102 ANAI VMS (II IIAII HAIA .i i| vi 11 I XAMI'I I KKI where P| is the total pressure ol 1 lit- reaelor, V is ilu- volume in the reactor, N| is (lie number of moles in (he reaelor, R is Ihe gas law eonslanl, and T is the temperature. Al the start of the reaction p£V = N'J.RT (3.90) where P" is the initial pressure of" the reactor and N" is the total number of moles in the reactor at the start of the reactions. Dividing equation (3.89) by equation (3.90) and rearranging yields P" (3.91) Substituting NT and N° from the stoichiometric table into equation (3.91) and rearranging yields = 1 + /Amol \ (3.92) Therefore, if one knows the pressure in a vessel as a function of time, one can calculate the conversion as a function of time. Figure 3.28 shows a plot of the pressure calculated from equation (3.92) as a function or conversion for various values of the Amol. Notice that the pressure always varies linearly with conversion. The pressure goes up when Amol is positive, while the pressure goes down when Amol is negative. Consequently, whenever Amol is nonzero, one can use the pressure to estimate the conversion. Examples 3.B and 3.C illustrate the use of the stoichiometric table to calculate the conversion as a function of time. 0.4 0.6 Conversion Figure 3.28 A plot ol (lit. |......mum vm-m-i .....w«........I. ulntn. I limn ,v|imiIi.,ii ii'i.'l Willi N^J I mill, N? 3 mol, and ■ - 1.0. a. 17 SOLVED EXAMPLES Kxamplc 3.A Fitting Data to Monod'i Law Table 3.A.I shows some data foi the |TOWth rate of Paramecium as a function ol the Paramecium concentration, lit the d.H.i In Monod's law (Monod |I942|). _ k[K2[par] fp ~ 1 + K2[par] Where |par] is the Paramecium concentration and k| and Ki arc constants, Solution There are two methods that people use to solve problems like tins . Rearranging the equations to get a linear lit and using least-squares . I hiing nonlinear least-squares i i'ii i, i ihe latter, but 1 wanted to give a picture of the former. Nun- are two versions of the linear plots: (3.A.1) . I iniweaver Burke plots . i adie l lofstee plots in ill. I ineweaver Burke method, one plots l/rale against [/concentration. Rearranging nil in.hi ( t.A. I) shows i = _L_ + _L rp k|K:|par| k. iii. n line, a plot of l/i> versus I/[par] should be a straight line. The intercept ihould Hie slope should be , „ . Once k| is determined from the mieieept, K can be l k,K. i a a. I A.I The rate of Paramecium reproduction as a function of the Paramecium i inn initiation I'......mi nun i urn i.nil.in.in. N/iiii' Kale, NA 111' llolll Paramecium Paramecium concentration, Kate. concentration, N/cm3 N/d in1 hour i N/cm3 Rata, N/d in' limn I • mi i 16 l(> 46 % 1 fl 12.8 16.6 46.4 16 1 I.M 8 1 23.2 19 59.2 47.4 ll/(. .' 17.6 20 62.4 55 II.' / M 16 1 23.8 62.4 V 1 '/ .' N 23.2 26 57.6 (ll 1 l(. ii Ii, 1 10 i 108.8 61 (. III ' l 1 32 11 80 /i I.M 1 1 1 M i .31,2 61 (. /•i lid ij ti II 8 (1 (. 109,6 76.4 IK. IM 63.2 ci: 103 .' ..... 11.......1 MiyilN I l').'/l Mil ANAI YM*. i H IIAII HAIA determined liuin iIk- slope Figure I A.I shows the plol this is not a wonderful linear relationship, Imi such .1 result is typical Next. I want to calculate the rate constants using a least squares procedure Table XA.2 shows the formulas used to do the least-squares lit. I listed the concentration in column A and the rate data in column B. Column C is one over the concentration; column D is one over the rate. I then used the SLOPE, INTERCEPT and RSQ Functions in Excel to calculate the slope of the line. Table 3.A.3 shows the numerical values in the plot. From the least-squares lit 0.194 [par] + 0.00711 (3.A.3) Comparison of equations (3.A.2) and (3.A.3) shows 1 0.00711 140.5, I (0.194 *k,) = 0.037. Figure 3.A.2 is a plot taking k| = 140.4 and K2 = 0.0366. The curve does not do a bad job of fitting the data, although there is a systematic error at high concentrations. This is typical. One can even get eases where the Lineweaver-Burke plot misses the trends in the data. We got the systematic error because we fit to 1 /rp. A plot of 1 /rp gives greater weight to data taken at small concentrations, and that is usually where the data are the least accurate. The Eadie-Hofstee plot avoids the difficulty at low concentrations by instead finding a way to linearize the data without calculating l/rp. Rearranging equation (3.A.1), we have rp(l+K2[par]) = klK2[parl Further rearrangement yields I par | k,K2-K2rp (3.A.4) (3.A.5) 0.11 i y 0.1 / / m 0.09 / 0.08 ■ / 0.07 - 0.06 ■ / - 0.05 - 0.04 ■ / u - 0.03 - *y - 0.02 - Jr m - 0.01 - 0 ' I I I 0 0.1 (I (, 0.2 0.3 0.4 0.5 1/Concentration Figure 3.A.1 A Lineweaver-Burke plot of tho clntii in lnhln I A I ......... i,„ nil- I iii.'W.-.iv.-i lloiki! plot Tabl«3.A.2 The formulas In th« ipc—drtwt tonntun^^a^M =C$1*C$2-$A13/ (1+C$2*$A13) =ABS(E11-B11 $F$1_ =ABS(E12-$B12) $F$1 =ABS(E13-$B13) "$F$1 _—---"T-~~ " ' I r i |Htff| th..t n«*tr« «r« from Eko.1 98. Exo.l 9S otvaa i ||Ult(. 106 ANAI CSC. l)l MA 11 HA I A 100 CD <3 DC I I 1 Data 1 1 1 1 1 ^- - -" - 1 -■ —■--- Lineweaver-Burke m m l l l i i i i i 50 0 0 10 20 30 40 50 60 70 80 90 Concentration Figure 3.A.2 The Lineweaver-Burke fit of the data in Table 3.A.1. Therefore a plot of rP/[par] versus rp should yield a straight line. Figure 3.A.3 shows the plot. In this case, the line does not fit the data very well. I did a least-squares lit through the data using a spreadsheet like that in Table 3.A.3. The results showed 'p [par] = 4.21 -0.0171rn (3.A.6) Comparing equation (3.A.6) and (3.A.5) shows K2 = 0.0171, k, =4.20/0.0171 = 246. Figure 3.A.4 shows how well the data actually fit's the line. Notice that there is still an error at high concentration. In this case the predicted rate is slightly too high. When you divide by |par], you give lower weight to the high concentration points. The result is that there are still errors. Still, the Eadie-Hofstee method fits better than the Lineweaver-Burke fit, even though R2 is 0.34 in Figure 3.A.3 while R: is 0.90 in Figure 3.A. 1. The last way to fit the data are with a nonlinear least-squares. The idea in nonlinear least-squares is to use the solver function of a spreadsheet to calculate the best values of the coefficients based on some criterion. A common criterion '.(I KM) ISO Hillo i II ttiU>n | il< ■! < iHlitt < liitil Hi I ill .In IAI SOI VI H I KAMI'I I '. 10/ I I_L 0 10 20 30 40 50 60 70 80 90 Concentration Figure 3.A.4 The Eadie-Hofstee fit of the data in Table 3.A.1. Nonlinear least squares i I I I_I_I-1-L_ 0 10 20 30 40 50 60 70 80 90 Concentration Figure 3.A.5 A nonlinear least-squares (it to the data in Table 3.A.1. minimize the total error, where the total error is defined b) Total error ^_J(ahs|i give you any uncertainty in the intercept, or the relative Important i i ilu ,|o|H' and intercept in fitting your data. If you lit some dala Y lo a function l(x), ilu ii It' is defined by R = I £[abs(Y f(x))]J dm_ £[abs(Y - Y)]2 (3.A.K) In n N is the average value of Y. When you use a leasl-squaies technique, you define ,.........mi thai you are tilling lo a variable Y. According lo how you define Y, you can iI.IIn ililleienl values ol R' The l.ineweavei Htiike, lathe llolslee.andiionlini.il , i .|,i.n, ■. methods define Y differently, as indicated in fable <■ A.6. The nel tesull is ......In \ allies ill R ' aie not eoinpaiable. In llie homework sel, we have an example where we lil Iwo dilleienl niodels lo n II i .In.i ncI I'hc Unit model lils llie dala lo two significant figures even though H r. in I he second model does not III llie data al all (ll soinelunes gels llie wrong lift).) Yi'l, K' In 0,73, litis example cleaily illusliales llie ideas thai l< does nol loll v«>u how well a (jiven model lils your (lulu. 1 K) ANAI Yí( H HAM IIAIA MIIVI Dl XAMľll Table 3.A.6 Tho values of Y and f(x) used to calculate R2 using the different methods Method Y f(x) Lineweaver- Burke 1 1 + K2|par| rp k|K2|par| Eadie-Hofstee rp [par] kiK2-K2rp Nonlinear least-squares rP k|K2|par| 1 + K2|parl One way around the difficulty is to define a uniform value of R2. For example, two definitions that we can use are R2= 1 Efabsfrp- k,K2[par] ) ]T[abs(rp - fp)]2 (3.A.9) data where rp is the average value of the rate. R2 = 1 E ilala L abs ^ rP [par] E dala abs Vlpar] k,K-> 1 + K2|par]y 'p [par] (3.A.10) where (rp/[par])av is the average value of (rp/[parj). Table 3.A.7 shows the values. Notice that the different definitions of R2 give wildly different values of R2. In the literature, people often use R2 as a measure of the goodness of fit and to distinguish between various models. One has to be very careful when doing that. After all, Table 3.A.7 shows that you can calculate wildly different values of R2 according to how you do the calculations. Still, in the literature people often ignore these complexities and naively use the R2 values that come out of their linear regression to assess how well their model fits the data. Readers can judge for themselves whether that is a valid approach. Example 3.B Tests of Statistical Significance: Analysis of Variance The next question is how can one really tell if one model fits a given data set belter than another. Table 3.A.7 The values of R2 calculated using the different methods Method R2 from linear repression R2 from equation 1 ? It '>) R- Írom equation 1 ' B 10) Lineweaver-Burke 0.901 OKI K 0.932 Eadie-Hofstee o til 0,190 0,323 Nonlinear leasl square! 0,903 II 0.33X 11« ubjm live of this example is tO [XOVtdC an objective lest Let's consult! Ihľ example lit I iiuiiplc LA (see Table 3.H.I). Which model fits best? Is the difference statistically significant? Mutton First, lei us see which model fits best. We do thai by calculating the varii..... ,,i |hi until .ui.l seeing which model has the lowest variance. The variance V, is defined I's ((experimental rate) - (calculated rate)) [Hunts __ (number of samples) - (number of independent parameters in model) ^nlisiiiiiiing iii equation (3.A.7) yields total error from equation LA.7 V, = number of samples — number of parameters (Ml I l i i ll 'i .....nportanl to calculate the variance as shown in (3.B.1) and not, foi example, the ... ..I one over (he rate. In order to use the statistical tests below, one will have i nine ili.it the error in the data follows what statisticians call a chi-squart u i .l,M,ihmi,m. II you calculate the errors in the rale, the errors usually do follow a %' ill lilliiilion. However, the errors in one over rate do not follow a x' distribution, As a ..... itllhough the Statistical measures below are meaningful for variances calculated v|| ,,|........ii ( l.B.I), Ihey are not meaningful for variances in one over the rale. I hi variances are easily calculated from the total errors in Table 3.A.5. For example. hi ih. nonlinear least-squares case 441" (32 points - 2 parameters) = 164 ......i,ui\. im Badie Hofstee 5648 V, = - = IK8 32-2 ( Lit,3) I I HI) i....... ill) Fits to the data in Example 3.A (!al< ulated Rate i itpe.....ental Nonlinear I east m. titration Rate Squares Lineweavei Burke Eadie i ľi iti i n ii 0 0 0 . in I K(,S ').(> S II i ti 1 • s 13.06 16.3« 11 4 2.3.2 K,(,o I7.s>4 13.73 3,2 17.6 21.07 .V I" 20.01 / M K. 1 urns 31.20 'K'IS N 32 30.71 n 82 "l fill N il i in /i 31J2 "1 fill .....H < K.I 119 IKK unit i '" I ■ !U II VI II I XAMI'I I ' I I I while lni llii" l.inewcawi lliiikc plot Vi = ^=3.5 32-2 (3.B.5) So the nonlinear least-squares method fits the data best. The next question is whether the difference is statistically significant. This is important, because one model could fit better, but the difference between the models could be within the noise in the data. Statisticians are still developing methods to test whether one model is better than another. It is easy to test what are called "nested models": two models that are the same except that one has one extra parameter. However, independent models are much harder to test. The Cox algorithm can do the testing rigorously. One can also do a test using a Bayesian maximum-likelihood or minimum-entropy algorithm. Both are beyond the scope of this book. In this problem we will provide an approximation that is not mathematically rigorous but has the advantage that it can be used for practical calculations. The method is based on a statistical test called the F test. The idea in the F test is to compute Finversc given by variance in weaker model variance in better model ' vH.6) So, if we want to compare the Lineweaver-Burkc and nonlinear least-squares methods, we calculate 315 Finversc = TTT =1.91 (3.B.7) I 04 Statistically, if the two variances are independent and the value of Firm;rsc is large enough, we can say llial ihe difference is statistically significant. Table 3.B.2 gives values of Finverse. In the table, Finverse is listed as a function of nf given by nf = number of data points — parameters in the model (3.B.8) If you want to see if one model fits rate data better than another does, you should always do an F test to see if the difference between two models is statistically significant. I want to say clearly that the F test is an approximation. It assumes that the two variances are independent, which is clearly not true. Still, it does give useful information even though the F test cannot be rigorously applied to this case. To read Table 3.B.2, if nf is 30, then Finverse must be 1.84 to have 95% confidence that one model is better than another and 2.39 to have 99% confidence that one model is better than another. We are between 95 and 99%, so we can say that we are between 95 and 99% and are certain that the nonlinear least-squares model fits the data better than does the Lineweaver-Burke model. One can calculate a more accurate value for the confidence using the FDIST function in Excel. FDIST calculates the probability that a given value of Finverse occurs by chance. So, 1-FDIST is the probability that it occurred by other than chance. %confidence = 1 - FDIST(Finverse, nf for better model, nf for worse model) (3.B.9) I used Excel to calculate 1 - FDIST(1.91, 30, 30) = 0.96 Tuhlo 3.B.2 Vnlimn of r,„».,.. lis n function of nf wlion both modoln hnwo the sumo value of nf Signilicuncc Level Ml ■Mr, >)v; 99% 'W.5% 1 39.86 161.5 4052 K..'i: ').() 19 99 199 ! 5.39 9.28 29.46 1/ 1 4.11 6.39 I5.l)8 23 S 3.45 5.05 10.97 14.94 6 3.05 4.28 8.42 11.07 / 2.78 3.79 6.99 H S'l 8 2.59 3.44 6.03 7.50 9 .',11 3.18 5.35 6.34 Hi 2.32 2.98 4.85 5.85 20 1.79 2.12 2.84 3.32 «i 1.61 1.84 2.39 2.63 10 1.51 1.69 2.11 2.3 50 1.44 1.60 1.95 2.1 ,,, I mn 'H.'i sure that Ihe nonlinear least-squares fit better than the Linewcavcr BUffcl i i..i I \i el also has a FINV function thai calculates Flmc,x via I...... FINV( I '/ \ KN k„n" 'if;)"1 (3.D.2) i h. ANAI YMMH HAM I lAI A Table 3.D.1 Formulas for th« spreadsheet for Essen's method A B C D E 1 2 Essen's Method 3 time cone first second third 4 ln(Ca0/Ca) (Ca0/Ca>-1 =(CA0/CA)"2-1 5 0 1 =LN(1/B5) =(1/B5-1) =((1/B5)"2-1) 6 1 0.91 =LN(1/B6) =(1/B6-1) =((1/B6)A2-1) 7 2 0.83 =LN(1/B7) =(1/B7-1) =((1/B7)"2-1) 8 3 0.77 =LN(1/B8) = (1/B8-1) =((1/B8)"2-1) 9 4 0.71 =LN(1/B9) =(1/B9-1) =((1/B9)"2-1) 10 5 0.67 =LN(1/B10) =(1/B10-1) =((1/B10)*2-1) 1 1 6 0.63 =LN(1/B11) =(1/B11-1) =((1/B11)"2-1) 12 7 0.59 =LN(1/B12) =(1/B12-1) =((1/B12)"2-1) 13 8 0.56 =LN(1/B13) =(1/B13-1 ) =((1/B13)"2-1) 14 9 0.53 =LN(1/B14) = (1/B14-1) =((1/B14)"2-1) 15 10 0.5 =LN(1/B15) =(1/B15-1) =((1/B15)"2-1) 16 11 0.48 =LN(1/B16) =(1/B16-1 ) =((1/B16)"2-1) 1 7 12 0.45 LN(1/B17) =(1/B17-1) =((1/B17)"2-1) 18 13 0.43 = LN(1/B18) =(1/B18-1) =((1/B18)"2-1) 19 14 0.42 =LN(1/B19) =(1/B19-1) = ((1/B19)"2-1) 20 14 0.4 =LN(1/B20) =(1/B20-1) = ((1/B20)"2-1) 21 16 0.38 = LN(1/B21 ) = (1/B21-1 ) =((1/B21)~2-1) 22 17 0.37 = LN(1/B22) =(1/B22-1) =((1/B22)"2-1) 23 Interce pt -Intercept ($C$5..$C$22, A$7..A$22) =Intercept ($D$5..$D$22, A$5..A$22) =Intercept ($E$5..$E$22, A$5..A$22) 24 Slope =slope ($C$5..$C$22, A$5..A$22) =slope ($D$5..$D$22, A$7..A$22,110) =slope ($E$7..$E$22, A$5..A$22) 25 Rho =RSQ ($C$5..$C$22, A$5..A$22) =RSQ ($D$5..$D$22, A$5..A$22) =RSQ ($E$5..$E$22, A$5..A$22) Substituting KN (3.D.3) KN should be a constant! Again, I used an Excel spreadsheet to do the calculations. When I solve this. I Inul ii useful to use the module macro in Excel to set up all of I ho ei|imlious The module rtUM 10 capability allows you to define your own functions, in VllUtl HANK i u vi 111 xAMl'l l 11/ i Oilc llli' Niimoilciil viilmm ttii Mix I •mini plots A H C 1) E Essen's Method 1 ime cone first second third ln(1/Ca) (Ca0/Ca)-1 (CA0/CA)"2-1 o 1 .00 0.000 0.000 n nun 1 0.91 0.094 0.099 0 . i'lill 2 0.83 0.186 0.20!, ii 452 3 0.77 0.261 0.299 i) (ill/ 4 0.71 0.342 0.408 n mil 5 0.07 0.400 0.493 i 2211 6 0.63 0.462 0.587 1 520 7 0.59 0 . 528 0.695 1 . h/:i 8 0.56 0.580 0.786 2 . 1 fill 9 0.53 0.635 0.887 2 . 560 10 0.50 0.693 1 .000 3 . 00(1 11 0.48 0.734 1 .083 3 . 340 12 0.45 0.799 1 .222 3 . (Kill 13 0.43 0.844 1 .326 ■1 . .11111 14 0.42 0.868 1 .381 1 (Kill 1 5 16 0.40 0.916 1 .500 5.250 0 . 38 0.960 1 .632 5.929 1 7 0.37 0.994 1 .703 6.305 Intercept 0.087 -0.005 0.477 Slope 0.057 0. i 00 0.373 Rho 0.984 0.999 n nil 1 I......in define your own functions by firsl inserting a module page into youi Workbook In my version of Excel you do thai by using die in.si-.ki macro moihii i lith ......eatc .i module, you can create your own functions by typing them onto the ......kilo page. i ihli 11 > * '.hows ihe module to calculate k,. l and k i, when- k i is given bj equation i.....i k i mikI k, are given b) equation 11.D.3). The Misi three hues define a fun< non i .n, ik, i Ihe lusi line defines kone as a public function. I called Iho function "kone", mil ki because ol a limitation of Excel, The word "public" in the function definition III n , the function available to Excel. The term "as variant" says thai the return typo is ......i "Variant" is a general return type that can be used for anything, fable Hi I gives H lť.1 ol olhei letuin types. I.....ond line ol the definition gives a value to "kone" accoidmg lo equation I *> I Ihe thiol line nils the liiiulnm lo return, Wo use log in I ho Inn, lion hi modules, log leliiins ill, MtUfVl log >>l a numboi while log 10 n linns n base 1(1 log 1 HI Af J Al V'.l'.i i| HAH HAIA !,( II VIII I XAMI'I I i i'l Tilble 3D.3 Modulo iihoiI to i • ky. k.i, whnro k|, k2, and k3 are defined by equation (3.D.2) Public Function kone(caO, ca, tau) As Variant kone = Log(caO/ca)/tau End Function tau) As Variant Public Function ktwo(caO, ca, ktwo = ((1#/ca)-(1#/ca0))/tau End Function Public Function kthree(caO, ca, tau) As Variant kthree = (C^/ca^-Cltf/caO^vtau/a End Function Table 3.D.4 Some Microsoft Excel/Visual BASIC return types Type Meaning Type Meaning As variant General return type (can be an integer, As Double Double-precision real, vector, matrix logical or text) real As single Single precision real As Integer Integer Table 3.D.3 also shows the function definitions to calculate k2 and k3 according to equation (3.50). The only thing that is weird in the definition is that the makes 1 a floatingpoint number. Once you type in the module, you can then use the function kone like any other function in Excel. So if you want to calculate kone with CaO = 1.0 Ca = 0.6 and tau = 3, you enter =kone(1, 0.6, 3) in a cell in your spreadsheet. I use these functions to calculate ki, k2, and k3 in my spreadsheet. Table 3.D.5 shows my spreadsheet. I listed time in column B and concentration in column C, and I wanted to calculate k) in column D, k2 in column E, and k3 in column F. I used the following steps to get the answer: 1. I named cell el CaO, and set it equal to 1. 2. I used the kone, ktwo, and kthree functions to calculate k,, k2, and k.i. For example, row 6 is for time=l. Ca0=l, Ca=c6, tau=b6, so kone calculates k| for CaO=ca0, ca=c6, tau=b6. I also included a spreadsheet (Table 3.D.6) that does not use the usei defined Function!, Table 3.D.6 shows the formulas used in the spreadsheet ( ohimn A is the tune, i hIiiiiiii B is the concentration, column C is k|, column l> is k >. and roh...... I is k, At lime .', CA = B33, ln(CA/CA) = lud/IUI) Also. ln(.',!) =kone(ca0,Ci2,Bl2) =ktwo(caO,C12,B12) =kthree(ca0,C12,B12) i>.',<; =kone(ca0,C13,B13) =ktwo(caO,C13,B13) =kthree(ca0,C13,B13) 0.!j3 =kone(ca0,C14,B14) =ktwo(caO,C14,B14) =kthree(caO,C14,B14) III 0.5 =kone(ca0,C15,B15) =ktwo(caO,C15,B15) =kthree(ca0,C15,B15) 0.48 =kone(caO,Ci6,Bl6) ktwo(caO,C16,BI6) =kthree(ca0,C16,B16) 0.45 =kone(caO,Cl7,B17) =ktwo(caO,C17,B17) =kthree(ca0,C17,B17) 0.43 =kone(ca0,C18,B18) =ktwo(caO,C18,B18) =kthree(ca0,C18,B18) I i 0.42 =kone(ca0,C19,Bl9) ktwo(c;H),C19,B19) =kthree(caO,C19,B19) 0.4 ii. :m =kone(caO,C20,B20) =ktwo(caO,C20,B20) =kthree(caO,C20,B20) =kone(caO,C21,B21) =ktwo(caO,C21 ,B21) -kthi-ee(c,](),c;'l ,h:'I ) II. 37 =kone(caO,C22,B22) =ktwo(ca0,C22,B22) =kthree(ca0,C22,B22) Notice that k; is constant while k| and k.i vary. Figure 3.18 also shows a plot of these i ,i i according to Van't Hoffs analysis, if k2 is constant, the reaction is second order; iii n hue, we conclude that the reaction is second-order. i..n,'//'. Method In Powell's method (sec formulas listed in Table I.D.N), one uses plots ol log of t versus In CA. One then shifts the curves left and right until things lit In ni\ spieadsheel. I defined a new variable sh i f I to do the calculations. I then varied b) hand. I put the top two points in the middles of the curves and then s.i" VI fin Ii , in vr hi best. Mn values foi Powell's method are shown in fable 3.D.9. \ plot ol the data in fable 3.D." shows that these data follow second older km, til . ,, I igurc I.D.I). i implc 1.K Using the Stoichiometric Table to Solve Stoichiometric Problems n ii the Stoichiometric table to do a problem from page I IS in leldei and Pun i m l I'l/H) Ethylene is being made via the dchydrogenation of ethane. ( I! (Ifi +H2 (I.E.I) inn, ill.it 11)11 inol/miniile are led into a How reactor. Analysis ol the evil stream in,In .it.-, that 40 mol/ni.....le ol hydrogen leave the rem lot ('iiletiliile the evil Mow rule. i/o ANAI YMM li MAM HAIA Table 3.D.6 Tho lormulas used lor Vim'l Holl» inotlioil A B C IJ 1 30 time cone kl k2 k3 31 0 1 ln(1/Ca)/t «Ca0/Ca)-1 )/t ((CA0/CA)*2-1 )/t/2 32 1 0.91 =LN(1/B32)/A32 =(1/B32-1 )/A32 =((1/B32)"2-11/A32/2 33 2 0.83 =LN(1/B33)/A33 = (1/B33-1)/A33 =((1/B33)"2-1)/A33/2 34 3 0.77 =LN(1/B34)/A34 =(1/B34-1)/A34 =((1/B34)"2-1 )/A34/2 35 4 0.71 =LN(1/B35)/A35 =(1/B35-1)/A35 =((1/B35)"2-1 )/A35/2 36 5 0.67 =LN(1/B36)/A36 =(1/B36-1)/A36 =((1/B36)"2-1 )/A36/2 37 6 0.63 =LN(1/B37)/A37 =(1/B37-1 )/A37 =((1/B37)"2-1 )/A37/2 38 7 0.59 =LN(1/B38)/A38 =(1/B38-1)/A38 =((1/B38)"2-1)/A38/2 39 8 0.56 =LN(1/B39)/A39 =(1/B39-1 )/A39 =((1/B39)"2-1)/A39/2 40 9 0.53 =LN(1/B40)/A40 =(1/B40-1 )/A40 =((1/B40)"2-1)/A40/2 41 10 0.5 =LN(1/B41)/A41 =(1/B41-1)/A41 =((1/B41 )"2-1 )/A41/2 42 11 0.48 =LN(1/B42)/A42 =(1/B42-1)/A42 =((1/B42)"2-1 )/A42/2 43 12 0.45 =LN(1/B43)/A43 =(1/B43-1)/A43 =((1/B43)"2-1)/A43/2 44 13 0.43 =LN(1/B44)/A44 =(1/B44-1)/A44 =((1/B44)"2-1)/A44/2 45 14 0.42 =LN(1/B45)/A45 =(1/845-1 )/A45 =((1/B45)"2-1)/A45/2 4 6 15 0.4 =LN(1/B46)/A46 = (1 /B46-1 )/A46 =((1/B46)"2-1)/A46/2 4/ 16 0.38 =LN(1/B47)/A47 = (1/B47-1 )/A47 = ((1/B47)"2-1)/A47/2 48 17 0.37 =LN(1/B48)/A48 =(1 /B48-1 )/A48 =((1/B48)"2-1)/A48/2 Table 3.E.1 Stoichiometric table for Example 3.E Species Flow-rate in Feed Change Moles Moles out of Reactor Ethane Ethylene Hydrogen F° ťc2h4 _fc2h6xc2h« (b/a)Kä2„6Xc2h6 (c/a)F£2H6Xc2h6 tcjH^1 ~xc2H6) Fc2h4 + (b/a)Fc 2H6Xc2Hft F„2 + (c/a)F«2H6Xc2H6 Table 3.E.2 The results after substituting data into Table 3.E.1 Species Flow Rate in Feed Change Moles Moles out of Reactor Ethane Ethylene Hydrogen 100 mol/minute 0 0 —100 mol/minute Xc2h6 100 mol/minute Xc2h6 100 mol/minute Xch6 100 mol/minute (1 - X, lU,, > 100 mol/minute Xc2h6 100 mol/minute Xc2h<, Solution Table 3.7 is a general stoichiometric table for the reaction. There are one reactant, two products, and no inerts in reaction (3.K.I). Therefore, the stoichiometlil table will be in the format of Table 3.E.I. In Table 3.7, a is the minus stoichiometric coefficient "I ( II,., b is the Stoichiometric coefficient of C2H4 and c is the stoichiometric coefficient ol ll> Substituting a b li. loo mol/minute, i,. P? 0 M.-I.K Tuhle \ I ! ! i( >l VI 1)1 XAMI'I I l/l i a.i. in/ ||„- iiiiiiii'iii.iI viiIimk, l,u Viiii'I Hull'-, in,•III,id Table 3.E.3 The results (in mol/minute) after substituting XC2h6 = 0 4 into Table 3.E.2 How Kale Change Moli-s mil Species feed Moles ol Reactoi Ethane 100 -40 60 Ethylene 0 1 HI in Hydrogen 1) 1 10 40 II C I) E F iui' cone k1 k2 I. i 1) 1 Im(1/Ca)/1 ((Ca0/Ca)-1)/t ((CA0/CA)'2-1)/t/2 1 ().!)! 0.094 0.099 0.104 a 0.83 0.093 0.102 o 1 M 3 0.77 0.087 0.100 li 114 4 0.71 0.086 0.102 II 1/3 _5 0.67 0.080 0.000 0.1/3 e 0.03 0.077 0.00H 0.127 7 0.59 0.075 0.000 0. 134 II 0.66 0.072 0.0OH 0 137 9 0.53 0.071 0.000 I). 14/ in 0.5 0.069 0.100 1). 11,0 1 t 0.48 0.067 0.098 0.152 12 0. 4b 0.067 0.102 0.164 13 0 . 43 0.065 0.102 o. I 70 14 0.42 0.062 0.099 ......' 1!) 0.4 0.061 0.100 0. 1 /', 1(1 0.38 0.060 0.102 0. Uli, 1 7 0.37 0.058 0.100 0.185 Nul K) mol/minute of hydrogen leave the reactor. Therefore | loo mol/hour]Xc,H6 = 4() mol/minute in X, ,,,„ =0.4 lil lnu X, i,,, into Table 3.E.2 yiclds Table 3.E.3. 1 ......3.1'' Chuchani and Mariin (1007) examined the pyrolysis ol inaleic <„n,<'(iiHOíhcooii . ('„ikciio i co | ii,o N) lomíme llu inalen ai id min a balch reactor and measining Ihe piessiue as ' Hun Ihe dala in Table *T I were oblained ai l.M) I ('. ulici correcling o mi lion < ii l ali ni.ilr Ihe min riilialiiin ol malí u ai lil as a linu lion ol linie ll'l Til Ihe dala lo a siuiple i.ilr i'i|iialiun ( I i..') (3.T I) li ni a lunt I li >i i Ich [I i,l, Table 3.D.8 Formulas used to analyze the data using Powell's method A B C D E F G H 52 Shift= -1 53 time cone log time In time +shift Ca/CaO Log Tau Log Tau Log Tau 54 0 1 First order Second Order Third Order 55 1 0.91 =L0G10(A55) =C55+SHIFT +B55/1 =L0G(LN(1/B55)) = L0G(1/B55-1) =L0G((1/B55"2-1 I/2) 56 2 0.83 =L0G10(A56) =C56+SHIFT +B56/1 =L0G(LN(1/B56)) = L0G(1/B56-1) = L0G((1/B56"2-1 ), 2. 57 3 0.77 =LOG10(A57) =C57+SHIFT +B57/1 =L0G(LN(1/B57)) = L0G(1/B57-1) = L0G((1/B57"2-1 i 2 58 4 0.71 =LOG10(A58) =C58+SHIFT +B58/1 =L0G(LN(1/B58)) = L0G(1/B58-1) = L0G((1/B58"2-1 I 2 59 5 0.67 =L0G10(A59) =C59+SHIFT +B59/1 =L0G(LN(1/B59)) =L0G(1/B59-1) = L0G((1/B59"2-1 i 2 60 6 0.63 =I_OG10(A60) =C60+SHIFT +B60/1 =L0G(LN(1/B60)) =L0G(1/B60-1) =L0G((1/B60*2-1 i 2 61 7 0.59 =L0G10(A61) =C61+SHIFT +B61 1 =L0G(LN(1/B61)) =L0G(1/B61-1) =L0G((1/B61"2-1 i 2 62 8 0.56 =L0G10(A62) =C62+SHIFT +B62/1 =L0G(LN(1/B62)) =L0G(1/B62-1) =L0G((1/B62"2-1 - 2 63 9 0.53 =L0G10(A63) =C63+SHIFT +B63/1 =L0G(LN(1/B63)) =L0G(1/B63-1) =L0G((1/B63"2-1 i 2 6-1 10 0.5 =L0G10(A64) =C64+SHIFT +B64/1 =L0G(LN(1/B64)) =L0G(1/B64-1) =L0G((1/B64"2-^ 2 65 11 0.48 =L0G10(A65) =C65+SHIFT +B65/1 =L0G(LN(1/B65)) =L0G(1/B65-1) =L0G((1/B65"2-" 2 6€ 67 12 0.45 =L0G10(A66) =C66+SHIFT +B66/1 =L0G(LN(1/B66)) = L0G> 1 B66-1 i = L0G((1 B66'2-" 2 13 0.43 =L0G10(A67) =C67+SHIFT +B67/1 =L0G(LN(1/B67)) =LOG(1 /B67-1) =L0G((1/B67"2-1 > 2 = = 14 0.42 =L0G10(A68) =C68+SHIFT +B68/1 =L0G(LN(1/B68)) = L0G(1/B68-1) = L0G((1/B68"2-' 2 15 0.4 =L0G10(A69) =C69+SHIFT +B69/1 =L0G(LN(1/B69)) = L0G(1/B69-1 ) = L0G((1 B69"2-" 2 _: 16 0.38 =L0G10(A70) =C70+SHIFT +B70/1 =L0G(LN(1/B70)) =L0G(1/B70-1) =LOG((1 B70'2-1 I 2 7" 17 0.37 =L0G10(A71) =C71+SHIFT +B71/1 =L0G(LN(1/B71)) =L0G(1/B71-1) =L0G((1 B7T2-1 i 2> z -. - - CaCaO o o o a 3 ° a I 1 o. a o -n' « § irst rder Q. i Shifted - jj ^/ O H a. ■ -\ yC 3. 2 a; a. S ■ m u \«: - ■ a. a - a u - M U a pc - z — — — o o □ CO o c — a 3 - - c CO c -J --c ~ -- — ■s. c - "1 o I S4 AM/VI i '.ľ . i íl MA 11 I IA I A M H VI HI hamm I 1?!) Solution Aeioidmg Id i'c|iiiillon ( * HK> Amol a (3.F.1) Solving 'It1 J \NlJ VAmol/ But N? =N^ a= 1, Amol = 2. Therefore X = 1 2 N V pO 1 I RT (1 -XA) = 52.1 torr /0.052 liter.atmX V mol-K / 1 atm 760 torr (1 -XA) CM = 1.4 x 10_3(1 -XA) Plugging in the numbers yields Table 3.F.2. I analyzed these data using the spreadsheet in Table 3.C.3. The results are given in Table 3.F.3. Notice that k, is constant up until about 70% conversion, which implies that the reaction is first-order in that conversion range (the side reaction starts at about 60% conversion). Table 3.F.1 Data for Example 3.F Pressure, atm t, minutes Pressure, atm t, minutes Pressure, atm t. minutes 52.1 0 110.9 5.2 124.7 9 72.8 1.5 II 4.4 6.0 134.7 40 92.5 3.0 121.6 7.0 139.8 51 Table 3.F.2 Concentration versus time for the data in Table 3.F.1 Time, Time, minutes XA Cm, mol/liter minutes XA Cm, mol/liter 0 0 1.40 x 10"3 9 0.667 4.66 x 10"4 1.5 0.199 1.12 x irr3 10 0.697 4.27 x 10~4 3 0.388 8.57 x 10~4 40 0.763 2.90 x 10 4 5.2 0.564 6.10 x I0~4 51 0.842 2.21 x 10~4 6 0.598 5.63 x 10"4 — — — I I.(t Develop a stoiclllonielili tnble lot the reaction CO I II < > > CO.. I II., ■,, I how lh.il the density is constant. Hltlullon following the analysis in Section 3.12.1, Table 3.C.I the stoichiometric tabic Im Ilm, HyHtcm becomes. ..... Ihc .......ber of moles is constant, the density is constant (lor an isotheini.il lit. ill gas i .....pie 3.11 limiting Keaetants i n Develop a stoichiometric table for the homogeneous gas-phase reaction: 2N02(g)+ i02(g): N203 lb) l)i vclop an expression for Ihc concentration of the oxygen as a function ol , (inversion of the nitrogen dioxide in a constant pressure reactor. ...... 1 I i Numerical Values in Tablo 3.F.2 A B C D E iii 1 ...... 1-Xa k1 K2 k3 111 0 I ln(1/Ca)/t uCa0/Ca)- o/t ((Ca0/Ca)"2- 1 1/1/2 t,' 1 .5 0 80 0.148 0 166 0. I (16 11 3.9 0 61 0.164 0 211 0.278 11 !i.2 0 44 0.160 0 246 0.410 in (i 0 40 0.152 0 248 0.432 m 7 0 33 0. 167 0 286 0.573 11 9 0 30 0.133 0 256 0.51,0 in 40 0 21 0.039 0 096 0.2/o m 51 0 16 0 . 036 0 104 0.383 ' idle iü.1 Stoichiometric table for Example 3.G i 11 II Ml. II. I.....I Initial Moles Change Moles N°H2 N°co2 < '< I; "II, -NcoXco -NC()XC() +N^0Xco +N0oXCO 0 Moles t Inreai ted n;',,(I Xco) K2 - NooXco Nľo2+N»()X(,, N li, I N;',,xn, Nľ„ I N',',. I N'/,,,, I N1,',. 126 ANAlYüüWII HAU HAI A M II VI I) I XAMI'I I ' Vit (c) II wc pul 1 mol öl mitogen dioxide and I mol of oxygen into a closed icacloi. will oxygen or nitrogen dioxide be used up first? We call the icaclanl thai is used up first the limiting reactant. Solution (a) Refer to Table 3.H.I. (b) The oxygen concentration: by definition co2 N02 where V is the volume. From the ideal-gas law PV = Nl01alRT (3111 ) (3.H.2) Substituting V from equation (3.H.2) into equation (3.H.1) and substiluiiii)' expressions for Nlola| and the concentration of N02 from the stoichiometric table yields / K, ~ ?XNq2N»0, \ P 2 V^+NnoJI-IXnoOJRT (c) The N02 is used up when XNn2 = 1. At that point No, = K2 ~ |XNo2N?,o2 = 1 - (j)(D(2) = 0.5 mol N02 is used up first. Therefore N02 is the limiting reactant. Example 3.1 Using the Equations in Section 3.13.2 for Design Ethyl ben/oate (E) Il made by reacting benzoyl chloride (B) with ethyl alcohol (A) (see Figure 3.1.1). Table 3.H.1 Stoichiometric table for Example 3.H Species Initial Moles Change Moles Moles Unreaeled N02 N° 1nn02 -XnotN^q, C(1-Xno:) Oj N° No2 -3XNo2Nno2 K2 - 3Xno2N°no2 N205 0 + |Xn02N(no2 jXnoiN^,,, Total > ("ii. Therefore, equation 3.77 applies. Rearranging equation 1.77 \ iekls I In 1 k, (Ml) mill k, k||C'A. Rearranging yields v It k, = - In t 111 I.I x 10 /minute (3.1,2) ('i, / 30 minutes \ I 0.3, hi .i Kii 0.95, ('n = (I - Xu) I mol/liter. Substituting into equation (3.1.1) yields In ( ! _l()9Sj = 250 minuics (3.1,3) I ■•'i|il<- l.J Multiple ReactantS In Section 3.13, we did a problem on the I il i Mil. i leaclion of hen/oquinone (B) and cyclopcnladiene (('I to yield an addiiel ...... dial you nou want to use dimethyl cyclopcnladiene ((") lor the lead ion \'lilni|' llie methyl groups increases k. Assume a 20% increase (I made up a iniiiilii i i i.....I mol/liter of C and 0.0K niol/liler of H are loaded into a well sluie.l hat! Ii Ii mi liii . o Determine the residence time needed to convert 95% of the hcn/oquinonC 10 ..liliu i i in Keminder, the reaction look 2.2 hours with cyclopcnladiene. Sulullnn liom equation (3.79) I dCh c„(V 1 1 li mi in brackets is constant. II kn goes up 20%, r goes down by 20%: 2.2 hours r = - —-— = |,B hours 1.2 upli » K Vim*) HofT riois for Mine < 'omplax Reactloni Bodi nitein and I und ' . s.....111.. I the tale ol the reaction II . II. 'Ulli 0.K.1) 12B ANAl Vratu li MAM liAIA .( ii vi ii i hamm r. liMl ill . i Imi.Ii reSCtOl. They did runs whcic 11 h ■ \ unveil known .iiihmiiiIs hI II ,inil Hi... anil measured the HBr concentration as a function oi time. The data listed in Table I.K.I were obtained at 301.3°C. How well do the data in Table 3.K.1 lit the rate equations (3.K.2) (3.K.3) rHBr = k2[H2][Br2] rHBl = kl5[H2][Br2]l/2 Solution I will use equation (3.30) to solve this problem. fCA -dCA -- = t (3.30) I am going to work in terms of the HBr concentration. According to equation (3.30) t = Inverting the limits yields 1hb, dlHBrJ CHBr (-rHBr) canr d[HBr] :°Br (rHBr) (3.K I. (3.K.51 Next, I need an expression for rHBr- Equations (3.K.2) and (3.K.3) give me expressions for riiHr in terms of the concentrations [H2], [Br2] in the reactor. In order to use the equations, I will need to know how [H2| and [Br2] are changing as the reaction proceeds. Next, I will use the stoichiometric table (Table 3.K.2) to develop an expression lot |H2] and [Br2] in terms of the initial concentration in the reactor and [HBr]. If we call [HBrl the amount of HBr that forms, then the change in the Br2 concentration is A[Br2l = ^[HBr] (3.K.6) Table 3.K.1 Bodenstein and Lund's [1904,1907] data for HBr production in a batch reactor Run l Run 2 Run 3 |H2]° = 0.5637 |Br2|° = 0.2947 |H2]° = 0.2281 [Br2]° = 0.1517 |H2|" = 0.3103 |Br2|" = 0.5069 Time, minutes |HBr| Time, minutes IHBrj Time, minutes 1 HBr] 0 0 0 0 0 0 14.5 0.0669 19.5 0.0322 15 (I.!)-!').1 24.5 0.0985 34.5 0.0527 35 0.1031 34.5 0.1262 54.5 0.0713 55 0.140d 49.5 0.1644 79.5 0.0912 NO 0.1752 79.5 0.2093 99.5 0.1040 102 0.In (3.K.2) works, then ruin = k2([H2]° - ().5|HBr|)(|Br:r - 0.5[HBr|) \,, Hiding to equation (3.K.5) I'm, d|HBr| I Ik 9) (1 K. 10) Hh lituting equation (3.K.9) into equation (3.K.10) and looking up the integral 111 all mi, i'i.il table yields k.i|ll ni)' lor k> yields 2 / [Br2]"([H2]° - 0.5[HBr)) \ I" IHr.. I") V 111. l(,([Br2]° - 0.5fHBrl)y t I K III 2 ! / |Hr,|"(|ll..|" ().5|IIH,|l\ M|ll.I" III. I" » " \ 111 • I". I M. 1" 0 -|IIHi|) / ( I K ,12) Meie is an Excel spreadsheet lo do Ihe call illations < oluinn A is ihe initial hydlOfM oin cull.ilion, column H is the initial horninu i oiu enllillloll, lohinin (' is the lin.il Mill 130 ANAIYNIMII HAH llAIA concentration, column I > is t, und column I is k. calculuicd trom equation (3.K. 111 A B C D E 3 [Hz]° í Br, r [ HBr] tau k. 4 0.5637 0.2947 0.0699 14.5 =2/D4/(A4-B4)'LN(B4/A4' (A4-0.5*C4)/(B4-0.5'C4)) 5 0.5637 0.2947 0.0985 24.5 =2/D5/(A5-B5)*LN(B5/A5' (A5-0.5*05)/(B5-0.5*05)) 6 0.5637 0.2947 0.1262 34.5 =2/D6/(A6-B6)*LN(B6/A6* (A6-0 . 5*C6) / (B6-0 . 5*06)) 7 0.5637 0.2947 0.1644 49.5 =2/D7/(A7-B7)*LN(B7/A7* (A7-0 . 5*07) / (B7-0 . 5*07) ) 8 0.5637 0.2947 0.2093 79.5 =2/D8/(A8-B8)*LN(B8/A8* (A8-0.5*C8)/(B8-0. 5*08)) 9 0.5637 0.2947 0.2306 99.5 =2/D9/(A9-B9)*LN(B9/A9* (A9-0.5*09)/(B9-0.5*09)) 10 0.5637 0.2947 0.2502 124.5 =2/D10/(A10-B10)*LN(B10/A10* (A1 0-0 . 5*C10) / (B10-0 . 5*010)) 11 0.5637 0.2947 0.2619 149.5 =2/D11/(A11-B11)*LN(B11/A11* (A11-0.5*011 )/(B11-0.5*011 )) 12 13 0.2281 0.1517 0.0322 19.5 =2/D13/(A13-B13)*LN(B13/A13* (A13-0.5*013)/(B13-0.5*013)) 14 0.2281 0.1517 0.0527 34.5 =2/D14/(A14-B14)*LN(B14/A14* (A14-0 . 5*C14) / (B14-0 . 5*014) ) 15 0.2281 0.1517 0.0713 54.5 =2/D15/ (A15-B15)*LN(B15/A15* (A15-0.5*C15)/(B15-0.5*015)) 16 __ 0.2281 0.1517 0.0912 79.5 =2/D16/(A16-B16)*LN(B16/A16* (A16-0.5*01 6)/(B16-0.5*016)) 0.2281 0.1517 0.104 99.5 =2/D17/(A17-B17)*LN(B17/A17* (A17-0.5*01 7)/(B17-0.5*017)) 18 0.2281 0.1517 0.1142 124.5 =2/D18/(A18-B18)*LN(B18/A18* (A18-0 . 5*018) / (B18-0. 5*018)) 17 0.2281 0.1517 0.1217 149.5 =2/D19/(A19-B19)*LN(B19/A19* (A19-0.5*C19)/(B19-0.5*019)) 20 0.2281 0.1517 0.1295 174.5 =2/D20/(A20-B20)*LN(B20/A20* (A20-0.5*020)/(B20-0.5*020)) 21 22 0.3103 0.5069 0.0492 15 =2/D22/(A22-B22)*LN(B22/A22* (A22-0. 5*022) / (B22-0. 5*022)) 23 0.3103 0.5069 0.1031 35 =2/D23/(A23-B23)*LN(B23/A23* (A?3 i).:./ (n:>;t o. i,-023) ) ( . u/1 I I mini OVIT laut ) '.( II VI III KAMI'I ľ. .'II .'/ •II A B c D E 1 0.3103 0.5069 0.1400 2/D24/(A24 B24 ) i N (H24/A24* (A24-0. 5*C24)/(B24-0. 5*C24)) o.:no3 0.5069 0.1752 80 =2/D25/(A25-B25)*LN(B25/A26* (A25-0. 5*025) / (B25-0. 5*C25)) 0.3103 0.5069 0.1963 102 =2/D26/(A26-B26)*LN(B26/A26* (A26-0.5*026)/(B26-0.5*C26)) 0.3103 0.5069 0.2179 125 =2/D27/(A27-B27)*LN(B27/A27* (A27-0.5*027)/(B27-0.5*027)) 0.3103 0.5069 0.236 155 =2/D28/(A28-B28)*LN(B28/A28* (A28-0.5*028)/(B28-0. 5*C28)) 0.3103 0.5069 0.2533 196 =2/D29/(A29-B29)*LN(B29/A29* (A29-0.5*029)/(B29-0. 5*C29)) i ľ n me die results: m i i I [Br2]°: k,, = x^[H°2] - [Br2]<> x ^arctan for [Br2]° > [H2]°: ki.5 TIM5 ,V/[H°]-[Br2]0/ t\/([Br2]° - [H2]")ü — arctan - 0.5[HBrJ [H°] - [Br2]<> (3.K.16) Table 3.K.3 The Excel module used to calculate k-i 5 Public Function kk(h20, br20, hbr, tau) As Variant If (br20 = h20) Then kk = 4#/tau*(1#/Sqr(br20-0.5*hbr)-1#/Sqr(br20)) Exit Function End If If (h20 > br20) Then kk = Atn(Sqr(br20)/Sqr(h20-br20)) kk = kk-Atn(Sqr(br20-0.5*hbr)/Sqr(h20-br20)) kk = kk*4#/tau/Sqr(ri20-br20) Exit Function End If x1 = Sqr(br20-0.5*hbr) + Sqr(br20-h20) x2 = Sqr(br20)-Sqr(br20-h20) x3 = Sqr(br20-0.5*hbr)-Sqr(br20-h20) x4 = Sqr(br20) + Sqr(br20 - h20) kk = 2/tau/Sqr(br20-h20)*LoQ((x1'x2)/(x3*x4 n End Function MM vi iii kami 'i i H 133 In /mill.I" 11 -N1111 *tt>*■ ^ 1 dur.-r in.ri)",i\ Hllii,.|")'" - (|Br21° - IH.il")1"! {(I Iii . I" ii »|lllli|)" 1 < I Hi . I" |nr I V |i|Hi.|.....■ I ([Br2]° |lf.|")"'M / I IK I /1 mm 1 used .1 spreadsheet to do the calculations: I hr. Function is so complicated that I found it easier to write an Excel module i" ■ 1I111 hue k| ., Table 3.K.3 is a printout of my module. I.ililc I.K.4 is the spreadsheet using the Excel module. I listed the values ol |ll..|" in III....... \ I Hi 11" 111 column B, |HBr| in column C, and r in column I >. I then used rtlj ......lion in calculate the values of k|.s in column E. 1 I show a speadsheel that calculates Ihc formulas by hand. 1 -ii.ii. 1 K 4 The formulas in the spreadsheet used to calculate k,., 111 A B c D E Br" [HBr] tau k_1 .5 1 !.(i37 0.2947 0.0699 14.5 =kk(A4,B4,C4,D4) 1 !)(i37 0.2947 0.0985 24.5 =kk(A5,B5,C5,D5) i. 5637 0.2947 0.1262 34 .5 =kk(A6,B6,C6,D6) 1 m;:i/ 0.2947 0.1644 49.5 =kk(A7,B7,C7,D7) 1 Mi:i7 0.2947 0.2093 79.5 =kk(A8,B8,C8,D8) 1 !,(i:i7 0.2947 0.2306 99.5 =kk(A9,B9,C9,D9) 1 m,:i/ 0.2947 0.2502 124.5 =kk(A10,B10,C10,D10) 1 m,:i/ 0.2947 0.2619 149.5 =kk(A11,B11,C11,D11) 1 .'i'mi 0.1517 0.0322 19.5 =kk(A13,B13,C13,D13) 1 ;';'in 0.1517 0.0527 34.5 =kk(A14,B14,C14,D14) 1 .vhi 0.1517 0.0713 54.5 =kk(A15,B15,C15,D15) 1 .';'iil 0.1517 0.0912 79.5 =kk(A16,B16,C16)D18) 1 .'.'H1 0.1517 0.104 99.5 =kk(A17,B17,C17,D17) 1 1 0.1517 0.1142 124.5 =kk(A18,B18,Cie,D18) 1 .'/ill 0.1517 0.1217 149.5 =kk(A19,B19,C19,D19) 1 .viu 0.1517 0.1295 174.5 =kk(A20,B20,C20,D20) 1 1103 0.5069 0.0492 15 =kk(A22,B22,C22,022) 1 M03 0.5069 0.1031 35 =kk(A23,B23,C23,D23) 1 1111:1 0.5069 0.1406 55 =kk(A24,B24,C24,D24) 1 1111:1 0.5069 0.1752 80 =kk(A25,B25,C25,02S) 1 1111:1 0.5069 0.1963 102 =kk(A26,B26,C26,02e) 1 1111:1 0.5069 0.2179 125 =kk(A27,B27,C27,D27) 11:1103 0.5069 0.236 155 kk(A2H,H28,C2ll,1)211) II. 3103 ii. Mil,9 0.25:1:1 10(1 kk(A29,H2!l,C29,l);MI) A B C D E 3 tau [H,] [Br2]° [HBr] ki.5 4 14.5 0.5637 0.2947 0.0699 =4 / D4 / SQRT (A4-B4) * (AT AN (SQRT(B4)/SQRT( A4-B4) ) -ATAN (SQRT(B4-0.5*C4)/SQRT(A4-B4))) 5 24.5 0.5637 0.2947 0.0985 =4/D5/SQRT(A5-B5)* (ATAN(SQRT(B5)/SQRT(A5-B5))-ATAN(SQRT(B5-0.5*C5)/SQRT(A5-B5))) 6 34.5 0.5637 0.2947 0.1262 =4/D6/SQRT(A6-B6)* (ATAN(SQRT(B6)/SQRT(A6-B6))-ATAN(SQRT(B6-0.5*C6)/SQRT(A6-B6))) 7 49.5 0.5637 0.2947 0.1644 =4/D7/SQRT(A7-B7)*(ATAN(SQRT(B7)/SQRT(A7-B7))-ATAN(SQRT(B7-0.5*C7)/SQRT(A7-B7))) 3 - 79.5 0.5637 0.2947 0.2093 =4/D8/SQRT(A8-B8)* (ATAN(SQRT(B8)/SQRT(A8-B8))-ATAN(SQRT(B8-0.5*C8)/SQRT(A8-B8))) 99.5 0.5637 0.2947 0.2306 =4/D9/SQRT(A9-B9)* (ATAN(SQRT(B9)/SQRT(A9-B9))-ATAN(SQRT(B9-0.5*C9)/SQRT(A9-B9))) 124.5 0.5637 0.2947 0.2502 =4/D10/SQRT(A10-B10)* (ATAN (SQRT(B10)/SQRT(A10-B10))-ATAN(SQRT(B10-0.5-C10)/SQRT(A10-B10))) 149.5 0.5637 0.2947 0.2619 =4/D11/SQRT(A11-B11)* (ATAN(SQRT(B11 )/SQRT(A11-B11))-ATAN(SQRT(B11-0.5*C11)/SQRT(A11-B11))) ■1 i A -_ C D E 13 19.5 0.2281 0 1517 0.0322 = 4 D1 3 SQRT (A1 3-B1 3)' (ATAN ( SQRT ( B3) 1 SQRT (A13-B13) ) —ATAN (SQRT (B13-0.5-C13)/SQRT(A13-B13))) 14 34.5 0.2281 0.1517 0.0527 =4/D14/SQRT(A14-B14)"(ATAN(SQRT(B14)/SQRT(A14-B14))-ATAN(SQRT(B14-0.5-C14)/SQRT(A14-B14))) ' 5 54.5 0.2281 0.1517 0.0713 =4/D15/SQRT(A15-B15)*(ATAN(SQRT(B15)/SQRT(A15-B15))-ATAN(SQRT(B15-0.5*C15)/SQRT(A15-B15))) 16 79.5 0.2281 0.1517 0.0912 =4/D16/SQRT(A16-B16)"(ATAN(SQRT(B16)/SQRT(A16-B16))—ATAN(SQRT'3"o— 0.5'C16)/SQRT(A16-B16))) * ** 99.5 0.2281 0.1517 0.104 =4/D17/SQRT(A17-B17)"(ATAN(SQRT(B17)/SQRT(A17-B17))—ATAN(SG- -0.5*C17)/SQRT(A17-B17))) ■ : -2-.5 0.2281 0.1517 0.1142 =4/D18/SQRT(A18-B18)"(ATAN(SQRT(B18)/SQRT(A18-B18))-ATAN(SQRT(Bi 6-0.5*C18)/SQRT(A18-B18))) • 9 149.5 0.2281 0.1517 0.1217 =4/D19/SQRT(A19-B19)*(ATAN(SQRT(B19)/SQRT(A19-B19))-ATAN(SQRT(B19-0.5-C19)/SQRT(A19-B19))) :: 174.5 0.2281 0.1517 0.1295 =4/D20/SQRT(A20-B20)4(ATAN(SQRT(B20)/SQRT(A20-B20))-ATAN(SQRT(B20-0.5-C20)/SQRT(A20-B20))) 11 15 0.3103 0.5069 0.0492 =2/022/SQRT(B22-A22) 'LN((SQRT(B22-0 .5"C22)+SQRT(B22-A22) )•{ SQRT( B22)-SQRT(B22-A22))/(SQRT(B22-0.5'C22)-SQRT(B22-A22))/(SQRT(B22)♦ SQRT(B22-A22))) w A B C D E 23 35 0.3103 0.5069 0.1031 =2/D23/SQRT(B23-A23) LN((SQRT(B23-0.5"C23)+SQRT(B23-A23)) (SORT( B23)—SQRT(B23-A23))/(SQRT(B23-0.5*C23) —SQRT(B23-A23))/(SORT(B23) + SQRT(B23-A23))) 24 55 0.3103 0.5069 0.1406 =2/D24/SQRT(B24-A24) *LN((SQRT(B24-0 .5*C24)+SQRT(B24-A24))*(SQRT( B24)-SQRT(B24-A24))/(SQRT(B24-0.5*C24)-SQRT(B24-A24))/(SQRT(B24) + SQRT(B24-A24))) 25 80 0.3103 0.5069 0.1752 =2/D25/SQRT(B25-A25) *LN ( (SQRT(B25-0.5*C25)+SQRT(B25-A25))*(SQRT( B25)-SQRT(B25-A25))/(SQRT(B25-0.5*C25)-SQRT(B25-A25))/(SQRT(B25) + SQRT(B25-A25))) 26 102 0.3103 0.5069 0.1963 =2/D26/SQRT(B26-A26)*LN((SQRT(B26-0.5*C26)+SQRT(B26-A26))*(SQRT( B26)-SQRT(B26-A26))/(SQRT(B26—0.5*C26)-SQRT(B26-A26))/(SQRT(B26)+ SQRT(B26-A26))) 125 0.3103 0.5069 0.2179 =2/D27/SQRT(B27-A27)*LN((SQRT(B27-0.5*C27)+SQRT(B27-A27))*(SQRT( B27)-SQRT(B27-A27))/(SQRT(B27-0.5*C27)-SQRT(B27-A27))/(SQRT(B27)+ SQRT(B27-A27))) ■ 155 0.3103 0.5069 0.236 =2/D28/SQRT(B28-A28 )*LN((SQRT(B28-0.5*C28)+SQRT(B28-A28))*(SQRT( B28)-SQRT(B28-A28))/(SQRT(B28-0.5*C28)-SQRT(B28-A28))/(SQRT(B28)* SQRT(B28-A28))) 196 0.3103 0.5069 0.2533 =2/D29/SQRT(B29-A29)*LN((SQRT(B29-0.5*C29)+SQRT(B29-A29))'(SQRT( B29)-SQRT(B29-A29))/(SQRT(B29-0.5*C29) -SQRT(B29-A29))/(SQRT(B29) + SQRT(B29-A29))) - — y — 2 = - S« — — - DC — 2 I i I —- y- — > y- > s. z a a n (ft - C z CO 0 - 11 c --1 X m - - - > a z D r z > _ > ± K - - . Bn W CO CO CO U CO 3 c c c CJ CO — CJ o o u — o o CO CO COOOOOC — — 2 C - 'J CO CT ~ - ~ ~ - ~ o — O B - - cc — r r