70 AIIAI cms i )l HAM HAIA Ml t IN« IIIAIA K 11 Ml'IHK .Al MAM I AW. '.INI tl I IIIAI.IANI'. 71 al a time, so it lakes many measurements l<> determine .1 1 .iu- , -quaiiou tic . the rttC .is a function of concentration). One can gel useful ilatu from a single run Willi 1.....idnccl measurement. Still, the advantage of a direct method is that one measures the rate equation He. the rate as a function of the rcactant concentration) directly. One does not have to make any assumptions about the form of the rale equation to gel an answer. Generally direct measurements are much easier to (it to a rate equation than indirect measurements because, in a direct measurement, one determines the rate equation directly, while in an indirect measurement, one needs to infer the rale equation by lilting a curve to the data. The latter process can introduce some degree of error. In my experience, when you are trying to determine kinetics for a new system, it is usually better to start with an indirect method. The indirect method gives you an approximate rale equation with a quick and easy experiment. That is often good enough. A direct method is required only when you need a precise rate equation. If you are designing a process where a 10% change in rate matters, you need to do direct rate measurements. If you can accept a 10% error and adjust the process accordingly, an indirect method will suffice. In my experience, direct measurements take 10-100 limes longer than indirect measurements, so direct methods are useful only when a high degree of accuracy is needed. 3.5 EXAMPLES OF DIRECT AND INDIRECT METHODS Indirect method! IK the most common kinetic measurements in the older literature. You uicasuic a concentration as a function of time and then lit the data to a rate equation. Hun 1 nicisuicnicnis aie harder. One has to find a way to measure the rate of reaction directly. Most direct method* involve differentiating a rate equation. However, that is not a necessity, for example, in the reactor in Figure 3.8, one could measure how much arsine Hows into the reactor and how much flows out. If one knows how many moles per hour of arsine Mow into the reactor and how many moles per hour How out of the reactor, one can calculate the rale from a mass balance: R As = — (Fin — Foul) Aw (3.9) where R,v is the rate of arsenic deposition per unit area, Aw is the area of the wafer. I „, is the flow rate of arsine inlo the reactor in mol/hour, and. F,„, is the How rate of arsine out of the reactor in mole/hour. Notice that one is still directly measuring the rate. Therefore, one would call the measurement a direct method or a differential method, even though you are not differentiating anything to get a direct measurement of the rate. Nole. however, that equation (3.9) applies only if the rate is constant across the wafer and there is no reaction anywhere else in the reactor. If there were a reaction somewhere else in the reactor, one would have to do analysis to eliminate those effects. One would call the measurement integral methods or indirect methods since one needs to do .111 analysis to determine the rale. Generally, one would only call .1 iiieasuieineui .1 dont determination of the rale equation when one can diiectK meastiie Ihe rale .is .1 function of concentration. If one has to do some analysis, n will be an indue. 1 method I In n ,ue many variations ol this idea lodav. most direct kinetic measurements are .....1 Continuously lllmd UUtk rcacloi (CSTKl figure t-l shows a diagram ol a 1 SI K Basically, you continuously Iced Hictanls into ihe reactor. Some ol ihe leaclanl .....Il I ul's lead, while olhei reaclanl molecules |iisl How through the reactor. You ihen ..... 1111 Ihe reactanl concentration in the mlel and outlet of the reactor. Ihe average >• >• linn 1.He \ \ m the reactor is given by 1 -III f „IUI r — A *-A rA = - ( \ KU ..... 1111 11 '""s llie icaeloi so that ihe mixlure stays well mixed. In that case, the tale is It 1 ml throughout Ihe reactor, so one can calculate Ihe rate directly from equation (3.10). s.....1" > impoilanl ilnecl method is Ihe method of initial rates In Ihe 1ml1.1l lale im lb,id. one runs the reaction in a batch reactor, as in an indirect measurement. However. IM in.ilwcs ihe lale differently. Consider Ihe data in figure 32. figure 3.2 show. .........lianou lune dala. Notice that according lo equation (2.4), the rale al any lime I iiľ Klope of the target lo the line. Therefore, one can use ihe slope lo gel a rale. In ílu initial rate method, one fits a (dashed) line lo Ihe initial pari of ihe concenli.il..... 1 11 ilu data Ihe rale is Ihe slope of Ihe line. One then changes the initial concentration, ■in,! c uei.iies ihe rale versus concentration data. The advantage of this approach is thai "n. 1 in ..peiale direct data quickly, by running several small reactors al once. I XAMPLES OF INDIRECT MEASUREMENTS .....1 measurements are generally made in a batch system. Most of ihe rale measure ..... m'u did in your chemistry lab were indirect measurements, for example, you mighl I. ." loaded some species into a beaker, measured Ihe concentration versus time, and In 1 1 hi 1.. .1 first-order or second-order rale law. Thai is an indirect measurement. The " 1 h fl......Ksis experiments that students sometimes do are also indirect measurements ..I .......• l»« I 'lie. 1 measurements require that you actually measure the rale. The measure ....... harder. They are seldom done in undergraduate labs. In Table 3.1. we briefly .....1,1 several techniques that one uses to measure rates of reaction. As an exercise, II.......1'' should go back and decide whether each method is a direct or indirect one 1 11 riNG DATA TO EMPIRICAL RATE LAWS: SINGLE REACTANTS M iln. point, we will be changing topics. We will assume that you have used eilhei a ''........ 'in indirect method lo measure the rale data for a given reaction. We will now dim lion how one Iiis the rale dala lo a rate equation. I In I'.'iieial Scheme will be to 1 I'li'-n.....B Ihe oidei ol the reaction. (In a complicated case, one also has lo de.........<■ die form ol ihe rale equation, 1 ' I 11 the . .instants I.....I I1.m is In ili'li'inuiic Ihe unlet ol I lie 1 em 110 72 ANAI Y'.l'.ul MAII DAIA anai .'.»'.« >l HAIA I MOM A I Ml M Ml NIIAI HIAC.IOII 73 3.8 ANALYSIS OF DATA FROM A DIFFERENTIAL REACTOR The easiest data to analyze arc data from a differential reactor. As noted above, in a differential reactor, one can measure the rate directly at a fixed concentration of reactants. One then varies the concentration of the reactants and measures the rate again. The data can then be fit to a rate equation using a regression technique. If the data are simple, a log-log plot of the rate versus concentration can be used to infer the order of the reaction. In a more complex case, one needs to assume a form for the rate equation and then fit the data to constants. In Chapters 4 and 5. we will discuss how one can guess a suitable rate form for the rate equation. However, the analysis is easy once the data are obtained. For example, copper and platinum have been suggested as components in memory chips. Steger and Masel (1994, 1998) have examined the rate that copper is etched in a mixture of oxygen and hexalluoropentanedione (CFjCOCHjCOCFj). Steger and Masel found that the main reaction is 02 + 2CF,COCH:i(rOCF, + Cu Cu(CF,COCHCOCF3 )2 + H20 (3.11) Steger and Masel loaded a copper disk into a flow reactor similar to the one shown in Figure 3.7. They then turned on the feed and measured the weight of the copper disk as a function of lime lo yield the dala in Figure 3.9. The data were then plugged into equation ( V 1(1) lo calculate a rale. Sieger and Masel then changed the 02 concentration in the reacloi and measured the rale again. Alter several runs, they were able to generate the dm shown iii Figure \ III The dala arc a little unusual in that Steger and Masel observed two different rale laws: one rale law when the copper is oxidized and a different rate law when (he COppet is reduced, Plllther, there is a change from one rate law to another as the oxygen pressure is increased in the reactor and the copper is oxidized. Still, the analysis ill ilus dala are simple. For example. Sieger and Masel used a regression technique to fit (he data on the unoxidized surface lo the following rate form: 1/2 1 + K2P£ • 3.12) 40 60 80 100 Time, minutos Figure 3.9 The weight of a copper disk as a function of time measured by Steuar and Mnnnl 0010 • Slope = 0.5 ■ Metallic color ■ Oxide color 10 100 Oxygen pressure, torr Hum" 3.10 The rate of copper etching as a function of the oxygen concentration. (Data ol Stognr and M«.l (1H98).| ......i how easy it is to do the analysis even though this is a very complex reaction I he technique can easily be generated. If one can measure a reaction in a differential ......... one can gel data for the rate as a function of the reaclanl concentration. One can tin n hi the rale data using a regression method. Detailed analysis of the dala arc given in I simple 3.F. The result is a straightforward analysis of rale data. i n 1 Pitfalls in the Analysis of Rate Data from a Differential Reactor ii.....ire a few pitfalls in analyzing the rale data from a differential reactoi • 11 is not uncommon for more than one rate equation to lit the measured kinetics « nihil the experimental uncertainties. One should not infer that the rale equation is .....eel just because the data fits. . I he quality of kinetic data varies with the equipment used and the method ol temperature measurement and control. Data taken on one apparatus are seldom directly comparable lo dala laken on a different apparatus. . h is not uncommon to observe 10-30% variations in rate taken in the same apparatus on different days. Usually, these variations can he traced lo variations in the temperature, pressure, or How rale in the reactor. . I he procedure used lo lil the dala can have a major effect on the values ol the parameters obtained in the dala analysis. . Hie quality of the regression coefficient (r) docs not tell you how well a model his mhii data, i imples \ A lo \.V illustrate this effect. In the examples, we lil the Paramecium dala in I i)-uie .' I lo two different rale equations: k,K,|par| I i K [pal | k,K2|par| I i K|par|v- Rate Rate (3.11) I VI4) nul lluee (lilleienl lilting procedures 74 au/u ,\r.....(All MAI A ANAI Yftlti «i| IIAM IIAIAIIIOMANINIKlItAI MfACIOM 76 Table 3.3 Results ol lour different ways In III III. r.ili. .Iii.. Rale Equation Filling Method Ka rJ Inl.ll I'.rmi (3.13) Lineweaver-Hurkc 141) 0.0370 0.901 9454 (3.13) Eadie-Hot'slee 246 0.0156 i) Hi 5647 (3.13) Nonlinear leasl 204 0.0221 ()."05 4919 squares (3.14) Nonlinear leasl — — 0.'X)S 4576 squares Table 3.3 illustrates the result of the analysis. Notice thai the rate constants vary by a factor of 2 according lo which fining procedure is used. The Eadie-Hofstee lilting method gives a terrible r bui a reasonable error. The Linewcaver-Burke method gives a great r, but a large error. Equations (3.13) and (3.14) lit the data almost equally well. The data show a small preference for equation (3.14). However, the analysis in Example 3.C shows thai the differences are not statistically significant. These examples illustrate the pitfalls in analy/.ing rate data. A rate equation can fit data poorly and still give a reasonable value of r. A rate equation can fit well and give a poor value of r2. Different models can fit the data within the statistical noise in the data. There is one other difficulty with direct measurements—the measurements are rather tedious. One needs lo make a separate measurement for each set of conditions. Generally, each point in Figure 3.10 Ux)k about a day lo take and a second day to reproduce, flic entire plot took over a month. Further, we needed several plots: one for each hexafluoropenlamedione concentration to determine a rate law. These measurements took months to complete. Also, differential reactors lend to be hard to operate. There were man) bad runs. One docs have to think about whether it is necessary to lake such difficult data when one is planning the experiment. 3.9 ANALYSIS OF RATE DATA FROM AN INTEGRAL REACTOR The alternative is to use an indirect method. The advantage of an indirect method is that one can generate a significant amount of data in a short lime. The disadvantage is that the data are harder to analyze than data from a differential method and the uncertainty in the data analysis is large with differential data. In the next several sections, we will discuss how one analyzes data from what is called an integral reaetor, which is a reactor you use to do both indirect and direct rate measurements. The simplest integral reactor is a batch reaetor. A batch reactor is a closed vessel like a flask with a stirrer. One loads the reactants into the batch reactor and measures the concentration of the reactants as a function of time. Typical data are shown in Figure 3.2. One then fits the data to a universal curve to infer the order of the reaction and the rate constant. We will derive an equation lor the performance of the batch reaction in Section 3.9.1 and use the equation in Section 3.10. The actual derivation is not important. One could skip to Section 3.10 without loss of continuity. 3.9.1 Derivation of the Concentration versus Time for Batch Reactions Generally, if one wants to use a batch reactor to determine kinetics, one loads the reaclanls into the reactor and runs the reaction while inciisiiiui(i llir > urn ciiliulniti vcisiis tunc (>nc ...... ",s ,hr » universal curve to mle, ., rate equation. In tins |«...... m "ill derive cM.,ess,ons lo, the imm-.sal unves. We will assume perfect mum,....... I.....in h oi the limn 'a = UA|" (3(IJ) W, will then derive a universal curve lor the concentration versus lime. We include Hie .....mion '"' Completeness. However, one can skip to equation (3.42) without loll ol I iiMlmuily < ons.de. a reaction A => B in Ihe batch reactor shown in Figure \ 11 lo. the purtx.se .....we will assume thai the concentration of A is the same everywhere in Ihe ........• ",u reactor is *»lcd *> »<> A evaporates, enters (he reactor, Ol leaves Ihi II .11 till II dice ,s no Row in or out of the reactor, ihcn all of (he A lhal is generated in Ihe " " 1.....IUSI accumulate in the reactor. Therefore (Kate of accumulation of A) Rate of generation of A where Rale of \ / moles of generation J = ( A reacted Of A / \ per unit time Nuhrttiluting the definition of rA into the right side of equation (3.17) yields Rale of v generation I = rAV or A / ..... v is ,Ik- volume of liquid in the batch reaclor. Similarly (Rate of accumulation of A) = di Where N\ is the number of moles of A in the batch reactor. ( I 16) (3.1/) I (.IK) (1.19) Figur« 3.11 Ah. 78 ANAI VMM M MAII HAIA ANAI Yliim il MAII HAIA I III IM AN INI 11 (HIAI IIIACIOM 77 Substituting equations (3IH> iiiul (VI1)) iiilu equation l t Idl shows dNA rAV = Integrating equation (3.20) yields "N^=fdt = ffi V rA J,, i i.2(» (3.21) where is the initial concentration of A, NA is the final concentration of A, and t is the reaction time. Inverting the limits in equation (3.21) yields rN* dNA j; n'. V(-rA) (3.22) It is often useful to work in terms of quantity called the conversion, XA, where XA is defined by y K - Na (3.23) I'liv.u ally. X\ is 11u* fraction of (he reaclant A that has been converted into products. < )n<- should memorize this definition before proceeding. Differentialing equation (3.23) and rearranging yields dNA « N^(-dXA) (3.24) Changing the integral in equation (3.22) from moles to conversion yields 'x* NA(-dXA) V(rA) (3.25) rearranging equation (3.25) yields V(-rA) (3.26) Note that rA is negative in equation (3.26), that is, (-rA) is positive. Consequently t works out to be positive in equation (3.26). Equations (3.22) and (3.26) are the key equations for batch reactors. Note that when a reaction proceeds, (-rA) decreases. I/V(—rA) increases as indicated in Figure 3.12. The time is proportional to the area under (he curve in Figure 3.12. Initially, it takes little lime to increase the conversion. However, as time proceeds, ihe curve in Figure 3.13 increases so it takes proportionally more lime lo gel lo highei conversion Another important idea is that the time (o gel lo a certain conversion is proportional to the area under Ihe curve in Figure ' I 2 Tins ideii i olli n used in n.i. loi design vU") a) IS 10 n-■-r 0% 20% 40% 60% eo% 100% Conversion Figure 3.12 A plot of ,, for a first-order reaction in a batch reactor V(-rA) 1.2 I o 8 g io 0.6 rr 0.4 0.2 (i i ' 1 Rate 1 1 ' 1 t Simple average Eq.(3.28) 0.2 0.4 0.6 Time (I H hfure 3.13 ( M versus C for a first-order reaction In a batch reactor. Notice that the r Neu we will derive an approximation for constant density systems. Examples ol loiisinni density systems include I I 11j11111 relictions in solution I (Ins phase reactions with no ncl change in moles, such as CO I II..O > ( <). I II.. I Hciu'tioiM in consiiiiii volume closed butch reiiclors 78 ANAI YfllflOl HAD DAI A For constant density, wc can remove v from the Integral m equation I L26): vJo -rA (3.27) (3.28) where CA is Ihe initial concentration of the reactant A in the reactor. It is also useful to transform equation (3.28) as follows. Note that if we divide equation (3.23) by V, and rearrange, wc find that for a constant-density system CA = <*(] -XA) (3.29) where CA is the concentration of A and, C"A is the initial concentration of A. Substituting equation (3.29) into equation (3.28) yields (3.30) when ( \ m the ii| >| <•■ linnl ol I lit.- inu-gial is ihe concenlraiion altera time, r. Switching ílu- limits in equation I I, <()) yields (3.31) Equations (3.28) and (3.31) form the basis of batch reactor design for constant-density systems. Notice that according to equation (3.31), the time is given as an integral of the rale equation, which is why ihis type of analysis is called an integral method. liquation (3.31) is different from equation (3.26) in that it considers a plol of l/(—rA). l/(—rA) decreases with increasing concentration as indicated in Figure 3.13. Note, however, that during the reaction you start with a high concentration of the reactants, and end up with a low concentration of reactions. Consequently, l/(—rA) increases as the reaction proceeds. The time needed to obtain to a given is proportional to the area under the curve in Figure 3.14. The idea that time is proportional to an area under the curve is quite important to reactor design. Next, it is useful to discuss how equation (3.31) relates to (he definition of a reaction rate in Chapter 2. Recall that in Chapter 2 we defined the reaction rale to be a rale in molecules per hour per reactor volume. Next, we want to demonstrate that equation (3.31) implies that Ihe production rale of species in die ic.u loi is piopmiioniil lo Ihe average rale ANAIVHIIIOI MAII IIAIAinoMANINIKIIIAI HIACIOH 79 In Ihe leiuloi limes the reacloi volume, where lite avciagc reaclion lale. i .x is given In 1 - fA 3 dCA (3.32) N..i. ih.n iluring a reaction in a batch reactor, rA is changing with lime as indicated In I tKlirc \ 13, The funclion rA is an average value of rA over Ihe course of the KM...... I*i rlorming ihe Integral in the denominator of equation (3.32) and rearranging yields reacloi. Ihe derivation is included because n.....\ ol out siudi-iiis need in i-i il .i)'.iin lluwevei, I am assuming ili.n most ol out icadcis have iiheadv seen the ili iIvmum Ihcivfoic, one could skip litis section wilhoiii loss ol ,.......mil) 80 ANAI YMMH MAM UAtA Let's begin by considering lusi unlet react.....s. lot ,i lirsi orilci reaction A > U rA = -k,CA (3.38) Substituting equation (3.38) into equation (3.31) and integrating yields (3.39) Substituting equation (3.29) into equation (3.39) yields k, \\-xJ (3.40) Similarly, for an nth-order reaction rA = -kn(CA)n (3.41) Substituting equation (3.41) into equation (3.31), integrating, and rearranging yields (3.42) where C'A is the initial A concentration; CA is the concentration at time, t; n is the order of the reaction: and, k„ is the rate constant. As an exercise, the reader may want to show thai equation (3.42) goes to equation (3.39) as n goes to 1.0. 3.9.3 Reversible Reactions The derivations in Section 3.9.1 were Tor irreversible reactions However, many reactions are reversible. Consider the reaction (3.43) which obeys From a mass balance (rA)=-k,[A] + k2[B] [B| = [A]0 - [A] Substituting (3.45) into (3.43) yields d|A| dt = (rA) = k,|A| ( k,t|A|„ |A|i (3.44) (3.45) 82 ANAlYSľlOI MAII IIA1A I II IIN<, IIAH HAIA KIIIH Al IIIIIAVlufl Hl The reaction rate goes to teto when lbe rate reaches equilibrium Ixi's define |A|,. as ihe concentration of A when the system reaches cquililii iiiiii U\ dchiuhou 0=-k,[A|c + k:(|A|„-|A|c) Subtracting equation (3.47) from (3.46) yields d[A] di = (k, +k2)([A|-|A]c) The solution of equation (3.48) is (k,+k,)'n /|A],,-[A]C\ V IA] - [AL J (3.47) (3.48) (3.49) which looks just like a first-order approach to equilibrium. Table 3.4 shows several other examples. Derivations are given in later sections. 3.10 QUALITATIVE BEHAVIOR Equations I 1.39), (3.42), and (3.49) are the key performance equations for nth-order kinetics In a batch reactor. Ii is interesting to plot these results as a function of lime. Figure 3.14 shows a plot ol ilir oiuccniialum as a function of lime calculated from equations (3.39) and (3.42) lor n - ',.1! All ol the curves are qualitatively the same. The concentration starts out varying linearly with time and then levels off. Generally, the curve levels off more quickly wuh larger values of n, namely, higher-order reactions. We plotted the results in two ways in Figure 3.14: first with k(CA) = 1. and then with k - I and C" = 1.5. Notice that all of the plots arc similar. The concentration of — < o econd order 0 0.5 1 Time Time Figure 3.14 A plot of the concentration as a function of tlma calculate) from aquations (3.38) and (3.42), 0.5 I Time i Igiiro 3.15 A replot of the data from Figure 3.14 as a function of ln(CA/CJi) and (CSJ/CA i 1 1.5 'i" ii.ii i.ml, A. initially decays quickly with time. However, the rate ol decay slow In ■ li i lenses, giving an exponential-looking curve in all cases. I he kC) point thai one needs lo draw from Figure 3.14 is that the concentration lime i i i lni a batch reactor lix>k qualitatively the same, independent of the order ol I he " '.....ii The concentration-time profiles are the same al low conversion independent i lín niilei of the reaction. There are some quantitative differences, especially at high .....vi iinn However, these differences arc often subtle, especially when (here is some ........i the data. Iheie are two ways lo look al Ibis result. On one hand, the performance ol a batch ......Ill's not vary lhal much with the order of the reaction, except at high conversion ■ 'ii ' i|iienily, for many purposes, you arc going to get the righl answer even if you IMtime the wrong order for Ihe reaction. On the other hand, il will be hard lo use a batch ■ in in determine ihe order of Ihe reaction because the resulls arc not going la \.hn MM......' Ii wilh Ihe reaction order. Consequently, there are some uncertainties in using i iii Ii n ii lni dala lo determine kinetics. i I I I 11 I INC RATE DATA TO IDEAL BEHAVIOR tlil Lssoiťs Method ••• ''" lid i.ilure, people say lhal one can gel some useliil information by rcploiiinr lbe limu in Plgure 3.14. Foi example, Figuře 3.15 shows a plot ol log (CA/C'A) versus luue "i.l |n " < \ > II veisus luue Noluc lhal a plot ol log (CA/0 is li,K')u •«»• " I"st ••'.....i. lion. wlnle a plul ol K A/CAI I is lighllv euived. In conlrasl, lbe plot ol 111 1 \l I | is line.ii loi a second oidei icaclioii and slightly cuiNcd loi a lusl oidei ii 1 1..... I heieloie. m ptinciple, one can dislinginsh Ivlween lusl and secouil oidei '........• l'\ piepaiiug hiincs hke lliose m Figuře t \*> nud secing which plot is ihe iiionI lnu ni fV*MI T ■ III I I M I IA I I I 'A I A I II I If)' . I [A I 1 I >A I A 11 ' II Ii AI IIIIIAVHIM ii', Bssen proposal thai one could )Tiiri.iii/i' iIil'sc nil.i1. lo determine the ordei <>i the rcuclion. The idea is u> 1. Measure the concentration versus lime in a batch reactor. 2. Fit the data to the hatch reactor equations [equation (3.39) for lirst-order. equation (3.42) for second-ordcr|. 3. Whichever fits best is assumed to be the correct rate equation lor the reaction. Note that according to equation (3.42), a plot of (CA)'~n versus time should be linear for n ^ 1 while according to equation (3.39), a plot of ln(CA/CA) versus time should he linear for n = I (i.e., a first-order reaction). Essen proposed that one can determine the order of a reaction by constructing plots of ln(CA/CA) and |(CA/CA)n_l - I] versus time for various values of n. and then doing some analysis to check the results. See Hareourt and Essen (1865, 1866. 1867). For example, in our undergraduate labs, we examine the reaction between Red Dye 4 and hydrogen peroxide to yield a yellow dye. Table 3.5 shows the concentration of Red Dye 4 as a function of time as measured in a batch reactor starting with equal dye and peroxide concentration. According to Essen, one should analyze this data by making a plot of ln(CA/CA) (for lirst-order reactions). (C"/CA)- 1 (for second-order reactions), and (CA/CA)2 — I (for third-order reactions), and see which plot is the most linear. The easiest way lo do this is lo put the data into a spreadsheet on a computer. One then uses the spreadsheet to plot ln(CA/CA). (CA/CA) - 1. (CA/CA)2 - 1 versus time lo see which plot is the most linear. Sample spreadsheets are given in Example 3.D. A linear regression process can be used to lit the rate constants. Figure 3.16 shows the plots. The data are actually second-order, hut you could not tell that from the figure. Notice that all of the plots look linear. If we exclude the first two points, Uien a plot of \n(C"A/CA) is linear with a regression coefficient of 0.981, a plot of (CA/CA) — 1 is linear with a regression coefficient of 0.999, while a plot of (C^/C/v)2 — 1 is linear with a regression coefficient of 0.984. A regression coefficient of 0.981 is as good as the measurements. There arc two ways to look at this result. First, all the regression coefficients are reasonable, which means that you cannot accurately determine the order of the reaction from the data. On the other hand, the plot of (CA/CA) is more linear than the other plots. According lo Figure 3.15, this is the expected behavior for a second-order reaction. Therefore, one might say that the data in Table 3.5 are as expected for a second-order reaction (i.e., n = 2). My own opinion is that such an assertion is not valid. In the supplementary material we note that a regression coefficient measures the uncertainty in the parameters obtained ii h o J O 2 - l ll'imi 0 5 10 15 0 5 10 15 0 5 10 15 Time, minutes Time, minutes Time, minutes Figure 3.16 An Essen plot of the data in Table 3.5. 0 10 20 30 40 Time, minutes Figure 3.17 Essen Plot ot the Data in Table 3.6. 50 Table 3.5 The concentration of dye as a function of time CA, mmol/liter t, minutes CA, mmol/liter t, minutes CA, mmol/liicr r. minutes I 0 0.63 6 0.45 12 0.91 1 0.59 7 0.43 13 0.83 2 0.56 8 0.42 14 0.77 3 0.53 9 0.40 15 0.71 4 0.50 Id 0.38 16 0.67 5 0.48 11 0.37 17 i.....i a linear lit lo the data assuming that the model is correct. It does not tell you how well a model works. In Example 3.A, we show that the regression coeflicicnl can v.iis by a factor of 2 according to how il is calculated. As ;i result il is not useful lo i omparc regression coefficients calculated lor different models; each regression oicllu icni i ' .iliulak'd using a different model (i.e., a lirst-order model vs. a second ordei model) As a result, comparisons of small differences in the regression cocflicicnts are rarel) in. ininglul. Still, that subtlety is often lost in the literature You often see people makinji the mistake of saying that a rale equation fits beltei because the regression coellii icnl is • losei to 1.0. That assertion is incorrect 86 ANAI YMM ll HA 11 HAIA I II I UJi i 11A 11 HAIA hUHIAI III IIAVH Hi 11/ I do not want to imply that the results in Figure 3.16 Mt useless I lie one advantage of Figure 3.16 is that one can calculate the rale constant lor the reaction from the slope of the line in Figure 3.16. According to equation (3.42). the slope of the curve is given by l>k„(CA)n-' (3.50) Slope = (n For the data in Figure 3.17. n = 2. Therefore, if we know the slope and CJA. we can calculate k„. Essen and Harcourt (1865) and Van't Hoff (1884) showed that one can use this method to analyze data for a wide variety of simple reactions. The advantage of this approach is that it is conceptually simple and it is easy to assess the goodness of the lit. The disadvantage of the method is thai it requires many plots, although that is not a problem if one has a spreadsheet available. 3.11.2 Van't Hoff's Method Van't Hoff (1884) used Essen's method to analyze a variety of rate data. He found that it usually worked. However, there were some examples where the method did not work well. Table 3.6 shows data for the reaction of chloroacetic acid and water at 100 K. Figure 3.17 shows a plot of ln(CA/CA) versus t. The plot is very linear, which suggests that the reaction is first-order. However. Van't Hoff examined the reaction in more detail and noted that if one rearranges equation (3.39). one can show that a lirst-ordcr reaction should obey (3.51) Table 3.6 shows values of k| computed in this way. Notice that k decreases slightly as the reaction proceeds. Schwab (1883) examined this effect in some detail, and proved that the rale constant varied with the product HCl concentration. Therefore, the reaction HOCH2COOH Table 3.6 Buchanan's (1871) data for the reaction CICH2COOH + H20 + HCl at 100°C / (C1CH2COOH|0\ k| - [ [ClCHjCOOH] )' [CICH2COOH], Time, hours grams/liter hour 1! 4 — 2 3.80 0.026 3 3.69 0.026 4 3.60 0.023 6 3.47 0.025 in 3.10 0.024 13 2.91 0.023 19 2.54 0.022 25 2.26 0.021 34.5 1.95 (ll).'l) 43 1.59 0.021 4* 1.39 0 021 is not ically hrsl ordei even though 11 plul ol lint 'A/l'Al versus 1 is lineal Kalhcr, Ihe reaction follows a complicated talc equation Van't Hull noted lhal one has lo be very careful in using the I-ascii plols to analyze .l.il.i because one can easily be fooled into thinking thai a reaction with a complex rale ■ quaiion follows simple first- or second-order kinetics. Van't Hoff asserted thai it was important lo check ihe results from Essen's analysis. To do the check, one computes 1 lioni equation (3.51) for n = 1, or k„ from k„ = I (n- l)t(CA)"- c° (3.52) lui 11 / I. One then makes a plot to cheek that kn is constant. k„ should always woik mil in he constant within experimental error. If k„ varies in any sort of systematic way. then iln reaction is not really first- or second-order. Instead, a complex rale equation will be iii riled to lit the rate data. In my experience, Van't Hoff's method is much more accurate than Essen's. I lind n U 1 lul to plot what I call a Van't Hoff plot, which is a plot of k„ versus t for various yulues of 11, and sec if it is constant. For example. Figure 3.18 is a Van'l Hull plot ol Ihe dye data in Table 3.5. Notice that k-> is constant, which means thai the reaction nails 11 mid-order. On the other hand. k| is not constant for the chloracetic acid data in table 1.6, which means that the reaction is not first- or second-order. We give lurlhei ' samples in the solved problems. In my experience. Van't Hoff plots are much nunc •• 1 ilmg than Essen plots. Surprisingly, though, you usually see Essen plols. not Van'l Hull plots, in most textbooks. Example 3.D illustrates this method more carefully, t>ne In mid carefully examine this example before proceeding. 111.3 Powell's Method I Inn- are many other ways to analyze kinetic data. A particularly powerful method ih.n cd mainly before computers were available is called Powell's method Powell's method is derived from equations (3.39) and (3.42). 0.2 0 15 0.1 ()()!, 1 1 ' Oxidation of 1 1 k3 > red dye k2 1 k, 1 5 10 15 Time, minutes 0.05 I 004 S c 5 ■V £ 0.03 0.02 , 1 1 1 1 Hydration of chloracetic acid 1 1s\ < ■ **** É ■ ■ •— . 1 . 1 . k, 10 20 30 40 Time, minutes Ml línům a til Vim) iiott fiotultiii. ■..........lablM 3.5 and 3.8. Hit ANAI Y'.ľ. i )l HAH DA IA nil MAI M III Ml till III lit) Rearranging equation (3.42) ami taking the logarithm of both sides yields log. {r-i[(S)"- = logl(1lk(C(;)'-1] + log11)(t) (3.53) Similarly, taking the logarithms of both sides of equation (3.39) yields I ^ ) ) - In(k) = ln(t) (3.54) Figure 3.1° shows a plot of (CA/CA) versus log (t) calculated from equation (3.53) for various values of k(C^)"~' and n. Notice that all of the curves calculated for a first-order reaction have an identical shape. They arc just shifted to the right or left of each other. In contrast, the curves for second-order reactions have a decidedly different shape. Consequently, one should be able to distinguish first-order reactions from second-order reactions by plotting CA/CA versus log (t) and looking at the general shape of the plot. The easiest way to use Powell's method is to employ a spreadsheet on a computer. The general procedure is to I. Make a plot of CA/CA versus log 10 (r). ' Program the spreadsheet lo calculate ideal curves for C"/CA versus log 10 (t) for n - j,1,2,3. in I lie equations (3.S3) and ( (.54) to calculate the values of log (t) for each value «1 Ca/CJ. do 11■ .ii logIM(k(C'A)" ') as a variable (i.e., SBSI). Set the variable initially to zero. 3. Vary loghl(k(CA)" 1) to see which curve fits. Example calculations are given in solved Example 3.D. Powell actually proposed his method before computers were available, and so he did noi use a spreadsheet. Instead, Powell used a set of universal curves for zero-, first-, second-, and third-order reactions to see how data lit. figure 3.20 is a suitable set of 1 0.8 o< 0.6 < 0.4 0.2 0 0.1 i i r t r i m i-r-1—i i i 11 i|-1-1 i nni|-1-1—i » t ttt]-1-1 i t t i m \ . ■ry \% x. V\ v* v ŕ" \% \ ^ \\ \ v* \\ \vv -1-1—L i i ml-1—i"**-1 ■ ■ *'1__r-».i_i . i ■ 11^^. ._ĺ i i i _, . , 1 1000 mooo 10 100 Time (log r) Figure 3.19 A plot of (Ca/CJ) versus log t calculated for a series of first order and second-order reactions. 0.1 1 10 100 Time (In t) I Igura 3.20 Howell's universal curves tor half , first-, second-, and third older loactiuim O 11 I 0.8 .. 06 (I ■! 0 2 -1 t 1 1—r-T 1 li--1-1-1—1—1 1 ? 1 j-1-1 i 1 1 1 1 1 IKs. # Original l^N. da,a \ 0, \ % \ • • Shitted data > A, 0. ^-i_--1 . Ä 1 J 1 _1_1_ .....' \\ 10 limn Figura 3.21 An illustration of a Powell plot for data taken for a second-order reaction. 1 .il curves. One can use these curves lo calculate the order of a reaction I In 1.......il si heme is lo I Make a plot of CA/CA versus In (r) as shown in Figure 3.20, making sure lhal the si ales 011 each of the axes are the same as in Figure 3.20. ' Slide the data to the left and right to see which curve in Figure 3.20 lils Ivsi 1..... I .' I illustrates this method. We look data anil plotted it on the Powell plot We linn In Hi .1 1 In- dala lo the right lo see which curve fits best. In this case, the second onlei hum in In ilri than the rest, so we conclude lhal the reaction is second-order. We do need In 1'.. Inn k ami make a table of k, versus r as described in Section 3.12 to calculate the nile 11 1 mi I sample I A gives additional examples ol analysis of data with Powells melhod 1 II HIE HALF LIFE METHOD xiiuiin 1 Indira 1 method to anal) ic rate data is called tbc half •Ufa method 1 he ball lift m. 1i1. m I was quili populai More the days ol computers We include it because the ideas .. . I11I 1 vi n il tin- method is mil used veiy mm Ii iinvmoie IK) Af JAl i' it' i t )| MAI I IIAI A IMI HAM I II I Ml m ■II The half-life, ri/:. is defined as the mm- m gel to 509 conversion, in the hall hie method, one makes a log-log plot of the hall-life versus time and calculates the order from the slope of the plot. Next, we will derive an expression for the hall-life as a function of CA for a first-Older reaction and an nth-order reaction. Rearranging equation (3.39) shows that for a lirst-order reaction I In C: ki yCA at 50% conversion (CA/CA)= 2. Therefore. tl/2, the half-life, is given by ln(2) Similarly, rearranging equation (3.42) shows that for an nth-order reaction: . n-l 1 (i - xA) (n- i)kn(CA)"~ XA = 0.5 at t|/i. Therefore, for an nth-order reaction: i = t r./2 = II one needs to know only an approximate rale constant, one does not have to M tUtll) mu in exactly 50'/í conversion. Rather, one can run to any fixed conversion and gel uselul ml.....i.ition. Nnw . il one needs to gel an exact rate constant, one needs lo add some extra work (>ne «ill have lo run the reaction versus time and calculate an exact lime lo 5091 comcisinii Note that one can calculate t|versus ('" from data for a single run figures t "■ and I ■ i Illustrate the process. I el's start the experiment at t = 0. If we start at r = 0, C'A ■ I mol/hiei. iheu i, . 1« I lit- tune to get to CA = 0.5 as shown in the left part of figure 3.23. Now imagine i niiug the experiment at t = 0.3. At r = 0.3, C*A = 0.9. Therefore, the hall hie is die inn. u lakes lo go from CA = 0.9 lo t'A = 0.45. Similarly, we can imagine slaiting the . .|......lenl at any value of (\ in figure 3.23 and calculate the hall hie lot thai value nl l , I lie result is thai one can construct a half-life plot from a single run I igiiie <24 shows data generated in this way. Notice that the hall hie decreases im. nl\ with ln(tA). The slope of the plot is -1.0, which implies thai I -n«-I.O im u J. In my experience, half-life plots were the easiest way lo analyze batch reactoi d.H.i Im Inn i nmpiilers were readily available. The biggest advantage of the half-life method II I lljmo I ľ I A.........Ii.il.......I Imw...........■ 1 'I ll .....i ill III" I.....I H.....ill" ilnlil "»l wz AN/VI »mm M HAH IIAIA I II I IN< I OA IA lni MI 'IHM Al HAH I AWV. Mllllll'll III Al'. I ANI ii in § 2 c E I 1-5 x Slop< 1—1.0 - * 0.4 0.5 0.6 0.7 0.8 0.9 1 Figure 3.24 The half-life as a function of Cp, for the data in Table 3.5 (with some extra points added). that there arc large differences hclwcen. for example, lirst- and second-order dala. Further, if one takes dala on a system that does not follow a simple rate equation, one knows it because the half life plot is curved on a log-log scale. Still, the half-life method is difficult lo automate, Consequently, hall-life plots are now rarely seen in the literature. Equations i 1,56) and ( t.5X) are very useful, however. One can use these equations to decide how lo tun experiments, or to estimate activation barriers from very little data. See Section 2.5,1 For more information, llieie are Othei methods in the literature, including the Ciugenheim method. These methods have largely disappeared as computers made their appearance. For an older review, see Roseveare i in use equation <3.f>2). we need an expression for C'n and (', in lenns ol X|( In tit. in \i section, wc will describe how lo gel It • til The Stoichiometric Table t nii.iilei a general reaction aA + bB-► eC + dD (3.63) hi It a h. c. and d arc the Stoichiometric coefficients for species A. H. (', and I). IV«|n'i lively Assume lhal moles of A and N',', moles of B are loaded into a closed ........ill vessel. Nexl. we will derive a relation hclwcen the concentrations of all ol the i i.i lunction of the concentration of a single species. A. To simplify Ihe algebia. II tliil to work in terms of a new variable, X.\. the fractional universum ol A. wheie H , i't dcllncd as n'; ......ills \ is the fraction of the reaclaiii A that has been converted into products. One I.....lil memorize this definition before proceeding. Ni'M. il is useful to calculate the concentrations ol each of the species in Ihe reaclot i i I.....Hun ol X\. logici I I WN| has a neat trick lhal makes il easier lo keep li.it k ol Ilu Mu nuk is called a stoichiometric table i it. inn hiomelric table is a table where the following information is presented dm ,i Imii h xysicm): i . 1111 ti 111 I The particular species i uhnuli .' Ihe niiinhei of moles ol each species initially present i ni....... 1 Ihe change in ihe niiinhei ol mules due lo reaclion I iihiiiiu I Ihe uumhei ol moles remaining in the rcacloi . 11 lor conversion XA lulili i / is a siou hiomelric lublc lot reaction I Wit) with some ol ihe mloimaiiou missing Ntnl, let's calculate the changes in moles ol each ol the spet ies as ,i linutími ol XA Aimiine thill we loud N" moles ol A anil N',', mules ol II into tin ie.n lm then we mil »4 ANA! , !■.....IAII MAIA Tnblo 3.7 The start of a stoichiometric i.ihlc lor reaction (3.63) Species A B C D Inerts Total liiin.ii Moles ( 'liangc in Moli' N'li N°c K ny = n a + n? + K + K + n? hnal Moli- N{, NÍ nt the reaction for a time. t. so that the conversion of species A is XA. From the definition of the conversion we obtain Moles of A converted = N°XA (3.65) From the sloichiomctry of the reaction, the number of moles of B converted is given by Similai l\ Moles of B converted = Moles of C* formed Moles of D formed NAXA (aKXA NAXA (3.66) (3.67) (3.68) ll is important to keep track of the sign. During the reaction, the concentrations of A and B decrease while the concentrations of C and D rise. Consequently, the change in moles of B is: l) XaN" while the change in moles of C is c=4-QxAN AN Therefore, the stoichiometric table is as given in Table 3.8: One can slightly simplify Table 3.8 by noting -a — b+c+d = 5wPv= Amo1 (3.69) (3.70) (3.71) where Amol is the change in moles when reaction (3.63) goes one time. Substituting equation (3.67) into Table 3.8 yields Table 3.9. 3.13.2 Derivation of the Performance Equation for Reaction (3.59) Now il is useful to go back and derive a performance equation loi readmit ( t SO) | el's go back to equation (3.62). Equation (3.62) was a performance equation loi lemlion ( * v»l III IINtl DAIA It) I MCIKU At HAH I AWS MOlllt'll HI At; I AN IS 95 i 'i 3.U The stoichiometric table for reaction (3.63) Imlial Moles \ II K - 1 Ng + II No + In. rtu N? 0 ii.i.ii N? <■ Change in Moles Final Moles Na*a (;)*»** (4)nSx. -a - b + c + d NA(I - XA) N'B-(")^XA Nc+-N,XA a N°D+^N«XA N? -a - h + c + d" i ...... 3.9 An alternative stoichiometric table for reaction (3.63) IpM lea Initial Moles Change in Moles Final Mules ii I I' iii. ll: total n°D N? -n^xa -m Na(1 -XA) n aA i...... Imlial Concentration (Ihsnge Final Concentration «",',X|. (c/hK-IJXi, i.i/loi 7,x„ <; 1 > XB. Therefore, the XB term in the numerator of the log term in equation (3.75) will be negligible. Similarly. in the denominator of equation (3.75) will be negligible. Consequently, when there is a large excess of cyclopeniadiene. equation (3.75) reduces to A comparison of equations (3.76) and (3.39) shows that equation (3.76) is a first-order rate equation with k, = kBC° (3.77) Physically, what is happening is that when C" » CB, the concentration of cyclopeniadiene does not change significantly during the reaction. The Cc is constant in the rate equation. Therefore, ihe reaction appears as if it were lirst-order. The other key limit is when C£ = C". Note that during reaction (3.59), ben/oquinone (B) and cyclopeniadiene (C) arc used up at the same rale Consequently, if iuilially CB= C£, then CB = Cc 1 1 /Xi '.I Will NIIAI III Al.llllN!. \ Ytrywhcic during Ihe reaction. Consequently. 1 In- late equation loi reaclion t 1 v>) will •. 1 lin i lo the equation loi a second oidei ic.u lion. One ..hi derive the same result by Idling (',', approach C" 111 equalion (3.X-I). and ...... I Hospital's nilc lo do the limit. I be result is I kBCB C„ n 19) I .|<■.....i> 1 I /')| is equivalent lo equation (3.42), with n = 2. Equations (3.75) and ( I /'>! .....id 10 plan experiments. Firsl. one swamps the reactor with one species, say. <'. <• 1.1 measures ihe concentration of B as a function of time, and uses Essen's method 01 I'.iiM II s method lo see if Ihe reaclion follows equalion (3.76). One then runs the reai.....1 nil equal H and C concentrations, and uses Powell's method or Essen's method lo see .1 ■. 1 ■ 1......n 1 I 79) works. II both equations lit the data, then one can he assured dial the I P (3X0) 1 we will derive equaiions for Ihe conccnlralions of A. B, and C versus 1 ass........y 'i.......nuns I and 2 are both lirst-order. II reaction I is First-order, then ii<\ di = rA = -k|CA 11 XI 1 In 1. 1 \ 1 s the concentration ol A. Solving equation (3.XI) yields CA = CAe k" I I N.'l ............I I and 2 are lirsl order, then dC, dt = k)CA - k2C| 13.X 11 In.....s die concentration ol I 1 omhiiiiiig equations ( VX2) and |IHI| and integrating yields Q. (e k" -c k") + C?C k. k 1 (3*84)