnO2 paste
(a) (b)
Figure 21-15 (a) The Leclanché cell is a dry cell that generates a potential difference of
about 1.6 volts. (b) Some commercial alkaline dry cells.
Addition of the half-reactions gives the overall cell reaction
Zn ϩ 2NH4
ϩ 88n Zn2ϩ ϩ 2NH3 ϩ H2 Ecell ϭ 1.6 V
As H2 is formed, it is oxidized by MnO2 in the cell. This prevents collection of H2 gas
on the cathode, which would stop the reaction.
H2 ϩ 2MnO2 88n 2MnO(OH)
The ammonia produced at the cathode combines with zinc ions and forms a soluble
compound containing the complex ions, [Zn(NH3)4]2ϩ.
Zn2ϩ ϩ 4NH3 88n [Zn(NH3)4]2ϩ
This reaction prevents polarization due to the buildup of ammonia, and it prevents
the concentration of Zn2ϩ from increasing substantially, which would decrease the cell
potential.
Alkaline dry cells are similar to Leclanché dry cells except that (1) the electrolyte is
basic (alkaline) because it contains KOH, and (2) the interior surface of the Zn container
is rough; this gives a larger surface area. Alkaline cells have a longer shelf life than ordinary
dry cells, and they stand up better under heavy use. The voltage of an alkaline cell
is about 1.5 volts. During discharge, the alkaline dry cell reactions are
Zn(s) ϩ 2OHϪ(aq) 88n Zn(OH)2(s) ϩ 2eϪ (anode)
2MnO2(s) ϩ 2H2O(ᐉ) ϩ 2eϪ 88n 2MnO(OH)(s) ϩ 2OHϪ(aq) (cathode)
Zn(s) ϩ 2MnO2(s) ϩ 2H2O(ᐉ) 88n Zn(OH)2(s) ϩ 2MnO(OH)(s) (overall)
21-22 Dry Cells 887
Outer steel
case
Steel outer
top
Tin-plated
inner top Sealing and
insulating
gasket
Zn (anode)
KOH saturated with ZnO in
absorbent material (electrolyte)
HgO mixed with
graphite (cathode)
Inner steel case
Gas vent
Barrier
The mercury battery of the type frequently used in watches, calculators, and hearing aids is
a primary cell. Although mercury in the water supply is known to cause health problems, no
conclusive evidence has been found that the disposal of household batteries contributes to
such problems. Nevertheless, manufacturers are working to decrease the amount of mercury
in batteries. In recent years, the amount of mercury in alkaline batteries decreased markedly;
at the same time, the life of such batteries has increased dramatically.
The buildup of reaction products at an
electrode is called polarization of the
electrode.
SECONDARY VOLTAIC CELLS
In secondary voltaic cells, or reversible cells, the original reactants can be regenerated.
This is done by passing a direct current through the cell in the direction opposite of the
discharge current flow. This process is referred to as charging, or recharging, a cell or
battery. The most common example of a secondary voltaic cell is the lead storage battery,
used in most automobiles.
THE LEAD STORAGE BATTERY
The lead storage battery is depicted in Figure 21-16. One group of lead plates contains
compressed spongy lead. These alternate with a group of lead plates that contain lead(IV)
oxide, PbO2. The electrodes are immersed in a solution of about 40% sulfuric acid. When
the cell discharges, the spongy lead is oxidized to lead ions, and the lead plates accumulate
a negative charge.
Pb 88n Pb2ϩ ϩ 2eϪ (oxidation)
The lead ions then react with hydrogen sulfate ions from the sulfuric acid to form insoluble
lead(II) sulfate. This begins to coat the lead electrode.
Pb2ϩ ϩ HSO4
Ϫ 88n PbSO4(s) ϩ Hϩ (precipitation)
Thus, the net process at the anode during discharge is
Pb ϩ HSO4
Ϫ 88n PbSO4(s) ϩ Hϩ ϩ 2eϪ (anode during discharge)
21-23
888 CHAPTER 21: Electrochemistry
During discharge: H2SO4
solution
(b)
PbO2(s) + 4H+
+ H+
(aq) + 2e– → Pb2+(aq) + 2H
(aq)
+ H+(aq)
2O
Pb2+
(aq) + HSO4
–
HSO4
–
(aq) → PbSO4(s)
PbO2 plate (cathode)
Pb(s) → Pb2+(aq) + 2e–
Pb2+(aq) + (aq) → PbSO4(s)
Pb plate (anode)
(a)
Figure 21-16 (a) A schematic representation of one cell of a lead storage battery. The
reactions shown are those taking place during the discharge of the cell. Alternate lead grids
are packed with spongy lead and lead(IV) oxide. The grids are immersed in a solution of
sulfuric acid, which serves as the electrolyte. To provide a large reacting surface, each cell
contains several connected grids of each type as in (b); for clarity, only one of each is shown
in (a). Such a cell generates a voltage of about 2 volts. Six of these cells are connected
together in series, so that their voltages add, to make a 12-volt battery.
The electrons travel through the external circuit and re-enter the cell at the PbO2 electrode,
which is the cathode during discharge. Here, in the presence of hydrogen ions, the
lead(IV) oxide is reduced to lead(II) ions, Pb2ϩ. These ions also react with HSO4
Ϫ ions
from the H2SO4 to form an insoluble PbSO4 coating on the lead(IV) oxide electrode.
PbO2 ϩ 4Hϩ ϩ 2eϪ 88n Pb2ϩ ϩ 2H2O (reduction)
Pb2ϩ ϩ HSO4
Ϫ 88n PbSO4(s) ϩ Hϩ (precipitation)
PbO2 ϩ 3Hϩ ϩ HSO4
Ϫ ϩ 2eϪ 88n PbSO4(s) ϩ 2H2O (cathode during discharge)
The net cell reaction for discharge and its standard potential are obtained by adding the
net anode and cathode half-reactions and their tabulated potentials. The tabulated E0
value for the anode half-reaction is reversed in sign because it occurs as oxidation during
discharge.
21-24 The Nickel–Cadmium (Nicad) Cell 889
E0
Pb ϩ HSO4
Ϫ 88n PbSO4(s) ϩ Hϩ ϩ 2eϪ Ϫ(Ϫ0.356 V)
PbO2 ϩ 3Hϩ ϩ HSO4
Ϫ ϩ 2eϪ 88n PbSO4(s) ϩ 2H2O ϩ1.685 V)
Pb ϩ PbO2 ϩ 2Hϩ ϩ 2HSO4
Ϫ 88n 2PbSO4(s) ϩ 2H2O E0
cell ϭ ϩ2.041 V)
2H2SO4











One cell creates a potential of about 2 volts. Automobile 12-volt batteries have six cells
connected in series. The potential declines only slightly during use, because solid reagents
are being consumed. As the cell is used, some H2SO4 is consumed, lowering its concen-
tration.
When a potential slightly greater than the potential the battery can generate is imposed
across the electrodes, the current flow can be reversed. The battery can then be recharged
by reversal of all reactions. The alternator or generator applies this potential when the
engine is in operation. The reactions that occur in a lead storage battery are summarized
as follows.
discharge
Pb ϩ PbO2 ϩ 2Hϩ ϩ 2HSO4
Ϫ 3::::::4 2PbSO4(s) ϩ 2H2O
charge
During many repeated charge–discharge cycles, some of the PbSO4 falls to the bottom
of the container and the H2SO4 concentration becomes correspondingly low. Eventually
the battery cannot be recharged fully. It can be traded in for a new one, and the lead can
be recovered and reused to make new batteries.
THE NICKEL–CADMIUM (NICAD) CELL
The nickel–cadmium (nicad) cell has gained widespread popularity because it can be
recharged. It thus has a much longer useful life than ordinary (Leclanché) dry cells. Nicad
batteries are used in electronic wristwatches, calculators, and photographic equipment.
The anode is cadmium, and the cathode is nickel(IV) oxide. The electrolytic solution
is basic. The “discharge” reactions that occur in a nicad battery are
Cd(s) ϩ 2OHϪ(aq) 88n Cd(OH)2(s) ϩ 2eϪ (anode)
NiO2(s) ϩ 2H2O(ᐉ) ϩ 2eϪ 88n Ni(OH)2(s) ϩ 2OHϪ(aq) (cathode)
Cd(s) ϩ NiO2(s) ϩ 2H2O(ᐉ) 88n Cd(OH)2(s) ϩ Ni(OH)2(s) (overall)
21-24
The decrease in the concentration of
sulfuric acid provides an easy method
for measuring the degree of discharge,
because the density of the solution
decreases accordingly. We simply
measure the density of the solution
with a hydrometer.
A generator supplies direct current (dc).
An alternator supplies alternating
current (ac), so a rectifier (an
electronic device) is used to convert
this to direct current for the battery.
This is one of the oldest and most
successful examples of recycling.
To see why a nicad battery produces a
constant voltage, write the Nernst
equation for its reaction. Look at Q.
The solid reaction product at each electrode adheres to the electrode surface. Hence, a
nicad battery can be recharged by an external source of electricity; that is, the electrode
reactions can be reversed. Because no gases are produced by the reactions in a nicad battery,
the unit can be sealed. The voltage of a nicad cell is about 1.4 volts, slightly less than that
of a Leclanché cell.
THE HYDROGEN–OXYGEN FUEL CELL
Fuel cells are voltaic cells in which the reactants are continuously supplied to the cell and
the products are continuously removed. The hydrogen–oxygen fuel cell (Figure 21-17)
already has many applications. It is used in spacecraft to supplement the energy obtained
from solar cells. Liquid H2 is carried on board as a propellant. The boiled-off H2 vapor
that ordinarily would be lost is used in a fuel cell to generate electrical power.
Hydrogen (the fuel) is supplied to the anode compartment. Oxygen is fed into the
cathode compartment. The diffusion rates of the gases into the cell are carefully regulated
for maximum efficiency. Oxygen is reduced at the cathode, which consists of porous
carbon impregnated with a finely divided Pt or Pd catalyst.
catalyst
O2 ϩ 2H2O ϩ 4eϪ 88888n 4OHϪ (cathode)
The OHϪ ions migrate through the electrolyte to the anode, an aqueous solution of a
base. The anode is also porous carbon containing a small amount of catalyst (Pt, Ag, or
CoO). Here H2 is oxidized to H2O.
H2 ϩ 2OHϪ 88n 2H2O ϩ 2eϪ (anode)
The net reaction is obtained from the two half-reactions.
O2 ϩ 2H2O ϩ 4eϪ 88n 4OHϪ (cathode)
2(H2 ϩ 2OHϪ 88n 2H2O ϩ 2eϪ) (anode)
2H2 ϩ O2 88n 2H2O (net cell reaction)
The net reaction is the same as the burning of H2 in O2 to form H2O, but combustion
does not actually occur. Most of the chemical energy from the formation of HXO
bonds is converted directly into electrical energy, rather than into heat energy as in
combustion.
When the H2/O2 fuel cell is used aboard spacecraft, it is operated at a high enough
temperature that the water evaporates at the same rate as it is produced. The vapor is
then condensed to pure water.
Current research is aimed at modifying the design of fuel cells to lower their cost.
Better catalysts would speed the reactions to allow more rapid generation of electricity
and produce more power per unit volume. The H2/O2 cell is nonpolluting; the only
substance released is H2O. Catalysts have been developed that allow sunlight to decompose
water into hydrogen and oxygen, which might be used to operate fuel cells, permitting
the utilization of solar energy.
Fuel cells have also been constructed using fuels other than hydrogen, such as methane
or methanol. Biomedical researchers envision the possibility of using tiny fuel cells to
operate pacemakers. The disadvantage of other power supplies for pacemakers, which are
primary voltaic cells, is that their reactants are eventually consumed so that they require
periodic surgical replacement. As long as the fuel and oxidizer are supplied, a fuel cell can
—in theory, at least—operate forever. Eventually, tiny pacemaker fuel cells might be operated
by the oxidation of blood sugar (the fuel) by the body’s oxygen at a metal electrode
implanted just below the skin.
21-25
890 CHAPTER 21: Electrochemistry
Rechargeable nicad batteries are
used to operate many electrical
devices.
The efficiency of energy conversion of
the fuel cell operation is 60–70% of
the theoretical maximum (based on
⌬G). This represents about twice the
efficiency that can be realized from
burning hydrogen in a heat engine
coupled to a generator.
21-25 The Hydrogen–Oxygen Fuel Cell 891
H2(g) + H2O(g)
2H2(g) ϩ O2(g) → 2H2O(ᐉ)
Voltmeter O2(g) + H2O(g)
eϪ
OHϪ
OHϪ
Anode
Porous graphite
plus catalyst
(a)
2H2(g) ϩ 4OHϪ
(aq) →
4H2O(ᐉ) ϩ 4eϪ
Oxidation
OHϪ
H2O
OHϪ
H2O
Kϩ
Kϩ
Kϩ
Kϩ
Container
eϪ
O2(g)H2(g)
Cathode
Porous graphite
plus catalyst
O2(g) ϩ 2H2O(ᐉ) ϩ 4eϪ
→
4OHϪ
(aq)
Reduction
Figure 21-17 (a) Schematic drawing of a hydrogen–oxygen fuel cell. (b) A hydrogen–
oxygen fuel cell that is used in spacecraft.
(b)
892 CHAPTER 21: Electrochemistry
Key Terms
Alkaline cell A dry cell in which the electrolyte contains KOH.
Ampere Unit of electric current; 1 ampere equals 1 coulomb per
second.
Anode The electrode at which oxidation occurs.
Cathode The electrode at which reduction occurs.
Cathode protection Protection of a metal against corrosion by
making it a cathode (attaching it to a sacrificial anode of a more
easily oxidized metal).
Cell potential Potential difference, Ecell, between reduction and
oxidation half-cells; may be at nonstandard conditions.
Concentration cell A voltaic cell in which the two half-cells are
composed of the same species but contain different ion con-
centrations.
Corrosion Oxidation of metals in the presence of air and
moisture.
Coulomb Unit of electric charge; the amount of charge that
passes a given point when 1 ampere of electric current flows
for 1 second.
Downs cell An electrolytic cell for the commercial electrolysis
of molten sodium chloride.
Dry cells Ordinary batteries (voltaic cells) for flashlights, radios,
and so on; many are Leclanché cells.
Electrochemistry The study of the chemical changes produced
by electric current and the production of electricity by chemical
reactions.
Electrode potentials Potentials, E, of half-reactions as reductions
versus the standard hydrogen electrode.
Electrodes Surfaces on which oxidation and reduction halfreactions
occur in electrochemical cells.
Electrolysis The process that occurs in electrolytic cells.
Electrolytic cell An electrochemical cell in which electrical
energy causes nonspontaneous redox reactions to occur.
Electrolytic conduction See Ionic conduction.
Electroplating Plating a metal onto a (cathodic) surface by
electrolysis.
Faraday An amount of charge equal to 96,485 coulombs; corresponds
to the charge on one mole of electrons, 6.022 ϫ 1023
electrons.
Faraday’s Law of Electrolysis The amount of substance that
undergoes oxidation or reduction at each electrode during electrolysis
is directly proportional to the amount of electricity that
passes through the cell.
Fuel cell A voltaic cell in which the reactants (usually gases)
are supplied continuously and products are removed continu-
ously.
Galvanic cell See Voltaic cell.
Half-cell The compartment in a voltaic cell in which the oxidation
or reduction half-reaction occurs.
Hydrogen–oxygen fuel cell A fuel cell in which hydrogen is
the fuel (reducing agent) and oxygen is the oxidizing agent.
Ionic conduction Conduction of electric current by ions through
a pure liquid or a solution.
Lead storage battery A secondary voltaic cell that is used in
most automobiles.
Leclanché cell A common type of dry cell.
Metallic conduction Conduction of electric current through a
metal or along a metallic surface.
Nernst equation An equation that corrects standard electrode
potentials for nonstandard conditions.
Nickel–cadmium cell (nicad battery) A dry cell in which the
anode is Cd, the cathode is NiO2, and the electrolyte is
basic.
Polarization of an electrode Buildup of a product of oxidation
or reduction at an electrode, preventing further reaction.
Primary voltaic cell A voltaic cell that cannot be recharged; no
further chemical reaction is possible once the reactants are
consumed.
Sacrificial anode A more active metal that is attached to a less
active metal to protect the less active metal cathode against
corrosion.
Salt bridge A U-shaped tube containing an electrolyte that connects
two half-cells of a voltaic cell.
Secondary voltaic cell A voltaic cell that can be recharged; the
original reactants can be regenerated by reversing the direction
of current flow.
Standard cell A cell in which all reactants and products are in
their thermodynamic standard states (1 M for solution species
and 1 atm partial pressure for gases).
Standard cell potential The potential difference, E0
cell, between
standard reduction and oxidation half-cells.
Standard electrochemical conditions 1 M concentration for
solution species, 1 atm partial pressure for gases, and pure solids
and liquids.
Standard electrode A half-cell in which the oxidized and
reduced forms of a species are present at unit activity: 1 M solutions
of dissolved species, 1 atm partial pressure of gases, and
pure solids and liquids.
Standard electrode potential By convention, the potential
(E0) of a half-reaction as a reduction relative to the standard
hydrogen electrode, when all species are present at unit
activity.
Standard hydrogen electrode (SHE) An electrode consisting
of a platinum electrode that is immersed in a 1 M Hϩ solution
and that has H2 gas bubbled over it at 1 atm pressure; defined
as the reference electrode, with a potential of exactly 0.0000 . . .
volt.
Voltage Potential difference between two electrodes; a measure
of the chemical potential for a redox reaction to occur.
Voltaic cell An electrochemical cell in which spontaneous chemical
reactions produce electricity; also called a galvanic cell.
Exercises 893
Exercises
Redox Review and General Concepts
*001. (a) Define oxidation and reduction in terms of electron
gain or loss. (b) What is the relationship between the
numbers of electrons gained and lost in a redox reaction?
(c) Do all electrochemical cells involve redox reactions?
*002. Define and illustrate (a) oxidizing agent and (b) reducing
agent.
*003. For each of the following unbalanced equations, (i) write
the half-reactions for oxidation and for reduction, and
(ii) balance the overall equation in acidic solution using
the half-reaction method.
(a) Hg2ϩ ϩ Cu 88n Hg ϩ Cu2ϩ
(b) MnO2 ϩ ClϪ 88n Mn2ϩ ϩ Cl2
(c) Sn2ϩ ϩ O2 88n Sn4ϩ ϩ H2O
*004. For each of the following unbalanced equations, (i) write
the half-reactions for oxidation and reduction, and
(ii) balance the overall equation using the half-reaction
method.
(a) FeS ϩ NO3
Ϫ 88n
NO ϩ SO4
2Ϫ ϩ Fe2ϩ (acidic solution)
(b) Cr2O7
2Ϫ ϩ Sn2ϩ 88n
Cr3ϩ ϩ Sn4ϩ (acidic solution)
(c) S2Ϫ ϩ Cl2 ϩ OHϪ 88n
SO4
2Ϫ ϩ ClϪ ϩ H2O (basic solution)
*005. (a) Compare and contrast ionic conduction and metallic
conduction. (b) What is an electrode? (c) What is an inert
electrode?
*006. Support or refute each of the following statements: (a) In
any electrochemical cell the positive electrode is the one
toward which the electrons flow through the wire. (b) The
cathode is the negative electrode in any electrochemical
cell.
*007. For each of the following unbalanced equations, (i) write
the half-reactions for oxidation and reduction, (ii) identify
the species that lose and the species that gain
electrons, and (iii) write the balanced net ionic equation
for the overall reaction.
(a) Zn(s) ϩ Cu2ϩ(aq) 88n Zn2ϩ(aq) ϩ Cu(s)
(b) NO2
Ϫ(aq) ϩ Cr2O7
2Ϫ(aq) 88n
NO3
Ϫ(aq) ϩ Cr3ϩ(aq) (acidic solution)
(c) N2O4(aq) ϩ BrϪ(aq) 88n
NO2
Ϫ(aq) ϩ Br3
Ϫ(aq) (basic solution)
*008. For each of the following unbalanced equations, (i) write
the half-reactions for oxidation and reduction, (ii) identify
the species that lose and the species that gain
electrons, and (iii) write the balanced net ionic equation
for the overall reaction.
(a) Ni(s) ϩ Cd2ϩ(aq) 88n Ni2ϩ(aq) ϩ Cd(s)
(b) MnO4
Ϫ(aq) ϩ Cu(s) 88n
Cu2ϩ(aq) ϩ Mn2ϩ(aq) (acidic solution)
(c) Cr(OH)3(s) ϩ IO3
Ϫ(aq) 88n
IϪ(aq) ϩ CrO4
2Ϫ(aq) (basic solution)
Electrolytic Cells: General Concepts
*009. (a) Solids such as potassium bromide, KBr, and sodium
nitrate, NaNO3, do not conduct electric current even
though they are ionic. Why? Can these substances be
electrolyzed as solids? (b) Support or refute the statement
that the Gibbs free energy change, ⌬G, is positive for any
electrolysis reaction.
*010. (a) Metallic magnesium cannot be obtained by electrolysis
of aqueous magnesium chloride, MgCl2.
Why? (b) There are no sodium ions in the overall cell
reaction for the electrolysis of aqueous sodium chloride.
Why?
*011. Consider the electrolysis of molten aluminum oxide,
Al2O3, dissolved in cryolite, Na3AlF6, with inert electrodes.
This is the Hall–Héroult process for commercial
production of aluminum (Section 22-6). The following
experimental observations can be made when current is
supplied:
(i) Silvery metallic aluminum is produced at one elec-
trode.
(ii) Oxygen, O2, bubbles off at the other electrode.
Diagram the cell, indicating the anode, the cathode, the
positive and negative electrodes, the half-reaction occurring
at each electrode, the overall cell reaction, and the
direction of electron flow through the wire.
*012. Do the same as in Exercise 11 for the electrolysis of
molten calcium chloride with inert electrodes. The observations
are
(i) Bubbles of pale green chlorine gas, Cl2, are produced
at one electrode.
(ii) Silvery white molten metallic calcium is produced at
the other electrode.
*013. Do the same as in Exercise 11 for the electrolysis of aqueous
potassium sulfate, K2SO4. The observations are
(i) Bubbles of gaseous hydrogen are produced at one electrode,
and the solution becomes more basic around that
electrode.
(ii) Bubbles of gaseous oxygen are produced at the other
electrode, and the solution becomes more acidic around
that electrode.
*014. Do the same as in Exercise 11 for the electrolysis of an
aqueous solution of copper(II) bromide, CuBr2. The
observations are
(i) One electrode becomes coated with copper metal, and
the color of the solution around that electrode fades.
(ii) Around the other electrode, the solution turns brown,
as bromine is formed and dissolves in water.
*015. (a) Write the equation for the half-reaction when H2O is
reduced in an electrochemical cell. (b) Write the equation
for the half-reaction when H2O is oxidized in an
electrochemical cell.
894 CHAPTER 21: Electrochemistry
Faraday’s Law
*016. What are (a) a coulomb, (b) electric current, (c) an
ampere, and (d) a faraday?
*017. Calculate the number of electrons that have a total charge
of 1 coulomb.
*018. For each of the following cations, calculate (i) the number
of faradays required to produce 1.00 mol of free metal
and (ii) the number of coulombs required to produce
1.00 g of free metal. (a) Cr3ϩ, (b) Hg2ϩ, (c) Hg2
2ϩ.
*019. For each of the following cations, calculate (i) the number
of faradays required to produce 1.00 mol of free metal
and (ii) the number of coulombs required to produce
1.00 g of free metal. (a) Mg2ϩ, (b) Al3ϩ, (c) Kϩ.
*020. In 500. min 3.00 g of copper is obtained by electrolysis
of a copper(I) nitrate solution. (a) How many amperes is
required for this experiment? (b) Using the same current
and time, what mass of copper would be obtained from a
copper(II) nitrate solution?
*021. A mass of 1.50 g of silver is plated from a silver nitrate
solution in 1.5 h. Calculate the (a) coulombs, (b) faradays,
and (c) amperes necessary for this process.
*022. Rhodium is an element that has the appearance of silver,
but does not tarnish like silver and, because it is very hard,
does not become worn or scratched like silver. What mass
of rhodium could be plated by electrolysis of a Rh(NO3)3
solution with a 0.75-A current for 10.0 min?
*023. Hydrogen may be the fuel of the future for automobiles
according to some experts. Hydrogen can be isolated from
water by electrolysis. Calculate the mass of hydrogen that
is released when 1.50 amperes is passed through salt water
for 8 h.
*024. The mass of silver deposited on a spoon during electroplating
was 0.855 mg. How much electric charge passed
through the cell?
*025. What mass of platinum could be plated onto a ring from
the electrolysis of a platinum(II) salt with a 0.385-A current
for 125.0 s?
*026. What mass of silver could be plated onto a spoon from
electrolysis of silver nitrate with a 2.50-A current for
25.0 min?
0*27. We pass enough current through a solution to plate out
one mole of nickel metal from a solution of NiSO4. In
other electrolysis cells, this same current plates out two
moles of silver from AgNO3 solution but liberates only
one-half mole of O2 gas. Explain these observations.
0*28. A current is passed through 500. mL of a solution of CaI2.
The following electrode reactions occur:
anode: 2IϪ 88n I2 ϩ 2eϪ
cathode: 2H2O ϩ 2eϪ 88n H2 ϩ 2OHϪ
After some time, analysis of the solution shows that
51.5 mmol of I2 has been formed. (a) How many
faradays of charge have passed through the solution?
(b) How many coulombs? (c) What volume of dry H2
at STP has been formed? (d) What is the pH of the
solution?
*029. The cells in an automobile battery were charged at a
steady current of 5.0 A for exactly 4 h. What masses of
Pb and PbO2 were formed in each cell? The overall reaction
is
2PbSO4(s) ϩ 2H2O(ᐉ) 88n
Pb(s) ϩ PbO2(s) ϩ 2H2SO4(aq)
0*30. The chemical equation for the electrolysis of a fairly concentrated
brine solution is
2NaCl(aq) ϩ 2H2O(ᐉ) 88n
Cl2(g) ϩ H2(g) ϩ 2NaOH(aq)
What volume of gaseous chlorine would be generated at
752 torr and 5°C if the process were 88% efficient and
if a current of 1.5 A flowed for 5.0 h?
*031. An electrolytic cell contains 40.0 mL of a 0.152 M solution
of FeCl3. A current of 0.620 A is passed through the
cell, causing deposition of Fe(s) at the cathode. What is
the concentration of Fe3ϩ(aq) in the cell after this current
has run for 20.0 min?
*032. Suppose 250. mL of a 0.433 M solution of CuCl2 is
electrolyzed. How long will a current of 0.75 A have to
run in order to reduce the concentration of Cu2ϩ to
0.167 M? What mass of Cu(s) will be deposited on the
cathode during this time?
*033. What is the charge on an ion of tin if 7.42 g of metallic
tin is plated out by the passage of 24,125 C through a
solution containing the ion?
0*34. Three electrolytic cells are connected in series; that is,
the same current passes through all three, one after
another. In the first cell, 1.20 g of Cd is oxidized to Cd2ϩ;
in the second, Agϩ is reduced to Ag; in the third, Fe2ϩ is
oxidized to Fe3ϩ. (a) Find the number of faradays passed
through the circuit. (b) What mass of Ag is deposited at
the cathode in the second cell? (c) What mass of Fe(NO3)3
could be recovered from the solution in the third cell?
Voltaic Cells: General Concepts
*035. What does voltage measure? How does it vary with time
in a primary voltaic cell? Why?
*036. (a) Why must the solutions in a voltaic cell be kept separate
and not allowed to mix? (b) What are the functions
of a salt bridge?
*037. A voltaic cell containing a standard Fe3ϩ/Fe2ϩ electrode
and a standard Ga3ϩ/Ga electrode is constructed, and the
circuit is closed. Without consulting the table of standard
reduction potentials, diagram and completely describe
the cell from the following experimental observations.
(i) The mass of the gallium electrode decreases, and the
gallium ion concentration increases around that elec-
Exercises 895
trode. (ii) The ferrous ion, Fe2ϩ, concentration increases
in the other electrode solution.
*038. Repeat Exercise 37 for a voltaic cell that contains standard
Co2ϩ/Co and Au3ϩ/Au electrodes. The observations
are: (i) Metallic gold plates out on one electrode, and the
gold ion concentration decreases around that electrode.
(ii) The mass of the cobalt electrode decreases, and
the cobalt(II) ion concentration increases around that
electrode.
*039. Appendix J lists selected reduction potentials in volts at
25°C. Why is it not necessary to list a mixture of reduction
and oxidation potentials?
0*40. In Section 4-8 we learned how to predict from the activity
series (Table 4-12) which metals replace which others
from aqueous solutions. From that table, we predict that
zinc will displace silver. The equation for this process is
Zn(s) ϩ 2Agϩ(aq) 88n Zn2ϩ(aq) ϩ 2Ag(s)
Suppose we set up a voltaic cell based on this reaction.
(a) What half-reaction would represent the reduction in
this cell? (b) What half-reaction would represent the oxidation?
(c) Which metal would be the anode? (d) Which
metal would be the cathode? (e) Diagram this cell.
*041. In a voltaic cell made with metal electrodes, is the more
active metal more likely to be the anode or the cathode?
Explain.
*042. When metallic copper is placed into aqueous silver
nitrate, a spontaneous redox reaction occurs. No electricity
is produced. Why?
*043. Assume that a voltaic cell utilizes the redox reaction
2Al(s) ϩ 3Cu2ϩ(aq) 88n
2Al3ϩ(aq) ϩ 3Cu(s) (acidic solution)
Potassium and nitrate ions may also be present. Draw this
voltaic cell, and label the anode, cathode, electron flow,
and ion flow.
*044. Assume that a voltaic cell, proposed as a method for the
purification of uranium, utilizes the redox reaction
3Mg(s) ϩ 2U3ϩ(aq) 88n
3Mg2ϩ(aq) ϩ 2U(s) (acidic solution)
Potassium and nitrate ions may also be present. Draw this
voltaic cell, and label the anode, cathode, electron flow,
and ion flow.
Standard Cell Potentials
*045. (a) What are standard electrochemical conditions?
(b) Why are we permitted to assign arbitrarily an electrode
potential of exactly 0 V to the standard hydrogen
electrode?
*046. What does the sign of the standard reduction potential
of a half-reaction indicate? What does the magnitude
indicate?
*047. (a) What are standard reduction potentials? (b) What
information is contained in tables of standard reduction
potentials (Tables 21-2 and 21-3, Appendix J)? How is the
information in such tables arranged?
*048. Standard reduction potentials are 2.9 V for F2(g)/FϪ,
0.8 V for Agϩ/Ag(s), 0.5 V for Cuϩ/Cu(s), 0.3 V for
Cu2ϩ/Cu(s), Ϫ0.4 V for Fe2ϩ/Fe(s), Ϫ2.7 V for
Naϩ/Na(s), and Ϫ2.9 V for Kϩ/K(s). (a) Arrange the oxidizing
agents in order of increasing strength. (b) Which
of these oxidizing agents will oxidize Cu under standardstate
conditions?
*049. Standard reduction potentials are 1.455 V for the
PbO2(s)/Pb(s) couple, 1.360 V for Cl2(g)/ClϪ, 3.06 V for
F2(g)/HF(aq), and 1.77 V for H2O2(aq)/H2O(ᐉ). Under
standard-state conditions, (a) which is the strongest oxidizing
agent, (b) which oxidizing agent(s) could oxidize
lead to lead(IV) oxide, and (c) which oxidizing agent(s)
could oxidize fluoride ion in an acidic solution?
*050. Arrange the following less commonly encountered metals
in an activity series from the most active to the least
active: radium [Ra2ϩ/Ra(s), E0 ϭ Ϫ2.9 V], rhodium
[Rh3ϩ/Rh(s), E0 ϭ 0.80 V], europium [Eu2ϩ/Eu(s), E0 ϭ
Ϫ3.4 V]. How do these metals compare in reducing
ability with the active metal lithium [Liϩ/Li(s), E0 ϭ
Ϫ3.0 V], with hydrogen, and with gold [Au3ϩ/Au(s),
E0 ϭ 1.5 V], which is a noble metal and one of the least
active of the metals?
*051. Arrange the following metals in an activity series from
the most active to the least active: nobelium [No3ϩ/No(s),
E0 ϭ Ϫ2.5 V], cobalt [Co2ϩ/Co(s), E0 ϭ Ϫ0.28 V], gallium
[Ga3ϩ/Ga(s), E0 ϭ Ϫ0.53 V], thallium [Tlϩ/Tl(s),
E0 ϭ Ϫ0.34 V], polonium [Po2ϩ/Po(s), E0 ϭ 0.65 V].
*052. Diagram the following cells. For each cell, write the balanced
equation for the reaction that occurs spontaneously,
and calculate the cell potential. Indicate the direction of
electron flow, the anode, the cathode, and the polarity (ϩ
or Ϫ) of each electrode. In each case, assume that the circuit
is completed by a wire and a salt bridge.
(a) A strip of magnesium is immersed in a solution that
is 1.0 M in Mg2ϩ, and a strip of silver is immersed in a
solution that is 1.0 M in Agϩ.
(b) A strip of nickel is immersed in a solution that is
1.0 M in Ni2ϩ, and a strip of silver is immersed in a solution
that is 1.0 M in Agϩ.
*053. Repeat Exercise 52 for the following cells.
(a) A strip of chromium is immersed in a solution that is
1.0 M in Cr3ϩ, and a strip of gold is immersed in a solution
that is 1.0 M in Au3ϩ.
(b) A strip of aluminum is immersed in a solution that is
1.0 M in Al3ϩ, and a strip of copper is immersed in a solution
that is 1.0 M in Cu2ϩ.
In answering Exercises 54–73, justify each answer by appropriate
calculations. Assume that each reaction occurs at standard electrochemical
conditions.
896 CHAPTER 21: Electrochemistry
*054. (a) Will Fe3ϩ oxidize Sn2ϩ to Sn4ϩ in acidic solution?
(b) Will dichromate ions oxidize fluoride ions to free fluorine
in acidic solution?
*055. (a) Will dichromate ions oxidize arsenous acid, H3AsO3,
to arsenic acid, H3AsO4, in acid solution? (b) Will
dichromate ions oxidize hydrogen peroxide, H2O2, to free
oxygen, O2, in acidic solution?
*056. (a) Will permanganate ions oxidize Cr3ϩ to Cr2O7
2Ϫ in
acidic solution? (b) Will sulfate ions oxidize arsenous acid,
H3AsO3, to arsenic acid, H3AsO4, in acid solution?
*057. Calculate the standard cell potential, E0
cell for the cell
described in Exercise 37.
*058. Calculate the standard cell potential, E0
cell, for the cell
described in Exercise 38.
*059. (a) Write the equation for the oxidation of Zn(s) by Br2(ᐉ).
(b) Calculate the potential of this reaction under standardstate
conditions. (c) Is this a spontaneous reaction?
*060. For each of the following cells, (i) write the net reaction
in the direction consistent with the way the cell is written;
(ii) write the half-reactions for the anode and cathode
processes; (iii) find the standard cell potential, E0
cell, at
25°C; and (iv) tell whether the standard cell reaction actually
occurs as given or in the reverse direction.
(a) Cr͉Cr3ϩ͉͉Sn2ϩ͉Sn
(b) Ag͉Agϩ͉͉Cd2ϩ͉Cd
*061. Repeat Exercise 60 for the following cells:
(a) Al͉Al3ϩ͉͉Ce4ϩ, Ce3ϩ͉Pt
(b) Zn͉Zn2ϩ͉͉Ni2ϩ͉Ni
*062. Which of the following reactions are spontaneous in
voltaic cells under standard conditions?
(a) H2(g) 88n Hϩ(aq) ϩ HϪ(aq)
(b) Zn(s) ϩ 4CNϪ(aq) ϩ Ag2CrO4(s) 88n
Zn(CN)4
2Ϫ(aq) ϩ 2Ag(s) ϩ CrO4
2Ϫ(aq)
(c) MnO2(s) ϩ 4Hϩ(aq) ϩ Sr(s) 88n
Mn2ϩ(aq) ϩ 2H2O(ᐉ) ϩ Sr2ϩ(aq)
(d) Cl2(g) ϩ 2H2O(ᐉ) ϩ ZnS(s) 88n
2HOCl(aq) ϩ H2S(aq) ϩ Zn(s)
*063. Consult a table of standard reduction potentials, and
determine which of the following reactions are spontaneous
under standard electrochemical conditions.
(a) Mn(s) ϩ 2Hϩ(aq) 88n H2(g) ϩ Mn2ϩ(aq)
(b) 2Al3ϩ(aq) ϩ 3H2(g) 88n 2Al(s) ϩ 6Hϩ(aq)
(c) 2Cr(OH)3(s) ϩ 6FϪ(aq) 88n
2Cr(s) ϩ 6OHϪ(aq) ϩ 3F2(g)
(d) Cl2(g) ϩ 2BrϪ(aq) 88n Br2(ᐉ) ϩ 2ClϪ(aq)
*064. Which of each pair is the stronger reducing agent? (a) Ag
or H2, (b) Sn or Pb, (c) Hg or Au, (d) ClϪ in acidic solution
or ClϪ in basic solution, (e) HCl or H2S, (f) Ag or
Au
*065. Which of each pair is the stronger oxidizing agent?
(a) Cuϩ or Agϩ, (b) Sn2ϩ or Sn4ϩ, (c) Fe2ϩ or Fe3ϩ,
(d) I2 or Br2, (e) MnO4
Ϫ in acidic solution or MnO4
Ϫ in
basic solution, (f) Hϩ or Cl2
0*66. The element ytterbium forms both 2ϩ and 3ϩ cations in
aqueous solution. E0 ϭ Ϫ2.797 V for Yb2ϩ/Yb(s), and
Ϫ2.267 V for Yb3ϩ/Yb(s). What is the standard-state
reduction potential for the Yb3ϩ/Yb2ϩ couple?
0*67. The standard reduction potential for Cuϩ to Cu(s) is
0.521 V, and for Cu2ϩ to Cu(s) it is 0.337 V. Calculate
the E0 value for the Cu2ϩ/Cuϩ couple.
*068. Consider a suggestion for purifying uranium without an
outside energy source by setting up a voltaic cell with the
reaction
3Mg(s) ϩ 2U3ϩ(aq) 88n
3Mg2ϩ(aq) ϩ 2U(s) (acidic solution)
The standard reduction potentials are Ϫ1.798 for the uranium
half-reaction and Ϫ2.714 for the magnesium
half-reaction. (a) Will this setup work spontaneously? (b)
Calculate the voltage produced by this cell as written.
*069. A reaction is proposed for a nickel–cadmium battery as
Ni(s) ϩ Cd2ϩ(aq) 88n Ni2ϩ(aq) ϩ Cd(s)
(a) Is the reaction spontaneous as written? (b) Calculate
the voltage produced by this voltaic cell as written.
*070. Selecting from half-reactions involving the following
species, write the spontaneous reaction that will produce
the voltaic cell with the highest voltage in an acidic solution:
Liϩ, Ba2ϩ, Ni2ϩ, H2O2, and F2.
*071. Propose the spontaneous reaction that will produce the
voltaic cell with the highest voltage output by choosing
from only reduction and oxidation potentials involving:
MnO4
Ϫ and MnO2, HgO and Hg, Ag2O and Ag, S and
S2Ϫ, and SO4
2Ϫ and SO3
2Ϫ.
*072. Tarnished silver is coated with a layer of Ag2S(s). The
coating can be removed by boiling the silverware in an
aluminum pan, with some baking soda or salt added to
make the solution conductive. Explain this from the point
of view of electrochemistry.
0*73. Describe the process of corrosion. How can corrosion of
an easily oxidizable metal be prevented if the metal must
be exposed to the weather?
Concentration Effects; Nernst Equation
For cell voltage calculations, assume that the temperature is 25°C
unless stated otherwise.
0*74. How is the Nernst equation of value in electrochemistry?
How would the Nernst equation be modified if we wished
to use natural logarithms, ln? What is the value of the
constant in the following equation at 25°C?
E ϭ E0 Ϫ ᎏ
con
n
stant
ᎏ ln Q
Exercises 897
*075. Identify all of the terms in the Nernst equation. What
part of the Nernst equation represents the correction factor
for nonstandard electrochemical conditions?
*076. By putting the appropriate values into the Nernst equation,
show that it predicts that the voltage of a standard
half-cell is equal to E0. Use the Zn2ϩ/Zn reduction halfcell
as an illustration.
*077. Calculate the potential associated with the following halfreaction
when the concentration of the cobalt(II) ion is
1.0 ϫ 10Ϫ3 M.
Co(s) 88n Co2ϩ ϩ 2eϪ
*078. Calculate the reduction potential for hydrogen ion in a
system having a perchloric acid concentration of 2.00 ϫ
10Ϫ4 M and a hydrogen pressure of 2.50 atm. (Recall that
HClO4 is a strong acid in aqueous solution.)
*079. The standard reduction potentials for the Hϩ/H2(g) and
O2(g), Hϩ/H2O(ᐉ) couples are 0.0000 V and 1.229 V,
respectively.
(a) Write the half-reactions and the overall reaction, and
calculate E0 for the reaction
2H2(g) ϩ O2(g) 88n 2H2O(ᐉ)
(b) Calculate E for the cell when the pressure of H2 is
3.00 atm and that of O2 is 1.30 atm.
*080. Consider the cell represented by the notation
Zn(s)͉ZnCl2(aq)͉͉Cl2(g, 1 atm); ClϪ(aq)͉C
Calculate (a) E0 and (b) E for the cell when the concentration
of the ZnCl2 is 0.25 mol/L.
*081. What is the concentration of Agϩ in a half-cell if the
reduction potential of the Agϩ/Ag couple is observed to
be 0.45 V?
*082. What must be the pressure of fluorine gas to produce a
reduction potential of 2.75 V in a solution that contains
0.38 M FϪ?
*083. Calculate the cell potential of each of the following electrochemical
cells at 25°C.
(a) Sn(s)͉Sn2ϩ(4.5 ϫ 10Ϫ3 M)͉͉Agϩ(0.110 M)͉Ag(s)
(b) Zn(s)͉Zn2ϩ(0.500 M)͉͉
Fe3ϩ (7.2 ϫ 10Ϫ6 M), Fe2ϩ(0.15 M)͉Pt
(c) Pt͉H2(1 atm)͉HCl(0.00880 M)͉Cl2(1 atm)͉Pt
*084. Calculate the cell potential of each of the following electrochemical
cells at 25°C.
(a) Pt͉H2(10.0 atm), Hϩ(1.00 ϫ 10Ϫ3 M)͉͉
Agϩ(0.00549 M)͉Ag(s)
(b) Pt͉H2(1.00 atm), Hϩ(pH ϭ 5.97)͉͉
Hϩ(pH ϭ 3.47), H2(1.00 atm)͉Pt
(c) Pt͉H2(0.0361 atm), Hϩ(0.0100 M)͉͉
Hϩ(0.0100 M), H2(5.98 ϫ 10Ϫ4 atm)͉Pt
0*85. Find the potential of the cell in which identical iron
electrodes are placed into solutions of FeSO4 of concentration
1.5 mol/L and 0.15 mol/L.
0*86. We construct a cell in which identical copper electrodes
are placed in two solutions. Solution A contains 0.75 M
Cu2ϩ. Solution B contains Cu2ϩ at some concentration
known to be lower than that in solution A. The potential
of the cell is observed to be 0.040 V. What is [Cu2ϩ]
in solution B?
0*87. We construct a standard copper–cadmium cell, close the
circuit, and allow the cell to operate. At some later time,
the cell voltage reaches zero, and the cell is “run down.”
(a) What will be the ratio of [Cd2ϩ] to [Cu2ϩ] at that
time? (b) What will be the concentrations?
0*88. Repeat Exercise 87 for a standard zinc–nickel cell.
0*89. The cell potential for the cell
Zn(s) ϩ 2Hϩ(_?_ M) 88n
Zn2ϩ (3.0 M) ϩ H2(g) (5.0 atm)
is observed to be 0.445 V. What is the pH in the Hϩ/H2
half-cell?
*090. We wish to produce a 0.375-volt concentration cell using
two hydrogen electrodes, both with hydrogen at a pressure
of one atmosphere. One of the solutions has a pH
of 1.5. Calculate the required pH of the other solution.
*091. A concentration cell prepared using two hydrogen electrodes,
both with the partial pressure of hydrogen being
one atmosphere, produces 0.450 volts. The pH of one
hydrogen electrode is 1.65; what is the pH of the other?
Relationships Among ⌬G0, E0
cell, and K
*092. How are the signs and magnitudes of E0
cell, ⌬G0, and K
related for a particular reaction? Why is the equilibrium
constant K related only to E0
cell and not to Ecell?
*093. In light of your answer to Exercise 92, how do you explain
the fact that ⌬G0 for a redox reaction does depend on the
number of electrons transferred, according to the equation
⌬G0 ϭ ϪnFE0
cell?
*094. Calculate E0
cell from the tabulated standard reduction
potentials for each of the following reactions in aqueous
solution. Then calculate ⌬G0 and K at 25°C from E0
cell.
Which reactions are spontaneous as written?
(a) MnO4
Ϫ ϩ 5Fe2ϩ 88n
Mn2 ϩ 5Fe3ϩ (acidic solution)
(b) 2Cuϩ 88n Cu2ϩ ϩ Cu(s)
(c) 3Zn(s) ϩ 2MnO4
Ϫ ϩ 4H2O 88n
2MnO2(s) ϩ 3Zn(OH)2(s) ϩ 2OHϪ
*095. Calculate ⌬G0 (overall) and ⌬G0 per mole of metal for
each of the following reactions from E0 values.
(a) Zinc dissolves in dilute hydrochloric acid to produce
a solution that contains Zn2ϩ, and hydrogen gas is
evolved.
898 CHAPTER 21: Electrochemistry
(b) Chromium dissolves in dilute hydrochloric acid to
produce a solution that contains Cr3ϩ, and hydrogen gas
is evolved.
(c) Silver dissolves in dilute nitric acid to form a solution
that contains Agϩ, and NO is liberated as a gas.
(d) Lead dissolves in dilute nitric acid to form a solution
that contains Pb2ϩ, and NO is liberated as a gas.
*096. Use tabulated reduction potentials to calculate the value
of the equilibrium constant for the reaction
Sn4ϩ ϩ 2Fe2ϩ 34 Sn2ϩ ϩ 2Fe3ϩ
*097. Use tabulated reduction potentials to calculate the equilibrium
constant for the reaction
2IϪ ϩ Cl2(g) 34 I2(s) ϩ 2ClϪ
*098. Using the following half-reactions and E0 data at 25°C:
PbSO4(s) ϩ 2eϪ 88n Pb(s) ϩ SO4
2Ϫ E0 ϭ Ϫ0.356 V
PbI2(s) ϩ 2eϪ 88n Pb(s) ϩ 2IϪ E0 ϭ Ϫ0.365 V
calculate the equilibrium constant for the reaction
PbSO4(s) ϩ 2IϪ 34 PbI2(s) ϩ SO4
2Ϫ
Practical Aspects of Electrochemistry
*099. Distinguish among (a) primary voltaic cells, (b) secondary
voltaic cells, and (c) fuel cells.
*100. Sketch and describe the operation of (a) the Leclanché
dry cell, (b) the lead storage battery, and (c) the hydrogen–
oxygen fuel cell.
*101. Why is the dry cell designed so that Zn and MnO2 do
not come into contact? What reaction might occur if they
were in contact? How would this reaction affect the usefulness
of the cell?
*102. People sometimes try to recharge dry cells, with limited
success. (a) What reaction would you expect at the zinc
electrode of a Leclanché cell in an attempt to recharge
it? (b) What difficulties would arise from the attempt?
*103. Briefly describe how a storage cell operates.
*104. How does a fuel cell differ from a dry cell or a storage
cell?
*105. Does the physical size of a commercial cell govern the
potential that it will deliver? What does the size affect?
Mixed Exercises
*106. Consider the electrochemical cell represented by
Mg(s)͉Mg2ϩ͉͉Fe3ϩ͉Fe(s). (a) Write the ion–electron equations
for the half-reactions and the overall cell equation.
(b) The standard reduction potential for Fe3ϩ/Fe(s) is
Ϫ0.036 V at 25°C. Determine the standard potential for
the reaction. (c) Determine E for the cell when the concentration
of Fe3ϩ is 10.0 mol/L and that of Mg2ϩ is 1.00 ϫ
10Ϫ3 mol/L. (d) If 150 mA is to be drawn from this cell
for a period of 20.0 min, what is the minimum mass for
the magnesium electrode?
*107. A sample of Al2O3 dissolved in a molten fluoride bath is
electrolyzed using a current of 1.20 A. (a) What is the
rate of production of Al in g/h? (b) The oxygen liberated
at the positive carbon electrode reacts with the carbon to
form CO2. What mass of CO2 is produced per hour?
*108. The “life” of a certain voltaic cell is limited by the amount
of Cu2ϩ in solution available to be reduced. If the cell
contains 30.0 mL of 0.165 M CuSO4, what is the maximum
amount of electric charge this cell could generate?
*109. A magnesium bar weighing 5.0 kg is attached to a buried
iron pipe to protect the pipe from corrosion. An average
current of 0.025 A flows between the bar and the pipe.
(a) What reaction occurs at the surface of the bar? of
the pipe? In which direction do electrons flow? (b) How
many years will be required for the Mg bar to be entirely
consumed (1 year ϭ 3.16 ϫ 107 s)? (c) What reaction(s)
will occur if the bar is not replaced after the time calculated
in part (b)?
*110. (a) Calculate the ratio of ion concentrations of Mn2ϩ and
Fe2ϩ necessary to produce a voltaic cell of 1.56 volts. The
electrodes are solid manganese and iron. (b) Draw this
voltaic cell. Indicate which electrode is the anode and
which is the cathode, as well as the direction of electron
flow.
*111. (a) Calculate the ratio of ion concentrations of Mg2ϩ and
Cu2ϩ necessary to produce a voltaic cell of 2.67 volts. The
electrodes are solid magnesium and solid copper. (b) Draw
this voltaic cell. Indicate which electrode is the anode and
which is the cathode, as well as the direction of electron
flow.
*112. The production of uranium metal from purified uranium
dioxide ore consists of the following steps:
UO2(s) ϩ 4HF(g) 88n UF4(s) ϩ 2H2O(ᐉ)
heat
UF4(s) ϩ 2Mg(s) 88n U(s) ϩ 2MgF2(s)
What is the oxidation number of U in (a) UO2, (b) UF4,
and (c) U? Identify (d) the reducing agent and (e) the substance
reduced. (f ) What current could the second
reaction produce if 0.500 g of UF4 reacted each minute?
(g) What volume of HF(g) at 25°C and 10.0 atm would
be required to produce 0.500 g of U? (h) Would 0.500 g
of Mg be enough to produce 0.500 g of U?
*113. Which of each pair is the stronger oxidizing agent?
(a) Hϩ or Cl2, (b) Ni2ϩ or Se in contact with acidic solution,
(c) Cr2O7
2Ϫ or Br2 (acidic solution).
*114. (a) Describe the process of electroplating. (b) Sketch and
label an apparatus that a jeweler might use for electro-
Exercises 899
plating silver onto jewelry. (c) A jeweler purchases highly
purified silver to use as the anode in an electroplating
operation. Is this a wise purchase? Why?
*115. The same quantity of electric charge that deposited 0.612 g
of silver was passed through a solution of a gold salt, and
0.373 g of gold was deposited. What is the oxidation state
of gold in this salt?
CONCEPTUAL EXERCISES
*116. Household electricity is an alternating current of 60 cycles
per second. Can a redox reaction be used to provide an
alternating current? Explain.
*117. Electroplating is performed by using a source of direct
current. (a) Why can’t an alternating current be used?
(b) What would happen if an alternating current were
used for electroplating?
*118. Figure 21-5 is a schematic diagram of the electrolytic cell
used to separate copper from the impurities, that is, zinc,
iron, silver, gold, and platinum. Can this process be used
to separate the metals left after the copper has been
removed? Explain.
*119. A zinc–copper cell like that shown in Figure 21-7 is constructed,
except that an inert platinum wire is used
instead of the salt bridge. Will the cell still produce a
potential?
BUILDING YOUR KNOWLEDGE
*120. An electrochemical cell was needed in which hydrogen
and oxygen would react to form water. (a) Using the following
standard reduction potentials for the couples
given, determine which combination of half-reactions
gives the maximum output potential:
E0 ϭ Ϫ0.828 V for H2O(ᐉ)/H2(g), OHϪ
E0 ϭ 0.0000 V for Hϩ/H2(g)
E0 ϭ 1.229 V for O2(g), Hϩ/H2O(ᐉ)
E0 ϭ 0.401 V for O2(g), H2O(ᐉ)/OHϪ
(b) Write the balanced equation for the overall reaction
in (a).
*121. (a) Given the following E0 values at 25°C, calculate Ksp
for cadmium sulfide, CdS.
Cd2ϩ(aq) ϩ 2eϪ 88n Cd(s) E0 ϭ Ϫ0.403 V
CdS(s) ϩ 2eϪ 88n Cd(s) ϩ S2Ϫ(aq) E0 ϭ Ϫ1.21 V0
(b) Evaluate ⌬G0 at 25°C for the process
CdS(s) 34 Cd2ϩ(aq) ϩ S2Ϫ(aq)
*122. Refer to tabulated reduction potentials. (a) Calculate Ksp
for AgBr(s). (b) Calculate ⌬G0 for the reaction
AgBr(s) 34 Agϩ(aq) ϩ BrϪ(aq)
*123. Under standard-state conditions, the following reaction
is not spontaneous:
BrϪ ϩ 2MnO4
Ϫ ϩ H2O(ᐉ) 88n
BrO3
Ϫ ϩ 2MnO2(s) ϩ 2OHϪ E0 ϭ Ϫ0.022 V
The reaction conditions are adjusted so that E ϭ 0.120 V
by making [BrϪ] ϭ [MnO4
Ϫ] ϭ 1.60 mol/L and [BrO3
Ϫ]
ϭ 0.60 mol/L. (a) What is the concentration of hydroxide
ions in this cell? (b) What is the pH of the solution
in the cell?
*124. Show by calculation that E0 ϭ Ϫ1.662 V for the reduction
of Al3ϩ to Al(s), regardless of whether the equation
for the reaction is written
(i) ᎏ1
3
ᎏAl3ϩ ϩ eϪ 88n ᎏ1
3
ᎏAl(s) ⌬G0 ϭ 160.4 kJ/mol
or
(ii) Al3ϩ ϩ 3eϪ 88n Al(s) ⌬G0 ϭ 481.2 kJ/mol
*125. We wish to fill a balloon with H2 at a pressure of 1.05 atm
and a temperature of 25°C. The volume of the balloon,
when filled, is 750 mL. How long must a current of 2.25 A
be passed through the cell in order to produce this amount
of H2 by electrolysis of water?
22 Metals I:
Metallurgy
OUTLINE
Metals
22-1 Occurrence of the Metals
Metallurgy
22-2 Pretreatment of Ores
22-3 Reduction to the Free Metals
22-4 Refining of Metals
Metallurgies of Specific Metals
22-5 Magnesium
22-6 Aluminum
22-7 Iron
22-8 Copper
22-9 Gold
OBJECTIVES
After you have studied this chapter, you should be able to
• Describe major sources of metals
• Describe some pretreatment techniques for ores
• Describe some reduction processes that produce free metals
• Describe some techniques for refining (purifying) metals
• Describe the specific metallurgies of five metals: magnesium, aluminum, iron,
copper, and gold
M
METALS
M
etals are widely used for structural purposes in buildings, cars, railroads, ships,
and aircraft. They also serve as conductors of heat and electricity. Medical and
nutritional research during recent decades has provided much insight into
important biological functions of metals. The metals Na, K, Ca, and Mg, as well as some
nonmetals (C, H, O, N, P, and S), are present in the human body in substantial quantities.
Many other metals are present in lesser quantities in the human body, but they are
essential to our well-being (see essay “Trace Elements and Life” in Chapter 23). In this
chapter, we shall study the occurrence of metals and examine processes for obtaining metals
from their ores.
1
2
3
4
5 6
7
8
910
11
12
1314
15
16
17
18
19
20
2122
23
A cluster of 21 minerals that are
present in ores, plus two samples of
native ores (free metals, Ag and Bi).
These are identified in the following
list.
1. Bornite (iridescent)—COPPER
2. Dolomite (pink)—MAGNESlUM
3. Molybdenite (gray)—MOLYBDENUM
4. Skutterudite (gray)—COBALT, NICKEL
5. Zincite (mottled red)—ZlNC
6. Chromite (gray)—CHROMIUM
7. Stibnite (top right, gray)—ANTIMONY
8. Gummite (yellow)—URANlUM
9. Cassiterite (rust, bottom right)—TIN
10. Vanadinite crystal on Goethite (red crystal)—
VANADIUM
11. Cinnabar (red)—MERCURY
12. Galena (gray)—LEAD
13. Monazite (white)—RARE EARTHS:
cerium, lanthanum, neodymium, thorium
14. Bauxite (gold)—ALUMINUM
15. Strontianite (white, spiny)—STRONTIUM
16. Cobaltite (gray cube)—COBALT
17. Pyrite (gold)—IRON
18. Columbinite (tan, gray stripe)—NIOBIUM,
TANTALUM
19. Native BISMUTH (shiny)
20. Rhodochrosite (pink)—MANGANESE
21. Rutile (shiny twin crystal)—TITANIUM
22. Native SILVER (filigree on quartz)
23. Pyrolusite (black, powdery)—MANGANESE
OCCURRENCE OF THE METALS
In our study of periodicity we learned that metallic character increases toward the left and
toward the bottom of the periodic table (Section 4-1) and that oxides of most metals are
basic (Section 6-8, part 2). The oxides of some metals (and metalloids) are amphoteric
(Section 10-6). In Section 13-17 we described metallic bonding and related the effectiveness
of metallic bonding to the characteristic properties of metals.
The properties of metals influence the kinds of ores in which they are found, and the
metallurgical processes used to extract them from their ores. Metals with negative standard
reduction potentials (active metals) are found in nature in the combined state. Those
with positive reduction potentials, the less active metals, may occur in the uncombined
free state as native ores. Examples of native ores are Cu, Ag, Au, and the less abundant
Pt, Os, Ir, Ru, Rh, and Pd. Cu, Ag, and Au are also found in the combined state.
Many “insoluble” compounds of the metals are found in the earth’s crust. Solids that
contain these compounds are the ores from which metals are extracted. Ores contain
minerals, comparatively pure compounds of the metals of interest, mixed with relatively
large amounts of gangue—sand, soil, clay, rock, and other material. Soluble compounds
are found dissolved in the sea or in salt beds in areas where large bodies of water have
evaporated. Metal ores can be classified by the anions with which the metal ions are
combined (Table 22-1 and Figure 22-1).
METALLURGY
Metallurgy is the commercial extraction of metals from their ores and the preparation of
metals for use. It usually includes several steps: (1) mining the ore, (2) pretreatment of
the ore, (3) reduction of the ore to the free metal, (4) refining or purifying the metal, and
(5) alloying, if necessary.
22-1
902 CHAPTER 22: Metals I: Metallurgy
Pronounce “gangue” as one syllable
with a soft final g.
The most widespread minerals are
silicates. But extraction of metals from
silicates is very difficult. Metals are
obtained from silicate minerals only
when there is no other more
economical alternative.
TABLE 22-1 Common Classes of Ores
Anion Examples and Names of Minerals
none (native ores) Au, Ag, Pt, Os, Ir, Ru, Rh, Pd, As, Sb, Bi, Cu
oxide hematite, Fe2O3; magnetite, Fe3O4; bauxite, Al2O3; cassiterite, SnO2;
periclase, MgO; silica, SiO2
sulfide chalcopyrite, CuFeS2; chalcocite, Cu2S; sphalerite, ZnS; galena, PbS;
iron pyrites, FeS2; cinnabar, HgS
chloride rock salt, NaCl; sylvite, KCl; carnallite, KClиMgCl2и6H2O
carbonate limestone, CaCO3; magnesite, MgCO3; dolomite, MgCO3иCaCO3;
rhodochrosite, MnCO3
sulfate gypsum, CaSO4и2H2O; epsom salts, MgSO4и7H2O; barite, BaSO4
silicate beryl, Be3Al2Si6O18; kaolinite, Al2(Si2O8)(OH)4; spodumene,
LiAl(SiO3)2
PRETREATMENT OF ORES
After being mined, many ores must be concentrated by removal of most of the gangue.
Most sulfides have relatively high densities and are denser than gangue. After pulverization,
the lighter gangue particles are removed by a variety of methods. One involves
22-2
22-2 Pretreatment of Ores 903
(Left) The pink mineral rhodochrosite is manganese carbonate, MnCO3. (Right) A pile of
manganese metal. Manganese is an important element in steels that must withstand great
shock, such as those used to construct rock crushers.
Sulfides
Oxides
Occur
uncombined
Halide salts
Phosphates
Silicates
C from coal,
B from borax
Carbonates
Key
Fe Co Ni Cu Zn Ga Ge As
H H He
Li Be B C N O F Ne
Na Mg Al Si P S Cl Ar
K Ca Sc Ti V Cr Mn Se Br Kr
Rb Sr Y Zr Nb Mo Ru Rh Pd Ag Cd In Sn Sb Te I Xe
RnBiPbTlHgAuPtIrOsReWTaHfLaBaCs
LuErHoDyTbGdEuSmPmNdPrCe Tm YbLanthanides
Figure 22-1 Major natural sources of the elements. The soluble halide salts are found in
oceans, salt lakes, brine wells, and solid deposits. Most helium is obtained from wells in the
United States and Russia. Most of the other noble gases are obtained from air.
blowing the lighter particles away via a cyclone separator (Figure 22-2a). These particles
can be sifted out through layers of vibrating wire mesh or on inclined vibration tables.
Hydrophilic (“water-loving”) particles are made wet by water; hydrophobic (“waterhating”)
particles are not made wet by water. These two kinds of particles are separated
in the flotation process. This method is particularly applicable to sulfides, carbonates,
and silicates, which are not “wetted” by water or else can be made water-repellent by treatment.
Their surfaces are easily covered by layers of oil or other flotation agents. A stream
of air is blown through a swirled suspension of such an ore in water and oil (or other
agent). Bubbles form in the oil on the mineral particles and cause them to rise to the
surface. The bubbles are prevented from breaking and escaping by a layer of oil and emulsifying
agent. A frothy ore concentrate forms at the surface. By varying the relative amounts
of oil and water, the types of oil additive, the air pressure, and so on, it is even possible
to separate one metal sulfide, carbonate, or silicate from another (Figure 22-2b).
Another pretreatment process involves chemical modification. This converts metal
compounds to more easily reduced forms. Carbonates and hydroxides may be heated to
drive off CO2 and H2O, respectively.
heat
CaCO3(s) 888n CaO(s) ϩ CO2(g)
heat
Mg(OH)2(s) 888n MgO(s) ϩ H2O(g)
Some sulfides are converted to oxides by roasting, that is, heating below their melting
points in the presence of oxygen from air. For example,
904 CHAPTER 22: Metals I: Metallurgy
Water and detergent
plus ore mixture
Light sulfide particles
in froth suspension
Froth
separation
Water and
detergent
recycle
Desired productRocky
material
Air
Air
(a) (b)
Crushed
ore
Air
Figure 22-2 (a) The cyclone separator enriches metal ores. Crushed ore is blown in at
high velocity. Centrifugal force takes the heavier particles, with higher percentages of metal,
to the wall of the separator. These particles spiral down to the collection bin at the bottom.
Lighter particles, not as rich in the metal, move into the center. They are carried out the
top in the air stream. (b) A representation of the flotation process for enrichment of copper
sulfide ore. The relatively light sulfide particles are suspended in the water–oil–detergent
mixture and collected as a froth. The denser material sinks to the bottom of the container.
Some sulfide minerals: iron pyrite
(fool’s gold), FeS2; black galena, PbS;
yellow orpiment, As2S3.
Panning for gold is possible because
of the high density of the native
metal.
heat
2ZnS(s) ϩ 3O2(g) 888n 2ZnO(s) ϩ 2SO2(g)
Roasting sulfide ores causes air pollution. Enormous quantities of SO2 escape into the
atmosphere (Section 6-8, part 4), where it causes great environmental damage (Figure
22-3). Federal regulations now require limitation of the amount of SO2 that escapes with
stack gases and fuel gases. Now most of the SO2 is trapped and used in the manufacture
of sulfuric acid (Section 24-12).
REDUCTION TO THE FREE METALS
The method used for reduction, or smelting, of metal ores to the free metals depends on
how strongly the metal ions are bonded to anions. When the bonding is strong, more
energy is required to reduce the metals. This makes reduction more expensive. The most
active metals usually have the strongest bonding.
The least reactive metals occur in the free state and thus require no reduction. Examples
include Au, Ag, and Pt. This is why gold and silver have been used as free metals
since prehistoric times. Some less active metals, such as Hg, can be obtained directly from
their sulfide ores by roasting. This reduces metal ions to the free metals by oxidation of
the sulfide ions.
heat
HgS(s) ϩ O2(g) 888n SO2(g) ϩ Hg(g)
cinnabar from air obtained as vapor;
later condensed
Roasting sulfides of the more active metals produces metal oxides, but no free metals.
heat
2NiS(s) ϩ 3O2(g) 888n 2NiO(s) ϩ 2SO2(g)
The resulting metal oxides are then reduced to free metals with coke or CO. If C must
be avoided, another reducing agent, such as H2, Fe, or Al, is used.
heat
SnO2(s) ϩ 2C(s) 888n Sn(ᐉ) ϩ 2CO(g)
heat
WO3(s) ϩ 3H2(g) 888n W(s) ϩ 3H2O(g)
22-3
22-3 Reduction to the Free Metals 905
Figure 22-3 Trees damaged by
acid rain and air pollution in the
southwestern United States.
Some of these environmental problems
are discussed in the Chemistry in Use
essay “Acid Rain.”
Rutile contains TiO2.
Coke is impure carbon.
The very active metals, such as Al and Na, are reduced electrochemically, usually from
their anhydrous molten salts. If H2O is present, it is reduced instead of the more active
metals. Tables 22-2 and 22-3 summarize reduction processes for some metal ions.
REFINING OF METALS
Metals obtained from reduction processes are almost always impure. Further refining
(purification) is usually required. This can be accomplished by distillation if the metal is
more volatile than its impurities, as in the case of mercury. Among the metals purified by
electrolysis (Sections 21-3 to 21-7) are Al (Section 22-6), Cu (Section 22-8), Ag, and Au
(Section 22-9).
Zone refining is often used when extremely pure metals are desired for such applications
as solar cells and semiconductors (Section 13-17). An induction heater surrounds a
bar of the impure solid and passes slowly from one end to the other (Figure 22-4). As it
passes, it melts portions of the bar, which slowly recrystallize as the heating element moves
away. The impurity does not fit into the crystal as easily as the element of interest, so
most of it is carried along in the molten portion until it reaches the end. Repeated passes
of the heating element produce a bar of high purity.
After refining, many metals are alloyed, or mixed with other elements, to change their
physical and chemical characteristics. In some cases, certain impurities are allowed to
remain during refining because their presence improves the properties of the metal. For
example, a small amount of carbon in iron greatly enhances its hardness. Examples of
alloys include brass (Cu and Zn), bronze (Cu and Sn), duralumin (Al, Cu, Mg, and Mn),
and stainless steel (Fe, Cu, Mn, plus others).
22-4
906 CHAPTER 22: Metals I: Metallurgy
TABLE 22-2 Reduction Processes for Some Metals
Metal Ion Typical Reduction Process
lithium, Liϩ
potassium, Kϩ
calcium, Ca2ϩ
Electrolysis of molten salt
sodium, Naϩ
magnesium, Mg2ϩ
aluminum, Al3ϩ
manganese, Mn2ϩ
zinc, Zn2ϩ
Reaction of oxide with coke (carbon)
chromium, Cr2ϩ, Cr3ϩ or carbon monoxide (CO)
iron, Fe2ϩ, Fe3ϩ
lead, Pb2ϩ
copper, Cu2ϩ
silver, Agϩ
Element occurring free, or easily obtained
by roasting the sulfide or oxide oremercury, Hg2ϩ
platinum, Pt2ϩ
gold, Auϩ
Increasingactivityofmetals
888888n88n88888nm888888m88m888888
Ultrapure silicon is produced by
zone refining.
The end containing the impurities can
be sliced off and recycled.
22-4 Refining of Metals 907
TABLE 22-3 Some Specific Reduction Processes
Metal Compound (ore) Reduction Process Comments
mercury HgS (cinnabar) Roast reduction; heating of ore in air
heat
HgS ϩ O2 888n Hg ϩ SO2
copper sulfides such as Cu2S Blowing of oxygen through purified Preliminary ore concentration and
(chalcocite) molten Cu2S purification steps required to
heat remove FeS impurities
Cu2S ϩ O2 888n 2Cu ϩ SO2
zinc ZnS (sphalerite) Conversion to oxide and reduction with Process also used for the production
carbon of lead from galena, PbS
heat
2ZnS ϩ 3O2 888n 2ZnO ϩ 2SO2
heat
ZnO ϩ C 888n Zn ϩ CO
iron Fe2O3 (hematite) Reduction with carbon monoxide
produced from coke
heat
2C (coke) ϩ O2 888n 2CO
heat
Fe2O3 ϩ 3CO 888n 2Fe ϩ 3CO2
titanium TiO2 (rutile) Conversion of oxide to halide salt and Also used for the reduction of UF4
reduction with an active metal obtained from UO2, pitchblende
heat
TiO2 ϩ 2Cl2 ϩ 2C 888n TiCl4 ϩ 2CO
heat
TiCl4 ϩ 2Mg 888n Ti ϩ 2MgCl2
tungsten FeWO4 (wolframite) Reduction with hydrogen Used also for molybdenum
heat
WO3 ϩ 3H2 888n W ϩ 3H2O
aluminum Al2O3иxH2O Electrolytic reduction (electrolysis) in
(bauxite) molten cryolite, Na3[AlF6], at
1000°C
heat
2Al2O3 888n 4Al ϩ 3O2
sodium NaCl (sea water) Electrolysis of molten chlorides Also for calcium, magnesium, and
heat other active metals in Groups IA
2NaCl 888n 2Na ϩ Cl2 and IIA
Impure
solid
Impurities are
concentrated
here
Molten
zone
Heater
Purified
solid
Figure 22-4 A representation of a zone refining apparatus.
METALLURGIES OF SPECIFIC METALS
The metallurgies of Mg, Al, Fe, Cu, and Au will be discussed as specific examples. The
order of increasing standard reduction potentials of these metals indicates the order of
increasing ease of reduction to the free metals.
Standard Reduction
Reduction Half-Reaction Potential E0, Volts
Mg2ϩ ϩ 2eϪ 88n Mg Ϫ2.37
Al3ϩ ϩ 3eϪ 88n Al Ϫ1.66
Fe2ϩ ϩ 2eϪ 88n Fe Ϫ0.44
Cu2ϩ ϩ 2eϪ 88n Cu ϩ0.337
Au3ϩ ϩ 3eϪ 88n Au ϩ1.50
MAGNESIUM
Magnesium occurs widely in carbonate ores, but most Mg comes from salt brines and
from the sea (Figure 22-5). Sea water is 0.13% Mg by mass. Because of its low density
(1.74 g/cm3), Mg is used in lightweight structural alloys for such items as automobile and
aircraft parts.
Magnesium ions are precipitated as Mg(OH)2 by addition of Ca(OH)2 (slaked lime) to
sea water. The slaked lime is obtained by crushing oyster shells (CaCO3), heating them
to produce lime (CaO), and then adding a limited amount of water (slaking).
22-5
908 CHAPTER 22: Metals I: Metallurgy
Mg and Al are active metals, Fe and
Cu are moderately active, and Au is
relatively inactive.
CaCO3→CaO + CO2 CaO + H2O→Ca(OH)2
Lime kilns Slaker
MgCl2 + Ca(OH)2→
Mg(OH)2 + CaCl2
Settling
tank
Mg(OH)2 + 2HCl→
MgCl2 + 2H2O
Filter
Mg(OH)2
HCl
Chlorine
gas
Dilute
MgCl2
soln.
Evaporators
MgCl2 soln.
Electrolytic cells
MgCl2→Mg + Cl2
Mg
Dryers Conc.
Hydrochloric
acid plant
Oyster
shells
CaCO3
Ocean water
INTAKE
FLUME
Strainers
Figure 22-5 Schematic diagram of an industrial plant for the production of magnesium
from the Mg2ϩ ions in sea water.
heat
CaCO3(s) 888n CaO(s) ϩ CO2(g) lime production
CaO(s) ϩ H2O(ᐉ) 888n Ca(OH)2(s) slaking lime
Ca(OH)2(s) ϩ Mg2ϩ(aq) 888n Ca2ϩ(aq) ϩ Mg(OH)2(s) precipitation
The last reaction occurs because Ksp for Mg(OH)2, 1.5 ϫ 10Ϫ11, is much smaller than that
for Ca(OH)2, 7.9 ϫ 10Ϫ6. The milky white suspension of Mg(OH)2 is filtered, and the
solid Mg(OH)2 is then neutralized with HCl to produce a MgCl2 solution. Evaporation
of the H2O leaves solid MgCl2, which is then melted and electrolyzed (Figure 22-6) under
an inert atmosphere to produce molten Mg and gaseous Cl2. The products are separated
as they are formed, to prevent recombination.
Magnesium is cast into ingots or alloyed with other light metals. The Cl2 by-product is
used to produce more HCl for neutralization of Mg(OH)2.
ALUMINUM
Aluminum is the most commercially important nonferrous metal. Its chemistry and uses
will be discussed in Section 23-7. Aluminum is obtained from bauxite, or hydrated
aluminum oxide, Al2O3иxH2O. Aluminum ions can be reduced to Al by electrolysis only
in the absence of H2O. First the crushed bauxite is purified by dissolving it in a concentrated
solution of NaOH to form soluble Na[Al(OH)4]. Then Al(OH)3иxH2O is
precipitated from the filtered solution by blowing in carbon dioxide to neutralize the unreacted
NaOH and one OHϪ ion per formula unit of Na[Al(OH)4]. Heating the hydrated
product dehydrates it to Al2O3.
22-6
ϩ ϩϩ2[Hϩ(aq) [Mg2ϩ(aq)ϩMg(OH)2(s) ClϪ(aq)] 2ClϪ(aq)] 2H2O
ϩMg(ᐉ)MgCl2(ᐉ) Cl2(g)
electrolysisevaporate solution,
then melt solid
22-6 Aluminum 909
Mg dipped out
Molten MgCl2
Steel cathode
Graphite anode
Chlorine gas
Inert
atmosphere
Molten Mg
–+
Figure 22-6 A cell for electrolyzing molten MgCl2. The magnesium metal is formed on
the steel cathode and rises to the top, where it is dipped off periodically. Chlorine gas is
formed around the graphite anode and is piped off.
A bed of limestone, CaCO3, along
the Verde River, Arizona.
Recall that Al2O3 is amphoteric.
Impurities such as oxides of iron,
which are not amphoteric, are left
behind in the crude ore.
Al2O3(s) ϩ 2NaOH(aq) ϩ 3H2O(ᐉ) 888n 2Na[Al(OH)4](aq)
2Na[Al(OH)4](aq) ϩ CO2(aq) 888n 2Al(OH)3(s) ϩ Na2CO3(aq) ϩ H2O(ᐉ)
heat
2Al(OH)3(s) 888n Al2O3(s) ϩ 3H2O(g)
The melting point of Al2O3 is 2045°C; electrolysis of pure molten Al2O3 would have
to be carried out at or above this temperature, with great expense. It can be done, however,
at a much lower temperature when Al2O3 is mixed with much lower-melting cryolite, a
mixture of NaF and AlF3 often represented as Na3[AlF6]. The molten mixture can be
electrolyzed at 1000°C with carbon electrodes. The cell used industrially for this process,
called the Hall–Héroult process, is shown in Figure 22-7.
The inner surface of the cell is coated with carbon or carbonized iron, which functions
as the cathode at which aluminum ions are reduced to the free metal. The graphite anode
is oxidized to CO2 gas and must be replaced frequently. This is one of the chief costs of
aluminum production.
(cathode) 4[Al3ϩ ϩ 3eϪ 88n Al(ᐉ)]
(anode) 3[C(s) ϩ 2O2Ϫ 88n CO2(g) ϩ 4eϪ]
(net reaction) 4Al3ϩ ϩ 3C(s) ϩ 6O2Ϫ 88n 4Al(ᐉ) ϩ 3CO2(g)
Molten aluminum is denser than molten cryolite, so it collects in the bottom of the cell
and is drawn off and cooled to a solid.
A more economical approach, the Alcoa chlorine process, has now been developed on
a commercial scale. The anhydrous bauxite is first converted to AlCl3 by reaction with
Cl2 in the presence of carbon. The AlCl3 is then melted and electrolyzed to give aluminum,
and the recovered chlorine is reused in the first step.
2Al2O3(s) ϩ 3C(coke) ϩ 6Cl2(g) 88n 4AlCl3(s) ϩ 3CO2(g)
2AlCl3(ᐉ) 88n 2Al(ᐉ) ϩ 3Cl2(g)
This process uses only about 30% as much electrical energy as the Hall–Héroult process.
The use of large amounts of electrical energy in electrolysis makes production of
aluminum from ores an expensive metallurgy. Methods for recycling used Al use less than
10% of the energy required to make new metal from bauxite by the Hall–Héroult process.
910 CHAPTER 22: Metals I: Metallurgy
For clarity, we have not shown waters
of hydration.
A mixture of compounds typically has
a lower melting point than any of the
pure compounds (Chapter 14).
Many consumer items are made of
aluminum.
Figure 22-7 (a) Schematic drawing
of a cell for producing aluminum by
electrolysis of a melt of Al2O3 in
Na3[AlF6]. The molten aluminum
collects in the container, which acts
as the cathode. (b) Casting molten
aluminum. Electrolytic cells used in
the Hall–Héroult process appear in
the background.
Iron cathode
Molten Al2O3
and cryolite
Carbon anode
+
Molten aluminum
(a)
–
(b)
Slag
Heated air ~ 900°C
Molten iron
Reducing
zone
Gasesrise
Solidsdescend
Hot gases used
to preheat air
Flue gas (CO, CO2)
Charge of ore, coke, and limestone
200°C
700°C
1200°C
2000°C
3Fe2O3 + CO → 2Fe3O4 + CO2
CaCO3 → CaO + CO2
Fe3O4 + CO → 3FeO + CO2
C + CO2 → 2CO
FeO + CO → Fe + CO2
Impure iron melts
Molten slag forms
Silicates and phosphates reduced
2C + O2 → 2CO
Processing of recycled Al now accounts for more than half of the production of this metal.
Recycling of aluminum drink cans is now so successful that more than two thirds of this
aluminum is used to produce new cans. The typical time elapsed from collection of cans
until new ones made from the same aluminum appear on supermarket shelves is only a
few weeks.
IRON
The most desirable iron ores contain hematite, Fe2O3, or magnetite, Fe3O4. As the available
supplies of these high-grade ores have dwindled, taconite, which is magnetite in very
hard silica rock, has become an important source of iron. The oxide is reduced in blast
furnaces (Figure 22-8) by carbon monoxide. Coke mixed with limestone (CaCO3) and
crushed ore is admitted at the top of the furnace as the “charge.” A blast of hot air from
the bottom burns the coke to carbon monoxide with the evolution of more heat.
22-7
22-7 Iron 911
In Fe3O4, the formal oxidation state of
iron is ϩᎏ8
3
ᎏ. Two thirds of the Fe atoms
are in the ϩ3 state and the other one
third are in the ϩ2 state. The formula
is sometimes written as Fe2O3 иFeO.
Figure 22-8 A schematic drawing
of a blast furnace for reduction of
iron ore.
heat
2C(s) ϩ O2(g) 888n 2CO(g) ϩ heat
Most of the oxide is reduced to molten iron by carbon monoxide, although some is reduced
directly by coke. Several stepwise reductions occur (see Figure 22-8), but the main reactions
for Fe3O4 can be summarized as follows:
Fe3O4 ϩ CO 88n 3FeO ϩ CO2
3(FeO ϩ CO 88n Fe ϩ CO2)
overall: Fe3O4(s) ϩ 4CO(g) 88n 3Fe(ᐉ) ϩ 4CO2(g)
Much of the CO2 reacts with excess coke to produce more CO to reduce the next incoming
charge.
heat
CO2(g) ϩ C(s) 888n 2CO(g)
The limestone, called a flux, reacts with the silica gangue in the ore to form a molten
slag of calcium silicate.
heat
CaCO3(s) 888n CaO(s) ϩ CO2(g)
limestone
heat
CaO(s) ϩ SiO2(s) 888n CaSiO3(ᐉ)
gangue slag
The slag is less dense than molten iron; it floats on the surface of the iron and protects
it from atmospheric oxidation. Both are drawn off periodically. Some of the slag is subsequently
used in the manufacture of cement.
The iron obtained from the blast furnace contains carbon, among other things. It is
called pig iron. If it is remelted, run into molds, and cooled, it becomes cast iron. This
is brittle because it contains much iron carbide, Fe3C. If all the carbon is removed, nearly
pure iron can be produced. It is silvery in appearance, quite soft, and of little use. If some
of the carbon is removed and other metals such as Mn, Cr, Ni, W, Mo, and V are added,
the mixture becomes stronger and is known as steel. There are many types of steel,
containing alloyed metals and other elements in various controlled proportions. Stainless
steels show high tensile strength and excellent resistance to corrosion. The most common
kind contains 14–18% chromium and 7–9% nickel.
Pig iron can also be converted to steel by burning out most of the carbon with O2 in
a basic oxygen furnace (Figure 22-9). Oxygen is blown through a heat-resistant tube
inserted below the surface of the molten iron. Carbon burns to CO, which subsequently
escapes and burns to CO2.
COPPER
Copper is so widely used, especially in its alloys such as bronze (Cu and Sn) and brass
(Cu and Zn), that it is becoming very scarce. The U.S. Bureau of Mines estimates that
the known worldwide reserves of copper ore will be exhausted during the first half of the
twenty-first century (Figure 22-10). It is now profitable to mine ores containing as little
as 0.25% copper. The increased use of fiber optics in place of copper in communications
cables may help to lessen the demand for this metal. The use of superconducting materials
in electricity transmission lines could eventually provide enormous savings.
22-8
912 CHAPTER 22: Metals I: Metallurgy
Iron ore is scooped in an open-pit
mine.
Reaction of a metal oxide (basic) with
a nonmetal oxide (acidic) forms a salt.
Considerable amounts of slag are also
used to neutralize acidic soil. If there
were no use for the slag, its disposal
would be a serious economical and
environmental problem.
Molten steel is poured from a basic
oxygen furnace.
22-8 Copper 913
Oxygen gun
Escaping
gases
Steel shell
CaO
wall lining
Water-cooled
hood
Oxygen
Iron ore,
scrap steel,
and molten iron
Figure 22-9 A representation of the basic oxygen process furnace. Much of the steel
manufactured today is refined by blowing oxygen through a furnace that is charged with
scrap and molten iron from a blast furnace. After the refined iron is withdrawn into a ladle,
alloying elements are added to produce the desired steel. The steel industry is one of the
nation’s largest consumers of oxygen.
Figure 22-10 An open-pit copper mine near Bagdad, Arizona.
The two main classes of copper ores are the mixed sulfides of copper and iron—such
as chalcopyrite, CuFeS2—and the basic carbonates, such as azurite, Cu3(CO3)2(OH)2, and
malachite, Cu2CO3(OH)2. Let’s consider CuFeS2 (or CuSиFeS). The copper compound
is separated from gangue by flotation (Figures 22-2b and 22-11) and then roasted to
remove volatile impurities. Enough air is used to convert iron(II) sulfide, but not
copper(II) sulfide, to the oxide.
heat
2CuFeS2(s) ϩ 3O2(g) 888n 2FeO(s) ϩ 2CuS(s) ϩ 2SO2(g)
The roasted ore is then mixed with sand (SiO2), crushed limestone (CaCO3), and some
unroasted ore that contains copper(II) sulfide in a reverberatory furnace at 1100°C. CuS
is reduced to Cu2S, which melts.
2CuS(s) ϩ O2(g) 88n Cu2S(ᐉ) ϩ SO2(g)
The limestone and silica form a molten calcium silicate glass. This dissolves iron(II) oxide
to form a slag less dense than the molten copper(I) sulfide, on which it floats.
heat
CaCO3(s) ϩ SiO2(s) 888n CaSiO3(ᐉ) ϩ CO2(g)
heat
CaSiO3(ᐉ) ϩ FeO(s) ϩ SiO2(s) 888n CaSiO3иFeSiO3(ᐉ)
The slag is periodically drained off. The molten copper(I) sulfide is drawn off into a
Bessemer converter, where it is again heated and treated with air. This oxidizes sulfide
ions to SO2 and reduces copper(I) ions to metallic copper. The overall process is
heat
Cu2S(ᐉ) ϩ O2(g) 888n 2Cu(ᐉ) ϩ SO2(g)
The impure copper is refined using an electrolytic cell like the one shown in Figure
22-12. Thin sheets of very pure copper are made to act as cathodes by connecting them
to the negative terminal of a dc generator. Chunks of impure copper connected to the
positive terminal function as anodes. The electrodes are immersed in a solution of
copper(II) sulfate and sulfuric acid. When the cell operates, Cu from the impure anodes
is oxidized and goes into solution as Cu2ϩ ions; Cu2ϩ ions from the solution are reduced
and plate out as metallic Cu on the pure Cu cathodes. Other active metals from the impure
914 CHAPTER 22: Metals I: Metallurgy
Figure 22-11 A copper ore being
enriched by flotation.
Mining low-gold ore in an open-pit
mine.
The steps of this process are described
in Exercise 36 at the end of the
chapter.
A sludge called anode mud collects
under the anodes. It contains such
valuable and difficult-to-oxidize
elements as Au, Pt, Ag, Se, and Te.
The separation, purification, and sale
of these elements reduce the cost of
refined copper.
(a) (b)
Two copper-containing minerals. (a) Malachite, Cu3(CO3)2(OH)2 or 2CuCO3 и Cu(OH)2, is
green. (b) azurite, Cu2CO3(OH)2 or CuCO3 и Cu(OH)2, is blue.
bars also go into solution after oxidation. They do not plate out onto the cathode bars of
pure Cu because of the far greater concentration of the more easily reduced Cu2ϩ ions
that are already in solution. Overall, there is no net reaction, merely a simultaneous transfer
of Cu from anode to solution and from solution to cathode.
Cu (impure) 88n Cu2ϩ ϩ 2eϪ (oxidation, anode)
Cu2ϩ ϩ 2eϪ 88n Cu (pure) (reduction, cathode)
Cu (impure) 88n Cu (pure) (no net reaction)
Nevertheless, the net effect is that small bars of very pure Cu and large bars of impure
Cu are converted into large bars of very pure Cu and small bars of impure Cu.
GOLD
Gold is an inactive metal (sometimes called a noble metal), so it occurs mostly in the
native uncombined state. It is sometimes found as gold telluride. Because of its high density,
metallic gold can be concentrated by panning. In this operation, gold-bearing sand and
gravel are gently swirled with water in a pan. The lighter particles spill over the edge, and
the denser nuggets of gold remain. Gold is concentrated by sifting crushed gravel in a
stream of water on a slightly inclined shaking table that contains several low barriers.
These impede the descent of the heavier gold particles but allow the lighter particles to
pass over. In a modern extension to this centuries-old process, the gold is then alloyed
with mercury and removed. The mercury is distilled away, leaving behind the pure gold.
22-9
22-9 Gold 915
+
Impure Cu anode
Pure Cu cathode
Electrolyte
H2SO4 + CuSO4
–
(a)
+
Pure Cu
Electrolyte now
contains Zn2+,
Fe2+
, other active
metal ions
“Anode sludge”
containing Ag,
Au, Pt
–
(b)
(c)
Figure 22-12 A schematic diagram of
the electrolytic cell used for refining
copper (a) before electrolysis and (b)
after electrolysis. (c) Commercial
electrolysis cells for refining copper.
The energy provided by the electric
generator forces a decrease in the
entropy of the system by separating
the Cu from its impurities in the
impure bars.
916 CHAPTER 22: Metals I: Metallurgy
Key Terms
Alloying Mixing of a metal with other substances (usually other
metals) to modify its properties.
Cast iron The brittle iron obtained when the pig iron from the
blast furnace is remelted, run into molds, and cooled; contains
much iron carbide, Fe3C.
Charge A sample of crushed ore as it is admitted to a furnace
for smelting.
Flotation A method by which hydrophobic (water-repelling)
particles of an ore are separated from hydrophilic (waterattracting)
particles in a metallurgical pretreatment process.
Flux A substance added to react with the charge, or a product of
its reduction, in metallurgy; usually added to lower a melting
point.
Gangue Sand, rock, and other impurities surrounding the mineral
of interest in an ore.
Hall–Héroult process A process in which a molten mixture of
Al2O3, NaF, and AlF3 is electrolyzed to obtain pure aluminum.
Metallurgy The overall processes by which metals are extracted
from ores.
Mineral A naturally occurring inorganic solid substance having
a definite chemical composition and characteristic crystalline
structure, color, and hardness; contains a metal that can be
extracted in a metallurgical process.
Native ore A metal that occurs in an uncombined or free state
in nature.
Ore A natural deposit containing a mineral of an element to be
extracted.
Pig iron The iron obtained from the blast furnace.
Refining Purifying of a substance.
Roasting Heating a compound below its melting point in the
presence of air.
Slag Unwanted material produced during smelting.
Smelting Chemical reduction of a substance at high temperature
in metallurgy.
Steel Iron alloyed with other metals, such as Mn, Cr, Ni, W, Mo,
and V, and sometimes with C and Si.
Zone refining A method of purifying a bar of metal by passing
it through an induction heater; this causes impurities to move
along in the melted portion.
Exercises
General Concepts
*01. List the chemical and physical properties that we usually
associate with metals.
*02. Define the term “metallurgy.” What does the study of metallurgy
include?
*03. What kinds of metals are most apt to occur in the uncombined
(native) state in nature?
*04. List the six anions (and their formulas) that are most often
combined with metals in ores. Give at least one example
of an ore of each kind. What anion is the most commonly
encountered?
*05. How does an ore differ from a mineral? Name the three
general categories of procedures needed to produce pure
metals from ores. Describe the purpose of each.
*06. Briefly describe one method by which gangue can be separated
from the desired mineral during the concentration
of an ore.
*07. Give the five general steps involved in extracting a metal
from its ore and converting the metal to a useful form.
Briefly describe the importance of each.
*08. Describe the flotation method of ore pretreatment. Are any
chemical changes involved?
*09. Smelting is the process by which a metal ore is reduced.
In this context, what does the term “reduced” mean?
*10. Which metals are likely to be found in the free state in
nature? Why are these metals in the free state in nature?
Describe the location of these elements on the periodic
table. Is there a pattern?
Gold is also recovered from the anode sludge from electrolytic purification of copper
Section 21-7. Gold is so rare that it is also obtained from very low-grade ores by the
cyanide process. Air is bubbled through an agitated slurry of the ore mixed with a solution
of NaCN. This causes slow oxidation of the metal and the formation of a soluble
complex compound.
4Au(s) ϩ 8CNϪ(aq) ϩ O2(g) ϩ 2H2O(ᐉ) 88n 4[Au(CN)2]Ϫ(aq) ϩ 4OHϪ(aq)
After filtration, free gold can then be regenerated by electrolytic reduction or by reduction
of [Au(CN)2]Ϫ with zinc.
Zn(s) ϩ 2[Au(CN)2]Ϫ(aq) 88n 2Au(s) ϩ Zn(CN)4
2Ϫ(aq)
Because of environmental concerns
about mercury toxicity, the cyanide
process is increasingly preferred. This
is not to suggest that mercury is more
toxic than cyanide. The problems
due to mercury are greater in that it
persists in the environment for a long
time, and mercury poisoning is
cumulative.
Exercises 917
*11. What kinds of ores are roasted? What kinds of compounds
are converted to oxides by roasting? What kinds are converted
directly to the free metals?
*12. Of the following compounds, which would you expect
to require electrolysis to obtain the free metals: KCl;
Cr2(SO4)3; Fe2O3; Al2O3; Ag2S; MgSO4? Why?
*13. At which electrode is the free metal produced in the electrolysis
of a metal compound? Why?
*14. Write the equation that describes the electrolysis of a brine
solution to form NaOH, Cl2, and H2. What mass of each
substance will be produced in an electrolysis cell for each
mole of electrons passed through the cell? Assume 100%
efficiency.
*15. The following equations represent reactions used in some
important metallurgical processes.
(a) Fe3O4(s) ϩ CO(g) 88n Fe(ᐉ) ϩ CO2(g)
(b) MgCO3(s) ϩ SiO2(s) 88n MgSiO3(ᐉ) ϩ CO2(g)
(c) Au(s) ϩ CNϪ ϩ H2O(ᐉ) ϩ O2(g) 88n
[Au(CN)2]Ϫ ϩ OHϪ
Balance the equations. Which one(s) represent reduction
to a free metal?
*16. Repeat Exercise 15 for
electrolysis
(a) Al2O3 (cryolite solution) 88888888n Al(ᐉ) ϩ O2(g)
(b) PbSO4(s) ϩ PbS(s) 88n Pb(ᐉ) ϩ SO2(g)
(c) TaCl5(g) ϩ Mg(ᐉ) 88n Ta(s) ϩ MgCl2(ᐉ)
*17. Suggest a method of obtaining manganese from an ore containing
manganese(III) oxide, Mn2O3. On what basis do
you make the suggestion?
*18. What is the purpose of utilizing the basic oxygen furnace
after the blast furnace in the production of iron?
*19. A blast furnace utilizes or produces five species that include
carbon. (a) List the five species that contain carbon. (b) Explain
the function of each step in the chain of events from
heating the ore to the production of steel.
*20. Magnesium hydroxide can be precipitated from sea water
by adding slaked lime, Ca(OH)2. (a) Why does this precipitation
occur? (b) Could this process be used to remove
sodium from sea water? Why?
*21. Describe the metallurgy of (a) copper and (b) magnesium.
*22. Describe the metallurgy of (a) iron and (b) gold.
*23. Briefly describe the Hall–Héroult process for the commercial
preparation of aluminum.
*24. (a) Compare the refining of aluminum by the Hall–
Héroult process and by the Alcoa chlorine process.
(b) Which process is less expensive and why? (c) The Alcoa
chlorine process is potentially more dangerous to the workers
than is the Hall–Héroult process. Why?
*25. A common iron ore is magnetite, Fe3O4. In this formula
for magnetite the oxidation number of iron is not an integer.
Write the formula in a form that includes only integer
oxidation numbers.
*26. Name some common minerals that contain iron. Write the
chemical formula for the iron compound in each. What is
the oxidation number of iron in each substance?
*27. What is steel? How does the hardness of iron compare with
that of steel?
*28. (a) What is the difference in composition between pure
iron and steel? (b) What is the name of the process by
which steel is produced?
*29. (a) What is an alloy? (b) Why are alloys produced? (c) How
are alloys produced?
*30. Describe and illustrate the electrolytic refining of Cu.
*31. Name the undesirable gaseous product formed during
the roasting of copper and other sulfide ores. Why is it
undesirable?
CONCEPTUAL EXERCISES
*32. One of the problems with roasting sulfide ores is the air
pollution caused by the gas that is produced. Describe some
consequences of releasing this gas into the atmosphere.
*33. Gold can be obtained by the separation of gold from the
anode sludge of the electrolytic purification of copper. (a)
Considering the nature of gold and copper, is this source
of gold to be expected? Explain. (b) Would finding silver
in the sludge also be expected? Explain.
BUILDING YOUR KNOWLEDGE
*34. The reaction
FeO(s) ϩ CO(g) 88n Fe(s) ϩ CO2(g)
takes place in the blast furnace at a temperature of 800 K.
(a) Calculate ⌬H0
800 for this reaction, using ⌬H0
f, 800 ϭ
Ϫ268 kJ/mol for FeO, Ϫ111 kJ/mol for CO, and
Ϫ394 kJ/mol for CO2. Is this a favorable enthalpy change?
(b) Calculate ⌬G0
800 for this reaction, using ⌬G0
f, 800 ϭ
Ϫ219 kJ/mol for FeO, Ϫ182 kJ/mol for CO, and
Ϫ396 kJ/mol for CO2. Is this a favorable free energy
change? (c) Using your values of ⌬H0
800 and ⌬G0
800, calculate
⌬S0
800.Native gold.
918 CHAPTER 22: Metals I: Metallurgy
*35. During the operation of a blast furnace, coke reacts with
the oxygen in air to produce carbon monoxide, which, in
turn, serves as the reducing agent for the iron ore. Assuming
the formula of the iron ore to be Fe2O3, calculate the
mass of air needed for each ton of iron produced. Assume
air to be 21% O2 by mass, and assume that the process is
93% efficient.
*36. The following reactions take place during the extraction
of copper from copper ore.
(a) 2Cu2S(ᐉ) ϩ 3O2(g) 88n 2Cu2O(ᐉ) ϩ 2SO2(g)
(b) 2Cu2O(ᐉ) ϩ Cu2S(ᐉ) 88n 6Cu(ᐉ) ϩ SO2(g)
Identify the oxidizing and reducing agents. Show that each
equation is correctly balanced by demonstrating that the
increase and decrease in oxidation numbers are equal.
*37. Assuming complete recovery of metal, which of the following
ores would yield the greater quantity of copper on
a mass basis? (a) an ore containing 3.60 mass % azurite,
Cu(OH)2и2CuCO3, or (b) an ore containing 4.95 mass %
chalcopyrite, CuFeS2.
*38. What mass of copper could be electroplated from a solution
of CuSO4, using an electric current of 2.50 A flowing
for 5.00 h? (Assume 100% efficiency.)
*39. (a) Calculate the weight, in pounds, of sulfur dioxide produced
in the roasting of 1 ton of chalcocite ore containing
10.3% Cu2S, 0.94% Ag2S, and no other source of sulfur.
(b) What weight of sulfuric acid can be prepared from the
SO2 generated, assuming 85% of it can be recovered from
stack gases, and 88% of that recovered can be converted
to sulfuric acid? (c) How many pounds of pure copper can
be obtained, assuming 78% efficient extraction and purification?
(d) How many pounds of silver can be produced,
assuming 81% of it can be extracted and purified?
*40. Forty-five pounds of Al2O3, obtained from bauxite, is
mixed with cryolite and electrolyzed. How long would a
0.900-A current have to be passed to convert all the Al3ϩ
(from Al2O3) to aluminum metal? What volume of oxygen,
collected at 775 torr and 125°C, would be produced
in the same period of time?
*41. Calculate the percentage of iron in hematite ore containing
61.5% Fe2O3 by mass. How many pounds of iron
would be contained in 1 ton of the ore?
*42. The roasting of cinnabar, HgS, is a process by which free
mercury is obtained.
heat
HgS ϩ O2 3:4 Hg ϩ SO2
Assuming that the roasting process is 100% efficient, what
mass, in tons, of SO2 is released into the air by the roasting
of 500. tons of cinnabar?
*43. The roasting of covellite, CuS, is a process by which free
copper is obtained.
heat
CuS ϩ O2 3:4 Cu ϩ SO2
Assuming that the roasting process is 91.5% efficient, what
mass, in tons, of SO2 is released into the air by the roasting
of 500. tons of covellite?
*44. Consider the separation of iron from hematite, Fe2O3, in
a blast furnace which has an overall efficiency of 98.0%
in order to leave a specific amount of carbon in the final
product.
heat
2C(coke) ϩ O2 3:4 2CO
heat
Fe2O3 ϩ 3CO 3:4 2Fe ϩ 3CO2
Calculate the mass of coke, in tons, necessary for the ultimate
reaction with 100. tons of hematite.
*45. Using data from Appendix K, calculate ⌬G0
298 for the following
reactions.
(a) Al2O3(s) 88n 2Al(s) ϩ ᎏ3
2
ᎏO2(g)
(b) Fe2O3(s) 88n 2Fe(s) ϩ ᎏ3
2
ᎏO2(g)
(c) HgS(s) 88n Hg(ᐉ) ϩ S(s)
Are any of the reactions spontaneous? Are the ⌬G0
298
values in line with what would be predicted based on the
relative activities of the metal ions involved? Do increases
in temperature favor these reactions?
*46. Find the standard molar enthalpies of formation of Al2O3,
Fe2O3, and HgS in Appendix K. Are the values in line with
what might be predicted, in view of the methods by which
the metal ions are reduced in extractive metallurgy?
23Metals II:
Properties and
Reactions
OUTLINE
The Alkali Metals (Group IA)
23-1 Group IA Metals: Properties
and Occurrence
23-2 Reactions of the Group IA
Metals
23-3 Uses of Group IA Metals and
Their Compounds
The Alkaline Earth Metals
(Group IIA)
23-4 Group IIA Metals: Properties
and Occurrence
23-5 Reactions of the Group IIA
Metals
23-6 Uses of Group IIA Metals and
Their Compounds
The Post-Transition Metals
23-7 Group IIIA: Periodic Trends
The d-Transition Metals
23-8 General Properties
23-9 Oxidation States
23-10 Chromium Oxides, Oxyanions,
and Hydroxides
OBJECTIVES
After you have studied this chapter, you should be able to
• Describe the properties and occurrence of the Group IA metals
• Describe some important reactions of the Group IA metals
• Describe some important uses of the Group IA metals and their compounds
• Describe the properties and occurrence of the Group IIA metals
• Describe some important reactions of the Group IIA metals
• Describe some important uses of the Group IIA metals and their compounds
• Identify the post-transition metals
• Describe some important periodic trends in the properties of Group IIIA metals and
some of their compounds
• Describe the important properties of aluminum and some of its compounds
• Identify the d-transition metals and describe some of their important properties and
their typical compounds
• Describe typical oxidation states of the transition metals
• Describe some oxides, oxyanions, and hydroxides of chromium
Metals and alloys have a wide
variety of uses. Strength, lightness,
and corrosion resistance are some
properties that are required of
metals used in aircraft construction.
II
n this chapter we will discuss some representative metals and some d-transition
metals. The representative elements are those in the A groups of the periodic table.
They have valence electrons in their outermost s and p atomic orbitals. Metallic character
increases from top to bottom within groups and from right to left within periods.
All the elements in Groups IA (except H) and IIA are metals. The heavier members of
Groups IIIA, IVA, and VA are called post-transition metals.
23-1 Group IA Metals: Properties and Occurrence 921
THE ALKALI METALS (GROUP IA)
GROUP IA METALS: PROPERTIES AND OCCURRENCE
The alkali metals are not found free in nature, because they are so easily oxidized. They
are most economically produced by electrolysis of their molten salts. Sodium (2.6% abundance
by mass) and potassium (2.4% abundance) are very common in the earth’s crust.
The other IA metals are quite rare. Francium consists only of short-lived radioactive
isotopes formed by alpha-particle emission from actinium (Section 26-4). Both potassium
and cesium also have natural radioisotopes. Potassium-40 is important in the potassium–
argon radioactive decay method of dating ancient objects (Section 26-12). The properties
of the alkali metals vary regularly as the group is descended (Table 23-1).
The free metals, except lithium, are soft, silvery, corrosive metals that can be cut with
a knife; lithium is harder. Cesium is slightly golden and melts in the hand (wrapped in
plastic because it is so corrosive). The relatively low melting and boiling points of the
alkali metals result from their fairly weak bonding forces. Each atom can furnish only one
electron for metallic bonding (Section 13-17). Because their outer electrons are so loosely
held, the metals are excellent electrical and thermal conductors. They ionize when irradiated
with low-energy light (the photoelectric effect). These effects become more
pronounced with increasing atomic size. Cesium is used in photoelectric cells.
The low ionization energies of the IA metals show that the single electron in the outer
shell is very easily removed. In all alkali metal compounds the metals exhibit the ϩ1 oxidation
state. Virtually all are ionic. The extremely high second ionization energies show that
removal of an electron from a filled shell is impossible by chemical means.
23-1
Hydrogen is included in Group IA in
the periodic table, but it is not a metal.
See the discussion of electrolysis of
sodium chloride in Section 21-3.
Alkali metals are excellent
conductors of electricity.
Li Be
Na Mg
K Ca
Rb Sr
IA IIA
Cs Ba
Fr Ra
Al
Ga
In Sn
Tl Pb
IIIA IVA
Bi
VA
We might expect the standard reduction potentials of the metal ions to become less
favorable (more negative) as the ionization energies for the metal atoms become more
favorable (less endothermic). The magnitude of the standard reduction potential of Li,
Ϫ3.05 volts, is unexpectedly large, however. The first ionization energy is the amount of
energy absorbed when a mole of gaseous atoms ionize. The standard reduction potential,
E0, indicates the ease with which aqueous ions are reduced to the metal (Section 21-14).
Thus, hydration energies must also be considered (Section 14-2). Because the Liϩ ion is
so small, its charge density (ratio of charge to size) is very high. It therefore exerts a
stronger attraction for polar H2O molecules than do the other IA ions. These H2O molecules
must be stripped off during the reduction process in a very endothermic manner,
making the E0 for the Liϩ/Li couple very negative (see Table 23-1).
The high charge density of Liϩ ion accounts for its ability to polarize large anions.
This gives a higher degree of covalent character in Li compounds than in other corresponding
alkali metal compounds. For example, LiCl is soluble in ethyl alcohol, a less
polar solvent than water; NaCl is not. Salts of the alkali metals with small anions are very
soluble in water, but salts with large and complex anions, such as silicates and aluminosilicates,
are not very soluble.
REACTIONS OF THE GROUP IA METALS
Many of the reactions of the alkali metals are summarized in Table 23-2. All are characterized
by the loss of one electron per metal atom. These metals are very strong reducing
agents. Reactions of the alkali metals with H2 and O2 were discussed in Sections 6-7 and
6-8, reactions with the halogens in Section 7-2; and reactions with water in Section 4-8,
part 2.
The high reactivities of the alkali metals are illustrated by their vigorous reactions with
water. Lithium reacts readily; sodium reacts so vigorously that the hydrogen gas it forms
may ignite; and potassium, rubidium, and cesium cause the H2 to burst into flames when
23-2
922 CHAPTER 23: Metals II: Properties and Reactions
TABLE 23-1 Properties of the Group IA Metals
Property Li Na K Rb Cs Fr
Outer electrons 2s1 3s1 4s1 5s1 6s1 7s1
Melting point (°C) 186 97.8 63.6 38.9 28.5 27
Boiling point (°C) 1347 904 774 688 678 677
Density (g/cm3) 0.534 0.971 0.862 1.53 1.87 —
Atomic radius (Å) 1.52 1.86 2.27 2.48 2.65 —
Ionic radius, Mϩ (Å) 0.90 1.16 1.52 1.66 1.81 —
Electronegativity 1.0 1.0 0.9 0.9 0.8 0.8
E0 (volts): Mϩ(aq) ϩ eϪ 88n M(s) Ϫ3.05 Ϫ2.71 Ϫ2.93 Ϫ2.93 Ϫ2.92 —
Ionization energies (kJ/mol)
M(g) 88n Mϩ(g) ϩ eϪ 520 496 419 403 377 —
Mϩ(g) 88n M2ϩ(g) ϩ eϪ 7298 4562 3051 2632 2420 —
⌬H0
hydration (kJ/mol): Mϩ(g) ϩ xH2O 88n Mϩ(aq) Ϫ544 Ϫ435 Ϫ351 Ϫ293 Ϫ264 —
Polarization of an anion refers to
distortion of its electron cloud. The
ability of a cation to polarize an anion
increases with increasing charge
density (ratio of charge to size) of
the cation.
the metals are dropped into water. The large amounts of heat evolved provide the activation
energy to ignite the hydrogen. The elements also react with water vapor in the air
or with moisture from the skin.
2K(s) ϩ 2H2O(ᐉ) 88n 2[Kϩ(aq) ϩ OHϪ(aq)] ϩ H2(g) ⌬H 0 ϭ Ϫ390.8 kJ/mol rxn
Alkali metals are stored under anhydrous nonpolar liquids such as mineral oil.
As is often true for elements of the second period, Li differs in many ways from the
other members of its family. Its ionic charge density and electronegativity are close to
those of Mg, so Li compounds resemble those of Mg in some ways. This illustrates the
diagonal similarities that exist between elements in successive groups near the top of the
periodic table.
IA IIA IIIA IVA
Li Be B C
Na Mg Al Si
Lithium is the only IA metal that combines with N2 to form a nitride, Li3N. Magnesium
readily forms magnesium nitride, Mg3N2. Both metals readily combine with carbon to
form carbides, whereas the other alkali metals do not react readily with carbon. The solubilities
of Li compounds are closer to those of Mg compounds than to those of other IA
compounds. The fluorides, phosphates, and carbonates of both Li and Mg are only slightly
soluble, but their chlorides, bromides, and iodides are very soluble. Both Li and Mg form
normal oxides, Li2O and MgO, when burned in air at 1 atmosphere pressure. The other
alkali metals form peroxides or superoxides.
The IA metal oxides are basic. They react with water to form strong bases.
Na2O(s) ϩ H2O(ᐉ) 88n 2[ Naϩ(aq) ϩ OHϪ(aq)]
K2O(s) ϩ H2O(ᐉ) 88n 2[ Kϩ(aq) ϩ OHϪ(aq)]
23-2 Reactions of the Group IA Metals 923
TABLE 23-2 Some Reactions of the Group IA Metals
Reaction Remarks
4M ϩ O2 88n 2M2O Limited O2
4Li ϩ O2 88n 2Li2O Excess O2 (lithium oxide)
2Na ϩ O2 88n Na2O2 (sodium peroxide)
M ϩ O2 88n MO2 M ϭ K, Rb, Cs; excess O2 (superoxides)
2M ϩ H2 88n 2MH Molten metals
6Li ϩ N2 88n 2Li3N At high temperature
2M ϩ X2 88n 2MX X ϭ halogen (Group VIIA)
2M ϩ S 88n M2S Also with Se, Te of Group VIA
12M ϩ P4 88n 4M3P
2M ϩ 2H2O 88n 2MOH ϩ H2 K, Rb, and Cs react explosively
2M ϩ 2NH3 88n 2MNH2 ϩ H2 With NH3(ᐉ) in presence of catalyst;
with NH3(g) at high temperature
(solutions also contain Mϩ ϩ solvated eϪ)
Sodium reacts vigorously with water.
2Na(s) ϩ 2H2O(ᐉ) 88n
2[Naϩ(aq) ϩ OH(aq)] ϩ H2(g)
The indicator, phenolphthalein, was
added to the water. As NaOH forms,
the solution turns pink.
USES OF GROUP IA METALS AND THEIR COMPOUNDS
Lithium, Li
Metallic lithium has the highest heat capacity of any element. It is used as a heat transfer
medium in experimental nuclear reactors. Extremely lightweight lithium–aluminum alloys
are used in aircraft construction. Lithium compounds are used in some lightweight dry
cells and storage batteries because they have very long lives, even in extreme temperatures.
LiCl and LiBr are very hygroscopic and are used in industrial drying processes and
air conditioning. Lithium compounds are used for the treatment of some types of mental
disorders (mainly manic depression).
Sodium, Na
Sodium is by far the most widely used alkali metal because it is so abundant. Its salts are
essential for life. The metal itself is used as a reducing agent in the manufacture of drugs
and dyes and in the metallurgy of such metals as titanium and zirconium.
heat
TiCl4(g) ϩ 4Na(ᐉ) 888n 4NaCl(s) ϩ Ti(s)
Highway lamps often incorporate Na arcs, which produce a bright yellow glow. A few
examples of the uses of sodium compounds are NaOH, called caustic soda, lye, or soda
lye (used for production of rayon, cleansers, textiles, soap, paper, and many polymers);
Na2CO3, called soda or soda ash, and Na2CO3и10H2O, called washing soda (also used
as a substitute for NaOH when a weaker base is acceptable); NaHCO3, called baking soda
23-3
924 CHAPTER 23: Metals II: Properties and Reactions
The yellowish glow of some highway
lamps is due to a sodium arc.
Mercury lamps give a bluish glow.
Spacings of energy levels are different for different alkali metals. The salts of the alkali
metals impart characteristic colors to flames: lithium (red), sodium ( yellow), and potassium
(violet).
(text continues on page 928)
The highly corrosive nature of both
lithium and sodium is a major
drawback to applications of the
pure metals.
23-3 Uses of Group IA Metals and Their Compounds 925
The Development of ScienceCCHEMISTRY IN USE
Trace Elements and Life
More than 300 years ago iron was the first trace element
shown to be essential in the human diet. An English physician,
Thomas Sedenham, soaked “iron and steel filings” in
cold Rhenish wine. He used the resulting solution to treat
patients suffering from chlorosis, now known to be an irondeficiency
anemia. Nearly 20 trace elements are now believed
to be required by humans. The discovery of the biological
functions of trace elements is an exciting and controversial
area of human nutrition research.
The trace elements can be classified into several categories
(see table). In 1989, the National Research Council recognized
that iron, iodine, zinc, selenium, copper, chromium,
manganese, and molybdenum were dietary essentials for
humans. Fluorine is also considered to be valuable for human
health, because of its benefits to the teeth and skeleton. These
nine trace elements are required by humans and other
animals because they are essential components in metalloenzymes
and hormones or because they promote health in a
specific tissue (such as fluorine in the teeth and skeleton).
The trace elements required by the human body in milligram
quantities include iron, zinc, copper, manganese, and fluorine.
Trace elements required in microgram (␮g) quantities
include iodine, selenium, chromium, and molybdenum.
Although probably required in ␮g quantities, no dietary recommendations
have been made for arsenic, nickel, silicon,
and boron even though there is evidence, primarily from animals,
that they are essential. There is only weak evidence that
cadmium, lead, lithium, tin, vanadium, and bromine are
essential for humans.
Iron deficiency is one of the most common nutrient deficiencies
in the world, occurring in up to 60% of the women,
infants, and children of some countries. Anemia, characterized
by a low concentration of hemoglobin in the blood or
by a low volume of packed red blood cells, is the usual symptom
of iron deficiency. Other symptoms include fatigue and
cognitive disorders. Up to 1% of the population may have
the genetic disease known as hereditary hemochromatosis,
which results in excess absorption of dietary iron, and can
lead to liver and heart damage. Concern about this disease
may eventually lead to routine screening so that persons with
this disorder can be treated before severe symptoms develop
and can avoid foods and supplements with large amounts of
iron. The recommended dietary allowance for iron for
women aged 23 to 50 years is 50% higher than that for men
in the same age group because of the iron lost in menstrual
bleeding.
Many dietary supplements include essential trace
elements.
(Box continues on next page)
926 CHAPTER 23: Metals II: Properties and Reactions
Iodine deficiency remains a major cause of mental retardation
and infant mortality and morbidity throughout the
world—even though iodine was shown to be essential for
human health nearly 100 years ago. More than 1 billion people
are believed to be at risk for iodine deficiency. In 1986,
the International Council for the Control of Iodine Deficiency
Disorders was established in an effort to improve
iodine nutrition and alleviate human suffering. This council
works closely with the World Health Organization, the
United Nations International Children’s Fund, and the
United Nations to alleviate iodine deficiency. Iodine is
required for the thyroid hormones, thyroxine and triiodothyronine,
that regulate the metabolic rate and O2
consumption of cells. Iodine is also intimately involved in the
control of growth and development, particularly during fetal
and infant life.
In the 1930s, zinc was discovered to be a dietary essential
in animals. Zinc deficiency was recognized as a potential public
health problem in the 1960s in Iran, where endemic
hypogonadism (delayed sexual development) and dwarfism
Dietary Trace Elements
Known to Be Essential Known or Suspected Functions
iron Hemoglobin, energy metabolism
iodine Thyroid hormones
zinc Enzymes, protein synthesis, cell division
copper Hemoglobin, bone, nerves, vascular system
selenium Enzymes, protect against oxidant stress
chromium Insulin action
manganese Enzymes, bone
molybdenum Enzymes, sulfur metabolism
fluorine Bones, teeth
Substantial Evidence
for Essentiality Known or Suspected Functions
arsenic Amino acid metabolism
boron Metabolism of calcium, magnesium, hormones
nickel Not known, suspected in some enzymes
silicon Bone and connective tissue
Weak Evidence
for Essentiality Known or Suspected Functions
bromine Not yet known
cadmium Not yet known
lead Not yet known
lithium Not yet known
tin Not yet known
vanadium Not yet known
23-3 Uses of Group IA Metals and Their Compounds 927
were discovered in adolescents consuming insufficient dietary
zinc. More than 200 zinc enzymes have been discovered. Zinc
is also important for the structure and function of biomembranes.
Loss of zinc from these membranes results in
increased susceptibility to oxidative damage, structural
strains, and alterations in specific receptor sites and transport
systems. Zinc also helps to stabilize the structures of RNA,
DNA, and ribosomes. Several transcription factors contain
“Zn fingers,” which are needed for the binding of these transcription
factors to the DNA. Thus, zinc is absolutely
necessary for adequate growth, protein synthesis, and cell
division. The best sources of zinc in the human diet are animal
foods such as meat, fish, poultry, and dairy products.
Copper was also shown to be essential in the early 1900s.
Copper is needed for the absorption and mobilization of iron,
so a deficiency of copper causes a type of anemia that is difficult
to distinguish from iron deficiency anemia. Copper is
also needed for the cardiovascular system, bone, brain, and
nervous system. Premature and malnourished infants are particularly
susceptible to developing copper deficiency, in part
because milk is a poor source of copper. Whole grains,
legumes, and nuts are the major dietary sources of copper.
Selenium was first suspected of being a dietary essential
in the 1950s. Selenium is considered to be an antioxidant
nutrient because it is present in enzymes that help protect
against toxic species of oxygen and free radicals. Selenium
deficiency is a major public health problem in certain parts
of China, where it increases the risk of heart disease, bone
and joint disorders, and liver cancer. Selenium is currently
under intensive investigation as a possible protector against
cancer. The content of selenium in foods is highly variable
and dependent on the selenium content of the soil. Generally
the best sources of selenium are muscle meats, cereals,
and grains.
The 1950s also saw the first evidence that chromium
might be a dietary essential. Chromium is believed to promote
the action of insulin and thus influences the metabolism
of carbohydrates, fats, and proteins. Reports of severe human
deficiency of chromium are rare and have been found primarily
in people receiving only intravenous feedings for
several months or years. Only a few laboratories in the world
can accurately measure the amount of chromium in foods and
body tissues because chromium is present in stainless steel,
which is ubiquitous in analytical laboratories and easily contaminates
biological samples.
Manganese and molybdenum are essential for enzymes in
humans and other animals, but a dietary deficiency of these
minerals is exceedingly rare in humans. Cobalt is essential
for vitamin B12, but the human body cannot make vitamin
B12 from cobalt and thus requires the preformed vitamin from
dietary sources. (It is possible to derive some vitamin B12 from
bacterial synthesis in the digestive tract.)
Efforts to discover whether other elements might be
essential intensified during the 1970s. Although it is believed
that arsenic, nickel, silicon, and boron are probably essential
to humans, it has been difficult to determine whether other
minerals have specific biological functions in humans or other
animals.
There are several reasons why it is difficult to establish
the essentiality of trace elements. Some elements, such as
arsenic and selenium, were first recognized for their extreme
toxicity, so it has been difficult to convince many health specialists
that a toxic element might also be a dietary essential
at low levels. Also, most trace elements are present in
extremely small amounts in diets and in tissues, and few laboratories
are equipped to prevent contamination of samples
and to measure these elements with the necessary precision.
Two factors have aided in the discovery of the roles of
many trace elements. One is the availability of two highly
sensitive analytical techniques, activation analysis and electrothermal
atomic absorption spectroscopy, that allow
detection of these elements in concentrations of only a few
parts per billion. The other is the use of special isolation
chambers that allow study of animals under carefully controlled
conditions, free of unwanted contaminants. The diets
fed to animals and their air supply must be carefully purified
to keep out even traces of unwanted elements, and their cages
must be made of plastics that contain no metals.
Our understanding of the biological functions of trace elements
is changing the way scientists think about diet and
health. For example, supplements of manganese, copper, and
zinc in combination with calcium have recently been shown
to improve human bone health to a greater extent than just
calcium alone. Silicon and boron are also believed to be
important for bone health. Deficiencies of selenium or copper
are suspected by some scientists of increasing the risk of
cancer or heart disease. Because chromium, copper, and zinc
influence glucose metabolism, future prevention and treatment
strategies for diabetes may involve these nutrients.
Mary Ann Johnson
College of Family and Consumer Sciences
University of Georgia
or bicarbonate of soda (used for baking and other household uses); NaCl (used as table
salt and as the source of all other compounds of Na and Cl); NaNO3, called Chile saltpeter
(a nitrogen fertilizer); Na2SO4, called salt cake, a by-product of HCl manufacture
(used for production of brown wrapping paper and cardboard boxes); and NaH (used for
synthesis of NaBH4, which is used to recover silver and mercury from waste water).
Other Group IA Metals
Like salts of Na (and probably Li), those of potassium are essential for life. KNO3,
commonly known as niter or saltpeter, is used as a potassium and nitrogen fertilizer.
Most other major industrial uses for K can be satisfied with the more abundant and
less expensive Na.
There are very few practical uses for the rare metals rubidium, cesium, and francium.
Cesium is used in some photoelectric cells (Section 5-11).
THE ALKALINE EARTH METALS (GROUP IIA)
GROUP IIA METALS: PROPERTIES AND OCCURRENCE
The alkaline earth metals are all silvery white, malleable, ductile, and somewhat harder
than their neighbors in Group IA. Activity increases from top to bottom within the group,
with Ca, Sr, and Ba being considered quite active. Each has two electrons in its highest
occupied energy level. Both electrons are lost in ionic compound formation, though not
as easily as the outer electron of an alkali metal. Compare the ionization energies in Tables
23-1 and 23-3. Most IIA compounds are ionic, but those of Be exhibit a great deal of
covalent character. This is due to the extremely high charge density of Be2ϩ. Compounds
of beryllium therefore resemble those of aluminum in Group IIIA (diagonal similarities).
The IIA elements exhibit the ϩ2 oxidation state in all their compounds. The tendency to
form 2ϩ ions increases from Be to Ra.
The alkaline earth metals show a wider range of chemical properties than the alkali
metals. The IIA metals are not as reactive as the IA metals, but they are much too reactive
to occur free in nature. They are obtained by electrolysis of their molten chlorides.
Calcium and magnesium are abundant in the earth’s crust, especially as carbonates and
sulfates. Beryllium, strontium, and barium are less abundant. All known radium isotopes
are radioactive and are extremely rare.
REACTIONS OF THE GROUP IIA METALS
Table 23-4 summarizes some reactions of the alkaline earth metals, which, except for stoichiometry,
are similar to the corresponding reactions of the alkali metals. Reactions with
hydrogen and oxygen were discussed in Sections 6-7 and 6-8.
Except for Be, all the alkaline earth metals are oxidized to oxides in air. The IIA oxides
(except BeO) are basic and react with water to give hydroxides. Beryllium hydroxide,
Be(OH)2, is quite insoluble in water and is amphoteric. Magnesium hydroxide, Mg(OH)2,
is only slightly soluble in water. The hydroxides of Ca, Sr, and Ba are strong bases.
Beryllium is at the top of Group IIA. Its oxide is amphoteric, whereas oxides of the
heavier members are basic. Metallic character increases from top to bottom within a group
23-5
23-4
928 CHAPTER 23: Metals II: Properties and Reactions
In Section 8-5 we found that gaseous
BeCl2 is linear. The Be atoms in BeCl2
molecules, however, act as Lewis acids.
In the solid state, the Cl atoms form
coordinate covalent bonds to Be,
resulting in a polymeric structure. In
such compounds, Be follows the octet
rule.
Cl Cl
Cl Cl
Be
Cl
Cl
Be Be
Amphoterism is the ability of a
substance to react with both acids
and bases (see Section 10-6).
and from right to left across a period. This results in increasing basicity and decreasing
acidity of the oxides in the same directions, as shown in the following table.
Group IA Group IIA Group IIIA
Li2O (basic) BeO (amphoteric) B2O3 (amphoteric)
Na2O (basic) MgO (basic) Al2O3 (amphoteric)
K2O (basic) CaO (basic) Ga2O3 (amphoteric)
In2O3 (basic)
Increasingmetalliccharacterofelements
Decreasingacidityofoxides
Increasingbasicityofoxides
Increasing metallic character of elements
Decreasing acidity of oxides
Increasing basicity of oxides
TABLE 23-3 Properties of the Group IIA Metals
Property Be Mg Ca Sr Ba Ra
Outer electrons 2s2 3s2 4s2 5s2 6s2 7s2
Melting point (°C) 1283 649 839 770 725 700
Boiling point (°C) 2484 1105 1484 1384 1640 1140
Density (g/cm3) 1.85 1.74 1.55 2.60 3.51 5
Atomic radius (Å) 1.12 1.60 1.97 2.15 2.22 2.20
Ionic radius, M2ϩ (Å) 0.59 0.85 1.14 1.32 1.49 —
Electronegativity 1.5 1.2 1.0 1.0 1.0 1.0
E0 (volts): M2ϩ(aq) ϩ 2eϪ 88n 2M(s) Ϫ1.85 Ϫ2.37 Ϫ2.87 Ϫ2.89 Ϫ2.90 Ϫ2.92
Ionization energies (kJ/mol)
M(g) 88n Mϩ(g) ϩ eϪ 899 738 599 550 503 509
Mϩ(g) 88n M2ϩ(g) ϩ eϪ 1757 1451 1145 1064 965 (979)
⌬H0
hydration (kJ/mol): M2ϩ(g) 88n M2ϩ(aq) — Ϫ1925 Ϫ1650 Ϫ1485 Ϫ1276 —
TABLE 23-4 Some Reactions of the Group IIA Metals
Reaction Remarks
2M ϩ O2 88n 2MO Very exothermic (except Be)
Ba ϩ O2 88n BaO2 Almost exclusively
M ϩ H2 88n MH2 M ϭ Ca, Sr, Ba at high temperatures
3M ϩ N2 88n M3N2 At high temperatures
6M ϩ P4 88n 2M3P2 At high temperatures
M ϩ X2 88n MX2 X ϭ halogen (Group VIIA)
M ϩ S 88n MS Also with Se, Te of Group VIA
M ϩ 2H2O 88n M(OH)2 ϩ H2 M ϭ Ca, Sr, Ba at 25°C; Mg gives MgO at high temperatures
M ϩ 2NH3 88n M(NH2)2 ϩ H2 M ϭ Ca, Sr, Ba in NH3(ᐉ) in presence of catalyst; NH3(g) with heat
Beryllium does not react with pure water even at red heat. It reacts with solutions of
strong bases to form the complex ion, [Be(OH)4]2Ϫ, and H2. Magnesium reacts with steam
to produce MgO and H2. Ca, Sr, and Ba react with water at 25°C to form hydroxides and
H2 (see Table 23-4). Group IIA compounds are generally less soluble in water than corresponding
IA compounds, but many are quite soluble.
USES OF GROUP IIA METALS AND THEIR COMPOUNDS
Beryllium, Be
Because of its rarity, beryllium has only a few practical uses. It occurs mainly as beryl,
Be3Al2Si6O18, a gemstone which, with appropriate impurities, may be aquamarine (blue)
or emerald (green). Because it is transparent to X-rays, “windows” for X-ray tubes are
constructed of beryllium. Beryllium compounds are quite toxic.
Magnesium, Mg
Metallic magnesium burns in air with such a brilliant white light that it is used in photographic
flash accessories and fireworks. It is very lightweight and is currently used in many
alloys for building materials. Like aluminum, it forms an impervious coating of oxide that
protects it from further oxidation. Given its inexhaustible supply in the oceans, it is likely
that many more structural uses will be found for it as the reserves of iron ores dwindle.
Calcium, Ca
Calcium and its compounds are widely used commercially. The element is used as a
reducing agent in the metallurgy of uranium, thorium, and other metals. It is also used
as a scavenger to remove dissolved impurities such as oxygen, sulfur, and carbon in molten
metals and to remove residual gases in vacuum tubes. It is a component of many alloys.
23-6
930 CHAPTER 23: Metals II: Properties and Reactions
Magnesium burns in air.
The tires on the Hummer vehicles
used in the Gulf War were made
blowout-proof by strong, lightweight
magnesium inserts. This is a
photograph of a civilian model.
Limestone is mainly calcium
carbonate.
A laboratory X-ray tube (left) and a close-up view of one of its windows (right). The
windows are made of beryllium metal.
Heating limestone, CaCO3, produces quicklime, CaO, which can then be treated with
water to form slaked lime, Ca(OH)2, an inexpensive base for which industry finds many
uses. When slaked lime is mixed with sand and exposed to the CO2 of the air, it hardens
to form mortar. Heating gypsum, CaSO4и2H2O, produces plaster of Paris, 2CaSO4иH2O.
heat
2[CaSO4и2H2O(s)] 888n 2CaSO4иH2O(s) ϩ 3H2O(g)
Strontium, Sr
Strontium salts are used in fireworks and flares, which show the characteristic red glow
of strontium in a flame. Strontium chloride is used in some toothpastes for persons with
sensitive teeth. The metal itself has no practical uses.
Barium, Ba
Barium is a constituent of alloys that are used for spark plugs because of the ease with
which it emits electrons when heated. It is used as a degassing agent for vacuum tubes. A
slurry of finely divided barium sulfate, BaSO4, is used to coat the gastrointestinal tract in
preparation for X-ray photographs because it absorbs X-rays so well. It is so insoluble that
it is not poisonous; all soluble barium salts are very toxic.
THE POST-TRANSITION METALS
The metals below the stepwise division of the periodic table in Groups IIIA through VA
are the post-transition metals. These include aluminum, gallium, indium, and thallium
from Group IIIA; tin and lead from Group IVA; and bismuth from Group VA. Aluminum
is the only post-transition metal that is considered very reactive. We will focus our discussion
on the metals of Group IIIA.
GROUP IIIA: PERIODIC TRENDS
The properties of the elements in Group IIIA (Table 23-5) vary less regularly down the
groups than those of the IA and IIA metals. The Group IIIA elements are all solids. Boron,
at the top of the group, is a nonmetal. Its melting point, 2300°C, is very high because it
crystallizes as a covalent solid. The other elements, aluminum through thallium, form
metallic crystals and have considerably lower melting points.
Aluminum, Al
Aluminum is the most reactive of the post-transition metals. It is the most abundant metal
in the earth’s crust (7.5%) and the third most abundant element. Aluminum is inexpensive
compared with most other metals. It is soft and can be readily extruded into wires or
rolled, pressed, or cast into shapes.
Because of its relatively low density, aluminum is often used as a lightweight structural
metal. It is often alloyed with Mg and some Cu and Si to increase its strength. Many
buildings are sheathed in aluminum, which resists corrosion by forming an oxide coating.
Pure aluminum conducts about two thirds as much electric current per unit volume as
copper, but it is only one third as dense (Al, 2.70 g/cm3; Cu, 8.92 g/cm3). As a result, a
mass of aluminum can conduct twice as much current as the same mass of copper.
23-7
23-7 Group IIIA: Periodic Trends 931
Calcium carbonate and calcium
phosphate occur in seashells and
animal bones.
This aluminum honeycomb material
is made by bonding aluminum foil
sheets to form hexagonal cells. It is
used to make sandwich construction
panels that have a very high
strength-to-weight ratio.
There are no true metals in Groups
VIA, VIIA, and VIIIA.
Aluminum is now used in electrical transmission lines and has been used in wiring in
homes. The latter use has been implicated as a fire hazard, however, due to the heat that
can be generated during high current flow at the junction of the aluminum wire and
fixtures of other metals.
Aluminum is a strong reducing agent.
Al3ϩ(aq) ϩ 3eϪ 88n Al(s) E0 ϭ Ϫ1.66 V
Aluminum is quite reactive, but a thin, very hard transparent film of Al2O3 forms when
Al comes into contact with air. This protects it from further oxidation. For this reason it
is even passive toward nitric acid, HNO3, a strong oxidizing agent. When the oxide coating
is sanded off, Al reacts vigorously with HNO3.
Al(s) ϩ 4HNO3(aq) 88n Al(NO3)3(aq) ϩ NO(g) ϩ 2H2O(ᐉ)
932 CHAPTER 23: Metals II: Properties and Reactions
TABLE 23-5 Properties of the Group IIIA Elements
Property B Al Ga In Tl
Outer electrons 2s22p1 3s23p1 4s24p1 5s25p1 6s26p1
Physical state (25°C, 1 atm) solid solid solid solid solid
Melting point (°C) 2300 660 29.8 156.6 303.5
Boiling point (°C) 2550 2367 2403 2080 1457
Density (g/cm3) 2.34 2.70 5.91 7.31 11.85
Atomic radius (Å) 0.85 1.43 1.35 1.67 1.70
Ionic radius, M3ϩ (Å) (0.20)* 0.68 0.76 0.94 1.03
Electronegativity 2.0 1.5 1.7 1.6 1.6
E0 (volts): M3ϩ(aq) ϩ 3eϪ 88n M(s) (Ϫ0.90)* Ϫ1.66 Ϫ0.53 Ϫ0.34 0.916
Oxidation states Ϫ3 to ϩ3 ϩ3 ϩ1, ϩ3 ϩ1, ϩ3 ϩ1, ϩ3
Ionization energies (kJ/mol)
M(g) 88n Mϩ(g) ϩ eϪ 801 578 579 558 589
Mϩ(g) 88n M2ϩ(g) ϩ eϪ 2427 1817 1971 1813 1961
M2ϩ(g) 88n M3ϩ(g) ϩ eϪ 3660 2745 2952 2692 2867
⌬H0
hydration (kJ/mol): M3ϩ(g) ϩ xH2O 88n M3ϩ(aq) — Ϫ4750 Ϫ4703 Ϫ4159 Ϫ4117
*For the covalent ϩ3 oxidation state.
Passive toward means “does not react
with.”
Aluminum wiring in a house.
The very negative enthalpy of formation of aluminum oxide makes Al a very strong
reducing agent for other metal oxides. The thermite reaction is a spectacular example
(Figure 23-1). It generates enough heat to produce molten iron for welding steel.
2Al(s) ϩ Fe2O3(s) 88n 2Fe(s) ϩ Al2O3(s) ⌬H0 ϭ Ϫ852 kJ/mol
Anhydrous Al2O3 occurs naturally as the extremely hard, high-melting mineral
corundum, which has a network structure. It is colorless when pure, but becomes colored
when transition metal ions replace a few Al3ϩ ions in the crystal. Sapphire is usually blue
and contains some iron and titanium. Ruby is red due to the presence of small amounts
of chromium.
Other Group IIIA Metals
Gallium is unusual in that it melts when held in the hand. It has the largest liquid state
temperature range of any element (29.8° to 2403°C). It is used in transistors and hightemperature
thermometers. Gallium-67 was one of the first artificially produced isotopes
to be used in medicine. It concentrates in inflamed areas and in certain melanomas.
Indium is a soft, bluish metal that is used in some alloys with silver and lead to make
good heat conductors. Most indium is used in electronics. Thallium is a soft, heavy metal
that resembles lead. It is quite toxic and has no important practical uses as a free metal.
23-7 Group IIIA: Periodic Trends 933
(a) (b) (c)
Figure 23-1 The thermite reaction. A mixture of Fe2O3 and aluminum powder was placed
in a clay pot with a piece of magnesium ribbon as a fuse. (a) The reaction was initiated by
lighting the magnesium fuse. (b) So much heat was produced by the reaction that the iron
melted as it was produced. (c) The molten iron dropped out of the clay pot and burned
through a sheet of iron that was placed under the pot.
Small amounts of different transition
metal ions give different colors to
sapphire, which is mostly aluminum
oxide, Al2O3.
Gallium metal melts below body temperature. ᭤
Periodic Trends
The atomic radii do not increase regularly as Group IIIA is descended (Table 6-1). The
atomic radius of Ga, 1.35 Å, is less than that of Al, 1.43 Å, which is directly above Ga.
The transition elements are located between calcium (IIA) and gallium (IIIA), strontium
(IIA) and indium (IIIA), and barium (IIA) and thallium (IIIA). The increase in nuclear
charge that accompanies filling of the (n Ϫ 1)d subshell results in the contraction of the
size of the atoms. This contraction is caused by the stronger attraction of the more highly
charged nuclei for the outer electrons. This causes the radii of Ga, In, and Tl to be smaller
934 CHAPTER 23: Metals II: Properties and Reactions
The Development of ScienceCCHEMISTRY IN USE
The Most Valuable Metal in the World
Imagine paying $8 million for one pound of aluminum!
Although aluminum currently costs less than $1.00 a pound,
it was considered the most valuable metal in 1827. Aluminum
was so cherished by royalty in the early to mid-1800s that
they alone ate with aluminum spoons and forks while their
lower-class guests dined with cheaper gold and silver service.
Aluminum is the most abundant metal in the earth’s crust
(7.5%); why was it originally so expensive?
Aluminum was first prepared by the following displacement
reaction.
AlCl3 ϩ 3K 88n Al ϩ 3KCl
Potassium was also expensive because it was made by passing
an electric current (from a voltaic cell) through molten
KCl. In addition to the great cost of energy required to melt
large quantities of KCl, copper and zinc (used in voltaic cells)
were also expensive metals in the early 1800s. Thus, the very
small amount of aluminum produced by this displacement
reaction was extremely expensive.
It was not practical to produce aluminum by passing an
electric current through molten Al2O3 because it has a high
melting point, 2000°C. This high temperature is difficult to
achieve and maintain; the components of most voltaic cells
melt below this temperature. Zinc melts at 420°C and copper
at 1083°C.
The cost of aluminum began to drop as the result of two
major advances in the late 1800s. The first came with the
invention of the electric generator, which could produce electricity
using steam or water. Electricity generated by steam
or water was quite inexpensive compared with electricity generated
by voltaic cells. Despite this cost reduction, aluminum
still cost more than $100,000 a pound. The second advance
took place in 1886, when chemists discovered that they could
lower the melting point of aluminum oxide by mixing it with
complex salts, such as Na3[AlF6]. Since 1886, the price of aluminum
has decreased markedly because of lower electricity
costs, improved production techniques, and recycling of discarded
aluminum products.
Although aluminum is no longer used in table services by
royalty, it is of inestimable value in energy conservation.
Around our homes we find energy-saving items such as aluminum
storm doors and windows, insulation backed with
aluminum foil, and aluminum siding. Because vehicle weight
significantly affects gas mileage, substituting aluminum for
heavier metals in cars, trucks, trains, and aircraft helps preserve
our petroleum supplies. Although the cost of aluminum
has decreased drastically, it is still a valuable metal because
of its ability to help us conserve energy and to improve our
standard of living at the same time.
Ronald DeLorenzo
Middle Georgia College
than would be predicted from the radii of B and Al. Atomic radii strongly influence other
properties. For example, Ga, In, and Tl are each much denser than the elements above
them due to their unusually small atomic radii.
The Group IIIA elements have the ns2np1 outer electron configuration. Aluminum
shows only the ϩ3 oxidation state in its compounds. The heavier metals (Ga, In, Tl) can
lose or share either the single p valence electron or the p and both s electrons to exhibit
the ϩ1 or ϩ3 oxidation state, respectively. In general the post-transition metals can exhibit
oxidation states of ( g Ϫ 2)ϩ and gϩ, where g ϭ periodic group number. As examples, TlCl
and TlCl3 both exist, as do SnCl2 and SnCl4. The stability of the lower state increases as
the groups are descended. This is called the inert s-pair effect because the two s electrons
remain nonionized, or unshared, for the ( g Ϫ 2)ϩ oxidation state. To illustrate, AlCl3
exists but not AlCl; TlCl3 is less stable than TlCl.
THE d-TRANSITION METALS
The term “transition elements” denotes elements in the middle of the periodic table. They
provide a transition between the “base formers” on the left and the “acid formers” on the
right. The term applies to both the d- and f-transition elements (d and f atomic orbitals
are being filled across this part of the periodic table). All are metals. We commonly use
the term “transition metals” to refer to the d-transition metals.
23-7 Group IIIA: Periodic Trends 935
The d-transition metals are located between Groups IIA and IIIA in the periodic table.
Strictly speaking, a d-transition metal must have a partially filled set of d orbitals. Zinc,
cadmium, and mercury (Group IIB) and their cations have completely filled sets of d
orbitals, but they are often discussed with d-transition metals because their properties are
similar. All of the other elements in this region have partially filled sets of d orbitals, except
the IB elements and palladium, which have completely filled sets. Some of the cations of
these latter elements have partially filled sets of d orbitals.
Sc Ti V Cr Mn Fe Co Ni Cu Zn
Y Zr Nb Mo Tc Ru Rh Pd Ag Cd
La Hf Ta W Re Os Ir Pt Au Hg
Ac Rf Db Sg Bh
IIIB IVB VB VIB VIIB VIIIB IB IIB
Hs Mt
Compare the radii and densities of
these elements with those of Group IA
and IIA metals in the same rows.
As is generally true, for each pair
of compounds, covalent character
is greater for the higher (more
polarizing) oxidation state of the metal.
Oxides of most nonmetals are acidic,
and oxides of most metals are basic
(except those having high oxidation
states).
GENERAL PROPERTIES
The following are general properties of transition elements.
1. All are metals.
2. Most are harder and more brittle and have higher melting points, boiling points,
and heats of vaporization than nontransition metals.
3. Their ions and their compounds are usually colored.
4. They form many complex ions (Chapter 25).
5. With few exceptions, they exhibit multiple oxidation states.
6. Many of them are paramagnetic, as are many of their compounds.
7. Many of the metals and their compounds are effective catalysts.
Some properties of 3d-transition metals are listed in Table 23-6.
OXIDATION STATES
Most transition metals exhibit more than one nonzero oxidation state. The maximum
oxidation state is given by a metal’s group number, but this is often not its most stable
oxidation state (Table 23-7).
The outer s electrons lie outside the d electrons and are always the first ones lost in
ionization. In the first transition series, scandium and zinc exhibit just one nonzero oxidation
state. Scandium loses its two 4s electrons and its only 3d electron to form Sc3ϩ. Zinc
loses its two 4s electrons to form Zn2ϩ.
3d 4s
Ϫ3eϪ
21Sc [Ar] __h __ __ __ __ __hg 888n 21Sc3ϩ [Ar]
Ϫ2eϪ
30Zn [Ar] __hg __hg __hg __hg __hg __hg 888n 30Zn2ϩ [Ar] 3d10
23-9
23-8
936 CHAPTER 23: Metals II: Properties and Reactions
Many transition metal ions are
highly colored. These flasks contain
(left to right), aqueous solutions of
Fe(NO3)3, Co(NO3)2, Ni(NO3)2,
Cu(NO3)2, and Zn(NO3)2. Colorless
Zn2ϩ ions differ from the others by
having completely filled 3d orbitals.
In the “building” of electron
configurations by the Aufbau Principle,
the outer s orbitals are occupied before
the inner d orbitals (Section 5-17).
TABLE 23-6 Properties of Metals in the First Transition Series
Properties Sc Ti V Cr Mn Fe Co Ni Cu Zn
Melting point (°C) Ϫ1541 Ϫ1660 Ϫ1890 Ϫ1850 Ϫ1244 Ϫ1535 Ϫ1495 Ϫ1453 Ϫ1083 Ϫ420
Boiling point (°C) Ϫ2831 Ϫ3287 Ϫ3380 Ϫ2672 Ϫ1962 Ϫ2750 Ϫ2870 Ϫ2732 Ϫ2567 Ϫ907
Density (g/cm3) Ϫ2.99 Ϫ4.54 Ϫ6.11 Ϫ7.18 Ϫ7.21 Ϫ7.87 Ϫ8.9 Ϫ8.91 Ϫ8.96 Ϫ7.13
Atomic radius (Å) Ϫ1.62 Ϫ1.47 Ϫ1.34 Ϫ1.25 Ϫ1.29 Ϫ1.26 Ϫ1.25 Ϫ1.24 Ϫ1.28 Ϫ1.34
Ionic radius, M2ϩ (Å) Ϫ— Ϫ0.94 Ϫ0.88 Ϫ0.89 Ϫ0.80 Ϫ0.74 Ϫ0.72 Ϫ0.69 Ϫ0.70 Ϫ0.74
Electronegativity Ϫ1.3 Ϫ1.4 Ϫ1.5 Ϫ1.6 Ϫ1.6 Ϫ1.7 Ϫ1.8 Ϫ1.8 Ϫ1.8 Ϫ1.6
E0 (V) for M2ϩ(aq) ϩ 2eϪ 88n M(s) Ϫ2.08* Ϫ1.63 Ϫ1.2 Ϫ0.91 Ϫ1.18 Ϫ0.44 Ϫ0.28 Ϫ0.25 ϩ0.34 Ϫ0.76
IE (kJ/mol) first Ϫ631 Ϫ658 Ϫ650 Ϫ652 Ϫ717 Ϫ759 Ϫ758 Ϫ757 Ϫ745 Ϫ906
IE (kJ/mol) second Ϫ1235 Ϫ1310 Ϫ1414 Ϫ1592 Ϫ1509 Ϫ1561 Ϫ1646 Ϫ1753 Ϫ1958 Ϫ1733
*For Sc3ϩ(aq) ϩ 3eϪ n Sc(s).
Titanium and nickel also have only a single nonzero oxidation state. All of the other
3d-transition metals exhibit at least two oxidation states in their compounds. For example,
cobalt can form Co2ϩ and Co3ϩ ions.
3d 4s 3d 4s
Ϫ2eϪ
27Co [Ar] __hg __hg __h __h __h __hg 888n 27Co2ϩ [Ar] __hg __hg __h __h __h __
Ϫ3eϪ
27Co [Ar] __hg __hg __h __h __h __hg 888n 27Co3ϩ [Ar] __hg __h __h __h __h __
The most common oxidation states of the 3d-transition elements are ϩ2 and ϩ3. The
elements in the middle of each series exhibit more oxidation states than those to the left
or right. As one moves down a group, higher oxidation states become more stable and
more common (opposite to the trend for representative elements). This is because the d
electrons are more effectively shielded from the nucleus as the group is descended and
are therefore more easily lost or more readily available for sharing. For example, cobalt
commonly exhibits the ϩ2 and ϩ3 oxidation states. Rh and Ir are just below Co. Their
common oxidation states are ϩ3 and ϩ4. The ϩ4 state is slightly more stable for Ir than
for the lighter Rh.
The oxides and hydroxides of lower oxidation states of a given transition metal are
basic. Those containing intermediate oxidation states tend to be amphoteric, and those
containing high oxidation states tend to be acidic. This is illustrated for the oxides and
hydroxides of Cr in the next section.
CHROMIUM OXIDES, OXYANIONS, AND HYDROXIDES
Typical of the metals near the middle of a transition series, chromium shows several oxidation
states. The most common are ϩ2, ϩ3, and ϩ6 (Table 23-8).
Oxidation–Reduction
The most stable oxidation state of Cr is ϩ3. Solutions of blue chromium(II) salts are easily
air-oxidized to chromium(III).
Cr3ϩ ϩ eϪ 88n Cr2ϩ E0 ϭ Ϫ0.41 V
23-10
TABLE 23-7 Nonzero Oxidation States of the 3d-Transition Metals*
IIIB IVB VB VIB VIIB VIIIB IB IIB
Sc Ti V Cr Mn Fe Co Ni Cu Zn
ϩ1 r
ϩ2 ϩ2 r ϩ2 ϩ2 ϩ2 ϩ2
ϩ3 ϩ3 r ϩ3 ϩ3 ϩ3 o
ϩ4 ϩ4 ϩ4 o
ϩ5 o
ϩ6 o
ϩ7 o
*Abbreviations: o ϭ oxidizing agent; r ϭ reducing agent.













Pentaamminechlorocobalt(III)
chloride, [Co(NH3)5(Cl)]Cl2, is a
compound that contains cobalt in
the ϩ3 oxidation state (left).
Hexaaquacobalt(II) chloride,
[Co(OH2)6]Cl2, contains cobalt in
the ϩ2 oxidation state (right).
Aqueous solutions of some
compounds that contain chromium.
(Left to right) Chromium(II) chloride
(CrCl2) is blue; chromium(III)
chloride (CrCl3) is green; potassium
chromate (K2CrO4) is yellow;
potassium dichromate (K2Cr2O7) is
orange.
Chromium(VI) species are oxidizing agents. Basic solutions containing chromate ions,
CrO4
2Ϫ, are weakly oxidizing. Acidification produces the dichromate ion, Cr2O7
2Ϫ, and
chromium(VI) oxide, both powerful oxidizing agents.
Cr2O7
2Ϫ ϩ 14Hϩ ϩ 6eϪ 88n 2Cr3ϩ ϩ 7H2O E0 ϭ ϩ1.33 V
Chromate–Dichromate Equilibrium
Red chromium(VI) oxide, CrO3, is the acid anhydride of two acids: chromic acid, H2CrO4,
and dichromic acid, H2Cr2O7. Neither acid has been isolated in pure form, although chromate
and dichromate salts are common. CrO3 reacts with H2O to produce strongly acidic
solutions containing hydrogen ions and (predominantly) dichromate ions.
2CrO3 ϩ H2O 88n [2Hϩ ϩ Cr2O7
2Ϫ] dichromic acid (red-orange)
From such solutions orange dichromate salts can be crystallized after adding a stoichiometric
amount of base. Addition of excess base produces yellow solutions from which
only yellow chromate salts can be obtained. The two anions exist in solution in a pHdependent
equilibrium.
ϩ6 ϩ6
2CrO4
2Ϫ ϩ 2Hϩ 34 Cr2O7
2Ϫ ϩ H2O Kc ϭ ϭ 4.2 ϫ 1014
yellow orange
[Cr2O7
2Ϫ]
ᎏᎏ
[CrO4
2Ϫ]2[Hϩ]2
Adding a strong acid to a solution that contains CrO4
2Ϫ/Cr2O7
2Ϫ ions favors the reaction
to the right and increases [Cr2O7
2Ϫ]. Adding a base favors the reaction to the left
and increases [CrO4
2Ϫ].
Cr
O
O
O
O
Cr
O
O
O
O Cr
O
O
O
dichromate ion, Cr2
O7
2Ϫ
chromate ion, CrO4
2Ϫ
2Ϫ
2Ϫ
TABLE 23-8 Some Compounds of Chromium
Ox. State Oxide Hydroxide Name Acidic/Basic Related Salt Name
ϩ2 CrO Cr(OH)2 chromium(II) CrCl2 chromium(II)
basic
black hydroxide anhydr. colorless chloride
aq. lt. blue
ϩ3 Cr2O3 Cr(OH)3 chromium(III) CrCl3 chromium(II)
amphoteric
green hydroxide anhydr. violet chloride
aq. green
KCrO2 potassium
green chromite
ϩ6 CrO3 H2CrO4 or chromic acid K2CrO4 potassium
weakly acidic
dk. red [CrO2(OH)2] yellow chromate
H2Cr2O7 or dichromic acid K2Cr2O7 potassium
acidic
[Cr2O5(OH)2] orange dichromate
Chromate ion, CrO4
2–
2–
Dichromate ion, Cr2O7
2–
2–
Dehydration of chromate or dichromate salts with concentrated H2SO4 produces CrO3.
Chromium(VI) oxide is a strong oxidizing agent. A powerful “cleaning solution” once used
for removing greasy stains and coatings from laboratory glassware was made by adding
concentrated H2SO4 to a concentrated solution of K2Cr2O7. The active ingredients are
CrO3, an oxidizing agent, and H2SO4, an excellent solvent.
Chromium(III) hydroxide is amphoteric (Section 10-6).
Cr(OH)3(s) ϩ 3Hϩ 88n Cr3ϩ ϩ 3H2O (rxn. with acids)
Cr(OH)3(s) ϩ OHϪ 88n Cr(OH)4
Ϫ (rxn. with bases)
Exercises 939
Key Terms
Alkali metals Group IA metals.
Alkaline earth metals Group IIA metals.
d-Transition metals Metals that have partially filled sets of d
orbitals; the B groups of the periodic table.
Diagonal similarities Chemical similarities of elements of
Period 2 to elements of Period 3 one group to the right; especially
evident toward the left of the periodic table.
Inert s-pair effect The tendency of the two outermost s electrons
to remain nonionized or unshared in compounds;
characteristic of the post-transition metals.
Post-transition metals Representative metals in the “p block.”
Representative metals Metals in the A groups in the periodic
table; their outermost electrons are in s and p orbitals.
Exercises
*01. How do the acidities or basicities of metal oxides vary with
oxidation numbers of the same metal?
*02. Discuss the general differences in electron configurations
of representative elements and d-transition metals.
*03. Compare the extents to which the properties of successive
elements across the periodic table differ for representative
elements and d-transition metals. Explain.
*04. Compare the metals and nonmetals with respect to
(a) number of outer-shell electrons, (b) electronegativities,
(c) standard reduction potentials, and (d) ionization
energies.
*05. How do the physical properties of metals differ from those
of nonmetals?
*06. Define “malleable” and “ductile.”
*07. (a) What is the meaning of the statement, “alkali metals
are corrosive?” (b) Are alkali metal ions corrosive? (c) Are
alkali metal ions an important part of your diet?
*08. Compare the alkali metals with the alkaline earth metals
with respect to (a) atomic radii, (b) densities, (c) first ionization
energies, and (d) second ionization energies.
Explain the comparisons.
*09. Summarize the chemical and physical properties of: (a) the
alkali metals and (b) the alkaline earth metals.
*10. (a) Compare and contrast the physical and chemical properties
of the Group IA metals with those of the Group IIA
metals. (b) Compare the periodic trends of the two groups.
*11. Write the general outer-electron configurations for atoms
of the IA and IIA metals. What oxidation state(s) would
you predict for these elements? What types of bonding
would you expect in most of the compounds of these
elements? Why?
*12. Write electron configurations (__hg notation) for (a) Mg,
(b) Mg2ϩ, (c) Na, (d) Naϩ, (e) Sn, (f ) Sn2ϩ, and (g) Sn4ϩ.
*13. Write electron configurations (__hg notation) for (a) K,
(b) Kϩ, (c) Sr, (d) Sr2ϩ, (e) Al, (f) Al3ϩ, and (g) Ga3ϩ.
*14. Are the elements in Groups IA and IIA found in the free
state in nature? What are the primary sources for these
elements?
*15. Describe some uses for (a) lithium and its compounds and
(b) sodium and its compounds.
*16. Where do the metals of Groups IA and IIA fall with respect
to H2 in the activity series? What does this tell us about
their reactivities with water and acids?
*17. Write chemical equations describing the reactions of O2
with each of the alkali and alkaline earth metals. Account
for differences within each family.
*18. Describe some uses for (a) calcium and its compounds and
(b) magnesium and its compounds.
*19. Write general equations for reactions of alkali metals with
(a) hydrogen, (b) sulfur, and (c) ammonia. Represent the
metal as M.
*20. Write general equations for reactions of alkali metals with
(a) water, (b) phosphorus, and (c) halogens. Represent the
metal as M and the halogen as X.
This cleaning solution must be used
with great caution because it is a
strong oxidizing agent and is
carcinogenic.
940 CHAPTER 23: Metals II: Properties and Reactions
*21. Write general equations for reactions of alkaline earth
metals with (a) hydrogen, (b) sulfur, and (c) ammonia. Represent
the metal as M.
*22. Write general equations for reactions of alkaline earth
metals with (a) water, (b) phosphorus, and (c) chlorine.
Represent the metal as M.
*23. What is meant by the term “diagonal relationships”?
*24. Give some illustrations of diagonal relationships in the
periodic table, and explain each.
*25. What is hydration energy? How does it vary for cations of
the alkali metals?
*26. How do hydration energies vary for cations of the alkaline
earth metals?
*27. How do the standard reduction potentials of the alkali
metal cations vary? Why?
*28. How do the standard reduction potentials of the alkaline
earth metal cations vary? Why?
*29. Why are the standard reduction potentials of lithium and
beryllium out of line with respect to group trends?
*30. Calculate ⌬H0 values at 25°C for the reactions of 1 mol of
each of the following metals with stoichiometric quantities
of water to form metal hydroxides and hydrogen. (a) Li,
(b) K, and (c) Ca. Rationalize the differences in these val-
ues.
*31. How are the d-transition metals distinguished from other
elements?
*32. What are the general properties of the d-transition met-
als?
*33. Why are trends in variations of properties of successive
d-transition metals less regular than trends among successive
representative elements?
*34. Write out the electron configurations for the following
species: (a) V; (b) Fe; (c) Cu; (d) Zn2ϩ; (e) Cr3ϩ; (f) Ni2ϩ;
(g) Ag; (h) Agϩ.
*35. Why do copper and chromium atoms have “unexpected”
electron configurations?
*36. Discuss the similarities and differences among elements of
corresponding A and B groups of the periodic table, IIIA
and IIIB for example.
*37. Copper exists in the ϩ1, ϩ2, and ϩ3 oxidation states.
Which is the most stable? Which would be expected to be
a strong oxidizing agent and which would be expected to
be a strong reducing agent?
*38. For a given transition metal in different oxidation states,
how does the acidic character of its oxides increase? How
do ionic and covalent character vary? Characterize a series
of metal oxides as examples.
*39. For different transition metals in the same oxidation state
in the same group (vertical column) of the periodic table,
how do covalent character and acidic character of their
oxides vary? Why? Cite evidence for the trends.
*40. Chromium(VI) oxide is the acid anhydride of which two
acids? Write their formulas. What is the oxidation state of
the chromium in these acids?
CONCEPTUAL EXERCISES
*41. Some of the oxides of the transition metals are amphoteric.
(a) What is meant by the term “amphoteric?” (b) Identify
five oxides that are amphoteric, five oxides that are basic,
and five oxides that are acidic.
*42. Calcium carbonate is a very common compound on this
planet. Name five common items that are composed primarily
of calcium carbonate.
*43. Magnesium and some of the magnesium alloys will burn
if heated to a high enough temperature. Some magnesium
alloys are used in construction and may have to be welded
at a high temperature. Suggest techniques or requirements
for welding these alloys so that they do not ignite.
*44. Some mental disorders are treated with medications that
contain lithium salts. Why is pure lithium unsuitable for
such purposes?
*45. Gallium thermometers are high-temperature thermometers.
(a) Why can’t they be used near room temperature?
(b) Propose a reason why mercury thermometers cannot
be used to measure temperatures in the upper range of gallium
thermometers.
BUILDING YOUR KNOWLEDGE
*46. How many grams of Co3O4 (a mixed oxide, CoOиCo2O3)
must react with excess aluminum to produce 190. g of
metallic cobalt, assuming 72.5% yield?
heat
3Co3O4 ϩ 8Al 888n 9Co ϩ 4Al2O3
*47. Calculate ⌬H0, ⌬S0, and ⌬G0 for the reaction of 1 mol of
Na with water to form aqueous NaOH and hydrogen.
*48. Calculate ⌬H0, ⌬S0, and ⌬G0 for the reaction of 1 mol of
Rb with water to form aqueous RbOH and hydrogen.
Compare the spontaneity of this reaction with that in Exercise
47.
*49. What is the ratio of [Cr2O7
2Ϫ] to [CrO4
2Ϫ] at 25°C in a
solution prepared by dissolving 1.5 ϫ 10Ϫ3 mol of sodium
chromate, Na2CrO4, in enough of an aqueous solution
buffered at pH ϭ 2.00 to produce 250. mL of solution?
Pure lithium.
24Some Nonmetals
and Metalloids
OO
nly about 20% of the elements are classified as nonmetals. With the exception
of H, they are in the upper right-hand corner of the periodic table. In this chapter
we shall consider the chemistry and properties of nonmetals and metalloids that
illustrate group trends and individuality of elements within groups of nonmetals.
OUTLINE
The Noble Gases (Group VIIIA)
24-1 Occurrence, Uses, and
Properties
24-2 Xenon Compounds
The Halogens (Group VIIA)
24-3 Properties
24-4 Occurrence, Production, and
Uses
24-5 Reactions of the Free Halogens
24-6 The Hydrogen Halides and
Hydrohalic Acids
24-7 The Oxoacids (Ternary Acids)
of the Halogens
Sulfur, Selenium, and Tellurium
24-8 Occurrence, Properties, and
Uses
24-9 Reactions of Group VIA
Elements
24-10 Hydrides of Group VIA
Elements
24-11 Group VIA Oxides
24-12 Oxoacids of Sulfur
Nitrogen and Phosphorus
24-13 Occurrence of Nitrogen
24-14 Hydrogen Compounds of
Nitrogen
24-15 Nitrogen Oxides
24-16 Some Oxoacids of Nitrogen
and Their Salts
24-17 Phosphorus
Silicon
24-18 Silicon and the Silicates
OBJECTIVES
After you have studied this chapter, you should be able to
• Describe the occurrence and use of the noble gases
• Describe compounds of the noble gases
• Describe the occurrence and production of the halogens
• Describe some important reactions and compounds of the halogens
• Describe the occurrence and production of sulfur, selenium, and tellurium
• Describe some important reactions and compounds of the heavier Group VIA
nonmetals
• Describe the occurrence and production of nitrogen and phosphorus
• Describe some important reactions of nitrogen and phosphorus
• Describe the occurrence and importance of silicon
• Describe a few important compounds of silicon
Recently mined sulfur awaits
shipment.
24-2 Xenon Compounds 943
THE NOBLE GASES (GROUP VIIIA)
OCCURRENCE, USES, AND PROPERTIES
The noble gases are very low-boiling gases. Except for radon, they can be isolated by fractional
distillation of liquefied air. Radon is collected from the radioactive disintegration
of radium salts. Table 24-1 gives the percentage of each noble gas in the atmosphere.
Helium is produced in the United States from some natural gas fields. This source was
discovered in 1905 by H. P. Cady and D. F. McFarland at the University of Kansas, when
they were asked to analyze a nonflammable component of natural gas from a Kansas gas
well. Uses of the noble gases are summarized in Table 24-2.
The noble gases are colorless, tasteless, and odorless. In the liquid and solid states the
only forces of attraction among the atoms are very weak dispersion forces. Polarizability
and interatomic interactions increase with increasing atomic size, and so melting and
boiling points increase with increasing atomic number. The attractive forces among He
atoms are so small that He remains liquid at 1 atmosphere pressure even at a temperature
of 0.001 K.
XENON COMPOUNDS
Until the early 1960s, chemists believed that the Group VIIIA elements would not combine
chemically with any elements. In 1962, Neil Bartlett (1932– ) and his research group at
the University of British Columbia were studying the powerful oxidizing agent PtF6. They
24-2
24-1
Br Kr
I Xe
At Rn
Cl Ar
F Ne
He
VIIA
PSi S
Se
Te
N
VIIIA
Radon is continually produced in small
amounts in the uranium radioactive
decay sequence (Section 26-11). Radon
gas is so unreactive that it eventually
escapes from the soil. Measurable
concentrations of radon, a radioactive
gas, have been observed in basements
of many dwellings.
A pressure of about 26 atmospheres is
required to solidify He at 0.001 K.
The noble gases are often called the
rare gases. They were formerly called
the “inert gases” because it was
incorrectly thought that they could not
enter into chemical combination.
TABLE 24-1 Percentages (by volume) of Noble Gases in the Atmosphere
He Ne Ar Kr Xe Rn
0.0005% 0.015% 0.94% 0.00011% 0.000009% Ϸ0%
accidentally prepared and identified O2
ϩPtF6
Ϫ by reaction of oxygen with PtF6. Bartlett
reasoned that xenon also should be oxidized by PtF6 because the first ionization energy
of O2 (1.31 ϫ 103 kJ/mol) is slightly larger than that of xenon (1.17 ϫ 103 kJ/mol). He
obtained a red crystalline solid initially believed to be XeϩPtF6
Ϫ but now known to be a
more complex compound.
Since Bartlett’s discovery, many other noble gas compounds have been made. All involve
very electronegative elements. Most are compounds of Xe, and the best characterized
compounds are xenon fluorides. Oxygen compounds are also well known. Reaction of Xe
with F2, an extremely strong oxidizing agent, in different stoichiometric ratios produces
xenon difluoride, XeF2; xenon tetrafluoride, XeF4; and xenon hexafluoride, XeF6, all colorless
crystals (Table 24-3).
All the xenon fluorides are formed in exothermic reactions. They are reasonably stable,
with XeXF bond energies of about 125 kJ/mol of bonds. For comparison, strong covalent
bond energies range from about 170 to 500 kJ/mol, whereas bond energies of
hydrogen bonds are typically less than 40 kJ/mol.
THE HALOGENS (GROUP VIIA)
The elements of Group VIIA are known as halogens (Greek, “salt formers”). The term
“halides” is used to describe their binary compounds. The heaviest halogen, astatine, is
an artificially produced element of which only short-lived radioactive isotopes are known.
944 CHAPTER 24: Some Nonmetals and Metalloids
TABLE 24-2 Uses of the Noble Gases
Noble Gas Use Useful Properties or Reasons
helium 1. Filling of observation balloons and other lighter- Nonflammable; 93% of lifting power of flammable
2. than-air craft hydrogen
2. He/O2 mixtures, rather than N2/O2, for deep-sea Low solubility in blood; prevents nitrogen narcosis
2. breathing and “bends”
3. Diluent for gaseous anesthetics Nonflammable, nonreactive
4. He/O2 mixtures for respiratory patients Low density, flows easily through restricted passages
5. Heat transfer medium for nuclear reactors Transfers heat readily; does not become radioactive;
chemically inert
6. Industrial applications, such as inert atmosphere Chemically inert
2. for welding easily oxidized metals
7. Liquid He used to maintain very low temperatures Extremely low boiling point
2. in research (cryogenics)
neon Neon signs Even at low Ne pressure, moderate electric current
causes bright orange-red glow; can be modified by
colored glass or mixing with Ar or Hg vapor
argon 1. Inert atmosphere for welding Chemically inert
2. Filling incandescent light bulbs Inert; inhibits vaporization of tungsten filament
and blackening of bulbs
krypton Airport runway and approach lights Gives longer life to incandescent lights than Ar, but
more expensive
xenon Xe and Kr mixture in high-intensity, short-exposure Both have fast response to electric current
photographic flash tubes
radon Radiotherapy of cancerous tissues Radioactive
Oxygen is second only to fluorine in
electronegativity.
Crystals of the noble gas compound
xenon tetrafluoride, XeF4.
PROPERTIES
The elemental halogens exist as diatomic molecules containing single covalent bonds.
Properties of the halogens show obvious trends (Table 24-4). Their high electronegativities
indicate that they attract electrons strongly. Many binary compounds that contain a
metal and a halogen are ionic.
The small fluoride ion (radius ϭ 1.19 Å) is not easily polarized (distorted) by cations,
whereas the large iodide ion (radius ϭ 2.06 Å) is. As a result, compounds containing IϪ
ions show greater covalent character than those containing FϪ ions. The properties of
ClϪ and BrϪ ions are intermediate between those of FϪ and IϪ.
The chemical properties of the halogens resemble one another more closely than do
those of elements in any other periodic group, with the exception of the noble gases and
possibly the Group IA metals. But their physical properties differ significantly. Melting
and boiling points of the halogens increase from F2 to I2. This follows their increase in
size and increase in ease of polarization of outer-shell electrons by adjacent nuclei, resulting
in greater intermolecular attractive (dispersion) forces. All halogens except astatine are
decidedly nonmetallic. They show the Ϫ1 oxidation number in most of their compounds.
Except for fluorine, they also exhibit oxidation numbers of ϩ1, ϩ3, ϩ5, and ϩ7.
24-3
24-3 Properties 945
TABLE 24-3 Xenon Fluorides
Preparation Reaction eϪ Pairs Xe Hybridization,
Compound (Molar ratio XeϺF2) Conditions Around Xe Molecular Geometry*
XeF2 1Ϻ1–3 400°C or irradiation 5 sp3d, linear
or elec. discharge
XeF4 1Ϻ5 Same as for XeF2 6 sp3d2, square planar
XeF6 1Ϻ20 300°C and 60 atm 7 sp3d3(?), exact geometry
or elec. discharge undetermined
*See Tables 8-3, 8-4.
TABLE 24-4 Properties of the Halogens
Property F Cl Br I At
Physical state (25°C, 1 atm) gas gas liquid solid solid
Color pale yellow yellow-green red-brown violet (g); black (s) —
Atomic radius (Å) 0.72 1.00 1.14 1.33 1.40
Ionic radius (XϪ) (Å) 1.19 1.67 1.82 2.06 —
Outer shell eϪ 2s22p5 3s23p5 4s24p5 5s25p5 6s26p5
First ionization energy (kJ/mol) 1681 1251 1140 1008 890
Electronegativity 4.0 3.0 2.8 2.5 2.1
Melting point (°C, 1 atm) Ϫ220 Ϫ101 Ϫ7.1 114 —
Boiling point (°C, 1 atm) Ϫ188 Ϫ35 59 184 —
XXX bond energy (kJ/mol) 155 242 193 151 —
OCCURRENCE, PRODUCTION, AND USES
The halogens are so reactive that they do not occur free in nature. The most abundant
sources of halogens are halide salts. A primary source of iodine is NaIO3. The halogens
are obtained by oxidation of the halide ions.
2XϪ 88n X2 ϩ 2eϪ
The order of increasing ease of oxidation is FϪ Ͻ ClϪ Ͻ BrϪ Ͻ IϪ Ͻ AtϪ.
Fluorine
Fluorine occurs in large quantities in the minerals fluorspar or fluorite, CaF2; cryolite,
Na3AlF6, and fluorapatite, Ca5(PO4)3F. It also occurs in small amounts in sea water, teeth,
bones, and blood. F2 is such a strong oxidizing agent that it has not been produced by
direct chemical oxidation of FϪ ions. The pale yellow gas is prepared by electrolysis of a
molten mixture of KF ϩ HF, or KHF2, in a Monel metal cell. This must be done under
anhydrous conditions because H2O is more readily oxidized than FϪ.
electrolysis
2KHF2 88888888n F2(g) ϩ H2(g) ϩ 2KF(s)
melt
In 1986, Carl O. Christe discovered that the strong Lewis acid SbF5 displaces the
weaker Lewis acid MnF4 from the hexafluoromanganate(IV) ion, [MnF6]2Ϫ. MnF4 is thermodynamically
unstable, and it decomposes into MnF3 and F2. Christe heated a mixture
of potassium hexafluoromanganate(IV) and antimony(V) fluoride in a passivated Teflon–
stainless steel container at 150°C for 1 hour. He obtained elemental F2 in better than 40%
yield. The overall reaction may be represented as
2K2MnF6(s) ϩ 4SbF5(ᐉ) 88n 4KSbF6(s) ϩ 2MnF3(s) ϩ F2(g)
Fluorine is used as a fluorinating agent. Many fluorinated organic compounds are stable
and nonflammable. They are used as refrigerants, lubricants, plastics (such as Teflon),
insecticides, and, until recently, aerosol propellants. Stannous fluoride, SnF2, is used in
toothpaste.
Chlorine
Chlorine (Greek chloros, “green”) occurs in abundance in NaCl, KCl, MgCl2, and CaCl2
in salt water and in salt beds. It is also present as HCl in gastric juices. The toxic, yellowish
green gas is prepared commercially by electrolysis of concentrated aqueous NaCl, in which
24-4
946 CHAPTER 24: Some Nonmetals and Metalloids
ϩ1 ϩ4 ϩ5 Ϫ1Ϫ1 ϩ1 ϩ5 ϩ3 Ϫ1 0Ϫ1
X
At2
Lewis dot formula
for a halogen molecule
X
F2 Cl2 Br2 I2
We often represent a halogen atom
as X, without specifying a particular
halogen.
Agϩ
IϪ
The diffuse cloud of the IϪ ion is
easily polarized by the small Agϩ
ion.
Monel metal is an alloy of Ni, Cu,
Al, and Fe. It is resistant to attack by
hydrogen fluoride.
The reactions of F2 with other
elements are dangerous because of the
vigor with which F2 oxidizes other
substances. They must be carried out
with extreme caution!
industrially important H2 and caustic soda (NaOH) are also produced (Section 21-4).
More than 26 billion pounds of chlorine was produced in the United States in 1997.
Chlorine is used to produce many commercially important products. Tremendous
amounts of it are used in extractive metallurgy and in chlorinating hydrocarbons to produce
a variety of compounds (such as polyvinyl chloride, a plastic). Chlorine is present as Cl2,
NaClO, Ca(ClO)2, or Ca(ClO)Cl in household bleaches as well as in bleaches for wood
pulp and textiles. Under carefully controlled conditions, Cl2 is used to kill bacteria in
public water supplies.
Bromine
Bromine (Greek bromos, “stench”) is less abundant than fluorine and chlorine. In the
elemental form it is a dense, freely flowing, corrosive, dark-red liquid with a brownish red
vapor at 25°C. It occurs mainly in NaBr, KBr, MgBr2, and CaBr2 in salt water, underground
salt brines, and salt beds. The major commercial source for bromine is deep brine
wells in Arkansas that contain up to 5000 parts per million (0.5%) of bromide.
Bromine is used in the production of silver bromide for light-sensitive eyeglasses and
photographic film; in the production of sodium bromide, a mild sedative; and in methyl
bromide, CH3Br, a soil fumigant that contributes to the destruction of the ozone layer.
Iodine
Iodine (Greek iodos, “purple”) is a violet-black crystalline solid with a metallic luster. It
exists in equilibrium with a violet vapor at 25°C. The element can be obtained from dried
seaweed or shellfish or from NaIO3 impurities in Chilean nitrate (NaNO3) deposits. It is
contained in the growth-regulating hormone thyroxine, produced by the thyroid gland.
“Iodized” table salt is about 0.02% KI, which helps prevent goiter, a condition in which
the thyroid enlarges. Iodine has been used as an antiseptic and germicide in the form of
tincture of iodine, a solution in alcohol.
The preparation of iodine involves reduction of iodate ion from NaIO3 with sodium
hydrogen sulfite, NaHSO3.
2IO3
Ϫ(aq) ϩ 5HSO3
Ϫ(aq) 88n 3HSO4
Ϫ(aq) ϩ 2SO4
2Ϫ(aq) ϩ H2O(ᐉ) ϩ I2(s)
Iodine is then purified by sublimation (Figure 13-16).
24-4 Occurrence, Production, and Uses 947
Chlorides occur in salt beds.
Bromine is a dark red, volatile liquid.
Iodine reacts with starch (as in this
potato) to form a deep-blue complex
substance.
More recently available is an aqueous
solution of an iodine complex of
polyvinylpyrrolidone, or “povidone.” It
does not sting when applied to open
wounds.
REACTIONS OF THE FREE HALOGENS
The free halogens react with most other elements and many compounds. For example,
all the Group IA metals react with all the halogens to form simple binary ionic compounds
(Section 7-2).
General Reaction Remarks
nX2 ϩ 2M 88n 2MXn All X2 with most metals (most vigorous reaction with
F2 and Group IA metals)
X2 ϩ nX2 88n 2XXЈn Formation of interhalogens (n ϭ 1, 3, 5, or 7); X is
larger than XЈ
X2 ϩ H2 88n 2HX With all X2.
3X2 ϩ 2P 88n 2PX3 With all X2, and with As, Sb, Bi replacing P
5X2 ϩ 2P 88n 2PX5 Not with I2; also Sb 88n SbF5, SbCl5;
As 88n AsF5; Bi 88n BiF5
X2 ϩ H2S 88n S ϩ 2HX With all X2
X2 ϩ 2XϪ 88n 2XЈϪ ϩ X2 F2 88n Cl2, Br2, I2
Cl2 88n Br2, I2
Br2 88n I2
The most vigorous reactions are those of F2, which usually oxidizes other species to
their highest possible oxidation states. Iodine is only a mild oxidizing agent (IϪ is a mild
reducing agent) and usually does not oxidize substances to high oxidation states. Consider
the following reactions of halogens with two metals that exhibit variable oxidation
numbers.
With Fe With Cu
2Fe ϩ 3F2 88n 2FeF3 (only)
2Fe ϩ 3Cl2 (excess) 88n 2FeCl3 Cu ϩ X2 88n CuX2 (X ϭ F, Cl, Br)
Fe ϩ Cl2 (lim. amt.) 88n FeCl2
Fe ϩ I2 88n FeI2 (only) 2Cu ϩ I2 88n 2CuI (only)
Fe3ϩ ϩ IϪ 88n Fe2ϩ ϩ ᎏ1
2
ᎏI2 Cu2ϩ ϩ 2IϪ 88n CuI ϩ ᎏ1
2
ᎏI2
THE HYDROGEN HALIDES AND HYDROHALIC ACIDS
The hydrogen halides are colorless gases that dissolve in water to give acidic solutions
called hydrohalic acids. The gases have piercing, irritating odors. The abnormally high
melting and boiling points of HF are due to its very strong hydrogen bonding (Figure
13-5).
24-6
24-5
948 CHAPTER 24: Some Nonmetals and Metalloids
ϩ3
ϩ3 ϩ2
ϩ1
ϩ1
ϩ2
ϩ2
Bromine reacts with powdered
antimony so vigorously that the
flask vibrates.
Iron and chlorine react to form
iron(III) chloride, FeCl3.
For instance, aqueous solutions
of hydrogen fluoride are called
hydrofluoric acid.
Hydrogen halides can be prepared by combination of the elements.
H2 ϩ X2 88n 2HX(g) X ϭ F, Cl, Br, I
The reaction with F2 to produce HF is explosive and very dangerous. The reaction
producing HCl does not occur significantly in the dark but occurs rapidly by a photochemical
chain reaction when the mixture is exposed to light. Light energy is absorbed
by Cl2 molecules, which break apart into very reactive chlorine atoms, which have unpaired
electrons (radicals). These subsequently attack H2 molecules and produce HCl molecules,
leaving hydrogen atoms (also radicals). The hydrogen radicals, in turn, attack Cl2 molecules
to form HCl molecules and chlorine radicals, and the process continues.
This chain reaction continues as long as there is a significant concentration of radicals.
Termination steps eliminate two radicals and can eventually terminate the reaction.
The reaction of H2 with Br2 is also a photochemical reaction. That of H2 with I2 is
very slow, even at high temperatures and with illumination.
All hydrogen halides react with H2O to produce hydrohalic acids that ionize.
The reaction is essentially complete for dilute aqueous solutions of HCl, HBr, and HI.
In dilute aqueous solutions; HF is a weak acid (Ka ϭ 7.2 ϫ 10Ϫ4). In concentrated solutions
more acidic dimeric (HF)2 units are present (Figure 24-1). They ionize as follows.
H H Hϩ ϪO Oϩ ϩH
H H
X X
Cl
H2
termination stepsCl2
HCl
ϩ
ϩ
ϩ
Cl Cl
H
H H
Cl2
H2
Cl2
2 Cl initiation step
chain propagation steps
HCl
HCl
Hϩϩ
ϩ ϩ Cl
Cl
H
h␯
24-6 The Hydrogen Halides and Hydrohalic Acids 949
A photochemical reaction is one in
which a species (usually a molecule)
interacts with radiant energy to
produce very reactive species. These
then undergo further reaction.
Concentrated solution (HF)2
Figure 24-1 Hydrogen bonding
(dashed lines) in dilute and
concentrated aqueous solutions
of hydrofluoric acid, HF.
Dilute solution
(HF)2(aq) ϩ H2O(ᐉ) 34 H3Oϩ(aq) ϩ HF2
Ϫ(aq) K Ϸ 5
The order of increasing acid strengths of the aqueous hydrohalic acids is:
HF ϽϽ HCl Ͻ HBr Ͻ HI
The only acid used in industry to a greater extent than HCl is H2SO4. Hydrochloric
acid is used in the production of metal chlorides, dyes, and many other commercially
important products. It is also used on a large scale to dissolve metal oxide coatings from
iron and steel prior to galvanizing or enameling.
Hydrofluoric acid is used in the production of fluorine-containing compounds and for
etching glass. The acid reacts with silicates, such as calcium silicate, CaSiO3, in the glass
to produce a very volatile and thermodynamically stable compound, silicon tetrafluoride,
SiF4.
CaSiO3(s) ϩ 6HF(aq) 88n CaF2(s) ϩ SiF4(g) ϩ 3H2O(ᐉ)
THE OXOACIDS (TERNARY ACIDS) OF THE HALOGENS
Table 24-5 lists the known oxoacids of the halogens, their sodium salts, and some trends
in properties. Only three oxoacids, HClO4, HIO3, and H5IO6, have been isolated in anhydrous
form. The others are known only in aqueous solution. In all these acids the H is
bonded through an O.
The only oxoacid of fluorine that has been prepared is unstable hypofluorous acid,
HOF. Aqueous hypohalous acids (except HOF) can be prepared by reaction of free halogens
(Cl2, Br2, I2) with cold water. The smaller the halogen, the farther to the right the
equilibrium lies.
X2 ϩ H2O 34 HX ϩ HOX (X ϭ Cl, Br, I)
hydrohalic acid hypohalous acid
Hypohalite salts can be prepared by reactions of the halogens with cold dilute bases.
24-7
950 CHAPTER 24: Some Nonmetals and Metalloids
0 Ϫ1 ϩ1
In HOF the oxidation states are
F ϭ Ϫ1, H ϭ ϩ1, O ϭ 0.
These reactions all involve
disproportionation of the halogen.
TABLE 24-5 Oxoacids of the Halogens and Their Salts
Thermal Oxidizing
Stability Oxidizing Power and
Oxidation Name and Acid Power Sodium Name Thermal Hydrolysis Nature of
State Acid of Acid Strength of Acid Salt of Salt Stability of Anion Halogen
ϩ1 HXO hypohalous NaXO sodium X ϭ F, Cl, Br, I
(HOX) acid (NaOX) hypohalite
ϩ3 HXO2 halous NaXO2 sodium X ϭ Cl, Br (?)
acid halite
ϩ5 HXO3 halic NaXO3 sodium X ϭ Cl, Br, I
acid halate
ϩ7 HXO4 perhalic NaXO4 sodium X ϭ Cl, Br, I
acid perhalate
ϩ7 H5XO6 paraperhalic several sodium X ϭ I only
acid types paraperhalates
Increase
Increases
Increases
Increase
X2 ϩ 2NaOH 88n NaX ϩ NaOX ϩ H2O (X ϭ Cl, Br, I)
sodium halide sodium hypohalite
The hypohalites are used as bleaching agents. Sometimes Cl2 is used as a bleach or as a
disinfectant, as in public water supplies. It reacts slowly with H2O to form HCl and HOCl.
The hypochlorous acid then decomposes into HCl and O radicals, which kill bacteria.
These oxygen radicals are very strong oxidizing agents. They are the effective bleaching
and disinfecting agent in aqueous solutions of Cl2 or hypochlorite salts.
Perchloric acid is the strongest of all common acids with respect to ionization. Hot,
concentrated perchloric acid is a very strong oxidizing agent that can explode in the presence
of reducing agents. Cold, dilute perchloric acid is only a weak oxidizing agent.
SULFUR, SELENIUM, AND TELLURIUM
OCCURRENCE, PROPERTIES, AND USES
Each Group VIA element is less electronegative than its neighboring halogen. Oxygen
and sulfur are clearly nonmetallic, but selenium is less so. Tellurium is usually classified
as a metalloid and forms metal-like crystals. Its chemistry is mostly that of a nonmetal.
Polonium is a metal. All 29 isotopes of polonium are radioactive.
Irregularities in the properties of elements within a given family increase toward the
middle of the periodic table. There are larger differences in the properties of the Group
VIA elements than in the properties of the halogens. The properties of elements in the
second period usually differ significantly from those of other elements in their families,
because second-period elements have no low-energy d orbitals. So, the properties of
oxygen are not very similar to those of the other Group VIA elements (Table 24-6). The
maximum number of electrons that O can have in its valence shell is eight. The heavier
Group VIA elements—S, Se, Te, and Po—have d orbitals in their valence shells; one or
24-8
Cl2 H2O HCl HOCl
HOCl
ϩ ϩ
HCl Oϩ
24-8 Occurrence, Properties, and Uses 951
0 Ϫ1 ϩ1
Solid household bleaches are usually
Ca(ClO)Cl. This is prepared by
reaction of Cl2 with Ca(OH)2.
Ca(OH)2 ϩ Cl2 88n
Ca(ClO)Cl ϩ H2O
The d orbitals do not occur until the
third energy level.
TABLE 24-6 Some Properties of Group VIA Elements
Property O S Se Te Po
Physical state (1 atm, 25°C) gas solid solid solid solid
Color colorless yellow red-gray to brass-colored, —
(very pale blue) black metallic luster
Outermost electrons 2s22p4 3s23p4 4s24p4 5s25p4 6s26p4
Melting point (1 atm, °C) Ϫ218 112 217 450 254
Boiling point (1 atm, °C) Ϫ183 444 685 990 962
Electronegativity 3.5 2.5 2.4 2.1 1.9
First ionization energy (kJ/mol) 1314 1000 941 869 812
Atomic radius (Å) 0.73 1.03 1.19 1.42 1.68
Ionic (2Ϫ) radius (Å) 1.26 1.70 1.84 2.07 —
Common oxidation states usually Ϫ2 Ϫ2, ϩ2, ϩ4, ϩ6 Ϫ2, ϩ2, ϩ4, ϩ6 Ϫ2, ϩ2, ϩ4, ϩ6 Ϫ2, ϩ6
more of these d orbitals can accommodate additional electrons to form up to six bonds.
Thus each of the Group VIA elements except O can bond covalently to as many as six
other atoms.
Sulfur
Sulfur makes up about 0.05% of the earth’s crust. It was one of the elements known to
the ancients. It was used by the Egyptians as a yellow coloring, and it was burned in some
religious ceremonies because of the unusual odor it produced; it is the “brimstone” of the
Bible. Alchemists tried to incorporate its “yellowness” into other substances to produce
gold.
Sulfur occurs as the free element—predominantly S8 molecules—and in metal sulfides
such as galena, PbS; iron pyrite, FeS2; and cinnabar, HgS. To a lesser extent, it occurs as
metal sulfates such as barite, BaSO4, and gypsum, CaSO4и2H2O, and in volcanic gases as
H2S and SO2.
Sulfur is found in much naturally occurring organic matter, such as petroleum and coal.
Its presence in fossil fuels causes environmental and health problems because many sulfurcontaining
compounds burn to produce sulfur dioxide, an air pollutant.
Much of the sulfur used in the United States is recovered from sulfur compounds in
natural gas and oil. During the oil-refining process, these compounds are reduced to
hydrogen sulfide, which is then oxidized to sulfur in the Claus furnace.
8H2S(g) ϩ 4O2(g) 88n S8(ᐉ) ϩ 8H2O(g)
Elemental sulfur is mined along the U.S. Gulf Coast by the Frasch process (Figure
24-2). Most of it is used in the production of sulfuric acid, H2SO4, the most important
of all industrial chemicals. Sulfur is used in the vulcanization of rubber and in the synthesis
of many important sulfur-containing organic compounds.
In each of the three physical states, elemental sulfur exists in many forms. The two
most stable forms of sulfur, the rhombic (mp 112°C) and monoclinic (mp 119°C) crystalline
modifications, consist of different arrangements of S8 molecules. These are
puckered rings containing eight sulfur atoms (Figure 2-3) and all SXS single bonds. Above
150°C, sulfur becomes increasingly viscous and darkens as the S8 rings break apart into
chains that interlock with one another through SXS bonds. The viscosity reaches a
maximum at 180°C, at which point sulfur is dark brown. Above 180°C, the liquid thins
as the chains are broken down into smaller chains. At 444°C, sulfur boils to give a vapor
containing S8, S6, S4, and S2 molecules.
952 CHAPTER 24: Some Nonmetals and Metalloids
᭣ Native sulfur. Elemental sulfur is deposited at the edges of some hot springs and geysers.
This formation surrounds Emerald Lake in Yosemite National Park.
Compressed air
Hot steam
Sulfur, air
and steam
Melted
sulfer
Figure 24-2 The Frasch process for mining sulfur. Three concentric pipes are used. Water
at about 170°C and a pressure of 100 lb/in2 (7 kg/cm2) is forced down the outermost pipe to
melt the sulfur. Hot compressed air is pumped down the innermost pipe. It mixes with the
molten sulfur to form a froth, which rises through the third pipe.
Selenium
Selenium is quite rare (9 ϫ 10Ϫ6% of the earth’s crust). It occurs mainly as an impurity
in sulfur, sulfide, and sulfate deposits. It is obtained from the flue dusts that result from
roasting sulfide ores and from the “anode mud” formed in the electrolytic refining of
copper. It is used as a red coloring in glass. The gray crystalline allotropic form of selenium
has an electrical conductivity that is very light-sensitive, so it is used in photocopy
machines and in solar cells.
Tellurium
Tellurium is even less abundant (2 ϫ 10Ϫ7% of the earth’s crust) than selenium. It occurs
mainly in sulfide ores, especially with copper sulfide, and as the tellurides of gold and
silver. It, too, is obtained from the “anode mud” from refining of copper. The element
forms brass-colored, shiny, hexagonal crystals having low electrical conductivity. It is added
to some metals, particularly lead, to increase electrical resistance and improve resistance
to heat, corrosion, mechanical shock, and wear.
REACTIONS OF GROUP VIA ELEMENTS
Some reactions of the Group VIA elements are summarized in the following table.
General Equation Remarks
xE ϩ yM 88n MyEx With many metals
zE ϩ MxEy 88n MxEyϩz Especially with S, Se
E ϩ H2 88n H2E Decreasingly in the series O2, S, Se, Te
E ϩ 3F2 88n EF6 With S, Se, Te, and excess F2
2E ϩ Cl2 88n E2Cl2 With S, Se (Te gives TeCl2); also with Br2
E2Cl2 ϩ Cl2 88n 2ECl2 With S, Se; also with Br2
E ϩ 2Cl2 88n ECl4 With S, Se, Te, and excess Cl2; also with Br2
E ϩ O2 88n EO2 With S (with Se, use O2 ϩ NO2)
24-9
(Above) Yellow crystalline sulfur contains S8 rings.
(Right) When molten sulfur is heated above 150°C, an amorphous
form of sulfur called “plastic sulfur” is formed as the S8 rings
break and the fragments link to form long chains.
E represents a Group VIA element.
HYDRIDES OF GROUP VIA ELEMENTS
All the Group VIA elements form covalent compounds of the type H2E (E ϭ O, S, Se,
Te, Po) in which the Group VIA element is in the Ϫ2 oxidation state. H2O is a liquid
that is essential for animal and plant life. H2S, H2Se, and H2Te are colorless, noxious,
poisonous gases. They are even more toxic than HCN. Egg protein contains sulfur, and
its decomposition forms H2S, giving off the odor of rotten eggs. H2Se and H2Te smell
even worse. Their odors are usually ample warning of the presence of these poisonous
gases.
Both the melting point and boiling point of water are very much higher than expected
by comparison with those of the heavier hydrides (Figure 13-5). This is a consequence of
hydrogen bonding in ice and liquid water (Section 13-2) caused by the strongly dipolar
nature of water molecules. The electronegativity differences between H and the other
VIA elements are much smaller than that between H and O, so no H-bonding occurs in
H2S, H2Se, or H2Te.
Aqueous solutions of hydrogen sulfide, selenide, and telluride are acidic; acid strength
increases as the group is descended: H2S Ͻ H2Se Ͻ H2Te. The same trend was observed
for increasing acidity of the hydrogen halides. The acid ionization constants are
H2S H2Se H2Te
H2E 34 Hϩ ϩ HEϪ Ka1: 1.0 ϫ 10Ϫ7 1.9 ϫ 10Ϫ4 2.3 ϫ 10Ϫ3
HEϪ 34 Hϩ ϩ E2Ϫ Ka2: 1.0 ϫ 10Ϫ19 Ϸ10Ϫ11 Ϸ1.6 ϫ 10Ϫ11
GROUP VIA OXIDES
Although others exist, the most important VIA oxides are the dioxides, which are acid
anhydrides of H2SO3, H2SeO3, and H2TeO3; and the trioxides, which are anhydrides of
H2SO4, H2SeO4, and H6TeO6. Let us consider SO2 and SO3 as examples.
Sulfur Dioxide, SO2
Sulfur dioxide is a colorless, poisonous, corrosive gas with a very irritating odor. Even in
small quantities, it causes coughing and nose, throat, and lung irritation. It is an angular
molecule with trigonal planar electronic geometry, sp2 hybridization at the S atom, and
resonance stabilization.
Sulfur dioxide is produced in reactions such as the combustion of sulfur-containing
fossil fuels and the roasting of sulfide ores.
2ZnS ϩ 3O2 88n 2ZnO ϩ 2SO2
SO2 is a waste product of these operations. In the past, it was released into the atmosphere
along with some SO3 produced by its reaction with O2. A more environmentally
friendly practice now is to trap SO2 and SO3 and use them to make H2SO4. Some coal
contains up to 5% sulfur, so both SO2 and SO3 are present in the flue gases when coal is
burned. No way has been found to remove all the SO2 from flue gases of power plants.
One way of removing most of the SO2 involves the injection of limestone, CaCO3, into
the combustion zone of the furnace. Here CaCO3 decomposes to lime, CaO. This then
24-11
24-10
954 CHAPTER 24: Some Nonmetals and Metalloids
H2S is a stronger acid than H2O.
The solubility of H2S in water is
approximately 0.10 mol/L at 25°C.
Here E represents S, Se, or Te.
؉4 Oxidation State
Formula Name
H2SO3 sulfurous acid
H2SeO3 selenous acid
H2TeO3 tellurous acid
؉6 Oxidation State
Formula Name
H2SO4 sulfuric acid
H2SeO4 selenic acid
H6TeO6 telluric acid
If the SO2 and SO3 are allowed to
escape into the atmosphere, they cause
highly acidic rain.
combines with SO2 to form calcium sulfite (CaSO3), an ionic solid, which is collected and
disposed of as solid waste
heat
CaCO3 888n CaO ϩ CO2 followed by CaO ϩ SO2 88n CaSO3
This process is called scrubbing, and a disadvantage of it is the formation of huge quantities
of solid waste (CaSO3, unreacted CaO, and by-products).
Catalytic oxidation is now used by the smelting industry to convert SO2 into SO3. This
is then dissolved in water to make solutions of H2SO4 (up to 80% by mass). The gases
containing SO2 are passed through a series of condensers containing catalysts to speed up
the reaction. In some cases the impure H2SO4 can be used in other operations in the same
plant.
Sulfur Trioxide, SO3
Sulfur trioxide is a liquid that boils at 44.8°C. It is the anhydride of H2SO4. It is formed
by the reaction of SO2 with O2. The reaction is very exothermic, but ordinarily very slow.
It is catalyzed commercially in the contact process by spongy Pt, SiO2, or vanadium(V)
oxide, V2O5, at high temperatures (400 to 700°C).
catalyst
2SO2(g) ϩ O2(g) 399999994 2SO3(g) ⌬H0 ϭ Ϫ197.6 kJ/mol ⌬S0 ϭ Ϫ188 J/Kиmol
The high temperature favors SO2 and O2 but allows the reaction to proceed much more
rapidly, so it is economically advantageous. The SO3 is then removed from the gaseous
reaction mixture by dissolving it in concentrated H2SO4 (95% H2SO4 by mass) to produce
polysulfuric acids—mainly pyrosulfuric acid, H2S2O7. This is called oleum, or fuming
sulfuric acid. The addition of fuming sulfuric acid to water produces commercial H2SO4.
SO3 ϩ H2SO4 88n H2S2O7 then H2S2O7 ϩ H2O 88n 2H2SO4
In the presence of certain catalysts, sulfur dioxide in polluted air reacts rapidly with O2
to form SO3. Particulate matter, or suspended microparticles, such as NH4NO3 and
elemental S, act as efficient catalysts.
24-11 Group VIA Oxides 955
Large amounts of SO2 and H2S are released
during volcanic eruptions.
The rose on the left is in an
atmosphere of sulfur dioxide, SO2.
Gaseous SO2 and aqueous solutions
of HSO3
Ϫ and SO3
2Ϫ ions are
used as bleaching agents. A similar
process is used to bleach wood pulp
before it is converted to paper.
The prefix pyro means “heat” or “fire.”
Pyrosulfuric acid may also be obtained
by heating concentrated sulfuric acid,
which results in the elimination of one
molecule of water from two molecules
of sulfuric acid.
2H2SO4 88n H2S2O7 ϩ H2O
OXOACIDS OF SULFUR
Sulfurous Acid, H2SO3
Sulfur dioxide readily dissolves in water to produce solutions of sulfurous acid, H2SO3.
The acid has not been isolated in anhydrous form.
H2O ϩ SO2 34 H2SO3
The acid ionizes in two steps in water
H2SO3 34 Hϩ ϩ HSO3
Ϫ Ka1 ϭ 1.2 ϫ 10Ϫ2
HSO3
Ϫ 34 Hϩ ϩ SO3
2Ϫ Ka2 ϭ 6.2 ϫ 10Ϫ8
When excess SO2 is bubbled into aqueous NaOH, sodium hydrogen sulfite, NaHSO3
is produced. This acid salt can be neutralized with additional NaOH or Na2CO3 to
produce sodium sulfite.
NaOH ϩ H2SO3 88n NaHSO3 ϩ H2O
NaOH ϩ NaHSO3 88n Na2SO3 ϩ H2O
The sulfite ion is pyramidal and has tetrahedral electronic geometry as predicted by the
VSEPR theory.
24-12
956 CHAPTER 24: Some Nonmetals and Metalloids
OH O
O
H
S
OH O
Ϫ
O
S O
O
O S
2Ϫ
Sulfurous acid, H2SO3 Hydrogen sulfite ion, HSO3
Ϫ Sulfite ion, SO3
2Ϫ
Sulfuric Acid, H2SO4
More than 40 million tons of sulfuric acid are produced annually worldwide. The contact
process is used for the commercial production of most sulfuric acid. The solution sold
commercially as “concentrated sulfuric acid” is 96–98% H2SO4 by mass and is about 18
molar H2SO4.
Pure H2SO4 is a colorless, oily liquid that freezes at 10.4°C and boils at 290 to 317°C
while partially decomposing to SO3 and water. There is some hydrogen bonding in solid
and liquid H2SO4.
Tremendous amounts of heat are evolved when concentrated sulfuric acid is diluted.
This illustrates the strong affinity of H2SO4 for water. H2SO4 is often used as a dehydrating
agent. Dilutions should always be performed by adding the acid to water to avoid
spattering the acid.
Sulfuric acid is a strong acid with respect to the first step of its ionization in water.
The second ionization occurs to a lesser extent (Example 18-17).
H2SO4 34 Hϩ ϩ HSO4
Ϫ Ka1 ϭ very large
HSO4
Ϫ 34 Hϩ ϩ SO4
2Ϫ Ka2 ϭ 1.2 ϫ 10Ϫ2
Pouring concentrated H2SO4 into
an equal volume of H2O liberates a
lot of heat—enough to raise the
temperature of the resulting solution
from room temperature to 121°C.
NITROGEN AND PHOSPHORUS
The Group VA elements provide a dramatic illustration of the vertical trends in metallic
properties. In this family, nitrogen and phosphorus are nonmetals, arsenic is predominantly
nonmetallic, antimony is more metallic, and bismuth is definitely metallic.
Properties of the Group VA elements are listed in Table 24-7.
Oxidation states of the VA elements range from Ϫ3 to ϩ5. Odd-numbered oxidation
states are favored. The VA elements form very few monatomic ions. Ions with a charge
of 3Ϫ occur for N and P, as in Mg3N2 and Ca3P2.
All of the Group VA elements show the Ϫ3 oxidation state in covalent compounds such
as NH3 and PH3. The ϩ5 oxidation state is found only in covalent compounds such as
phosphorus pentafluoride, PF5; nitric acid, HNO3; and phosphoric acid, H3PO4; and in
polyatomic ions such as NO3
Ϫ and PO4
3Ϫ. Each Group VA element exhibits the ϩ3 oxidation
state in one of its oxides, for instance, N2O3 and P4O6. These are acid anhydrides of
nitrous acid, HNO2, and phosphorous acid, H3PO3; both are weak acids. No other element
exhibits more oxidation states than nitrogen (Table 24-8).
24-12 Oxoacids of Sulfur 957
O
O
O
OH
H
S
O
O
O
OH S
Ϫ
O
O
O
O S
2Ϫ
Sulfuric acid, H2SO4 Hydrogen sulfate ion, HSO4
Ϫ Sulfate ion, SO4
2Ϫ
TABLE 24-7 Properties of the Group VA Elements
Property N P As Sb Bi
Physical state (1 atm, 25°C) gas solid solid solid solid
Color colorless red, white, black yellow, gray yellow, gray gray
Outermost electrons 2s22p3 3s23p3 4s24p3 5s25p3 6s26p3
Melting point (°C) Ϫ210 44 (white) 813 (gray, 28 atm)* 631 (gray) 271
Boiling point (°C) Ϫ196 280 (white) sublimes 613 1750 1560
Atomic radius (Å) 0.75 1.10 1.20 1.40 1.50
Electronegativity 3.0 2.1 2.1 1.9 1.8
First ionization energy (kJ/mol) 1402 1012 947 834 703
Oxidation states Ϫ3 to ϩ5 Ϫ3 to ϩ5 Ϫ3 to ϩ5 Ϫ3 to ϩ5 Ϫ3 to ϩ5
*Sublimes at lower pressures.
OCCURRENCE OF NITROGEN
Nitrogen, N2, is a colorless, odorless, tasteless gas that makes up about 75% by mass and
78% by volume of the atmosphere. Nitrogen compounds form only a minor portion of
the earth’s crust, but all living matter contains nitrogen. The primary natural inorganic
deposits of nitrogen are very localized. They consist mostly of KNO3 and NaNO3. Most
sodium nitrate is mined in Chile.
24-13
958 CHAPTER 24: Some Nonmetals and Metalloids
TABLE 24-8 Oxidation States of Nitrogen and Examples
؊3 ؊2 ؊1 0 ؉1 ؉2 ؉3 ؉4 ؉5
NH3 N2H4 NH2OH N2 N2O NO N2O3 NO2 N2O5
ammonia hydrazine hydroxylamine nitrogen dinitrogen nitrogen dinitrogen nitrogen dinitrogen
oxide oxide trioxide dioxide pentoxide
NH4
ϩ NH2Cl H2N2O2 HNO2 N2O4 HNO3
ammonium chloramine hyponitrous nitrous dinitrogen nitric acid
ion acid acid tetroxide
NH2
Ϫ NO2
Ϫ NO3
Ϫ
amide ion nitrite ion nitrate ion
Every protein contains nitrogen in
each of its fundamental amino acid
units.
Industrial
fixation
Via electrical
storms and
rain
Atmospheric nitrogen
Nitrogen fixation
by cyanobacteria
Denitrifying bacteria
in anaerobic mud
Urea
Decomposers
Amino acids
Ammonia
Nitrogen-fixing
bacteria in roots
and soil
NO3
Ϫ and NH4
ϩ
in soil water
Amino acid
synthesis
Runoff
from
land
A schematic representation of the nitrogen cycle.
The extreme abundance of N2 in the atmosphere and the low relative abundance of
nitrogen compounds elsewhere are due to the chemical inertness of N2 molecules. This
results from the very high bond energy of the NmN bond (945 kJ/mol).
Although N2 molecules are relatively unreactive, nature provides mechanisms by which
N atoms are incorporated into proteins, nucleic acids, and other nitrogenous compounds.
The nitrogen cycle is the complex series of reactions by which nitrogen is slowly but
continually recycled in the atmosphere (our nitrogen reservoir), lithosphere (earth), and
hydrosphere (water).
When N2 and O2 molecules collide near a bolt of lightning, they can absorb enough
electrical energy to produce molecules of NO. An NO molecule is quite reactive because
it contains one unpaired electron. NO reacts readily with O2 to form nitrogen dioxide,
NO2. Most NO2 dissolves in rainwater and falls to the earth’s surface. Bacterial enzymes
reduce the nitrogen in a series of reactions in which amino acids and proteins are produced.
These are then used by plants, eaten by animals, and metabolized. The metabolic products
are excreted as nitrogenous compounds such as urea, (NH2)2CO, and ammonium
salts such as NaNH4HPO4. These can also be enzymatically converted to ammonia, NH3,
and amino acids.
Nitrogen is converted directly into NH3 in another way. Members of the class of plants
called legumes (including soybeans, alfalfa, and clover) have nodules on their roots. Within
the nodules live bacteria that produce an enzyme called nitrogenase. These bacteria extract
N2 directly from air trapped in the soil and convert it into NH3. The ability of nitrogenase
to catalyze this conversion, called nitrogen fixation, at ordinary temperatures and
pressures with very high efficiency is a marvel to scientists. They must resort to very
extreme and costly conditions to produce NH3 from nitrogen and hydrogen (the Haber
process, Section 17-7).
Ammonia is the source of nitrogen in many fertilizers. Unfortunately, nature does not
produce NH3 and related plant nutrient compounds rapidly enough to provide an adequate
food supply for the world’s growing population. Commercial synthetic fertilizers have
helped to lessen this problem, but at great cost for the energy that is required to produce
them.
HYDROGEN COMPOUNDS OF NITROGEN
We have already described ammonia and some of its reactions. Please review Sections
17-7 and 18-4.
Liquid ammonia (bp Ϫ33.4°C) is used as a solvent for some chemical reactions. It is
hydrogen bonded, just as H2O is, but NH3 is a much more basic solvent. Its weak autoionization
produces the ammonium ion, NH4
ϩ, and the amide ion, NH2
Ϫ. This is similar
to H2O, which ionizes to produce some H3Oϩ and OHϪ ions.
NH3(ᐉ) ϩ NH3(ᐉ) 34 NH4
ϩ ϩ NH2
Ϫ K ϭ 10Ϫ35
base2 acid1 acid2 base1
Many ammonium salts are known. Most are very soluble in water. They can be prepared
by reactions of ammonia with acids. Reaction with nitric acid gives ammonium nitrate.
NH3(aq) ϩ [Hϩ(aq) ϩ NO3
Ϫ(aq)] 88n [NH4
ϩ ϩ NO3
Ϫ(aq)] ammonium nitrate
Amines are organic compounds that are structurally related to ammonia. We think of
them as being derived from NH3 by the replacement of one or more hydrogens with
24-14
24-14 Hydrogen Compounds of Nitrogen 959
Nitrogen is sold as compressed gas in
cylinders. The boiling point of N2 is
Ϫ195.8°C (Ϫ320°F). N2 is obtained by
fractional distillation of liquid air.
Root nodules on soybeans.
Great advances have been made in
cattle breeding in recent decades.
Semen from superior bulls can be
collected and stored in liquid
nitrogen for 30 years or more.
organic groups (Sections 18-4 and 27-12). All involve sp3-hybridized N. All are weak bases
because of the unshared pair of electrons on N.
960 CHAPTER 24: Some Nonmetals and Metalloids
Ϫ3 ϩ5 ϩ1
Ammonia, NH3 Methylamine, CH3NH2 Dimethylamine, (CH3)2NH Trimethylamine, (CH3)3
NITROGEN OXIDES
Nitrogen forms several oxides, in which it exhibits positive oxidation states of 1 to 5 (Table
24-8). All have positive free energies of formation, owing to the high dissociation energy
of N2 and O2 molecules. All are gases except N2O5, a solid that melts at 30.0°C.
Dinitrogen Oxide (؉1 Oxidation State)
Molten ammonium nitrate undergoes autooxidation–reduction (decomposition) at 170 to
260°C to produce dinitrogen oxide, also called nitrous oxide.
heat
NH4NO3(ᐉ) 888n N2O(g) ϩ 2H2O(g) ϩ heat
At higher temperatures, explosions occur, producing N2, O2, and H2O.
heat
2NH4NO3(ᐉ) 888n 2N2(g) ϩ O2(g) ϩ 4H2O(g) ϩ heat
Dinitrogen oxide supports combustion because it produces O2 when heated.
heat
2N2O(g) 888n 2N2(g) ϩ O2(g) ϩ heat
The molecule is linear but unsymmetrical, with a dipole moment of 0.17 D.
Nitrogen Oxide (؉2 Oxidation State)
The first step of the Ostwald process (Section 24-16) for producing HNO3 from NH3 is
used for the commercial preparation of nitrogen oxide, NO.
catalyst
4NH3(g) ϩ 5O2(g) 88888n 4NO(g) ϩ 6H2O(g)
heat
NO is not produced in large amounts in nature under usual conditions. It is formed by
direct reaction of N2 and O2 in electrical storms.
24-15
Some dentists use N2O for its mild
anesthetic properties. It is also known
as laughing gas because of its side
effects.
Dinitrogen oxide, or nitrous oxide,
N2O, mp Ϫ90.8°C, bp Ϫ88.8°C.
NO is a colorless gas that condenses at Ϫ152°C to a blue liquid. Gaseous NO is paramagnetic
and contains one unpaired electron per molecule.
Its unpaired electron makes nitric oxide very reactive. Molecules that contain unpaired
electrons are called radicals. NO reacts with O2 to form NO2, a brown, corrosive gas.
2NO(g) ϩ O2(g) 88n 2NO2(g) ϩ heat
In recent years, researchers have learned that very low concentrations of NO have important
roles in a variety of physiological processes, not only in humans, but in such a variety
of animals as barnacles, chickens, trout, and fruit flies. It has recently been found to be
involved in transmission of signals by nerves, blood clotting, local control of blood flow,
the immune system’s ability to kill tumor cells and bacteria, and possibly even memory.
Nitrogen Dioxide and Dinitrogen Tetroxide (؉4 Oxidation State)
Nitrogen dioxide is formed by reaction of NO with O2. It is prepared in the laboratory
by heating heavy metal nitrates.
heat
2Pb(NO3)2(s) 888n 2PbO(s) ϩ 4NO2(g) ϩ O2(g)
The NO2 molecule is angular. It is represented by resonance structures.
Because each NO2 molecule contains one unpaired electron, NO2 readily dimerizes to
form colorless, diamagnetic dinitrogen tetroxide, N2O4, at low temperatures.
2NO2(g) 34 N2O4(g) ⌬H0 ϭ Ϫ57.2 kJ/mol rxn
brown colorless
The N2O4 molecule is also represented by resonance structures.
N
O
O
N
O
O
N
O
O
N
O
O
N
O
O
N
O
O
N
O
O
N
O
O
N
O
O
nitrogen dioxide
N
O
O
N O N O
24-15 Nitrogen Oxides 961
Nitrogen oxide, or nitric oxide, NO,
mp Ϫ163.6°C, bp Ϫ151.8°C, bond
distance (1.15 Å) intermediate
between NmO (1.06 Å) and NUO
(1.20 Å).
Nitrogen dioxide, NO2, mp
Ϫ11.20°C, one unpaired electron,
bond length 1.197 Å; brown gas.
Dinitrogen tetroxide, N2O4, bp 21.2°C, no unpaired electrons,
NXN bond length 1.64 Å, NXO bond length 1.17Å,
OXNXO bond angle 126°.
962 CHAPTER 24: Some Nonmetals and Metalloids
The EnvironmentCCHEMISTRY IN USE
Nitrogen Oxides and Photochemical Smog
Nitrogen oxides are produced in the atmosphere by natural
processes. Human activities contribute only about 10% of all
the oxides of nitrogen (collectively referred to as NOx) in the
atmosphere, but the human contribution occurs mostly in
urban areas, where the oxides may be present in concentrations
a hundred times greater than in rural areas.
Just as NO is produced naturally by the reaction of N2
and O2 in electrical storms, it is also produced by the same
reaction at the high temperatures of internal combustion
engines and furnaces.
N2(g) ϩ O2(g) 34 2NO(g) ⌬H0 ϭ ϩ180 kJ/mol rxn
At ordinary temperatures the reaction does not occur to a
significant extent. Because it is endothermic, it is favored by
high temperatures. Even in internal combustion engines and
furnaces, the equilibrium still lies far to the left, so only small
amounts of NO are produced and released into the atmosphere.
Even very small concentrations of nitrogen oxides
cause serious problems, however.
The NO radicals react with O2 to produce NO2 radicals.
Both NO and NO2 are quite reactive, and they do considerable
damage to plants and animals. NO2 reacts with H2O in
the air to produce corrosive droplets of HNO3 and more NO.
3NO2 ϩ H2O 88n NO ϩ 2HNO3 (nitric acid)
The HNO3 may be washed out of the air by rainwater
(acid rain), or it may react with traces of NH3 in the air to
form solid NH4NO3, a particulate pollutant.
HNO3 ϩ NH3 88n NH4NO3
This situation occurs in all urban areas, but the problem is
worse in warm, dry climates, which are conducive to lightinduced
(photochemical) reactions. Here ultraviolet (UV)
radiation from the sun produces damaging oxidants. The
brownish hazes that often hang over such cities as Los Angeles,
Denver, and Mexico City are due to the presence of
brown NO2. Problems begin in the morning rush hour as
NO is exhausted into the air. The NO combines with O2 to
form NO2. Then, as the sun rises higher in the sky, NO2
absorbs UV radiation and breaks down into NO and oxygen
radicals.
UV
NO2 88n NO ϩ O
The extremely reactive O radicals combine with O2 to produce
O3 (ozone).
O ϩ O2 88n O3
Ozone is a powerful oxidizing agent that damages rubber,
plastic materials, and all plant and animal life. It also reacts
with hydrocarbons from automobile exhaust and evaporated
gasoline to form secondary organic pollutants such as aldehydes
and ketones (Section 27-11). The peroxyacyl nitrates
(PANs), perhaps the worst of the secondary pollutants, are
especially damaging photochemical oxidants that are very
irritating to the eyes and throat.
Catalytic converters in automobile exhaust systems reduce
emissions of oxides of nitrogen.
R C
O O
O
O O N R ϭ hydrocarbon chain or ring
nitrate
acyl
peroxy
ϩ4 ϩ2 ϩ5
Photochemical smog casts a haze over urban or industrial areas; its
severity depends on the weather.
SOME OXOACIDS OF NITROGEN AND THEIR SALTS
The main oxoacids of nitrogen are nitrous acid, HNO2, and nitric acid, HNO3.
Nitrous Acid (؉3 Oxidation State)
Nitrous acid, HNO2, is unstable and cannot be isolated in pure form. It is prepared as a
pale blue solution when H2SO4 reacts with cold aqueous sodium nitrite. Nitrous acid is
a weak acid (Ka ϭ 4.5 ϫ 10Ϫ4). It acts as an oxidizing agent toward strong reducing agents
and as a reducing agent toward very strong oxidizing agents.
Lewis formulas for nitrous acid and the nitrite ion follow.
N
O O
N
O O
ϪϪ
nitrite ion
H N
O O
nitrous acid
24-16
24-16 Some Oxoacids of Nitrogen and Their Salts 963
Nitric Acid (؉5 Oxidation State)
Pure nitric acid, HNO3, is a colorless liquid that boils at 83°C. Light or heat causes it to
decompose into NO2, O2, and H2O. The presence of the NO2 in partially decomposed
aqueous HNO3 causes its yellow or brown tinge. The Lewis structure of nitric acid is
HNO3 is commercially prepared by the Ostwald process. At high temperatures, NH3
is catalytically converted to NO, which is cooled and then air-oxidized to NO2. Nitrogen
dioxide reacts with H2O to produce HNO3 and some NO. The NO produced in the third
step is then recycled into the second step. More than 18 billion pounds of HNO3 was
produced in the United States in 1997.
Nitric acid is very soluble in water (Ϸ16 mol/L). It is a strong acid and a strong oxidizing
agent.
4NH3(g)
2NO(g)
3NO2(g) H2O(ᐉ) 2[Hϩ ϩ ϩNO3
Ϫ] NO(g)ϩ
ϩ O2(g) 2NO2(g)
5O2(g) 4NO(g) 6H2O(g)ϩ ϩ
recycle
Pt
1000°C
cool
H O
O
O
N
O
H O
O
N
Nitrogen also forms hyponitrous acid,
H2N2O2, in which N is in the ϩ1
oxidation state, as well as hyponitrite
salts such as Na2N2O2.
Nitric acid, HNO3, mp Ϫ42°C; bp
83°C; bond lengths NXO (terminal)
1.22 Å, NXO (central) 1.41 Å.
NaNO2 and NaNO3 as Food Additives
The brown color of “old” meat is the result of oxidation of blood and is objectionable to
many consumers. Nitrites and nitrates are added to food to retard this oxidation and also
to prevent growth of botulism bacteria. Nitrate ions, NO3
Ϫ, are reduced to NO2
Ϫ ions,
which are then converted to NO. This in turn reacts with the brown oxidized form of the
heme in blood. This reaction keeps meat red longer. Controversy has arisen, however,
concerning the possibility that nitrites combine with amines under the acidic conditions
in the stomach to produce carcinogenic nitrosoamines.
PHOSPHORUS
Phosphorus is always combined in nature. Phosphorus is present in all living organisms—as
organophosphates and in calcium phosphates such as hydroxyapatite,
Ca5(PO4)3(OH), and fluorapatite, Ca5(PO4)3F, in bones and teeth. It also occurs in these
and related compounds in phosphate minerals, which are mined mostly in Florida and
North Africa.
Industrially, the element is obtained from phosphate minerals by heating them at 1200
to 1500°C in an electric arc furnace with sand (SiO2) and coke.
24-17
RЈ
R
N N O (R and RЈ ϭ organic groups)
nitroso
group
amine
group
964 CHAPTER 24: Some Nonmetals and Metalloids
Nitric acid, HNO3, reacts with protein-containing materials such as this feather, staining
them yellow. Perhaps you have spilled nitric acid on your skin and seen it turn yellow.
Copper (left beaker) and zinc (right
beaker) react with concentrated nitric
acid.
White phosphorus reacts with air, so
it is stored under water. It contains
tetrahedral P4 molecules.
heat
2Ca3(PO4)2 ϩ 6SiO2 ϩ 10C 888n 6CaSiO3 ϩ 10CO ϩ P4
calcium phosphate calcium silicate
(phosphate rock) (slag)
Vaporized phosphorus is condensed to a white solid (mp ϭ 44.2°C, bp ϭ 280.3°C)
under H2O to prevent oxidation. Even when kept under H2O, white phosphorus slowly
converts to the more stable red phosphorus allotrope (mp ϭ 597°C; sublimes at 431°C).
Red phosphorus and tetraphosphorus trisulfide, P4S3, are used in matches. They do not
burn spontaneously, yet they ignite easily when heated by friction. Both white and red
phosphorus are insoluble in water.
The largest use of phosphorus is in fertilizers. Phosphorus is an essential nutrient, and
nature’s phosphorus cycle is very slow owing to the low solubility of most natural phosphates.
Phosphate fertilizers are therefore essential. To increase the solubility of the natural
phosphates, they are treated with H2SO4 to produce “superphosphate of lime,” a mixture
of two salts. This solid is pulverized and applied as a powder.
This reaction represents the most popular use of sulfuric acid, the industrial chemical
produced in largest quantity.
SILICON
Silicon is a shiny, blue-gray, high-melting, brittle metalloid. It looks like a metal, but it is
chemically more like a nonmetal. It is second only to oxygen in abundance in the earth’s
crust, about 87% of which is composed of silica (SiO2) and its derivatives, the silicate
minerals. The crust is 26% Si, compared with 49.5% O. Silicon does not occur free in
nature. Pure silicon crystallizes with a diamond-type structure, but the Si atoms are less
closely packed than C atoms. Its density is 2.4 g/cm3 compared with 3.51 g/cm3 for
diamond.
Ca3(PO4)2 ϩ ϩ ϩ и2H2SO4 4H2O [Ca(H2PO4)2 2(CaSO4 2H2O)]
evaporate
phosphate rock calcium
dihydrogen
phosphate
calcium
sulfate
dihydrate
superphosphate of lime
24-17 Phosphorus 965
The tips of “strike anywhere” matches contain
tetraphosphorus trisulfide and red phosphorus.
Friction converts kinetic energy into heat, which
initiates a spontaneous reaction.
P4S3(s) ϩ 8O2(g) 88n P4O10(s) ϩ 3SO2(g)
Superphosphate fertilizer.
Pure silicon is used in solar cells to
collect energy from the sun.
SILICON AND THE SILICATES
Elemental silicon is usually prepared by the high-temperature reduction of silica (sand)
with coke. Excess SiO2 prevents the formation of silicon carbide.
heat
SiO2(s, excess) ϩ 2C(s) 888n Si(s) ϩ 2CO(g)
Reduction of a mixture of silicon and iron oxides with coke produces an alloy of iron
and silicon known as ferrosilicon. It is used in the production of acid-resistant steel alloys,
such as “duriron,” and in the “deoxidation” of steel. Aluminum alloys for aircraft are
strengthened with silicon.
Elemental silicon is used to make silicone polymers. Its semiconducting properties
(Section 13-17) are used in transistors and solar cells.
The biggest chemical differences between silicon and carbon are that (1) silicon does
not form stable double bonds, (2) it does not form very stable SiXSi bonds unless the
silicon atoms are bonded to very electronegative elements, and (3) it has vacant 3d orbitals
in its valence shell into which it can accept electrons from donor atoms. The SiXO single
bond is the strongest of all silicon bonds and accounts for the stability and prominence
of silica and the silicates.
Silicon dioxide (silica) exists in two familiar forms in nature: quartz, small chips of
which occur in sand; and flint (Latin silex), an uncrystallized amorphous type of silica.
Silica is properly represented as (SiO2)n because it is a polymeric solid of SiO4 tetrahedra
sharing all oxygens among surrounding tetrahedra (Figure 13-32c). For comparison, solid
carbon dioxide (dry ice) consists of discrete OUCUO molecules, as does gaseous CO2.
Some gems and semiprecious stones such as amethyst, opal, agate, and jasper are crystals
of quartz with colored impurities.
Most of the crust of the earth is made up of silica and silicates. The natural silicates
comprise a large variety of compounds. The structures of all these are based on SiO4 tetrahedra,
with metal ions occupying spaces between the tetrahedra. The extreme stability of
the silicates is presumably due to the donation of extra electrons from O into vacant 3d
orbitals of Si. In many common minerals, called aluminosilicates, Al atoms replace some
Si atoms with very little structural change. Because an Al atom has one less positive charge
in its nucleus than Si does, it is also necessary to introduce a univalent ion, such as Kϩ or
Naϩ.
The physical characteristics of the silicates are often suggested by the arrangement of
the SiO4 tetrahedra. A single-chain silicate, diopside [CaMg(SiO3)2]n, and a double-chain
silicate, asbestos [Ca2Mg5(Si4O11)2(OH2)]n, occur as fibrous or needle-like crystals. Talc,
[Mg3Si4O10(OH)2]n, a silicate with a sheet-like structure, is flaky. Micas are sheet-like
aluminosilicates with about one of every four Si atoms replaced by Al. Muscovite mica is
[KAl2(AlSi3O10)(OH)2]n. Micas occur in thin sheets that are easily peeled away from one
another.
The clay minerals are silicates and aluminosilicates with sheet-like structures. They
result from the weathering of granite and other rocks. The layers have enormous “inner
surfaces” that can absorb large amounts of H2O. Clay mixtures often occur as minute
platelets with a very large total surface area. When wet, the clays are easily shaped. When
heated to high temperatures, they lose H2O; when fired in a furnace, they become very
rigid.
Fused sodium silicate, Na2SiO3, and calcium silicate, CaSiO3, are the major components
of the glass used in such things as drinking glasses, bottles, and window panes. Glass
is a hard, brittle material that has no fixed composition or regular structure. Because it
24-18
966 CHAPTER 24: Some Nonmetals and Metalloids
Si is much larger than C. As a result,
SiXSi bonds are too long to permit
the effective pi bonding that is
necessary for multiple bonds.
Natural quartz crystals, SiO2.
“Asbestos” refers to a group of
impure magnesium silicate minerals.
As you can see, asbestos is a fibrous
material. When inhaled, these fibers
are highly toxic and carcinogenic.
has no regular structure, it does not break evenly along crystal planes, but breaks to form
rounded surfaces and jagged edges. The basic ingredients are produced by heating a
mixture of Na2CO3 and CaCO3 with sand until it melts, at about 700°C.
heat
[CaCO3 ϩ SiO2](ᐉ) 888n CaSiO3(ᐉ) ϩ CO2(g)
heat
[Na2CO3 ϩ SiO2](ᐉ) 888n Na2SiO3(ᐉ) ϩ CO2(g)
The resulting “soda–lime” glass is clear and colorless (if all CO2 bubbles escape and if
the amounts of reactants are carefully controlled).
Exercises 967
Key Terms
Chain initiation step The first step in a chain reaction; produces
reactive species (such as radicals) that then propagate the
reaction.
Chain propagation step An intermediate step in a chain reaction;
in such a step one or more reactive species is consumed,
and another reactive species is produced.
Chain reaction A reaction in which reactive species, such as radicals,
are produced in more than one step. Consists of an
initiation step, one or more propagation steps, and one or more
termination steps.
Chain termination step The combination of reactive species
(such as radicals) which terminates the chain reaction.
Contact process An industrial process by which sulfur trioxide
and sulfuric acid are produced from sulfur dioxide.
Frasch process A method by which elemental sulfur is mined or
extracted. Sulfur is melted with superheated water (at 170°C
under high pressure) and forced to the surface of the earth as
a slurry.
Haber process An industrial process for the catalyzed production
of ammonia from N2 and H2 at high temperature and
pressure.
Halogens Group VIIA elements; F, Cl, Br, I, and At. The free
elements exist as diatomic molecules.
Nitrogen cycle The complex series of reactions by which nitrogen
is slowly but continually recycled in the atmosphere,
lithosphere, and hydrosphere.
Noble gases Group VIIIA elements; He, Ne, Ar, Kr, Xe, and
Rn.
Ostwald process An industrial process for the production of
nitrogen oxide and nitric acid from ammonia and oxygen.
PANs Abbreviation for peroxyacyl nitrates, photochemical oxidants
in smog.
Particulate matter Finely divided solid particles suspended in
polluted air.
Photochemical oxidants Photochemically produced oxidizing
agents capable of causing damage to plants and animals.
Photochemical smog A brownish smog occurring in urban areas
that receive large amounts of sunlight; caused by photochemical
(light-induced) reactions among nitrogen oxides, hydrocarbons,
and other components of polluted air that produce
photochemical oxidants.
Radical An atom or group of atoms that contains one or more
unpaired electrons (usually very reactive species).
Exercises
The Noble Gases
*01. (a) Write the Lewis dot representations of the noble
gases.
(b) Why are the noble gases so unreactive?
*02. Why were the noble gases among the last elements to be
discovered?
*03. List some of the uses of the noble gases and reasons for
the uses.
*04. Arrange the noble gases in order of increasing (a) atomic
radii, (b) melting points, (c) boiling points, (d) densities,
and (e) first ionization energies.
*05. Explain the order of increasing melting and boiling points
of the noble gases in terms of polarizabilities of the atoms
and forces of attraction between them.
*06. What gave Neil Bartlett the idea that compounds of xenon
could be synthesized? Which noble gases are known to
form compounds? With which elements are the noble gas
atoms bonded?
*07. Describe the bonding and geometry in XeF2, XeF4, and
XeF6.
0*8. Xenon(VI) fluoride can be produced by the combination
of xenon(IV) fluoride with fluorine. Write a chemical equation
for this reaction. What mass of XeF6 could be
produced from 2.75 g of XeF4 and excess fluorine?
The Halogens
*09. Write the electron configuration for each halogen atom.
Write the Lewis symbol for a halogen atom, X. What is
968 CHAPTER 24: Some Nonmetals and Metalloids
the usual oxidation state of the halogens in binary compounds
with metals, semiconducting elements, and most
nonmetals?
*10. Write the Lewis structure of a halogen molecule, X2.
Describe the bonding in the molecule. What is the trend
of bond length and strength going down the family from
F2 to I2?
*11. What types of intermolecular forces are found in molecular
halogens? What is the trend in these forces going down
the group from F2 to I2? Describe the physical state of each
molecular halogen at room temperature and pressure.
*12. List the halogens in order of increasing (a) atomic radii,
(b) ionic radii, (c) electronegativities, (d) melting points,
(e) boiling points, and (f) standard reduction potentials.
*13. Write the equations describing the half-reactions and net
reaction for the electrolysis of molten KF/HF mixtures. At
which electrodes are the products formed? What is the purpose
of the HF?
*14. Carl O. Christe’s preparation of F2 did not involve direct
chemical oxidation. Explain this statement.
*15. Write equations describing general reactions of the free
halogens, X2, with (a) Group IA (alkali) metals, (b) Group
IIA (alkaline earth) metals, and (c) Group IIIA metals. Represent
the metals as M.
*16. Write balanced equations for any reactions that occur in
aqueous mixtures of (a) NaI and Cl2, (b) NaCl and Br2,
(c) NaI and Br2, (d) NaBr and Cl2, and (e) NaF and I2.
*17. An aqueous solution contains either NaBr or a mixture of
NaBr and NaI. Using only aqueous solutions of I2, Br2,
and Cl2 and a small amount of CH2Cl2, describe how you
might determine what is in the unknown solution.
*18. Write equations illustrating the tendency of FϪ to stabilize
high oxidation states of cations and the tendency of IϪ
to stabilize low oxidation states. Why is this the case?
*19. Why are the free halogens more soluble in water than most
nonpolar molecules?
*20. Distinguish between hydrogen bromide and hydrobromic
acid.
*21. Refer to Figure 13-5. What is the order of decreasing boiling
points of the hydrogen halides? Why is the HF “out
of line?”
*22. Describe the effect of hydrofluoric acid on glass.
*23. What is the acid anhydride of perchloric acid?
*24. Name the following compounds: (a) KBrO3; (b) KOBr;
(c) NaClO4; (d) NaClO2; (e) HOBr; (f ) HBrO3; (g) HIO3;
(h) HClO4.
*25. Write the Lewis formulas and structures of the four ternary
acids of chlorine.
*26. Write equations describing reactions by which the following
compounds can be prepared: (a) hypohalous acids of
Cl, Br, and I (in solution with hydrohalic acids); (b) hypohalite
salts; (c) chlorous acid; (d) perchloric acid.
*27. What is the order of increasing acid strength of the ternary
chlorine acids? Explain the order.
*28. Choose the strongest acid from each group: (a) HOCl,
HOBr, HOI; (b) HOCl, HClO2, HClO3, HClO4; (c) HOI,
HBrO3, HClO4. Explain your choices.
Group IIIA Elements
*29. Write abbreviated electron configurations for atomic oxygen,
selenium, and polonium.
*30. Write out the electron configurations of oxide, sulfide, and
selenide ions.
*31. Characterize the Group VIA elements with respect to color
and physical state under normal conditions.
*32. The Group VIA elements, except oxygen, can exhibit oxidation
states ranging from Ϫ2 to ϩ6, but not Ϫ3 or ϩ7.
Why?
*33. Sulfur, selenium, and tellurium are all capable of forming
six-coordinate compounds such as SF6. Give two reasons
why oxygen cannot be the central atom in such sixcoordinate
molecules.
*34. For the following species, draw (i) diagrams that show the
hybridization of atomic orbitals and (ii) three-dimensional
structures that show all hybridized orbitals and outermost
electrons. (iii) Determine the oxidation state of the Group
VIA element (other than oxygen) in each species. (a) H2S;
(b) SF6; (c) SF4; (d) SO2; (e) SO3.
*35. Repeat Exercise 34 for (a) SeF6, (b) SO3
2Ϫ, (c) SO4
2Ϫ,
(d) HSO4
Ϫ, and (e) thiosulfate ion, S2O3
2Ϫ (one S is central
atom).
*36. Write equations for the reactions of (a) S, Se, and Te with
excess F2; (b) O2, S, Se, and Te with H2; (c) S, Se, and Te
with O2.
*37. Write equations for the reactions of (a) S and Te with
HNO3; (b) S and Se with excess Cl2; (c) S and Se with Na,
Ca, and Al.
*38. Discuss the acidity of the aqueous Group VIA hydrides,
including the relative values of acid ionization constants.
What is primarily responsible for the order of increasing
acidities in this series?
*39. Compare the structures of the dioxides of sulfur, selenium,
tellurium, and polonium. How do they relate to the metallic
or nonmetallic character of these elements?
*40. Write equations for reactions of
(a) NaOH with sulfuric acid (1Ϻ1 mole ratio)
(b) NaOH with sulfuric acid (2Ϻ1 mole ratio)
(c) NaOH with sulfurous acid (1Ϻ1 mole ratio)
(d) NaOH with sulfurous acid (2Ϻ1 mole ratio)
(e) NaOH with selenic acid, H2SeO4 (1Ϻ1 mole ratio)
(f ) NaOH with selenic acid, (2Ϻ1 mole ratio)
(g) NaOH with tellurium dioxide (1Ϻ1 mole ratio)
(h) NaOH with tellurium dioxide (2Ϻ1 mole ratio)
*41. How much sulfur dioxide could be produced from complete
combustion of 1.00 ton of coal containing 5.85%
sulfur?
Exercises 969
*42. What mass of H2SO4 could be produced in the following
process if 1.00 ton of FeS2 is used? The unbalanced equations
for the process are
FeS2(s) ϩ O2(g) 88n Fe2O3(s) ϩ SO2(g)
SO2(g) ϩ O2(g) 88n SO3(g)
SO3(g) ϩ H2SO4(ᐉ) 88n H2S2O7(ᐉ)
H2S2O7(ᐉ) ϩ H2O(ᐉ) 88n H2SO4(aq)
*43. Common copper ores in the western United States contain
the mineral chalcopyrite, CuFeS2. Assuming that a
commercially useful ore contains 0.291 mass % Cu and
that all the sulfur ultimately appears in the smelter stack
gases as SO2, calculate the mass of sulfur dioxide generated
by the conversion of 1.00 ton of the ore.
*44. A gaseous mixture at some temperature in a 1.00-L vessel
originally contained 1.00 mol SO2 and 5.00 mol O2. Once
equilibrium conditions were attained, 77.8% of the SO2
had been converted to SO3. What is the value of the equilibrium
constant (Kc) for this reaction at this temperature?
Nitrogen and Phosphorus
*45. Characterize each of the Group VA elements with respect
to normal physical state and color.
*46. Write out complete electron configurations for the atoms
of the Group VA elements; nitride ion, N3Ϫ; and phosphide
ion, P3Ϫ.
*47. Compare and contrast the properties of (a) N2 and P4;
(b) HNO3 and H3PO4; (c) N2O3 and P4O6.
*48. Describe the natural nitrogen cycle.
*49. List natural sources of nitrogen and phosphorus and at least
two uses for each.
*50. Discuss the effects of temperature, pressure, and catalysts
on the Haber process for the production of ammonia. (You
may wish to consult Section 17-7.)
*51. Determine the oxidation states of nitrogen in the following:
(a) NO3
Ϫ; (b) NO2
Ϫ; (c) N2H4; (d) NH3; (e) NH2
Ϫ.
*52. Determine the oxidation states of nitrogen in the following:
(a) N2; (b) NO; (c) N2O4; (d) HNO3; (e) HNO2.
*53. Draw three-dimensional structures showing all outer-shell
electrons, describe molecular and ionic geometries, and
indicate hybridization (except for N3Ϫ) at the central element,
for the following species: (a) N2, (b) N3Ϫ; (c) NH3,
(d) NH4
ϩ; (e) NH2
Ϫ, amide ion.
*54. Draw three-dimensional structures showing all outer-shell
electrons, describe molecular and ionic geometries, and
indicate hybridization at the central element for the following
species: (a) NH2Br, bromamine; (b) HN3, hydrazoic
acid; (c) N2O2; (d) NO2
ϩNO3
Ϫ, solid nitronium nitrate;
(e) HNO3; (f) NO2
Ϫ.
*55. Draw three-dimensional structures showing all outer-shell
electrons for the following species: (a) P4; (b) P4O10;
(c) As4O6; (d) H3PO4; (e) AsO4
3Ϫ.
*56. Write formula unit equations for the following: (a) thermal
decomposition of potassium azide, KN3; (b) reaction
of gaseous ammonia with gaseous HCl; (c) reaction of
aqueous ammonia with aqueous HCl; (d) thermal decomposition
of ammonium nitrate at temperature above 260°C;
(e) reaction of ammonia with oxygen in the presence of red
hot platinum catalyst; (f ) thermal decomposition of nitrous
oxide (dinitrogen oxide), N2O; (g) reaction of NO2 with
water.
*57. Write the formula unit equation for the preparation of
“superphosphate of lime.”
*58. Write two equations illustrating the ability of ammonia to
function as a Lewis base.
*59. In liquid ammonia would sodium amide, NaNH2, be
acidic, basic, or neutral? Would ammonium chloride,
NH4Cl, be acidic, basic, or neutral? Why?
*60. Which of the following molecules have a nonzero
dipole moment—that is, are polar molecules? (a) NH3;
(b) NH2Cl; (c) NO; (d) NH2OH; (e) HNO3.
*61. Describe with equations the Ostwald process for the production
of nitrogen oxide, NO, and nitric acid.
*62. Why is NO so reactive?
*63. Write a Lewis formula for NO2. Would you predict that
it is very reactive? How about N2O4 (dimerized NO2)?
*64. At room temperature, a sample of NO2 gas is brown.
Explain why this sample loses its color as it is cooled.
*65. Discuss the problem of NOx emissions with respect to air
pollution. Use equations to illustrate the important reac-
tions.
*66. What are the acid anhydrides of (a) nitric acid, HNO3;
(b) nitrous acid, HNO2; (c) phosphoric acid, H3PO4; and
(d) phosphorous acid, H3PO3?
*67. Calcium phosphate (phosphate rock) is not applied directly
as a phosphorus fertilizer. Why?
*68. Discuss the use of sodium nitrite as a meat preservative.
CONCEPTUAL EXERCISES
*69. Not much is known about astatine because it is very rare,
is radioactive, and decays very quickly. Would you predict
the chemical and physical properties of astatine to be more
Industrial smokestack.
970 CHAPTER 24: Some Nonmetals and Metalloids
like those of a metal or nonmetal? Defend your answer on
the basis of astatine’s location on the periodic table.
*70. Elemental chlorine is obtained by the electrolysis of molten
NaCl (Downs cell). Elemental fluorine is obtained by the
electrolysis of KHF2 in a cell made of Monel metal (a stainless
steel alloy). Both of these processes are dangerous.
Why?
*71. Acid rain consists of acids formed by the reaction of acid
anhydrides and water. List four acid anhydrides commonly
involved in acid rain formation and the sources of each.
*72. Sulfur deposits are mined by the Frasch process. Explain
why free sulfur is not found on the earth’s surface.
*73. (a) On the basis of electronic structures, explain why most
metal halides are ionic. (b) On the basis of your explanation,
predict which halide compounds are likely to be
covalent.
*74. (a) Which of the halogens is the most active chemically
and (b) which of the halogens is most likely to be reduced
from the free state? (c) Which of the halogens is most likely
to act as an oxidizing agent? (d) Which of the halogens is
least likely to be an effective oxidizing agent? (e) Which of
the halogens is likely to be a liquid under room conditions?
(f ) Which of the halogens is found free in nature?
*75. The nitrogen cycle indicates that electrical storms (lightning)
are a source of nitrogen compounds. (a) What is the
function of the lightning? (b) What compounds of nitrogen
are produced during electrical storms? (c) What is the
effect of these compounds on the rain water?
BUILDING YOUR KNOWLEDGE
*76. Standard enthalpies of formation are Ϫ402 kJ/mol for
XeF6(s) and Ϫ261.5 kJ/mol for XeF4(s). Calculate ⌬H0
rxn
at 25°C for the preparation of XeF6 from XeF4 and F2(g).
*77. The average atomic mass of N is 14.0067 amu. There are
two isotopes which contribute to this average: 14
07N
(14.00307 amu) and 15
07N (15.00011 amu). Calculate the
percentage of 15
07N atoms in a sample of naturally occurring
nitrogen.
*78. The NmN and the NXN bond energies are listed in
Tables 15-2 and 15-3. Predict whether four gaseous nitrogen
atoms would form two gaseous nitrogen molecules or
a gaseous tetrahedral molecule similar to P4, basing your
prediction on the amount of energy released as the molecules
are formed. Repeat the calculations for phosphorus
using 485 kJ/mol for PmP and 201 kJ/mol for PXP.
*79. Commercial concentrated HNO3 contains 69.5 mass %
HNO3 and has a density of 1.42 g/mL. What is the molarity
of this solution? What volume of the concentrated acid
should you use to prepare 10.0 L of dilute HNO3 solution
with a concentration of 6.00 M?
*80. What is the total mass of silicon in the crust of the earth?
Assume that the radius of the earth is 6400 km, the crust
is 50 km thick, the density of the crust is 3.5 g/cm3, and
25.7 mass % of the crust is silicon.
*81. How many grams of xenon oxide tetrafluoride, XeOF4, and
how many liters of HF at STP could be prepared, assuming
complete reaction of 10.5 g of xenon tetrafluoride,
XeF4, with a stoichiometric quantity of water according to
the equation below?
6XeF4(s) ϩ 8H2O(ᐉ) 88n
2XeOF4(ᐉ) ϩ 4Xe(g) ϩ 16HF(g) ϩ 3O2(g)
*82. Argon crystallizes at Ϫ235°C in a face-centered cubic unit
cell with a ϭ 5.43 Å. Determine the apparent radius of an
argon atom in the solid.
*83. A reaction mixture contained 100. g of K2MnF6 and 174 g
of SbF5. Fluorine was produced in 38.3% yield. How many
grams of F2 were produced? What volume is this at STP?
25Coordination
Compounds
C
OUTLINE
25-1 Coordination Compounds
25-2 Ammine Complexes
25-3 Important Terms
25-4 Nomenclature
25-5 Structures
Isomerism in Coordination
Compounds
25-6 Structural (Constitutional)
Isomers
25-7 Stereoisomers
Bonding in Coordination
Compounds
25-8 Crystal Field Theory
25-9 Color and the Spectrochemical
Series
OBJECTIVES
After you have studied this chapter, you should be able to
• Recognize coordination compounds
• Recognize metals that form soluble ammine complexes in aqueous solutions and write
the formulas for common ammine complexes
• Use the terminology that describes coordination compounds
• Apply the rules for naming coordination compounds
• Recognize common structures of coordination compounds
• Describe various kinds of structural (constitutional) isomerism and distinguish among
structural isomers
• Recognize stereoisomers
• Describe the crystal field theory of bonding in coordination compounds
• Explain the origin of color in complex species
• Use the spectrochemical series to explain colors of a series of complexes
C
oordination compounds are found in many places on the earth’s surface. Every
living system includes many coordination compounds. They are also important
components of everyday products as varied as cleaning materials, medicines, inks,
and paints. A list of important coordination compounds appears to be endless because new
ones are discovered every year.
COORDINATION COMPOUNDS
In Section 10-10 we discussed Lewis acid–base reactions. A base makes available a share
in an electron pair, and an acid accepts a share in an electron pair, to form a coordinate
25-1
Red blood cells (1200x). The red
blood cells that transport O2
throughout our bodies contain
hemoglobin, a coordination
compound.
Covalent bonds in which the shared
electron pair is provided by one atom
are called coordinate covalent bonds.
covalent bond. Such bonds are often represented by arrows that point from the electron
pair donor (Lewis base) to the acceptor (Lewis acid).
Electron configurations of the elements of the three d-transition series are given in
Table 25-1 and in Appendix B. Most d-transition metal ions have vacant d orbitals
that can accept shares in electron pairs. Many act as Lewis acids by forming coordinate
covalent bonds in coordination compounds (coordination complexes, or complex ions).
Complexes of transition metal ions or molecules include cationic species
(e.g., [Cr(OH2)6]3ϩ, [Co(NH3)6]3ϩ, [Ag(NH3)2]ϩ), anionic species (e.g., [Ni(CN4)]2Ϫ,
[MnCl5]3Ϫ), and neutral species (e.g., [Fe(CO)5], [Pt(NH3)2Cl2]). Many complexes are very
stable, as indicated by their low dissociation constants, Kd (Section 20-6 and Appendix I).
We now understand from molecular orbital theory that all substances have some vacant
orbitals—they are potential Lewis acids. Most substances have unshared pairs of electrons
—they are potential Lewis bases.
Sn
Cl
Cl
Cl2 ϩ
Ϫ
Cl
Cl
Cl
Cl
Cl
Cl Cl
Cl
2Ϫ
Sn
hexachlorostannate(IV) iontin(IV) chloride,
a Lewis acid
chloride ion,
a Lewis base
H
H
H
N H
H
H
N
Cl Cl
Cl
B B Cl
Cl
ϩ
Cl
ammonia,
a Lewis base
boron trichloride,
a Lewis acid
The red arrows represent coordinate
covalent bonds. These arrows do not
imply that two SnXCl bonds are
different from the others. Once
formed, all the SnXCl bonds in
the [SnCl6]2Ϫ ion are alike.
25-1 Coordination Compounds 973
TABLE 25-1 Ground State Electron Configurations of d-Transition Metals
Period 4 Period 5 Period 6
21Sc [Ar]3d14s2
39Y [Kr]4d15s2
57La [Xe]5d16s2
22Ti [Ar]3d24s2
40Zr [Kr]4d25s2
72Hf [Xe]4f 145d26s2
23V [Ar]3d34s2
41Nb [Kr]4d45s1
73Ta [Xe]4f 145d36s2
24Cr [Ar]3d54s1
42Mo [Kr]4d55s1
74W [Xe]4f 145d46s2
25Mn [Ar]3d54s2
43Tc [Kr]4d55s2
75Re [Xe]4f 145d56s2
26Fe [Ar]3d64s2
44Ru [Kr]4d75s1
76Os [Xe]4f 145d66s2
27Co [Ar]3d74s2
45Rh [Kr]4d85s1
77Ir [Xe]4f 145d76s2
28Ni [Ar]3d84s2
46Pd [Kr]4d10
78Pt [Xe]4f 145d96s1
29Cu [Ar]3d104s1
47Ag [Kr]4d105s1
79Au [Xe]4f 145d106s1
30Zn [Ar]3d104s2
48Cd [Kr]4d105s2
80Hg [Xe]4f 145d106s2
We often write water as OH2 rather
than H2O when we want to emphasize
that oxygen is the donor atom.
Several of the apparent irregularities in
these electron configurations can be
explained by the special stability of
half-filled and filled sets of d orbitals
(Section 5-17).
Many important biological substances are coordination compounds. Hemoglobin and
chlorophyll are two examples (Figure 25-1). Hemoglobin is a protein that carries O2 in
blood. It contains iron(II) ions bound to large porphyrin rings. The transport of oxygen
by hemoglobin involves the coordination and subsequent release of O2 molecules by the
Fe(II) ions. Chlorophyll is necessary for photosynthesis in plants. It contains magnesium
ions bound to porphyrin rings. Vitamin B12 is a large complex of cobalt. Coordination
compounds have many practical applications in such areas as medicine, water treatment,
soil and plant treatment, protection of metal surfaces, analysis of trace amounts of metals,
electroplating, and textile dyeing.
Bonding in transition metal complexes was not understood until the pioneering research
of Alfred Werner (1866–1919), a Swiss chemist who received the Nobel Prize in chemistry
in 1913. Great advances have been made since in the field of coordination chemistry,
but Werner’s work remains the most important contribution by a single researcher.
Prior to Werner’s work, the formulas of transition metal complexes were written with
dots, CrCl3и6H2O, AgClи2NH3, just like double salts such as iron(II) ammonium sulfate
hexahydrate, FeSO4и(NH4)2SO4и6H2O. The properties of solutions of double salts are
the properties expected for solutions made by mixing the individual salts. However, a solution
of AgClи2NH3, or more properly [Ag(NH3)2]Cl, behaves differently from either a
solution of (very insoluble) silver chloride or a solution of ammonia. The dots have been
called “dots of ignorance,” because they signified that the mode of bonding was unknown.
Table 25-2 summarizes the types of experiments Werner performed and interpreted to
lay the foundations for modern coordination theory.
Werner isolated platinum(IV) compounds with the formulas that appear in the first
column of Table 25-2. He added excess AgNO3 to solutions of carefully weighed amounts
of the five salts. The precipitated AgCl was collected by filtration, dried, and weighed.
Double salts are ionic solids resulting
from the cocrystallization of two salts
from the same solution into a single
structure. In the example given, the
solid is produced from an aqueous
solution of iron(II) sulfate, FeSO4,
and ammonium sulfate, (NH4)2SO4.
974 CHAPTER 25: Coordination Compounds
Fe
CH2
N N
N N
CH2
CH2CH2
COOHCOOH
CH3CH3
CHCH3 CH2
CH
CH2
CH3
Porphyrin ring system + iron = heme group
(shown as disks at left)(a)
Mg
C
N N
N N
CH2
CH2
COOCH3
COOC20H39
CH3CH3
CH2CH3CH3
CH
CH2
CH3
C
HC
O
Chlorophyll a(b)
Fe
Fe
Fe
Fe
Figure 25-1 (a) A model of a hemoglobin molecule (MW ϭ 64,500 amu). Individual atoms
are not shown. The four heme groups in a hemoglobin molecule are represented by disks.
Each heme group contains one Fe2ϩ ion and porphyrin rings. A single red blood cell
contains more than 265 million hemoglobin molecules and more than 1 billion Fe2ϩ ions.
(b) The structure of chlorophyll a, which also contains a porphyrin ring with a Mg2ϩ ion at
its center. Chlorophyll is necessary for photosynthesis. The porphyrin ring is the part of the
molecule that absorbs light. The structure of chlorophyll b is slightly different.
He determined the number of moles of AgCl produced. This told him the number of ClϪ
ions precipitated per formula unit. The results are in the second column. Werner reasoned
that the precipitated ClϪ ions must be free (uncoordinated), whereas the unprecipitated
ClϪ ions must be bonded to Pt so they could not be precipitated by Agϩ ions. He also
measured the conductances of solutions of these compounds of known concentrations. By
comparing these with data on solutions of simple electrolytes, he found the number of
ions per formula unit. The results are shown in the third column. Piecing the evidence
together, he concluded that the correct formulas are the ones listed in the last two columns.
The NH3 and ClϪ within the brackets are bonded by coordinate covalent bonds to the
Lewis acid, Pt(IV) ion.
The charge on a complex is the sum of its constituent charges.
We can use this relationship to determine or confirm the charge on a complex species.
For example, the charge on [Pt(NH3)6]4ϩ can be calculated as
Charge ϭ [charge on Pt(IV)] ϩ 6 ϫ (charge on NH3)
ϭ (4ϩ) ϩ 6 ϫ (0) ϭ 4ϩ
The charge on [Pt(NH3)4Cl2]2ϩ is
Charge ϭ [charge on Pt(IV)] ϩ 4 ϫ (charge on NH3) ϩ 2 ϫ (charge on ClϪ)
ϭ (4ϩ) ϩ 4 ϫ (0) ϩ 2 ϫ (1Ϫ) ϭ 2ϩ
AMMINE COMPLEXES
Ammine complexes contain NH3 molecules bonded to metal ions. Because the ammine
complexes are important compounds, we will describe some of them briefly.
Most metal hydroxides are insoluble in water, and so aqueous NH3 reacts with nearly
all metal ions to form insoluble metal hydroxides, or hydrated oxides.
2[(NH3(aq) ϩ H2O(ᐉ) 34 NH4
ϩ(aq) ϩ OHϪ(aq)]
Cu2ϩ(aq) ϩ 2OHϪ(aq) 88n Cu(OH)2(s)
Cu2ϩ(aq) ϩ 2NH3(aq) ϩ 2H2O(ᐉ) 88n Cu(OH)2(s) ϩ 2NH4
ϩ(ag)
25-2
The conductance of a solution of an
electrolyte is a measure of its ability to
conduct electricity. It is related to the
number of ions and the charges on
ions in solution.
25-2 Ammine Complexes 975
TABLE 25-2 Interpretation of Experimental Data by Werner
Moles AgCl Number of Ions
Precipitated per per Formula Unit
Formula Formula Unit (based on conductance) True Formula Ions/Formula Unit
PtCl4и6NH3 4 5 [Pt(NH3)6]Cl4 [Pt(NH3)6]4ϩ 4 ClϪ
PtCl4и5NH3 3 4 [Pt(NH3)5Cl]Cl3 [Pt(NH3)5Cl]3ϩ 3 ClϪ
PtCl4и4NH3 2 3 [Pt(NH3)4Cl2]Cl2 [Pt(NH3)4Cl2]2ϩ 2 ClϪ
PtCl4и3NH3 1 2 [Pt(NH3)3Cl3]Cl [Pt(NH3)3Cl3]ϩ 2 ClϪ
PtCl4и2NH3 0 0 [Pt(NH3)2Cl4] no ions
Colors of coordination compounds
depend on which metals and
ligands are present. From left: the
[Ni(NH3)6]2ϩ ion is purple; the
[Ni(H2O)6]2ϩ ion is green; the
[Cu(H2O)4]2ϩ ion is light blue; and
the [Cu(NH3)4]2ϩ ion is deep blue.
Similarly,
Cr3ϩ ϩ 3NH3 ϩ 3H2O 88n Cr(OH)3(s) ϩ 3NH4
ϩ
In general terms we can represent this reaction as
Mnϩ ϩ nNH3 ϩ nH2O 88n M(OH)n(s) ϩ nNH4
ϩ
where Mnϩ represents all of the common metal ions except the cations of the strong bases
(Group IA cations and the heavier members of Group IIA: Ca2ϩ, Sr2ϩ, and Ba2ϩ).
The hydroxides of some metals and some metalloids are amphoteric (Section 10-6).
Aqueous NH3 is a weak base (Kb ϭ 1.8 ϫ 10Ϫ5), so the [OHϪ] is too low to dissolve
amphoteric hydroxides to form hydroxo complexes.
However, several metal hydroxides do dissolve in an excess of aqueous NH3 to form
ammine complexes. For example, the hydroxides of copper and cobalt are readily soluble
in an excess of aqueous ammonia solution.
Co(OH)2 (a blue compound that
turns gray quickly) dissolves in
excess aqueous NH3 to form
[Co(NH3)6]2ϩ ions (yellow-orange).
976 CHAPTER 25: Coordination Compounds
TABLE 25-3 Common Metal Ions That Form Soluble Complexes with an
Excess of Aqueous Ammoniaa
Insoluble Hydroxide Formed Complex Ion Formed
Metal Ion by Limited Aq. NH3 by Excess Aq. NH3
Co2ϩ Co(OH)2 [Co(NH3)6]2ϩ
Co3ϩ Co(OH)3 [Co(NH3)6]3ϩ
Ni2ϩ Ni(OH)2 [Ni(NH3)6]2ϩ
Cuϩ CuOH 88n ᎏ1
2
ᎏCu2Ob [Cu(NH3)2]ϩ
Cu2ϩ Cu(OH)2 [Cu(NH3)4]2ϩ
Agϩ AgOH 88n ᎏ1
2
ᎏAg2Ob [Ag(NH3)2]ϩ
Zn2ϩ Zn(OH)2 [Zn(NH3)4]2ϩ
Cd2ϩ Cd(OH)2 [Cd(NH3)4]2ϩ
Hg2ϩ Hg(OH)2 [Hg(NH3)4]2ϩ
aThe ions of Rh, Ir, Pd, Pt, and Au show similar behavior.
bCuOH and AgOH are unstable and decompose to the corresponding oxides.
Cu(OH)2 (light blue) dissolves in excess aqueous NH3 to form [Cu(NH3)4]2ϩ ions (deep blue).
Cu(OH)2(s) ϩ 4NH3 34 [Cu(NH3)4]2ϩ ϩ 2OHϪ
Co(OH)2(s) ϩ 6NH3 34 [Co(NH3)6]2ϩ ϩ 2OHϪ
Interestingly, the metal hydroxides that exhibit this behavior are derived from the 12 metals
of the cobalt, nickel, copper, and zinc families. All the common cations of these metals
except Hg2
2ϩ (which disproportionates) form soluble complexes in the presence of excess
aqueous ammonia (Table 25-3).
IMPORTANT TERMS
The Lewis bases in coordination compounds may be molecules, anions, or (rarely) cations.
They are called ligands (Latin ligare, “to bind”). The donor atoms of the ligands are the
atoms that donate shares in electron pairs to metals. In some cases it is not possible to
identify donor atoms, because the bonding electrons are not localized on specific atoms.
Some small organic molecules such as ethylene, H2CUCH2, bond to a transition metal
through the electrons in their double bonds. Examples of typical simple ligands are listed
in Table 25-4.
Ligands that can bond to a metal through only one donor atom at a time are monodentate
(Latin dent, “tooth”). Ligands that can bond simultaneously through more than one
donor atom are polydentate. Polydentate ligands that bond through two, three, four, five,
or six donor atoms are called bidentate, tridentate, quadridentate, quinquedentate, and hexadentate,
respectively. Complexes that consist of a metal atom or ion and polydentate ligands
are called chelate complexes (Greek chele, “claw”).
The coordination number of a metal atom or ion in a complex is the number of
donor atoms to which it is coordinated, not necessarily the number of ligands. The coordination
sphere includes the metal or metal ion (called the central atom) and its
ligands, but no uncoordinated counterions. For example, the coordination sphere of
hexaamminecobalt(III) chloride, [Co(NH3)6]Cl3, is the hexaamminecobalt(III) ion,
[Co(NH3)6]3ϩ. These terms are illustrated in Table 25-5.
25-3
TABLE 25-4 Typical Simple Ligands with Their Donor Atoms Shaded
Molecule Name Name as Ligand Ion Name Name as Ligand
ammonia ammine chloride chloro
water aqua fluoride fluoro
carbon monoxide carbonyl cyanide cyanoa
phosphine phosphine hydroxide hydroxo
nitrogen oxide nitrosyl nitrite nitrob
aNitrogen atoms can also function as donor atoms, in which case the ligand name is “isocyano.”
bOxygen atoms can also function as donor atoms, in which case the ligand name is “nitrito.”
NH3
OH2
C O
PH3
N O
Cl
Ϫ
F
Ϫ
C N Ϫ
OHϪ
N
O Ϫ
O
TABLE 25-5 Some Ligands and Coordination Spheres (complexes)
Coordination Oxidation Coordination
Ligand(s) Classification Sphere Number of M Number of M
NH3 monodentate [Co(NH3)6]3ϩ ϩ3 6
ammine hexaamminecobalt(III)
bidentate ϩ3 6
BrϪ monodentate
bromo
H2NXCH2XCH2XNH2 bidentate ϩ2 4
ethylenediamine (en)
tridentate ϩ3 6
hexadentate ϩ3 6
H2N CH2 CH2 NH2
(or N N)
ethylenediamine (en)
H2
N
N
H2
CH2
3
3ϩ
CH2
Co
[Co(en)3]3ϩ
tris(ethylenediamine)cobalt(III)
ion
H2
N
N
H2
CH2
0
CH2
Br2Cu
[Cu(en)Br2]
dibromoethylenediamine-
copper(II)
H2N NH2(CH2)2 N
H
(CH2)2
(or N N)N
diethylenetriamine (dien)
H2
N
N
H2
CH2 2
3ϩ
CH2
CH2
CH2
Fe
[Fe(dien)2]3ϩ
bis(diethylenetriamine)iron(III)
ion
H
N
N
N
(CH2)2
C
O
CH2
ethylenediaminetetraacetato (edta)
C
CC
O O
O O
CH2
CH2CH2
Ϫ
O
Ϫ
O
Ϫ
O
Ϫ
Co
ON
N O
Ϫ
H2C O
O
O
C
H2C
H2
C
CC
C
H2
H2C
H2C O
O
O
C
[Co(edta)]Ϫ
(ethylenediaminetetraacetato)cobaltate(III)
ion





NOMENCLATURE
The International Union of Pure and Applied Chemistry (IUPAC) has adopted a set of
rules for naming coordination compounds. The rules are based on those originally devised
by Werner.
1. Cations are always named before anions, with a space between their names.
2. In naming the coordination sphere, ligands are named in alphabetical order. The
prefixes di ϭ 2, tri ϭ 3, tetra ϭ 4, penta ϭ 5, hexa ϭ 6, and so on specify the number
of each kind of simple (monodentate) ligand. For example, in dichloro, the “di” indicates
that two ClϪ ions act as ligands. For complicated ligands (polydentate chelating
agents), other prefixes are used, including: bis ϭ 2, tris ϭ 3, tetrakis ϭ 4,
pentakis ϭ 5, and hexakis ϭ 6. The names of complicated ligands are enclosed in
parentheses. The numeric prefixes are not used in alphabetizing. When a prefix
denotes the number of substituents on a single ligand, as in dimethylamine,
NH(CH3)2, it is used to alphabetize ligands.
3. The names of anionic ligands end in the suffix -o. Examples are FϪ, fluoro; OHϪ,
hydroxo; O2Ϫ, oxo; S2Ϫ, sulfido; CO3
2Ϫ, carbonato; CNϪ, cyano; SO4
2Ϫ, sulfato;
NO3
Ϫ, nitrato; S2O3
2Ϫ, thiosulfato.
4. The names of neutral ligands are usually unchanged. Four important exceptions are
NH3, ammine; H2O, aqua; CO, carbonyl; and NO, nitrosyl.
5. Some metals exhibit variable oxidation states. The oxidation number of such a metal
is designated by a Roman numeral in parentheses following the name of the complex
ion or molecule.
6. The suffix “-ate” at the end of the name of the complex signifies that it is an anion.
If the complex is neutral or cationic, no suffix is used. The English stem is usually
used for the metal, but where the naming of an anion is awkward, the Latin stem
is substituted. For example, “ferrate” is used rather than “ironate,” and “plumbate”
rather than “leadate” (Table 25-6).
The following examples illustrate these rules.
K2[Cu(CN)4] potassium tetracyanocuprate(II)
[Ag(NH3)2]Cl diamminesilver chloride
[Cr(OH2)6](NO3)3 hexaaquachromium(III) nitrate
[Co(en)2Br2]Cl dibromobis(ethylenediamine)cobalt(III) chloride
[Ni(CO)4] tetracarbonylnickel(0)
Na[Al(OH)4] sodium tetrahydroxoaluminate
[Pt(NH3)4][PtCl6] tetraammineplatinum(II) hexachloroplatinate(IV)
[Cu(NH3)2(en)]Br2 diammine(ethylenediamine)copper(II) bromide
Na2[Sn(OH)6] sodium hexahydroxostannate(IV)
[Co(en)3](NO3)3 tris(ethylenediamine)cobalt(III) nitrate
K4[Ni(CN)2(ox)2] potassium dicyanobis(oxalato)nickelate(II)
[Co(NH3)4(OH2)Cl]Cl2 tetraammineaquachlorocobalt(III) chloride
25-4
The term ammine (two m’s) signifies
the presence of ammonia as a ligand.
It is different from the term amine
(one m), which describes some organic
compounds (Section 27-12) that are
derived from ammonia.
25-4 Nomenclature 979
TABLE 25-6 Names for
Some Metals
in Complex
Anions
Name* of Metal
Metal in Complex Anions
aluminum aluminate
antimony antimonate
chromium chromate
cobalt cobaltate
copper cuprate
gold aurate
iron ferrate
lead plumbate
manganese manganate
nickel nickelate
platinum platinate
silver argentate
tin stannate
zinc zincate
*Stems derived from Latin names for metals
are shown in italics.
The abbreviation ox represents the
oxalate ion (COO)2
2Ϫ or C2O4
2Ϫ.
C C
OO
O O
ϪϪ
The oxidation state of aluminum is not
given because it is always ϩ3.
980 CHAPTER 25: Coordination Compounds
TABLE 25-7 Idealized Geometries for Various Coordination Numbers
Coordination Number Geometry Examples Model
2 [Ag(NH3)2]ϩ
[Ag(NH3)2]ϩ
⌴⌴⌴⌴⌴⌴⌴⌴⌴⌴⌴⌴⌴⌴
[Cu(CN)2]Ϫ
[Zn(CN)4]2Ϫ
[Cd(NH3)4]2ϩ
[Cd(NH3)4]2ϩ4
[Ni(CN)4]2Ϫ
4
[Cu(OH2)4]2ϩ
[Pt(NH3)2Cl2]
[Cu(OH2)4]2ϩ
[Fe(CO)5]
[CuCl5]3Ϫ
5
[Fe(CO)5]
[Ni(CN)5]3Ϫ
[MnCl5]3Ϫ
5
[Fe(CN)6]4Ϫ
[Fe(OH2)6]2ϩ
6 [Fe(CN)6]4Ϫ
L M L
linear
L
M
LL
L
tetrahedral
M
LL
L L
square planar
trigonal bipyramidal
L
L
L
L
L
M
L
L
LL
L
M
square pyramidal
L
octahedral
L
LL
L
L
M
[MnCl5]3Ϫ
STRUCTURES
The structures of coordination compounds are governed largely by the coordination
number of the metal. Many have structures similar to the simple molecules and ions we
studied in Chapter 8. Unshared pairs of electrons in d orbitals usually have only small
influences on geometry because they are not in the outer shell. Table 25-7 summarizes
the geometries for common coordination numbers.
Both tetrahedral and square planar geometries are common for complexes with coordination
number 4. For coordination number 5, the trigonal bipyramidal structure and
the square pyramidal structure are both common. Transition metal complexes with coordination
numbers as high as 7, 8, and 9 are known. The geometries tabulated are ideal
geometries. Actual structures are sometimes distorted, especially if the ligands are not all
the same. The distortions are due to compensations for the unequal electric fields generated
by the different ligands.
ISOMERISM IN COORDINATION COMPOUNDS
Isomers are different compounds that have the same molecular formula; they have the
same number and kinds of atoms arranged differently. The term isomers comes from the
Greek word meaning “equal weights.” Because their structures are different, isomers have
different physical and chemical properties. Here we shall restrict our discussion of isomerism
to that caused by different arrangements of ligands about central metal ions.
There are two major classes of isomers: structural (constitutional) isomers and stereoisomers.
For coordination compounds, each can be further subdivided as follows.
Structural Isomers Stereoisomers
1. ionization isomers 1. geometric (positional) isomers
2. hydrate isomers 2. optical isomers
3. coordination isomers
4. linkage isomers
25-5
25-5 Structures 981
Some coordination compounds. Starting at the top
left and moving clockwise:
[Cr(CO)6] (white), CO is the ligand.
K3{Fe[(COO)2]3} (green), (COO)2
2Ϫ
(oxalate ion) is the ligand.
[Co(H2NXCH2XCH2XNH2)3]I3 (yelloworange),
ethylenediamine is the ligand.
[Co(NH3)5(OH2)]Cl3 (red), NH3 and H2O
are ligands.
K3[Fe(CN)6] (red-orange), CNϪ is the ligand.
(A drop of water fell on this sample.)
Differences between structural isomers involve either more than one coordination sphere
or different donor atoms on the same ligand. They contain different atom-to-atom bonding
sequences. Simple stereoisomers of coordination compounds involve only one coordination
sphere and the same ligands and donor atoms. Before considering stereoisomers, we will
describe the four types of structural isomers.
STRUCTURAL (CONSTITUTIONAL) ISOMERS
Ionization (Ion–Ion Exchange) Isomers
These isomers result from the interchange of ions inside and outside the coordination
sphere. For example, red-violet [Co(NH3)5Br]SO4 and red [Co(NH3)5SO4]Br are ionization
isomers.
In structure A the SO4
2Ϫ ion is free and is not bound to the cobalt(III) ion. A solution of
A reacts readily with a solution of barium, BaCl2, to precipitate BaSO4, but does not react
readily with AgNO3. In structure B the SO4
2Ϫ ion is bound to the cobalt(III) ion and so
it does not react with BaCl2 in aqueous solution. The BrϪ ion is free, however, and a solution
of B reacts with AgNO3 to precipitate AgBr. Equimolar solutions of A and B also have
different electrical conductivities. The sulfate solution, A, conducts electric current better
because its ions have 2ϩ and 2Ϫ charges rather than 1ϩ and 1Ϫ. Other examples of this
type of isomerism include
[Pt(NH3)4Cl2]Br2 and [Pt(NH3)4Br2]Cl2
[Pt(NH3)4SO4](OH)2 and [Pt(NH3)4(OH)2]SO4
[Co(NH3)5NO2]SO4 and [Co(NH3)5SO4]NO2
[Cr(NH3)5SO4]Br and [Cr(NH3)5Br]SO4
Hydrate Isomers
In some crystalline complexes, water can be inside and outside the coordination sphere. For
example, when treated with excess AgNO3(aq), solutions of the following three hydrate
isomers yield three, two, and one mole of AgCl precipitate, respectively, per mole of
complex.
Br
NH3
NH3
H3N
H3N NH3
2ϩ
[Co(NH3)5Br]SO4
pentaamminebromocobalt(III)
sulfate
A
SO4
2Ϫ
[Co(NH3)5SO4]Br
pentaamminesulfatocobalt(III)
bromide
B
BrϪ
Co
OSO3
NH3
NH3
H3N
H3N NH3
ϩ
Co
25-6
982 CHAPTER 25: Coordination Compounds
Isomers such as those shown here may
or may not exist in the same solution in
equilibrium. Such isomers are formed
by different reactions.
We write the sulfato ligand as OSO3
to emphasize that it is coordinated
through a lone pair on one of the O
atoms.
Recall that BaSO4 (Ksp ϭ 1.1 ϫ 10Ϫ10)
and AgBr (Ksp ϭ 3.3 ϫ 10Ϫ13) are
classified as insoluble in H2O, whereas
BaBr2 is soluble and Ag2SO4 (Ksp ϭ
1.7 ϫ 10Ϫ5) is moderately soluble.
In A:
Ba2ϩ ϩ SO4
2Ϫ 88n BaSO4(s)
Agϩ ϩ SO4
2Ϫ 88n no rxn
In B:
Agϩ ϩ BrϪ 88n AgBr(s)
Ba2ϩ ϩ BrϪ 88n no rxn
Coordination Isomers
Coordination isomerism can occur in compounds containing both complex cations and
complex anions. Such isomers involve exchange of ligands between cation and anion, that
is, between coordination spheres.
Linkage Isomers
Certain ligands can bind to metal ions in more than one way. Examples of such ligands
are cyano, XCNϪ, and isocyano, XNCϪ; nitro, XNO2
Ϫ, and nitrito, XONOϪ. The
donor atoms are on the left in these representations.
Cl
Cl
NH3
H3N
H3N NH3
2ϩ 2Ϫ
[Pt(NH3)4][PtCl6],
tetraammineplatinum(II) hexachloroplatinate(IV)
Pt
ClCl
Cl Cl
Pt
Cl
Cl
NH3
H3N
H3N NH3
2ϩ 2Ϫ
[Pt(NH3)4Cl2][PtCl4],
tetraamminedichloroplatinum(IV) tetrachloroplatinate(II)
Pt
ClCl
Cl Cl
Pt
25-6 Structural (Constitutional) Isomers 983
OH2
OH2
OH2
H2O
H2O OH2
3ϩ 2ϩ
[Cr(OH2)6]Cl3
hexaaquachromium(III)
chloride
3 ClϪCr
[Cr(OH2)4Cl2]Cl · 2H2O
tetraaquadichlorochromium(III)
chloride dihydrate
ClϪ
,
2H2O
Cl
Cl
OH2
H2O
H2O OH2
ϩ
Cr
[Cr(OH2
)5
Cl]Cl2
· H2
O
pentaaquachlorochromium(III)
chloride hydrate
2 ClϪ
,
H2O
Cl
OH2
OH2
H2O
H2O OH2
Cr
violet blue-green green
STEREOISOMERS
Compounds that contain the same atoms and the same atom-to-atom bonding sequences,
but that differ only in the spatial arrangements of the atoms relative to the central atom,
are stereoisomers. Complexes with only simple ligands can exist as stereoisomers only if
they have coordination number 4 or greater. The most common coordination numbers
among coordination complexes are 4 and 6, and so they will be used to illustrate stereoiso-
merism.
Geometric (cis–trans) Isomers
In geometric isomers, or cis-trans isomers, of coordination compounds, the same ligands
are arranged in different orders within the coordination sphere. Geometric isomerism
occurs when atoms or groups of atoms can be arranged on two sides of a rigid structure.
Cis means “adjacent to” and trans means “on the opposite side of.” Cis- and transdiamminedichloroplatinum(II)
are shown below.
25-7
NH3
H3N
N O
O
H3N
H3N NH3
2ϩ 2ϩ
[Co(NH3)5NO2]Cl2
pentaamminenitrocobalt(III)
chloride
yellow, stable in acids
2 ClϪ
[Co(NH3)5ONO]Cl2
pentaamminenitritocobalt(III)
chloride
red, decomposes in acids
2 ClϪ
NH3
H3N
Co
H3N
H3N NH3
Co
O
ON
Pentaamminenitrocobalt(III) ion,
[Co(NH3)5NO2]2ϩ
984 CHAPTER 25: Coordination Compounds
Pentaamminenitritocobalt(III) ion,
[Co(NH3)5ONO]2ϩ
A complex with coordination number
2 or 3 that contains only simple ligands
can have only one spatial arrangement.
Try building models to see this.
ClXPtXCl angle Ϸ 90° ClXPtXCl angle Ϸ 180°
NXPtXN angle Ϸ 90° NXPtXN angle Ϸ 180°
NH3
Cl
ClH3N
trans
dark yellow
Pt
NH3
Cl
Cl NH3
cis
pale yellow
Pt
cis-diamminedichloroplatinum(II),
cis-[Pt(NH3)2Cl2]
25-7 Stereoisomers 985
In the cis isomer, the chloro groups are closer to each other (on the same side of the
square) than they are in the trans isomer. The ammine groups are also closer together in
the cis complex.
Geometric isomerism is possible for octahedral complexes. For example, complexes of
the type MA4B2 can exist in two isomeric forms. Consider as an example the complex ion
[Co(NH3)4Cl2]ϩ. The two like ligands (ClϪ) can be either cis or trans to each other. These
two complexes are different colors: solutions and salts of the cis isomer are violet and those
of the trans isomer are green.
NH3
Cl
Cl NH3
NH3
NH3
ϩ
cis-tetraamminedichlorocobalt(III) ion,
cis-[Co(NH3)4Cl2]
ϩ
Co
A stereoview of cis-tetraamminedichlorocobalt(III) ion, cis-[Co(NH3)4Cl2]ϩ. The two chloro
ligands are adjacent to each other.
trans-diamminedichloroplatinum(II),
trans-[Pt(NH3)2Cl2]
The cis isomer has been used
successfully in cancer chemotherapy.
The trans isomer has no such activity.
Cis–trans isomerism is not possible for
tetrahedral complexes, in which all
angles are (ideally) 109.5°.
ϩ
trans-tetraamminedichlorocobalt(III) ion,
trans-[Co(NH3)4Cl2]ϩ
Co
NH3
NH3
NH3
Cl
H3N Cl
A stereoview of trans-tetraamminedichlorocobalt(III) ion, trans-[Co(NH3)4Cl2]ϩ. The
two chloro ligands are across from each other.
986 CHAPTER 25: Coordination Compounds
Figure 25-2 The dichlorobis(ethylenediamine)cobalt(III) ion, [Co(en)2Cl2]ϩ,
exists as a pair of cis–trans isomers. Ethylenediamine is represented as .
Cis-dichlorobis(ethylenediamine)cobalt(III) perchlorate, [Co(en)2Cl2]ClO4, is purple.
Trans-dichlorobis(ethylenediamine)cobalt(III) chloride, [Co(en)2Cl2]Cl, is green.
N N
Co
Cl
Cl
N
N
N
N
+
cis-isomer
Co
Cl
N
Cl
N
N
N
+
trans-isomer
Complexes involving bidentate ligands, such as ethylenediamine, can also exhibit this
kind of isomerism, as shown in Figure 25-2.
Octahedral complexes with the general formula MA3B3 can exhibit another type of
geometric isomerism, called mer–fac isomerism. This can be illustrated with the complex
ion [Pt(NH3)3Cl3]ϩ (see Table 25-2). In one isomer the three similar ligands (e.g., the ClϪ
ligands) lie at the corners of a triangular face of the octahedron; this is called the fac isomer
(for facial).
In the other isomer, the three similar ligands lie in the same plane; this is called the
mer isomer (so called because the plane is analogous to a meridian on a globe).
Complexes of the type [MA2B2C2] can exist in several isomeric forms. Consider as an
example [Cr(OH2)2(NH3)2Br2]ϩ. First, the members of all three pairs of like ligands may
be either trans to each other (A) or cis to each other (B).
cis-diammine-cis-diaqua-cisdibromochromium(III)
ion
B
ϩ
Cr
OH2
NH3
NH3
Br
Br OH2
ϩ
trans-diammine-trans-diaqua-transdibromochromium(III)
ion
A
Cr
OH2
OH2
BrH3N
Br NH3
A stereoview of mer–triammine-trichloroplatinum(IV) ion, mer-[Pt(NH3)3Cl3]ϩ. The
three chloro ligands are all one meridian (in a plane), and the three ammine ligands
are all on a perpendicular meridian (a plane perpendicular to the first).
Pt
NH3
Cl
ClCl
H3N NH3
fac-triamminetrichloroplatinum(IV) ion,
fac-[Pt(NH3)3Cl3]ϩ
ϩ
A stereoview of fac-triamminetrichloroplatinum(IV) ion, fac-[Pt(NH3)3Cl3]ϩ. The three
chloro ligands are all on one triangular face of an octahedron, and the three ammine
ligands are all on the opposite triangular face.
Pt
NH3
Cl
NH3
Cl
H3N Cl
mer-triamminetrichloroplatinum(IV) ion,
mer-[Pt(NH3)3Cl3]ϩ
ϩ
Cl
Then, members of one of the pairs may be trans to each other, but the members of the
other two pairs are cis.
There is no trans–trans–cis isomer.
Why?
988 CHAPTER 25: Coordination Compounds
Optical isomers of cis-diammine-cis-diaqua-cis-dibromochromium(III) ion
B-I B-II
Rotate by
180Њ
ϩ
Cr
OH2
NH3
NH3
Br
Br OH2
ϩ
Cr
OH2
NH3
BrH3N
H2O Br
B-II؅
ϩ
Cr
OH2
NH3
OH2
Br
Br NH3
cis-diammine-trans-diaquacis-dibromochromium(III)
ion
C
ϩ
Cr
OH2
OH2
NH3
Br
Br NH3
trans-diammine-cis-diaquacis-dibromochromium(III)
ion
D
ϩ
Cr
NH3
NH3
OH2
Br
Br OH2
cis-diammine-cis-diaquatrans-dibromochromium(III)
ion
E
ϩ
Cr
Br
Br
OH2
H3N
H3N OH2
In C, only the two H2O ligands are in a trans arrangement; in D, only the NH3 ligands
are trans; and in E, only the two BrϪ ligands are trans.
Further interchange of the positions of the ligands produces no new geometric isomers.
However, one of the five geometric isomers (B) can exist in two distinct forms called optical
isomers.
Optical Isomers
The cis-diammine-cis-diaqua-cis-dibromochromium(III) geometrical isomer (B) exists in
two forms that bear the same relationship to each other as left and right hands. They are
nonsuperimposable mirror images of each other and are called optical isomers or enan-
tiomers.
Complexes B-I and B-II are mirror images of each other. To see that they are not identical,
imagine rotating B-II about its vertical axis by 180° (B-IIЈ), so that the two BrϪ
ligands are at the left, as they are in B-I. Then B-IIЈ has the OH2 ligand at the right front
position and the NH3 ligand at the right rear position, which is not the same as in B-I.
No rotation of B-II makes it identical to B-I. Thus these two arrangements are nonsuperimposable
mirror images of each other. Stereo images of these two complexes appear in
Figure 25-3.
An object that is not superimposable
with its mirror image is said to be
chiral.
Optical isomers interact with polarized light in different ways. Separate equimolar solutions
of each rotate a plane of polarized light (see Figures 25-4 and 25-5) by equal amounts
but in opposite directions. One solution is dextrorotatory (rotates to the right) and the
other is levorotatory (rotates to the left). Optical isomers are called dextro and levo isomers.
The phenomenon by which a plane of polarized light is rotated is called optical activity.
It can be measured with a device called a polarimeter (Figure 25-5) or with more sophisticated
instruments. A single solution containing equal amounts of the two isomers is a
racemic mixture. This solution does not rotate a plane of polarized light. The equal and
opposite effects of the two isomers exactly cancel. To exhibit optical activity, the dextro
and levo isomers must be separated from each other. This is done by one of a number of
chemical or physical processes broadly called optical resolution.
Alfred Werner was also the first person
to demonstrate optical activity in an
inorganic compound (not of biological
origin). This demonstration silenced
critics of his theory of coordination
compounds, and in his opinion, it was
his greatest achievement. Louis Pasteur
had demonstrated the phenomenon of
optical activity many years earlier in
organic compounds of biological
origin.
25-7 Stereoisomers 989
Figure 25-3 The optical isomers
of the cis-diammine-cis-diaqua-cisdibromochromium(III)
ion.
990 CHAPTER 25: Coordination Compounds
Sodium
lamp
Polarizing filter
Polarizing filter
oriented parallel
to first
Polarizing filter
oriented perpendicular
to first
No light is
transmitted
Undiminished
intensity
“Plane” of
polarized light
Monochromatic
light
Polarizing
filter
Light
source
Sample
cell
Detector
␪
␪
Figure 25-4 Light from a lamp or from the sun consists of electromagnetic waves that
vibrate in all directions perpendicular to the direction of travel. Polarizing filters absorb all
waves except those that vibrate in a single plane. The third polarizing filter, with a plane of
polarization at right angles to the first, absorbs the polarized light completely.
Figure 25-5 The plane of polarization of plane-polarized light is rotated through an angle
(␪) as it passes through an optically active medium. Species that rotate the plane to the right
(clockwise) are dextrorotatory, and those that rotate it to the left are levorotatory.
BONDING IN COORDINATION COMPOUNDS
Bonding theories for coordination compounds should be able to account for structural
features, colors, and magnetic properties. The earliest accepted theory was the valence
bond theory (Chapter 8). It can account for structural and magnetic properties, but it
offers no explanation for the wide range of colors of coordination compounds. The crystal
field theory gives satisfactory explanations of color as well as of structure and magnetic
properties for many coordination compounds. We will therefore discuss only this more
successful theory in the remainder of this chapter.
CRYSTAL FIELD THEORY
Hans Bethe (1906– ) and J. H. van Vleck (1899–1980) developed the crystal field theory
between 1919 and the early 1930s. It was not widely used, however, until the 1950s. In
its original form, it assumed that the bonds between ligand and metal were completely
ionic.
In a metal ion surrounded by other atoms, the d orbitals are at higher energy than they
are in an isolated metal ion. If the surrounding electrons were uniformly distributed about
the metal ion, the energies of all five d orbitals would increase by the same amount (a
spherical crystal field). Because the ligands approach the metal ion from different directions,
they affect different d orbitals in different ways. Here we illustrate the application of these
ideas to complexes with coordination number 6 (octahedral crystal field).
25-8
Modern ligand field theory is based on
crystal field theory. It attributes partial
covalent character and partial ionic
character to bonds. It is a more
sophisticated theory, beyond the
scope of this text.
25-8 Crystal Field Theory 991
3ϩ
The two optical isomers of the tris(ethylenediamine)cobalt(III) ion
Rotate by
180Њ
Co
N
N
NN
N N
N
N
3ϩ
Co
N
N
NN
N N
3ϩ
Co
N
N
N
N
Another pair of optical isomers follows, each of which contains three molecules of
ethylenediamine, a bidentate ligand.
The dx2
–y2, and dz2 orbitals are directed along a set of mutually perpendicular x, y, and
z axes (p. 212). As a group, these orbitals are called eg orbitals. The dxy, dyz, and dxz
orbitals, collectively called t2g orbitals, lie between the axes. The ligand donor atoms
approach the metal ion along the axes to form octahedral complexes. Crystal field theory
proposes that the approach of the six donor atoms (point charges) along the axes sets up
an electric field (the crystal field). Electrons on the ligands repel electrons in eg orbitals
on the metal ion more strongly than they repel those in t2g orbitals (Figure 25-6). This
removes the degeneracy of the set of d orbitals and splits them into two sets, the eg set at
higher energy and the t2g set at lower energy.
The energy separation between the two sets is called the crystal field splitting energy,
⌬octahedral, or ⌬oct. It is proportional to the crystal field strength of the ligands—that is, how
strongly the ligand electrons repel the electrons on the metal ion.
The d electrons on a metal ion occupy the t2g set in preference to the higher energy
eg set. Electrons that occupy the eg orbitals are strongly repelled by the relatively close
approach of ligands. The occupancy of these orbitals tends to destabilize octahedral
complexes.
As always, electrons occupy the orbitals in the arrangement that results in the lowest
energy. The electron pairing energy, P, is the expenditure of energy that is necessary to
pair electrons by bringing two negatively charged particles into the same region of space.
We must compare this with the crystal field splitting energy, ⌬oct, the difference in energy
between t2g and eg. Weak field ligands such as FϪ cause only small splitting energies,
whereas strong field ligands such as CNϪ give large values of ⌬oct.
If the splitting energy, ⌬oct, is smaller than the pairing energy, P, the electrons occupy
all five nondegenerate orbitals singly before pairing. After all d orbitals are half-filled,
dxy dxz dyz dx2–y2 dz2
free metal ion
(no crystal field)
metal ion in
spherical
crystal field
dx2–y2 dz2 }eg
dxy dxz dyz
metal ion in
octahedral
crystal field
}t2g
⌬oct (not to scale)
Recall that degenerate orbitals are
orbitals of equal energy.
992 CHAPTER 25: Coordination Compounds
Figure 25-6 Effects of the
approach of ligands on the energies
of d orbitals on the metal ion. In an
octahedral complex, the ligands (L)
approach the metal ion (M) along
the x, y, and z axes, as indicated by
the blue arrows. (a) The orbitals of
the eg type—dx
2
–y
2 (shown here)
and dz
2 —point directly toward the
incoming ligands, so electrons in
these orbitals are strongly repelled.
(b) The orbitals of the t2g type—dxy
(shown here), dxz and dyz —do not
point toward the incoming ligands,
so electrons in these orbitals are less
strongly repelled.
M
L
L
L
L
L
L
(b) t2g type orbitals (dxy shown here)
L
L
L
L
L
L
M
(a) eg type orbitals (dx2 – y2 shown here)
Typical values of ⌬oct are between 100
and 400 kJ/mol.
additional electrons then pair with electrons in the t2g set. Such a complex would have
the same number of unpaired electrons on the metal atom or ion as when the metal is
uncomplexed; this is called a high spin complex. But if the splitting energy is greater
than the pairing energy, the electrons will be at lower energy if they pair in the t2g orbital
before any electrons occupy the higher energy eg orbitals. Such a complex could have
fewer unpaired electrons on the metal atom than when the metal is uncomplexed, so it is
called a low spin complex.
Let us now describe the hexafluorocobaltate(III) ion, [CoF6]3Ϫ, and the hexacyanocobalt(III)
ion, [Co(CN)6]3Ϫ. Both contain the d6 Co3ϩ ion. The [CoF6]3Ϫ ion is a
paramagnetic complex, whereas [Co(CN)6]3Ϫ is a diamagnetic complex. We will focus our
attention on the d electrons.
The Co3ϩ ion has six electrons (four unpaired) in its 3d orbitals.
3d
Co3ϩ [Ar] __hg __h __h __h __h
Magnetic measurements indicate that [CoF6]3Ϫ also has four unpaired electrons per ion.
So there must be four electrons in t2g orbitals and two in eg orbitals. Fluoride, FϪ, is a
weak field ligand, so the crystal field splitting energy is very small (⌬oct Ͻ P) and electron
pairing is unfavorable. Thus [CoF6]3Ϫ is a high spin complex.
On the other hand, [Co(CN)6]3Ϫ is diamagnetic, so all six d electrons must be paired
in the t2g orbitals. Cyanide ion, CNϪ, is a strong field ligand, which generates a large
crystal field splitting energy (⌬oct Ͼ P), making electron pairing more favorable;
[Co(CN)6]3Ϫ is a low spin complex.
The difference in configurations between [CoF6]3Ϫ and [Co(CN)6]3Ϫ is due to the relative
magnitudes of the crystal field splitting, ⌬oct, caused by the different crystal field
strengths of FϪ and CNϪ. The CNϪ ion interacts with vacant metal orbitals more strongly
than the FϪ ion does. As a result, the crystal field splitting generated by the close approach
of six CNϪ ions (strong field ligands) to the metal ion is greater than that generated by
the approach of six FϪ ions (weak field ligands).
A diamagnetic species has no unpaired
electrons. A paramagnetic species has
unpaired electrons, and the strength
of the paramagnetism depends on the
number of unpaired electrons. See
Chapter 5.
25-8 Crystal Field Theory 993
᭤[Co(OH2)6]2ϩ ions are pink (bottom). A limited amount of aqueous ammonia produces
Co(OH)2, a blue compound that quickly turns gray (middle). Co(OH)2 dissolves in excess
aqueous ammonia to form [Co(NH3)6]2ϩ ions, which are yellow-orange (top).
[CoF6]3Ϫ is a high spin complex.
[Co(CN)6]3Ϫ is a low spin complex.
Co3ϩ ion in
spherical
crystal field
}eg
[CoF6]3Ϫ
}t2g
⌬oct ϭ small; weak crystal
field due to FϪ
(not to scale)
3d, free Co3ϩ ion
Co3ϩ ion in
spherical
crystal field
}eg
[Co(CN)6]3Ϫ
}t2g
⌬oct ϭ large; strong crystal
field due to CNϪ
(not to scale)
3d, free Co3ϩ ion
⌬oct for [Co(CN)6]3Ϫ Ͼ ⌬oct for [CoF6]3Ϫ
Low spin configurations exist only for octahedral complexes having metal ions with d4,
d5, d6, and d7 configurations. For d1–d3 and d8–d10 ions, only one possibility exists. In
these cases, the configuration is designated as high spin. All dn possibilities are shown in
Table 25-8.
COLOR AND THE SPECTROCHEMICAL SERIES
A substance appears colored because it absorbs light that corresponds to one or more of
the wavelengths in the visible region of the electromagnetic spectrum (4000 to 7000 Å)
and transmits or reflects the other wavelengths. Our eyes are detectors of light in the
visible region, and so each wavelength in this region appears as a different color. A combination
of all wavelengths in the visible region is called “white light”; sunlight is an example.
The absence of all wavelengths in the visible region is blackness.
In Table 25-9 we show the relationships among colors absorbed and colors transmitted
or reflected in the visible region. The first column displays the wavelengths absorbed. The
spectral color is the color associated with the wavelengths of light absorbed by the sample.
When certain visible wavelengths are absorbed from incoming “white” light, the light not
absorbed remains visible to us as transmitted or reflected light. For instance, a sample that
absorbs orange light appears blue. The complementary color is the color associated with
the wavelengths that are not absorbed by the sample. The complementary color is seen
when the spectral color is removed from white light.
Most transition metal compounds are colored, a characteristic that distinguishes them
from most compounds of the representative elements. In transition metal compounds, the
d orbitals in any one energy level of the metals are not degenerate. No longer do all have
the same energy, as they do in isolated atoms. They are often split into two sets of orbitals
separated by energies, ⌬oct, that correspond to wavelengths of light in the visible region.
25-9
A convenient way to describe
d-transition metal ions is to indicate
the number of nonbonding electrons
in d orbitals.
994 CHAPTER 25: Coordination Compounds
TABLE 25-8 High and Low Spin Octahedral Configurations
dn Examples High Spin Low Spin dn Examples High Spin Low Spin
d1 Ti3ϩ __ __ eg same as high spin d6 Fe2ϩ, Ru2ϩ, Pd4ϩ, __h __h eg
__ __ eg
Rh3ϩ, Co3ϩ
__h __ __ t2g
__hg __h __h t2g
__hg __hg __hg t2g
d2 Ti2ϩ, V3ϩ __ __ eg same as high spin d7 Co2ϩ, Rh2ϩ __h __h eg
__h __ eg
__h __h __ t2g
__hg __hg __h t2g
__hg __hg __hg t2g
d3 V2ϩ, Cr3ϩ __ __ eg same as high spin d8 Ni2ϩ, Pt2ϩ __h __h eg same as high spin
__h __h __h t2g
__hg __hg __hg t2g
d4 Mn3ϩ, Re3ϩ __h __ eg
__ __ eg d9 Cu2ϩ __hg __h eg same as high spin
__h __h __h t2g
__hg __h __h t2g
__hg __hg __hg t2g
d5 Mn2ϩ, Fe3ϩ __h __h eg
__ __ eg d10 Zn2ϩ, Agϩ, Hg2ϩ __hg __hg eg same as high spin
Ru3ϩ
__h __h __h t2g
__hg __hg __h t2g
__hg __hg __hg t2g
The absorption of visible light causes electronic transitions between orbitals in these sets.
Table 25-10 gives the colors of some transition metal nitrates in aqueous solution. Solutions
of representative (A group) metal nitrates are colorless.
One transition of a high spin octahedral Co(III) complex is depicted as follows.
Ground State Excited State Energy of Light Absorbed
eg
__h __h Absorption __hg __h _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
h88888888888n
of light ⌬E ϭ h␯ depends on ⌬oct
t2g
__hg __h __h __h __h __h _ _ _ _ _ _ _ _ _ _ _ _ _ _ _g
The frequency (␯), and therefore the wavelength and color, of the light absorbed are
related to ⌬oct.1 This, in turn, depends on the crystal field strength of the ligands. So the
colors and visible absorption spectra of transition metal complexes, as well as their
magnetic properties, provide information about the strengths of the ligand–metal inter-
actions.
A color wheel shows colors and their
complementary colors. For example,
green is the complementary color of
red. The data in Table 25-9 are
given for specific wavelengths. Broad
bands of wavelengths are shown in
this color wheel.
25-9 Color and the Spectrochemical Series 995
TABLE 25-9 Complementary Colors
Spectral Color Complementary Color
Wavelength Absorbed (Å) (color absorbed) (color observed)
4100 violet lemon yellow
4300 indigo yellow
4800 blue orange
5000 blue-green red
5300 green purple
5600 lemon yellow violet
5800 yellow indigo
6100 orange blue
6800 red blue-green
TABLE 25-10 Colors of
Aqueous
Solutions of
Some
Transition
Metal
Nitrates
Transition Color of
Metal Ion Aq. Solution
Cr3ϩ Deep blue
Mn2ϩ Pale pink
Fe2ϩ Pale green
Fe3ϩ Orchid
Co2ϩ Pink
Ni2ϩ Green
Cu2ϩ Blue
1The numerical relationship between ⌬oct and the wavelength, ␭, of the absorbed light is found by
combining the expressions E ϭ h␯ and ␯ ϭ c/␭, where c is the speed of light. For one mole of a complex,
⌬oct ϭ ENA ϭ ᎏ
hc
␭
NA
ᎏ where NA is Avogadro’s number
Adding a Co(II) salt to molten glass
gives the glass a deep blue color.
Orange
Red
Blue
Purple
Yellow
Green
Planck’s constant is
h ϭ 6.63 ϫ 10Ϫ34 Jиs.
By interpreting the visible spectra of many complexes, it is possible to arrange common
ligands in order of increasing crystal field strengths.
IϪ Ͻ BrϪ Ͻ ClϪ Ͻ FϪ Ͻ OHϪ Ͻ H2O Ͻ (COO)2
2Ϫ Ͻ NH3 Ͻ en Ͻ NO2
Ϫ Ͻ CNϪ
This arrangement is called the spectrochemical series. Strong field ligands, such as CNϪ,
produce large crystal field splittings and usually produce low spin complexes, where
possible. Weak field ligands, such as ClϪ, produce small crystal field splittings and high
spin complexes. Low spin complexes usually absorb higher-energy (shorter-wavelength)
light than do high spin complexes. The colors of several six-coordinate Cr(III) complexes
are listed in Table 25-11.
In [Cr(NH3)6]Cl3, the Cr(III) is bonded to six ammonia ligands, which produce a relatively
high value of ⌬oct. This causes the [Cr(NH3)6]3ϩ ion to absorb relatively high energy
visible light in the blue and violet regions. Thus, we see yellow-orange, the complementary
color.
Water is a weaker field ligand than ammonia, and therefore ⌬oct is less for [Cr(OH2)6]3ϩ
than for [Cr(NH3)6]3ϩ. As a result, [Cr(OH2)6]Br3 absorbs lower energy (longer wavelength)
light. This causes the reflected and transmitted light to be higher energy bluish
gray, the color that describes [Cr(OH2)6]Br3.
Increasing crystal field strength
The colors of complex compounds that contain a given metal depend on the ligands. The
yellow compound at the left is a salt that contains [Co(NH3)6]3ϩ ions. In the next three
compounds, left to right, one NH3 ligand in [Co(NH3)6]3ϩ has been replaced by NCSϪ
(orange), H2O (red), and ClϪ (purple). The green compound at the right is a salt that
contains [Co(NH3)4Cl2]ϩ ions.
996 CHAPTER 25: Coordination Compounds
TABLE 25-11 Colors of Some Chromium(III) Complexes
[Cr(OH2)6]Cl3 violet [Cr(NH3)6]Cl3 yellow
[Cr(OH2)6]Br3 bluish gray [Cr(NH3)5Cl]Cl2 purple
[Cr(OH2)4Cl2]Cl green [Cr(NH3)4Cl2]Cl violet
[Cr(OH2)4Br2]Br green [Cr(CON2H4)6][SiF6]3 green
We see the light that is transmitted
(passes through the sample) or that is
reflected by the sample.
Different anions sometimes cause
compounds containing the same
complex cation to have different
colors. For example, two differentcolored
compounds in this table both
contain the [Cr(OH2)6]3ϩ ion.
Exercises 997
Key Terms
Ammine complexes Complex species that contain ammonia
molecules bonded to metal ions.
Central atom The atom or ion to which the ligands are bonded
in a complex species.
Chelate A ligand that utilizes two or more donor atoms in bonding
to metals.
cis–trans isomerism See Geometric isomerism.
Complementary color The color associated with the wavelengths
of light that are not absorbed—that is, the color
transmitted or reflected.
Coordinate covalent bond A covalent bond in which both
shared electrons are donated by the same atom; a bond between
a Lewis base and a Lewis acid.
Coordination compound or complex A compound containing
coordinate covalent bonds between electron pair donors and a
metal.
Coordination isomers Isomers involving exchange of ligands
between a complex cation and a complex anion of the same
coordination compound.
Coordination number The number of donor atoms coordinated
to a metal.
Coordination sphere The metal ion and its coordinated ligands,
but not any uncoordinated counterions.
Crystal field theory A theory of bonding in transition metal
complexes in which ligands and metal ions are treated as point
charges; a purely ionic model. Ligand point charges represent
the crystal (electric) field perturbing the metal’s d orbitals that
contain nonbonding electrons.
⌬oct The energy separation between eg and t2g sets of metal d
orbitals caused by octahedral complexation of ligands.
Dextrorotatory Describes an optically active substance that
rotates the plane of plane-polarized light to the right; also called
dextro.
Donor atom A ligand atom whose electrons are shared with a
Lewis acid.
eg orbitals A set of dx2–y2 and dz2 orbitals; those d orbitals within
a set with lobes directed along the x, y, and z axes.
Enantiomers Stereoisomers that differ only by being nonsuperimposable
mirror images of each other, like left and right hands;
also called optical isomers.
Geometric isomerism Occurs when atoms or groups of atoms
can be arranged in different ways on two sides of a rigid structure;
also called cis–trans isomerism. In geometric isomers of
coordination compounds, the same ligands are arranged in different
orders within the coordination sphere.
High spin complex The crystal field designation for a complex
in which all t2g and eg orbitals are singly occupied before any
pairing occurs.
Hydrate isomers Isomers of crystalline complexes that differ in
terms of the presence of water inside or outside the coordination
sphere.
Ionization isomers Isomers that result from interchange of ions
inside and outside the coordination sphere.
Isomers Different compounds that have the same formula.
Levorotatory Refers to an optically active substance that
rotates the plane of plane-polarized light to the left; also called
levo.
Ligand A Lewis base in a coordination compound.
Linkage isomers Isomers in which a particular ligand bonds to
a metal ion through different donor atoms.
Low spin complex The crystal field designation for a complex
in which pairing occurs to fill the t2g orbitals before any electrons
occupy the eg orbitals.
Optical activity The ability of one of a pair of optical isomers
to rotate the plane of polarized light.
Optical isomers See Enantiomers.
Pairing energy The energy required to pair two electrons in the
same orbital.
Plane-polarized light Light waves in which all the electric vectors
are oscillating in one plane.
Polarimeter A device used to measure optical activity.
Polydentate Describes ligands with more than one donor
atom.
Racemic mixture An equimolar mixture of optical isomers that
is, therefore, optically inactive.
Spectral color The color associated with the wavelengths of
light that are absorbed.
Spectrochemical series An arrangement of ligands in order of
increasing ligand field strength.
Square planar complex A complex in which the metal atom or
ion is in the center of a square plane, with a ligand donor atom
at each of the four corners.
Stereoisomers Isomers that differ only in the way in which
atoms are oriented in space; they include geometric and optical
isomers.
Strong field ligand A ligand that exerts a strong crystal or ligand
electric field and generally forms low spin complexes with
metal ions when possible.
Structural (constitutional) isomers (Applied to coordination
compounds.) Isomers whose differences involve more than a
single coordination sphere or else different donor atoms; they
include ionization isomers, hydrate isomers, coordination isomers,
and linkage isomers.
t2g orbitals A set of dxy, dyz, and dxz orbitals; those d orbitals
within a set with lobes bisecting the x, y, and z axes.
Weak field ligand A ligand that exerts a weak crystal or ligand
field and generally forms high spin complexes with
metals.
998 CHAPTER 25: Coordination Compounds
Exercises
Basic Concepts
*01. What property of transition metals allows them to form
coordination compounds easily?
*02. Suggest more appropriate formulas for NiSO4и6H2O,
Cu(NO3)2и4NH3, and Co(NO3)2и6NH3.
*03. What are the two constituents of a coordination complex?
What type of chemical bonding occurs between these con-
stituents?
*04. Define the term “coordination number” for the central
atom or ion in a complex. What values of the coordination
numbers for metal ions are most common?
*05. Distinguish among the terms ligands, donor atoms, and
chelates.
*06. Identify the ligands and give the coordination number
and the oxidation number for the central atom or ion
in each of the following: (a) [Co(NH3)2(NO2)4]Ϫ;
(b) [Cr(NH3)5Cl]Cl2; (c) K4[Fe(CN)6]; (d) [Pd(NH3)4]2ϩ.
*07. Repeat Exercise 6 for (a) Na[AuCl2]; (b) [Ag(NH3)2]ϩ;
(c) [Pt(NH3)2Cl4]; (d) [Co(en)3]3ϩ
*08. What is the term given to the phenomenon of ring formation
by a ligand in a complex? Describe a specific
example.
*09. Describe the experiments of Alfred Werner on the compounds
of the general formula PtCl4иnNH3, where n ϭ 2,
3, 4, 5, 6. What was his interpretation of these experiments?
*10. For each of the compounds of Exercise 9, write formulas
indicating the species within the coordination sphere. Also
indicate the charges on the complex ions.
*11. Write a structural formula showing the ring(s) formed by
a bidentate ligand such as ethylenediamine with a metal
ion such as Fe3ϩ. How many atoms are in each ring? The
formula for this complex ion is [Fe(en)3]3ϩ.
Ammine Complexes
*12. Write net ionic equations for reactions of solutions of
the following transition metal salts in water with a limited
amount of aqueous ammonia: (It is not necessary to
show the ions as hydrated.) (a) CuCl2; (b) Zn(NO3)2;
(c) Fe(NO3)3; (d) Co(NO3)2; (e) Ni(NO3)2.
*13. Write net ionic equations for the reactions of the insoluble
products of Exercise 12 with an excess of aqueous ammonia,
if a reaction occurs.
*14. Which of the following insoluble metal hydroxides will
dissolve in an excess of aqueous ammonia? (a) Zn(OH)2;
(b) Cr(OH)3; (c) Fe(OH)2; (d) Ni(OH)2; (e) Cd(OH)2.
*15. Write net ionic equations for the reactions in Exercise 14.
Naming Coordination Compounds
*16. Give systematic names for the following compounds.
(a) [Fe(CO)5]
(b) Na2[Co(OH2)2(OH)4]
(c) [Ag(NH3)2]Br
(d) [Cr(en)3](NO3)3
(e) [Pt(NH3)4(NO2)2] F2
(f ) K2[Cu(CN)4]
(g) [Ni(NH3)4(H2O)2](NO3)2
(h) Na[Al(H2O)2(OH)4]
(i) [Co(NH3)4Cl2][Cr(C2O4)2]
*17. Give systematic names for the following compounds.
(a) Na[Au(CN)2]
(b) [Ni(CO)4]
(c) [CoCl6]3Ϫ
(d) [Co(H2O)6]3ϩ
(e) Na2[Pt(CN)4]
(f ) K[Cr(NH3)2(OH)2Cl2]
(g) [Mo(NCS)2(en)2]ClO4
(h) [Mn(NH3)2(OH2)3(OH)]SO4
(i) [Co(en)2Cl2]I
*18. Give systematic names for the following compounds.
(a) [Ag(NH3)2]Cl
(b) [Fe(en)3]PO4
(c) [Co(NH3)6]SO4
(d) [Co(NH3)6]2(SO4)3
(e) [Pt(NH3)4]Cl2
(f ) (NH4)2[PtCl4]
(g) [Co(NH3)5SO4]NO2
*19. Give systematic names for the following compounds.
(a) K2[PdCl6]
(b) (NH4)2[PtCl6]
(c) [Co(en)3]Cl2
(d) [Co(en)3]Cl3
(e) [Cr(en)2(NH3)2](NO3)2
(f ) [Co(NH3)4Cl2]Cl
(g) (NH4)3[CuCl5]
*20. Write formulas for the following.
(a) diamminedichlorozinc
(b) tin(IV) hexacyanoferrate(II)
(c) tetracyanoplatinate(II) ion
(d) potassium hexachlorostannate(IV)
(e) tetraammineplatinum(II) ion
(f ) hexaamminenickel(II) bromide
(g) tetraamminecopper(II) pentacyanohydroxoferrate(III)
(h) diaquadicyanocopper(II)
(i) potassium hexachloropalladate(IV)
*21. Write formulas for the following compounds.
(a) trans-diamminedinitroplatinum(II)
(b) rubidium tetracyanozincate
(c) triaqua-cis-dibromochlorochromium(III)
(d) pentacarbonyliron(0)
(e) sodium pentacyanocobaltate(II)
(f ) hexammineruthenium(III) tetrachloronickelate(II)
(g) sodium tetracyanocadmate
(h) hexaamminecobalt(III) chloride
Exercises 999
Structures of Coordination Compounds
*22. Write formulas, and provide names for three complex
cations in each of the following categories.
(a) cations coordinated to only monodentate ligands
(b) cations coordinated to only bidentate ligands
(c) cations coordinated to two bidentate and two monodentate
ligands
(d) cations coordinated to one tridentate ligand, one
bidentate ligand, and one monodentate ligand
(e) cations coordinated to one tridentate ligand and three
monodentate ligands
*23. Provide formulas and names for three complex anions that
fit each description in Exercise 22.
*24. How many geometric isomers can be formed by complexes
that are (a) octahedral MA2B4 and (b) octahedral MA3B3?
Name any geometric isomers that can exist. Is it possible
for any of these isomers to show optical activity (exist as
enantiomers)? Explain.
*25. Write the structural formulas for (a) two isomers of
[Pt(NH3)2Cl2], (b) four isomers (including linkage isomers)
of [Co(NH3)3(NO2)3], and (c) two isomers (including ionization
isomers) of [Pt(NH3)3Br]Cl.
*26. Determine the number and types of isomers that would be
possible for each of the following complexes.
(a) tetraamminediaquachromium(III) ion
(b) triamminetriaquachromium(III) ion
(c) tris(ethylenediamine) chromium(III) ion
(d) dichlorobis(ethylenediamine)platinum(IV) chloride
(e) diamminedibromodichlorochromate(III) ion
*27. Indicate whether the complexes in each pair are identical
or isomers.
en
Cl(b) ϩ
enCr
NH3
ClH3N
Cl
ϩ
Cr
NH3
H3N
Cl
(a)
Pt
Cl
NH3
BrH3N
Br Cl
Pt
Cl
NH3
BrH3N
Br Cl
*28. Distinguish between constitutional (structural) isomers and
stereoisomers.
*29. Distinguish between an optically active complex and a
racemic mixture.
*30. Write the formula for a potential ionization isomer of each
of the following compounds. Name each one.
(a) [Co(NH3)4Br2]Br
(b) [Ni(en)2(NO2)2]Cl2
(c) [Fe(NH3)5CN]SO4
*31. Write the formula for a potential hydrate isomer of each
of the following compounds. Name each one.
(a) [Cu(OH2)4]Cl2 (b) [Ni(OH2)5Br]BrиH2O
*32. Write the formula for a potential coordination isomer of
each of the following compounds. Name each one.
(a) [Co(NH3)6][Cr(CN)6] (b) [Ni(en)3][Cu(CN)4]
*33. Write the formula for a potential linkage isomer of each
of the following compounds. Name each one.
(a) [Co(en)2(NO2)2]Cl (b) [Cr(NH3)5(CN)](CN)2
Crystal Field Theory
*34. Describe clearly what ⌬oct is. How is ⌬oct actually
measured experimentally? How is it related to the spectrochemical
series?
*35. On the basis of the spectrochemical series, determine
whether the following complexes are low spin or high
spin.
(a) [CrCl4Br2]3Ϫ
(b) [Co(en)3]3ϩ
(c) [Fe(OH2)6]3ϩ
(d) [Fe(NO2)6]3Ϫ
(d)
en
3ϩ
en
Co
NH3
NH3
en
3ϩ
en
Co
NH3
NH3
H3N
en
(c)
en
3ϩ 3ϩ
en
Co
NH3
NH3
en
Co
NH3
1000 CHAPTER 25: Coordination Compounds
*36. On the basis of the spectrochemical series, determine
whether the following complexes are low spin or high spin.
(a) [Cu(OH2)6]2ϩ
(b) [MnF6]3Ϫ
(c) [Co(CN)6]3Ϫ
(d) [Cr(NH3)6]3ϩ
*37. Write out the electron distribution in t2g and eg orbitals for
the following in an octahedral field.
Metal Ions Ligand Field Strength
V2ϩ weak
Mn2ϩ strong
Mn2ϩ weak
Ni2ϩ weak
Cu2ϩ weak
Fe3ϩ strong
Cuϩ weak
Ru3ϩ strong
*38. Write formulas for two complex ions that would fit into
each of the categories of Exercise 37. Name the complex
ions you list.
*39. Determine the electron distribution in (a) [Co(CN)6]3Ϫ, a
low spin complex ion, and (b) [CoF6]3Ϫ, a high spin complex
ion.
CONCEPTUAL EXERCISES
*40. How can you predict whether a complex is high or low
spin? How can you determine whether a complex is high
or low spin?
*41. Define the terms diamagnetic and paramagnetic. What feature
of electronic structure is directly related to these
properties?
*42. What are three common metal coordination numbers
found in coordination chemistry? Give the shape usually
associated with each of the common coordination numbers.
*43. Several isomers are possible for [Co(en)(NH3)(H2O)2Cl]2ϩ.
(Two of the isomers have optical isomers.) How many isomers
can you draw? Draw the structure of an isomer that
does not have an optical isomer. Draw the structure of an
isomer that would have an optical isomer. (Hint: “en” is
not large enough to bind at trans locations.)
BUILDING YOUR KNOWLEDGE
Values of equilibrium constants are listed in Appendix I.
*44. Calculate (a) the molar solubility of Zn(OH)2 in pure water,
(b) the molar solubility of Zn(OH)2 in 0.25 M NaOH solution,
and (c) the concentration of [Zn(OH)4]2Ϫ ions in the
solution of (b).
*45. The yellow complex oxalatorhodium compound
K3[Rh(C2O4)3] can be prepared from the wine-red complex
compound K3[RhCl6] by boiling a concentrated aqueous
solution of K3[RhCl6] and K2C2O4 for two hours and then
evaporating the solution until the product crystallizes.
heat
K3[RhCl6](aq) ϩ 3K2C2O4(aq) 888n
K3[Rh(C2O4)3](s) ϩ 6KCl(aq)
What is the theoretical yield of the oxalato complex if
1.150 g of the chloro complex is heated with 4.95 g of
K2C2O4? In an experiment, the actual yield was 0.88 g.
What is the percent yield?
*46. Consider the formation of the triiodoargentate(I) ion.
Agϩ ϩ 3IϪ 88n [AgI3]2Ϫ
Would you expect an increase or decrease in the entropy
of the system as the complex is formed? The standardstate
absolute entropy at 25°C is 72.68 J/K mol for Agϩ,
111.3 J/K mol for IϪ, and 253.1 J/K mol for [AgI3]2Ϫ. Calculate
⌬S0 for the reaction and confirm your prediction.
*47. Molecular iodine reacts with IϪ to form a complex ion.
I2(aq) ϩ IϪ 34 [I3]Ϫ
Calculate the equilibrium constant for this reaction given
the following data at 25°C.
I2(aq) ϩ 2eϪ 88n 2IϪ E0 ϭ 0.535 V
[I3]Ϫ ϩ 2eϪ 88n 3IϪ E0 ϭ 0.5338 V
*48. Calculate the pH of a solution prepared by dissolving 0.30
mol of tetraamminecopper(II) chloride, [Cu(NH3)4]Cl2, in
water to give 1.0 L of solution. Ignore hydrolysis of Cu2ϩ.
*49. Use the following standard reduction potential data to
answer the questions.
Co3ϩ ϩ eϪ 34 Co2ϩ
E0 ϭ 1.808 V
Co(OH)3(s) ϩ eϪ 34 Co(OH)2(s) ϩ OHϪ(aq)
E0 ϭ 0.17 V
[Co(NH3)6]3ϩ ϩ eϪ 34 [Co(NH3)6]2ϩ
E0 ϭ 0.108 V
[Co(CN)6]3Ϫ ϩ eϪ 34 [Co(CN)5]3Ϫ ϩ CNϪ
E0 ϭ Ϫ0.83 V
O2(g) ϩ 4Hϩ(10Ϫ7 M) ϩ 4eϪ 34 2H2O(ᐉ)
E0 ϭ 0.815 V
2H2O(ᐉ) ϩ 2eϪ 34 H2(g) ϩ 2OHϪ
E0 ϭ Ϫ0.828 V
Which cobalt(III) species among those listed would oxidize
water? Which cobalt(II) species among those listed
would be oxidized by water? Explain your answers.
26Nuclear
Chemistry
OUTLINE
26-1 The Nucleus
26-2 Neutron–Proton Ratio and
Nuclear Stability
26-3 Nuclear Stability and Binding
Energy
26-4 Radioactive Decay
26-5 Equations for Nuclear
Reactions
26-6 Neutron-Rich Nuclei (Above
the Band of Stability)
26-7 Neutron-Poor Nuclei (Below
the Band of Stability)
26-8 Nuclei with Atomic Number
Greater Than 83
26-9 Detection of Radiation
26-10 Rates of Decay and Half-Life
26-11 Disintegration Series
26-12 Uses of Radionuclides
26-13 Artificial Transmutations of
Elements
26-14 Nuclear Fission
26-15 Nuclear Fission Reactors
26-16 Nuclear Fusion
OBJECTIVES
After you have studied this chapter, you should be able to
• Describe the makeup of the nucleus
• Describe the relationships between neutron–proton ratio and nuclear stability
• Tell what is meant by the band of stability
• Calculate mass deficiency and nuclear binding energy
• Describe the common types of radiation emitted when nuclei undergo radioactive decay
• Write and balance equations that describe nuclear reactions
• Predict the different kinds of nuclear reactions undergone by nuclei, depending on their
positions relative to the band of stability
• Describe methods for detecting radiation
• Understand half-lives of radioactive elements
• Carry out calculations associated with radioactive decay
• Interpret disintegration series
• Tell about some uses of radionuclides, including the use of radioactive elements for
dating objects
• Describe some nuclear reactions that are induced by bombardment of nuclei with
particles
• Tell about nuclear fission and some of its applications, including nuclear reactors
• Tell about nuclear fusion and some prospects for and barriers to its use for the
production of energy
Positron emission tomography
(PET) allows mapping of tissues.
This PET scan shows the
distribution of radioactive glucose in
a healthy human brain.
CC
hemical properties are determined by electron distributions and are only indirectly
influenced by atomic nuclei. Until now, we have discussed ordinary chemical
reactions, so we have focused on electron configurations. Nuclear reactions involve
changes in the composition of nuclei. These extraordinary processes are often accompanied
by the release of tremendous amounts of energy and by transmutations of elements.
Some differences between nuclear reactions and ordinary chemical reactions follow.
Nuclear Reaction Ordinary Chemical Reaction
1. Elements may be converted from 1. No new elements can be produced.
2. one to another.
2. Particles within the nucleus are 2. Only the electrons participate.
2. involved.
3. Tremendous amounts of energy are 3. Relatively small amounts of energy are
2. released or absorbed. 2. released or absorbed.
4. Rate of reaction is not influenced by 4. Rate of reaction depends on factors such
2. external factors. 2. as concentration, temperature, catalyst,
2. and pressure.
Medieval alchemists spent years trying to convert other metals into gold without
success. Years of failure and the acceptance of Dalton’s atomic theory early in the nineteenth
century convinced scientists that one element could not be converted into another.
Then, in 1896 Henri Becquerel discovered “radioactive rays” (natural radioactivity)
coming from a uranium compound. Ernest Rutherford’s study of these rays showed that
atoms of one element may indeed be converted into atoms of other elements by spontaneous
nuclear disintegrations. Many years later it was shown that nuclear reactions initiated
by bombardment of nuclei with accelerated subatomic particles or other nuclei can also
transform one element into another—accompanied by the release of radiation (induced
radioactivity).
Becquerel’s discovery led other researchers, including Marie and Pierre Curie, to
discover and study new radioactive elements. Many radioactive isotopes, or radioisotopes,
now have important medical, agricultural, and industrial uses.
Nuclear fission is the splitting of a heavy nucleus into lighter nuclei. Nuclear fusion
is the combination of light nuclei to produce a heavier nucleus. Huge amounts of energy
are released when these processes occur. These processes could satisfy a large portion of
our future energy demands. Current research is aimed at surmounting the technological
problems associated with safe and efficient use of nuclear fission reactors and with the
development of controlled fusion reactors.
THE NUCLEUS
In Chapter 5 we described the principal subatomic particles (Table 26-1). Recall that the
neutrons and protons together constitute the nucleus, with the electrons occupying essentially
empty space around the nucleus. The nucleus is only a minute fraction of the total
volume of an atom, yet nearly all the mass of an atom resides in the nucleus. Thus, nuclei
are extremely dense. It has been shown experimentally that nuclei of all elements have
approximately the same density, 2.4 ϫ 1014 g/cm3.
26-1
26-1 The Nucleus 1003
Marie Sklodowska Curie (1867–
1934) is the only person to have
been honored with Nobel Prizes in
both physics and chemistry. In 1903,
Pierre (1859–1906) and Marie Curie
and Henri Becquerel (1852–1908)
shared the prize in physics for the
discovery of natural radioactivity.
Marie Curie also received the 1911
Nobel Prize in chemistry for her
discovery of radium and polonium
and the compounds of radium. She
named polonium for her native
Poland. Marie’s daughter, Irene
Joliot-Curie (1897–1956), and
Irene’s husband, Frederick Joliot
(1900–1958), received the 1935
Nobel Prize in chemistry for the
first synthesis of a new radioactive
element.
In Chapter 5, we represented an atom
of a particular isotope by its nuclide
symbol. Radioisotopes are often called
radionuclides.
If enough nuclei could be gathered
together to occupy one cubic
centimeter, the total weight would
be about 250 million tons!
From an electrostatic point of view, it is amazing that positively charged protons can
be packed so closely together. Yet many nuclei do not spontaneously decompose, so they
must be stable. In the early twentieth century when Rutherford postulated the nuclear
model of the atom, scientists were puzzled by such a situation. Physicists have since
detected many very short-lived subatomic particles (in addition to protons, neutrons, and
electrons) as products of nuclear reactions. Well over 100 have been identified. A discussion
of these many particles is beyond the scope of a chemistry text. Furthermore their
functions are not entirely understood, but it is now thought that they help to overcome
the proton–proton repulsions and to bind nuclear particles (nucleons) together. The
attractive forces among nucleons appear to be important over only extremely small
distances, about 10Ϫ13 cm.
NEUTRON–PROTON RATIO AND NUCLEAR STABILITY
The term “nuclide” is used to refer to different atomic forms of all elements. The term
“isotope” applies only to different forms of the same element. Most naturally occurring
nuclides have even numbers of protons and even numbers of neutrons; 157 nuclides fall
into this category. Nuclides with odd numbers of both protons and neutrons are least
common (there are only four), and those with odd–even combinations are intermediate
in abundance (Table 26-2). Furthermore, nuclides with certain “magic numbers” of protons
and neutrons seem to be especially stable. Nuclides with a number of protons or a number
of neutrons or a sum of the two equal to 2, 8, 20, 28, 50, 82, or 126 have unusual stability.
Examples are 4
2He, 16
08O, 42
20Ca, 88
38Sr, and 208
082Pb. This suggests an energy level (shell) model
for the nucleus similar to the shell model of electron configurations.
Figure 26-1 is a plot of the number of neutrons (N) versus number of protons (Z) for
the stable nuclides (the band of stability). For low atomic numbers, the most stable
nuclides have equal numbers of protons and neutrons (N ϭ Z). Above atomic number 20,
the most stable nuclides have more neutrons than protons. Careful examination reveals
an approximately stepwise shape to the plot due to the stability of nuclides with even
numbers of nucleons.
26-2
1004 CHAPTER 26: Nuclear Chemistry
The nuclide symbol for an element
(Section 5-7) is
A
ZE
where E is the chemical symbol for the
element, Z is its atomic number, and A
is its mass number.
TABLE 26-1 Fundamental Particles of Matter
Particle Mass Charge
Electron (eϪ) 0.00054858 amu 1Ϫ
Proton (p or pϩ) 1.0073 amu 1ϩ
Neutron (n or n0) 1.0087 amu none
TABLE 26-2 Abundance of Naturally Occurring Nuclides
Number of protons even even odd odd
Number of neutrons even odd even odd
Number of such nuclides 157 52 50 4
NUCLEAR STABILITY AND BINDING ENERGY
Experimentally, we observe that the masses of atoms other than 1
1H are always less than
the sum of the masses of their constituent particles. We now know why this mass deficiency
occurs. We also know that the mass deficiency is in the nucleus of the atom and has nothing
to do with the electrons; however, because tables of masses of isotopes include the electrons, we
shall also include them.
The mass deficiency, ⌬m, for a nucleus is the difference between the sum of the
masses of electrons, protons, and neutrons in the atom (calculated mass) and the actual
measured mass of the atom.
⌬m ϭ (sum of masses of all eϪ, pϩ, and n0) Ϫ (actual mass of atom)
For most naturally occurring isotopes, the mass deficiency is only about 0.15% or less of
the calculated mass of an atom.
EXAMPLE 26-1 Mass Deficiency
Calculate the mass deficiency for chlorine-35 atoms in amu/atom and in g/mol atoms. The
actual mass of a chlorine-35 atom is 34.9689 amu.
Plan
We first find the numbers of protons, electrons, and neutrons in one atom. Then we determine
the “calculated” mass as the sum of the masses of these particles. The mass deficiency is
the actual mass subtracted from the calculated mass. This deficiency is commonly expressed
either as mass per atom or as mass per mole of atoms.
26-3
120
100
80
60
40
20
20 40 60
1:1 ratio
Neutron-poor
nuclei
Neutron-rich
nuclei
1.2
80
Number of protons
Numberofneutrons
1.4
1.5
Figure 26-1 A plot of the number
of neutrons versus the number of
protons in stable nuclei. As atomic
number increases, the N/Z ratio (the
decimal fractions) of the stable nuclei
increases. The stable nuclei are
located in an area known as the band
of stability. Most radioactive nuclei
occur outside this band.
Do you remember how to find the
numbers of protons, neutrons, and
electrons in a specified atom? Review
Section 5-7.
Solution
Each atom of 35
17Cl contains 17 protons, 17 electrons, and (35 Ϫ 17) ϭ 18 neutrons. First we
sum the masses of these particles.
protons: 17 ϫ 1.0073 amu ϭ 17.124 amu m88 (masses from Table 26-1)
electrons: 17 ϫ 0.00054858 amu ϭ 0.0093 amu
neutrons: 18 ϫ 1.0087 amu ϭ 18.157 amu
sum ϭ 35.290 amu m88 calculated mass
Then we subtract the actual mass from the “calculated” mass to obtain ⌬m.
⌬m ϭ 35.290 amu Ϫ 34.9689 amu ϭ 0.321 amu mass deficiency
(in one atom)
We have calculated the mass deficiency in amu/atom. Recall (Section 5-9) that 1 gram is
6.022 ϫ 1023 amu. We can show that a number expressed in amu/atom is equal to the same
number in g/mol of atoms.
ϭ ϫ ϫ
ϭ 0.321 g/mol of 35Cl atoms m88 (mass deficiency in a mole of Cl atoms)
You should now work Exercises 14a and 16a, b.
What has happened to the mass represented by the mass deficiency? In 1905, Einstein
set forth the Theory of Relativity. He stated that matter and energy are equivalent. An
obvious corollary is that matter can be transformed into energy and energy into matter.
The transformation of matter into energy occurs in the sun and other stars. It happened
on earth when controlled nuclear fission was achieved in 1939 (Section 26-14). The reverse
transformation, energy into matter, has not yet been accomplished on a large scale.
Einstein’s equation, which we encountered in Chapter 1, is E ϭ mc2. E represents the
amount of energy released, m the mass of matter transformed into energy, and c the speed
of light in a vacuum, 2.997925 ϫ 108 m/s (usually rounded off to 3.00 ϫ 108 m/s).
A mass deficiency represents the amount of matter that would be converted into energy
and released if the nucleus were formed from initially separate protons and neutrons. This
energy is the nuclear binding energy, BE. It provides the powerful short-range force
that holds the nuclear particles (protons and neutrons) together in a very small volume.
We can rewrite the Einstein relationship as
BE ϭ (⌬m)c2
Specifically, if 1 mole of 35Cl nuclei were to be formed from 17 moles of protons and 18
moles of neutrons, the resulting mole of nuclei would weigh 0.321 gram less than the
original collection of protons and neutrons (Example 26-1).
Nuclear binding energies may be expressed in many different units, including kilojoules/mole
of atoms, kilojoules/gram of atoms, and megaelectron volts/nucleon. Some
useful equivalences are
6.022 ϫ 1023 atoms
ᎏᎏᎏ
1 mol 35Cl atoms
1 g
ᎏᎏ
6.022 ϫ 1023 amu
0.321 amu
ᎏᎏ
atom
_?_ g
ᎏ
mol
1006 CHAPTER 26: Nuclear Chemistry
1 megaelectron volt (MeV) ϭ 1.60 ϫ 10Ϫ13 J and 1 joule (J) ϭ 1 kgиm2/s2
Let’s use the value of ⌬m for 35Cl atoms to calculate their nuclear binding energy.
EXAMPLE 26-2 Nuclear Binding Energy
Calculate the nuclear binding energy 35Cl in (a) kilojoules per mole of Cl atoms, (b) kilojoules
per gram of Cl atoms, and (c) megaelectron volts per nucleon.
Plan
The mass deficiency that we calculated in Example 26-1 is related to the binding energy by
the Einstein equation.
Solution
The mass deficiency is 0.321 g/mol ϭ 3.21 ϫ 10Ϫ4 kg/mol.
(a) BE ϭ (⌬m)c2 ϭ ϫ (3.00 ϫ 108 m/s)2 ϭ 2.89 ϫ 1013
ϭ 2.89 ϫ 1013 J/mol 35Cl atoms ϭ 2.89 ϫ 1010 kJ/mol of 35Cl atoms
(b) From Example 26-1, the actual mass of 35Cl is
or
We use this mass to set up the needed conversion factor.
BE ϭ ϫ ϭ 8.26 ϫ 108 kJ/g 35Cl atoms
(c) The number of nucleons in one atom of 35Cl is 17 protons ϩ 18 neutrons ϭ 35 nucleons.
1 mol 35Cl atoms
ᎏᎏᎏ
34.9689 g 35Cl atoms
2.89 ϫ 1010 kJ
ᎏᎏᎏ
mol of 35Cl atoms
34.9689 g
ᎏᎏ
mol 35Cl atoms
34.9689 amu
ᎏᎏ35Cl atom
kgиm2/s2
ᎏᎏ
mol 35Cl atoms
3.21 ϫ 10Ϫ4 kg
ᎏᎏ
mol 35Cl atoms
26-3 Nuclear Stability and Binding Energy 1007
BE ϭ ϫ ϫ ϫ ϫ
ϭ 8.57 MeV/nucleon
1 35Cl atom
ᎏᎏ
35 nucleons
1 mol 35Cl atoms
ᎏᎏᎏ
6.022 ϫ 1023 35Cl atoms
1 MeV
ᎏᎏ
1.60 ϫ 10Ϫ13 J
1000 J
ᎏ
kJ
2.89 ϫ 1010 kJ
ᎏᎏᎏ
mol of 35Cl atoms
You should now work Exercises 14 and 16.
The nuclear binding energy of a mole of 35Cl nuclei, 2.89 ϫ 1013 J/mol, is an enormous
amount of energy—enough to heat 6.9 ϫ 107 kg ( Ϸ 76,000 tons) of water from
0°C to 100°C! Stated differently, this is also the amount of energy that would be required
to separate 1 mole of 35Cl nuclei into 17 moles of protons and 18 moles of neutrons. This
has never been done.
Figure 26-2 is a plot of average binding energy per gram of nuclei versus mass number.
It shows that nuclear binding energies (per gram) increase rapidly with increasing mass
number, reach a maximum around mass number 50, and then decrease slowly. The nuclei
with the highest binding energies (mass numbers 40 to 150) are the most stable. Large
amounts of energy would be required to separate these nuclei into their component
The mass number, Z, is equal to the
number of nucleons in one atom.
Some unstable radioactive nuclei do
emit a single proton, a single neutron,
or other subatomic particles as they
decay in the direction of greater
stability. None decomposes entirely
into elementary particles.
neutrons and protons. Even though these nuclei are the most stable ones, all nuclei are
stable with respect to complete decomposition into protons and neutrons because all
(except 1H) nuclei have mass deficiencies. In other words, the energy equivalent of the
loss of mass represents an associative force that is present in all nuclei except 1H. It must
be overcome to separate the nuclei completely into their subatomic particles.
RADIOACTIVE DECAY
Nuclei whose neutron-to-proton ratios lie outside the stable region undergo spontaneous
radioactive decay by emitting one or more particles or electromagnetic rays or both. The
type of decay that occurs usually depends on whether the nucleus is above, below, or to
the right of the band of stability (Figure 26-1). Common types of radiation emitted in
decay processes are summarized in Table 26-3.
The particles can be emitted at different kinetic energies. In addition, radioactive decay
often leaves a nucleus in an excited (high-energy) state. Then the decay is followed by
gamma ray emission.
(excited nucleus) 88n Z
ME* 88n Z
ME ϩ 0
0␥
The energy of the gamma ray (h␯) is equal to the energy difference between the ground
and excited nuclear states. This is like the emission of lower energy electromagnetic radiation
that occurs as an atom in its excited electronic state returns to its ground state
(Section 5-12). Studies of gamma ray energies strongly suggest that nuclear energy levels
are quantized just as are electronic energy levels. This adds further support for a shell
model for the nucleus.
The penetrating abilities of the particles and rays are proportional to their energies.
Beta particles and positrons are about 100 times more penetrating than the heavier and
slower-moving alpha particles. They can be stopped by a ᎏ1
8
ᎏ-inch-thick (0.3 cm) aluminum
26-4
1008 CHAPTER 26: Nuclear Chemistry
10
8
6
4
2
0
0 50 100
Mass number
Bindingenergypergram(unitsof108
kJ/g)
150 200 250
Figure 26-2 Plot of binding energy per gram versus mass number. Very light and very
heavy nuclei are relatively unstable.
Recall that the energy of
electromagnetic radiation is
E ϭ h␯, where h is Planck’s constant
and ␯ is the frequency.
plate. They can burn skin severely but cannot reach internal organs. Alpha particles have
low penetrating ability and cannot damage or penetrate skin. They can damage sensitive
internal tissue if inhaled, however. The high-energy gamma rays have great penetrating
power and severely damage both skin and internal organs. They travel at the speed of
light and can be stopped by thick layers of concrete or lead.
26-4 Radioactive Decay 1009
TABLE 26-3 Common Types of Radioactive Emissions
Type and Symbola Identity Mass (amu) Charge Velocity Penetration
beta (␤Ϫ, Ϫ0
Ϫ1␤, Ϫ0
Ϫ1e) electron 0.00055 1Ϫ Յ90% speed of light low to moderate,
depending on energy
positronb (Ϫ0
ϩ1␤, Ϫ0
ϩ1e) positively charged 0.00055 1ϩ Յ90% speed of light low to moderate,
electron depending on energy
alpha (␣, 4
2␣, 4
2He) helium nucleus 4.0026 2ϩ Յ10% speed of light low
proton (1
1p, 1
1H) proton, hydrogen 1.0073 1ϩ Յ10% speed of light low to moderate,
nucleus depending on energy
neutron (1
0n) neutron 1.0087 0ϩ Յ10% speed of light very high
gamma (0
0␥) ray high-energy electro- 0 0ϩ speed of light high
magnetic radiation
such as X-rays
aThe number at the upper left of the symbol is the number of nucleons, and the number at the lower left is the number of positive charges.
bOn the average, a positron exists for only about a nanosecond (1 ϫ 10Ϫ9 second) before colliding with an electron and being converted into the corresponding amount
of energy.
A technician cleans lead glass blocks that form part of the
giant OPAL particle detector at CERN, the European center
for particle physics near Geneva, Switzerland.
Robotics technology is used to manipulate
highly radioactive samples safely.
EQUATIONS FOR NUCLEAR REACTIONS
In chemical reactions, atoms in molecules and ions are rearranged, but matter is neither
created nor destroyed, and atoms are not changed into other atoms. In earlier chapters,
we learned to write balanced chemical equations to represent chemical reactions. Such
equations must show the same total number of atoms of each kind on both sides of the
equation and the same total charge on both sides of the equation. In a nuclear reaction,
a different kind of transformation occurs, one in which a proton can change into a neutron,
or a neutron can change into a proton, but the total number of nucleons remains the same.
This leads to two requirements for the equation for a nuclear reaction:
1. The sum of the mass numbers (the left superscript in the nuclide symbol) of the
reactants must equal the sum of the mass numbers of the products.
2. The sum of the atomic numbers (the left subscript in the nuclide symbol) of the
reactants must equal the sum of the atomic numbers of the products; this maintains
charge balance.
Because such equations are intended to describe only the changes in the nucleus, they do
not ordinarily include ionic charges (which are due to changes in the arrangements of
electrons). In the following sections, we will see many examples of such equations for
nuclear reactions.
NEUTRON-RICH NUCLEI (ABOVE THE BAND
OF STABILITY)
Nuclei in this region have too high a ratio of neutrons to protons. They undergo decays
that decrease the ratio. The most common such decay is beta emission. A beta particle is
an electron ejected from the nucleus when a neutron is converted into a proton.
1
0n 88n 1
1p ϩ Ϫ0
Ϫ1␤
Thus, beta emission results in an increase of one in the number of protons (the atomic
number) and a decrease of one in the number of neutrons, with no change in mass number.
Examples of beta particle emission are
228
088Ra 88n 228
089Ac ϩ Ϫ0
Ϫ1␤ and 14
06C 88n 14
07N ϩ Ϫ0
Ϫ1␤
The sum of the mass numbers on each side of the first equation is 228, and the sum of
the atomic numbers on each side is 88. The corresponding sums for the second equation
are 14 and 6, respectively.
NEUTRON-POOR NUCLEI (BELOW THE BAND
OF STABILITY)
Two types of decay for nuclei below the band of stability are positron emission or electron
capture (K capture). Positron emission is most commonly encountered with
artificially radioactive nuclei of the lighter elements. Electron capture occurs most often
with heavier elements.
26-7
26-6
26-5
1010 CHAPTER 26: Nuclear Chemistry
A positron has the mass of an electron but a positive charge. Positrons are emitted
when protons are converted to neutrons.
1
1p 88n 1
0n ϩ ϩ0
ϩ1␤
Thus, positron emission results in a decrease by one in atomic number and an increase by
one in the number of neutrons, with no change in mass number.
38
19K 88n 38
18Ar ϩ ϩ0
ϩ1␤ and 15
08O 88n 15
07N ϩ ϩ0
ϩ1␤
The same effect can be accomplished by electron capture (K capture), in which an electron
from the K shell (n ϭ 1) is captured by the nucleus.
106
047Ag ϩ ϩ0
Ϫ1e 88n 106
046Pd and 37
18Ar ϩ ϩ0
Ϫ1e 88n 37
17Cl
Some nuclides, such as 22
11Na, undergo both electron capture and positron emission.
22
11Na ϩ ϩ0
Ϫ1e 88n 22
10Ne (3%) and 22
11Na 88n 22
10Ne ϩ ϩ0
ϩ1␤ (97%)
Some of the neutron-poor nuclei, especially the heavier ones, increase their neutron-toproton
ratios by undergoing alpha emission. Alpha particles are helium nuclei, 4
2He,
consisting of two protons and two neutrons. Alpha emission also results in an increase of
the neutron-to-proton ratio. An example is the alpha emission of lead-204.
204
082Pb 88n 200
080Hg ϩ 4
2␣
NUCLEI WITH ATOMIC NUMBER GREATER THAN 83
All nuclides with atomic number greater than 83 are beyond the band of stability and are
radioactive. Many of these decay by alpha emission.
226
088Ra 88n 222
086Rn ϩ 4
2␣ and 210
084Po 88n 206
082Pb ϩ 4
2␣
The decay of radium-226 was originally reported in 1902 by Rutherford and Soddy. It
was the first transmutation of an element ever observed. A few heavy nuclides also decay
by beta emission, positron emission, and electron capture.
Some isotopes of uranium (Z ϭ 92) and elements of higher atomic number, the
transuranium elements, also decay by spontaneous nuclear fission. In this process a heavy
nuclide splits into nuclides of intermediate mass and neutrons.
252
098Cf 88n 142
056Ba ϩ 106
042Mo ϩ 4 1
0n
DETECTION OF RADIATION
Photographic Detection
Emanations from radioactive substances affect photographic plates just as ordinary visible
light does. Becquerel’s discovery of radioactivity resulted from the unexpected exposure
of such a plate, wrapped in black paper, by a nearby enclosed sample of a uraniumcontaining
compound, potassium uranyl sulfate. After a photographic plate has been
developed and fixed, the intensity of the exposed spot is related to the amount of radiation
that struck the plate. Quantitative detection of radiation by this method is difficult
and tedious.
26-9
26-8
26-9 Detection of Radiation 1011
Electron capture by the nucleus differs
from an atom gaining an electron to
form an ion.
␣-particles carry a double positive
charge, but charge is usually not
shown in nuclear reactions.
The only stable nuclide with atomic
number 83 is 209
083Bi.
Detection by Fluorescence
Fluorescent substances can absorb high-energy radiation such as gamma rays and subsequently
emit visible light. As the radiation is absorbed, the absorbing atoms jump to excited
electronic states. The excited electrons return to their ground states through a series of
transitions, some of which emit visible light. This method may be used for the quantitative
detection of radiation, using an instrument called a scintillation counter.
Cloud Chambers
The original cloud chamber was devised by C. T. R. Wilson (1869–1959) in 1911. A
chamber contains air saturated with vapor. Particles emitted from a radioactive substance
ionize air molecules in the chamber. Cooling the chamber causes droplets of liquid to
condense on these ions. The paths of the particles can be followed by observing the foglike
tracks produced. The tracks may be photographed and studied in detail. Figures 26-3
and 26-4 show a cloud chamber and a cloud chamber photograph, respectively.
Gas Ionization Counters
A common gas ionization counter is the Geiger–Müller counter (Figure 26-5). Radiation
enters the tube through a thin window. Windows of different stopping powers can
be used to admit only radiation of certain penetrating powers.
1012 CHAPTER 26: Nuclear Chemistry
Dry ice
Figure 26-3 A cloud chamber. The
emitter is glued onto a pin stuck into
a stopper that is mounted on the
chamber wall. The chamber has
some volatile liquid in the bottom
and rests on dry ice. The cool
air near the bottom becomes
supersaturated with vapor. When an
emission speeds through this vapor,
ions are produced. These ions serve
as “seeds” about which the vapor
condenses, forming tiny droplets,
or fog.
The Geiger counter can detect only
␤ and ␥ radiation. The ␣-particles
cannot penetrate the walls or window
of the tube.
collision
Figure 26-4 A historic cloud
chamber photograph of alpha
tracks in nitrogen gas. The
forked track was shown to be
due to a speeding proton
(going off to the left) and an
isotope of oxygen (going off to
the right). It is assumed that
the ␣-particle struck the
nucleus of a nitrogen atom at
the point where the track
forks.
RATES OF DECAY AND HALF-LIFE
Radionuclides have different stabilities and decay at different rates. Some decay nearly
completely in a fraction of a second and others only after millions of years. The rates of
all radioactive decays are independent of temperature and obey first-order kinetics. In
Section 16-3 we saw that the rate of a first-order process is proportional only to the
concentration of one substance. The rate law and the integrated rate equation for a firstorder
process (Section 16-4) are
rate of decay ϭ k[A] and ln΂ᎏ
A
A
0
ᎏ΃ϭ akt
Here A represents the amount of decaying radionuclide of interest remaining after some
time t, and A0 is the amount present at the beginning of the observation. The k is the rate
constant, which is different for each radionuclide. Each atom decays independently of the
others, so the stoichiometric coefficient a is always 1 for radioactive decay. We can therefore
drop it from the calculations in this chapter and write the integrated rate equation as
ln ΂ᎏ
A
A
0
ᎏ΃ϭ kt
Because A0/A is a ratio, A0 and A can represent either molar concentrations of a reactant
or masses of a reactant. The rate of radioactive disintegrations follows first-order kinetics,
so it is proportional to the amount of A present; we can write the integrated rate equation
in terms of N, the number of disintegrations per unit time:
ln ΂ᎏ
N
N
0
ᎏ΃ϭ kt
26-10
26-10 Rates of Decay and Half-Life 1013
Gas molecules Path of a
single
radiation
Window
+
–
e–
+
Figure 26-5 The principle of operation of a gas ionization counter. The center wire is
positively charged, and the shell of the tube is negatively charged. When radiation enters
through the window, it ionizes one or more gas atoms. The electrons are attracted to the
central wire, and the positive ions are drawn to the shell. This constitutes a pulse of electric
current, which is amplified and displayed on the meter or other readout.
A sample of carnotite, a uranium
ore, shown with a Geiger–Müller
counter.
In nuclear chemistry, the decay rate is usually expressed in terms of the half-life, t1/2,
of the process. This is the amount of time required for half of the original sample to react.
For a first-order process, t1/2 is given by the equation
t1/2 ϭ ϭ
The isotope strontium-90 was introduced into the atmosphere by the atmospheric
testing of nuclear weapons. Because of the chemical similarity of strontium to calcium, it
now occurs with Ca in measurable quantities in milk, bones, and teeth as a result of its
presence in food and water supplies. It is a radionuclide that undergoes beta emission with
a half-life of 28 years. It may cause leukemia, bone cancer, and other related disorders. If
we begin with a 16-␮g sample of 90
38Sr, 8 ␮g will remain after one half-life of 28 years.
After 56 years, 4 ␮g will remain; after 84 years, 2 ␮g; and so on (Figure 26-6).
Similar plots for other radionuclides all show the same shape of exponential decay
curve. About ten half-lives (280 years for 90
38Sr) must pass for radionuclides to lose 99.9%
of their radioactivity.
EXAMPLE 26-3 Rate of Radioactive Decay
The “cobalt treatments” used in medicine to arrest certain types of cancer rely on the ability
of gamma rays to destroy cancerous tissues. Cobalt-60 decays with the emission of beta particles
and gamma rays, with a half-life of 5.27 years.
60
27Co 88n 60
28Ni ϩ Ϫ0
Ϫ1␤ ϩ 0
0␥
How much of a 3.42-␮g sample of cobalt-60 remains after 30.0 years?
0.693
ᎏ
k
ln 2
ᎏ
k
1014 CHAPTER 26: Nuclear Chemistry
0 20 40 60 80 100 120
0
4
8
12
16
1 half-
life
2 half-
lives
3 halflives
4 half-
lives
Time (years)
Massof90
Sr(g)␮
Figure 26-6 The decay of a 16-␮g sample of 90
38Sr.
In 1963, a treaty was signed by the
United States, the Soviet Union, and
the United Kingdom prohibiting the
further testing of nuclear weapons in
the atmosphere. Since then, strontium-
90 has been disappearing from the air,
water, and soil according to the curve
in Figure 26-6. So the treaty has
largely accomplished its aim up to
the present.
Gamma rays destroy both cancerous
and normal cells, so the beams of
gamma rays must be directed as nearly
as possible at only cancerous tissue.
Plan
We determine the value of the specific rate constant, k, from the given half-life. This value is
then used in the first-order integrated rate equation to calculate the amount of cobalt-60
remaining after the specified time.
Solution
We first determine the value of the specific rate constant.
t1/2 ϭ so k ϭ ϭ ϭ 0.131 yϪ1
This value can now be used to determine the ratio of A0 to A after 30.0 years.
ln ΂ᎏ
A
A
0
ᎏ΃ϭ kt ϭ 0.131 yϪ1(30.0 y) ϭ 3.93
Taking the inverse ln of both sides, ᎏ
A
A
0
ᎏ ϭ 51.
A0 ϭ 3.42 ␮g, so
A ϭ ᎏ
A
51
0
ᎏ ϭ ᎏ
3.4
5
2
1
␮g
ᎏ ϭ 0.067 ␮g 60
27Co remains after 30.0 years.
You should now work Exercise 54.
DISINTEGRATION SERIES
Many radionuclides cannot attain nuclear stability by only one nuclear reaction. Instead,
they decay in a series of disintegrations. A few such series are known to occur in nature.
Two begin with isotopes of uranium, 238U and 235U, and one begins with 232Th. All three
of these end with a stable isotope of lead (Z ϭ 82). Table 26-4 outlines in detail the 238U,
235U, and 232Th disintegration series, showing half-lives. For any particular decay step,
the decaying nuclide is called the parent nuclide, and the product nuclide is the daughter.
Uranium-238 decays by alpha emission to thorium-234 in the first step of one series.
Thorium-234 subsequently emits a beta particle to produce protactinium-234 in the
second step. The series can be summarized as shown in Table 26-4a. The net reaction for
the 238U series is
238
092U 88n 206
082Pb ϩ 8 4
2He ϩ 6 Ϫ0
Ϫ1␤
“Branchings” are possible at various points in the chain. That is, two successive decays
may be replaced by alternative decays, but they always result in the same final product.
There are also decay series of varying lengths starting with some of the artificially produced
radionuclides (Section 26-13).
26-11
0.693
ᎏ
5.27 y
0.693
ᎏ
t1/2
0.693
ᎏ
k
26-11 Disintegration Series 1015
᭤ The ␥-radiation from 60Co is used to treat cancers near the surface of the body.
1016 CHAPTER 26: Nuclear Chemistry
TABLE 26-4 Emissions and Half-Lives of Members of Natural Radioactive Series*
(a) 238U Series (b) 235U Series (c) 232Th Series
*Abbreviations are y, year; d, day; m, minute; and s, second. Less prevalent decay branches are shown in blue.
U ␣238
92
Th ␤234
90
24.1 d
Pa ␤234
91
6.75 h
U ␣234
92
2.47 ϫ 105 y
Th ␣230
90
8.0 ϫ 104 y
Ra ␣226
88
1.60 ϫ 103 y
Rn ␣222
86
3.82 d
Po ␣218
84
3.05 m
Pb ␤214
82
26.8 m
␤
0.04%
At218
85␣
2 s
Bi ␣214
83
19.7 m
Tl ␤210
81
1.32 m
␤
Pb ␤210
82
20.4 y
Bi ␣210
83
5.01 d
Tl ␤206
81
4.19 m
␤
≈100%
Po210
84␣
138 d
Pb206
82
99.96%
Po214
84␣
1.6 ϫ 10Ϫ4 s
4.51 ϫ 109 y
Pb207
82
Rn ␣219
86
4.00 s
Po ␣215
84
1.78 ϫ 10Ϫ3 s
Pb ␤211
82
36.1 m
␤
5 ϫ 10Ϫ4%
At215
85␣
10Ϫ4 s
Bi ␣211
83
2.16 m
Tl ␤207
81
4.79 m
␤
99.7%
Po211
84␣
0.52 s
U ␣235
92
Th ␤231
90
25.5 h
Pa ␣231
91
3.25 ϫ 104 y
Ac ␣227
89
21.6 y
Fr ␤223
87
22 m
␤
98.8%
Th227
90␣
18.2 d
Ra ␣223
88
11.4 d
7.1 ϫ 108 y
Pb208
82
Po216
84
Pb ␤212
82
10.6 h
␤
0.014%
At216
85␣
3 ϫ 10Ϫ4 s
Bi ␣212
83
60.6 m
Tl ␤208
81
3.10 m
␤
66.3%
Po212
84␣
3.0 ϫ 10Ϫ7 s
Th ␣232
90
Ra ␤228
88
6.7 y
Ac ␤228
89
6.13 h
Th ␣228
90
1.91 y
Ra ␣224
88
3.64 d
Rn ␣220
86
55.3 s
␣
0.14 s
1.41 ϫ 1010 y
USES OF RADIONUCLIDES
Radionuclides have practical uses because they decay at known rates. Some applications
make use of the radiation that is continuously emitted by radionuclides.
Radioactive Dating
The ages of articles of organic origin can be estimated by radiocarbon dating. The
radioisotope carbon-14 is produced continuously in the upper atmosphere as nitrogen
atoms capture cosmic-ray neutrons.
14
07N ϩ 1
0n 88n 14
06C ϩ 1
1H
The carbon-14 atoms react with oxygen molecules to form 14CO2. This process continually
supplies the atmosphere with radioactive 14CO2, which is removed from the
atmosphere by photosynthesis. The intensity of cosmic rays is related to the sun’s activity.
As long as this remains constant, the amount of 14CO2 in the atmosphere remains constant.
14CO2 is incorporated into living organisms just as ordinary 12CO2 is, so a certain fraction
of all carbon atoms in living substances is carbon-14. This decays with a half-life of
5730 years.
14
06C 88n 14
07N ϩ Ϫ0
Ϫ1␤
After death, the plant no longer carries out photosynthesis, so it no longer takes up 14CO2.
Other organisms that consume plants for food stop doing so at death. The emissions from
the 14C in dead tissue then decrease with the passage of time. The activity per gram of
carbon is a measure of the length of time elapsed since death. Comparison of ages of
ancient trees calculated from 14C activity with those determined by counting rings indicates
that cosmic ray intensity has varied somewhat throughout history. The calculated
ages can be corrected for these variations. The carbon-14 technique is useful only for
dating objects less than 50,000 years old. Older objects have too little activity to be dated
accurately.
The potassium–argon and uranium–lead methods are used for dating older objects.
Potassium-40 decays to argon-40 with a half-life of 1.3 billion years.
40
19K ϩ Ϫ0
Ϫ1e 88n 40
18Ar
Because of its long half-life, potassium-40 can be used to date objects up to 1 million years
old by determination of the ratio of 40
19K to 40
18Ar in the sample. The uranium–lead method
is based on the natural uranium-238 decay series, which ends with the production of stable
lead-206. This method is used for dating uranium-containing minerals several billion years
old because this series has an even longer half-life. All the 206Pb in such minerals is assumed
to have come from 238U. Because of the very long half-life of 238
092U, 4.5 billion years, the
amounts of intermediate nuclei can be neglected. A meteorite that was 4.6 billion years
old fell in Mexico in 1969. Results of 238U/206Pb studies on such materials of extraterrestrial
origin suggest that our solar system was formed several billion years ago.
EXAMPLE 26-4 Radiocarbon Dating
A piece of wood taken from a cave dwelling in New Mexico is found to have a carbon-14
activity (per gram of carbon) only 0.636 times that of wood cut today. Estimate the age of the
wood. The half-life of carbon-14 is 5730 years.
26-12
26-12 Uses of Radionuclides 1017
In 1992, hikers in the Italian Alps
found the remains of a man who had
been frozen in a glacier for about
4000 years. This discovery is
especially important because of the
unusual preservation of tissues,
garments, and personal belongings.
Radiocarbon dating is used to
estimate the ages of archaeological
finds such as these.
In recent decades atmospheric testing
of nuclear warheads has also caused
fluctuations in the natural abundance
of 14C.
Gaseous argon is easily lost from
minerals. Measurements based on the
40K/40Ar method may therefore not be
as reliable as desired.
In Table 26-4(a) we see that the first
step, the decay of 238U, is the slowest
step (longest half-life). We learned in
Section 16-7 that the slowest step is
the rate-determining step.
Plan
As we did in Example 26-3, we determine the specific rate constant k from the known halflife.
The time required to reach the present fraction of the original activity is then calculated
from the first-order decay equation.
Solution
First we find the first-order specific rate constant for 14C.
t1/2 ϭ or k ϭ ϭ ϭ 1.21 ϫ 10Ϫ4 yϪ1
The present 14C activity, N (disintegrations per unit time), is 0.636 times the original activity,
N0.
N ϭ 0.636 N0
We substitute into the first-order decay equation
ln ΂ᎏ
N
N
0
ᎏ΃ϭ kt
ln ΂ᎏ
0.63
N
6
0
N0
ᎏ΃ϭ (1.21 ϫ 10Ϫ4 yϪ1)t
We cancel N0 and solve for t.
ln ΂ ΃ϭ (1.21 ϫ 10Ϫ4 yϪ1)t
0.452 ϭ (1.21 ϫ 10Ϫ4 yϪ1)t or t ϭ 3.74 ϫ 103y (or 3740 y)
You should now work Exercises 58 and 62.
EXAMPLE 26-5 Uranium–Lead Dating
A sample of uranium ore is found to contain 4.64 mg of 238U and 1.22 mg of 206Pb. Estimate
the age of the ore. The half-life of 238U is 4.51 ϫ 109 years.
Plan
The original mass of 238U is equal to the mass of 238U remaining plus the mass of 238U that
decayed to produce the present mass of 206Pb. We obtain the specific rate constant, k, from
the known half-life. Then we use the ratio of original 238U to remaining 238U to calculate the
time elapsed, with the aid of the first-order integrated rate equation.
Solution
First we calculate the amount of 238U that must have decayed to produce 1.22 mg of 206Pb,
using the isotopic masses.
_?_ mg 238U ϭ 1.22 mg 206Pb ϫ ᎏ
2
2
0
3
6
8
m
m
g
g
2
2
0
3
6
8
P
U
b
ᎏ ϭ 1.41 mg 238U
Thus, the sample originally contained 4.64 mg ϩ 1.41 mg ϭ 6.05 mg of 238U.
We next evaluate the specific rate (disintegration) constant, k.
t1/2 ϭ so k ϭ ϭ ϭ 1.54 ϫ 10Ϫ10 yϪ1
Now we calculate the age of the sample, t.
0.693
ᎏᎏ
4.51 ϫ 109 y
0.693
ᎏ
t1/2
0.693
ᎏ
k
1
ᎏ
0.636
0.693
ᎏ
5730 y
0.693
ᎏ
t1/2
0.693
ᎏ
k
1018 CHAPTER 26: Nuclear Chemistry
ln ΂ᎏ
A
A
0
ᎏ΃ϭ kt
ln ΂ᎏ
6
4
.
.
0
6
5
4
m
m
g
g
ᎏ΃ϭ (1.54 ϫ 10Ϫ10 yϪ1)t
ln 1.30 ϭ (1.54 ϫ 10Ϫ10 yϪ1)t
ϭ t or t ϭ 1.70 ϫ 109 years
The ore is approximately 1.7 billion years old.
You should now work Exercise 76.
Medical Uses
The use of cobalt radiation treatments for cancerous tumors was described in Example
26-3. Several other nuclides are used as radioactive tracers in medicine. Radioisotopes
of an element have the same chemical properties as stable isotopes of the same element,
so they can be used to “label” the presence of an element in compounds. A radiation
detector can be used to follow the path of the element throughout the body. Modern
computer-based techniques allow construction of an image of the area of the body where
the radioisotope is concentrated. Salt solutions containing 24Na can be injected into the
bloodstream to follow the flow of blood and locate obstructions in the circulatory system.
Thallium-201 tends to concentrate in healthy heart tissue, whereas technetium-99 concentrates
in abnormal heart tissue. The two can be used together to survey damage from
heart disease.
Iodine-123 concentrates in the thyroid gland, liver, and certain parts of the brain. This
radioisotope is used to monitor goiter and other thyroid problems, as well as liver and
brain tumors. One of the most useful radioisotopes in medical applications in recent years
is an isotope of technetium, an element that does not occur naturally on earth. This isotope,
99mTc, is produced by the decay of 99Mo.
99Mo 88n 99mTc ϩ Ϫ0
Ϫ1␤
The “m” in the superscript of 99mTc stands for “metastable.” This isotope is formed at a high
energy, and then slowly decays with a half-life of 6.0 hours by emitting gamma radiation.
99mTc 88n 99Tc ϩ ␥
The “Chemistry in Use” feature on the textbook Web site describes some medical applications
of 99mTc.
Another form of imaging that uses positron emitters (Section 26-6) is positron emission
tomography (PET). Isotopes commonly used in this technique are short-lived
positron emitters such as 11C (t1/2 ϭ 20.4 min), 13N (t1/2 ϭ 9.98 min), 15O (t1/2 ϭ 2.05
min), and 18F (t1/2 ϭ 110 min). The appropriate isotope is incorporated into a chemical
that is normally taken up by the tissues that are being investigated, for instance carbon
dioxide or glucose including 11C or water including 15O. This radioactive chemical can
then be administered by inhalation or injection. The patient is then placed into a cylindrical
gamma ray detector. When these radioisotopes decay, the emitted positron quickly
encounters an electron and reacts in a matter–antimatter annihilation, to give off two
gamma rays in opposite directions.
Ϫ0
ϩ1␤ ϩ Ϫ0
Ϫ1␤ 88n 2␥
0.262
ᎏᎏ
1.54 ϫ 10Ϫ10 yϪ1
26-12 Uses of Radionuclides 1019
This isotope of iodine is also used
in the treatment of thyroid cancer.
Because of its preferential absorption
in the thyroid gland, it delivers
radiation where it is needed.
A scan of the radiation released by
radioactive iodine concentrated in
thyroid tissue gives an image of the
thyroid gland.
The directions of emission of millions of such pairs of gamma rays detected over several
minutes allow for a computer reconstruction of an image of the tissue containing the
positron emitter.
The energy produced by the decay of plutonium-238 is converted into electrical energy
in heart pacemakers. The relatively long half-life of the isotope allows the device to be
used for 10 years before replacement.
Agricultural Uses
The pesticide DDT is toxic to humans and animals repeatedly exposed to it. DDT persists
in the environment for a long time. It concentrates in fatty tissues. The DDT once used
to control the screwworm fly was replaced by a radiological technique. Irradiating the
male flies with gamma rays alters their reproductive cells, sterilizing them. When great
numbers of sterilized males are released in an infested area, they mate with females, who,
of course, produce no offspring. This results in the reduction and eventual disappearance
of the population.
Labeled fertilizers can also be used to study nutrient uptake by plants and to study the
growth of crops. Gamma irradiation of some foods allows them to be stored for longer
periods without spoiling. For example, it retards the sprouting of potatoes and onions. In
1999, the FDA approved gamma irradiation of red meat as a way to curb food-borne
illnesses. In addition to significantly reducing levels of Listeria, Salmonella, and other
bacteria, such irradiation is currently the only known way to completely eliminate the
dangerous strain of Escherichia coli bacteria in red meat. Absorption of gamma rays
by matter produces no radioactive nuclides, so foods preserved in this way are not
radioactive.
Industrial Uses
There are many applications of radiochemistry in industry and engineering. When great
precision is required in the manufacture of strips or sheets of metal of definite thicknesses,
the penetrating powers of various kinds of radioactive emissions are utilized. The thickness
of the metal is correlated with the intensity of radiation passing through it. The flow
1020 CHAPTER 26: Nuclear Chemistry
The procedure works because the
female flies mate only once. In an area
highly populated with sterile males, the
probability of a “productive” mating is
very small.
A weak radioactive source such as
americium is used in some smoke
detectors. Radiation from the source
ionizes the air to produce a weak
current. Smoke particles interrupt
the current flow by attracting the
ions. This decrease in current
triggers the alarm.
Moderate irradiation with gamma rays from radioactive isotopes has kept the strawberries at
the right fresh for 15 days, while those at the left are moldy. Such irradiation kills mold
spores but does no damage to the food. The food does not become radioactive.
of a liquid or gas through a pipeline can be monitored by injecting a sample containing
a radioactive substance. Leaks in pipelines can also be detected in this way.
In addition to the 238Pu-based heart pacemaker already mentioned, lightweight,
portable power packs that use radioactive isotopes as fuel have been developed for other
uses. Polonium-210, californium-242, and californium-244 have been used in such generators
to power instruments for space vehicles and in polar regions. These generators can
operate for years with only a small loss of power.
Research Applications
The pathways of chemical reactions can be investigated using radioactive tracers. When
radioactive 35S2Ϫ ions are added to a saturated solution of cobalt sulfide in equilibrium
with solid cobalt sulfide, the solid becomes radioactive. This shows that sulfide ion
exchange occurs between solid and solution in the solubility equilibrium.
CoS(s) 34 Co2ϩ(aq) ϩ S2Ϫ(aq) Ksp ϭ 8.7 ϫ 10Ϫ23
Photosynthesis is the process by which the carbon atoms in CO2 are incorporated into
glucose, C6H12O6, in green plants.
sunlight
6CO2 ϩ 6H2O 888888888n C6H12O6 ϩ 6O2
chlorophyll
The process is more complex than the net equation implies; it actually occurs in many
steps and produces a number of intermediate products. By using labeled 14CO2, we can
identify the intermediate molecules. They contain the radioactive 14C atoms.
ARTIFICIAL TRANSMUTATIONS OF ELEMENTS
The first artificially induced nuclear reaction was carried out by Rutherford in 1915. He
bombarded nitrogen-14 with alpha particles to produce an isotope of oxygen and a proton.
14
07N ϩ 4
2␣ 88n 1
1H ϩ 17
08O
Such reactions are often indicated in abbreviated form, with the bombarding particle and
emitted subsidiary particles shown parenthetically between the parent and daughter nuclei.
14
07N (4
2␣, 1
1p) 17
08O
Several thousand different artificially induced reactions have been carried out with
bombarding particles such as neutrons, protons, deuterons (2
1H), alpha particles, and other
small nuclei.
Bombardment with Positive Ions
A problem arises with the use of positively charged nuclei as projectiles. For a nuclear
reaction to occur, the bombarding nuclei must actually collide with the target nuclei, which
are also positively charged. Collisions cannot occur unless the projectiles have sufficient
kinetic energy to overcome coulombic repulsion. The required kinetic energies increase
with increasing atomic numbers of the target and of the bombarding particle.
Particle accelerators called cyclotrons (atom smashers) and linear accelerators have
overcome the problem of repulsion. A cyclotron (Figure 26-7) consists of two hollow, Dshaped
electrodes called “dees.” Both dees are in an evacuated enclosure between the poles
of an electromagnet. The particles to be accelerated are introduced at the center in the
26-13
26-13 Artificial Transmutations of Elements 1021
The first cyclotron was constructed
by E. O. Lawrence (1901–1958) and
M. S. Livingston at the University of
California in 1930.
gap between the dees. The dees are connected to a source of high-frequency alternating
current that keeps them oppositely charged. The positively charged particles are attracted
toward the negative dee. The magnetic field causes the path of the charged particles to
curve 180° to return to the space between the dees. Then the charges are reversed on the
dees, so the particles are repelled by the first dee (now positive) and attracted to the second.
This repeated process is synchronized with the motion of the particles. They accelerate
along a spiral path and eventually emerge through an exit hole oriented so that the beam
hits the target atoms (Figure 26-8).
1022 CHAPTER 26: Nuclear Chemistry
Strong
magnetic field Oscillating
circuit
Small negative
electrode
Hollow D-shaped
electrode
(–)
High-energy beam
Figure 26-7 Schematic representation of a cyclotron.
The path of the particle is initially
circular because of the interaction
of the particle’s charge with the
electromagnet’s field. As the particle
gains energy, the radius of the path
increases, and the particle spirals
outward.
Figure 26-8 A beam of protons (bright blue stream) from a cyclotron at the Argonne
National Laboratory. Nuclear reactions take place when protons and other atomic particles
strike the nuclei of atoms.
In a linear accelerator the particles are accelerated through a series of tubes within an
evacuated chamber (Figure 26-9). The odd-numbered tubes are at first negatively charged
and the even ones positively charged. A positively charged particle is attracted toward the
first tube. As it passes through that tube, the charges on the tubes are reversed so that the
particle is repelled out of the first tube (now positive) and toward the second (negative)
tube. As the particle nears the end of the second tube, the charges are again reversed. As
this process is repeated, the particle is accelerated to very high velocities. The polarity is
changed at constant frequency, so subsequent tubes are longer to accommodate the
increased distance traveled by the accelerating particle per unit time. The bombardment
target is located outside the last tube. If the initial polarities are reversed, negatively
charged particles can also be accelerated. The longest linear accelerator, completed in
1966 at Stanford University, is about 2 miles long. It is capable of accelerating electrons
to energies of nearly 20 GeV.
Many nuclear reactions have been induced by such bombardment techniques. At the
time of development of particle accelerators, there were a few gaps among the first 92
elements in the periodic table. Particle accelerators were used between 1937 and 1941 to
synthesize three of the four “missing” elements: numbers 43 (technetium), 85 (astatine),
and 87 (francium).
26-13 Artificial Transmutations of Elements 1023
To
electrical
power
Odd
Even
0
To
vacuum
pump
Target
1 2 3 4 5 6
= •
Figure 26-9 Diagram of an early type of linear accelerator. An alpha emitter is placed in
the container at the left. Only those ␣-particles that happen to be emitted in line with the
series of accelerating tubes can escape.
The first linear accelerator was built
in 1928 by a German physicist, Rolf
Wideroe.
One gigaelectron volt (GeV) ϭ 1 ϫ
109 eV ϭ 1.60 ϫ 10Ϫ10 J. This is
sometimes called 1 billion electron
volts (BeV) in the United States.
An aerial view of the particle
accelerator dedicated in 1978 at
the Fermi National Accelerator
Laboratory (Fermilab), near Batavia,
Illinois. This proton accelerator, 4
miles in circumference, accelerates
protons to energies of 1 trillion
electron volts. Construction of a
vastly larger accelerator known as
the superconducting supercollider, or
SSC, was begun near Waxahachie,
Texas. However, this project was
abandoned in 1994 when the U.S.
Congress voted to discontinue
funding for it.
96
42Mo ϩ 2
1H 88n 97
43Tc ϩ 1
0n
209
083Bi ϩ 4
2␣ 88n 210
085At ϩ 3 1
0n
230
090Th ϩ 1
1p 88n 223
087Fr ϩ 2 4
2␣
Many hitherto unknown, unstable, artificial isotopes of known elements have also been
synthesized so that their nuclear structures and behavior could be studied.
Neutron Bombardment
Neutrons bear no charge, so they are not repelled by nuclei as positively charged projectiles
are. They do not need to be accelerated to produce bombardment reactions. Neutron
beams can be generated in several ways. A frequently used method involves bombardment
of beryllium-9 with alpha particles.
9
4Be ϩ 4
2␣ 88n 12
06C ϩ 1
0n
Nuclear reactors (Section 26-15) are also used as neutron sources. Neutrons ejected in
nuclear reactions usually possess high kinetic energies and are called fast neutrons. When
they are used as projectiles they cause reactions, such as (n, p) or (n, ␣) reactions, in which
subsidiary particles are ejected. The fourth “missing” element, number 61 (promethium),
was synthesized by fast neutron bombardment of neodymium-142.
142
060Nd ϩ 1
0n 88n 143
061Pm ϩ Ϫ0
Ϫ1␤
Slow neutrons (“thermal” neutrons) are produced when fast neutrons collide with
moderators such as hydrogen, deuterium, oxygen, or the carbon atoms in paraffin. These
neutrons are more likely to be captured by target nuclei. Bombardments with slow
neutrons can cause neutron-capture (n, ␥) reactions.
200
080Hg ϩ 1
0n 88n 201
080Hg ϩ 0
0␥
Slow neutron bombardment also produces the 3H isotope (tritium).
6
3Li ϩ 1
0n 88n 3
1H ϩ 4
2␣ (n, ␣) reaction
E. M. McMillan (1907–1991) discovered the first transuranium element, neptunium, in
1940 by bombarding uranium-238 with slow neutrons.
238
092U ϩ 1
0n 88n 239
092U ϩ 0
0␥
239
092U 88n 239
093Np ϩ Ϫ0
Ϫ1␤
Several additional elements have been prepared by neutron bombardment or by bombardment
of the nuclei so produced with positively charged particles. Some examples are
238
092U ϩ 1
0n 88n 239
092U ϩ 0
0␥
239
092U 88n 239
093Np ϩ Ϫ0
Ϫ1␤
· plutonium
239
093Np 88n 239
094Pu ϩ Ϫ0
Ϫ1␤
239
094Pu ϩ 4
2␣ 88n 242
096Cm ϩ 1
0n curium
246
096Cm ϩ 12
06C 88n 254
102No ϩ 4 1
0n nobelium
252
098Cf ϩ 10
05B 88n 257
103Lr ϩ 5 1
0n lawrencium
1024 CHAPTER 26: Nuclear Chemistry
Fast neutrons move so rapidly that
they are likely to pass right through
a target nucleus without reacting.
Hence, the probability of a reaction is
low, even though the neutrons may be
very energetic.
NUCLEAR FISSION
Isotopes of some elements with atomic numbers above 80 are capable of undergoing fission
in which they split into nuclei of intermediate masses and emit one or more neutrons.
Some fissions are spontaneous; others require that the activation energy be supplied by
bombardment. A given nucleus can split in many different ways, liberating enormous
amounts of energy. Some of the possible fissions that can result from bombardment
of fissionable uranium-235 with fast neutrons follow. The uranium-236 is a short-lived
intermediate.
160
062Sm ϩ 72
30Zn ϩ 4 1
0n ϩ energy
146
057La ϩ 87
35Br ϩ 3 1
0n ϩ energy
235
092U ϩ 1
0n 88n 236
092U 888n 141
056Ba ϩ 92
36Kr ϩ 3 1
0n ϩ energy
144
055Cs ϩ 90
37Rb ϩ 2 1
0n ϩ energy
144
054Xe ϩ 90
38Sr ϩ 2 1
0n ϩ energy
Recall that the binding energy is the amount of energy that must be supplied to the
nucleus to break it apart into subatomic particles. Figure 26-10 is a plot of binding energy
per nucleon versus mass number. It shows that atoms of intermediate mass number have
the highest binding energies per nucleon; therefore, they are the most stable. Thus, fission
is an energetically favorable process for heavy atoms, because atoms with intermediate
masses and greater binding energies per nucleon are formed.
Which isotopes of which elements undergo fission? Experiments with particle accelerators
have shown that every element with an atomic number of 80 or more has one or
more isotopes capable of undergoing fission, provided they are bombarded at the right
energy. Nuclei with atomic numbers between 89 and 98 fission spontaneously with long
half-lives of 104 to 1017 years. Nuclei with atomic numbers of 98 or more fission spontaneously
with shorter half-lives of a few milliseconds to 60.5 days. One of the natural decay
modes of the transuranium elements is via spontaneous fission. In fact, all known nuclides
26-14
26-14 Nuclear Fission 1025
888888n
888888n
8888n
8888n
The term “nucleon” refers to a nuclear
particle, either a neutron or a proton.
1 MeV ϭ 1.60 ϫ 10Ϫ13 J
10
9
8
He4
2
O16
8 Kr
fission
84
36Fe56
26
7
6
5
4
3
2
1
0 20 40 60 80 100 120 140 160 180 200 220 240
Mass number
Bindingenergypernucleon,MeV
fusion
U238
92
O18
8
B10
5
Li7
3
H2
1 Figure 26-10 Variation in nuclear
binding energy with atomic mass.
The most stable nucleus is 56
26Fe,
with a binding energy of 8.80 MeV
per nucleon.
with mass numbers greater than 250 do this because they are too big to be stable. Most
nuclides with mass numbers between 225 and 250 do not undergo fission spontaneously
(except for a few with extremely long half-lives). They can be induced to undergo fission
when bombarded with particles of relatively low kinetic energies. Particles that can supply
the required activation energy include neutrons, protons, alpha particles, and fast electrons.
For nuclei lighter than mass 225, the activation energy required to induce fission
rises very rapidly.
In Section 26-2 we discussed the stability of nuclei with even numbers of protons and
even numbers of neutrons. We should not be surprised to learn that both 233U and 235U
can be excited to fissionable states by slow neutrons much more easily than 238U, because
they are less stable. It is so difficult to cause fission in 238U that this isotope is said to be
“nonfissionable.”
Typically, two or three neutrons are produced per fission reaction. These neutrons can
collide with other fissionable atoms to repeat the process. If sufficient fissionable material,
the critical mass, is contained in a small enough volume, a sustained chain reaction
can result. If too few fissionable atoms are present, most of the neutrons escape and no
chain reaction occurs. Figure 26-11 depicts a fission chain reaction.
In an atomic bomb, two subcritical portions of fissionable material are brought together
to form a critical mass. A nuclear fission explosion results. A tremendous amount of heat
energy is released, as well as many radionuclides whose effects are devastating to life and
the environment. The radioactive dust and debris are called fallout.
NUCLEAR FISSION REACTORS
In a nuclear fission reactor, the fission reaction is controlled by inserting materials to
absorb some of the neutrons so that the mixture does not explode. The energy that is
produced can be safely used as a heat source in a power plant.
26-15
1026 CHAPTER 26: Nuclear Chemistry
Kr
n
U
Ba
Sr
Xe
n
n
n
n
n
n
n
n
n
92
36
Kr
92
36
Kr
92
36
235
92
U235
92
U235
92
141
56
Ba
141
56
Ba
141
56
U235
92
144
54
90
38
Figure 26-11 A self-propagating nuclear chain reaction. A stray neutron induces a single
fission, liberating more neutrons. Each of them induces another fission, each of which is
accompanied by release of two or three neutrons. The chain continues to branch in this way,
very quickly resulting in an explosive rate of fission.
The launching of the nuclear
submarine Hyman G. Rickover into
the Thames River in Connecticut
(August 27, 1983).
Light Water Reactors
Most commercial nuclear power plants in the United States are “light water” reactors,
moderated and cooled by ordinary water. Figure 26-12 is a schematic diagram of a light
water reactor plant. The reactor core at the left replaces the furnace in which coal, oil,
or natural gas is burned in a fossil fuel plant. Such a fission reactor consists of five main
components: (1) fuel, (2) moderator, (3) control rods, (4) cooling system, and (5) shielding.
Fuel
Rods of U3O8 enriched in uranium-235 serve as the fuel. Unfortunately, uranium ores
contain only about 0.7% 235
092U. Most of the rest is nonfissionable 238
092U. The enrichment
is done in processing and reprocessing plants by separating gaseous 235UF6 from 238UF6,
prepared from the ore. Separation by diffusion is based on the slower rates of diffusion
of heavier gas molecules (Section 12-14). Another separation procedure uses the ultra-
centrifuge.
A potentially more efficient method of enrichment would involve the use of sophisticated
tunable lasers to ionize 235
092U selectively and not 238
092U. The ionized 235
092U could then
be made to react with negative ions to form another compound, easily separated from the
mixture. For this method to work, we must construct lasers capable of producing radiation
monochromatic enough to excite one isotope and not the other—a difficult challenge.
Moderator
The fast neutrons ejected during fission are too energetic to be absorbed efficiently by
other nuclei. Thus, they must be slowed by collisions with atoms of comparable mass that
do not absorb them, called moderators. The most commonly used moderator is ordinary
water, although graphite is sometimes used. The most efficient moderator is helium, which
26-15 Nuclear Fission Reactors 1027
Molten sodium
or
Liquid water
under high
pressure
(carries heat
to steam generator)
Nuclear
reactor
Control rod
Uranium fuel
Pump
Steam
generator
Steam Steam turbine
(generates electricity)
River100° F80° F
Pump
Condenser
(steam from turbine
is condensed
by river water)
P
P
P
+
–
Pump
Figure 26-12 A schematic diagram
of a light water reactor plant. This
design includes two closed loops of
water. The water that carries heat
from the reactor to the steam
generator is in a closed loop and is
not released to the environment.
Uranium is deposited on the
negative electrode in the
electrorefining phase of fuel
reprocessing. The crystalline mass
is about 97% LiCl and KCl. The
remaining 3% uranium chloride is
responsible for the amethyst color.
slows neutrons but does not absorb them all. The next most efficient is “heavy water”
(deuterium oxide, 2
1H2O or 2
1D2O). This is so expensive that it has been used chiefly in
research reactors. A Canadian-designed power reactor that uses heavy water is more
neutron-efficient than light water reactors. This design is the basis of the many reactors
in Canada.
Control Rods
Cadmium and boron are good neutron absorbers.
10
05B ϩ 1
0n 88n 7
3Li ϩ 4
2␣
The rate of a fission reaction is controlled by the use of movable control rods, usually
made of cadmium or boron steel. They are automatically inserted in or removed from
spaces between the fuel rods. The more neutrons absorbed by the control rods, the fewer
fissions occur and the less heat is produced. Hence, the heat output is governed by the
control system that operates the rods.
Cooling System
Two cooling systems are needed. First, the moderator itself serves as a coolant for the
reactor. It transfers fission-generated heat to a steam generator. This converts water to
steam. The steam then goes to turbines that drive generators to produce electricity.
Another coolant (river water, sea water, or recirculated water) condenses the steam from
the turbine, and the condensate is then recycled into the steam generator.
The danger of meltdown arises if a reactor is shut down quickly. The disintegration of
radioactive fission products still goes on at a furious rate, fast enough to overheat the fuel
elements and to melt them. So it is not enough to shut down the fission reaction. Efficient
cooling must be continued until the short-lived isotopes are gone and the heat from
their disintegration is dissipated. Only then can the circulation of cooling water be stopped.
The 1979 accident at Three Mile Island, near Harrisburg, Pennsylvania, was due to
stopping the water pumps too soon and the inoperability of the emergency pumps. A
combination of mechanical malfunctions, errors, and carelessness produced the overheating
that damaged the fuel assembly. It did not and could not explode, although melting
of the core material did occur. The 1986 accident at Chernobyl, in the USSR, involved
a reactor of a very different design and was far more serious. The effects of that disaster
will continue for decades.
Shielding
It is essential that people and the surrounding countryside be adequately shielded from
possible exposure to radioactive nuclides. The entire reactor is enclosed in a steel containment
vessel. This is housed in a thick-walled concrete building. The operating personnel
are further protected by a so-called biological shield, a thick layer of organic material
made of compressed wood fibers. This absorbs the neutrons and beta and gamma rays
that would otherwise be absorbed in the human body.
Breeder Reactors
The possibility of shortages in the known supply of 235
092U has led to the development of
breeder reactors, which can manufacture more fuel than they use. A breeder reactor is
1028 CHAPTER 26: Nuclear Chemistry
The neutrons are the worst problem
of radiation. The human body contains
a high percentage of H2O, which
absorbs neutrons very efficiently. A
new weapon, the neutron bomb,
produces massive amounts of neutrons
and so is effective against people, but
it does not produce the long-lasting
radiation of the fission atomic bomb.
designed not only to generate electrical power but also to maximize neutron capture in
the core by 238
092U. The fuel of a typical breeder reactor consists of the abundant but nonfissionable
isotope 238
092U mixed with 235
092U or 239
094Pu, which produce neutrons when they
undergo fission. Some of these neutrons are absorbed by 238
092U to form 239
092U. This unstable
uranium isotope soon leads, after two steps of beta emission to 239
094Pu.
␤ decay ␤ decay
238
092U ϩ 1
0n 88n 239
092U 888888888888n 239
093Np 888888888888n 239
094Pu
t1/2 ϭ 23.4 min t1/2 ϭ 2.35 days
This fissionable 239
094Pu can then be used as fuel in a reactor.
For every 235
092U or 239
094Pu nucleus that undergoes fission, more than one neutron is
captured by 238
092U to produce 239
094Pu. Thus, the breeder reactor can produce more fissionable
material than it consumes. After about 7 years, enough 239
094Pu can be collected to fuel
a new reactor and to refuel the original one.
Nuclear Power: Hazards and Benefits
Controlled fission reactions in nuclear reactors are of great use and even greater potential.
The fuel elements of a nuclear reactor have neither the composition nor the extremely
compact arrangement of the critical mass of a bomb. Thus, no possibility of nuclear
explosion exists. However, various dangers are associated with nuclear energy generation.
The possibility of “meltdown” has been discussed with respect to cooling systems
in light water reactors. Proper shielding precautions must be taken to ensure that the
radionuclides produced are always contained within vessels from which neither they nor
their radiations can escape. Long-lived radionuclides from spent fuel must be stored
underground in heavy, shock-resistant containers until they have decayed to the point
that they are no longer biologically harmful. As examples, strontium-90 (t1/2 ϭ 28 years)
and plutonium-239 (t1/2 ϭ 24,000 years) must be stored for 280 years and 240,000 years,
respectively, before they lose 99.9% of their activities. Critics of nuclear energy contend
that the containers could corrode over such long periods, or burst as a result of earth
tremors, and that transportation and reprocessing accidents could cause environmental
contamination with radionuclides. They claim that river water used for cooling is returned
to the rivers with too much heat (thermal pollution), thus disrupting marine life. (It
should be noted, though, that fossil fuel electric power plants cause the same thermal
pollution for the same amount of electricity generated.) The potential for theft also exists.
Plutonium-239, a fissionable material, could be stolen from reprocessing plants and used
to construct atomic weapons.
Proponents of the development of nuclear energy argue that the advantages far
outweigh the risks. Nuclear energy plants do not pollute the air with oxides of sulfur,
nitrogen, carbon, and particulate matter, as fossil fuel electric power plants do. The big
advantage of nuclear fuels is the enormous amount of energy liberated per unit mass of
fuel. At present, nuclear reactors provide about 22% of the electrical energy consumed
in the United States. In some parts of Europe, where natural resources of fossil fuels
are scarcer, the utilization of nuclear energy is higher. For instance, in France and
Belgium, more than 80% of electrical energy is produced from nuclear reactors. With
rapidly declining fossil fuel reserves, it appears likely that nuclear energy and solar energy
will become increasingly important. Intensifying public concerns about nuclear power,
however, may mean that further growth in energy production using nuclear power in
the United States must await technological developments to overcome the remaining
hazards.
26-15 Nuclear Fission Reactors 1029
Nuclear waste may take centuries to
decompose, so we cannot afford to
take risks in its disposal. Suggested
approaches include casting it into
ceramics, as shown here, to eliminate
the possibility of the waste dissolving
in ground water. The encapsulated
waste could then be deposited in
underground salt domes. Located in
geologically stable areas, such salt
domes have held petroleum and
compressed natural gas trapped
for millions of years. The political
problems of nuclear waste disposal
are at least as challenging as the
technological ones. Refer to the
Chemistry in Use box entitled
“Managing Nuclear Wastes” in
this chapter.
1030 CHAPTER 26: Nuclear Chemistry
The EnvironmentCCHEMISTRY IN USE
Managing Nuclear Wastes
Some may consider a career in managing nuclear waste as
being just about the worst job anyone would ever want, but
hundreds of technically trained people have spent years working
to solve the problems associated with nuclear power. The
major part of the continuing challenge is political. Nuclear
power plants generate about 23% of the electricity in the
United States. Most of the high-level nuclear waste (HLW)
that is generated from nuclear power plants—in the form of
spent nuclear fuel (SNF)—is generated where many people
live, in the eastern half of the United States. The safest place
for a repository is away from people, in a dry, remote location,
probably in the western United States, where there are
fewer people (and fewer votes!).
SNF constitutes about half of the HLW in the United
States. The other half comes from the construction and existence
of nuclear weapons. All HLW is a federal responsibility.
About 90% of the radioactivity in nuclear waste is from HLW.
The largest volume of nuclear waste is low-level waste (LLW)
and that is mostly the responsibility of the state (or group of
states) in which it is generated. LLW is rather awkwardly
defined, being everything that is neither HLW nor defense
waste and consists of wastes from hospitals; pharmaceutical
labs; research labs; and the moon suits, tools, and the like
from nuclear power plants. In the eastern United States, most
of the LLW is in the form of the plastic beads that make up
the ion-exchange resins used in nuclear power plants to clean
various loops of water used in power production.
Plutonium wastes from the Los Alamos National Laboratory
in northern New Mexico were trucked for the first
time to the federal Waste Isolation Pilot Plant in Carlsbad
in March 1999. The 600 pounds (270 kg) of waste consisted
of plutonium-contaminated clothing and metal cans, packed
in boxes and stainless steel containers. Most of the material
was from the laboratory’s manufacture of nuclear batteries
used in NASA’s deep space probes and will be buried in the
depository carved out of ancient salt caverns about half a mile
(0.8 km) below ground.
Most current attention is focused on SNF for two reasons.
It is highly radioactive and it can be seen as a “local”
problem because it is made where electric customers live.
Europe has reprocessing plants to recover the unused fissionable
material for new fuel, but the United States
disallowed the practice in the 1970s. This partially explains
why spent fuel rods have been piling up at U.S. nuclear plants.
Research has focused on Yucca Mountain, Nevada, at the
western edge of the National Test Site, for its suitability as a
nuclear waste repository for SNF and some defense waste.
Many political leaders of Nevada strongly oppose this plan,
and they seriously question that nuclear waste can be safely
kept out of the human environment for 10,000 years, as is
required under the federal Nuclear Waste Policy Act.
The numbers describing SNF are barely comprehensible
to most people. The volume of all existing SNF could cover
a large football stadium to a depth of 4 or 5 feet, but no sensible
person would want to confine that much heat and
radioactivity to one place. Another description is the 70,000
metric tons of SNF generated to date in power plants, a
figure that means little unless one understands thousandkilogram
quantities and knows the density of fission products.
The plans for Yucca Mountain, should it be found to
be a suitable site, will hold in its many miles of tunnels and
caverns, all the SNF so far generated and that expected to be
generated in the next few years.
The SNF portion of HLW can be understood by
chemists who see in it nearly every element on the periodic
chart of the elements. After a 235U nucleus undergoes fission
and releases its excess nuclear binding energy, it leaves
a pair of new atoms. These fission products are like newly
born forms of the elements that are already well known and,
like newborns, are unstable until they mature. There are
about 1000 isotopes of about 100 different elements in SNF,
and most are radioactive. They decay into stable elements
at different rates, giving off alpha, beta, and gamma emissions.
It will take about 7000 years until the SNF will be
only as radioactive as the rocks and minerals that make up
our planet.
These fission products are housed in long titanium rods,
each about the diameter of a pencil, that constitute the fuel
assembly in a nuclear power plant. Workers wearing gloves
can handle fuel assemblies before fissioning occurs. But after
removal from a nuclear reactor, the fuel assembly is stored
in a cooling pool of water beside the reactor for at least 10
years. If the power plant has a small cooling pool, on-site
storage of the oldest fuel assemblies occurs in specially constructed
concrete casks until the federal government takes
ownership and finds a suitable place for it. Fuel rod consolidation
is sometimes practiced to save space because much of
the space in a fuel assembly was present so power plant water
26-15 Nuclear Fission Reactors 1031
could easily pass and pick up the heat generated by the fission
process.
Other options that have been considered for HLW include
outer space ejection and burial in deep ocean trenches. The
consensus worldwide is that deep geological isolation is the
best option. The United States leads in studying a specific
site, Yucca Mountain. In other countries, even those generating
a larger percentage of their power with nuclear power,
the small volumes awaiting burial allow them more time to
choose a location.
For the student who likes a challenge, nuclear waste management
is a good one. From the cost-benefit analyses of all
the ways we make and use electrical power to the way the
wastes are handled, one can find an issue or a career. Here
are a few key issues to study and discuss:
• Transportation of the waste to its repository. Should it be
done by rail or by truck? Should there be public notification
of the time of transport? Are there response
measures in place in case of an accident?
• The site’s seismicity. Will there be significant volcanic or
seismic activity near the site in the next 10,000 years?
• Hydrology. Is there enough evidence to ensure that
radionuclides will not seep into groundwater to any
significant degree?
• Public education. Should conservation be taught, and
should teachers promote or discourage the role of
nuclear power in our nation’s power mix?
• Other technical options. Should one investigate nuclear
physics options that might transmute the long-lived
radioisotopes into ones with shorter half-lives regardless
of the costs?
• Weapons disarmament. Should the plutonium from
“disarmed” nuclear weapons eventually be turned into
nuclear fuel or made useless immediately and buried
with other HLW?
For more information, see the following Web sites:
The Department of Energy, http://www.doe.gov
Yucca Mountain studies project, http://www.ymp.gov
The American Nuclear Society, http://www.ans.org
See also Radwaste magazine.
Donald H. Williams
Professor of Chemistry
Hope College
Nuclear waste
receiving building
Hoist
building
Filter
house
Administration
building
1,000 feet
Rock strata
Steel waste
canisters containing
spent nuclear fuel
Tunnel network
Mine
level
Ventilation
shaft
Waste
shaftShaft
for people,
material
Tunnel
In the United States, permanent storage sites for high-level radioactive
wastes will probably be deep underground in rock formations. Shown is
the kind of nuclear waste facility designed for Yucca Mountain, which
would be a three-square-mile complex of interconnected tunnels located in
dense volcanic rock 305 meters (1000 feet) beneath the mountain.
NUCLEAR FUSION
Fusion, the joining of light nuclei to form heavier nuclei, is favorable for the very light
atoms. In both fission and fusion, the energy liberated is equivalent to the loss of mass
that accompanies the reactions. Much greater amounts of energy per unit mass of reacting
atoms are produced in fusion than in fission.
Spectroscopic evidence indicates that the sun is a tremendous fusion reactor consisting
of 73% H, 26% He, and 1% other elements. Its major fusion reaction is thought to involve
the combination of a deuteron, 2
1H, and a triton 3
1H, at tremendously high temperatures
to form a helium nucleus and a neutron.
2
1H ϩ 3
1H 88n 4
2He ϩ 1
0n ϩ energy
Thus, solar energy is actually a form of fusion energy.
Fusion reactions are accompanied by even greater energy production per unit mass of
reacting atoms than are fission reactions. They can be initiated only by extremely high
temperatures, however. The fusion of 2
1H and 3
1H occurs at the lowest temperature of any
fusion reaction known, but even this is 40,000,000 K! Such temperatures exist in the sun
and other stars, but they are nearly impossible to achieve and contain on earth. Thermonuclear
bombs (called fusion bombs or hydrogen bombs) of incredible energy have
been detonated in tests but, thankfully, never in war. In them the necessary activation
energy is supplied by the explosion of a fission bomb.
It is hoped that fusion reactions can be harnessed to generate energy for domestic
power. Because of the tremendously high temperatures required, no currently known
structural material can confine these reactions. At such high temperatures all molecules
dissociate and most atoms ionize, resulting in the formation of a new state of matter called
26-16
1032 CHAPTER 26: Nuclear Chemistry
The deuteron and triton are the nuclei
of two isotopes of hydrogen, called
deuterium and tritium. Deuterium
occurs naturally in water. When the
D2O is purified as “heavy water,” it
can be used for several types of
chemical analysis.
Our sun supplies energy to the earth
from a distance of 93,000,000 miles.
Like other stars, it is a giant nuclear
fusion reactor. Much of its energy
comes from the fusion of deuterium,
2
1H, producing helium, 4
2He.
The explosion of a thermonuclear (hydrogen)
bomb releases tremendous amounts of energy. If
we could learn how to control this process, we
would have nearly limitless amounts of energy.
a plasma. A very high temperature plasma is so hot that it melts and decomposes anything
it touches, including structural components of a reactor. The technological innovation
required to build a workable fusion reactor probably represents the greatest challenge ever
faced by the scientific and engineering community.
Recent attempts at the containment of lower temperature plasmas by external magnetic
fields have been successful, and they encourage our hopes. Fusion as a practical energy
source, however, lies far in the future at best. The biggest advantages of its use would be
that (1) the deuterium fuel can be found in virtually inexhaustible supply in the oceans;
and (2) fusion reactions would produce only radionuclides of very short half-life, primarily
tritium (t1/2 ϭ 12.3 years), so there would be no long-term waste disposal problem. If
controlled fusion could be brought about, it could liberate us from dependence on uranium
and fossil fuels.
26-16 Nuclear Fusion 1033
Nuclear fusion provides the energy of our
sun and other stars. Development of
controlled fusion as a practical source of
energy requires methods to initiate and
contain the fusion process. Here a very
powerful laser beam has initiated a fusion
reaction in a 1-mm target capsule that
contained deuterium and tritium. In a 0.5picosecond
burst, 1013 neutrons were
produced by the reaction 2
1H ϩ 3
1H 88n
4
2He ϩ 1
0n.
Plasmas have been called the fourth
state of matter.
The plasma in a fusion reactor must
not touch the walls of its vacuum
vessel, which would be vaporized.
In the Tokamak fusion test reactor,
the plasma is contained within a
magnetic field shaped like a
doughnut. The magnetic field is
generated by D-shaped coils around
the vacuum vessel.
1034 CHAPTER 26: Nuclear Chemistry
Key Terms
Alpha particle (␣) A particle that consists of two protons and
two neutrons; identical to a helium nucleus.
Artificial transmutation An artificially induced nuclear reaction
caused by bombardment of a nucleus with subatomic particles
or small nuclei.
Band of stability A band containing stable (nonradioactive)
nuclides in a plot of number of neutrons versus number of protons
(atomic number).
Beta particle (␤) An electron emitted from the nucleus when a
neutron decays to a proton and an electron.
Binding energy (nuclear binding energy) The energy equivalent
(E ϭ mc2) of the mass deficiency of an atom.
Breeder reactor A fission reactor that produces more fissionable
material than it consumes.
Chain reaction A reaction that, once initiated, sustains itself and
expands.
Cloud chamber A device for observing the paths of speeding
particles as vapor molecules condense on the ionized air molecules
in their tracks.
Control rods Rods of materials such as cadmium or boron steel
that act as neutron absorbers (not merely moderators), used in
nuclear reactors to control neutron fluxes and therefore rates
of fission.
Critical mass The minimum mass of a particular fissionable
nuclide, in a given volume, that is required to sustain a nuclear
chain reaction.
Cyclotron A device for accelerating charged particles along a
spiral path.
Daughter nuclide A nuclide that is produced in a nuclear decay.
Electron capture Absorption of an electron from the first
energy level (K shell) by a proton as it is converted to a neutron;
also K capture.
Fast neutron A neutron ejected at high kinetic energy in a
nuclear reaction.
Fluorescence Absorption of high-energy radiation by a substance
and the subsequent emission of visible light.
Gamma ray (␥) High-energy electromagnetic radiation.
Geiger-Müller counter A type of gas ionization counter used
to detect radiation.
Half-life of a radionuclide The time required for half of a given
sample to undergo radioactive decay.
Heavy water Water containing deuterium, a heavy isotope of
hydrogen, 2
1H.
Linear accelerator A device used for accelerating charged particles
along a straight-line path.
Mass deficiency The amount of matter that would be converted
into energy if an atom were formed from constituent
particles.
Moderator A substance such as hydrogen, deuterium, oxygen,
or paraffin capable of slowing fast neutrons upon collision.
Nuclear binding energy The energy equivalent of the mass
deficiency; energy released in the formation of an atom from
subatomic particles.
Nuclear fission The process in which a heavy nucleus splits into
nuclei of intermediate masses and one or more neutrons are
emitted.
Nuclear fusion The combination of light nuclei to produce a
heavier nucleus.
Nuclear reaction A reaction involving a change in the composition
of a nucleus; it can evolve or absorb an extraordinarily
large amount of energy.
Nuclear reactor A system in which controlled nuclear fission
reactions generate heat energy on a large scale. The heat energy
is subsequently converted into electrical energy.
Nucleons Particles comprising the nucleus; protons and neu-
trons.
Nuclides Different atomic forms of all elements (in contrast to
isotopes, which are different atomic forms of a single element).
Parent nuclide A nuclide that undergoes nuclear decay.
Plasma A physical state of matter that exists at extremely high
temperatures, in which all molecules are dissociated and most
atoms are ionized.
Positron A nuclear particle with the mass of an electron but
opposite charge.
Radiation High-energy particles or rays emitted in nuclear decay
processes.
Radioactive dating A method of dating ancient objects by determining
the ratio of amounts of a parent nuclide and one of its
decay products present in an object and relating the ratio to the
object’s age via half-life calculations.
Radioactive tracer A small amount of radioisotope that replaces
a nonradioactive isotope of the element in a compound whose
path (e.g., in the body) or whose decomposition products are
to be monitored by detection of radioactivity; also called a
radioactive label.
Radioactivity The spontaneous disintegration of atomic nuclei.
Radioisotope A radioactive isotope of an element.
Radionuclide A radioactive nuclide.
Scintillation counter A device used for the quantitative detection
of radiation.
Slow neutron A fast neutron slowed by collision with a moder-
ator.
Thermonuclear energy Energy from nuclear fusion reactions.
Transuranium elements The elements with atomic numbers
greater than 92 (uranium); none occurs naturally and all must
be prepared by nuclear bombardment of other elements.
Exercises 1035
Exercises
Nuclear Stability and Radioactivity
*01. Define and compare nuclear fission and nuclear fusion.
Briefly describe current uses of nuclear fission and fusion.
*02. Differentiate between natural and induced radioactivity.
Use the periodic table to identify the locations of those
elements that are the result of induced radioactivity.
*03. How do nuclear reactions differ from ordinary chemical
reactions?
*04. What is the equation that relates the equivalence of matter
and energy? What does each term in this equation
represent?
*05. What is mass deficiency? What is binding energy? How
are the two related?
*06. What are nucleons? What is the relationship between the
number of protons and the atomic number? What is the
relationship among the number of protons, the number of
neutrons, and the mass number?
*07. Define the term “binding energy per nucleon.” How can
this quantity be used to compare the stabilities of nuclei?
*08. Describe the general shape of the plot of binding energy
per nucleon against mass number.
*09. (a) Briefly describe a plot of the number of neutrons against
the atomic number (for the stable nuclides). Interpret the
observation that the plot shows a band with a somewhat
step-like shape. (b) Describe what is meant by “magic numbers”
of nucleons.
*10. Potassium, Z ϭ 19, has a series of naturally occurring isotopes:
39K, 40K, 41K. Identify the isotope(s) of potassium
that is/are most likely to be stable and which tend to decay.
*11. Platinum, Z ϭ 78, has a series of naturally occurring isotopes:
190Pt, 192Pt, 194Pt, 195Pt, 196Pt, and 198Pt. Identify
the isotope(s) of platinum that is/are most likely to be stable
and which tend to decay.
*12. Indicate the type of emission and the decay product
predicted for each unstable isotope listed in Exercise 10.
*13. Indicate the type of emission and the decay product
predicted for each unstable isotope listed in Exercise 11.
*14. The actual mass of a 62Ni atom is 61.9283 amu. (a) Calculate
the mass deficiency in amu/atom and in g/mol for
this isotope. (b) What is the nuclear binding energy in
kJ/mol for this isotope?
*15. The actual mass of a 108Pd atom is 107.90389 amu. (a) Calculate
the mass deficiency in amu/atom and in g/mol for
this isotope. (b) What is the nuclear binding energy in
kJ/mol for this isotope?
*16. Calculate the following for 64Zn (actual mass ϭ 63.9291
amu). (a) mass deficiency in amu/atom; (b) mass deficiency
in g/mol; (c) binding energy in J/atom; (d) binding energy
in kJ/mol; (e) binding energy in MeV/nucleon.
*17. Calculate the following for 49Ti (actual mass ϭ 48.94787
amu). (a) mass deficiency in amu/atom; (b) mass deficiency
in g/mol; (c) binding energy in J/atom; (d) binding energy
in kJ/mol; (e) binding energy in MeV/nucleon.
*18. Calculate the nuclear binding energy in kJ/mol for each of
the following (a) 127
053I; (b) 81
35Br; (c) 35
17Cl. The atomic masses
are 126.9044 amu, 80.9163 amu, and 34.96885 amu,
respectively.
*19. Repeat Exercise 18 for (a) 36
16S, (b) 39
19Kr, and (c) 24
12Mg.
Their respective atomic masses are 35.96709 amu,
38.96371 amu, and 23.98504 amu. Which of these nuclides
has the greatest binding energy per nucleon?
*20. Compare the behaviors of ␣, ␤, and ␥ radiation (a) in an
electrical field, (b) in a magnetic field, and (c) with respect
to ability to penetrate various shielding materials, such as
a piece of paper and concrete. What is the composition of
each type of radiation?
*21. Why are ␣-particles that are absorbed internally by the
body particularly dangerous?
*22. Name some radionuclides that have medical uses, and list
the uses.
*23. Describe how radionuclides can be used in (a) research,
(b) agriculture, and (c) industry.
*24. Name and describe four methods for detection of radia-
tion.
*25. Describe how (a) nuclear fission and (b) nuclear fusion generate
more stable nuclei.
*26. What evidence exists to support the theory that nucleons
are arranged in “shells” or energy levels within the
nucleus?
Nuclear Reactions
*27. Consider a radioactive nuclide with a neutron/proton ratio
that is larger than those for the stable isotopes of that
element. What mode(s) of decay might be expected for this
nuclide, and why?
*28. Repeat Exercise 27 for a nuclide with a neutron/proton
ratio that is smaller than those for the stable isotopes.
*29. Calculate the neutron/proton ratio for each of the following
radioactive nuclides, and predict how each of the
nuclides might decay: (a) 13
05B (stable mass numbers for B
are 10 and 11); (b) 92
38Sr (stable mass numbers for Sr are
between 84 and 88); (c) 192
082Pb (stable mass numbers for Pb
are between 204 and 208).
*30. Repeat Exercise 29 for (a) 193
079 Au (stable mass number for
Au is 197), (b) 189
075Re (stable mass numbers for Re are 185
and 187), and (c) 137
059Pr (stable mass number for Pr is
141).
*31. Write the symbols for the daughter nuclei in the following
radioactive decays (␤ refers to an eϪ).
Ϫ␤ Ϫ␣
(a) 125Sn 888n (d) 147Sm 888n
Ϫn Ϫp
(b) 13C 888n (e) 184Ir 888n
Ϫ␥ ϩ␤
(c) 11B 888n (f ) 40
19K 888n
*32. Predict the kind of decays you would expect for the following
radionuclides. (a) 60
27Co (n/p ratio too high); (b) 20
11Na
(n/p ratio too low); (c) 222
086Rn; (d) 67
29Cu; (e) 238
092U; (f) 11
06C.
*33. What are nuclear bombardment reactions? Explain the
shorthand notation used to describe bombardment
reactions.
*34. Fill in the missing symbols in the following nuclear
bombardment reactions.
(a) 23
11Na ϩ _?_ 88n 23
12Mg ϩ 1
0n
(b) 59
27Co ϩ 1
0n 88n 56
25Mn ϩ _?_
(c) 96
42Mo ϩ 4
2He 88n 100
043Tc ϩ _?_
(d) 209
083Bi ϩ _?_ 88n 210
084Po ϩ 1
0n
(e) 238
092U ϩ 16
08O 88n _?_ ϩ 5 1
0n
*35. Fill in the missing symbols in the following nuclear
bombardment reactions.
(a) 232
090Th ϩ _?_ 88n 240
096Cm ϩ 4 1
0n
(b) _?_ ϩ 1
1H 88n 29
14Si ϩ 0
0␥
(c) 0?
26Mg ϩ _?_ 88n 0?
26Al ϩ 1
0n
(d) 40
18Ar ϩ _?_ 88n 0?
43K ϩ 1
1H
*36. Write the symbols for the daughter nuclei in the following
nuclear bombardment reactions. (a) 60
28Ni (n, p); (b) 98
42Mo
(1
0n, ␤); (c) 35
17Cl (p, ␣).
*37. Write the symbols for the daughter nuclei in the following
nuclear bombardment reactions. (a) 20
10Ne (␣, ␥); (b) 15
07N
( p, ␣); (c) 10
05B (n, ␣).
*38. Write the nuclear equation for each of the following bombardment
processes. (a) 14
07N (␣, p) 17
08O; (b) 106
046Pd (n, p)
106
045Rh; (c) 23
11Na (n, ␤Ϫ)X. Identify X.
*39. Repeat Exercise 38 for the following. (a) 113
048Cd (n, ␥) 114
048Cd;
(b) 6
3Li (n, ␣) 3
1H; (c) 2
1H (␥, p)X. Identify X.
*40. Write the shorthand notation for each of the following
nuclear reactions.
(a) 6
3Li ϩ 1
0n 88n 4
2He ϩ 3
1H
(b) 31
15P ϩ 2
1H 88n 32
15P ϩ 1
1H
(c) 238
092U ϩ 1
0n 88n 239
093Np ϩ Ϫ0
Ϫ1e
*41. Repeat Exercise 40 for the following.
(a) 253
099Es ϩ 4
2He 88n 256
101Md ϩ 1
0n
(b) 27
13Al ϩ 1
0n 88n 26
13Al ϩ 2 1
0n
(c) 37
17Cl ϩ 1
1H 88n 1
0n ϩ 37
18Ar
*42. Write the nuclear equations for the following processes.
(a) 63
28Ni undergoing ␤Ϫ emission; (b) two deuterium ions
undergoing fusion to give 3
2He and a neutron; (c) a nuclide
being bombarded by a neutron to form 7
3Li and an ␣particle
(identify the unknown nuclide); (d) 14
07N being
bombarded by a neutron to form three ␣-particles and an
atom of tritium.
*43. Write the nuclear equations for the following processes.
(a) 220
086Rn undergoing ␣ decay; (b) 110
049In undergoing
positron emission; (c) 127
053I being bombarded by a proton
to form 121
054Xe and seven neutrons; (d) tritium and deuterium
undergoing fusion to form an ␣-particle and a
neutron; (e) 95
42Mo being bombarded by a proton to form
95
43Tc and radiation (identify this radiation).
*44. “Radioactinium” is produced in the actinium series from
235
092U by the successive emission of an ␣-particle, a ␤Ϫparticle,
an ␣-particle, and a ␤Ϫ-particle. What are the
symbol, atomic number, and mass number for “radioac-
tinium?”
*45. An alkaline earth element (Group IIA) is radioactive. It
undergoes decay by emitting three ␣-particles in succession.
In what periodic table group is the resulting element
found?
*46. A nuclide of element rutherfordium, 257
104Rf, is formed by
the nuclear reaction of californium-249 and carbon-12,
with the emission of four neutrons. This new nuclide
rapidly decays by emitting an ␣-particle. Write the equation
for each of these nuclear reactions.
*47. Supply the missing information to each of the following
equations.
(a) 53
24Cr ϩ 4
2He 88n 1
0n ϩ _?_
(b) 187
075Re ϩ ␤ 88n _?_
(c) 243
095Am ϩ 1
0n 88n 00?
244Cm ϩ _?_ ϩ ␥
(d) 35
17Cl ϩ _?_ 88n 32
16S ϩ 4
2He
1036 CHAPTER 26: Nuclear Chemistry
Exercises 1037
*48. Supply the missing information to each of the following
reactions.
(a) 0?
14N ϩ _?_ 88n 0?
17O ϩ p
(b) 235
092U ϩ 1
0n 88n 137
054? ϩ 2 1
0n ϩ _?_
(c) 241
095Am ϩ 12
0?C 88n 4 1
0n ϩ _?_
*49. Describe how (a) cyclotrons and (b) linear accelerators
work.
Rates of Decay
*50. What does the half-life of a radionuclide represent? How
do we compare the relative stabilities of radionuclides in
terms of half-lives?
*51. Why must all radioactive decays be first order?
*52. Describe the process by which steady-state (constant) ratios
of carbon-14 to (nonradioactive) carbon-12 are attained in
living plants and organisms. Describe the method of radiocarbon
dating. What factors limit the use of this method?
*53. The half-life of 19
08O is 29 s. What fraction of the isotope
originally present would be left after 10.0 s?
*54. The half-life of 11
06C is 20.3 min. How long will it take for
95.0% of a sample to decay? How long will it take for
99.5% of the sample to decay?
*55. The activity of a sample of tritium decreased by 5.5% over
the period of a year. What is the half-life of 3
1H?
*56. A very unstable isotope of beryllium, 8Be, undergoes ␣
emission with a half-life of 0.07 fs. How long does it take
for 99.99% of a 1.0-␮g sample of 8Be to undergo decay?
*57. The 14
06C activity of an artifact from a burial site was
8.6/minиg C. The half-life of 14
06C is 5730 years, and the
current 14
06C activity is 15.3/minиg C (that is, 15.3 disintegrations
per minute per gram of carbon). How old is the
artifact?
*58. A piece of wood from a burial site was analyzed using 14
06C
dating and was found to have an activity of 11.8/minиg C.
Using the data given in Exercise 57 for 14
06C dating, determine
the age of this piece of wood.
*59. Analysis of an ant found in a piece of amber provided 14.0
disintegrations of 14
06C/minиg C, whereas a living ant of the
same species produces 17.0 disintegrations per minute.
Calculate the approximate age of the fossilized ant. The
half-life of carbon-14 is 5730 years.
*60. A skeleton was found in the woods and the police would
like to place the approximate time of death. A sample of
bone from the skeleton produces 12.0 disintegrations per
min per g of C. Bone of recent origin produces 14.0 disintegrations
per minиper g C. Calculate the approximate
age of the bone sample. The half-life of carbon-14 is 5730
years.
*61. Strontium-90 is one of the harmful radionuclides that
results from nuclear fission explosions. It decays by beta
emission with a half-life of 28 years. How long would it
take for 99.99% of a given sample released in an atmospheric
test of an atomic bomb to disintegrate?
*62. Carbon-14 decays by beta emission with a half-life of 5730
years. Assuming a particular object originally contained
7.50 ␮g of carbon-14 and now contains 0.80 ␮g of carbon-
14, how old is the object?
Fission and Fusion
*63. Briefly describe a nuclear fission process. What are the two
most important fissionable materials?
*64. What is a chain reaction? Why are nuclear fission processes
considered chain reactions? What is the critical mass of a
fissionable material?
*65. Where have continuous nuclear fusion processes been
observed? What is the main reaction that occurs in such
sources?
*66. The reaction that occurred in the first fusion bomb was
7
3Li ( p, ␣) X. (a) Write the complete equation for the
process, and identify the product, X. (b) The atomic masses
are 1.007825 amu for 1
1H, 4.00260 amu for ␣, and
7.01600 amu for 7
3Li. Find the energy for the reaction in
kJ/mol.
*67. Summarize how an atomic bomb works, including how the
nuclear explosion is initiated.
*68. Discuss the pros and cons of the use of nuclear energy
instead of other, more conventional types of energy based
on fossil fuels.
*69. Describe and illustrate the essential features of a light water
fission reactor.
*70. How is fissionable uranium-235 separated from nonfissionable
uranium-238?
*71. Distinguish between moderators and control rods of
nuclear reactors.
*72. What are the major advantages and disadvantages of fusion
as a potential energy source compared with fission? What
is the major technological problem that must be solved to
permit development of a fusion reactor?
Mixed Exercises
*73. Calculate the binding energy, in kJ/mol of nucleons, for
the following isotopes. (a) 15
08O with a mass of 15.00300 amu;
(b) 16
08O with a mass of 15.99491 amu; (c) 17
08O with a mass
of 16.99913 amu; (d) 18
08O with a mass of 17.99915 amu;
(e) 19
08O with a mass of 19.0035 amu. Which of these would
you expect to be most stable?
*74. The first nuclear transformation (discovered by Rutherford)
can be represented by the shorthand notation 14
07N
(␣, p) 17
08O. (a) Write the corresponding nuclear equation
for this process. The respective atomic masses are 14.00307 amu
1038 CHAPTER 26: Nuclear Chemistry
for 14
07N, 4.00260 amu for 4
2He, 1.007825 amu for 1
1H, and
16.99913 amu for 17
08O. (b) Calculate the energy change of
this reaction in kJ/mol.
*75. A proposed series of reactions (known as the carbon–
nitrogen cycle) that could be important in the very hottest
region of the interior of the sun is
12C ϩ 1H 88n A ϩ ␥
A 88n B ϩ ϩ0
ϩ1e
B ϩ 1H 88n C ϩ ␥
C ϩ 1H 88n D ϩ ␥
D 88n E ϩ ϩ0
ϩ1e
E ϩ 1H 88n 12C ϩ F
Identify the species labeled A–F.
*76. The ultimate stable product of 238U decay is 206Pb. A certain
sample of pitchblende ore was found to contain 238U
and 206Pb in the ratio 67.8 atoms 238UϺ32.2 atoms 206Pb.
Assuming that all the 206Pb arose from uranium decay, and
that no U or Pb has been lost by weathering, how old is
the rock? The half-life of 238U is 4.51 ϫ 109 years.
*77. Potassium-40 decays to 40Ar with a half-life of 1.3 ϫ 109
years. What is the age of a lunar rock sample that contains
these isotopes in the ratio 33.4 atoms 40KϺ66.6 atoms 40Ar?
Assume that no argon was originally in the sample and that
none has been lost by weathering.
CONCEPTUAL EXERCISES
*78. Both nuclear and conventional power plants produce
wastes to which the environment is sensitive. Discuss these
wastes and the current problems created by the wastes.
*79. If the earth is 4.5 ϫ 109 years old and the amount of
radioactivity in a sample becomes smaller with time, how
is it possible for there to be any radioactive elements left
on earth that have half-lives of less than a few million years?
BUILDING YOUR KNOWLEDGE
*80. Show by calculation which reaction produces the larger
amount of energy per atomic mass unit of material react-
ing.
fission: 235
092U ϩ 1
0n 88n 94
40Zr ϩ 140
058Ce ϩ 6 Ϫ0
Ϫ1e ϩ 2 1
0n
fusion: 2 2
1H 88n 3
1H ϩ 1
1H
The atomic masses are 235.0439 amu for 235
092U; 93.9061
amu for 94
40Zr; 139.9053 amu for 140
058Ce; 3.01605 amu for
3
1H; 1.007825 amu for 1
1H; 2.0140 amu for 2
1H.
*81. What would be the volume of helium, measured at STP,
generated from the decay of the 8Be sample in Exercise
56?
27Organic Chemistry I:
Formulas, Names,
and Properties
OUTLINE
Saturated Hydrocarbons
27-1 Alkanes and Cycloalkanes
27-2 Naming Saturated
Hydrocarbons
Unsaturated Hydrocarbons
27-3 Alkenes
27-4 Alkynes
Aromatic Hydrocarbons
27-5 Benzene
27-6 Other Aromatic Hydrocarbons
27-7 Hydrocarbons: A Summary
Functional Groups
27-8 Organic Halides
27-9 Alcohols and Phenols
27-10 Ethers
27-11 Aldehydes and Ketones
27-12 Amines
27-13 Carboxylic Acids
27-14 Some Derivatives of
Carboxylic Acids
27-15 Summary of Functional
Groups
Fundamental Classes of Organic
Reactions
27-16 Substitution Reactions
27-17 Addition Reactions
27-18 Elimination Reactions
27-19 Polymerization Reactions
OBJECTIVES
After you have studied this chapter, you should be able to
• Describe saturated hydrocarbons (alkanes and cycloalkanes)—their structures and their
nomenclature
• Describe unsaturated hydrocarbons (alkenes and alkynes)—their structures and their
nomenclature
• Draw and name constitutional isomers of alkanes
• Draw and name constitutional and geometric isomers of alkenes
• Describe and name some aromatic hydrocarbons—benzene, condensed aromatics, and
substituted aromatic compounds
• Describe some common functional groups and name compounds from the following
classes
halides
alcohols and phenols
ethers
aldehydes and ketones
amines
carboxylic acids and some of their derivatives
• Recognize examples of the three fundamental classes of organic reactions: substitution,
addition, and elimination
• Recall some uses of three fundamental classes of organic reactions and some compounds
that are prepared by each class of reaction
• Recognize some common polymers and the reactions by which they are formed; identify
the monomers from which they are formed
Waxes are high-molecular-weight
straight-chain alkanes, alcohols, and
carboxylic acids.
OO
rganic chemistry is the chemistry of compounds that contain CXC or CXH
bonds. Why is one entire branch of chemistry devoted to the behavior of the
compounds of just one element? The answer is twofold: (1) There are many more
compounds that contain carbon than there are compounds that do not (more than 11
million organic compounds have been identified), and (2) the molecules containing carbon
can be so much larger and more complex (a methane molecule contains five atoms per
molecule and DNA contains tens of billions of atoms per molecule).
Originally the term “organic” was used to describe compounds of plant or animal origin.
“Inorganic” was used to describe compounds from minerals. In 1828, Friedrich Wöhler
(1800–1882) synthesized urea by boiling ammonium cyanate with water.
This disproved the theory that held that organic compounds could be made only by living
things. Today many organic compounds are manufactured from inorganic materials.
We encounter organic chemistry in every aspect of our lives. All life is based on a
complex interrelationship of thousands of organic substances—from simple compounds
such as sugars, amino acids, and fats to vastly more complex ones such as the enzymes
that catalyze life’s chemical reactions and the huge DNA molecules that carry genetic
information from one generation to the next. The food we eat (including many additives);
the clothes we wear; the plastics and polymers that are everywhere; our life-saving medicines;
the paper on which we write; our fuels; many of our poisons, pesticides, dyes, soaps
and detergents—all involve organic chemistry.
We normally think of petroleum and natural gas as fuel sources, but most synthetic
organic materials are also derived from these two sources. More than half of the top 50
commercial chemicals are organic compounds derived in this way. Indeed, petroleum and
natural gas may one day be more valuable as raw materials for organic synthesis than as
fuel sources. If so, we may greatly regret our delay in developing alternative energy sources
while burning up vast amounts of our petroleum and natural gas deposits as fuels.
A carbon atom has four electrons in its outermost shell with ground state configuration
1s22s22p2. The C atom can attain a stable configuration by forming four covalent
bonds. As we saw in Chapter 8, each C atom can form single, double, or triple bonds by
utilizing various hybridizations. The bonding of carbon is summarized in Table 27-1 using
examples that we saw in Chapters 7 and 8. Carbon is unique among the elements in the
extent to which it forms bonds between like atoms and in the diversity of compounds that
are formed. Bonds between atoms of the same element rarely occur for any element except
carbon. Carbon atoms form long chains, branched chains, and rings that may also have
chains attached to them. A tremendous variety of carbon-containing compounds is known.
Although millions of organic compounds are known, the elements they contain are
very few: C and H; often N, O, S, P, or a halogen; and sometimes another element. The
great number and variety of organic compounds are a result of the many different arrangements
of atoms, or structures, that are possible. The chemical and physical properties of
organic compounds are related to the structures of their molecules. Thus, the basis for
organizing and understanding organic chemistry is our understanding of structure and
bonding.
In this text we shall give only an introduction to organic chemistry. In this chapter we
organize organic compounds into the most common classes or “families” according to
their structural features and learn to name various types of compounds. In this chapter
and the next, we will present a few typical reactions that organic substances undergo.
O
CH2NNH4OCN NH2
urea
(an organic compound)
H2O
boil
ammonium cyanate
(an inorganic compound)
Urea, H2NXCOXNH2, is the
principal end product of metabolism
of nitrogen-containing compounds in
mammals. It is eliminated in the urine.
An adult man excretes about 30 grams
of urea in 24 hours.
Organic Chemistry I: Formulas, Names, and Properties 1041
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 11.2, Carbon–Carbon Bonds.
Organic molecules are based on a framework of carbon–carbon and carbon–hydrogen
bonds. Many compounds contain only the two elements C and H; they are called hydrocarbons.
Hydrocarbons that contain a delocalized ring of pi bonds such as the benzene
ring (Section 9-6) are called aromatic hydrocarbons. Those that do not contain such
delocalized systems are called aliphatic hydrocarbons. Aliphatic hydrocarbons that
contain only sigma (␴) bonds (i.e., only single bonds) are called saturated hydrocarbons.
Those that contain both sigma and pi (␲) bonds (i.e., double, triple, or delocalized bonds)
are called unsaturated hydrocarbons. These classifications are diagrammed in Figure
27-1. The first seven sections of this chapter are devoted to the study of hydrocarbons.
A functional group is a special arrangement of atoms within an organic molecule that
is responsible for some characteristic chemical behavior of the compound. Different molecules
that contain the same functional groups have similar chemical behavior. We shall
follow the study of hydrocarbons with a presentation of some important characteristic
functional groups.
1042 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
TABLE 27-1 Hybridization of Carbon in Covalent Bond Formation
Hybridization and Orbitals Used by Bonds Formed by
Resulting Geometry Each C Atom Each C Atom Example
sp3, tetrahedral four sp3 hybrids four ␴ bonds ethane
sp2, trigonal planar three sp2 hybrids, three ␴ bonds,
ethylene
H H
HH
C C
H
H H
H H
C C H
one p orbital one ␲ bond
sp, linear two sp hybrids, two ␴ bonds,
acetylene H C C Htwo p orbitals two ␲ bonds
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 11.3, Hydrocarbons.
Hydrocarbons
Aliphatic
hydrocarbons
Aromatic
hydrocarbons
Alkanes Cycloalkanes Alkenes Alkynes
Saturated Unsaturated
Figure 27-1 Classification of hydrocarbons.
SATURATED HYDROCARBONS
ALKANES AND CYCLOALKANES
The saturated hydrocarbons, or alkanes, are compounds in which each carbon atom
is bonded to four other atoms. Cycloalkanes are saturated hydrocarbons that
contain carbon atoms joined to form a ring.
Alkanes
Each H atom in a saturated hydrocarbon is bonded to only one C atom. Saturated hydrocarbons
contain only single bonds. Petroleum and natural gas are composed mostly of
saturated hydrocarbons.
In Section 8-7 we examined the structure of the simplest alkane, methane, CH4. We
saw that methane molecules are tetrahedral with sp3 hybridization at carbon (Figure
27-2).
Ethane, C2H6, is the next simplest saturated hydrocarbon. Its structure is quite similar
to that of methane. Two carbon atoms share a pair of electrons. Each carbon atom also
shares an electron pair with each of three hydrogen atoms. Both carbon atoms are sp3
hybridized (Figure 27-3). Propane, C3H8, is the next member of the family (Figure 27-4).
27-1
The term saturated comes from early
studies in which chemists tried to
add hydrogen to various organic
substances. Those to which no more
hydrogen could be added were called
saturated, by analogy with saturated
solutions.
27-1 Alkanes and Cycloalkanes 1043
(a)
CH4 or H C
H
H
H
H
H
H
C
H
1s
sp3
(b)
(c)
(d)
C
H
HH
H
(e)
Figure 27-2 Representations of a molecule of methane, CH4. (a) The condensed and
Lewis formulas for methane. (b) The overlap of the four sp3 carbon orbitals with the s
orbitals of four hydrogen atoms forms a tetrahedral molecule. (c) A ball-and-stick model,
(d) a space-filling model of methane, and (e) a three-dimensional representation that uses
the wedged line to indicate a bond coming forward and a dashed line to represent a bond
projecting backward.
Two different compounds have the formula C4H10 but different structures and, hence,
different properties. Such isomers result when two molecules contain the same atoms
bonded together in different orders. The structures of these two isomeric C4H10 alkanes
are shown in Figure 27-5. These two structures correspond to the two ways in which a
hydrogen atom can be removed from a propane molecule and replaced by a XCH3 group.
If a XCH3 replaces an H on either of the end carbon atoms, the result is butane. It is
considered to be a straight-chain hydrocarbon because there is no branching. Though
somewhat misleading, the term “straight chain” is widely used. The carbon chains are
linear only in the structural formulas that we write. They are actually zigzag due to the
tetrahedral bond angles at each carbon and are sometimes further kinked or twisted. Think
of such a chain of carbon atoms as continuous. We can trace a single path from one terminal
carbon to the other and pass through every other C atom without backtracking.
If the XCH3 group replaces an H from the central carbon atom of propane, the result
is the branched-chain hydrocarbon 2-methylpropane, or isobutane. This is the simplest
branched-chain hydrocarbon.
The formulas of the alkanes can be written in general terms as CnH2nϩ2, where n is
the number of carbon atoms per molecule. The first five members of the series are
Iso ϭ “same”; mer ϭ “part.” As we
saw in Sections 25-6 and 25-7, isomers
are substances that have the same
numbers and kinds of atoms arranged
differently. Isomerism in organic
compounds is discussed more
systematically in Chapter 28.
1044 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
CH3CH3 or H C
H
H
H
H
C H
(a) (b)
(c)
C
H
H
H
H
C
H
H
(d)
Figure 27-3 Models of ethane, C2H6. (a) The condensed and Lewis formulas for ethane.
(b) A ball-and-stick model, (c) a space-filling model of ethane, and (d) a 3-D representation.
Figure 27-4 Ball-and-stick and
space-filling models and a threedimensional
representation of
propane, C3H8.
C C C HH
H
H H
H
H
H
CH4 C2H6 C3H8 C4H10 C5H12
Number of C atoms ϭ n ϭ 1 2 3 4 5
Number of H atoms ϭ 2n ϩ 2 ϭ 4 6 8 10 12
The formula of each alkane differs from the next by CH2, a methylene group.
A series of compounds in which each member differs from the next by a specific number
and kind of atoms is called a homologous series. The properties of members of such a
series are closely related. The boiling points of the lighter members of the straight-chain
saturated hydrocarbon series are shown in Figure 27-6. As the molecular weights of the
straight-chain hydrocarbons increase, their boiling points also increase regularly. Properties
such as boiling point depend on the forces between molecules (Chapter 13). Carbon–
carbon and carbon–hydrogen bonds are essentially nonpolar and are arranged tetrahedrally
around each C atom. As a result, saturated hydrocarbons are nonpolar molecules,
and the only significant intermolecular forces are dispersion forces (Section 13-2). These
forces, which are due to induced dipoles, become stronger as the sizes of the molecules
and the number of electrons in each molecule increase. Thus, trends such as those depicted
in Figure 27-6 are due to the increase in effectiveness of dispersion forces.
Some systematic method for naming compounds is necessary. The system in use today
is prescribed by the International Union of Pure and Applied Chemistry (IUPAC). The
names of the first 20 straight-chain alkanes are listed in Table 27-2. You should become
familiar with at least the first ten. The names of the alkanes starting with pentane have
prefixes (from Greek) that give the number of carbon atoms in the molecules. All alkane
names have the -ane ending.
Constitutional isomers (structured isomers) have the same molecular formula but
differ in the order in which their atoms are attached.
We have seen that there are two saturated C4H10 hydrocarbons. For the C5 hydrocarbons,
there are three possible arrangements of the atoms. These three different C5H12
alkanes are examples of constitutional isomers.
Figure 27-5 Ball-and-stick models
of the two isomeric C4H10
hydrocarbons, butane,
CH3CH2CH2CH3, and
methylpropane, CH3CHCH3.
A
CH3
27-1 Alkanes and Cycloalkanes 1045
Ϫ100
Ϫ200
0
200
300
400
100
2 4 6 8 10 12 14 16 18 20
Gases
Liquids
Solids
Number of carbon atoms in alkane
Boilingpoint,ЊC
Figure 27-6 A plot of normal
boiling point versus the number of
carbon atoms in straight-chain,
saturated hydrocarbons.
The number of constitutional isomers increases rapidly as the number of carbon atoms
in saturated hydrocarbons increases. There are five isomeric C6H14 alkanes (Table 27-3).
Table 27-4 displays the number of isomers of some saturated hydrocarbons (alkanes). Most
of the isomers have not been prepared or isolated. They probably never will be.
As the degree of branching increases for a series of molecules of the same molecular
weight, the molecules become more compact. A compact molecule can have fewer points
of contact with its neighbors than more extended molecules do. As a result, the total
induced dipole forces (dispersion forces) are weaker for branched molecules, and the
boiling points of such compounds are lower. This pattern can be seen in the data in Table
27-3 and in Example 27-1.
EXAMPLE 27-1 Isomeric Alkanes
Write the Lewis formula of each of the three isomers having the molecular formula of C5H12.
The boiling points of the three isomers are 9.5°C, 27.9°C, and 36.1°C. Match each compound
with its boiling point.
Plan
We first write the carbon skeleton of the straight-chain isomer (no branching). Then we remove
one carbon from the straight-chain structure and reposition it to form the skeleton for a
branched isomer. We check to see whether branching is possible at other carbon atoms. Then
1046 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
TABLE 27-2 Some Straight-Chain Hydrocarbons (Alkanes)
Molecular Normal Normal State at Room
Formula IUPAC Name bp (°C) mp (°C) Temperature
CH4 methane Ϫ161 Ϫ184
C2H6 ethane Ϫ88 Ϫ183
gas
C3H8 propane Ϫ42 Ϫ188
C4H10 butane ϩ0.6 Ϫ138
C5H12 pentane 36 Ϫ130
C6H14 hexane 69 Ϫ94
C7H16 heptane 98 Ϫ91
C8H18 octane 126 Ϫ57
C9H20 nonane 150 Ϫ54
C10H22 decane 174 Ϫ30
liquid
C11H24 undecane 194.5 Ϫ25.6
C12H26 dodecane 214.5 Ϫ9.6
C13H28 tridecane 234 Ϫ6.2
C14H30 tetradecane 252.5 ϩ5.5
C15H32 pentadecane 270.5 10
C16H34 hexadecane 287.5 18
C17H36 heptadecane 303 22.5
C18H38 octadecane 317 28
solid
C19H40 nonadecane 330 32
C20H42 eicosane 205.(at 15 torr) 36.7
we remove a second carbon from the unbranched isomer and, if possible, form additional
branching, and so on. Finally, we add hydrogen atoms to each skeleton to complete the
structure.
Solution
The three compounds are:
The compound with the most branching would be expected to be the one with the lowest
boiling point, and the straight-chain isomer would be expected to have the highest boiling
point.
CH3C(CH3)3, bp ϭ 9.5°C; CH3CH2CH(CH3)2, bp ϭ 27.9°C;
and CH3CH2CH2CH2CH3, bp ϭ 36.1°C.
(Models of these compounds are shown in the margin.)
You should now work Exercise 15.
H
H H
H H
C C
H H
H
C C H Hsame as
H H
H
C C
H H
H H
C C H
H
H HC
H
H HC
H
H H
H H
C C
H H
H H
C C
H
H
C H, H
H
H
C
H
H
C
H
H C H
H H
H
C H, and
C
CH3CH2CH2CH2CH3, CH3CH2CH(CH3)2,
and CH3C(CH3)3.
TABLE 27-3 Isomeric C6H14 Alkanes
Normal
IUPAC Name Formula bp (°C)
hexane CH3CH2CH2CH2CH2CH3 68.7
2-methylpentane 60.3
3-methylpentane 63.3
2,2-dimethylbutane 49.7
2,3-dimethylbutane 58.0
CH3 CH3
CH3CH CHCH3
CH3
CH3
CH3CH2CCH3
CH3
CH3CH2CHCH2CH3
CH3
CH3CH2CH2CHCH3
TABLE 27-4 Numbers of
Possible
Constitutional
Isomers of
Alkanes
Formula Isomers
C7H16 9
C8H18 18
C9H20 35
C10H22 75
C11H24 159
C12H26 355
C13H28 802
C14H30 1,858
C15H32 4,347
C20H42 366,319
C25H52 36,797,588
C30H62 4,111,846,763
Cycloalkanes
The cyclic saturated hydrocarbons, or cycloalkanes, have the general formula CnH2n. The
cycloalkanes (and other ring compounds that we will encounter later) are often shown in
simplified skeletal form in which each intersection of two lines represents a C atom; we
mentally add enough H atoms to give each carbon atom four bonds. The first four unsubstituted
cycloalkanes and their simplified representations are
In some of these structures, the bond angles are somewhat distorted from the ideal
tetrahedral angle of 109.5°, the most severe distortions being 60° in cyclopropane and 90°
in cyclobutane. As a result, these rings are said to be “strained,” and these two compounds
are unusually reactive for saturated hydrocarbons. Cyclopentane is stable with a nearly
flat ring, because the bond angles in a regular pentagon (108°) are near the tetrahedral
angle (109.5°).
Cyclohexane is not stable as a flat ring. The bond angles of the flat ring would be 120°,
not the 109.5°, of an sp3-hybridized carbon. To avoid the strain, the ring “puckers” and
becomes nonplanar (Figures 27-7 and 27-8).
CH2
CH2
CH2
CH2
H2C
H2C
CH2
CH2H2C
H2C CH2H2CCH2
CH2
H2C
CH2
CH2
H2C
cyclohexanecyclopropane cyclobutane cyclopentane
Noncyclic compounds are sometimes
referred to as acyclic compounds.
1048 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
NAMING SATURATED HYDROCARBONS
It is important to realize that many compounds (and their names) were so familiar to
chemists before the development of the IUPAC system (beginning about 1890) that they
continued to be called by their common, or “trivial,” names. In this and the next chapter,
IUPAC names of compounds appear in blue type, and their common alternative names
are shown in black type.
27-2
Figure 27-7 Stereoview of a nonplanar form of cyclohexane.
Figure 27-8 A model showing a
nonplanar arrangement of cyclohexane.
In skeletal representations, each
intersection represents a carbon with
any attached hydrogens. (See Section
27-3 for more details on writing
skeletal representations.)
The IUPAC naming system is based on the names of the unbranched hydrocarbons
given in Table 27-2 and their higher homologues. To name a branched-chain hydrocarbon,
we first find the longest chain of carbon atoms and use the root name that corresponds
to that hydrocarbon. We then indicate the positions and kinds of substituents attached to
the chain. Alkyl group substituents attached to the longest chain are thought of as fragments
of hydrocarbon molecules obtained by the removal of one hydrogen atom. We give
them names related to the parent hydrocarbons from which they are derived. Other alkyl
groups are named similarly (Table 27-5). We use the general symbol R to represent any
alkyl group. The cycloalkyl groups derived from the first four cycloalkanes are called cyclopropyl,
cyclobutyl, cyclopentyl, and cyclohexyl, respectively.
cyclopropyl cyclobutyl cyclopentyl cyclohexyl
27-2 Naming Saturated Hydrocarbons 1049
TABLE 27-5 Some Alkanes and the Related Alkyl Groups
Parent Hydrocarbon Alkyl group, —R
CH4, methane XCH3, methyl
C2H6, ethane XC2H5, ethyl
C3H8, propane XC3H7, propyl XC3H7, isopropyl
C4H10, butane XC4H9, butyl XC4H9, sec-butyl
(read as “secondary butyl”)
C4H10, methylpropane XC4H9, t-butyl XC4H9, isobutyl
(common name, isobutane) (read as “tertiary butyl”)
CH2
CH3
CH CH3
CH3
CH3
C C(CH3)3CH3 orH
CH3
CH3
C CH(CH3)3CH3 or
H H
H H
C C
H
CH
H
H
C H
H H
H H
C C
H
H
C
H
H
C H CH2(CH2)2CH3or
H H
H H
C C
H
H
CH
H
H
C H CH3(CH2)2CH3or
H H
H
C CH
H
H
C H
H H
H H
C C
H
H
C H CH2CH2CH3or
H H
H H
C CH
H
H
C H CH3CH2CH3or
H H
H H
C C H CH2CH3orH
H H
H H
C C H CH3CH3or
H
H
C CH3H orH
H
H
C CH4H or
Summary of IUPAC Rules for Naming Alkanes
1. Find the longest chain of C atoms. Choose the base name that describes the
number of C atoms in this chain, with the ending -ane (see Table 27-2). The
longest chain may not be obvious at first if branches of different sizes are present.
2. Number the C atoms in this longest chain beginning at the end nearest the first
branching. If necessary, go to the second branch closest to an end, and so on,
until a difference is located. If there is branching at equal distances from both
ends of the longest chain, begin numbering at the end nearest the branch that
is first in alphabetical order.
3. Assign the name and position number to each substituent. Arrange the
substituents in alphabetical order. Hyphenated prefixes, such as tert- and sec-, are
not used in alphabetization of the substituents.
4. Use the appropriate prefix to group like substituents: di- ϭ 2, tri- ϭ 3, tetra- ϭ
4, penta- ϭ 5, and so on. Don’t consider these prefixes when alphabetizing
attached groups.
5. Write the name as a single word. Use hyphens to separate numbers and letters
(plus some hyphenated prefixes) and commas to separate numbers. Don’t leave
any spaces.
Let us name the following compound.
(CH3)2CHCH2CH2CH2CH3 or (CH3)2CH(CH2)3CH3
We follow Rules 1 and 2 to number the carbon atoms in the longest chain.
The methyl group is attached to the second carbon atom in a six-carbon chain, so the
compound is named 2-methylhexane.
The following examples further illustrate the rules of nomenclature.
EXAMPLE 27-2 Naming Alkanes
Give the IUPAC name of the compound represented by the structural formula
CCH3 (CH3)3C(CH2)2CH3CH2 CH2 CH3
CH3
CH3
or
CHCH3 CH2CH2 CH2 CH3
CH3
1 2 3 4 5 6
H
HH
C
H
H
CH C
H H
H H
C C
H
H
C H or
H
H C H
Formerly the rule was to name the
substituents in order of increasing
complexity. This was sometimes
difficult to determine. With the use
of computers in literature searches, it
became necessary to adopt the more
definitive alphabetization of the names
of substituents.
1050 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
Parentheses are used to conserve space.
Formulas written with parentheses
must indicate unambiguously the
structure of the compound. The
parentheses here indicate that the
CH3 group is attached to the C next
to it.
Remember that the (dash) formulas
indicate which atoms are bonded
to each other. They do not show
molecular geometry. In this molecule,
each C atom is tetrahedrally bonded to
four other atoms.
It is incorrect to name the compound
5-methylhexane because that violates
Rule 2.
Plan
We first find the longest carbon chain and number it to give the substituent encountered first
the smallest possible number. Then we name the substituents as in Table 27-5 and specify the
number of each as indicated in Rule 4.
Solution
The longest chain contains five carbons, so this compound is named as a derivative of pentane.
There are two methyl groups, both at carbon number 2. The IUPAC name of this compound
is 2,2-dimethylpentane.
EXAMPLE 27-3 Naming Alkanes
Give the IUPAC name of the compound represented by the structural formula
Plan
The approach is the same as in Example 27-2. We should be aware that the longest carbon
chain might not be written in a straight line.
Solution
The longest chain contains six carbons, so this compound is named as a derivative of hexane.
There are three methyl substituents: two at carbon number 2 and one at carbon number 4.
The IUPAC name of this compound is 2,2,4-trimethylhexane.
You should now work Exercises 13 and 14.
EXAMPLE 27-4 Writing Formulas from Names
Write the structure for 4-tert-butyl-2,5-dimethylheptane.
Plan
The root name “heptane” indicates that there is a seven-carbon chain.
CXCXCXCXCXCXC
The names and numbers of the substituents tell us where to attach the alkyl groups.
CCH3 CH2 CH CH3
CH3
CH3 CH2
CH3
CCH3 CH2 CH CH2 CH3
CH3
CH3 CH3
or
(rewritten to emphasize the six-C chain)
1 2 3 4
5
6
1 2 3 4 5 6
CCH3 CH2 CH CH3
CH3
CH3 CH2
CH3
CCH3 CH2 CH2 CH3
CH3
CH3
1 2 3 4 5
27-2 Naming Saturated Hydrocarbons 1051
Solution
Then we fill in enough hydrogens to saturate each C atom and arrive at the structure
EXAMPLE 27-5 Writing Formulas from Names
Write the structure for the compound 2-cyclopropyl-3-ethylpentane.
Plan
The root name “pentane” tells us that the structure is based on a five-carbon chain. We place
the substituents at the positions indicated by the numbers in the name.
Solution
You should now work Exercise 16.
The names of substituted cycloalkanes are derived analogously to those of alkanes.
(1) The base name is determined by the number of carbon atoms in the ring using the
same base name as the alkane with the addition of cyclo in front. (2) If only one substituent
is attached to the ring, no “location number” is required because all positions in a
cycloalkane are equivalent. (3) Two or more functional groups on the ring are identified
by location numbers, which should be assigned sequentially to the ring carbons in the
order that gives the smallest sum of location numbers.
CH3
CH2CH3
CH3
CH3
CH3
1
2
34
5
1,3-dimethylcyclopentane
(not 1,4-dimethylcyclopentane)
1-ethyl-1-methylcyclopropanemethylcyclobutane
CHCH3 CH CH2 CH3
CH2 CH2
CH
CH2
CH3
1 2 3 4 5
where the symbol represents
the cyclopropyl group,
CCH3 CH3
CH2
CH3
CH CH3CH
CH3 CH3
CH2CH3 CH
CCH3 CH3
CH3
CH3
C C CC C C
CH3 CH3
1 2 3 4 5 6 7
1,3-dimethylcyclopentane can exist as
two different isomers; one with both
methyl groups on the same side of the
ring (cis-), another with the methyl
groups on opposite sides of the plane
of the ring (trans-). (See Section 27-3.)
1052 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
Recall that the cycloalkanes may also
be represented by the general formula
CnH2n.
UNSATURATED HYDROCARBONS
There are three classes of unsaturated hydrocarbons: (1) the alkenes and their cyclic counterparts,
the cycloalkenes; (2) the alkynes and the cycloalkynes; and (3) the aromatic
hydrocarbons.
ALKENES
The simplest alkenes contain one carbon–carbon double bond, CUC, per molecule. The
general formula for noncyclic alkenes is CnH2n. The simplest alkene is C2H4, ethene,
which is usually called by its common name, ethylene.
The bonding in ethylene was described in Section 8-13. The hybridization (sp2) and
bonding at other double-bonded carbon atoms are similar. Both carbon atoms in C2H4
are located at the centers of trigonal planes. Rotation about CUC double bonds does not
occur significantly at room temperature. Compounds that have the general formula
(XY)CUC(XY) therefore exist as a pair of cis–trans isomers. Figure 27-9 shows the cis–
trans isomers of 1,2-dichloroethene. The existence of compounds with different arrangements
of groups on the opposite sides of a bond with restricted rotation is called geometric
isomerism. This cis–trans isomerism can occur across double bonds in alkenes and across
single bonds in rings.
Two shared electron pairs draw the atoms closer together than a single electron pair
does. Thus, carbon–carbon double bonds are shorter than CXC single bonds, 1.34 Å
versus 1.54 Å. The physical properties of the alkenes are similar to those of the alkanes,
but their chemical properties are quite different.
The root for the name of each alkene is derived from the alkane having the same
number of C atoms as the longest chain containing the double bond. In the trivial
(common) system of nomenclature, the suffix -ylene is added to the characteristic root. In
systematic (IUPAC) nomenclature, the suffix -ene is added to the characteristic root.
27-3
Four CXH ␴ bonds, one CXC ␴
bond (green, hatched), and one CXC
␲ bond (tan, hatched) in the planar
C2H4 molecule.
27-3 Alkenes 1053
Figure 27-9 Two isomers of 1,2-dichloroethene are possible because rotation about the
double bond is restricted. This is an example of geometric isomerism. A ball-and-stick model
and a space-filling model are shown for each isomer. (a) The cis isomer. (b) The trans
isomer.
(b)
CC
ClH
HCl
trans-1,2-dichloroethene
mp –50°C, bp 47.5°C
(a)
CC
ClCl
HH
cis-1,2-dichloroethene
mp –80.5°C, bp 60.3°C
one bond
H
H H
H
CC σ
σ
σ σ
σ
π
Our Daily LivesCCHEMISTRY IN USE
Petroleum
Petroleum, or crude oil, was discovered in the United States
(Pennsylvania) in 1859 and in the Middle East (Iran) in 1908.
It has been found in many other locations since these initial
discoveries and is now pumped from the ground in many
parts of the world. Petroleum consists mainly of hydrocarbons.
Small amounts of organic compounds containing
nitrogen, sulfur, and oxygen are also present. Each oil field
produces petroleum with a particular set of characteristics.
Distillation of petroleum produces several fractions.
Gasoline is much in demand, so higher hydrocarbons (C12
and up) are “cracked” to increase the amount of gasoline that
can be made from a barrel of petroleum. The hydrocarbons
are heated, in the absence of air and in the presence of a catalyst,
to produce a mixture of smaller alkanes that can be used
in gasoline. This process is called catalytic cracking.
The octane number (rating) of a gasoline indicates how
smoothly it burns and how much engine “knock” it produces.
(Engine knock is caused by premature detonation of fuel in
the combustion chamber.) 2,2,4-trimethylpentane, isooctane,
has excellent combustion properties and was arbitrarily assigned
an octane number of 100. Heptane, CH3(CH2)5CH3,
has very poor combustion properties and was assigned an octane
number of zero.
isooctane (octane number ϭ 100)
CCH3 CH2 CH CH3
CH3
CH3
CH3
Mixtures of these two were prepared and burned in test
engines to establish the octane scale. The octane number of
such a mixture is the percentage of isooctane in it. Gasolines
burned in standard test engines are assigned octane numbers
based on the compression ratio at which they begin to knock.
A 90-octane fuel produces the same amount of knock as the
90% isooctane/10% heptane mixture. Branched-chain compounds
produce less knock than straight-chain compounds.
The octane numbers of two isomeric hexanes are
CH3(CH2)4CH3
hexane (octane number ϭ 25)
(CH3)3CCH2CH3
2,2-dimethylbutane (octane number ϭ 92)
Petroleum Fractions
Fraction* Principal Composition Distillation Range (°C)
natural gas C1–C4 below 20
bottled gas C5–C6 20–60
gasoline C4–C12 40–200
kerosene C10–C16 175–275
fuel oil, diesel oil C15–C20 250–400
lubricating oils C18–C22 above 300
paraffin C23–C29 mp 50–60
asphalt — viscous liquid
(“bottoms fraction”)
coke — solid
*Other descriptions and distillation ranges have been used, but all are similar.
CH3CH2CH2CH2CH2CH2CH3
heptane (octane number ϭ 0)
Petroleum refinery towers.
Summary of IUPAC Rules for Naming Alkenes and Cycloalkenes
1. Locate the C atoms in the longest C chain that contains the double bond. Use the
base name prefix with the ending -ene.
2. Number the C atoms of this chain sequentially beginning at the end nearer the
double bond. Insert the number describing the position of the double bond (indicated
by its first carbon location) before the base name. (This is necessary only
for chains of four or more C atoms, because only one position is possible for a
double bond in a chain of two or three carbon atoms.)
3. In naming alkenes, the double bond takes positional precedence over substituents
on the carbon chain. The double bond is assigned the lowest possible number.
4. To name compounds with possible geometric isomers, consider the two largest
groups within the carbon chain that contains the double bond—these are indicated
as part of the base name. The isomer in which the largest groups at each
end of the CUC are located on opposite sides is called trans. If the largest groups
are on the same side, the molecule is referred to as cis. Insert the prefix cis- or
trans- just before the number of the double bond to indicate whether the largest
groups are on the same or opposite sides, respectively, of the double bond.
5. For cycloalkenes, the double bond is assumed to be between C atoms 1 and 2,
so no position number is needed to describe it.
Some illustrations of this naming system follow.
The following two names illustrate the application of Rule 3.
4-Methyl-2-pentene can exist as either of two isomers. Their names illustrate the application
of Rule 4.
CC
CH3 CH(CH3)2
HH
4-methyl-cis-2-pentene
CC
CH3
CH(CH3)2
H
H
4-methyl-trans-2-pentene
1
2 3
4 5 1
2 3
4 5
CH
CH3
CH CH3CHCH2
CH3
CH2 CH3CCH3
4-methyl-2-pentene2-methyl-1-butene
1234 4321 5
CH2 CH2 CH3 CH2CH CH3CH2 CH2CH
ethene
(ethylene)
systematic:
trivial:
propene
(propylene)
1-butene
(butylene)
CH3CH CHCH3 CH3
CH3
CH2C
2-butene
methylpropene
(isobutylene)
1234
1 2 3 4
The prefix trans- means “across” or
“on the other side of.” As a reminder
of this terminology, think of words
such as “transatlantic.”
27-3 Alkenes 1055
There are two isomers of 2-butene,
cis-2-butene and trans-2-butene.
Some alkenes, called polyenes, have two or more carbon–carbon double bonds per
molecule. The suffixes -adiene, -atriene, and so on are used to indicate the number of
(CUC) double bonds in a molecule.
1 2 3 4 4 3 2 1
CH2UCHXCHUCH2 CH3XCHUCUCH2
1,3-butadiene 1,2-butadiene
1,3-Butadiene and similar molecules that contain alternating single and double bonds are
described as having conjugated double bonds. Such compounds are of special interest
because of their polymerization reactions (Section 27-19).
1056 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
EXAMPLE 27-6 Naming Alkenes
Give the IUPAC names of the following two alkenes.
Plan
For each compound, we first find the longest chain that includes the double bond, and then
number it beginning at the end nearer the double bond (Rules 1 and 2). Then we specify the
identities and positions of substituents in the same way we did for alkanes. In (a) we specify
the geometric isomer by locating the two largest groups attached to the double-bonded carbons
and then describing their relationship using the cis–trans terminology.
Solution
(a)
The longest such chain contains five atoms and has a double bond beginning at atom 2; thus,
the compound is named as a derivative of 2-pentene. Now we must apply Rule 4. The two
CC
CH3 CH2CH3
CH3H
1
2 3
4 5
CC
CH3 CH2CH3
CH3H
(a)
CC
CH3CH2 CH2CH3
CH2CH3H
(b)
Problem-Solving Tip: When Is the Location of a Substituent or
a Multiple Bond Not Included in an
IUPAC Name?
If only one location is possible for a substituent or double bond, the number that indicates
the location can be left out of the compound’s name. If there is doubt, put the
number in the name. For example, “2-methyl-1-propene” is named simply methylpropene.
There are no other locations possible for the substituent or the double bond.
The name without the numbers is the correct name. Both numbers are necessary to name
3-methyl-1-butene, however.
CHCH
CH3
CH2CCH3
CH3
CH2CH3
methylpropene 3-methyl-1-butene
largest groups in the chain are the terminal XCH3 (carbon 1) and the XCH2CH3 (carbons 4
and 5); these are on the same side of the double bond, so we name the compound as a derivative
of cis-2-pentene. The only substituent is the methyl group at carbon 3. The full name of
the compound is 3-methyl-cis-2-pentene.
(b) There are two choices for the longest chain, and either one would have an ethyl substituent.
We could number either chain from the other end and still have the double bond starting at
carbon 3; we number from the end that gives the carbon bearing the ethyl group the lowest
possible position number, 3. Carbon 3 has two equivalent substituents, so geometric isomerism
is not possible, and we do not use the cis–trans terminology. The name is 3-ethyl-3-hexene.
You should now work Exercise 20.
EXAMPLE 27-7 Naming Alkenes
Name the following alkene.
Plan
As we did in Example 27-6(b), we first find the longest chain that includes the double bond,
and then number it beginning at the end nearer the double bond (Rules 1 and 2). Then we
specify the identities and positions of substituents in the same way we did for alkanes.
Solution
There are two choices for the longest chain, and either one would have an ethyl substituent.
Carbon 3 has two equivalent substituents, so geometric isomerism is not possible, and we do
not use the cis-trans terminology. The name is 3-ethyl-2-pentene.
You should now work Exercise 22.
The cycloalkenes are represented by the general formula CnH2n–2. Two cycloalkenes and
their skeletal representations are
CH2
CH2
CH2
CH
H2C
HC
oror
CH2
CH2
CH
H2C
HC
cyclohexenecyclopentene
CC
CH3 CH2CH3
CH2CH3H
1234
5
CC
CH3 CH2CH3
CH2CH3H
54
32
1
or
CC
CH3 CH2CH3
CH2CH3H
CC
CH3CH2 CH2CH3
CH2CH3H
1234
56
CC
CH3CH2 CH2CH3
CH2CH3H
12
34
56
or
One would not number the carbons as:
Rule 1 requires that we use the longest
chain that includes both carbons of the
double bond.
CH3 CH2CH3
CH2CH3H
C C 4 5
3
2 1
27-3 Alkenes 1057
EXAMPLE 27-8 Cycloalkenes
Write the skeletal representation of 3-methylcyclohexene.
Plan
We draw the ring of the specified size with one double bond in the ring. We number the ring
so the double bond is between atoms 1 and 2. Then we add the designated substituents at the
indicated positions.
Solution
We number the six-membered ring so that the double bond is between atoms 1 and 2.
A methyl group is attached at carbon 3; the correct structure is
Remember that each intersection of two lines represents a carbon atom; there are enough H
atoms at each C atom to make a total of four bonds to carbon.
You should now work Exercise 21.
ALKYNES
The alkynes, or acetylenic hydrocarbons, contain one or more carbon–carbon triple
bonds, XCmCX. The noncyclic alkynes with one triple bond per molecule have the
general formula CnH2nϪ2. The bonding in all alkynes is similar to that in acetylene (Section
27-4
1
3
6 4
5
2
CH3
1 3
6 4
5
2
1058 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
Problem-Solving Tip: How to Draw Skeletal Representations
The rules for drawing skeletal representations follow.
1. A carbon atom is assumed to be at each intersection of two lines and at the end of
each line. All carbon–carbon bonds are shown as lines. A carbon atom might be shown
for clarity.
2. Because carbon has four bonds, we mentally supply hydrogens to give each carbon
its four bonds.
3. Atoms other than hydrogen and carbon are shown. The following skeletal representation
is that of 3-chloro-4-ethyl-cis-3-heptene.
and its formula is C9H17Cl.
Cl
8-14). Triply bonded carbon atoms are sp-hybridized. The triply bonded atoms and their
adjacent atoms lie on a straight line (Figure 27-10).
Alkynes are named like the alkenes except that the suffix -yne is added to the characteristic
root. The first member of the series is commonly called acetylene. Its molecular
formula is C2H2. It is thermodynamically unstable, decomposing explosively to C(s) and
H2(g) at high pressures. It may be converted into ethene (also known as ethylene) and
then to ethane by the addition of hydrogen. The condensed formulas for several small
alkynes are
Figure 27-10 Models of acetylene,
HXCmCXH.
27-4 Alkynes 1059
CCH3CH CH CH
ethyne
(acetylene)
propyne 1-butyne
CCH3 CH2 CH
CCH3 CH3C CCH3
CH3
CH CH
3-methyl-1-butyne
2-butyne
The triple bond takes positional precedence over substituents on the carbon chain. It is
assigned the lowest possible number in naming.
Summary of IUPAC Rules for Naming Alkynes
Alkynes are named like the alkenes except for the following two points.
1. The suffix -yne is added to the characteristic root.
2. Because the linear arrangement about the triple bond does not lead to geometric
isomerism, the prefixes cis- and trans- are not used.
EXAMPLE 27-9 Alkynes
Write the structural formula of 5,5-dimethyl-2-heptyne.
Plan
The structure is based on a seven-carbon chain with a triple bond beginning at carbon 2. We
add methyl groups at the positions indicated.
Solution
1 2 3 4 5 6 7
CXCmCXCXCXCXC
Two methyl groups are attached to carbon 5, and sufficient hydrogens are added to complete
the bonding at each C atom.
The carbon chain is numbered to give the triple bond the lowest possible numbers even though
that results in higher position numbers for the methyl substituents.
You should now work Exercise 28.
CH3C
CH3
CH3
CCH2CCH2CH3
1 75 6432
EXAMPLE 27-10 Naming Alkynes
Give the IUPAC name of the compound
Plan
We find the longest chain and number it to give the triple bond the lowest possible number.
Then we specify the substituent(s) by name and position number.
Solution
The longest continuous carbon chain that includes the triple bond has six C atoms. There are
two ways in which we could choose and number such a chain, and in both the triple bond
would be between C atoms 3 and 4.
Because we also want the methyl substituent to have the lowest possible number, we choose
the first possibility. The name of the compound is 2-methyl-3-hexyne.
You should now work Exercise 30.
AROMATIC HYDROCARBONS
Originally the word “aromatic” was applied to pleasant-smelling substances. The word
now describes benzene, its derivatives, and certain other compounds that exhibit similar
chemical properties. Some have very foul odors because of substituents on the benzene
ring. On the other hand, many fragrant compounds do not contain benzene rings.
Steel production at one time required large amounts of coke. This was prepared by
heating bituminous coal to high temperatures in the absence of air. This process also favors
production of coal gas and coal tar. Because of the enormous amount of coal converted to
coke, coal tar was produced in large quantities. It served as a source of aromatic compounds.
For each ton of coal converted to coke, about 5 kg of aromatic compounds were obtained.
Today petroleum refining is the major source of aromatic hydrocarbons.
Early research on the reactions of the aromatic hydrocarbons led to methods for
preparing a great variety of dyes, drugs, flavors, perfumes, and explosives. More recently,
large numbers of polymeric materials, such as plastics and fabrics, have been prepared
from these compounds.
BENZENE
Benzene is the simplest aromatic hydrocarbon. By studying its reactions, we can learn a
great deal about aromatic hydrocarbons. Benzene was discovered in 1825 by Michael
Faraday when he fractionally distilled a by-product oil obtained in the manufacture of
illuminating gas from whale oil.
27-5
CH3CHC CCH2CH3
CH3
CH3CHC CCH2CH3
CH3
1 5 6432 6 2 1345
CH3CHC CCH2CH3
CH3
The main components of coal gas
are hydrogen (Ϸ 50%) and methane
(Ϸ 30%).
Distillation of coal tar produces a
variety of aromatic compounds.
1060 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
Elemental analysis and determination of its molecular weight showed that the molecular
formula for benzene is C6H6. The formula suggests that it is highly unsaturated. But
its properties are quite different from those of alkenes and alkynes.
The facts that only one monosubstitution product is obtained in many reactions and
that no addition products can be prepared show conclusively that benzene has a symmetrical
ring structure. Stated differently, every H atom is equivalent to every other H atom,
and this is possible only in a symmetrical ring structure (a).
The debate over the structure and bonding in benzene raged for at least 30 years. In
1865, Friedrich Kekulé (1829–1896) suggested that the structure of benzene was intermediate
between two structures [part (b) of the preceding structures] that we now call
resonance structures. We often represent benzene as
All 12 atoms in a benzene molecule lie in a plane. This suggests sp2 hybridization of
each carbon. The six sp2-hybridized C atoms lie in a plane, and the unhybridized p orbitals
extend above and below the plane. Side-by-side overlap of the p orbitals forms pi orbitals
(Figure 9-11). The electrons associated with the pi bonds are delocalized over the entire
benzene ring (Figure 27-11a,b).
H
H
H
H
H
H
or, more simply The structure of benzene is described
in detail in Section 9-6 in terms of
MO theory.
27-5 Benzene 1061
C
C
C
C
C
C
H
H
H
H
H
H
(skeleton only)
(a)
C
C
C
C
C
C
H
H
H
H
H
H
C
C
C
C
C
C
H
H
H
H
H
H
(b)
A computer-generated model of a
molecule of benzene, C6H6. A balland-stick
model is shown inside a
representation of the molecular
surface.
Figure 27-11 (a) A model of the benzene molecule, C6H6, and (b) its electron distribution.
(c) A model of toluene, C6H5CH3. This is a derivative of benzene in which one H atom has
been replaced by an alkyl group.
(a) (c)
CH3
(b)
OTHER AROMATIC HYDROCARBONS
Benzene molecules bearing alkyl substituents are called alkylbenzenes. The simplest of
these is methylbenzene (common name, toluene), shown in Figure 27-11c. The dimethylbenzenes
are called xylenes. Three different compounds (Table 27-6) have the formula
C6H4(CH3)2 (see margin). These three xylenes are constitutional isomers. In naming these
(as well as other disubstituted benzenes), we use prefixes ortho- (abbreviated o-), meta(m-),
or para- ( p-) to refer to relative positions of substituents on the benzene ring. The
ortho- prefix refers to two substituents located on adjacent carbon atoms; for example,
1,2-dimethylbenzene is o-xylene. The meta- prefix identifies substituents on C atoms 1
and 3, so 1,3-dimethylbenzene is m-xylene. The para- prefix refers to substituents on
C atoms 1 and 4, so 1,4-dimethylbenzene is p-xylene.
Summary of Rules for Naming Derivatives of Benzene
1. If there is only one group on the ring, no number is needed to designate its posi-
tion.
2. If there are two groups on the ring, we use the traditional designations.
ortho- or o- for 1,2-disubstitution
meta- or m- for 1,3-disubstitution
para- or p- for 1,4-disubstitution
3. If there are three or more groups on the ring, location numbers are assigned to
give the minimum sum of numbers.
Examples are
When an H atom is removed from a benzene molecule, C6H6, the resulting group,
C6H5X or , is called “phenyl.” Sometimes we name mixed alkyl-aromatic
hydrocarbons on that basis.
CC
CH3 CH3
H
2
1
3
4
phenylcyclohexane 2-phenyl-cis-2-butene
H
H
CH2CH3
H
H
H
ethylbenzene
H
H
CH2CH3
CH2CH3
H
H
m-diethylbenzene, or
1,3-diethylbenzene
H
H
CH2CH3
CH2CH3
CH2CH3
H
1,2,4-triethylbenzene
27-6
1062 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
CH3
CH3
4
5
6
3
2
1
para-xylene
bp ϭ 138ЊC
mp ϭ 13ЊC
CH3
CH3
4
5
6
3
2
1
CH3
CH3
4
5
6
3
2
1
ortho-xylene
bp ϭ 144ЊC
mp ϭ Ϫ27ЊC
meta-xylene
bp ϭ 139ЊC
mp ϭ Ϫ54ЊC
Another class of aromatic hydrocarbons consists of “condensed” or “fused-ring”
aromatic systems. The simplest of these are naphthalene, anthracene, and phenanthrene.
No hydrogen atoms are attached to the carbon atoms that are involved in fusion of aromatic
rings, that is, carbon atoms that are members of two or more aromatic rings.
The traditional name is often used as part of the base name in naming an aromatic
hydrocarbon and its derivatives. You should know the names and structures of the aromatic
hydrocarbons discussed thus far: benzene, toluene, the three xylenes, naphthalene,
anthracene, and phenanthrene.
Distillation of coal tar provides four volatile fractions as well as the pitch that is used
for surfacing roads and in the manufacture of “asphalt” roofing (Figure 27-12). Eight
aromatic hydrocarbons obtained in significant amounts by efficient fractional distillation
of the “light oil” fraction are those in Table 27-6.
H
H
H
H
H
H
H
H H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H H
phenanthrene, C14H10anthracene, C14H10naphthalene, C10H8
27-6 Other Aromatic Hydrocarbons 1063
TABLE 27-6 Aromatic Hydrocarbons from Coal Tar
Name Formula Normal bp (°C) Normal mp (°C) Solubility
benzene C6H6 80 ϩ6
toluene C6H5CH3 111 Ϫ95
o-xylene C6H4(CH3)2 144 Ϫ27
m-xylene C6H4(CH3)2 139 Ϫ54 All insoluble
p-xylene C6H4(CH3)2 138 ϩ13 in water
naphthalene C10H8 218 ϩ80
anthracene C14H10 342 ϩ218
phenanthrene C14H10 340 ϩ101
Naphthalene.
HYDROCARBONS: A SUMMARY
Hydrocarbons contain only carbon and hydrogen. They can be subdivided into various
groups. To assist you in organizing what you have studied so far in this chapter, take a
few minutes and study Figure 27-13.
FUNCTIONAL GROUPS
The study of organic chemistry is greatly simplified by considering hydrocarbons as parent
compounds and describing other compounds as derived from them. In general, an organic
molecule consists of a skeleton of carbon atoms with special groups of atoms attached to
that skeleton. These special groups of atoms are often called functional groups because
they represent the most common sites of chemical reactivity (function). The only functional
groups that are possible in hydrocarbons are double and triple (i.e., pi) bonds. Atoms
other than C and H are called heteroatoms, the most common being O, N, S, P, and
the halogens. Most functional groups contain one or more heteroatoms.
In the next several sections we will introduce some common functional groups that
contain heteroatoms and learn a little about the resulting classes of compounds. We will
continue to represent alkyl groups with the symbol RX. We commonly use that symbol
to represent either an aliphatic (e.g., alkyl) or an aromatic (e.g., an aryl such as phenyl)
group. When we specifically mean an aryl group, we will use the symbol ArX.
ORGANIC HALIDES
Almost any hydrogen atom in a hydrocarbon can be replaced by a halogen atom to give
a stable compound. Table 27-7 shows some organic halides and their names.
27-8
27-7
As you study the following sections,
you may wish to refer to the summary
in Section 27-15.
1064 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
1000 lb of coal tar
distilled yields:
Light oil
redistilled
yields approx.:
16 lb Benzene
2.5 lb Toluene
0.3 lb Xylene
Middle oil
extracted
with NaOH
and redistilled:
Phenol
and
cresols
together
about
20 lb
Naphthalene
40–60 lb
Heavy oil
yields
impure cresols
and phenols
Green oil
yields
anthracene
and
phenanthrene
5–20 lb
Pitch
500–
600 lb
Figure 27-12 Fractions obtained from coal tar.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 11.5, Functional Groups.
In the IUPAC naming system, the organic halides are named as halo- derivatives of the
parent hydrocarbons. The prefix halo- can be fluoro-, chloro-, bromo-, or iodo-. Simple alkyl
chlorides are sometimes given common names as alkyl derivatives of the hydrogen halides.
For instance, the IUPAC name for CH3CH2XCl is chloroethane; it is commonly called
ethyl chloride by analogy to HXCl, hydrogen chloride.
A carbon atom can be bonded to as many as four halogen atoms, so an enormous
number of organic halides can exist. Completely fluorinated compounds are known as
fluorocarbons or sometimes perfluorocarbons. The fluorocarbons are even less reactive
than hydrocarbons. Saturated compounds in which all H atoms have been replaced by
some combination of Cl and F atoms are called chlorofluorocarbons or sometimes freons.
These compounds were widely used as refrigerants and as propellants in aerosol cans.
However, the release of chlorofluorocarbons into the atmosphere has been shown to be
quite damaging to the earth’s ozone layer. Since January 1978, the use of chlorofluorocarbons
in aerosol cans in the United States has been banned, and efforts to develop both
controls for existing chlorofluorocarbons and suitable replacements continue. The production
and sale of freons have been banned in many countries.
Freon is a DuPont trademark for
certain chlorofluorocarbons; other
companies’ related products are
known by other names. Typical freons
are trichlorofluoromethane,
CFCl3 (called Freon-11), and
dichlorodifluoromethane, CF2Cl2
(called Freon-12).
27-8 Organic Halides 1065
Hydrocarbons
(contain C, H only)
(Sections 27-1 to 27-6)
Aliphatic hydrocarbons
Contain C–C single,
double, and triple bonds,
but no aromatic rings
(Sections 27-1 to 27-4)
Saturated hydrocarbons
Contain only C–C
and
C–H single bonds
(Sections 27–1 and 27–2)
Alkanes, CnH2n+2
Contain no rings
methane, ethane,
hexane,
2-methylhexane,
2,2-dimethylpentane
cyclopropane,
cyclopentane,
cyclohexane
Acyclic alkenes, CnH2n
ethylene, 1-butene,
2-methyl-1-butene
Cycloalkenes, CnH2n–2
cyclopentene,
cyclohexene
Cycloalkanes, CnH2n
Contain rings
Alkenes
Contain one or more
C=C double bonds
(Section 27-3)
Polyenes
Contain more than one
C=C per molecule—
e.g., 1,3-butadiene
acetylene, 1-butyne,
3-methyl-1-butyne
benzene, naphthalene,
anthracene, toluene,
p-xylene
Alkynes, CnH2n–2
Contain one or more
CϵC triple bonds
(Section 27-4)
Unsaturated hydrocarbons
Contain one or more
C=C double or
CϵC triple bonds
(Sections 27-3 and 27-4)
Aromatic hydrocarbons
(Sections 27-5 to 27-6)
SOME EXAMPLES
Figure 27-13 A classification of
hydrocarbons.
The simplest phenol is called phenol.
The most common member of a class
of compounds is often called by the
class name. Salt, sugar, alcohol, and
phenol are examples.
ALCOHOLS AND PHENOLS
Alcohols and phenols contain the hydroxyl group (XOXH) as their functional group.
Alcohols may be considered to be derived from saturated or unsaturated hydrocarbons
by the replacement of at least one H atom by a hydroxyl group. The properties of alcohols
result from a hydroxyl group attached to an aliphatic carbon atom, XCXOXH.
Ethanol (ethyl alcohol) is the most common example (Figure 27-14).
When a hydrogen atom on an aromatic ring is replaced by a hydroxyl group (Figure
27-15), the resulting compound is known as a phenol. Such compounds behave more like
acids than alcohols. Alternatively, we may view alcohols and phenols as derivatives of water
in which one H atom has been replaced by an organic group.
H
H H
H H
C C O HO HO HH
phenolwater ethanol
27-9
TABLE 27-7 Some Organic Halides
Normal
Formula Structural Formula bp (°C) IUPAC Name Common Name
CH3Cl Ϫ23.8 chloromethane methyl chloride
CH2Cl2 40.2 dichloromethane methylene chloride
CHCl3 61 trichloromethane chloroform
CCl4 76.8 tetrachloromethane carbon tetrachloride
CHCl2Br 90 bromodichloromethane —
H
H
H
C Cl
Cl
H
H
C Cl
Cl
Cl
H C Cl
Cl
Cl
Cl
C Cl
Cl
Br
H C Cl
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 11.6, Reactions of Alcohols.
Indeed, this is a better view. The structure of water was discussed in Section 8-9. The
hydroxyl group in an alcohol or a phenol is covalently bonded to a carbon atom, but the
OXH bond is quite polar. The oxygen atom has two unshared electron pairs, and the
CXOXH bond angle is nearly 104.5°.
The presence of a bonded alkyl or aryl group changes the properties of the XOH
group. Alcohols are so very weakly acidic that they are thought of as neutral compounds.
Phenols are weakly acidic.
Many properties of alcohols depend on whether the hydroxyl group is attached to a
carbon that is bonded to one, two, or three other carbon atoms.
Primary alcohols contain one R group; secondary alcohols contain two R groups;
and tertiary alcohols contain three R groups bonded to the carbon atom to which
the XOH group is attached.
TABLE 27-7 (continued)
Normal
Formula Structural Formula bp (°C) IUPAC Name Common Name
(CH3)2CHI 89.5 2-iodopropane isopropyl iodide
CH3ClCUCHCH2CH2Cl 40 2,5-dichloro-cis-2- —
pentene; trans
isomer is also
possible
C5H7Cl 25 3-chlorocyclopentene is —
shown; other isomers
are also possible
C6H5I 118 iodobenzene phenyl iodide
C6H4ClBr 204 1-bromo-2-chloro- o-bromochlorobenzene
benzene is shown;
other isomers are
also possible
C14H9Cl5 1,1,1-trichloro-2,2- DDT
bis(p-chlorophenyl)–
ethane
H H
I H
C CH
H
H
C H
Cl
ClCH3 CH2C
H
H
H
C C
Cl
2
1
5 4
3
I
Cl
Br
Cl Cl
Cl
ClCl
C
C
H
Figure 27-14 Models of ethanol
(also called ethyl alcohol or grain
alcohol), CH3CH2OH.
1068 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
Figure 27-15 Models of phenol,
C6H5OH.
Representing alkyl groups as R, we can illustrate the three classes of alcohols. The R
groups may be the same or different.
Nomenclature
The systematic name of an alcohol consists of the root name plus an -ol ending. A numeric
prefix indicates the position of the XOH group on a chain of three or more carbon atoms.
EXAMPLE 27-11 Formulas of Alcohols
There are four constitutional isomers of the saturated acyclic four-carbon alcohols with one
XOH per molecule. Write the structural formula of each, and identify each as primary,
secondary, or tertiary. Name each isomer.
Plan
The carbon skeleton can be either a straight chain, CXCXCXC, or branched, CXCXC.
A
C
Each skeleton has two different types of carbons to which the XOH group can be attached.
Solution
You should now work Exercise 46.
The polyhydric alcohols contain more than one XOH group per molecule. Those
containing two OH groups per molecule are called glycols. Important examples of polyhydric
alcohols include
CCH3
CH3
CH3
OH
2-methyl-2-propanol
tertiary butyl alcohol
3؇:
CH3CH2CHCH3
OH
2-butanol
secondary butyl alcohol
2؇:
CHCH3
CH3
CH2OH
2-methyl-1-propanol
isobutyl alcohol
1؇:CH3CH2CH2CH2OH
1-butanol
butyl alcohol
1؇:
R
RЈ
H
C OH
a secondary (2Њ) alcohol
R
H
H
C OH
a primary (1Њ) alcohol
R
RЈ
RЉ
C OH
a tertiary (3Њ) alcohol
In writing organic structures, we often
use primes when we wish to specify
that the alkyl groups might be
different, for example, R, RЈ, RЉ.
27-9 Alcohols and Phenols 1069
CH3 OH
methanol
methyl alcohol
(wood alcohol)
CH2CH2CH3 OH
1-propanol
propyl alcohol
(a primary alcohol)
CH3CHCH3
OH
2-propanol
isopropyl alcohol
(a secondary alcohol)
CH2CH3CCH3
CH3
OH
2-methyl-2-butanol
t-pentyl alcohol
(a tertiary alcohol)
Acyclic compounds contain no rings.
Polyhydric alcohols are used in
permanent antifreeze and in
cosmetics.
Phenols are usually referred to by their common names. Examples are
As you might guess, cresols occur in “creosote,” a wood preservative.
Physical Properties
The hydroxyl group, XOH, is quite polar, whereas alkyl groups, R, are nonpolar. The
properties of alcohols depend on two factors: (1) the number of hydroxyl groups per molecule
and (2) the size of the nonpolar portion of the molecule.
The low-molecular-weight alcohols are soluble in water in all proportions (miscible).
Beginning with the four butyl alcohols, solubility in water decreases rapidly with increasing
molecular weight. This is because the nonpolar parts of such molecules are much larger
than the polar parts. Many polyhydric alcohols are very soluble in water because they
contain two or more polar XOH groups per molecule.
Table 27-8 shows that the boiling points of unbranched primary alcohols increase, and
their solubilities in water decrease, with increasing molecular weight. The boiling points
H
H H
H H
C C O H
polar
part
nonpolar
part
ROH:
CH3
OH
OH
OH
OH
OH
hydroquinone o-cresol
CH3
CH3
OH
m-cresol
OH
p-cresolresorcinol
OH OH
CH2 CH2
1,2-ethanediol
ethylene glycol
(the major ingredient in
permanent antifreeze)
OH OH
CH CH2CH3
1,2-propanediol
propylene glycol
OH OH
CH CH2
OH
CH2
1,2,3-propanetriol
glycerine or glycerol
(a moisturizer in cosmetics)
The o-, m-, and p- notation was
introduced in Section 27-6.
1070 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
TABLE 27-8 Physical Properties of Primary Alcohols
Normal Solubility in H2O
Name Formula bp (°C) (g/100 g at 20°C)
methanol CH3OH 65 miscible
ethanol CH3CH2OH 78.5 miscible
1-propanol CH3CH2CH2OH 97 miscible
1-butanol CH3CH2CH2CH2OH 117.7 7.9
1-pentanol CH3CH2CH2CH2CH2OH 137.9 2.7
1-hexanol CH3CH2CH2CH2CH2CH2OH 155.8 0.59
of alcohols are much higher than those of the corresponding alkanes (see Table 27-2)
because of the hydrogen bonding of the hydroxyl groups.
Most phenols are solids at 25°C. Phenols are only slightly soluble in water unless they
contain other functional groups that interact with water.
Uses
Many alcohols and phenols have considerable commercial importance. Methanol,
CH3OH, was formerly produced by the destructive distillation of wood and is sometimes
called wood alcohol. It is now produced in large quantities from carbon monoxide and
hydrogen. It is very toxic and causes permanent blindness when taken internally.
Ethanol, CH3CH2OH, also known as ethyl alcohol or grain alcohol, can be prepared
by fermentation of sugar. The most ancient written literature refers to beverages that were
obviously alcoholic!
Many simple alcohols are important raw materials in the industrial synthesis of polymers,
fibers, explosives, plastics, and pharmaceutical products. Phenols are widely used in
the preparation of plastics and dyes. Solutions of phenols are used as antiseptics and disinfectants.
Some uses of polyhydric alcohols depend on their relatively high boiling points.
For instance, glycerine is used as a wetting agent in cosmetic preparations. Ethylene glycol
(bp 197°C), which is miscible with water, is used in commercial permanent antifreeze.
ETHERS
When the word “ether” is mentioned, most people think of the well-known anesthetic,
diethyl ether. There are many ethers. Their uses range from artificial flavorings to refrigerants
and solvents. An ether is a compound in which an O atom is bonded to two organic
groups.
Alcohols are considered derivatives of water in which one H atom has been replaced by
an organic group. Ethers may be considered derivatives of water in which both H atoms
have been replaced by organic groups.
HXOXH RXOXH RXOXRЈ
water alcohol ether
The similarity is only structural, however, because ethers are only slightly polar and are
chemically rather unreactive. In fact, their physical properties are similar to those of the
corresponding alkanes; for example, CH3OCH3 is like CH3CH2CH3.
Three kinds of ethers are known: (1) aliphatic, (2) aromatic, and (3) mixed. Common
names are used for ethers in most cases.
C CO
27-10
Fermentation is an enzymatic process
carried out by certain kinds of bacteria.
27-10 Ethers 1071
H3C CH3O
methoxymethane
dimethyl ether
(an aliphatic ether)
H3C CH2CH3O
methoxyethane
methyl ethyl ether
(an aliphatic ether)
methoxybenzene
methyl phenyl ether
anisole
(a mixed ether)
phenoxybenzene
diphenyl ether
(an aromatic ether)
O CH3 O
When the benzene ring is a
substituent, it is called a phenyl group
(XC6H5).
Models of diethyl ether (top) and
methyl phenyl ether (bottom).
Diethyl ether is a very low boiling liquid (bp 35°C). Dimethyl ether is a gas that is used
as a refrigerant. The aliphatic ethers of higher molecular weights are liquids, and the
aromatic ethers are liquids and solids.
Even ethers of low molecular weight are only slightly soluble in water. Diethyl ether
is an excellent solvent for organic compounds. It is widely used to extract organic
compounds from plants and other natural sources.
ALDEHYDES AND KETONES
Aldehydes and ketones contain the carbonyl group, . In aldehydes, at least one
H atom is bonded to the carbonyl group. Ketones have two carbon atoms bonded to a
carbonyl group. Models of formaldehyde (the simplest aldehyde) and acetone (the simplest
ketone) are shown in Figure 27-16.
Aldehydes are usually called by their common names. These are derived from the name
of the acid with the same number of C atoms (Table 27-9). The systematic (IUPAC) name
is derived from the name of the parent hydrocarbon. The suffix -al is added to the characteristic
stem. The carbonyl group takes positional precedence over most other
substituents.
Formaldehyde has long been used as a disinfectant and as a preservative (including
embalming fluid) for biological specimens. Its main use is in the production of certain
plastics and in binders for plywood. Many important natural substances are aldehydes and
ketones. Examples include sex hormones, some vitamins, camphor, and the flavorings
extracted from almonds and cinnamon. Aldehydes contain a carbon–oxygen double bond,
so they are very reactive compounds. As a result, they are valuable in organic synthesis,
particularly in the construction of carbon chains.
O
CR H
aliphatic
aldehyde
O
CR R
aliphatic
ketone
O
CAr H
O
CAr Ar
aromatic
ketone
O
CAr R
mixed
ketone
aromatic
aldehyde
C O.
27-11
The simplest aldehyde, formaldehyde,
O
B
HXCXH, has two H atoms and no
alkyl or aryl groups.
1072 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
Figure 27-16 (a) Models of formaldehyde, HCHO, the simplest aldehyde. (b) Models of
acetone, CH3XCOXCH3, the simplest ketone.
(a) (b)
The simplest ketone is called acetone. Other simple, commonly encountered ketones
are usually called by their common names. These are derived by naming the alkyl or aryl
groups attached to the carbonyl group.
The systematic names for ketones are derived from their parent hydrocarbons. The
suffix -one is added to the characteristic stem.
The ketones are excellent solvents. Acetone is very useful because it dissolves most
organic compounds yet is miscible with water. Acetone is widely used as a solvent in the
manufacture of lacquers, paint removers, explosives, plastics, drugs, and disinfectants.
Some ketones of high molecular weight are used extensively in blending perfumes. Structures
of some naturally occurring aldehydes and ketones are
CH3 CH2
O
C CH3
butanone
CH3 CH2
O
C CH CH2 CH3
CH3
4-methyl-3-hexanone
1 2 3 4 1 2 3 4 5 6
O
C
O
C CH3O
cyclohexanone acetophenone
(methyl phenyl ketone)
benzophenone
(diphenyl ketone)
CH3 CH3
O
C CH3 CH2CH3
O
C CH3CH2 CH2CH3
O
C
diethyl ketonemethyl ethyl ketoneacetone
27-11 Aldehydes and Ketones 1073
TABLE 27-9 Properties of Some Simple Aldehydes
Common Name Formula Normal bp (°C)
formaldehyde (methanal) Ϫ21
acetaldehyde (ethanal) 20.2
propionaldehyde (propanal) 48.8
benzaldehyde 179.5
H
O
C H
CH3
O
C H
O
CH3CH2C H
O
C H
Steroid molecules have similar
molecular shapes but different
biochemical functions. Progesterone
(top), a female sex hormone, and
testosterone (bottom), a male sex
hormone. Both are ketones.
AMINES
The amines are derivatives of ammonia in which one or more hydrogen atoms have been
replaced by alkyl or aryl groups. Many low-molecular-weight amines are gases or lowboiling
liquids (Table 27-10). Amines are basic compounds (Table 18-6, Section 28-4).
The aliphatic amines of low molecular weight are soluble in water. Aliphatic diamines of
fairly high molecular weight are soluble in water because each molecule contains two
highly polar XNH2 groups that form hydrogen bonds with water.
The odors of amines are quite unpleasant; many of the malodorous compounds that
are released as fish decay are low-molecular-weight amines. Amines of high molecular
weight are nonvolatile, so they have little odor. One of the materials used to manufacture
nylon, hexamethylenediamine, is an aliphatic amine. Many aromatic amines are used to
prepare organic dyes that are widely used in industrial societies. Amines are also used to
produce many medicinal products, including local anesthetics and sulfa drugs.
Amines are widely distributed in nature in the form of amino acids and proteins, which
are found in all higher animal forms, and in alkaloids, which are found in most plants.
27-12
CCH3CH
CH2
(CH2)12
muscone
(musk deer, used
in perfumes)
testosterone
(male sex hormone)
camphor
O
O
H3C
OH
O
H3C
H3C
CH3
CH3
CH
benzaldehyde
(almonds)
cinnamaldehyde
(cinnamon)
vanillin
(vanilla)
O
CH CH CH
O
HO
OCH3
CH
O
Ammonia acts as a Lewis base because
there is one unshared pair of electrons
on the N atom (Section 10-10).
1074 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
TABLE 27-10 Boiling Points of Ammonia and Some Amines
Name Formula Boiling Point (°C)
ammonia NH3 Ϫ33.4
methylamine CH3NH2 Ϫ6.5
dimethylamine (CH3)2NH 7.4
trimethylamine (CH3)3N 3.5
ethylamine CH3CH2NH2 16.6
aniline C6H5NH2 184
ethylenediamine H2NCH2CH2NH2 116.5
pyridine C5H5N 115.3
pyrrole C4H5N 129.8
Some of these substances are fundamental building blocks of animal tissue, and minute
amounts of others have dramatic physiological effects, both harmful and beneficial. Countless
other biologically important substances, including many vitamins, antibiotics, and other
drugs, contain amino groups, XNR2 (where R can represent an H, alkyl, or aryl group).
Structure and Nomenclature
There are three classes of amines, depending on whether one, two, or three hydrogen
atoms have been replaced by organic groups. They are called primary, secondary, and
tertiary amines, respectively.
The systematic names of amines are based on consideration of the compounds as derivatives
of ammonia. Amines of more complex structure are sometimes named as derivatives
of the parent hydrocarbon with the term amino- used as a prefix to describe XNH2.
Aniline is the simplest aromatic amine. Many aromatic amines are named as derivatives
of aniline.
Heterocyclic amines contain nitrogen as a part of the ring, bound to two carbon
atoms. Many of these amines are found in coal tar and a variety of natural products. Some
aromatic and heterocyclic amines are called by their common names.
N N
N
N
N
NN
N
H
N
H
pyrrole
(secondary)
quinoline
(tertiary)
pyridine
(tertiary)
purine pyrimidine
N
H H
Cl
p-chloroaniline
(primary)
N,N-dimethylaniline
(tertiary)
aniline
(primary)
N
H H
Br
Br
3,4-dibromo-N-methylaniline
(secondary)
N
H CH3
N
H3C CH3
H
H NH2
NH2
CH2CH3H H
C C
H H
H H
C C H
2-aminobutane or sec-butylamine 1-amino-3-ethylcyclohexane
N
NH3
H
H
ammonia
N
RNH2
CH3 H
H
methylamine
(a primary amine)
N
R2NH
CH3 CH3
H
dimethylamine
(a secondary amine)
N
R3N
CH3 CH3
CH3
trimethylamine
(a tertiary amine)
H
N
N
CH2
CH2
CH
strychnine
O O
C
N
N
CH3
nicotine
27-12 Amines 1075
Genes, the units of chromosomes that carry hereditary characteristics, are long stretches
of double helical deoxyribonucleic acid, or DNA. DNA is composed of four nucleotide
bases: adenine, guanine, cytosine, and thymine. The first two are modified purines, and
the latter two are modified pyrimidines. The sequence of these building blocks in DNA
acts as a code for the order of amino acids in the proteins of an organism. The DNA in
each cell of an organism contains the instructions for making the complete organism (see
Section 28-10).
CARBOXYLIC ACIDS
Compounds that contain the carboxyl group, , are acidic. They are called
carboxylic acids. Their general formula is RXCOOH. Most are weak acids. They are
much stronger acids than most phenols, however. Carboxylic acids are named systematically
by dropping the terminal -e from the name of the parent hydrocarbon and adding
-oic acid (Table 27-11). Many older names survive, however, and organic acids are often
called by common names. In aromatic acids the carboxyl group is attached to an aromatic
ring (Figure 27-17).
Organic acids occur widely in natural products, and many have been known since
ancient times. Their common (trivial) names are often derived from a Greek or Latin
word that indicates the original source (see Table 27-11).
The names of modified carboxylic acids are often derived from the trivial names of the
acids. Positions of substituents are sometimes indicated by lowercase Greek letters, beginning
with the carbon adjacent to the carboxyl carbon, rather than by numbering the carbon
chain.
C O H
O
27-13
N
N
N
N
H
NH2
N
N
N
N
H
O
adenine
N
N
H
NH2
thymine
H
H2N
N
H
O N
H
O
CH3
O
cytosineguanine
Aliphatic carboxylic acids are
sometimes referred to as fatty acids
because many have been obtained from
animal fats.
1076 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
TABLE 27-11 Some Aliphatic Carboxylic Acids
Formula Common Name IUPAC Name
HCOOH formic acid methanoic acid
CH3COOH acetic acid ethanoic acid
CH3CH2COOH propionic acid propanoic acid
CH3CH2CH2COOH butyric acid butanoic acid
CH3CH2CH2CH2CH2COOH caproic acid hexanoic acid
CH3(CH2)10COOH lauric acid dodecanoic acid
CH3(CH2)14COOH palmitic acid hexadecanoic acid
CH3(CH2)16COOH stearic acid octadecanoic acid
In the general formulas for alcohols
and ethers, R cannot be H. In
carboxylic acids, R can be H.
Formic acid was obtained by
distillation of ants (L. formica, “ant”);
acetic acid occurs in vinegar (L.
acetum, “vinegar”); butyric acid in
rancid butter (L. butyrum, “butter”);
and stearic acid in animal fats (Gr.
stear, “beef suet”). Caproic acid is one
of the so-called “goat acids.” Its odor
is responsible for the name.
A space-filling model of a portion of
the DNA double helical structure.
EXAMPLE 27-12 Isomeric Carboxylic Acids
Write the structural formula of each of the four carboxylic acids having the molecular formula
of C5H10O2. Give the IUPAC name of each.
Plan
We first write the structural formula of the carboxylic acid that does not have branching. Then
we remove one carbon from the straight-chain structure and reposition it to form branched
isomers. We check to see whether branching is possible at more than one location. We find
that branching is possible at two different locations. Remove a second carbon from the
unbranched isomer and, if possible, from additional branching. The two removed carbons can
be added as methyl branches. Each isomer is named by naming the longest chain that contains
the carboxylic acid group. The final -e of the name is changed to -oic acid. Names for branches
appear at the beginning of the name.
Solution
Isomer 1.
Isomer 2.
Isomer 3.
Isomer 4.
You should now work Exercise 72a, b.
O
CCH3C OH
CH3
CH3
dimethylpropanoic acid
CH3CHCH2C
O
OH
CH3
3-methylbutanoic acid
CH3CH2CH
O
C OH
CH3
2-methylbutanoic acid
OH
O
CH3CH2CH2CH2C pentanoic acid
C
O
OH
CH
Br
CH3
␣␤
2-bromopropanoic acid
␣( -bromopropionic acid)
C
O
OH
OH
CH2 CH2 CH2
␣␤␥
4-hydroxybutanoic acid
␥( -hydroxybutyric acid)
Derivatives of ␣-hydroxy carboxylic
acids are currently used in some
cosmetic preparations.
H
C
H
H
H
(a) acetic acid
(an aliphatic acid)
(b) benzoic acid
(an aromatic acid)
C
O
O
H
C
O
O
Figure 27-17 (a) Models of acetic
acid. (b) Models of benzoic acid.
Some carboxylic acid molecules contain more than one XCOOH group (Table 27-12).
These acids are nearly always called by their common names. For example, oxalic acid,
found in rhubarb, spinach, and tomatoes, is a dicarboxylic acid; citric acid, found in citrus
fruits, is a tricarboxylic acid.
Aromatic acids are called by their common names or named as derivatives of benzoic
acid, which is considered the “parent” aromatic acid.
Although many carboxylic acids occur in the free state in nature, many occur as amides
or esters (Section 27-14). ␣-Amino acids are substituted carboxylic acids with the general
structure
OH
C
O OH
C
Cl CH3
CH3
O OH
C
O
2,4-dimethylbenzoic acidp-chlorobenzoic acidbenzoic acid
These two derivatives of acetic acids
act as herbicides (weed killers) by
overstimulating the plant’s growth
system.
1078 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
TABLE 27-12 Some Naturally Occurring Organic Acids
Name Source
acetic acid CH3XCOOH vinegar
citric acid citrus fruits
lactic acid sour milk
malic acid apples, watermelons,
grape juice, wine
oxalic acid HOOCXCOOH rhubarb, spinach, tomatoes
quinic acid cranberries
tartaric acid grape juice, wine
OH
COOH
CHOOC CH2 COOHCH2
OH
CH COOHCH3
OH
CH COOHHOOC CH2
C
CC
CH2 COOH
C CH2 OHHO
H
OHH
OHH
OH
CH
OH
CH COOHHOOC
CH2
Cl
COCl OH
O
CH2
CH3
COCl OH
O
Crystals of glycine viewed under
polarized light. Glycine, the simplest
amino acid, has the structure shown
in the text, with R ϭ H.
27-14 Some Derivatives of Carboxylic Acids 1079
where R can be either an alkyl or an aryl group. ␣-Amino acids are the components of
proteins, which make up the muscle and tissue of animals. Many other acids are important
in the metabolism and synthesis of fats by enzyme systems. Acetic acid (the acid in
vinegar) is the end product in the fermentation of most agricultural products. It is the
fundamental unit used by living organisms in the biosynthesis of such widely diverse classes
of natural products as long-chain fatty acids, natural rubber, and steroid hormones.
SOME DERIVATIVES OF CARBOXYLIC ACIDS
Three important classes of acid derivatives are formed by the replacement of the hydroxyl
group by another atom or group of atoms. Each of these derivatives contains an acyl group.
Aromatic compounds of these types (with R ϭ aryl groups) are also encountered
frequently.
Acyl Halides (Acid Halides)
The acyl halides, sometimes called acid halides, are structurally related to carboxylic
acids by the replacement of the XOH group by a halogen, most often XCl. They are
usually named by combining the stems of the common names of the carboxylic acids with
the suffix -yl and then adding the name of the halide ion. Examples are
Acid halides are very reactive and have not been observed in nature.
Esters
Esters can be thought of as the result of removing one molecule of water from a carboxylic
acid and an alcohol. Removing a molecule of water from acetic acid and ethyl alcohol
gives ethyl acetate.
CH3C
O
HOOH CH3C
O
OCH2CH3 H2OCH2CH3
ethyl alcohol ethyl acetateacetic acid
ϩ ϩ
ClC
O
ClC
O
FCCH3CH2CH2CH3
O
benzoyl chloridebutyryl fluorideacetyl chloride
O
CR Cl
an acyl chloride
(an acid chloride)
O
CR NH2
an amide
O
CR RO
an ester
27-14
OHC
OH
NH2
CR
As we will see in Section 28-6, one
method of forming esters involves
acid-catalyzed reaction of an alcohol
with a carboxylic acid.
1080 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
The fragment is derived from acetic acid, the parent acid; the OXCH2CH3
fragment is derived from ethanol, the parent alcohol. Models of ethyl acetate, an ester,
are shown in Figure 27-18.
Esters are nearly always called by their common names. These consist of, first, the
name of the alkyl group in the alcohol, and then the name of the anion derived from the
acid.
Because of their inability to form hydrogen bonds, esters tend to be liquids with boiling
points much lower than those of carboxylic acids of similar molecular weights.
Most simple esters are pleasant-smelling substances. They are responsible for the flavors
and fragrances of most fruits and flowers and many of the artificial fruit flavorings that
are used in cakes, candies, and ice cream (Table 27-13). Esters of low molecular weight
are excellent solvents for nonpolar compounds. Ethyl acetate is an excellent solvent that
gives many nail polish removers their characteristic odor.
Fats (solids) and oils (liquids) are esters of glycerol and aliphatic acids of high molecular
weight. “Fatty acids” are all organic acids whose esters occur in fats and oils. Fats and
oils have the general formula
OCH3C
O
methyl benzoate
OCH3C
O
phenyl acetate
OC(CH3)3CH3CH2C
O
t-butyl propionate
C
O
CH3
TABLE 27-13 Some Common Esters
Ester Formula Odor of
isoamyl acetate CH3COOC5H11 bananas
ethyl butyrate C3H7COOC2H5 pineapples
amyl butyrate C3H7COOC5H11 apricots
octyl acetate CH3COOC8H17 oranges
isoamyl isovalerate C4H9COOC5H11 apples
methyl salicylate C6H4(OH)(COOCH3) oil of wintergreen
methyl anthranilate C6H4(NH2)(COOCH3) grapes
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 11.7, Fats and Oils.
Figure 27-18 Models of ethyl acetate, an ester.C O CH2CH3,CH3
O
Apple blossoms.
27-14 Some Derivatives of Carboxylic Acids 1081
Naturally occurring fats and oils are mixtures of many different esters. Milk fat, lard,
and tallow are familiar, important fats. Soybean oil, cottonseed oil, linseed oil, palm oil,
and coconut oil are examples of important oils.
The triesters of glycerol are called glycerides. Simple glycerides are esters in which all
three R groups are identical. Two examples are
Waxes are esters of fatty acids and alcohols other than glycerol. Most are derived from
long-chain fatty acids and long-chain monohydric alcohols. Both usually contain even
numbers of carbon atoms. Beeswax is largely C15H31COOC30H61; carnauba wax contains
C25H51COOC30H61. Both are esters of myricyl alcohol, C30H61OH.
Fats are solid esters of glycerol and (mostly) saturated acids. Oils are liquid esters that
are derived primarily from unsaturated acids and glycerol. The acid portion of a fat usually
contains an even number of carbon atoms, often 16 or 18. Some acids that occur frequently
in fats and oils are
butyric CH3CH2CH2COOH
lauric CH3(CH2)10COOH
myristic CH3(CH2)12COOH
palmitic CH3(CH2)14COOH
stearic CH3(CH2)16COOH
oleic CH3(CH2)7CHUCH(CH2)7COOH
linolenic CH3CH2CHUCHCH2CHUCHCH2CHUCH(CH2)7COOH
ricinoleic CH3(CH2)5CHOHCH2CHUCH(CH2)7COOH
Figure 27-19a shows a model of stearic acid, a long-chain saturated fatty acid.
Amides
Amides are thought of as derivatives of organic acids and ammonia, primary amines, or
secondary amines. Amides contain the functional group. They are named asC N
O
H2C (CH2)14CH3
(CH2)14CH3
(CH2)14CH3
CO
O
O
O
HC C
O
H2C C
O
H2C (CH2)16CH3
(CH2)16CH3
(CH2)16CH3
CO
O
O
O
HC C
O
H2C C
O
glyceryl tristearate
(stearin)
glyceryl tripalmitate
(palmitin)
CH2CR O
O
O
O
CHCR
O
CH2CR
O
C
O
RThe fatty acid portions, , may be saturated or unsaturated.
The R’s may be in the same or different groups.
Most natural fatty acids contain even
numbers of carbon atoms because they
are synthesized in the body from twocarbon
acetyl groups.
Honeybees produce the wax to build
their honeycombs.
Glycerol is
CH2
CHHO
HO
HO
CH2
derivatives of the corresponding carboxylic acids, the suffix -amide being substituted for
-ic acid or -oic acid in the name of the parent acid. The presence of alkyl or aryl substituents
attached to nitrogen is designated by prefixing the letter N and the name of the substituent
to the name of the unsubstituted amide.
Proteins are complex amides of high molecular weight. Some synthetic fibers are also
polyamides (Sections 27-19 and 28-9).
Figure 27-19 Models of longchain
fatty acids. The saturated fatty
acids (a) are linear and tend to pack
like sticks of wood to form solid
masses in blood vessels, thereby
constricting them. The trans
unsaturated fatty acids have a slight
Z-shaped kink in the chain, but are
also essentially linear molecules. By
contrast, cis unsaturated fatty acids
(b) are bent and so do not pack as
well as linear structures and do not
collect in blood vessels as readily.
Many natural vegetable fats and oils
contain esters of cis unsaturated fatty
acids, or polyunsaturated fatty acids.
Health problems associated with
saturated fatty acids can be decreased
by eating less animal fat, butter, and
lard. Problems due to trans fatty
acids are reduced by avoiding
processed vegetable fats.
1082 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
O
CH3C NH2 C NH2
O
CH3C
CH3
CH3N
O
CH3C
H
N
O
N-phenylacetamide
(acetanilide)
N,N-dimethylacetamidebenzamideacetamide
Crystals of the amide known as
acetaminophen (Tylenol) viewed
under polarized light. The structure
of acetaminophen is
C N OH
H
CH3
O
(a) (b)
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 11.8, Amino Acids and Proteins.
27-14 Some Derivatives of Carboxylic Acids 1083
Our Daily LivesCCHEMISTRY IN USE
Butter, Margarine, and trans Fats
Humans had consumed butter for thousands of years before
France introduced the world to margarine in the late 1860s;
by 1874, margarine had reached the United States. Margarine
consumption has increased rapidly, due in part to an increased
risk of heart disease that has been associated with consumption
of butter.
Both butter and margarine are primarily fats, but they
contain different kinds of fats. The main ingredient in butter
is cream, the concentrated fat from milk, whereas the main
ingredient in margarine is vegetable oil, the concentrated fat
from plants. Although it is widely accepted that animal fats
pose a greater risk for heart disease than vegetable oils, most
vegetable oils used in margarine and processed foods are
modified by hydrogenation.
The hydrogenation process adds hydrogen atoms to
unsaturated vegetable oils; this causes vegetable oils to solidify,
which makes them creamier and prolongs their shelf life.
Unfortunately, the hydrogenation process changes about
40% of the naturally occurring cis molecules of vegetable oil
into trans isomers; the resulting fat molecules are referred to
as trans fats. Oils produced in nature contain only cis isomers
and are considered healthful, but the trans fats have been
linked to many thousands of deaths due to heart disease. Diets
high in hydrogenated vegetable oil may be as unhealthy as
those high in saturated animal fats.
To understand the different effects produced by animal
fats, vegetable oils, and trans fats, we need to know a little
about blood cholesterol. Medical specialists recognize several
types of cholesterol, two of which are HDL (high-density
lipoproteins) and LDL (low-density lipoproteins). Because
high levels of HDL reduce the risk of heart disease, HDL
has become known as “good” cholesterol; because high levels
of LDL increase the risk for heart disease, LDL has
become known as “bad” cholesterol. In the average U.S.
adult, the total amount of cholesterol (including HDL and
LDL) circulating in the blood is about 200 mg per 100 mL
of serum.
A diet high in animal fats usually increases blood cholesterol
levels beyond 200 mg per 100 mL of serum, thereby
increasing the risk for heart disease. When animal fats
increase the total blood cholesterol level, however, they
increase the levels of both HDL and LDL. In contrast, vegetable
oils decrease the total blood cholesterol level, which
makes them healthier foods than other fats. Trans fats (from
hydrogenated vegetable oils) also decrease the total blood
cholesterol level, giving the appearance of being healthy, but
unfortunately they accomplish this reduction by simultaneously
reducing the desirable HDL levels and increasing the
undesirable LDL levels.
The damage done by hydrogenated vegetable oils is particularly
disturbing because margarine is one of the largest
sources of calories in the U.S. diet; approximately 40% of
margarine fat is the trans isomer. Hydrogenated vegetable oils
are widely used in shortening, cookies, crackers, chips, and
other processed foods. Ironically, most of these foods are
touted as being healthy because they contain no cholesterol
or saturated fats.
What can we do to protect our health? For many years,
nutritionists have been telling us what appears to be the “best
solution”: reduce your total fat intake to 20% of your daily
calories, and reduce your consumption of animal fats by substituting
vegetable oils. Whenever possible, use vegetable oils
that have not been hydrogenated.
Ronald DeLorenzo
Middle Georgia College
Blood pressure check.
Figure 27-20 Summary of some functional groups and classes of organic compounds.
1084 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
FUNCTIONAL GROUP CLASS OF COMPOUNDS SOME EXAMPLES
C C
Organic
compounds
with functional
groups
C C
X
(Xϭ F, Cl, Br, I)
OH
C O C
N
O
C OH
O
C OR
O
C N
O
C Y
Alkenes
(Section 27-3)
Alkynes
(Section 27-4)
Organic halides
(Section 27-8)
Alcohols, R OH
(Section 27-9)
Phenols, Ar OH
(Section 27-9)
Ethers
(Section 27-10)
Amines
(Section 27-12)
Carboxylic acids
(Section 27-13)
Esters
(Section 27-14)
Amides
(Section 27-14)
Aldehydes (Y ϭ H)
(Section 27-11)
Ketones (Y ϭ R or Ar)
(Section 27-11)
ethylene, butene,
cyclohexene
acetylene, propyne
methyl chloride,
chloroform
ethyl alcohol, methyl alcohol,
2-butanol
phenol, m-cresol,
diethyl ether,
methyl phenyl ether,
diphenyl ether
methylamine, aniline,
dimethylamine, pyridene
formic acid, acetic acid
benzoic acid
ethyl acetate,
glyceryl tristearate
N-methylacetamide,
acetanilide,
various polyamides
formaldehyde, acetaldehyde
benzaldehyde
acetone, cyclohexanone,
camphor
O
C X
(X halogen)
Acyl halides
(Section 27-14) acetyl chloride
SUMMARY OF FUNCTIONAL GROUPS
Some important functional groups and the corresponding classes of related compounds
are summarized in Figure 27-20.
FUNDAMENTAL CLASSES OF ORGANIC REACTIONS
Organic compounds display very different abilities to react, ranging from the limited reactivity
of hydrocarbons and fluorocarbons to the great variety of reactions undergone by
the millions of organic molecules that contain several functional groups. Reactivity
depends on structure. We can usually predict the kinds of reactions a compound can
undergo by identifying the functional groups it contains. But the electronic and structural
features that are near a functional group can also affect its reactivity. One of the fascinations
of organic chemistry is our ability to “fine-tune” both physical and chemical
properties by making small changes in structure. The successes of this approach are innumerable,
including the development of fuels and their additives or alternatives, the
improvement of pharmaceuticals to enhance their effectiveness and minimize their ill
effects, and the development of polymers and plastics with an incredible variety of properties
and uses.
In the following sections of this chapter we will present a few of the kinds of reactions
that organic compounds undergo. A topic of such vast scope as reactivity of organic
compounds can be made manageable only if we divide our study of this field into subtopics.
Nearly all organic transformations involve at least one of three fundamental classes of
reactions. The following three sections will address substitution, addition, and elimination
reactions. We will also look at some reaction sequences that combine reaction steps
from more than one of the fundamental classes.
SUBSTITUTION REACTIONS
In a substitution reaction an atom or a group of atoms attached to a carbon atom
is removed, and another atom or group of atoms takes its place. No change occurs
in the degree of saturation at the reactive carbon atom.
The saturated hydrocarbons (alkanes and cycloalkanes) are chemically rather inert
materials. For many years they were known as paraffins because they undergo few reactions.
They do not react with such powerful oxidizing agents as potassium permanganate
and potassium dichromate. They do react with the halogens, however, with oxygen when
ignited, and with concentrated nitric acid. As expected, members of a homologous series
(see Section 27-1) have similar chemical properties. If we study the chemistry of one of
these compounds, we can make predictions about the others with a fair degree of certainty.
The saturated hydrocarbons can react without a big disruption of the molecular structure
only by displacement, or substitution of one atom for another. At room temperature,
chlorine and bromine react very slowly with saturated straight-chain hydrocarbons. At
27-16
27-15
27-16 Substitution Reactions 1085
higher temperatures, or in the presence of sunlight or other sources of ultraviolet light,
H atoms in the hydrocarbon can be replaced easily by halogen atoms. These substitution
reactions are called halogenation reactions.
The reaction is represented as
Many organic reactions produce more than a single product. For example, the chlorination
of CH4 may produce several other products in addition to CH3Cl, as the following
equations show.
When a hydrocarbon has more than one C atom, its reaction with Cl2 is more complex.
The first step in the chlorination of ethane gives the product that contains one Cl atom
per molecule.
When a second hydrogen atom is replaced, a mixture of the two possible products is
obtained.
Cl Cl H
H
H
Cϩ HClϩ
heat or
UV
ethane chloroethane
(ethyl chloride)
bp ϭ 13.1ЊC
H
C Cl
H
H H
H
H
C
H
C
H
C ClCl
Cl
Cl Cl
H
Cl C Cl
Cl
Cl
ϩ HClϩ
tetrachloromethane
(carbon tetrachloride)
bp ϭ 76.8ЊC
H
C ClClCl Cl
H
Cl
H
C Cl
Cl
ϩ HClϩ
trichloromethane
(chloroform)
bp ϭ 61ЊC
H
H
C ClCl ClH Cl
H
H
C Clϩ HClϩ
dichloromethane
(methylene chloride)
bp ϭ 40.2ЊC
H
H
H
C Cl ClH H
H
H
C Clϩ HClϩ
heat or
UV
methane chlorine chloromethane
(methyl chloride)
bp ϭ Ϫ23.8ЊC
Note that only one half of the chlorine
atoms occur in the organic product.
The other half form hydrogen
chloride, a commercially valuable
compound.
1086 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
Substitution is the most common kind of reaction of an aromatic ring. Halogenation,
with chlorine or bromine, occurs readily in the presence of iron or anhydrous iron(III)
chloride (a Lewis acid) catalyst.
The equation is usually written in condensed form as
Aromatic rings can undergo nitration, substitution of the nitro group XNO2, in a
mixture of concentrated nitric and sulfuric acids at moderate temperatures.
The explosive TNT (2,4,6-trinitrotoluene) is manufactured by the nitration of toluene
in three steps.
ϩ
H2SO4
ϩ3HNO3 3H2O
nitric acid 2,4,6-trinitrotoluene (TNT)toluene
NO2
NO2
CH3
O2N
CH3
ϩ
conc.
H2SO4
50Њ
ϩHNO3 H2O
nitric acid nitrobenzene
NO2
ϩ
Cl
ϩCl2 HCl
FeCl3
benzene chlorobenzene
H
H
H H
HH
ϩ
Cl
H
H H
HH
ϩCl2 HCl
FeCl3
Cl Cl
H
H
H
C
ϩ HClϩ
heat or
UV
1,1-dichlorethane
bp ϭ 57ЊC
C Cl
Cl
H
H
H
C
1,2-dichloroethane
bp ϭ 84ЊC
C ClCl
H
H
H
H
H
C
H
C Cl
H
The product mixture does not contain
equal numbers of moles of the
dichloroethanes so we do not show a
stoichiometrically balanced equation.
Because reactions of saturated
hydrocarbons with chlorine can
produce many products, the reactions
are not always as useful as might be
desired.
27-16 Substitution Reactions 1087
A model of a molecule of TNT.
When iron is added as a catalyst, it
reacts with chlorine to form iron(III)
chloride, which is the true catalyst.
The H2SO4 is both a catalyst and a
dehydrating agent. The H in the
product H2O comes from the
hydrocarbon; the OH comes from
HNO3.
Groups other than hydrogen can be substituted by other atoms or groups of atoms.
For example, the bromine atom of an alkyl bromide can be replaced with a hydroxyl group
to form an alcohol. The mechanism of this type of reaction is often studied in introductory
organic chemistry courses.
Alcohols react with common inorganic oxyacids to produce inorganic esters. For
instance, nitric acid reacts with alcohols to produce nitrates by substitution of nitrate,
XONO2, for hydroxyl, XOH.
CH3CH2OH ϩ HNO3 88n CH3CH2XONO2 ϩ H2O
ethanol nitric acid ethyl nitrate
The substitution reaction of nitric acid with glycerol produces the explosive nitroglycerine.
Alfred Nobel’s (1833–1896) discovery in 1866 that this very sensitive material could
be made into a “safe” explosive by absorbing it into diatomaceous earth or wood meal led
to his development of dynamite.
Cold, concentrated H2SO4 reacts with alcohols to form alkyl hydrogen sulfates. The
reaction with lauryl alcohol is an important industrial reaction.
CH3(CH2)10CH2XOHo ϩ HOSO3Ho 88n CH3(CH2)10CH2XOSO3H ϩ H2O
1-dodecanol (lauryl alcohol) sulfuric lauryl hydrogen sulfate
acid
The neutralization reaction of an alkyl hydrogen sulfate with NaOH then produces the
sodium salt of the alkyl hydrogen sulfate.
CH3(CH2)10CH2XOSO3H ϩ NaϩOHϪ 88n CH3(CH2)10CH2XOSO3
ϪNaϩ ϩ H2O
sodium lauryl sulfate (a detergent)
Sodium salts of the alkyl hydrogen sulfates that contain about 12 carbon atoms are excellent
detergents. They are also biodegradable. (Soaps and detergents were discussed in
Section 14-18.)
ADDITION REACTIONS
An addition reaction involves an increase in the number of atoms or groups attached
to carbon. The molecule becomes more nearly saturated.
27-17
ϩ ϩ
H2SO4
3HNO3
glycerol glyceryl trinitrate
(nitroglycerine)
H2C
H2C
OH
HC OH HC ONO2 3H2O
H2C ONO2
H2C ONO2
OH
ϩ ϩCH3 OHϪ BrϪCH3CH
Br
CH3 CH3CH
OH
Simple inorganic esters may be
thought of as compounds that contain
one or more alkyl groups covalently
bonded to the anion of a ternary
inorganic acid. Unless indicated, the
term “ester” refers to organic esters.
1088 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
Interestingly, nitroglycerine (“nitro”) is
taken by some who have heart disease.
It acts as a vasodilator (dilates the
blood vessels) to decrease arterial
tension.
Nobel’s brother had been killed and
his father physically disabled in a
nitroglycerine explosion in 1864.
Nobel willed $9.2 million to establish
a fund for annual prizes in physics,
chemistry, medicine, literature, and
peace. The prizes were first awarded
in 1901.
The principal reactions of alkenes and alkynes are addition reactions rather than substitution
reactions. For example, contrast the reactions of ethane and ethylene with Cl2.
ethane: CH3XCH3 ϩ Cl2 88n CH3XCH2Cl ϩ HCl (substitution, slow)
ethylene: (addition, rapid)
Carbon–carbon double bonds are reaction sites and so represent functional groups. Most
addition reactions involving alkenes and alkynes proceed rapidly at room temperature. By
contrast, many substitution reactions of the alkanes require catalysts and high tempera-
tures.
Bromine adds readily to the alkenes to give dibromides. The reaction with ethylene is
The addition of Br2 to alkenes is used as a simple qualitative test for unsaturation.
Bromine, a dark red liquid, is dissolved in a nonpolar solvent. When an alkene is added,
the solution becomes colorless as the Br2 reacts with the alkene to form a colorless
compound. This reaction may be used to distinguish between alkanes and alkenes.
Hydrogenation is an extremely important addition reaction of the alkenes. Hydrogen
adds across double bonds at elevated temperatures, under high pressures, and in the presence
of an appropriate catalyst (finely divided Pt, Pd, or Ni).
catalyst
CH2UCH2 ϩ H2 888888n CH3XCH3
heat
Some unsaturated hydrocarbons are converted to saturated hydrocarbons in the manufacture
of high-octane gasoline and aviation fuels. Unsaturated vegetable oils can also be
converted to solid cooking fats (shortening) by hydrogenation of most of the carbon–
carbon double bonds present (Figure 27-21). Most of the double bonds that remain are
the trans- isomer.
The hydration reaction (addition of water) is another very important addition reaction
of alkenes. It is used commercially for the preparation of a wide variety of alcohols from
petroleum by-products. Ethanol, the most important industrial alcohol, is produced industrially
by the hydration of ethylene from petroleum, using H2SO4 as a catalyst.
H2SO4
H2CUCH2 ϩ H2O 888888n H3CXCH2OH
3H2
Ni catalyst
heat
olein (an oil, liquid) stearin (a fat, solid)
H2COC(CH2)7CH CH(CH2)7CH3
O
H2COC(CH2)16CH3
O
O
O
HCOC(CH2)16CH3
H2COC(CH2)16CH3
CH(CH2)7CH3HCOC(CH2)7CH
O
CH(CH2)7CH3H2COC(CH2)7CH
O
ϩH2C CH2 Br2 CH2
Br Br
CH2 1,2-dibromoethane
(ethylene dibromide)
ϩH2C CH2 Cl2 CH2
Cl Cl
CH2
Figure 27-21 Hydrogenation of
the olefinic double bonds in
vegetable oil converts it to solid fat.
27-17 Addition Reactions 1089
Think of H2O as HOH.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 11.4, Hydrocarbons and
Addition Reactions.
One of the commercially most important addition reactions of the alkenes forms polymers.
This reaction will be discussed in Section 27-19.
The alkynes contain two pi bonds, both of which are sources of electrons, and they are
more reactive than the alkenes. The most common reaction of the alkynes is addition
across the triple bond. The reactions with hydrogen and with bromine are typical.
Other unsaturated bonds can also undergo addition reactions. Probably the most
O
B
important example is the carbonyl group, XCX. Because of the availability of unshared
pairs of electrons on the oxygen atom, the products can undergo a wide variety of subsequent
reactions. For example, HCN adds to the CUO bond of acetone.
This reaction is a key early step in the production of the transparent plastic known as
Plexiglas or Lucite.
ELIMINATION REACTIONS
An elimination reaction involves a decrease in the number of atoms or groups
attached to carbon. The degree of unsaturation increases.
An elimination reaction, dehydrohalogenation, can occur for chloro-, bromo- and
iodoalkanes. In such a reaction, the halogen, X, from one C atom and a hydrogen from
an adjacent C atom are eliminated. A double bond between two carbon atoms is formed;
the molecule becomes more unsaturated. The net reaction is the transformation of an alkyl
halide (or haloalkane) into an alkene. Dehydrohalogenation reactions usually require a
strong base such as sodium hydroxide, NaOH.
HH
H H
C CH H NaOH H2O NaBr
H
H
C
Br
bromoethane ethene
(ethylene)
H
C ϩ ϩ ϩ
27-18
NaOH(aq)
O
CH3 CH3 CH3 CH3C
OH
CN
C HCNϩ
H H
HH
H H
C C C C
H2
H H
BrH
Br H
C C C C
Br2
1,2-dibromoethene
1090 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
A related reaction is dehydration, in which an alcohol is converted into an alkene and
water by the elimination of HX and XOH from adjacent carbon atoms. The dehydration
of an alcohol to form an alkene can be considered the reverse of the hydration of an
alkene to form an alcohol (Section 27-17). Dehydration reactions are catalyzed by acids.
Such simple elimination reactions are relatively rare. Elimination reactions, however,
frequently occur as individual steps in more complex reaction sequences. Many more elimination
reactions are encountered in courses in organic chemistry.
POLYMERIZATION REACTIONS
A polymer is a large molecule that is a high-molecular-weight chain of small molecules.
The small molecules that are linked to form polymers are called monomers. Typical polymers
consist of hundreds or thousands of monomer units and have molecular weights up
to thousands or millions.
Polymerization is the combination of many small molecules (monomers) to form
large molecules (polymers).
Polymers are divided into two classes: natural and synthetic. Important biological molecules
such as proteins, nucleic acids, and polysaccharides (starches and the cellulose in
wood and cotton) are natural polymers. Natural rubber and natural fibers such as silk and
wool are also natural polymers. Familiar examples of synthetic polymers include plastics
such as polyethylene, Teflon, and Lucite (Plexiglas) and synthetic fibers such as nylon,
Orlon, and Dacron. In this section we will describe some processes by which polymers
are formed from organic compounds.
27-19
Concentrated sulfuric acid is an
excellent dehydrating agent. Here it
removes water from sucrose, a sugar
with the formula C12H22O11.
Dehydration of sucrose produces
(mostly) carbon.
27-19 Polymerization Reactions 1091
The word fragment -mer means “part.”
Recall that isomers are compounds that
are composed of the same (iso) parts
(mers). A monomer is a “single part”; a
large number of monomers combine to
form a polymer, “many parts.”
Addition Polymerization
Polymerization is an important addition reaction (Section 27-17) of the alkenes. Polymers
formed by this kind of reaction are called addition polymers. The formation of polyethylene
is an important example. In the presence of appropriate catalysts (a mixture of
aluminum trialkyls, R3Al, and titanium tetrachloride, TiCl4), ethylene polymerizes into
chains containing 800 or more carbon atoms.
catalyst
nCH2UCH2 888888n X( CH2XCH2 )Xn
ethylene polyethylene
The polymer may be represented as CH3(CH2XCH2)nCH3, where n is approximately
400. Polyethylene is a tough, flexible plastic. It is widely used as an electrical insulator
and for the fabrication of such items as unbreakable refrigerator dishes, plastic cups, and
squeeze bottles. Polypropylene is made by polymerizing propylene, CH3XCHUCH2, in
much the same way. Teflon is made by polymerizing tetrafluoroethylene in a similar
reaction.
catalyst
nCF2UCF2 888888n X( CF2XCF2 )Xn
heat
tetrafluoroethylene Teflon
The molecular weight of Teflon is about 2 ϫ 106. Approximately 20,000 CF2UCF2 molecules
polymerize to form a single giant molecule. Teflon is a very useful polymer. It does
not react with concentrated acids and bases or with most oxidizing agents. It does not
dissolve in most organic solvents.
Natural rubber is obtained from the sap of the rubber tree, a sticky liquid called latex.
Rubber is a polymeric hydrocarbon formed in the sap by the combination of about 2000
molecules of 2-methyl-1,3-butadiene, commonly called isoprene. The molecular weight
of rubber is about 136,000.
When natural rubber is warmed, it flows and becomes sticky. To eliminate this problem,
vulcanization is used. This is a process in which sulfur is added to rubber and the mixture
is heated to approximately 140°C. Sulfur atoms combine with some of the double bonds
in the linear polymer molecules to form bridges that bond one rubber molecule to another.
This cross-linking by sulfur atoms converts the linear polymer into a three-dimensional
polymer. Fillers and reinforcing agents are added during the mixing process to increase
the durability of rubber and to form colored rubber. Carbon black is the most common
reinforcing agent. Zinc oxide, barium sulfate, titanium dioxide, and antimony(V) sulfide
are common fillers.
Some synthetic rubbers are superior to natural rubber in some ways. Neoprene is a
synthetic elastomer (an elastic polymer) with properties quite similar to those of natural
rubber. The basic structural unit is 2-chloro-1,3-butadiene, commonly called chloroprene,
which differs from isoprene in having a chlorine atom rather than a methyl group at carbon
2 of the 1,3-butadiene chain.
CH
chloroprene neoprene (a synthetic rubber)
nCH2 CH2
Cl
C
Cl
C(CH2 CH CH2)n
polymerization
isoprene natural rubber
2nCH2 CH2
CH3
C CH (CH2 CH2
CH3
C CH CH
CH3
CH2 CH2)C n
Teflon is a trade name owned by
DuPont, a company that has developed
and manufactured many fluorinated
polymers.
1092 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 11.9, Addition Polymerization.
Numerous other polymers are elastic
enough to be called by the generic
name “rubber.”
Many cooking utensils with
“nonstick” surfaces are coated with a
polymer such as Teflon.
Neoprene is less affected by gasoline and oil and is more elastic than natural rubber. It
resists abrasion well and is not swollen or dissolved by hydrocarbons. It is widely used to
make hoses for oil and gasoline, electrical insulation, and automobile and refrigerator
parts.
When two different monomers are mixed and then polymerized, copolymers are
formed. Depending on the ratio of the two monomers and the reaction conditions, the
order of the units can range from quite regular (e.g., alternating) to completely random.
In this way, polymers with a wide variety of properties can be produced. The most important
rubber produced in the largest amount in the United States is styrene butadiene
rubber (SBR), a polymer of styrene with butadiene in a 1Ϻ3 molecular ratio.
Dihydric alcohols contain two XOH
groups per molecule. They are often
called glycols.
27-19 Polymerization Reactions 1093
The double bonds in SBR can be cross-linked by vulcanization as described for natural
rubber. SBR is used primarily for making tires. Other copolymers are used to make car
bumpers, body and chassis parts, wire insulation, sporting goods, sealants, and caulking
compounds.
Some addition polymers and their uses are listed in Table 27-14.
Condensation Polymerization
Some polymerization reactions are based on condensation reactions, in which two molecules
combine by splitting out or eliminating a small molecule. For such a polymer to be formed,
each monomer must have two functional groups, one on each end. A polymer formed in
this way is called a condensation polymer. There are many useful condensation polymers,
based on a wide variety of bifunctional molecules.
Polyesters (short for “polymeric esters”) are condensation polymers that are formed
when dihydric alcohols react with dicarboxylic acids. An ester linkage is formed at each end
of each monomer molecule to build up large molecules. A useful polyester is prepared
from terephthalic acid and ethylene glycol.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 11.10, Condensation
Polymerization.
HH
H
C3 C
CH2 CH2
C
ϩ
H
H H
C C
HH
H
H H
C C polymerization
H H
H
CH2
H H
CH2)
CH2
CH2(CH2
n
HO OH HOC
O
H HOCH2CH2O
HOCH2CH2O
ϪH2O
n
C
O
C
O
C
O
C
O
HOCH2CH2OC
O
OH H HOCH2CH2O
terephthalic acid
polyethylene terephthalate (also known as PET)
(Dacron, Mylar)
terephthalic acidethylene glycol ethylene glycol
1094 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
TABLE 27-14 Some Important Addition Polymers
Polymer Production,
Polymer Name Tons/Year in
(some trade names) Some Uses United States Monomer Name Monomer Formula
polyethylene electrical insulation; .14 million ethylene
(Polythene) toys and molded
objects; bags;
squeeze bottles
polypropylene bottles; films; lab 6.7 million propylene
(Herculon, equipment, toys;
Vectra) packaging film;
filament for rope,
webbing,
carpeting; molded
auto and appliance
parts
polyvinyl chloride pipe, siding, gutters; 7.4 million vinyl chloride
(PVC) floor tiles
polyacrylonitrile acrylic fibers for 95,000 acrylonitrile
(Orlon, Acrilan) carpets, clothing,
knitwear
polystyrene molded toys, dishes, 3.3 million styrene
(Styrene, kitchen equipment;
Styrofoam, insulating foam,
Styron) e.g., ice chests;
rigid foam
packaging
poly(methyl meth- high-quality 507,000 methyl methacrylate
acrylate) transparent
(Plexiglas, objects; waterLucite)
based paints;
contact lenses
polybutadiene automotive tire tread, 610,000 1,3-butadiene
hoses, belts; metal
can coatings
ethylene-propylene appliance parts; auto 324,000 ethylene, propylene formulas given above
copolymer hoses, bumpers,
body and chassis
parts; coated
fabrics
SBR copolymer tires 1 million styrene, 1,3-butadiene formulas given above
H CH3
H
C C
O
C OCH3
H H
H
C C H
H H
C C
H H
HH
C C
H H
CH3H
C C
H H
ClH
C C
H H
CNH
C C
H H
H
C C
More than 2 million tons of this polymer is produced annually in the United States.
Dacron, the fiber produced from this polyester, accounts for approximately 50% of all
synthetic fibers. It absorbs very little moisture, and its properties are nearly the same
whether it is wet or dry. Additionally, it possesses exceptional elastic recovery properties
so it is used to make “permanent-press” fabrics. This polyester can also be made into films
of great strength (e.g., Mylar), which can be rolled into sheets 1/30 the thickness of a
human hair. Such films can be magnetically coated to make audio- and videotapes.
The polymeric amides, polyamides, are an especially important class of condensation
polymers. Nylon is the best known polyamide. It is prepared by heating anhydrous hexamethylenediamine
with anhydrous adipic acid, a dicarboxylic acid. This substance is often
called nylon 66 because the parent diamine and dicarboxylic acid each contain six carbon
atoms.
Polymers have a wide range of
properties and uses. Mylar sheet
polymer (another polyethylene
terephthalate) is used to protect
documents, such as this photocopy
of a later version of the Declaration
of Independence (top). Another kind
of polymer coating is used to make
shatterproof covers on glass items
such as fluorescent light bulbs
(bottom).
27-19 Polymerization Reactions 1095
Molten nylon is drawn into threads (Figure 27-22). After cooling to room temperature,
these can be stretched to about four times their original length. The “cold drawing”
process orients the polymer molecules so that their long axes are parallel to the fiber axis.
At regular intervals there are NXH,O hydrogen bonds that cross-link adjacent chains to
give strength to the fiber.
Petroleum is the ultimate source of both adipic acid and hexamethylenediamine. We do
not mean that these compounds are present in petroleum, only that they are made from
it. The same is true for many other industrial chemicals. The cost of petroleum is an important
factor in our economy because so many products are derived from petroleum.
adipic acid
hexamethylenediamine
(CH2)6H2N NH2
heat
ϪH2O
(CH2)4HO OH
O
C
O
C
ϩ nylon 66
(a polyamide)
NH NH NHC C Cn
O
(CH2)4 (CH2)6
OO
A patch made of Dacron polymer is used to close a defect in a human heart.
The molecular weight of the polymer
varies from about 10,000 to about
25,000. It melts at about 260–270°C.
Some types of natural condensation polymers play crucial roles in living systems.
Proteins (Section 28-9) are polymeric chains of L-amino acids linked by peptide bonds. A
peptide bond is formed by the elimination of a molecule of water between the amino
group of one amino acid and the carboxylic acid group of another.
Nylon is formed at the interface
where hexamethylenediamine (in the
lower water layer) and adipyl
chloride (a derivative of adipic acid,
in the upper hexane layer) react.
The nylon can be drawn out and
wound on a stirring rod. In industry,
adipic acid is used in place of adipyl
chloride.
1096 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
Figure 27-22 Fibers of synthetic polymers are made by extrusion of the molten material
through tiny holes, called spinnerets. After cooling, nylon fibers are stretched to about four
times their original length to orient the polymer molecules.
H
H
H
O
N NC C OH C C OH OH ϩϩ H2OH N C C N C CH
R H RЈ
H O
H R
H O
H RЈ
H O
peptide
bond
When this process is carried out repeatedly, a large molecule called a polypeptide is
formed.
H
H
H
O
N N NC C C C C C N C C
R H R
H O
H R
H O
H R
H O
N C C N C C
H R
H O
H R
H O
N C C OH
H R
H O
peptide bonds
Spinneret
Nylon fibers
Pump
Pressure
N2
285ЊC
Nylon chips
Proteins are discussed in greater detail in Section 28-9.
Key Terms 1097
Key Terms
Acid halide See Acyl halide.
Acyl group The group of atoms remaining after removal of an
XOH group of a carboxylic acid.
Acyl halide A compound derived from a carboxylic acid by
replacing the XOH group with a halogen (X), usually XCl;
general formula is also called acid halide.
Addition reaction A reaction in which two atoms or groups of
atoms are added to a molecule, one on each side of a double or
triple bond. The number of groups attached to carbon increases,
and the molecule becomes more nearly saturated.
Alcohol A hydrocarbon derivative in which an H attached to a
carbon atom not part of an aromatic ring has been replaced by
an XOH group.
Aldehyde A compound in which an alkyl or aryl group and
a hydrogen atom are attached to a carbonyl group; general
formula is
Aliphatic hydrocarbons Hydrocarbons that do not contain aromatic
rings.
Alkanes See Saturated hydrocarbons.
Alkenes Unsaturated hydrocarbons that contain a carbon–
carbon double bond.
Alkyl group A group of atoms derived from an alkane by the
removal of one hydrogen atom.
Alkylbenzene A compound containing an alkyl group bonded
to a benzene ring.
Alkynes Unsaturated hydrocarbons that contain a carbon–
carbon triple bond.
Amide A compound containing the group.
Amine A compound that can be considered a derivative of
ammonia, in which one or more hydrogens are replaced by alkyl
or aryl groups.
Amino acid A compound containing both an amino group and
a carboxylic acid group.
Amino group The XNH2 group.
Aromatic hydrocarbons Benzene and similar condensed ring
compounds; contain delocalized rings of electrons.
Aryl group The group of atoms remaining after a hydrogen atom
is removed from an aromatic system.
Carbonyl group The group.
Carboxylic acid A compound containing a group.
Condensation polymer A polymer that is formed by a condensation
reaction.
C O H
O
C
O
C N
O
R C
O
H; R may be H.
R C
O
X;
Condensation reaction A reaction in which a small molecule,
such as water or hydrogen chloride, is eliminated and two molecules
are joined.
Conjugated double bonds Double bonds that are separated
from each other by one single bond, as in CUCXCUC.
Constitutional isomers Compounds that contain the same
numbers of the same kinds of atoms but that differ in the order
in which their atoms are bonded together. Also known as structural
isomers.
Copolymer A polymer formed from two different compounds
(monomers).
Cycloalkanes Cyclic saturated hydrocarbons.
Dehydration The reaction in which HX and XOH are eliminated
from adjacent carbon atoms to form water and a more
unsaturated bond.
Dehydrohalogenation An elimination reaction in which a
hydrogen halide, HX (X ϭ Cl, Br, I), is eliminated from a
haloalkane. A CUC double bond is formed.
Elimination reaction A reaction in which the number of groups
attached to carbon decreases. The degree of unsaturation in the
molecule increases.
Ester A compound of the general formula
where R and RЈ may be the same or different and may be either
aliphatic or aromatic. R may be H. RЈ cannot be H.
Ether A compound in which an oxygen atom is bonded to two
alkyl or two aryl groups, or one alkyl and one aryl group.
Fat A solid triester of glycerol and (mostly) saturated fatty acids.
Fatty acid A long-chain aliphatic acid; many can be obtained
from animal fats.
Functional group A group of atoms that represents a potential
reaction site in an organic compound.
Geometric isomers Compounds with different arrangements of
groups on the opposite sides of a bond with restricted rotation,
such as a double bond; for example, cis–trans isomers of certain
alkenes.
Glyceride A triester of glycerol.
Heterocyclic amine An amine in which nitrogen is part of a
ring.
Homologous series A series of compounds in which each member
differs from the next by a specific number and kind of
atoms.
Hydration reaction A reaction in which the elements of water,
H and OH, add across a double or triple bond.
Hydrocarbon A compound that contains only carbon and
hydrogen.
Hydrogenation The reaction in which hydrogen adds across a
double or triple bond.
Isomers Different compounds that have the same molecular
formula.
C OR RЈ,
O
1098 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
Exercises
Basic Ideas
*001. (a) Give a brief definition of organic chemistry. (b) Can life,
as currently known, exist without organic compounds? (c)
What happened to the “vital force” theory?
*002. How is carbon’s tendency to bond to other carbon atoms
unique among the elements?
*003. (a) Why do many organic compounds have low melting
and boiling points? (b) Some organic compounds, like
table sugar, do not have a boiling point. Suggest a reason
why.
*004. (a) What are the principal sources of organic compounds?
(b) Some chemists argue that the ultimate source of all
natural occurring organic compounds is carbon dioxide.
Could this be possible? Hint: Think about the origins of
coal, natural gas, and petroleum.
*005. List 10 “everyday” uses of organic compounds. Choose
examples from at least five different classes of compounds.
Aliphatic Hydrocarbons
*006. (a) Define and compare the structures of alkanes, alkenes,
and alkynes. (b) Write the general formula of each of these
hydrocarbon families. (c) What is a saturated hydrocarbon?
*007. (a) What are hydrocarbons? (b) What are the cycloalkanes?
(c) Write the general formula for acyclic
alkanes.
*008. Describe the bonding in and the geometry of molecules
of the following alkanes: (a) methane; (b) ethane;
(c) propane; (d) butane. How are the formulas for these
compounds similar? How are they different?
*009. (a) What are straight-chain hydrocarbons? (b) What are
branched-chain hydrocarbons? (c) Cite three examples of
each.
*010. (a) What is a homologous series? (b) Provide specific
examples of compounds that are members of a homologous
series. (c) What is a methylene group? (d) How does
each member of a homologous series differ from compounds
that come before and after it in the series?
(e) Name three homologous series that are also aliphatic
hydrocarbons.
*011. (a) What are alkyl groups? (b) Draw structures for and
write the names of the first five straight-chain alkyl
groups. (c) What is the origin of the names for alkyl
groups?
*012. Could a substance with the molecular formula C3H8 be
a cycloalkane? Could C3H8 be a branched alkane having
Ketone A compound in which a carbonyl group is bound to two
alkyl or two aryl groups, or to one alkyl and one aryl group.
Oil A liquid triester of glycerol and unsaturated fatty acids.
Organic chemistry The chemistry of substances that contain
carbon–carbon or carbon–hydrogen bonds.
Peptide bond A bond formed by elimination of a molecule of
water between the amino group of one amino acid and the carboxylic
acid group of another.
Phenol A hydrocarbon derivative that contains an XOH group
bound to an aromatic ring.
Pi bond A chemical bond formed by the side-to-side overlap of
atomic orbitals.
Polyamide A polymeric amide.
Polyene A compound that contains more than one double bond
per molecule.
Polyester A polymeric ester.
Polyhydric alcohol An alcohol that contains more than one
XOH group.
Polymerization The combination of many small molecules
(monomers) to form large molecules (polymers).
Polymers Large molecules formed by the combination of many
small molecules (monomers).
Polypeptide A polymer composed of amino acids linked by
peptide bonds.
Primary alcohol An alcohol with no or one R group bonded to
the carbon bearing the XOH group.
Primary amine An amine in which one H atom of ammonia has
been replaced by an organic group.
Protein A naturally occurring polymeric chain of L-amino acids
linked together by peptide bonds.
Saturated hydrocarbons Hydrocarbons that contain only single
bonds. They are also called alkanes or paraffin hydrocarbons.
Secondary alcohol An alcohol with two R groups bonded to the
carbon bearing the XOH group.
Secondary amine An amine in which two H atoms of ammonia
have been replaced by organic groups.
Sigma bond A chemical bond formed by the end-to-end overlap
of atomic orbitals.
Structural isomers See Constitutional isomers.
Substitution reaction A reaction in which an atom or a group
of atoms attached to a carbon atom is replaced by another atom
or group of atoms. No change occurs in the degree of saturation
at the reactive carbon.
Tertiary alcohol An alcohol with three R groups bonded to the
carbon bearing the XOH group.
Tertiary amine An amine in which three H atoms of ammonia
have been replaced by organic groups.
Unsaturated hydrocarbons Hydrocarbons that contain double
or triple carbon–carbon bonds.
Vulcanization The process in which sulfur is added to rubber
and heated to 140°C, to cross-link the linear rubber polymer
into a three-dimensional polymer.
Exercises 1099
a methyl group as a substituent attached to the longest
chain?
*013. Name the following compound by the IUPAC system:
CH3CH(CH3)CH2CH3. Draw a constitutional isomer of
this compound, and give its correct IUPAC name.
*014. Give the IUPAC name of each of the following
compounds.
(a) CH3CH2CH2CH(CH3)CH2CH3
(b) CH3CH2CH(CH2CH3)CH2CH3
(c) CH3CH(CH2CH2CH3)CH2CH3
(d) CH3CH2CH2CH(CH3)2
(e) CH3CH(CH3)CH2CH2CH3
*015. Write structural formulas for the isomeric saturated
hydrocarbons having the molecular formula C6H14.
Which one would have the highest boiling point?
*016. Write structural formulas for four constitutional isomers
of C7H14 that contain a cyclopropyl substituent. Name
each by the IUPAC system.
*017. Write the structural formula for 2,2-dimethylpropane.
*018. Write the IUPAC name for each of the following.
*019. Write structural formulas for 1,1,2-trimethylcyclohexane,
isopropylcyclobutane, and sec-butylcyclohexane.
*020. (a) How does the general formula for the alkenes differ
from the general formula for the alkanes? (b) Why are
the general formulas identical for alkenes and cycloalkanes
that contain the same number of carbon atoms?
*021. (a) What are cycloalkenes? (b) What is their general formula?
(c) Provide the structural formulas and names of
three examples.
*022. Describe the bonding at each carbon atom in (a) ethene,
(b) propene, (c) 1-butene, and (d) 2-methyl-2-butene.
*023. (a) What are geometric isomers? (b) Why is rotation
around a double bond not possible at room temperature?
(c) What do cis and trans mean? (d) Draw structures for
cis- and trans-3-methyl-3-hexene. How should their melting
and boiling points compare?
*024. How do carbon–carbon single bond lengths and carbon–
carbon double bond lengths compare? Why?
*025. (a) What are alkynes? (b) What other name is used to
describe them? (c) What is the general formula for
alkynes? (d) How does the general formula for alkynes
compare with the general formula for cycloalkenes?
Why?
*026. Describe the bonding and geometry associated with the
triple bond of alkynes.
*027. Write the structural formulas of the following
compounds: (a) 1-butyne; (b) 2-methylpropene;
(c) 2-ethyl-3-methyl-1-butene; (d) 3-methyl-1-butyne.
CH3
CH3
CH3
CH3
CH3
CH3
(a) (b) (c)
*028. Write the structural formulas of the following
compounds: (a) 3-hexyne; (b) 1,3-pentadiene;
(c) 3,3-dimethylcyclobutene; (d) 3,4-diethyl-1-
hexyne.
*029. Write the IUPAC names for the following compounds.
*030. Repeat Exercise 29 for
(c) CH3CmCCH3
*031. Each of the hydrocarbon families has members that
are cyclic hydrocarbons. (a) What is a cyclic hydrocarbon?
(b) Write the structural formula of one example
of each of the following: (i) cycloalkene; (ii) cycloalkane;
and (iii) cycloalkyne.
Aromatic Hydrocarbons
*032. (a) What distinguishes an aromatic compound from other
cyclic compounds? (b) What is the principal source of
aromatic hydrocarbons?
*033. (a) The structure for benzene is sometimes drawn as a
hexagon enclosing a circle; what is the meaning of the
circle? (b) Write the structural formulas of the three most
common aromatic hydrocarbons.
*034. (a) What is a phenyl group? (b) How many isomeric
monophenylnaphthalenes are possible?
*035. There are only three isomeric trimethylbenzenes. Write
their structural formulas, and name them.
*036. (a) How many isomeric dibromobenzenes are possible?
(b) What names are used to designate these isomers?
*037. Write the structural formulas for the following
compounds: (a) p-difluorobenzene; (b) ethylbenzene;
(c) 1,3,5-tribromobenzene; (d) 1,3-diphenylbutane;
(e) p-chlorotoluene.
CH3CH2CHCH2CH3(f)
CH2CH2CH3
CH3CH2CHCH3(e)
CH2CH3
CH3CHCH2CH3(b)
CH3
CH3
CH3C(d) CHCH3
CH3
(a)
(d)CH2(b) C(CH3)2
CH3
CH3CHCHCH2CH3(c)
CH3
CH3
CH3CCH2CH2CH3
CH3
(a)
1100 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
*038. Write the IUPAC names for the following compounds.
Alkyl and Aryl Halides
*039. (a) Write the general representation for the formula of an
alkyl halide. (b) How does this differ from the representation
for the formula of an aryl halide?
*040. Write the IUPAC names for the following halides.
(c) CH2ClCH2Cl
*041. Write the structural formulas for the following:
(a) 2,2-dichloropentane; (b) 3-bromo-1-butene;
(c) 1,2-dichloro-2-fluoropropane; (d) p-dichloro-
benzene.
*042. Name the following.
*043. The compound 1,2-dibromo-3-chloropropane (DBCP)
was used as a pesticide in the 1970s. Recently, agricultural
workers have claimed that exposure to DBCP made
them sterile. Write the structural formula for this com-
pound.
(d)
Cl
Br
Cl
Br
Cl
(b)
CH3
Cl
(c)
I
Br
Br
I
(a)
Cl
Cl
Cl
Cl H
C CCl(d) Cl
CH3
CHCH3(b) CH2Cl
(a) CH2CH2Cl
(d)
CH3
CH3
CH3
CH3
H3C(b)
CH3
CH3
H3C
(c)
CH2CH3
CH2CH3
CH3
H3C
(a)
CH3
CH2CH3
Alcohols and Phenols
*044. (a) What are alcohols and phenols? (b) How do they
differ? (c) Why can alcohols and phenols be viewed as
derivatives of hydrocarbons? as derivatives of water?
*045. (a) Distinguish among primary, secondary, and tertiary
alcohols. (b) Write names and formulas for three alcohols
of each type.
*046. Write structural formulas for and write the IUPAC names
of the eight (saturated) alcohols that contain five carbon
atoms and one XOH group per molecule. Which ones
may be classified as primary alcohols? secondary alcohols?
tertiary alcohols?
*047. (a) What are glycols? (b) Write the structural formula of
three examples. (c) Why are glycols more soluble in water
than monohydric alcohols that contain the same number
of carbon atoms?
*048. Refer to Table 27-8, and explain the trends in boiling
points and solubilities of alcohols in water.
*049. Why are methyl alcohol and ethyl alcohol called wood
alcohol and grain alcohol, respectively?
*050. Why are most phenols only slightly soluble in water?
*051. Write the structural formula for each of the following
compounds: (a) 2-butanol; (b) cyclohexanol;
(c) 1,4-butanediol.
*052. Give the IUPAC name for each of the following com-
pounds.
*053. Which of the following compounds are phenols? Give the
IUPAC name for each compound.
*054. Write the structural formulas for the following:
(a) p-iodophenol; (b) 4-nitrophenol (the nitro group is
XNO2); (c) m-nitrophenol.
(e) OH(c)
OH
OH
(d)
OH
(b) OH
CH3
CH3(d) C OH
CH3
CH3
CH3(b) C CH2CH2OH
CH3
CH3
CH2 CH2(c) CH2
OH OH
CH3CH2CHCH2OH(a)
CH3
(a) CH2CH2OH
Exercises 1101
Ethers
*055. Distinguish among aliphatic ethers, aromatic ethers, and
mixed ethers.
*056. Briefly describe the bonding around the oxygen atom in
dimethyl ether. What intermolecular forces are found in
this ether?
*057. What determines whether an ether is “symmetrical” or
“asymmetrical”?
0*58. Write the structural formulas for the following: (a) methoxymethane;
(b) 2-ethoxypropane; (c) 1,3-dimethoxybutane;
(d) ethoxybenzene; (e) methoxycyclobutane.
*059. Give the IUPAC name for the following ethers.
(a) CH3XOXCH2CH2CH3
Amines
*060. (a) What are amines? (b) Why are amines described as
derivatives of ammonia?
*061. Write the general representation for the formula for a
compound that is (a) a primary amine, (b) a secondary
amine, (c) a tertiary amine. Is (CH3)2CHNH2 a secondary
amine? Give a reason for your answer.
*062. Name the following amines.
*063. The stench of decaying proteins is due in part to two
compounds whose structures and common names are
CH3CH2CH2CH2 CH2CH2CH2CH3(d) N
CH2CH2CH2CH3
(c) NHCH3
(b) NH2O2N
(The —NO2 substituent is called “nitro-.”)
CH3(a) CH2
CH3 CH2
NH
(d) CH3O
(c) CH2CH3O
CH3 CH3CH(b) O
CH3
H2NCH2CH2CH2CH2NH2 putrescine
H2NCH2CH2CH2CH2CH2NH2 cadaverine
Name these compounds as amino-substituted alkanes.
Aldehydes and Ketones
*064. (a) List several naturally occurring aldehydes and ketones.
(b) What are their sources? (c) What are some uses of
these compounds?
*065. (a) Distinguish between aldehydes and ketones. (b) Cite
three examples (each) of aliphatic and aromatic aldehydes
and ketones by drawing structural formulas, and give
the IUPAC names of the compounds.
0*66. Write the structural formulas for the following:
(a) 2-methylbutanal; (b) propanal; (c) o-methoxybenzaldehyde;
(d) butanone; (e) bromopropanone;
(f ) 3-heptanone.
0*67. Name the following compounds.
Carboxylic Acids and Their Derivatives
*068. (a) What are carboxylic acids? (b) Write structural formulas
for and write the IUPAC names of five carboxylic
acids.
*069. (a) Why are aliphatic carboxylic acids sometimes called
fatty acids? Cite two examples.
*070. (a) What are acyl chlorides, or acid chlorides? (b) Write
structural formulas for four acid chlorides, and give their
IUPAC names.
*071. (a) What are esters? (b) Write structural formulas for four
esters, and give their IUPAC names.
*072. Write the structural formulas for the following:
(a) methylpropanoic acid; (b) 3-bromobutanoic acid;
(c) p-nitrobenzoic acid; (d) potassium benzoate; (e) 2aminopropanoic
acid.
*073. (a) What are fats? What are oils? (b) Write the general
formulas for fats and oils.
*074. (a) What are glycerides? Distinguish between simple glycerides
and mixed glycerides. (b) Write the IUPAC names
and structural formulas for three simple glycerides.
*075. Write the IUPAC names and structural formulas for some
acids that occur in fats and oils (as esters).
*076. What are waxes?
(d) C CH2 CH3
O
(b) O
H C CH2 CH(c)
Br
Br O
CH3CHCH2CH(a)
CH3
O
1102 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
*077. Give the IUPAC names for the following esters.
*078. Name the following esters.
Reactions of Organic Molecules
*079. (a) When an addition reaction occurs, the molecule
becomes more nearly saturated. Explain. (b) When an
elimination reaction occurs, the molecule becomes more
unsaturated. Explain.
*080. (a) What is a substitution reaction? (b) What is a halogenation
reaction? (c) What is an addition reaction?
*081. (a) Describe the reaction of methane with chlorine
in ultraviolet light. (b) Write equations that show structural
formulas for all compounds that can be formed by
reaction (a). (c) Name all compounds in these equations.
(d) Why are the halogenation reactions of the larger alkanes
of limited value?
*082. (a) Describe the reaction of ethane with chlorine in ultraviolet
light. (b) Write equations that show structural
formulas for all compounds that can be formed by reaction
(a). (c) Give the IUPAC names for all compounds in
these equations.
*083. Which of the following compounds could undergo
addition reactions? (a) propane; (b) 1,3-butadiene;
(c) cyclopentene; (d) acetone.
*084. Most reactions of the alkanes that do not disrupt the carbon
skeleton are substitution reactions, whereas the
alkenes are characterized by addition to the double bond.
What does this statement mean?
*085. How can bromination be used to distinguish between
alkenes and alkanes?
*086. Write equations for two reactions in which alkenes
undergo addition reactions with halogens. Give the
IUPAC names for all compounds.
*087. When a substitution reaction occurs, there is no change
in saturation. Explain.
*088. (a) What is hydrogenation? (b) Why is it important?
(c) Write equations for two reactions that involve hydrogenation
of alkenes. (d) Give the IUPAC names for all
compounds in part (c).
CH3CH2CH2COCH2CH3(b)
O
(a) C O
O
CH3C OCH2CH3(a)
O
CH3COCH3(b)
O
*089. Describe two qualitative tests that can be used to distinguish
between alkenes and alkanes. (b) Cite some specific
examples. (c) What difference in reactivity is the basis for
the qualitative distinction between alkanes and alkenes?
*090. (a) Why are alkynes more reactive than alkenes? (b) What
is the most common kind of reaction that alkynes
undergo? (c) Write equations for three such reactions.
(d) Give the IUPAC names for all compounds in part (c).
*091. (a) What is the most common kind of reaction that the
benzene ring undergoes? (b) Write equations for the reaction
of benzene with chlorine in the presence of an iron
catalyst and for the analogous reaction with bromine.
*092. Write equations to illustrate both aromatic and aliphatic
substitution reactions of toluene using (a) chlorine, (b)
bromine, and (c) nitric acid.
*093. Classify each reaction as substitution, addition, or elimi-
nation.
(a) CH3CH2Br ϩ CNϪ 88n CH3CH2CN ϩ BrϪ
(b) CH3CHCH2Br ϩ Zn 88n
A
Br CH3CHUCH2 ϩ ZnBr2
H2SO4
(c) C6H6 ϩ HNO3 888888n C6H5NO2 ϩ H2O
*094. Describe a simple test to distinguish between the two isomers
2-pentene and cyclopentane.
Polymers
*095. (a) What is a polymer? (b) What is the term for the smaller
molecule that serves as the repeating unit making
up a polymer? (c) What are typical molecular weights of
polymers?
*096. (a) What is polymerization? (b) Write equations for three
polymerization reactions.
*097. (a) Give an example of a condensation reaction. (b) What
is the essential feature of monomers used in condensation
polymerizations?
0*98. The examples of condensation polymers given in the text
are all copolymers, that is, they contain two different
monomers. Is it possible for a single monomer to polymerize
so as to form a condensation homopolymer? If so,
suggest an example. If not, explain why not.
*099. (a) What is a monomer? (b) Name three polymers commonly
found in a classroom and their use or function.
*100. (a) What is a copolymer? (b) What is a condensation polymer?
(c) Name three polymers that are addition polymers.
*101. Poly(vinyl alcohol) has a relatively high melting point,
258°C. How would you explain this behavior? A segment
of the polymer is
CH2CHCH2CHCH2CHCH2CHCH2CH
OH OH OH OH OH
. . . . . .
Exercises 1103
*102. What changes could be made in the structures of polymer
molecules that would increase the rigidity of a
polymer and raise its melting point?
*103. Methyl vinyl ketone, can be polymerized
by addition polymerization. The addition
reaction involves only the CUC bond. Write the
structural formula of a four-unit segment of this
polymer.
*104. (a) What is rubber? (b) What is vulcanization? (c) What
is the purpose of vulcanizing rubber? (d) What are fillers
and reinforcing agents? (e) What is their purpose?
*105. (a) What is an elastomer? (b) Cite a specific example.
(c) What are some of the advantages of neoprene compared
with natural rubber?
*106. (a) What are polyamides? (b) What kind of reaction forms
polyamides?
*107. (a) What are polyesters? (b) What is Dacron? (c) How
is Dacron prepared? (d) What is Mylar? (e) Is it reasonable
to assume that a polyester can be made from
propylene glycol and terephthalic acid? If so, sketch its
structure.
*108. Suppose the following glycol is used with terephthalic
acid to form a polyester. Write the structure of the polymer,
showing two repeating units.
*109. Write the structural formula of the monomer used in
preparation of each of the following polymers.
*110. Is it possible to produce a copolymer by addition polymerization?
If so, give an example. If not, explain why
not.
*111. (a) What is nylon? (b) How is it prepared?
*112. Common nylon is called nylon 66. (a) What does
this mean? (b) Write formulas for two other possible
nylons.
C
CH3
CH2 C
CH3
CH3 CH3
CH2 CH2C
CH3
CH3
(b) . . . . . .
C
CH3
CH2 C
CH3
CH2 CH2C
CH3
(a) . . .. . .
CH2OHHOCH2
CH3CCH CH2,
O
Mixed Exercises
*113. Identify the class of organic compounds (ester, ether,
ketone, and so on) to which each of the following belongs.
*114. Identify the class of organic compounds (ester, ether,
ketone, etc.) to which each of the following belongs.
*115. Identify and name the functional groups in each of the
following.
*116. Identify and name the functional groups in the
following.
(a) CH3CH2 C CH2 C CH2CH3
O O
(d) CH2CH3N
CH3
(c)
O C
OH
NO2
CH2CH2OH
(b) OCH2 CH C CH2CH2CH2CH3
O
(a) C OHHO
O
C
O
(c) CH2CH2CCl
O
(b) CH2O CH
O
(a) COH
O
CH3CCH2
O
(d)O(b)
CH2CH2OH(a) OH(c)
(d) CNH2
O
1104 CHAPTER 27: Organic Chemistry I: Formulas, Names, and Properties
(c) dioxane (also known as 1,4-dioxin)
(d) morphine
(e) epinephrine (adrenaline)
*117. Identify and name the functional groups in each of the
following.
(a) morpholine
(b) citric acid
(c) coniine (from the hemlock plant; the poison that
Socrates drank)
N H
CH2CH2CH3
C OH
O
H2C
CC OH
O
HO
C OH
O
H2C
N HO
CH CH3CH2NHO
HO
OH H
HO OHO
N
CH3
OO
CHCH2 CH3
OCH3
HO N
OH H
(b)
(d) glucose (a simple sugar, also known as dextrose)
(e) vitamin C (also called ascorbic acid)
*118. Name the following compounds.
(a) CH3CH(CH3)CH2CH2OH
(f ) (CH3CH2)3N
*119. Write the structural formulas for the following compounds:
(a) p-bromotoluene; (b) cyclohexanol; (c)
2-methoxy-3-methylbutane; (d) diethylamine; (e) ochlorophenol;
(f ) 1,4-butanediol.
*120. Give the IUPAC names for the following
compounds.
(b) CH3CH2CH2CH2NH2
CH3C
H3C O
COH
CH3
(f )CH3CH2CH2CH2CH
O
(c)
CH3CH2CHCH3
OCH3
(e)
(d) OCH3CH2CHCH2OH
CH3
(a)
NH2
Br Br
Br
(h)CCH3 CH2
Cl
(d)
O(g)CHCH3 CH3
NH2
(c)
(b)
OH
CH3
BrBr(e)
C
H
OH
OH
OO
HO
HOCH2
CCCCCCH
H H HOH OH
H
O
HH OH OH OH
Exercises 1105
CONCEPTUAL EXERCISES
*121. Suggest why the C6 ring in cyclohexane is not planar.
*122. Can an aromatic compound also be a saturated com-
pound?
BUILDING YOUR KNOWLEDGE
*123. (a) How do the melting points and boiling points of the
alkanes vary with molecular weight? (b) Do you expect
them to vary in this order? (c) Why or why not? Use
intermolecular forces to explain your answer.
*124. Write a Lewis formula for butanone.
*125. (a) What are resonance structures? (b) Draw resonance
structures for benzene. (c) What do we mean when we
say that the electrons associated with the ␲ bonds in benzene
are delocalized over the entire ring?
*126. Lidocaine has replaced Novocain (procaine) as the
favored anesthetic in dentistry. What functional group do
the two compounds have in common?
H2N
O
COCH2CH2N(CH2CH3)2
Novocain
NH
CH3 O
CH3
CCH2N(CH2CH3)2
lidocaine
28 Organic Chemistry II:
Shapes, Selected
Reactions, and
Biopolymers
A
OUTLINE
Shapes of Organic Molecules
28-1 Constitutional Isomers
28-2 Stereoisomers
28-3 Conformations
Selected Reactions
28-4 Reactions of Brønsted–Lowry
Acids and Bases
28-5 Oxidation–Reduction Reactions
28-6 Formation of Carboxylic
Acid Derivatives
28-7 Hydrolysis of Esters
Biopolymers
28-8 Carbohydrates
28-9 Polypeptides and Proteins
28-10 Nucleic Acids
OBJECTIVES
After you have studied this chapter, you should be able to
• Describe the differences between constitutional isomers and stereoisomers
• Distinguish between isomers and conformations
• Describe some common organic acids and bases and their relative strengths
• Recall some reactions that involve oxidation and reduction of organic compounds
• Recall reactions that form carboxylic acid derivatives
• Recall reactions in which esters and amides are formed
• Write equations in which carbonyl groups are used to produce other functional groups
• Write equations showing the hydrolysis of esters (saponification)
• Describe the structure of carbohydrates
• Describe the structure of proteins
• Describe the structure of nucleic acids
SHAPES OF ORGANIC MOLECULES
A
s we learned in Chapters 25 and 27, the chemical and physical properties of a
substance depend on the arrangements, as well as the identities, of its atoms.
Isomers are substances that have the same number and kind of atoms—that is, the
same molecular formula—but with different structural formulas. Because their structures
are different, isomers are different substances and have different properties.
Capsaicin, the substance responsible
for the “hot” in peppers, has phenol,
ester, and amide functional groups.
In addition to acting as an
inflammatory agent, it can act as a
pain reliever, presumably by
blocking nerve endings involved in
the transmission of pain.
Isomers can be broadly divided into two major classes: constitutional isomers and
stereoisomers. In Chapter 25 we discussed isomerism in coordination compounds, and in
Chapter 27 we learned about some isomeric organic compounds. In this chapter we will
take a more systematic look at some three-dimensional aspects of organic structures—a
subject known as stereochemistry (“spatial chemistry”).
CONSTITUTIONAL ISOMERS
Constitutional (or structural) isomers differ in the order in which their atoms are
bonded together.
In our studies of hydrocarbons in Sections 27-1 through 27-7, we saw some examples of
constitutional isomerism. Recall that there are three isomers of C5H12.
28-1
1108 CHAPTER 28: Organic Chemistry II: Shapes, Selected Reactions, and Biopolymers
H
H H
H H
C C
H H
H H
C C
H
H
C H H H
H H
H H
C C
H
H
C
CH3
H
C H H
H
H
C
H
H
C
CH3
CH3
C
pentane, C5H12 methylbutane
isopentane, C5H12
dimethylpropane
neopentane, C5H12
CH2UCHCHUCHCH2CH3 CH2UCHCH2CHUCHCH3 CH2UCHCH2CH2CHUCH2
1,3-hexadiene, C6H10 1,4-hexadiene, C6H10 1,5-hexadiene, C6H10
H
H H
OH OH
C C
H
H
C H
1,2-propanediol, C3H8O2
H
H H
OH OH
C C
H
H
C H
1,3-propanediol, C3H8O2
Cl
Cl
Cl
Cl
o-dichlorobenzene, C6H4Cl2 m-dichlorobenzene, C6H4Cl2
Cl
Cl
p-dichlorobenzene, C6H4Cl2
These three isomers differ in the lengths of their base chains but not in the functional
groups present (i.e., only alkyl groups are present in this case). As a result, they differ
somewhat in their melting and boiling points but differ only very slightly in the reactions
they undergo.
In one kind of constitutional isomerism, the compounds have the same number and
kind of functional groups on the same base chain or the same ring but in different positions.
Isomers that vary only in the location of groups of atoms usually have very similar chemical
and physical properties differing mainly in physical properties such as melting and
boiling points. The following groups of isomers are examples in which the type of functional
group does not change.
Sometimes the different order of arrangements of atoms results in different functional
groups. Some examples of this type of constitutional isomerism follow.
An alcohol and an ether:
An aldehyde and a ketone:
STEREOISOMERS
In stereoisomers the atoms are linked together in the same atom-to-atom order,
but their arrangements in space are different.
There are two types of stereoisomers: geometric and optical.
Geometric Isomers
Geometric isomers (or cis–trans isomers) differ only in the spatial orientation of groups
about a plane or direction. Two geometric isomers have the same molecular formula, the
same functional groups, the same base chain or ring, and the same order of attachment
of atoms; they differ in orientation either (1) around a double bond or (2) across the ring
in a cyclic compound. If the larger groups are on opposite sides of the ring or the double
bond, the designation trans appears in the name; if they are on the same side, the designation
is cis. We learned in Section 27-3 about the geometric isomerism associated with
the double bond in alkenes such as the 1,2-dichloroethenes (Figure 27-9). Similarly, two
or more substituents can be either on the same side or on opposite sides of the ring, as
shown in Figures 28-1 and 28-2. This kind of isomerism is possible when substituents
have replaced an H from a XCH2X unit in a ring. Because substituents on an aromatic
ring are bonded in the plane of the ring, such substitutions do not lead to geometric
isomerism.
28-2
H
H
C C
H
H
C
H
H
C
H
H
butanal, C4H8O
(butyraldehyde)
H
H
C
H
H
H
C
H H
C
O
C H
butanone, C4H8O
(methyl ethyl ketone)
O
H
H H
C C
HH
OH
ethanol, C2H6O
H
H
C O
H
H H
C H
methoxymethane, C2H6O
(dimethyl ether)
28-2 Stereoisomers 1109
Figure 28-1 Models of (a) cis-dichlorocyclopropane and (b) trans-dichlorocyclopropane.
(a)
(b)
Figure 28-2 Models of (a) cis-1chloro-3-methylcyclopentane,
(b)
trans-1-chloro-3-methylcyclopentane, and
(c) cis-1-chloro-3-methylcyclopentane
(stereoview).
1110 CHAPTER 28: Organic Chemistry II: Shapes, Selected Reactions, and Biopolymers
Optical Isomers
Many objects are mirror images of each other and cannot be superimposed. Your two
hands are a familiar example of this; each hand is a nonsuperimposable mirror image of
the other (Figure 28-3). “Superimposable” means that if one object is placed over the
other, the positions of all parts will match.
An object that is not superimposable with its mirror image is said to be chiral (from
the Greek word cheir, meaning “hand”); an object that is superimposable with its mirror
image is said to be achiral. Examples of familiar objects that are chiral are a screw, a
propeller, a foot, an ear, and a spiral staircase; examples of common objects that are achiral
are a plain cup with no decoration, a pair of eyeglasses, and a sock.
Optical isomers that are nonsuperimposable mirror images of each other (chiral)
are called enantiomers of one another. Enantiomers can exist in two forms that
bear the same relationship to each other as do left and right hands.
As an example of this, consider first the two models of bromochloromethane, CH2BrCl,
shown in Figure 28-4. They are mirror images of each other, and they can be superimposed.
Thus, this molecule is achiral and is not capable of optical isomerism. Now consider
(b)(a)
(c)
We speak of a screw or a propeller as
being “right-handed” or “left-handed.”
bromochloroiodomethane, CHBrClI (Figure 28-5). This molecule is not superimposable
with its mirror image, so it is chiral, and the two forms are said to be enantiomers of each
other. Any compound that contains four different atoms or groups bonded to the same
carbon atom is chiral; that is, it exhibits optical isomerism. Such a carbon is said to be
asymmetric (meaning “without symmetry”). Most simple chiral molecules contain at least
one asymmetric carbon atom, although there are other ways in which molecular chirality
can occur.
Stereoisomers have the same type and number of atoms, connected in the same order,
but arranged differently in space. Optical isomers (enantiomers) and geometric isomers
are subgroups of stereoisomers. They differ, however, in that geometric isomers have
different physical and chemical properties, whereas optical isomers have physical properties
that are identical (e.g., melting point, boiling point, and density). Optical isomers also
undergo the same chemical reactions, except when they interact with other chiral
compounds. Consequently, their properties in biological systems may be very different.
They also often exhibit different solubilities in solvents that are composed of chiral
molecules.
Figure 28-3 Mirror images. Place your left hand in front of a mirror; you will observe that
it looks like your right hand. We say that the two hands are mirror images of each other;
each hand is in every way the “reverse” of the other. Now try placing one hand directly over
the other; they are not identical. Hence, they are nonsuperimposable mirror images. Each
hand is a chiral object.
28-2 Stereoisomers 1111
Isomers
Stereoisomers
Optical
(enantiomers)
(D- or L-)
Constitutional
(structural)
Geometric
(cis- or trans-)
Mirror plane
Figure 28-4 Models of two mirror-image forms of bromochloromethane, CH2BrCl. The
two models are the same (superimposable), so they are achiral. CH2BrCl does not exhibit
optical isomerism.
Figure 28-5 (a, b) Models of the
two mirror-image forms of
bromochloroiodomethane,
CHBrClI. (Colors: Cl, green; Br, red;
I, purple.) (c) The same model as in
(a), turned so that H and I point the
same as in (b); however, the Br and
Cl atoms are not in the same
positions in (b) and (c). The two
models in (a) and (b) cannot be
superimposed on each other no
matter how we rotate them, so they
are chiral. These two forms of
CHBrClI represent different
compounds that are optical isomers of
each other.
1112 CHAPTER 28: Organic Chemistry II: Shapes, Selected Reactions, and Biopolymers
Optical isomers also differ from each other in one notable physical property: They
interact with polarized light in different ways. The main features of this subject were
presented in Chapter 25. Separate equimolar solutions of two optical isomers rotate a
plane of polarized light (Figures 25-4 and 25-5) by equal amounts but in opposite directions.
One of the optical isomers is designated as the D-isomer and its nonsuperimposable
mirror image is designated as the L-isomer. The phenomenon in which a plane of polarized
light is rotated by samples of either isomer is called optical activity. It can be measured
with a polarimeter (Figure 25-5). A racemic mixture is a single sample containing equal
amounts of the two optical isomers of a compound. Such a solution does not rotate a plane
of polarized light because the equal and opposite effects of the two isomers exactly cancel.
The isomers must be separated from each other to exhibit optical activity.
One very important way in which optical isomers differ chemically from one another
is in their biological activities. ␣-Amino acids have the general structure
where R represents any of a number of common substituents. The central carbon atom
has four atoms or groups bonded to it. In ␣-amino acids the four groups are all different,
so each amino acid can exist as two optical isomers. The exception is glycine, in which
H2N C
H
R
COOH
Stereoview of L-phenylalanine.
(a) (c)(b)
R ϭ H. Figure 28-6 shows this mirror-image relationship for optical isomers of phenylalanine,
in which R ϭ XCH2C6H5. All naturally occurring phenylalanine in living systems
is the L-form. In fact, only one isomer of each of the various optically active amino acids
is found in proteins.
CONFORMATIONS
The conformations of a compound differ from one another in the extent of rotation about
one or more single bonds. The CXC single bond length, 1.54 Å, is relatively independent
of the structure of the rest of the molecule.
Rotation about a single CXC bond is possible; in fact, at room temperature it occurs
very rapidly. In drawings and models, like those in Figure 28-7, it might at first appear
that rotation about a single bond would be unrestricted. The electrons of the groups
28-3
As we saw in Section 27-3, rotation
does not occur around carbon–carbon
double bonds at room temperature.
28-3 Conformations 1113
H
C
H H
H
C
H H
H
C
H H
C
H H
H
Staggered
conformation
Eclipsed
conformation
(a) (b)
Figure 28-7 Two possible conformations of ethane. (a) Staggered. (b) Eclipsed. Rotation of
one CH3 group about the CXC single bond, as shown by the curved arrows, converts one
conformation to the other.
Figure 28-6 Models of the two optical isomers of phenylalanine. The naturally occurring
phenylalanine in all living systems is the form shown on the left.
Figure 28-8 Two staggered
conformations of butane, C4H10.
1114 CHAPTER 28: Organic Chemistry II: Shapes, Selected Reactions, and Biopolymers
attached to the carbon atoms do repel one another somewhat, however. At room temperature
ethane molecules possess sufficient energy to cause rapid rotation about the single
carbon–carbon bond from one conformation to another. The staggered conformation of
ethane is slightly more stable (of lower energy) than the eclipsed conformation (see Figure
28-7); in the eclipsed conformation there is a slightly stronger repulsion between H atoms
on adjacent C atoms.
Consider the two conformations of butane shown in Figure 28-8. Again, staggered
conformations are slightly more stable than eclipsed ones. At room temperature many
conformations are present in a sample of butane.
Take care to distinguish between conformational differences and isomerism. The two
forms of butane shown in Figure 28-8 are not isomers of each other. Either form can be
converted to the other by rotation about a single bond, which is a very easy process that
does not involve breaking any bonds. By contrast, at least one chemical bond would have
to be broken and then re-formed to convert one isomer to another. This is most obvious
with isomerism in which a conversion would change the order of attachment of the atoms.
It is also true for geometric isomers that differ in orientation about a double bond. To
convert such a cis isomer to a trans isomer, it would be necessary to rotate part of the
molecule about the double bond. Such a rotation would move the p orbitals out of the
parallel alignment that is necessary to form the pi component of the double bond (Section
8-13). The breaking of this pi bond is quite costly in terms of energy; it occurs only with
the input of energy in the form of heat or light.
We saw in Section 27-1 that cyclohexane adopts a folded or “puckered” form. Cyclohexane
actually exists in two forms called the chair and twist boat forms (Figure 28-9). The
chair form is the more stable of the two because, on the average, the hydrogens (or other
substituents) are farther from one another than in the boat form. Chair and boat cyclohexane,
however, are not different compounds. Either form is easily converted into the
other by rotation around single bonds without breaking any bonds, and the two forms
cannot be separated. Thus, they are different conformations of cyclohexane.
SELECTED REACTIONS
Organic compounds can have quite different abilities to react, ranging from the limited
reactivity of hydrocarbons and fluorocarbons to the great variety of reactions undergone
by the thousands of organic molecules that contain several functional groups. Reactivity
depends on structure. We can usually predict the kinds of reactions a compound can
undergo by identifying the functional groups it contains.
In Chapter 27 we saw three fundamental classes of organic reactions and we associated
the fundamental reactions with specific functional groups. In this chapter we will present
some additional types of organic reactions.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 11.5, Functional Groups.
In the Brønsted–Lowry description, an
acid is a proton donor and a base is a
proton acceptor. Review the terminology
of conjugate acid–base pairs in Section
10-4.
28-4 Reactions of Brønsted–Lowry Acids and Bases 1115
REACTIONS OF BRØNSTED–LOWRY ACIDS AND BASES
Many organic compounds can act as weak Brønsted–Lowry acids or bases. Their reactions
involve the transfer of Hϩ ions, or protons (Section 10-4). Like similar reactions of
inorganic compounds, these acid–base reactions of organic acids and bases are usually fast
and reversible. Consequently, we can discuss the acidic or basic properties of organic
compounds in terms of equilibrium constants (Section 18-4).
Some Organic Acids
O
B
The most important organic acids contain carboxyl groups, XCXOXH. They are called
carboxylic acids (Section 27-13). They ionize slightly when dissolved in water, as illustrated
with acetic acid.
28-4
CH3COOH CH3COOϪH2O H3Oϩ Ka 1.8 10Ϫ5ϩϩ ϭ ϭ ϫ
[CH3COOϪ][H3Oϩ]
[CH3COOH]
acid1 base2 acid2base1
Figure 28-9 The two conformations of cyclohexane. (a) Chair, (b) stereoview of the chair
conformation, (c) twist boat and (d) stereoview of the twist boat conformation.
(a)
(c)
(b)
(d)
Sources of some naturally occurring
carboxylic acids.
1116 CHAPTER 28: Organic Chemistry II: Shapes, Selected Reactions, and Biopolymers
Acetic acid is 1.3% ionized in 0.10 M solution. Regardless of the lengths of the chains,
the acid strengths of the monocarboxylic acids are approximately the same, with Ka values
(in water) in the range 10Ϫ5 to 10Ϫ4. Their acid strengths increase dramatically when electronegative
substituents are present on the ␣-carbon atom (Ka values in water range from
10Ϫ3 to 10Ϫ1). Compare acetic acid and the three substituted acetic acids in Table 28-1.
There are two main reasons for this increase: (1) the electronegative substituents pull electron
density from the carboxylic acid group, and (2) the more electronegative substituents
help to stabilize the resulting carboxylate anion by spreading the negative charge over
more atoms.
The alcohols are so very weakly acidic that they do not react with strong bases. They
have about the same acid strength as water (see Table 28-1), and some of their reactions
are analogous to those of water.
The reactive metals react with alcohols to form alkoxides with the liberation of
hydrogen.
2CH3CH2XOH ϩ 2Na 88n H2 ϩ 2[Naϩ ϩ CH3CH2OϪ]
ethanol sodium ethoxide
(an alkoxide)
Alkoxides are strong bases that react with water (hydrolyze) to form the parent alcohol
and a metal hydroxide.
[Naϩ ϩ CH3CH2OϪ] ϩ HXOH 88n CH3CH2OH ϩ [Naϩ ϩ OHϪ]
sodium ethoxide ethanol
Phenols react with metallic sodium to produce phenoxides; the reactions are analogous
to those of alcohols. Because phenols are more acidic than alcohols, their reactions
are more vigorous.
ϩ ϩ ϩ2Na NaϩH2
OϪOH
2 2
sodium phenoxidephenol
In Section 18-3 we defined
pKa ϭ Ϫlog Ka
When Ka goes up by a factor of 10,
pKa goes down by one unit. We see
that the stronger an acid, the lower its
pKa value.
This is similar to the reaction of water
with active metals.
2HXOH ϩ 2Na 88n
H2 ϩ 2[Naϩ ϩ OHϪ]
TABLE 28-1 Ka and pKa Values of Some Carboxylic Acids
Name Formula Ka pKa
formic acid HCOOH 1.8 ϫ 10Ϫ40 3.74
acetic acid CH3COOH 1.8 ϫ 10Ϫ50 4.74
propanoic acid CH3CH2COOH 1.4 ϫ 10Ϫ50 4.85
monochloroacetic acid ClCH2COOH 1.5 ϫ 10Ϫ30 2.82
dichloroacetic acid Cl2CHCOOH 5.0 ϫ 10Ϫ20 1.30
trichloroacetic acid Cl3CCOOH 2.0 ϫ 10Ϫ10 0.70
benzoic acid C6H5COOH 6.3 ϫ 10Ϫ50 4.20
phenol* C6H5OH 1.3 ϫ 10Ϫ10 9.89
ethanol* CH3CH2OH Ϸ10Ϫ18 Ϸ18.
*Phenol and ethanol are not carboxylic acids. Phenol is weakly acidic compared with carboxylic acids, whereas ethanol
is even weaker than water.
Reaction of sodium metal with
ethanol gives sodium ethoxide and
hydrogen.
28-4 Reactions of Brønsted–Lowry Acids and Bases 1117
We saw in Section 18-9 that salts derived from strong acids and weak bases react with
water (hydrolyze) to give acidic solutions. The example given there involved the acidic
character of the ammonium ion.
NH4
ϩ ϩ H2O 34 NH3 ϩ H3Oϩ Ka ϭ ᎏ
[NH
[N
3]
H
[H
4
ϩ
3O
]
ϩ]
ᎏ ϭ 5.6 ϫ 10Ϫ10
acid1 base2 base1 acid2
Similar hydrolysis reactions occur with organic ammonium salts.
RNH3
ϩ ϩ H2O 34 RNH2 ϩ H3Oϩ Ka ϭ ᎏ
[RN
[R
H
N
2]
H
[H
3
ϩ
3O
]
ϩ]
ᎏ
acid1 base2 base1 acid2
We recall from Chapter 18 that the relationship Kw ϭ KaKb describes the strengths of any
conjugate acid–base pair in aqueous solution. For instance, we can use this relationship
for the CH3NH3
ϩ/CH3NH2 pair (we obtain Kb for methylamine, CH3NH2, from
Appendix G).
Ka(CH3NH3
ϩ
)ϭ ᎏ
Kb(C
K
H
w
3NH2)
ᎏ ϭ ᎏ
1
5
.
.
0
0
ϫ
ϫ
1
1
0
0
Ϫ
Ϫ
1
4
4
ᎏ ϭ 2.0 ϫ 10Ϫ11
In summary, we can rank the acid strengths of these classes of organic species.
carboxylic acids Ͼ phenols Ͼ substituted ammonium ions Ͼ alcohols
Our discussion has emphasized water solutions of these acids. Many organic compounds
are soluble in numerous other solvents. The properties of acids and bases in other solvents
depend on solvent properties such as polarity, acidity or basicity, and polarizability.
Some Organic Bases
The most important organic bases are the amines. When dissolved in water, they are
partially converted to substituted ammonium ions. This equilibrium is defined as shown
in the following equations for the ionization of primary, secondary, and tertiary amines.
base1 acid2 acid1 base2
primary, 1° RNH2 ϩ H2O 34 RNH3
ϩ ϩ OHϪ Kb ϭ ᎏ
[RN
[
H
R
3
N
ϩ
H
][O
2]
HϪ]
ᎏ.
secondary, 2° R2NH ϩ H2O 34 R2NH2
ϩ ϩ OHϪ Kb ϭ ᎏ
[R2N
[
H
R2
2
N
ϩ]
H
[O
]
HϪ]
ᎏ.
tertiary, 3° R3N ϩ H2O 34 R3NHϩ ϩ OHϪ Kb ϭ ᎏ
[R3N
[
H
R
ϩ
3N
][O
]
HϪ]
ᎏ.
Most low-molecular-weight aliphatic amines are somewhat stronger bases than ammonia.
Table 28-2 shows that aliphatic amines are much stronger bases than aromatic and heterocyclic
amines. The basicities of amines often decrease roughly in the order tertiary Ͼ
secondary Ͼ primary. Other structural factors and solvation effects, however, may
outweigh this tendency, especially with tertiary amines.
Several antibacterial mouthwashes
that contain a quaternary pyridinium
chloride salt.
The weaker a base, the stronger its
conjugate acid (Section 10-4).
The formula of a quaternary
ammonium salt is R4NCl.
pKb ϭ Ϫlog Kb
The weaker a base, the higher its pKb
value.
OXIDATION–REDUCTION REACTIONS
Oxidation of an organic molecule usually corresponds to increasing its oxygen content
or decreasing its hydrogen content. Reduction of an organic molecule usually corresponds
to decreasing its oxygen content or increasing its hydrogen content.
For example, the oxygen content increases when an aldehyde is converted to a carboxylic
acid, so this process is an oxidation.
Converting a primary alcohol to an aldehyde or a secondary alcohol to a ketone is also
an oxidation; the hydrogen content decreases.
oxidation
R C
O
RЈ (ketone)H
RЈ
CR2° alcohol
OH
Hydrogen content decreases
oxidation
R C
O
H (aldehyde)H
H
CR1° alcohol
OH
Oxygen content increases
oxidation
R C
O
OHHCR
O
28-5
We can rank the base strengths of common organic bases as:
alkoxides Ͼ aliphatic amines Ϸ phenoxides Ͼ carboxylates Ϸ aromatic amines Ϸ heterocyclic amines
TABLE 28-2 Basicities of Ammonia and Some Amines in Water
Name Formula Kb pKb
ammonia NH3 1.8 ϫ 10Ϫ5 4.74
methylamine CH3NH2 5.0 ϫ 10Ϫ4 3.30
dimethylamine (CH3)2NH 7.4 ϫ 10Ϫ4 3.13
trimethylamine (CH3)3N 7.4 ϫ 10Ϫ5 4.13
ethylamine CH3CH2NH2 4.7 ϫ 10Ϫ4 3.33
aniline C6H5NH2 4.2 ϫ 10Ϫ10 9.38
ethylenediamine H2NCH2CH2NH2 8.5 ϫ 10Ϫ5 (Kb1) 4.07
pyridine C5H5N 1.5 ϫ 10Ϫ9 8.82
The notation RЈ emphasizes that
the two R groups may be the same
(R ϭ RЈ, e.g., in the formation of
acetone, or different
(R RЈ, e.g., in the formation of
methyl ethyl ketone,
CH3 CH2CH3).C
O
CH3 CH3)C
O
We could describe the oxidation and
reduction of organic compounds in
terms of changes in oxidation numbers,
just as we did for inorganic compounds
in Sections 4-4 through 4-5. Formal
application of oxidation number rules
to organic compounds often leads
to fractional oxidation numbers for
carbon. For organic species, the
descriptions in terms of increase or
decrease of oxygen or hydrogen are
usually easier to apply.
28-5 Oxidation–Reduction Reactions 1119
Research & TechnologyCCHEMISTRY IN USE
Chemical Communication
The geometries of molecules play important roles in chemical
reactivity. Molecular geometry is particularly important
in a group of substances known as pheromones. Pheromones
are chemicals used for communication between members of
the same species. Pheromone activity has been observed in
many forms of life, from insects to humans, and pheromone
research is being done at many scholarly institutions.
If you’ve ever observed lines of ants moving in opposite
directions, you have observed the influence of pheromones
on insect behavior. When an ant finds food, it immediately
heads toward its nest while secreting 9-oxy-2-decenoic acid
from an abdominal gland. When other ants cross this acid
trail, they compulsively follow it to the food source and carry
the nourishment back to their nest. Soon, many ants will be
following the acid trail and reinforcing it with their own
9-oxy-2-decenoic acid secretions. Eventually, the food source
becomes exhausted, trail reinforcement stops, and the acid
trail evaporates. Ants are so dependent on the acid trail that
if a part of it were wiped away the ants following the trail
in both directions would come to a complete stop. They
wouldn’t know where to go.
Perhaps an even more impressive example than the total
dependence on chemical communication by ants is demonstrated
with the so-called “death pheromone.” Immediately
upon the death of an ant, fellow ants continue to groom the
dead ant and treat it as if it were still living. This attention
continues until the dead ant’s body produces the death pheromone,
10-octadecenoic acid. On sensing this pheromone,
colleagues carry the dead ant to the nearest garbage site.
Interestingly, if 10-octadecenoic acid is applied to a living
ant, the living ant is similarly dumped into the garbage. The
discarded ant will quickly return only to be carried off again,
and this process continues until the death pheromone
evaporates.
Because pheromones are used by female insects to indicate
their state of fertile readiness, pheromones have proven to be
an effective weapon in controlling some crop-damaging
insects. For example, when a specific mating pheromone is
applied to crops, male cotton bollworms and female tobacco
budworms compulsively mate with one another. Because of
physical incompatibilities, their bodies become interlocked
and both insects eventually die. Less drastic uses of
pheromones to control crop damage involve baiting traps
with sex pheromones to lure and trap male insects. Trapping
males eventually slows reproduction, and the insect population
may decrease to controllable levels. Some of these sex
pheromones are so powerful that a single drop has the potential
of attracting millions of males. In fact, some male insects
can detect a single molecule of female pheromone from a
great distance and then successfully seek out and find the
female.
Chemical communication is not confined to the insect
world. Female dogs secrete the chemical p-hydroxybenzoate
to attract males. Just like the ants and cotton bollworms, who
are dependent on detecting chemicals for their actions, male
dogs will attempt to mate with various objects to which phydroxybenzoate
has been applied.
When we examine the molecular structures and functional
groups of known pheromones, we find that they have little
in common. Some pheromones contain stereoisomers, and
some insects can distinguish between the stereoisomers. The
structures of pheromones play vital roles in their activity. Part
of the structure is an upper limit of about 20 carbon atoms,
a limit probably imposed by Graham’s Law. Most pheromones
must travel through the air; those with low molecular
weights are often more volatile. Scientists suspect that the
physical motions of pheromone molecules, which are also a
function of molecular structure, play an important role in the
communication mechanism.
Ronald DeLorenzo
Middle Georgia College
Aldehydes are easily oxidized to
carboxylic acids. They must therefore
be removed from the reaction mixture
as soon as they are formed. Aldehydes
have lower boiling points than the
alcohols from which they are formed,
so the removal of aldehydes is easily
accomplished.
1120 CHAPTER 28: Organic Chemistry II: Shapes, Selected Reactions, and Biopolymers
An aldehyde can be oxidized to a carboxylic acid.
In each of these “oxidation” reactions, something else must act as the oxidizing agent
(which is reduced). These oxidizing agents are often inorganic species such as dichromate
ions, Cr2O7
2Ϫ. The reverse of each of the preceding reactions is a reduction of the organic
molecule. In this reverse reaction, the reducing agent (the substance that is oxidized) is
often an inorganic compound.
Let us look at a few important types of organic oxidations and reductions.
Oxidation of Alcohol
Aldehydes can be prepared by the oxidation of primary alcohols. The reaction mixture is
heated to a temperature slightly above the boiling point of the aldehyde so that the aldehyde
distills out as soon as it is formed. Potassium dichromate in the presence of dilute
sulfuric acid is the common oxidizing agent.
Ketones can be prepared by the oxidation of secondary alcohols. Ketones are not as
susceptible to oxidation as are aldehydes.
Aldehydes and ketones can be prepared by a catalytic process that involves passing
alcohol vapors and air over a copper gauze or powder catalyst at approximately 300°C.
Here the oxidizing agent is O2.
H
OH
K2Cr2O7
dil. H2SO4
O
cyclooctanol cyclooctanone
CH3CH3 CH
OH
K2Cr2O7
dil. H2SO4
2-propanol
(isopropyl alcohol)
acetone
CH3 C
O
CH3
CH3CH2CH2CH2OH
K2Cr2O7
dil. H2SO4
1-butanol
bp ϭ 117.5°C
butanal
(butyraldehyde)
bp ϭ 75.7°C
H
O
CH3CH2CH2C
CH3OH H HC
O
K2Cr2O7
dil. H2SO4
methanol
bp ϭ 65°C
methanal
(formaldehyde)
bp ϭ Ϫ21°C
Oxygen content increases
oxidation
R C
O
OHHCR
O
Ketones are not as easily oxidized as
aldehydes, because oxidation of a
ketone requires the breaking of a
carbon–carbon bond. Thus, it is not
as important that they be quickly
removed from the reaction mixture.
These two reactions can also be
described as a type of elimination
reaction (Section 27-18). Two
hydrogen atoms are eliminated to
form a CUO double bond.
Organic reactions are sometimes
written in extremely abbreviated form.
This is often the case when a variety of
common oxidizing or reducing agents
will accomplish the desired conversion.
28-5 Oxidation–Reduction Reactions 1121
Formaldehyde is quite soluble in water; the gaseous compound can be dissolved in water
to give a 40% solution.
Acetaldehyde can be prepared by the similar oxidation of ethanol.
Oxidation of tertiary alcohols is difficult because the breaking of a carbon–carbon bond
is required. Such oxidations are of little use in synthesis.
Reduction of Carbonyl Compounds
Reduction of a variety of compounds that contain the carbonyl group provides synthetic
methods to produce primary and secondary alcohols. A common, very powerful reducing
agent is lithium aluminum hydride, LiAlH4; other reducing agents include sodium in
alcohol and sodium borohydride, NaBH4.
Oxidation of Alkylbenzenes
Unsubstituted aromatic hydrocarbons (Sections 27-5, 27-6) are quite resistant to oxidation
by chemical oxidizing agents. The reactions of strong oxidizing agents with
alkylbenzenes illustrate the stability of the benzene ring system. Heating toluene with a
basic solution of KMnO4 results in a nearly 100% yield of benzoic acid. The ring itself
remains intact; only the nonaromatic portion of the molecule is oxidized.
carboxylic acid
reduction
R C
O
R CH2OHOH
ester 1° alcohol(s)
1° alcohol
aldehyde
reduction
R C
O
R CH2OHH 1° alcohol
reduction
R C
O
R CH2OH ϩORЈ RЈ OH
ketone 2° alcohol(s)
reduction
R C
O
R CH
OH
RЈ RЈ
H
ydrogen content incre
ases2CH3CH2OH O2ϩ ϩ 2H2O2CH3 C
O
H
300°C
Cu
acetaldehydeethanol
2CH3OH O2ϩ ϩ 2H2O2H C
O
H
300°C
Cu
formaldehydemethanol
1122 CHAPTER 28: Organic Chemistry II: Shapes, Selected Reactions, and Biopolymers
Two such alkyl groups on an aromatic ring are oxidized to give a diprotic acid, as the
following example illustrates.
Terephthalic acid is used to make “polyesters,” an important class of polymers (Section
27-19).
Combustion of Organic Compounds
The most extreme oxidation reactions of organic compounds occur when they burn in
O2. Such combustion reactions (Section 6-8) are highly exothermic. When the combustion
takes place in excess O2, the products are CO2 and H2O. Examples of alkane combustions
are
methane: CH4(g) ϩ 2O2(g) 88n CO2(g) ϩ 2H2O(ᐉ) ⌬H0 ϭ Ϫ891 kJ
octane: 2C8H18(ᐉ) ϩ 25O2(g) 88n 16CO2(g) ϩ 18H2O(ᐉ) ⌬H0 ϭ Ϫ1.090 ϫ 104 kJ
The heat of combustion is the amount of energy liberated per mole of hydrocarbon
burned. Heats of combustion are assigned positive values by convention (Table 28-3) and
are therefore equal in magnitude, but opposite in sign, to ⌬H0 values for combustion reactions.
The combustion of hydrocarbons produces large volumes of O2 and H2O in addition
to large amounts of heat. The rapid formation of these gases at high temperature and
pressure drives the pistons or turbine blades in internal combustion engines.
CH3
CH3
(oxidation)
p-xylene
COOH
COOH
terephthalic acid
CH3 (2) HCl(aq)
(1) heat, OHϪ, KMnO4
toluene
C
O
OH
benzoic acid
Recall that ⌬H0 is negative for an
exothermic process.
Acetylene is produced by the slow
addition of water to calcium carbide.
CaC2(s) ϩ 2H2O(ᐉ) 88n
HCmCH(g) ϩ Ca(OH)2(s)
The light of one kind of headlamp
used by miners and cave explorers is
given off by the combustion of
acetylene.
TABLE 28-3 Heats of Combustion of Some Alkanes
Heat of Combustion
Hydrocarbon kJ/mol J/g
methane CH4 891 55.7
propane C3H8 2220 50.5
pentane C5H12 3507 48.7
octane C8H18 5450 47.8
decane C10H22 6737 47.4
ethanol* C2H5OH 1372 29.8
*Not an alkane; included for comparison only.
Hexane, C6H14, an alkane, burns
cleanly in air to give CO2 and H2O
(top). 1-Hexene, C6H12, an alkene,
burns with a flame that contains soot
(middle). Burning o-xylene, an
aromatic hydrocarbon, produces
large amounts of soot (bottom).
28-6 Formation of Carboxylic Acid Derivatives 1123
In the absence of sufficient oxygen, partial combustion of hydrocarbons occurs. The
products may be carbon monoxide (a very poisonous gas) or carbon (which deposits on
spark plugs, in the cylinder head, and on the pistons of automobile engines). Many modern
automobile engines now use microcomputer chips and sensors to control the air supply
and to optimize the fuel/O2 ratio. The reactions of methane with insufficient oxygen are
2CH4 ϩ 3O2 88n 2CO ϩ 4H2O and CH4 ϩ O2 88n C ϩ 2H2O
All hydrocarbons undergo similar reactions.
The alkenes, like the alkanes, burn in excess oxygen to produce carbon dioxide and water
in exothermic reactions.
CH2UCH2(g) ϩ 3O2 (g, excess) 88n 2CO2(g) ϩ 2H2O(ᐉ) ⌬H0 ϭ Ϫ1387 kJ
When an alkene (or any other unsaturated organic compound) is burned in air, a yellow,
luminous flame is observed, and considerable soot (unburned carbon) is formed. This reaction
provides a qualitative test for unsaturation. Saturated hydrocarbons burn in air without
forming significant amounts of soot.
Acetylene lamps are charged with calcium carbide. Very slow addition of water produces
acetylene, which is burned as it is produced. Acetylene is also used in the oxyacetylene
torch for welding and cutting metals. When acetylene is burned with oxygen, the flame
reaches temperatures of about 3000°C.
Like other hydrocarbons, the complete combustion of aromatic hydrocarbons, such as
benzene, releases large amounts of energy.
2C6H6(ᐉ) ϩ 15O2(g) 88n 12CO2(g) ϩ 6H2O(ᐉ) ⌬H0 ϭ Ϫ6548 kJ
Because they are so unsaturated, aromatic hydrocarbons burn in air with a yellow, sooty
flame.
FORMATION OF CARBOXYLIC ACID DERIVATIVES
The carboxylic acid derivatives introduced in Section 27-14 can be formed by substitution
of another group in place of XOH in the carboxyl group. The acyl halides (acid halides)
are usually prepared by treating acids with PCl3, PCl5, or SOCl2 (thionyl chloride). In
general terms, the reaction of acids with PCl5 may be represented as
The acyl halides are much more reactive than their parent acids. Consequently, they are
often used in reactions to introduce an acyl group into another molecule.
When an organic acid is heated with an alcohol, an equilibrium is established with the
resulting ester and water. The reaction is catalyzed by traces of strong inorganic acids,
such as a few drops of concentrated H2SO4.
C
O
OH PCl5 CH3CH3 ϩ C
O
Cl HCl(g) POCl3ϩ ϩ
acetic acid acetyl chloride
R C
O
OH PCl5ϩ R C
O
Cl HCl(g) POCl3ϩ ϩ
acid phosphorus
pentachloride
an acyl chloride
(an acid chloride)
phosphorus
oxychloride
28-6
Many experiments have shown
conclusively that the OH group from
the acid and the H from the alcohol
are the atoms that form water
molecules.
1124 CHAPTER 28: Organic Chemistry II: Shapes, Selected Reactions, and Biopolymers
In general terms, the reaction of an organic acid and an alcohol may be represented as
(R and RЈ may be the same or different groups.)
Reactions between acids and alcohols are usually quite slow and require prolonged
boiling (refluxing). The reactions between most acyl halides and most alcohols, however,
occur very rapidly without requiring the presence of an acid catalyst.
Amides are usually not prepared by the reaction of an amine with an organic acid. Acyl
halides react readily with primary and secondary amines to produce amides. The reaction
of an acyl halide with a primary or secondary amine produces an amide and a salt of the
amine.
HYDROLYSIS OF ESTERS
Because most esters are not very reactive, strong reagents are required for their reactions.
Esters can be hydrolyzed by refluxing with solutions of strong bases.
The hydrolysis of esters in the presence of strong bases is called saponification (soap-
making).
In general terms, the hydrolysis of esters may be represented as
ϩNaϩOHϪ CR
O
OϪNaϩϩR RЈ RЈOHC
O
O
ester salt of an acid alcohol
heat
ϩ CH3CH2OHNaϩOHϪϩCH3 CH2CH3C
O
O CH3C
O
OϪNaϩ
ethyl acetate sodium acetate ethanol
heat
28-7
ϩ ϩ2CH3NH2 CH3NH3
ϩClϪCH3 ClC
O
CH3
CH3
C N
HO
methylamine
(a primary amine)
acetyl chloride
(an acyl halide)
N-methylacetamide
(an amide)
methylammonium
chloride (a salt)
C
O
Cl CH3CH3 CH3CH2 CH2CH3ϩ O H C
O
O HClϩ
acetyl chloride ethyl alcohol ethyl acetate
C
O
OH RЈ RR ϩ O H C
O
O H2OϩRЈ
acid alcohol ester
C
O
OH CH3CH2 CH3CH3 ϩ OH C
O
O H2OϩCH2CH3
acetic acid ethyl alcohol ethyl acetate, an ester
Hϩ, heat
In this preparation, one half of the
amine is converted to an amide and
the other half to a salt.
Tripalmitin, a triglyceride.
American pioneers prepared their
soaps by boiling animal fat with an
alkaline solution obtained from the
ashes of hardwood. The resulting
soap could be “salted out” by adding
sodium chloride, making use of the
fact that soap is less soluble in a salt
solution than in water.
28-7 Hydrolysis of Esters 1125
Like other esters, fats and oils (Section 27-14) can be hydrolyzed in strongly basic solution
to produce salts of the acids and the trihydric alcohol glycerol. The resulting sodium
salts of long-chain fatty acids are soaps. In Section 14-18 we described the cleansing action
of soaps and detergents.
3NaϩOHϪ OϪNaϩϩ ϩ
C
O
(CH2)16CH3O
glyceryl tristearate
(a fat)
sodium stearate
(a soap)
glycerol
(glycerine)
H2C
C
O
(CH2)16CH3OH2C
C
O
(CH2)16CH3 3CH3(CH2)16C
H2O
heat
HC OHO
O H2C OH
H2C OH
HC
BIOPOLYMERS
As we study the molecules and ions in living organisms such as humans, we learn that
substances with a huge range of sizes and complexities are involved. These include small
molecules and ions such as sodium ions, trace metals, ketones, alcohols, and carboxylic
acids. Steroids, hormones, vitamins, fats, and oils are somewhat larger. As we continue up
the scale of sizes of substances present in living systems, we encounter the mammoth polymers
(biopolymers, or polymers found in biological systems). Concepts discussed in the
first 27 chapters and the first half of this chapter can be applied to large molecules and
ions as well as small ones. In the following sections we will look briefly at three classes of
biopolymers (carbohydrates, proteins, and nucleic acids), keeping in mind that almost all
of the chemical concepts used in these discussions were introduced earlier in this text.
See the Saunders Interactive
General Chemistry CD-ROM,
Screen 11.7, Functional Groups 2:
Fats and Oils.
The word saccharide comes from the
Latin saccharum, for “sugar” or
“sweet.”
1126 CHAPTER 28: Organic Chemistry II: Shapes, Selected Reactions, and Biopolymers
CARBOHYDRATES
Carbohydrates have the general formula Cn(H2O)m, thus the name hydrated carbons or,
more commonly, carbohydrates. Monosaccharides, or simple sugars, are the monomers
from which more complex carbohydrates are constructed. The most important simple
sugars contain either five or six carbon atoms. Monosaccharides that contain an aldehyde
group are called aldoses, and those that contain a ketone group are called ketoses. The
most commonly encountered monosaccharide is glucose, C6H12O6, also known as dextrose
or blood sugar. Figure 28-10a shows glucose as it is often written to emphasize its aldehyde
functional group. We can see from this formula that carbons 2, 3, 4, and 5 in glucose
each have four different groups and are, therefore, chiral. Each different monosaccharide
has a unique arrangement of atoms about its chiral atoms. Some important monosaccharides
are shown in Table 28-4.
28-8
(d)
Figure 28-10 The cyclization of
glucose to form either the ␣- or the ␤cyclic
form. (a) The linear representation
of the aldehyde form of glucose. (b) A
coiled representation that shows the
aldehyde function group near the XOH
group of carbon 5. (c) The two cyclic
forms of glucose. (d) A stereoview of the
␣-form of glucose.
OH
CH O
1
CH OH
2
CHO H
3
CH
4
CH OH
5
CH2OH
6
(b)
(c)
(a)
H
HO C
HO C
C
C
C
H
H
H
H
O
O
H
OH
CH2OH
6
4
2
5
3
1
HO C
H
HO
H
C
H
C
C
C
OH
OH
O
H
CH2OH
3
5
6
4
2
1
α
-glucoseα
HO C
H
HO
H
C
H
C
C
C
OH
OH
O
H
CH2OH
3
5
6
4
2
1
H
β
-glucoseβ
Besides their aldehyde or ketone form, saccharides can also exist in a cyclic form. Figure
28-10 shows the interconversion of glucose between its linear form and its cyclic form.
Figure 28-10b shows the molecule rotated into a conformation that brings the hydroxyl
group of carbon 5 near the aldehyde carbon. In this conformation, a reversible reaction
occurs in which a hydrogen is moved from the hydroxyl group to the carbonyl oxygen,
and an ether bond is formed between the two carbons (Figure 28-10c). During the formation
of the ether bond, a ring is also formed and carbon atom 1 becomes chiral. This new
chiral center can exist as either of two isomers; the resulting stereoisomers are referred
to as ␣- or ␤-forms.
Most naturally occurring carbohydrates contain more than one monosaccharide unit.
Complex carbohydrates are formed when two or more monosaccharides are linked. This
linkage between two monosaccharides occurs by elimination of a molecule of water and
the formation of an ether bond between two of the cyclic structures, as shown in Figure
28-11. The newly formed CXO bond linking the rings is called the glycosidic bond.
Figure 28-11 illustrates the linkage of ␣-glucose and ␤-fructose to form sucrose. Fructose
is a ketone-containing carbohydrate that exists in the cyclic form shown. A disaccharide
is a molecule consisting of two monosaccharides joined together by a glycosidic bond.
Sucrose is a disaccharide; two other disaccharides, each involving only glucose units, are
shown in Figure 28-12. In such a molecule, the CXOXC linkage between the monosaccharide
units involves carbon number 1 or 2 of one monosaccharide and carbon number
1, 4, or 6 of the other monosaccharide. The arrangement about the carbon number 1 or
2 is what is described as the ␣- or ␤-linkage. Using these notations, we describe maltose
as consisting of two glucose units joined by an ␣-1,4-linkage. Lactose has a ␤-1,4-linkage.
Table 28-5 lists some of the common sugars.
28-8 Carbohydrates 1127
TABLE 28-4 Some Important Monosaccharides
Five Carbon Atom Monosaccharides
Six Carbon Atom Monosaccharides
ribose
CH OH
CHO
CH OH
CH OH
CH2OH
arabinose
CHO H
CHO
CH OH
CH OH
CH2OH
ribulose
C O
CH OH
CH OH
CH2OH
CH2OH
glucose
CH OH
CHO
CHO H
CH OH
CH OH
CH2OH
mannose
CHO H
CHO
CHO H
CH OH
CH OH
CH2OH
galactose
CH OH
CHO
CHO H
CHO H
CH OH
CH2OH
fructose
C O
CHO H
CH OH
CH OH
CH2OH
CH2OH
Figure 28-11 Formation of
sucrose by the condensation of
an ␣-glucose molecule with a
␤-fructose molecule.
H
HO C
HO C
C
C
C
H
H
H
O
H
OH
CH2OH
3
OH
4
6
5 2
1
H H
OH H
OH
O
OHHOH2C
C
C
C
C
CH2OH
5
4 3
2
1
6
H
HO C
HO C
C
C
C
H
H
H
O
H
CH2OH
OH 1
H H
OH H
OH
O
HOH2C
C
C
C
C
CH2OH
2
O
-glucoseα
-fructoseβ sucrose
– H2O
+ H2O
+ H2O
H
HO C
HO C
C
C
C
H
H
H
O
H
O
CH2OH
OH
1 H
C
HO C
C
C
C
H
H
H
O
H
OH
CH2OH
OH
4
α
α
HO
H C
HO C
C
C
C
H
H
H
O
H O
CH2OH
OH 1
H
C
HO C
C
C
C
H
OH
H
O
H
H
CH2OH
OH
4 ββ
lactose ( -form)β
maltose ( -form)α
TABLE 28-5 Sweetness of Common Sugars Relative to
Sucrose
Sweetness Relative
Substance to Sucrose at 1.00
lactose (milk sugar, a disaccharide) 0.16
galactose (a monosaccharide in milk sugar) 0.32
maltose (a disaccharide used in beer making) 0.33
glucose (dextrose, a common monosaccharide) 0.74
sucrose (table sugar, a disaccharide) 1.00
fructose (fruit sugar, a monosaccharide) 1.74
Figure 28-12 Two disaccharides:
maltose and lacrose. Each involves
two glucose units. Maltose has
␣-1,4-linkage and lactose has a
␤-1,4-linkage.
A carbohydrate that contains three monosaccharides is called a trisaccharide, and so
on. Carbohydrates that contain from four to ten monosaccharides are usually called by
the general term oligosaccharides. Carbohydrates that contain larger numbers of monosaccharide
units are called polysaccharides.
Amylose, a type of starch found in plants, is a polysaccharide. It consists of a linear
combination of several thousand glucose units joined by ␣-1,4-linkages. In our bodies we
store glucose in a form known as amylopectin, or glycogen. Glycogen is similar to amylose;
it typically contains about 30,000 glucose monomers with the majority linked via an
␣-1,4-linkage but, unlike amylose, it contains occasional branches via ␣-1,6-linkages. In
Figure 28-13 each sphere represents a glucose unit. All links are ␣-1,4 except where
branching takes place; branching utilizes an ␣-1,6-linkage. Our digestive systems contain
enzymes (see Sections 16-9, 28-9) that can break this biopolymer, glycogen, into its
monomeric glucose units as we need energy to drive other biological processes.
Cellulose, the structural biopolymer found in plants, contains the ␤-1,4-linkage of
glucose units. Our digestive systems are incapable of breaking the ␤-1,4-linkage, so we
cannot utilize cellulose as a source of stored glucose. Some animals such as termites and
cattle are able to digest cellulose.
Figure 28-13 Amylopectin, or glycogen.
28-8 Carbohydrates 1129
Bacteria in the digestive tracts of termites
and cattle produce an enzyme that allows
them to digest the cellulose in wood and
grass.
POLYPEPTIDES AND PROTEINS
Two amino acids can react to form an amide bond (see Section 27-19). Compounds formed
by the linking of small numbers of amino acids are called peptides, and the amide bond
linking the amino acid monomers is called the peptide bond.
28-9
TABLE 28-6 The 20 Common Amino Acids Found in Proteins
Nonpolar R Groups
alanine (Ala)
phenylalanine (Phe)
glycine (Gly)
proline (Pro)
isoleucine (Ile)
tryptophan (Trp)
leucine (Leu)
methionine (Met)
valine (Val)
Polar, Neutral R Groups
asparagine (Asn) serine (Ser)
glutamine (Gln) threonine (Thr)
Polar, Acidic R Groups Polar, Basic R Groups
aspartic acid (Asp)
arginine (Arg)
glutamic acid (Glu)
cysteine (Cys)
histidine (His)
lysine (Lys)
tyrosine (Tyr)
CH3CHCOOH
NH2
HCHCOOH
NH2
CH3CH2CHCHCOOH
CH3
NH2
(CH3)2CHCH2CHCOOH
NH2
CH3SCH2CH2CHCOOH
NH2
H2NCCH2CHCOOH
NH2
O
H2NCCH2CH2CHCOOH
NH2
O
HOOCCH2CHCOOH
NH2
HOOCCH2CH2CHCOOH
NH2
HSCH2CHCOOH
NH2
HO CH2CHCOOH
NH2
CH2CHCOOH
NH2
COOH
N
H
CH2CHCOOH
NH2
N
H
(CH3)2CHCHCOOH
NH2
HOCH2CHCOOH
NH2
CH3CHCHCOOH
NH2
OH
H2NCNHCH2CH2CH2CHCOOH
NH2
NH
CH2CHCOOH
N
N
H NH2
H2NCH2CH2CH2CH2CHCOOH
NH2
Prefixes di-, tri-, and so on are used to indicate the number of amino acid monomers that
are joined to form the peptide. The compound indicated in the preceding formula is a
dipeptide.
Twenty different ␣-amino acids (see Section 27-13) are found widely in nature. Each
amino acid is commonly given a three-letter abbreviation. These acids differ in the R
groups attached to the ␣-carbon (the carbon adjacent to the XCOOH group). Differences
in R groups lead to differences in the properties of the ␣-amino acids. The amino
acids are usually classified by their R groups, or side chains, according to two criteria. The
first classification depends on whether the R groups are polar or nonpolar, the second
describes whether the R group contains an acidic or basic group. The 20 common amino
acids are shown, according to these classifications, in Table 28-6.
Many small peptides are known to have physiological significance. Aspartame, a
common artificial sweetener, is a dipeptide. Glutathione, a tripeptide, acts as a scavenging
agent for harmful oxidizing agents believed to cause cancer. Enkephalins, which are naturally
occurring analgesics (pain relievers) that occur in the brain, are pentapeptides. The
antibiotics gramicidin-S and tyrocidine A are cyclic decapeptides (i.e., they contain ten
amino acids linked into a ring by peptide bonds) produced by a bacterium.
H2NCHCNHCHCOOCH3
O
CH2
PhCOOH
CH2
H2NCHCH2CH2CNHCHCNHCH2COOH
O
CH2SHCOOH
O
aspartame
(Asp-Phe-OCH3)
glutathione
(Gln-Cys-Gly)
28-9 Polypeptides and Proteins 1131
ϩϩ OHH N C OHC
H
R
H2O
O
peptide
bond
H
H N C OHC
H
RЈ
O
H
H N C NC
H
R
O
C
O
C
H
RЈH H
Space-filling models of aspartame
(top) and glutathione (bottom).
Tyr-Gly-Gly-Phe-Leu
leucine enkephalin
Tyr-Gly-Gly-Phe-Met
methionine enkephalin
A protein is a biopolymer, of molecular weight typically 5000 g/mol or greater,
consisting of one or more polypeptide chains. The ␣-amino acids that occur in nearly all
proteins are the L optical isomer.
An enzyme is a protein that catalyzes a biological reaction.
Each protein consists of one or more polypeptide chains with a unique sequence of
amino acids. The order of these monomers is termed the primary structure of the protein.
The many possible interactions between different amino acids in the polypeptide chain
cause each protein molecule to adopt a characteristic shape, usually depending on its environment.
The secondary structure of a protein is the arrangement in space of the
polypeptide backbone, without reference to the conformations of the side chains (R
groups). Myoglobin is the oxygen-storing protein in muscles. Its amino acid sequence
(primary structure) and a three-dimensional representation of its secondary structure are
shown in Figure 28-14. The tertiary structure of a protein describes the overall shape
of the protein, including the side chains and any other nonpeptide components of the
protein. Myoglobin is referred to as a globular protein. Some proteins consist of multiple
polypeptide units called subunits. These subunits are held to one another by noncovalent
interactions such as hydrogen bonding, dipole–dipole interactions, or ionic attractions, to
form the quaternary structure of the protein.
Some organisms such as bacteria also
produce D-amino acids.
1132 CHAPTER 28: Organic Chemistry II: Shapes, Selected Reactions, and Biopolymers
COOH
NH2
Val-Leu-Ser-Glu-Gly-Glu-Trp-Gln-Leu-Val-Leu-His-Val-Trp-Ala-Lys-Val-
Glu-Ala-Asp-Val-Ala-Gly-His-Gly-Gln-Asp-Ile-Leu-Ile-Arg-Leu-Phe-Lys-
Ser-His-Pro-Glu-Thr-Leu-Glu-Lys-Phe-Asp-Arg-Phe-Lys-His-Leu-Lys-
Thr-Glu-Ala-Glu-Met-Lys-Ala-Ser-Glu-Asp-Leu-Lys-Lys-His-Gly-Val-Thr-
Val-Leu-Thr-Ala-Leu-Gly-Ala-Ile-Leu-Lys-Lys-Lys-Gly-His-His-Glu-Ala-
Glu-Leu-Lys-Pro-Leu-Ala-Gln-Ser-His-Ala-Thr-Lys-His-Lys-Ile-Pro-Ile-Lys-
Tyr-Leu-Glu-Phe-Ile-Ser-Glu-Ala-Ile-Ile-His-Val-Leu-His-Ser-Arg-His-
Pro-Gly-Asn-Phe-Gly-Ala-Asp-Ala-Gln-Gly-Ala-Met-Asn-Lys-Ala-Leu-
Glu-Leu-Phe-Arg-Lys-Asp-Ile-Ala-Ala-Lys-Tyr-Lys-Glu-Leu-Gly-Tyr-Gln-Gly
Figure 28-14 The amino acid sequence and the three-dimensional representation of
myoglobin, the oxygen-storing protein in muscles.
The folding and coiling patterns of a protein often affect its biological activity, for
instance, its effectiveness as an enzyme. Forces such as heat and ionic environment
can alter the pattern of folding and coiling, causing the enzyme to become inactive, or
denatured.
Proteins make up more than 50% of the dry weight of animals and bacteria. They
perform many important functions in living organisms, a few of which are indicated in
Table 28-7. Each protein carries out a specific biochemical function. Each is a polypeptide
with its own unique sequence of amino acids. The amino acid sequence of a protein
determines exactly how it folds up in a three-dimensional conformation and how it
performs its precise biochemical task.
28-9 Polypeptides and Proteins 1133
A model of the protein
ribonuclease A in stereoview.
The different colors represent
regions within the folding and
coiling patterns.
A “ribbon” model of the protein calmodulin. In this
type of model, the ribbon represents the
polypeptide chain. This protein coordinates with
Ca2ϩ ions (white spheres) and aids in transporting
them in living systems.
NUCLEIC ACIDS
We have seen that polysaccharides are polymers composed of a single type of monomer
(carbohydrates), as are proteins (amino acids). The third type of biopolymer is more
complex. Nucleic acids use three very different types of monomers: the phosphate group,
one of two simple carbohydrate units (deoxyribose or ribose), and selected organic bases
(Figure 28-15). A typical segment of the resulting polymer is shown in Figure 28-16.
There are two types of nucleic acids: ribonucleic acid (RNA) and deoxyribonucleic acid
(DNA). One of the ways in which they differ is in the carbohydrate that they contain.
28-10
1134 CHAPTER 28: Organic Chemistry II: Shapes, Selected Reactions, and Biopolymers
TABLE 28-7 Some Functions of Proteins
Example Function
Enzymes
amylase converts starch to glucose
DNA polymerase I repairs DNA molecule
trans aminase transfers amino group from one amino acid to another
Structural Proteins
viral coat proteins outer covering of virus
keratin hair, nails, horns, hoofs
collagen tendons, cartilage
Hormones
insulin, glucagon regulate glucose metabolism
oxytocin regulates milk production in female mammals
vasopressin increases retention of water by kidney
Contractile Proteins
actin thin contractile filaments in muscle
myosin thick filaments in muscle
Storage Proteins
casein a nutrient protein in milk
ferritin stores iron in spleen and egg yolk
Transport Proteins
hemoglobin carries O2 in blood
myoglobin carries O2 in muscle
serum albumin carries fatty acids in blood
cytochrome c transfers electrons
Immunological Proteins
␥-globulins form complexes with foreign proteins
Toxins
neurotoxin blocker of nerve function in cobra venom
ricin nerve toxin in South American frog (most toxic substance
known—0.000005 g is fatal to humans)
The two carbohydrates in nucleic acids are ribose (contained in RNA) and 2-deoxyribose
(contained in DNA); these carbohydrates differ in the presence or absence of an oxygen
atom on carbon 2.
The bases found in nucleic acids are shown in Figure 28-17. Each type of nucleic acid
contains two pyrimidine bases and two purine bases. The two purine bases adenine (A)
and guanine (G) and the pyrimidine base cytosine (C) are found in both RNA and DNA.
O OH
HO
H
H
H
H
H
2
4
3
1
HOCH2
5
2-deoxyribose
O OH
OHHO
H
H
H
H
HOCH2
5
2
4
3
1
ribose
28-10 Nucleic Acids 1135
–O
O
P
O–
O CH2 O
C
H
C
H
H
C
OH
H
C
H
H
H
N
N N
N
NH2
base
carbohydrate group
phosphate group
HO P O
O
CH2
C
H
H
O
C
O
H
C
C
H
H
P
O
CH2
C
OHO
O
C
H
C
H C
H H
HO
P
O
OHO
Base
Repeating
unit along
DNA
chain
Base
N
N
N
N
H
NH2
NH2 N N
H
N
HN
O
N
H
HN
O
O
NH2
O N
H
N
O
N
H
O
CH3
HN
Purine bases Pyrimidine bases
adenine (A)
DNA
RNA
guanine (G)
DNA
RNA
uracil (U)
RNA
cytosine (C)
DNA
RNA
thymine (T)
DNA
Figure 28-16 A short segment of
DNA (deoxyribonucleic acid).
Figure 28-15 The three kinds of units that are polymerized to form nucleic acids.
Figure 28-17 The bases found in DNA and RNA.
In RNA, the other pyrimidine base is uracil (U), whereas DNA contains thymine (T).
Specific pairs of these bases hydrogen bond to each other in particularly favorable
arrangements, so they always occur in distinctive hydrogen-bonded base pairs. Adenine
and thymine (in DNA) or uracil (in RNA) each form two hydrogen bonds, so they are
always paired together; similarly, guanine and cytosine each form three hydrogen bonds,
so they always pair to each other (Figure 28-18). The resulting complementary basepairing
links two strands of the polymer in a very specific order to form a double helix.
This order is the key to nucleic acid function. Figure 28-19 shows a short segment of the
double strand of DNA. In that figure, we see that adenine is paired with thymine and
guanine is paired with cytosine. In RNA, adenine forms base pairs with uracil instead of
with thymine.
DNA is one of the largest molecules known. Human DNA is estimated to have up to
three billion base pairs, resulting in a molecular weight in the tens of billions. RNA is
smaller and more variable, with molecular weights of 20,000 to 40,000. Genetic information
for each organism is stored in its DNA as a result of the locations and sequence
of the base pairs. The information is replicated when the strands unravel and new complementary
strands are formed. The genetic information of the DNA is used to guide the
many syntheses that occur in a living cell. For example, for the synthesis of a specific
protein, a specific portion of the DNA double helix unwinds. Then an RNA molecule is
built using the sequence of base pairs of the DNA as a pattern. The RNA then migrates
from the cell’s nucleus to the location where the protein is to be synthesized. There the
base sequence of the RNA determines the sequence of amino acids to be used in the
protein construction. With its many possible sequences of the same base pairs to transmit
information to control synthesis of other molecules, DNA contains a vast amount of
genetic information.
A stereoview of an idealized model of DNA.
1136 CHAPTER 28: Organic Chemistry II: Shapes, Selected Reactions, and Biopolymers
Figure 28-18 The two kinds of hydrogen-bonded base-pairing that occur in DNA. In
RNA, adenine forms a base pair with uracil instead of with thymine.
sugar
sugar
adenine
thymine
H
N
O
N N
N
N
N N
O
H
H
CH3
•••
•••
sugar
sugar
guanine
cytosine
N
N
N N
N
O
N N
O
H
H•••
•••
H
N
H•••
H
Key Terms 1137
Key Terms
Achiral Describes an object that can be superimposed with its
mirror image.
Aldose A monosaccharide that contains an aldehyde group.
Base pairing The complementary hydrogen bonding of cytosine
with guanine and of adenine with thymine (in DNA) or with
uracil (in RNA).
Biopolymer Polymers found in biological systems.
Brønsted–Lowry acid A species that acts as a proton donor.
Brønsted–Lowry base A species that acts as a proton acceptor.
Carbohydrate A biopolymer made up of saccharide units.
Chiral Describes an object that cannot be superimposed with its
mirror image.
Conformation One specific geometry of a molecule. The conformations
of a compound differ from one another only by
rotation about single bonds.
Constitutional isomers Compounds that contain the same
number of the same kinds of atoms but that differ in the order
in which their atoms are bonded together. Also known as structural
isomers.
2-Deoxyribose The carbohydrate found in DNA.
Dipeptide A compound in which two amino acid monomers
have joined to form the peptide.
Disaccharide A molecule consisting of two monosaccharides
joined together by a glycosidic bond.
DNA Deoxyribonucleic acid; a nucleic acid consisting of phosphate,
2-deoxyribose, and the four bases adenine, cytosine,
guanine, and thymine.
Enantiomers See Optical isomers.
Enzyme A protein that catalyzes a biological reaction.
Geometric isomers Compounds with different arrangements of
groups on the opposite sides of a bond with restricted rotation,
such as a double bond or a single bond in a ring; for example,
cis–trans isomers of certain alkenes.
Glycosidic bond A bond linking monomers in a polysaccharide.
Ketose A monosaccharide that contains a ketone group.
Monomers The small molecules from which polymers are formed.
Monosaccharide Simple sugars (monomers) from which more
complex carbohydrates are constructed.
Nucleic acid A biopolymer consisting of repeating units of
ribose or deoxyribose, phosphate, and selected bases.
Oligosaccharide A molecule consisting of four to ten monosaccharides
joined together by glycosidic bonds.
Optical isomers Molecules that are nonsuperimposable mirror
images of each other, that is, that bear the same relationship to
each other as do left and right hands; also called enantiomers.
Oxidation (as applied to organic compounds) The increase of
oxygen content or the decrease of hydrogen content of an
organic molecule.
Peptide bond A bond formed by elimination of a molecule of
water between the amino group of one amino acid and the carboxylic
acid group of another.
Peptides Compounds formed by the linking of small numbers
of amino acids.
Polymerization The combination of many small molecules
(monomers) to form large molecules (polymers).
Polymers Large molecules formed by the combination of many
small molecules (monomers).
Polypeptide A polymer composed of amino acids linked by peptide
bonds.
Figure 28-19 A short segment of the double strand
of a DNA molecule.C
A
G
G
T
C
T A
TA
GC
CG
TA
C G
A T
CG
CG
G
A T
AT
A T
Ribose The carbohydrate found in RNA.
RNA Ribonucleic acid, a nucleic acid consisting of phosphate,
ribose, and the four bases adenine, cytosine, guanine, and uracil.
Saponification The hydrolysis of esters in the presence of strong
bases.
Soap The sodium salt of a long-chain fatty acid.
Stereoisomers Isomers in which the atoms are linked together
in the same atom-to-atom order, but with different arrangements
in space. See Geometric isomers, Optical isomers.
Trisaccharide A molecule consisting of three monosaccharides
joined together by glycosidic bonds.
Polysaccharide Carbohydrates that contain more than ten
monosaccharide units.
Primary structure The sequence of the monomeric units in proteins,
carbohydrates, and so on.
Protein A naturally occurring polymeric chain of L-amino acids
linked together by peptide bonds.
Racemic mixture A single sample containing equal amounts of
the two enantiomers (optical isomers) of a compound; does not
rotate the plane of polarized light.
Reduction (as applied to organic compounds) The decrease of
oxygen content or the increase of hydrogen content of an
organic molecule.
1138 CHAPTER 28: Organic Chemistry II: Shapes, Selected Reactions, and Biopolymers
Exercises
Geometries of Organic Molecules
*01. Distinguish between two types of constitutional isomerism.
Give an example of each type.
*02. Distinguish between two types of stereoisomerism. Give
an example of each type.
*03. Which of the following compounds can exist as cis and
trans isomers? Draw them. (a) 2,3-dimethyl-2-butene;
(b) 2-chloro-2-butene; (c) dichlorobenzene; (d) 1,1-di-
chlorocyclobutane.
*04. Which of the following compounds can exist as cis and trans
isomers? Draw them. (a) 1-butene; (b) 2-bromo-1-butene;
(c) 2-bromo-2-butene; (d) 1,2-dichlorocyclopentane.
*05. Distinguish between conformations and isomers.
*06. What is optical isomerism? What are enantiomers?
*07. Which of the following compounds would exhibit optical
isomerism?
(d) CH3CHUCHCH3
*08. Draw three-dimensional representations of the enantiomeric
pairs in Exercise 7.
*09. Write formulas and names for the isomers of (a) bromochlorobenzene,
(b) trifluorobenzene, and (c) chlorotoluene.
What kind of isomerism is illustrated by each of
these sets of compounds?
*10. Write formulas and names for the isomers of (a) propanol,
(b) pentachloropropene, and (c) chlorofluoroethane.
*11. Draw structural representations of two conformations of
butane. Indicate the one having the lower potential
energy.
*12. Draw structural representations of two conformations of
cyclohexane. Give the name commonly applied to each.
CH3CHCH3
OH
(b)
HO
Cl
H
(c) CCH3CHCH2CH3
Br
(a)
Selected Reactions
*13. (a) List the four most acidic organic functional groups.
(b) List the four common organic functional groups that
are bases.
*14. Why are aqueous solutions of amines basic? Show, with
equations, how the dissolution of an amine in water is
similar to the dissolution of ammonia in water.
*15. Show that the reaction of amines with inorganic acids such
as HCl are similar to the reactions of ammonia with inorganic
acids.
*16. What are the equilibrium concentrations of the species
present in a 0.10 M solution of aniline? Kb ϭ 4.2 ϫ 10Ϫ10
C6H5NH2(aq) ϩ H2O(ᐉ) 34 C6H5NH3
ϩ ϩ OHϪ
*17. Which solution would be more acidic: a 0.10 M solution
of aniline hydrochloride, C6H5NH3Cl (Kb ϭ 4.2 ϫ 10Ϫ10
for aniline, C6H5NH2), or a 0.10 M solution of methylamine
hydrochloride, CH3NH3Cl (Kb ϭ 5.0 ϫ 10Ϫ4 for
methylamine, CH3NH2)? Justify your choice.
*18. Choose the compound that is the stronger acid in each
set.
(d) orOH OH
(c) orOH
O
C OH
CH3CH2OH(b) or OH
CH3CH2CH2OH(a) or
O
C OH
*19. (a) What are alkoxides? (b) What do we mean when we say
that the low-molecular-weight alkoxides are strong bases?
*20. (a) Write equations for the reactions of three alcohols with
metallic sodium. (b) Name all compounds in these equations.
(c) Are these reactions similar to the reaction of
metallic sodium with water? How?
*21. Which physical property of aldehydes is used to advantage
in their production from alcohols?
*22. The Kb for lidocaine (a monobasic organic compound,
MW ϭ 234.3) is 7.0 ϫ 10Ϫ6. What is the pH of a 1.5 percent
solution of lidocaine? The density of the solution is
1.00 g/mL.
*23. How are the terms “oxidation” and “reduction” often used
in organic chemistry? Classify the following changes
as either oxidation or reduction: (a) CH4 to CH3OH;
(b) CH2UCH2 to CH3XCH3; (c) CH3CH2CHO to
CH3CH2CH3;
*24. Classify the following changes as either oxidation or reduction:
(a) CH3OH to CO2 and H2O; (b) CH2CH2
to CH3CHO; (c) CH3COOH to CH3CHO;
(d) CH3CHUCH2 to CH3CH2CH3.
*25. Write equations to illustrate the oxidation of the following
aromatic hydrocarbons by potassium permanganate in
basic solution: (a) toluene; (b) ethylbenzene; (c) 1,2-di-
methylbenzene.
*26. (a) Do you expect toluene to produce soot as it burns?
Why? (b) Would you expect the flames to be blue or yel-
low?
*27. Describe the preparation of three aldehydes from alcohols,
and write appropriate equations. Name all reactants and
products.
*28. Describe the preparation of three ketones from alcohols,
and write appropriate equations. Name all reactants and
products.
*29. An ester is the product of the reaction of an alcohol with
a carboxylic acid and is the organic version of an inorganic
salt. What is an inorganic ester?
*30. (a) Write equations for the formation of three inorganic
esters. (b) Name the inorganic ester formed in each case.
*31. (a) What is nitroglycerine? (b) Write the chemical equation
that shows the preparation of nitroglycerine.
(c) List two important uses for nitroglycerine. Are they
similar?
*32. Write equations for the formation of three different esters,
starting with an acid and an alcohol in each case. Name all
compounds.
*33. Write equations for the formation of three different esters,
starting with a different acid chloride and a different alcohol
in each case. Name all compounds.
(d) toCH3
O
C OH
*34. Write equations for the hydrolysis of (a) methyl acetate,
(b) ethyl formate, (c) butyl acetate, and (d) octyl acetate.
Name all products.
*35. (a) What is saponification? (b) Why is this kind of reaction
called saponification?
Biopolymers
*36. What is necessary if a molecule is to be capable of polymerization?
Name three types of molecules that can
polymerize.
*37. A cellulose polymer has a molecular weight of 750,000.
Estimate the number of units of the monomer, ␤-glucose
(C6H12O6) in this polymer. This polymerization reaction
can be represented as
xC6H12O6 88n cellulose ϩ (x Ϫ 1)H2O
*38. Write structural formulas of two monosaccharides that
contain six carbon atoms.
*39. Write the structural formula of a trisaccharide containing
only ␣-1,4-linked glucose units.
*40. Write two structural formulas of fructose, one in its cyclic
form and one in its straight-chain form.
*41. Describe the structure of a natural amino acid molecule.
What kind of isomerism do most amino acids exhibit?
Why?
*42. How are the amino acid units in a polypeptide joined
together? What are the links called?
*43. Consider only two amino acids:
Write the structural formulas for the dipeptides that could
be formed containing one molecule of each amino acid.
*44. How many different dipeptides can be formed from the
three amino acids A, B, and C? Write the sequence of
amino acids in each. Assume that an amino acid could occur
more than once in each dipeptide.
*45. How many different tripeptides can be formed from the
three amino acids A, B, and C? Write the sequence of
amino acids in each. Assume that an amino acid could occur
more than once in each tripeptide.
*46. Aspartame (trade name NutraSweet) is a methyl ester of a
dipeptide:
Write the structural formulas of the two amino acids that
are combined to make aspartame (ignoring optical iso-
merism).
CH2
COOCH3
NH2CH NH
O
C CH
CH2COOH
NH2 COOH
H
C
RЈ
NH2 COOH
H
C
R
and
Exercises 1139
*55. (a) Name three biopolymers found in animals. (b) Name
at least one biopolymer that is composed of identical
monomers.
*56. Name the three types of monomers that are found in DNA.
Give a structural formula of an example of each.
CONCEPTUAL EXERCISES
*57. Nylon is decomposed by acids, but polyethylene is not.
Suggest an explanation for this difference in behavior.
*58. How does the heat of combustion of ethyl alcohol compare
with the heats of combustion of low-molecular-weight
saturated hydrocarbons on a per-mole basis and on a pergram
basis?
BUILDING YOUR KNOWLEDGE
*59. A laboratory procedure calls for oxidizing 2-propanol to
acetone using an acidic solution of K2Cr2O7. An insufficient
amount of K2Cr2O7 is on hand, however, so the
laboratory instructor decides to use an acidic solution of
KMnO4 instead. What mass of KMnO4 is required to carry
out the same amount of oxidation as 1.00 g of K2Cr2O7?
*60. The chemical equation for the water gas reaction is
C(s) ϩ H2O(g) 34 CO(g) ϩ H2(g)
At 1000 K, the value of Kp for this reaction is 3.2. When
we treat carbon with steam and allow the reaction to reach
equilibrium, the partial pressure of water vapor is observed
to be 15.6 atm. What are the partial pressures of CO and
H2 under these conditions?
*61. (a) In aqueous solution, acetic acid exists mainly in the
molecular form (Ka ϭ 1.8 ϫ 10Ϫ5). Calculate the freezing
point depression for a 0.10 molal aqueous solution
of acetic acid, ignoring any ionization of the acid.
Kf ϭ 1.86°C/molal for water. (b) In nonpolar solvents such
as benzene, acetic acid exists mainly as dimers
as a result of hydrogen bonding. Calculate the freezing
point depression for a 0.10 molal solution of acetic acid in
benzene. Kf ϭ 5.12°C/molal for benzene. Assume complete
dimer formation.
*62. What is the pH of a 0.10 M solution of sodium benzoate?
Ka ϭ 6.3 ϫ 10Ϫ5 for benzoic acid, C6H5COOH. Would
this solution be more or less acidic than a 0.10 M solution
of sodium acetate? Ka ϭ 1.8 ϫ 10Ϫ5 for acetic acid,
CH3COOH.
CH3 CH3C C
O O
O OH
H
*47. Write the structural formula of the carbohydrate found in
deoxyribonucleic acid (DNA). How does this carbohydrate
differ from the one found in ribonucleic acid (RNA)?
*48. Write structural formulas of the four bases found in DNA.
Circle at least one functional group on each that gives the
compound its base properties.
*49. Draw a short segment of the DNA biopolymer that contains
three each of the phosphate, carbohydrate, and base groups.
*50. In the double-strand form of DNA the base groups are
paired. What base is found paired (hydrogen bonded) with
thymine?
Mixed Exercises
*51. Identify the major products of each reaction.
*52. Identify the major products of each reaction.
*53. Write a structural formula for each of the following
compounds:
(a) CH3OH (d) CH3OCH3
(b) CH3COOH (e) CH3CHCHCH2CH3
(c) CH3CHOHCH3
*54. Write a structural formula for each of the following
compounds:
(a) CH3ONO2 (d) ClCH2CHCHCl
(b) CH3CHCH2 (e) CH3CH2CH2ONa
(c) CH3CH2CH2OH
O
C OH
OCCH3
O
(c)
NaOH(aq)
heat
(b)
O
CH3CH2COCH3
KOH(aq)
heat
(a) CH2OH ϩ Na
(c)
O
CH3CH
MnO4
Ϫ
Hϩ
(b)
O
C OH ϩ CH3OH
H2SO4
heat
CH3
2NaOH
O
C OH
OH
ϩ(a)
1140 CHAPTER 28: Organic Chemistry II: Shapes, Selected Reactions, and Biopolymers
APPENDIX A
SOME MATHEMATICAL OPERATIONS
In chemistry we frequently use very large or very small numbers. Such numbers are conveniently
expressed in scientific, or exponential, notation.
SCIENTIFIC NOTATION
In scientific notation, a number is expressed as the product of two numbers. By convention,
the first number, called the digit term, is between 1 and 10. The second number, called
the exponential term, is an integer power of 10. Some examples follow.
10000 ϭ 1 ϫ 104 24327 ϭ 2.4327 ϫ 104
1000 ϭ 1 ϫ 103 7958 ϭ 7.958 ϫ 103
100 ϭ 1 ϫ 102 594 ϭ 5.94 ϫ 102
10 ϭ 1 ϫ 101 98 ϭ 9.8 ϫ 101
1 ϭ 1 ϫ 100
1/10 ϭ 0.1 ϭ 1 ϫ 10Ϫ1 0.32 ϭ 3.2 ϫ 10Ϫ1
1/100 ϭ 0.01 ϭ 1 ϫ 10Ϫ2 0.067 ϭ 6.7 ϫ 10Ϫ2
1/1000 ϭ 0.001 ϭ 1 ϫ 10Ϫ3 0.0049 ϭ 4.9 ϫ 10Ϫ3
1/10000 ϭ 0.0001 ϭ 1 ϫ 10Ϫ4 0.00017 ϭ 1.7 ϫ 10Ϫ4
The exponent of 10 is the number of places the decimal point must be shifted to give the
number in long form. A positive exponent indicates that the decimal point is shifted right
that number of places. A negative exponent indicates that the decimal point is shifted left.
When numbers are written in standard scientific notation, there is one nonzero digit to the
left of the decimal point.
7.3 ϫ 103 ϭ 73 ϫ 102 ϭ 730 ϫ 101 ϭ 7300
4.36 ϫ 10Ϫ2 ϭ 0.436 ϫ 10Ϫ1 ϭ 0.0436
0.00862 ϭ 0.0862 ϫ 10Ϫ1 ϭ 0.862 ϫ 10Ϫ2 ϭ 8.62 ϫ 10Ϫ3
In scientific notation the digit term indicates the number of significant figures in the
number. The exponential term merely locates the decimal point and does not represent
significant figures.
Addition and Subtraction
In addition and subtraction all numbers are converted to the same power of 10, and the
digit terms are added or subtracted.
(4.21 ϫ 10Ϫ3) ϩ (1.4 ϫ 10Ϫ4) ϭ (4.21 ϫ 10Ϫ3) ϩ (0.14 ϫ 10Ϫ3) ϭ 4.35 ϫ 10Ϫ3
(8.97 ϫ 104) Ϫ (2.31 ϫ 103) ϭ (8.97 ϫ 104) Ϫ (0.231 ϫ 104) ϭ 8.74 ϫ 104
A-1
Recall that, by definition,
(any base)0 ϭ 1.
Multiplication
The digit terms are multiplied in the usual way, the exponents are added algebraically,
and the product is written with one nonzero digit to the left of the decimal.
(4.7 ϫ 107)(1.6 ϫ 102) ϭ (4.7)(1.6) ϫ 107ϩ2 ϭ 7.52 ϫ 109 ϭ 7.5 ϫ 109
(8.3 ϫ 104)(9.3 ϫ 10Ϫ9) ϭ (8.3)(9.3) ϫ 104Ϫ9 ϭ 77.19 ϫ 10Ϫ5 ϭ 7.7 ϫ 10Ϫ4
Division
The digit term of the numerator is divided by the digit term of the denominator, the exponents
are subtracted algebraically, and the quotient is written with one nonzero digit to
the left of the decimal.
ᎏ
8
2
.
.
4
0
ϫ
ϫ
1
1
0
0
7
3
ᎏ ϭ ᎏ
8
2
.
.
4
0
ᎏ ϫ 107Ϫ3 ϭ 4.2 ϫ 104
ᎏ
8.
3
4
.
1
8
2
1
ϫ
ϫ
1
1
0
0
Ϫ
9
3
ᎏ ϭ ᎏ
8
3
.4
.8
1
1
2
ᎏ ϫ 10[9Ϫ(Ϫ3)] ϭ 0.45292 ϫ 1012 ϭ 4.53 ϫ 1011
Powers of Exponentials
The digit term is raised to the indicated power, and the exponent is multiplied by the
number that indicates the power.
(1.2 ϫ 103)2 ϭ (1.2)2 ϫ 103ϫ2 ϭ 1.44 ϫ 106 ϭ 1.4 ϫ 106
(3.0 ϫ 10Ϫ3)4 ϭ (3.0)4 ϫ 10Ϫ3ϫ4 ϭ 81 ϫ 10Ϫ12 ϭ 8.1 ϫ 10Ϫ11
Electronic Calculators To square a number: (1) enter the number and (2) touch the (x2)
button.
(7.3)2 ϭ 53.29 ϭ 53 (two sig. figs.)
To raise a number y to power x: (1) enter the number; (2) touch the (yx) button; (3) enter
the power; and (4) touch the (=) button.
(7.3)4 ϭ 2839.8241 ϭ 2.8 ϫ 103 (two sig. figs.)
(7.30 ϫ 102)5 ϭ 2.0730716 ϫ 1014 ϭ 2.07 ϫ 1014 (three sig. figs.)
Roots of Exponentials
The exponent must be divisible by the desired root if a calculator is not used. The root
of the digit term is extracted in the usual way, and the exponent is divided by the desired
root.
͙2ෆ.5ෆ ϫෆ 1ෆ0ෆ5ෆ ϭ ͙2ෆ5ෆ ϫෆ 1ෆ0ෆ4ෆ ϭ ͙2ෆ5ෆ ϫ ͙1ෆ0ෆ4ෆ ϭ 5.0 ϫ 102
͙
3
2ෆ.7ෆ ϫෆ 1ෆ0ෆϪෆ8ෆ ϭ ͙
3
2ෆ7ෆ ϫෆ 1ෆ0ෆϪෆ9ෆ ϭ ͙
3
2ෆ7ෆ ϫ ͙
3
1ෆ0ෆϪෆ9ෆ ϭ 3.0 ϫ 10Ϫ3
Electronic Calculators To extract the square root of a number: (1) enter the number and (2)
touch the (͙xෆ) button.
͙2ෆ3ෆ ϭ 4.7958315 ϭ 4.8 (two sig. figs.)
To extract some other root: (1) enter the number y; (2) touch the (INV) and then the ( yx)
button; (3) enter the root to be extracted, x; and (4) touch the (=) button.
A-2 APPENDIX A: Some Mathematical Operations
Two significant figures in answer.
Two significant figures in answer.
These instructions are applicable to
most calculators. If your calculator has
other notation, consult your
calculator’s instruction booklet.
Three significant figures in answer.
On some models, this function is
performed by the ͙
x
yෆ button.
LOGARITHMS
The logarithm of a number is the power to which a base must be raised to obtain the
number. Two types of logarithms are frequently used in chemistry: (1) common logarithms
(abbreviated log), whose base is 10, and (2) natural logarithms (abbreviated ln), whose base
is e ϭ 2.71828. The general properties of logarithms are the same no matter what base is
used. Many equations in science were derived by the use of calculus, and these often
involve natural (base e) logarithms. The relationship between log x and ln x is as follows.
ln x ϭ 2.303 log x
Finding Logarithms The common logarithm of a number is the power to which 10
must be raised to obtain the number. The number 10 must be raised to the third power
to equal 1000. Therefore, the logarithm of 1000 is 3, written log 1000 ϭ 3. Some examples
follow.
Number Exponential Expression Logarithm
1000 103 3
100 102 2
10 101 1
1 100 0
1/10 ϭ 0.1 10Ϫ1 Ϫ1
1/100 ϭ 0.01 10Ϫ2 Ϫ2
1/1000 ϭ 0.001 10Ϫ3 Ϫ3
To obtain the logarithm of a number other than an integral power of 10, you must use
either an electronic calculator or a logarithm table. On most calculators, you do this by
(1) entering the number and then (2) pressing the (log) button.
log 7.39 ϭ 0.8686444 ϭ 0.869
log 7.39 ϫ 103 ϭ 3.8686 ϭ 3.869
log 7.39 ϫ 10Ϫ3 ϭ Ϫ2.1314 ϭ Ϫ2.131
The number to the left of the decimal point in a logarithm is called the characteristic, and
the number to the right of the decimal point is called the mantissa. The characteristic only
locates the decimal point of the number, so it is usually not included when counting significant
figures. The mantissa has as many significant figures as the number whose log was
found.
To obtain the natural logarithm of a number on an electronic calculator, (1) enter the
number and (2) press the (ln) or (ln x) button.
ln 4.45 ϭ 1.4929041 ϭ 1.493
ln 1.27 ϫ 103 ϭ 7.1468 ϭ 7.147
Finding Antilogarithms Sometimes we know the logarithm of a number and must find
the number. This is called finding the antilogarithm (or inverse logarithm). To do this on a
calculator, we (1) enter the value of the log; (2) press the (INV) button; and (3) press the
(log) button.
log x ϭ 6.131; so x ϭ inverse log of 6.131 ϭ 1.352 ϫ 106
log x ϭ Ϫ1.562; so x ϭ inverse log of Ϫ1.562 ϭ 2.74 ϫ 10Ϫ2
A-2
A-2 Logarithms A-3
On some calculators, the inverse log is
found as follows:
1. enter the value of the log
2. press the (2ndF) (second function)
button
3. press (10x)
ln 10 ϭ 2.303
To find the inverse natural logarithm, we (1) enter the value of the ln; (2) press the (INV)
button; and (3) press the (ln) or (ln x) button.
ln x ϭ 3.552; so x ϭ inverse ln of 3.552 ϭ 3.49 ϫ 101
ln x ϭ Ϫ1.248; so x ϭ inverse ln of Ϫ1.248 ϭ 2.87 ϫ 10Ϫ1
Calculations Involving Logarithms
Because logarithms are exponents, operations involving them follow the same rules as the
use of exponents. The following relationships are useful.
log xy ϭ log x ϩ log y or ln xy ϭ ln x ϩ ln y
log ᎏ
x
y
ᎏ ϭ log x Ϫ log y or ln ᎏ
x
y
ᎏ ϭ ln x Ϫ ln y
log x y ϭ y log x or ln xy ϭ y ln x
log ͙
y
xෆ ϭ log x1/y ϭ ᎏ
1
y
ᎏ log x or ln ͙
y
xෆ ϭ ln x1/y ϭ ᎏ
1
y
ᎏ ln x
QUADRATIC EQUATIONS
Algebraic expressions of the form
ax2 ϩ bx ϩ c ϭ 0
are called quadratic equations. Each of the constant terms (a, b, and c) may be either
positive or negative. All quadratic equations may be solved by the quadratic formula.
x ϭ
If we wish to solve the quadratic equation 3x2 Ϫ 4x Ϫ 8 ϭ 0, we use a ϭ 3, b ϭ Ϫ4, and
c ϭ Ϫ8. Substitution of these values into the quadratic formula gives
x ϭ ϭ
ϭ ϭ
The two roots of this quadratic equation are
x ϭ 2.4 and x ϭ Ϫ1.1
As you construct and solve quadratic equations based on the observed behavior of
matter, you must decide which root has physical significance. Examination of the equation
that defines x always gives clues about possible values for x. In this way you can tell which
is extraneous (has no physical significance). Negative roots are often extraneous.
When you have solved a quadratic equation, you should always check the values you
obtained by substitution into the original equation. In the preceding example we obtained
x ϭ 2.4 and x ϭ Ϫ1.1. Substitution of these values into the original quadratic equation,
3x2 Ϫ 4x Ϫ 8 ϭ 0, shows that both roots are correct. Such substitutions often do not give
a perfect check because some round-off error has been introduced.
4 Ϯ 10.6
ᎏ
6
4 Ϯ ͙1ෆ1ෆ2ෆ
ᎏᎏ
6
4 Ϯ ͙1ෆ6ෆ ϩෆ 9ෆ6ෆ
ᎏᎏ
6
Ϫ(Ϫ4) Ϯ ͙(Ϫෆ4ෆ)2ෆ Ϫෆ 4ෆ(3ෆ)(ෆϪෆ8ෆ)ෆ
ᎏᎏᎏᎏ
2(3)
Ϫb Ϯ ͙b2ෆ Ϫෆ 4ෆacෆ
ᎏᎏ
2a
A-3
A-4 APPENDIX A: Some Mathematical Operations
On some calculators, the inverse
natural logarithm is found as follows:
1. enter the value of the ln
2. press the (2ndF) (second function)
button
3. press (ex)