Entropy, S
Natural processes have a preferred direction of progress, systems tend to progress in the direction of increasing entropy
Entropy
= a measure of the number of ways in which a system may be arranged
= a measure of accessible energy levels
= a measure of thermal effects of reversible processes
Reversible processes = a small change in conditions could reverse the direction
Irreversible processes 1
Spontaneous Processes
Proceed in a given direction without being driven by an outside source of energy
Increase entropy S of Universe
Proceed in a direction towards states with highest probabilities
Lead to disipitation of energy
2
Spontaneous Processes
Expansion of gas Heat transfer
AS = R In Vfin/Vinit AS = Cp ln x2/Ti
(1 mol of ideal gas)
Second Law of Thermodynamics
Entropy of universe increases
Spontaneous processes increase entropy of universe
ASuniv = ASsyst + ASsurr
ASuniv > 0 Spontaneous processes
ASuniv < 0 Process does not proceed in given direction
AS j = 0 Equilibrium
To establish spontaneity of a process, we need to know ASsyst and AS
surr
Third Law of Thermodynamics
Entropy of ideal crystal at 0 K equlas zero
• ideal crystal does not exist
• 0 K cannot be attained
Reference state - perfect ordering
- motions, vibrations, rotations ceased
S = kln W
W = number of microstates of a system
AtOK   W=1,S = 0
Boltzmann Equation
S = kln W
k = R/NA = 1.38066 1023J K1
Ludwig Edward Boltzmann
W = number of microstates of a system
It is possible to establish value of S for
a given state
(in contrast to H or U)
On October 5,1906 committed suicide in Duinu by Trieste
Boltzmann Equation
T = 0
Winit = 1 S = 0
k = Boltzmann constant = 1.3807 10~23 J K~1 W = number of microstates of a system
Wfin = ?
In WB = S/k = 41 /1.3807 10"23 = 1024
Standard Entropy
S° = Standard molar entropy of a substance látky at 298 K and 1 bar
(increase of S on heating a substance from 0 K to 298 K) S° = AS = S (298 K) - S (0 K) J moh1 K"1
Standard Entropies S° at 298 K and 1 bar
Substance	S°, J K-1 mol1	Substance	S°, J K"1 mol1
S8(g)	431	H20 (g)	189
SF6(g)	292	H20 (I)	70
02(g)	205	H20 (s)	41
C02(g)	248	CaC03(s)	93
CO(g)	198	CaO(s)	40
H2 (g)	131	Sn (s) white	52
CH3OH (g)	240	Sn (s) gray	44
CH3OH (I)	127	C(s) graphite	6
C2H5OH (I)	161	C(s) diamond 2	
10
Standard Entropies S°
Dissolution
Substance   S°, J K1 moM
CH3OH (I) CH3OH (aq) NH4CI (s) NH4CI (aq)
127 133
168
12
Standard Entropies S°
Molecular mass, number of atoms in a molecule, number of vibrations and rotations
Substance
f2 (g) ci2 (g)
Br2 (g)
i2(g)
S°, J K1 mo I1 203 223
Heavy molecules - energy levels close spaced, more available states
13
Standard Entropies S°
Substance     S°, J K1 mol"
Chemical composition
More complicated molecules
NaCI (s) MgCI2 (s) AICI3 (s)
74 90 167
14
Standard Entropies S°
Reaction Entropy
AS0 = y n     S°      - Y n S°
r prod     prod react react
Products - Reactants
CH4(g) + 2 02(g) -> C02(g) + 2 H20(1) AS0 = [2(69.9) + 213.6] - [182.6 + 2(205.0)]
= -242.8 J K
AS°r < 0   for reactions:
Formation of solids or liquids from gases
Total number of moles of gases decresas
AS°r > 0 for reactions:
Formation of gases from solids or liquids
Total number of moles of gases incresas
Second Law of Thermodynamics
Entropy of universe increases
Spontaneous processes increase entropy of universe
ASuniv = ASsyst + ASsurr
ASuniv > 0 Spontaneous processes
ASuniv < 0 Process does not proceed in given direction
AS j = 0 Equilibrium
To establish spontaneity of a process, we need to know ASsyst and AS
surr
Heat Exchange between System and
Surroundings
■V At p = const
■fcg^^     Heat (surr) = - AH (system)
I       Given off (+) Lost (-)
Removed (—) Absorbed (+)
We can establish AS.
AH
< 0 exo
> 0 endo
AS
surr
> 0 incr
At 298 K
Sb406(s) + 6C(s) -> 4Sb(s) + 6C02(g) AH = 778 kj ASL1PP = -AH/T = -778 kj / 298 K = -2.6 kj K1
(aj_lb)_IS^Ml
Transfer of the same amount of heat at lower temperature increases relatively more entropy of surroundings - cold surroundings are more ordered and then more disturbs
Reaction Entropy
2Fe(s) + 3H20(g) -> Fe203(s) + 3H2(g) AS°r = [S°(Fe203(s) + 3S°H2(g)] - [2S°Fe(s) + 3S°H20(g)]
AS° = -141.5 J K1
Is this reaction spontaneous at 298 K, is AS°univ > 0?
ASuniv _ ASsyst + ASsurr
AS°r = AS°syst =-141.5 J K-1
Reaction Spontaneity
AS0surr = -AH°syst/T = -AH°r/T
AH° = AH°f(Fe203(s)) + 3AH°f(H2(g)) - 2AH°f(Fe (s)) - 3 AH°f(H20(g)) = -100 kJ
AS°surr = - AH°syst/T = 336 J K'
AS univ - AS syst + AS surr
= -141.5 + 336= 194.0 J K-1
Reaction is spontaneous at 298 K, AS°univ > 0
Entropy of Phase Transitions
H20(l) ±5 H20(g) at 373 K
H20(s) t; H20(l) at 273 K
23
Entropy of Phase Transitions
H20(l) ±5 H20(g) pri 373 K
Phase Transitions are equilibrium processes at which AS°univ = 0
AS°syst = S°(H20(g)) - S°(H20(l)) =
195.9 J K"1-86.6 J K"1 = 109.1 JK-1
AS°surr = -AHvap/ T = -40.7 kJ/373 K = -109.1 J K1
AS°univ = AS°syst + AS°surr = 0
25
Spontaneous Processes and Gibbs Energy
Reaction is spontaneous when ASuniv > 0
ASUniv - ASsyst + ASsurr - ASsyst   AHsyst/T > 0
Multiply by   - T
AH - TASsyst < 0
Multiply by -1 reverse unequal.
AG ■ Gibbs Free Energy (= - TASuniv)
AG = AH^, - TASs;sl When AG is negative, reaction is spontaneous
Gibbs Free Energy
1. AG is a state function
2. AG° Gibbs Free Energy at standard cond. -298 K
-1 bar for gases
-1 mol I-1 concentration
3. AG° tabulated
1/202 (g) + N2(g) ^ N20 (g)
AG°f(N20) = 104.18 kJ moh1
Reactants are more stabile than products Kinetic factors of N20 stability
27
Standard Gibbs Free Energy of Formation
AG°f calculated from AH°f and S° C(graphite) + 02(g) S C02(g)
AH°f = AHr° = - 393.5 kJ mol1
AS° = S° (C02(g)) - S° (C(graphite)) - S° (02(g)) AS° = 213.60 - 5.74 - 205.00 = 2.86 J K 1 mol1
AG°f = AH°f-TAS°f
AG°f = -393.5 - (298)(2.86) = - 394.360 J K 1 mol1
28
Standard Gibbs Free Energy of Formation
(at 25 °C)
AG°f
Substance	AG °„ kJ moľ1
NH3	-16.45
C02	- 394.4
N02	+ 51.3
H20 (g)	- 228.6
H20 (1)	- 237.1
C6H6	+ 124.3
C2H5OH	-174.8
AgCI	-109.8
CaC03	- 1128.8 29
AG°r Calculated from AGf°
AG°r = 2 nprod AG°f (prod) - 2 nreact AG°f (react)
aA + bB ^ cC + dD
AG0 = cAG°f (C ) + dAG°f (D) - aAG°f (A) - bAG°f (B)
3NO(g) ±5 N20(g) + N02(g)      AG0 = ?
AG0 = AG°f(N20)+ AG°f(N02)-3AG°f(NO) AG°r = 104.18 + 51.29 - 3(86.55) = -104 kJ mol1
Influence of Temperature on AG0	
AG0 = AH0 - T AS0	
AH0 +   AS0 +       AG0 Negative at high T	
AH0 +   AS0 -       AG0 Positive at all T	
AH0 -   AS0 +       AG0 Negative at all T	
AH0 -   AS0 -        AG0 Negative at low T	
	31
Chemical Equilibria A
In laboratory
NaXO* + CaCU     CaCCX + 2 NaCi
Natron on banks of salt lakes in Egypt CaCCX + 2 NaCI    NaXCX + CaCU
C. L. Berthollet A (       . .
,a-,ac a An excess of a product can reverse the
(1748-1822) r u       i *■
v '   course of chemical reaction
Reversibile reaction Na2C03 + CaCI2 ^
CaCO, + 2 NaCI
L
Reaction Quotient Q
Unequilibrium
concentrations
powered to stoichiometric
coefficients
Reversibile rea
[A] = [B] = 1 M [C] = [D] = 0
[A] = [B] = 0 [C] = [D] = 1 M
Q - Reaction At the start of reaction:
quotient
Q = 0/1 0
How far a reaction
proceeded from A complete reaction:
reactants to products Q _ 1/0 ^
(for a = b = c = d = 1)
Chemical Composition and AG
One of the most important equations in chemistry !
AG = AG0 + RT InQ Q = Reaction quotient
3NO(g) ±5 N20(g) + N02(g)   AG0 = -104 kJ mol1
NO = 0.3 atm ; N70 = 2 atm ; NO, = 1 atm Which direction ?
AG = AG0 + RT InQ = -104.0 + (8.314 J K 1 mol1)(298 K) In (74.1)
AG = - 93.3 kJ mol-1 Reaction is spontaneous to the right
more NO decomposes to products 34
aA + bB ±5 cC + dD
Pure
rcac rants
Chemical Composition and AG
AG = AG°r + RT InQ
Direction of spontaneous reaction
AG!
AG°r < 0
Pure products
AG
1
Pure
reactants
Pure products
T
H
a
AG0
Pure
reactants
Q = K
At equilibrium AG = 0
AG0 and Equilibrium Constant K
AG = AG0 + RT InQ At equilibrium AG = 0 and Q = K
AG0 = - RT InK   II -—
aA + bB ^ cC + dD
Reaction Quotient Q and Equilibrium Constant K
Q = K. System at equilibrium, no change.
Q > K. Concentrations of products are larger then equilibrium concentrations. A part of the products must convert back to reactants to attain equilibrium. Reaction shift to the left.
Q < K. Concentrations of reactants are larger then equilibrium concentrations. A part of the reactants must react to products to attain equilibrium. Reaction shift to the right. 37
AG0 and Equilibrium Constant K	
3NO(g) ±5 N2Q(g) + NQ2(g)   AG0 = -104 kJ mol1	
	■
	
k [no2][n2o\\	
	38
AG0 = - RT InK
AG0	K	At equilibrium (AG = 0)
<0	> 1	More products
>0	< 1	More reactants
= 0	= 1	
Equilibrium Constant K
K is a function only of temperature Pure phases (I, s) do not influence equilibrium Concentration of solvent is not considered K is dimensionless
Concentrations related to standard state 1 M
40
Guldberg-Waage Law
1864 Law of mass action
aA + bB ±5 cC + dD
Cato Maximilian Guldberg (1836-1902) Peter Waage (1833-1900) 41
Guldberg-Waage Law	
cC + dD ±5 aA + bB	
Reverse reaction, Knew = 1/ K	
ncC + ndD ±z naA + nbB	
Multiply equation by a constant   Knew = (K)n	
Sum of chemical equations	
Knew = K1 X K2	42
Guldberg-Waage Law	
2 N02 i=» 2 NO + 02 Kj	
2 S02 + 02 ^ 2 S03 K2	
N02 + S02 i+ NO + S03    K3 = ?	
K3 = (Kx x K2)1/2 = a/Kj x K2	
	43
Attaining Chemical Equilibrium
H2 + l2 ^2 HI 2 HI -> H2 + l2
concentration						
		H2 + 12 -> 2 HI		_                      2HI^H2 +	2	
		HI				
						
0		^2, 12 equilibrium ^ state		\~^2/ '2 equilibrium ^tate		
	time ->		time   -> 1			
	1 Equilibrium concentrations			1 Equilibrium concentrations		
LeChatelier's Principle
Reversibile reactions
If a chemical system at equilibrium experiences a change in concentration, temperature, volume, or partial pressure, then the equilibrium shifts to counteract the imposed change and a new equilibrium is established.
Henri LeChatelier (1850-1936)
. Addition of H
time
Concentrations
K is const.    C02 (g) + H2 (g) i;   H20 (g) + CO (g)
Trapping water, shift to right
NaCI (s) + H2S04(l) ±5 Na2S04(s) + HCI (g)f Gaseous HCI escapes, shift to right
H2(g) + l2(g)^2HI(g)
Addition of inert N2, does not take part in reaction, no change in number of moles, no shift
46
Transfer of Oxygen and C02
hemoglobin Fe2+ highspin
^ oxyhemoglobin Fe2+ lowspin
02-hemoglobin ±+ hemoglobin ^ CO-hemoglobin
H20 + C02 + C032- tř 2HCO3-
47
Pressure
Reactions with changing number of moles of gases 2 N02(g) ^ N204(g)
Ang = (nprod " nreact) = 1 - 2 = -1
V halved, p 2x bigger Q = 72 Kp M____
Increased pressure shifts reaction to the right
compress
shift
New
equilibrium
48
Pressure m
■
N2(g) + 3 H2(g) ±5 2 NH3(g) AH = -92 kJ moh1    j V
• the reaction is exothermic \
• decreasing number of moles of gases *
Fritz Haber
According to LeChatelier, the yield will be at (i868-1934) maximum at high pressure and low Mn . _ ,n,0
a   K NP in Chemistry 1918
temperature
At low temperature, the reaction is slow uses Fe catalyst to speed up Conditions
20-100 MPa and 400-600 °C
wmfi
i/
Pressure
K is con
H2 P NH3
N2(g) + 3 H2(g) ±5 2 NH3(g)
50
Equilibrium Constant K pV=nRT  p = (n/V)RT=cRT p = Pj + p2 + p3 +.....Partial pressures N2(g) + 3 H2(g) ±5 2 NH3(g)		
	„ [NH3]2	51
		
Equilibrium Constant K
jA + kB ±5 IC + mD
K = Kc (RT)An
A/7 = (/ + m) - (j + k)
(c<2 x i?r)i(cD x j?rr
C*V)(Jfc*)
Heterogeneous Equilibria
CaC03(s) ±5 CaO(s) + C02(g)
K = [C02][CaO] / [CaC03] =        [C02] = p(C02)
Activity (concentration) of pure liquids and solids is constant and does not appear in K.
[CaO] = [CaC03] = const.   Addition does not change K
53
Heterogeneous Equilibria
2H20(/) ±5 2H2(g) + 02(g) K=[H2]2[02] Kp = p2(H2)p(02)
2H20(g) ±5 m2(g) + 02(g)
K = [H2]2[02] / [H20]2     Kv = p2(H2) p(02) / p2(H20)
Temperature
K changes with
Compare K at T1 and T2   (K-, and K2)
van't Hoff equation
55
Temperature
cooling
heating
2 N02 (g) ±5 N204 (g) AH° = - 63 kJ mor
Exothermic reactions shift to right
- T2 < T,
on cooling = K incr., K2 >
Heat is a product in exothermic reaction 2N02(g) ^N204(g) + Q
	
	
Equilibria in Exothermic Reactions
N2(g) + 3 H2(g) ^ 2 NH3(g)    AH° = - 92 kJ moM Yield decreases with increasing T
Temperature
CaC03(s) ±5 CaO(s) + C02(g)   AH° = 556 kJ moh1
Endothermic reactions shift to right on heating T2>T1
K incr, K2 > ^ , K = p(C02)
Heat is a reactant in endothermic reactions CaC03(s) + Q ±5 CaO(s) + C02(g)
Equilibrium Concentrations
H2(0) + F2(fli) *? 2HF(g)
K= 1.15 102 = [HF]2/[H2][F2] [H2]0=1.00/W [F2]0=2.00M [HF]0 = 0
Initial
Change
Equilibrium
H2(flf) 1.00
-x
1.00-x
F2(0) 2.00
- x
2.00-x
2HF(^) 0
+2x 2x
K = 1.15 102 = [HF]2 / [H2][F2] = (2x)2 / (1.00 - x)(2.00 - x)
59
Equilibrium Concentrations
x, 2 = [- b + (b2 - 4ac),/2] / 2a
x1 = 2.14 mol h1 a x2 = 0.968 mol h1 Use
x2= 0.968 mol h1
[H2]= 1.000 M- 0.968 yW= 3.2 10"2yW [F2] = 2.000 yW-0.968 yW= 1.032 M [HF] = 2 (0.968 /W) = 1.936 M
60