M ANA! YMMII MA 11 tlAIA (0 x 05 0.6 0.7 0.8 0.9 t Figure 3.24 The half-life as a (unction of CA for the data in Table 3.5 (with sum.....tt.i i.....ii1. i.l.lmll thiii there are large differences between, for example, first- and sccond-onler data I milier, J if one takes data on a system (hat does not follow a simple rate equation, one know.* II because the half-life plot is curved on a log-log scale. Still, the half-life method is dillli nil to automate. Consequently, half-life plots arc now rarely seen in the literature. Equations (3.56) and (3.58) arc very useful, however. One can use these equation! Ill decide how to run experiments, or to estimate activation barriers from vei\ Imli iImIM See Section 2.5.1 for more information. There are other methods in the literature, including the Gugenheim method llii'tf methods have largely disappeared as computers made their appearance hot an ilkM review, sec Roseveare (1931). 3.13 FITTING DATA TO EMPIRICAL RATE LAWS: MULTIPLE REACTANT8 The results in Section 3.7 are useful only in the ease of a reaction where a single nm..... is converted into products. In a more typical reaction, two reactants, lot e\ample, t ittitl B. react to form products. One needs a more complex analysis to consider those ■ nsi>« In the work that follows, we will derive an expression for the conversion veisus nine uf a system with multiple reactions. However, we need some further information in i For example, consider the Diels-Alder reaction of ben/.oquinone (Hi and i vi l<■i«»un dicne (C) to yield an adduct (A): O Id I I Ml| It is useful to consider what data we need to analyze rule data lot ic.ii lion i l v»i Nc»|, we will derive an expression lot lite conversion ol l>eii/iH|inn......is ,i liuulioii ol tltttf IIIIINilllAIA ll)| MI'IIIICAI IIAII IAWI1 MIH HIM I HIAi.i.< III Ihn stoichiometric table for reaction (3.83) '«|ir< les A H (' li lllrlls lot III hull il Mold ( I.....)•<' »' Moles I nul NI..I n". N" n(;xa -(;)n»aXa + (c-)n';xa o a - b + c + d N'A(I -XA) NÍ (")N'AXA Nc+C-N + VxA ("*""||+c + j)wbaXa N + (a)N^ K i jNJXa |i Nll + (^xA in. o. N? \ ** / 0 N? lnlill +n.■>> ,li..vv. .i iTiiri.il stoicliionielric table, while Table 3.11 shows the specilic slou liiomeliic i ilili lin the ease considered here. ,ii, uniting results from fable 3.11 into equation (3.60) yields <-r„> - K.(CbHCc) = KB(CB(I - X»)H(Cg - X»C,)) (3.72) \......hug In equation (3.30) T = C ( r„) (3.73) Initial ( 'oiicrnliali.nl Clí * ii i ll.llllT 1.10 The stoichiometric table for the reaction bB + cC aA t|H'i lev II C final ( '.iiiienli.ilii.il <;x„ n/lnC^X,, lil/hll |',\|1 Cb(I Xb> C1,' (i/hK^X,, < A i i,i/lm Km MAM t »AI A Table 3.11 The stoichiometric tnble lor the reaction C l 11 : A Species Initial CuucenUMkn ( li.tiiec ľuuil (niicenlriilion B C A C°c C°A -C°BXB CbXb Cg(l - XH) Č£-C&Xb C°A+C°BXB Substituting equation (3.72) into equation (3.73) yields o (kBc£(l-XB)(Cc-XBCB)) (3.74) The integral equation (3.74) looks imposing. Fortunately, I was able to look it up in Gradshleyn and Ry/hik (1965). '.I Ulil N1IAI Ml Al.HON! ' »«'i\whcie dining tin- icaclion (•..iiscqiicnllv. the tale equation loi icaclion ( I V)> will ii •In. .• i<> ihe equation loi a second oulei icaclion. One can derive the same result by letting CB approach C|' In equation (3.84). and UMiiti I'Hospital's rule to do the limit. The result is (3.79) I qihiiion ( * /'>) is equivalent to equation (3.42), with n = 2. Equations (3.75) and ( * P9) .....led to plan experiments. First, one swamps the reactor with one species. s.i\ < .•..I measures the concentration of B as a function of lime, and uses Fssen's method "i Ms method to see il the reaction follows equation (3.76). One then urns the reaction .>iili equal It and (' concentrations, and uses Powell's method or Fssen's method to see 'i ■ .pi.it..... I 1.79) works. II both equations lil the data, then one can be assured llial the .....lil.i lollows eqiialion (3.60). If eilher equation (3.76) or (3.7')) fails, one Mfdl .i nunc complex procedure to lil the data. Such procedures are beyond the scope ol the II ii inn here. X ^ Ill i - X B \ ) _ klt<('( (/'/,) (3.75) Fquaiion (3.75) has two interesting limits: the limit where C£ y>> Cn and the limit where C" = Cg. The first limit, called the "swamping limit." corresponds to running the reaction with a huge excess of cyclopcntadiene (C). If C*('. is in large excess, C^/Cg X> 1 > XB. Therefore, the XB term in the numerator of the log term in equation (3.75) will be negligible. Similarly. CB in the denominator of equation (3.75) will be negligible. Consequently, when there is a large excess of cyclopentadiene, equation (3.75) reduces to I . / I T = kBC In (l -XB) (3.76) 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 cyclopentadiene does not change significantly during the reaction. The Cc is constant in the rale equation. Therefore, the reaction appears as if it were first-order. The other key limit is when CB = C£. Note that during reaction (3.59), hcn/oquinonc (B) and cyclopentadiene (C) are used up al the same Mile. Consequently, it initially CB= C£, then C, - Co I t /K) 'il I QUI N1IAI REACTIONS I hi Ii In one oilier example ill.it we will need to consider later in this book, which is id. hum where a reaclant A is lirst converted into an intermediate. I, and then into ,i I'i.mIii. t, p: A=^=>l=i=>P (3,80) N. \ [. we will derive equations for the concentrations of A. B. and C versus t assuming ........i. linns I and 2 are both lirst-order. II n .ii lion I is lirst-order. then dC* dt = rA = -k|CA (t HI i ..... 1 v 's I he concentration of A. Solving equation (3.XI) yields CA=CVk" i \ X2) II n .u umis I nud 2 are lirst-order. then dCi dl Ma k,C, 11.83) rX " ( i is the conceiilratioii of I. ......billing equations ( \ HI) and ( VX3) and integrating yields C, M* (e k" e k"> i f/c k" kj - ki (3.84) HB anai Y'iľiOl mam iiaia SI (Jl II NMAI MI AI:I Ii iN'. 09 where ('" is ilu- initial concentration <>l U We could integrate anil calculate the Concentration Of P. However, il is much easier in calculate the concentration of P from a mass balance. Cp = c^ + c,; + c?-cAc, (3.85) Equations (3.82)-(3.84) are the key equations lor sequential reactions. Figure 3.25 shows a plot of equation CA, C,, and C„ versus k,t for various values of k2/k, calculated from equations (3.82), (3.84), and (3.85). We have plotted the data as a function of kit, to eliminate the k, dependence. The concentration of the reaclant A decreases exponentially with time as was shown in Figure 3.14. However, in Figure 3.26, we plot the data as a function of k,t. All of the CA data fall on a universal curve, independent of k and k2. ,.....enlialinu ol the inlcimcdliilc shows quite different bchavim I he iniciiiicili.iie ........h i .m. hi Increases, icaches a maximum, ami (hen declines again. Hie height nl ilie IMHlmum ilecie.ises as k../k, incicascs. Physically, what is happening is lhal m the iinli.il ,. m .,i ih. rem Hon, tin- reactutl a is being convened into the Intermediate, >«» initially the .........trillion ol the mteimediate rises. However, as the concentration nl the intenncdiiile i.uiliľ. up. the rate Of conversion of the intermediate into the product ľ begins to hocOfM ,m,,„,iIiiiiI Eventually, the rale of destruction of the intermediate through reaction ' gets i.. In largei than the rale of production of the intermediate through reaction I. At thai die concentration <>( the intermediate falls again. Ini Inline reference, M is useful to plot two other quantities: q, = cA - cA ( I NCI ■mil Figure 3.25 A plot of equations (3.82)-(3.85) with CJJ = 1. Cf = 0. 1.0 [—r ' I 1 I 1 i 1 I i I i I C, kj/k, =1.1 .kjj/k, =5.0__ kg/k, = 10.0 0123456 70 123456 70 1234567 M k,r k,t Figure 3.26 A plot of C, and C* versus k, I lor k,/H, - 1 1, 8.0, 10, 100. (3.87) lhal equation (3.85) reduces to equation (3.86) when Q = C? = 0. Therefore, CJ m ,,|u.iii,Mi (3,86) will he the product concentration il the intermediate concentration || in eligible. Ci is more complicated, C| is the concentration ol the intermediate ........i, would calculate if the derivative on the left side of equation (3.82) were ., i,. I Inure I 27 shows a plot of the CJ and C{ versus k,T. We have multiplied ('} and i , b\ k/k| to make the figure easier to sec. Notice that when k./k, > 10, C'| is | i distinguishable from C|, except at the very start of the reaction. CJ is h.ucl\ .......nubiiblc from CP, That will be very Important for the discussion m the ttexl I 0 0 4 jlf I..II n pli.l ill ('.,. (I.....«I uml (tilniiUl"«l v«.n» k,, lliglil ,i|il.....I k .1lllnnn) mul k I " •111« k|i 100 ANAIVülüOI HAU IIAIA mi am iiiini 4 ha ii iiaia i iii im vi ii iimi ani i i'i ii ! .m ii ii i ii iani ii !t 101 Table 3.12 Summary of key i:oi»:o|)(s Two methods to measure rates: direct and indirect Direct: high accuracy; many runs needed Indirect: lower accuracy: fewer experiments Methods lo analyze indirect data Essen Construct plots of MC^/Ca). (C^/CA)n- 1 See if lineal Van't Hoff Calculate k|, tj, kj See if constant Van'l HolT—easier and more accurate Methods In analyze direct data I easi squares Nonlinear I east-squares Nonlinear least-squares easier and more accurate Key equations for indirect data N? r Jo V(-rA) JO -t"A Je' -rA C, ! (n-l)kn(t*)n-1 (3.26) (3.28) (3.31) (3.39) (3.42) Chapter, because it means thai one can calculate an accurate value of the concentration ol ihc iniermediatc by assuming ihc derivative in equation (3.82) lo be zero whenever k2/k, > 10. 3.15 SUMMARY Table 3.I2 summarizes the key results from this chapter. In this chapter, wc reviewed some basic kinetic analysis. We discussed the order of a reaction and the activation energy, and showed how those quantities could be measured. We then briefly discussed planning experiments and noted that a key decision in planning an experiment is to decide between direct and indirect methods. Direct methods require harder experiments, but lite data are easier to analyze. Indirect methods require easy experiments, but the analysis of the rate data are less certain. We also discussed how one analyzes data from batch reactors. We expect that most of these concepts are already familiar to our readers. Still, we recommend that the reader review the solved examples at the end of this chapter before going on to Chapier 4. 3.16 SUPPLEMENTAL MATERIAL: MEASURING RATE DATA FROM VOLUME AND PRESSURE CHANGES There is one other topic thai I wanted lo include in the ti'M. I'M u did mil icallv Ml. which is to discuss how one uses prcssiiie ami vol......• ihanges in imIci tate dala. As noted previously, kinetic measurements llisl became popului d......y tin laitei hall of il.......eteenlh century. Al llie lime, il was haul 10 make kmelu ineasiiiciiieiils None ni (he spectroscopic lechniquei desnibcd in Chapter I had been invented Accurate nm iiil.alames were uol generally available. Analytic techniques were primitive As a nli people needed to lind a series ol (neks to measure ihc concentration changes. A i minium expcriinenl was to run a gas-phase reaction in a closed vessel ami measure llie i Klinge in the pressure of ihc gas as a function of lime. I in example, in 1883, Van't Hoff examined the decomposition of ursine, specifically, .......... (I'll, on a glass vessel. Today, one would examine that reaction with a ........balance as described in Section 3.5. In 1883, however, no one had a balance lhal i rnsiiive enough to measure how much arsenic was deposited. Van't I loll had to lnul iimiiliei method to make the measurements. He loaded Ihc arsine into a closed Mask ami .......lured the pressure as a function of lime. Note that reaction (3.5) converts 2 mol <'l ...... gas to 3 niol of hydrogen gas, plus I mol of solid arsenic. II 2 mol are convened ...... I mol hi a fixed-size vessel, the pressure in the vessel will go up \ .hi i Moll noted lhal if one measures the pressure as a function of lime, one can Ina k calculate the amount of arsine that was converted and therefore calculate a rate Ii .......ai.il in sec this type ol measurement loday. However, there are many examples ol Um (\pe in i he literature. Consequently, it is useful to derive a series of equations to see In in one can use the pressure in a system lo calculate llie Concentrations of all of llie i....... ami I he amount of gas readmit that is converted. I he derivation will siarl with the stoichiometric table we derived previously lor the o in Hull aA + bB->cC-fdD (3.88) ......i b, e. and d are the stoichiometric coeflicienls for species A. H, C. and I). i ■ i lively. Assume that N" N1^ moles of B are loaded into a closed isothermal vessel ■ we will derive a relation between the number of moles of A in the vessel and llie |in.....ure in Ihc vessel. Recall from the discussion in Section 3.13.1 lhal the stoichiometric table fin this i. ii in hi is given by a lable such as 'fable 3.13 (see also Table 3.7). Ni-nt, we will calculate the pressure in the system as a function of Ihc initial piesstue v......ling lo the ideal-gas law: P,V = N, KT (3.89) :i 13 A stoichiometric table for reaction (3.87) 11 ii Mal Moles Change in Moles A o i' ..... lnul INli Nf n'J o*- Ns(A,;;")x. Pinal Moles NJ(I-Xa) n;> i S;s, n'|' n>n<;(a7')x* 102 ANAI YSISOI HAH DAIA .Ml VI I > I "AMI 'I I III I where PT is the total pressure of the reactor, V is the volume in the reactor. N, is the number of moles in the reactor, r is the gas law constant, and T is the temperature. At the start of the reaction P£v = n^rt (3.90) where P^ is the initial pressure of the reactor and Nlr 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 _PHn°t (3.91) Substituting nt and n'-j from the stoichiometric table into equation (3.91) and rearranging yields PT _ (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. I'igure 3.2X shows a plot of the pressure calculated from equation (3.92) as a function or conversion for various values of the Atnol. Notice that the pressure always varies linearly with conversion. Ihe 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. I:\amples 3.IS and 3.C illustrate the use of the stoichiometric table to calculate the conversion as a function of time. 0.4 06 Conversion Figure 3.28 A plot <.l Hie i»..■•-•.............■.....,u„................„i,„i i„„„ ™,,,„ii,„, , i u;'| wild N, ') riiul. .in, I .i III 3.17 80LVED EXAMPLES I wimple 3.A lillinu Dala 10 MOOOd'l law Table 3 A I shows some .l.il.i loi Hi. growth i .tit- ol Paramecium as a function ol the Paramecium concenlralion. In the d.u.i i,. Mounds law (Monod 11942|). , t \ I i k|K2lparl 'p " I + K2(parl • i„ ,, |pai | is the Paramecium concentration and k, and K2 are constants. Solution There are two methods that people use to solve problems like this . Rearranging the equations to get a linear lit and using least -square* . Doing nonlinear least-squares I |.n I, i the latter, but I wanted to give a picture of Ihe former. I here are IWO versions of the linear plots: . I meweavei Hurke plots . I ...In- llolslee plots ....... I meweaver Hurke method, one plots l/rale against I/concenlralion Kc.iii.iiipnr . iiiniiiou I VA. I) shows ! = _L_ + 1 (3A2) r,, k|K.,lpar| k, li,. ,, line. ,i ploi ol I/i,. versus l/[par| should be- a straight line. The intercept should ' I he slope should be ' . Once k, is determined from the intercept, K ..in 1« V i k i K j i A I The rate of Paramecium reproduction as a function of the Paramecium i inn initiation _ HI, m 1 Mini ľaraiiieciuni ľ.lľ.llllll llllll minium. Kate. 1 llIKľllll.lllľll Rule. coiK.'iilriili.in. Kiile. N/im1 hour N/cm' N/tcni' hour) NA.ni' N/kín' how i 10.4 16 36 46 M tf. I ] B H, (, lr. 1 l'i 1 l.'l K 1 23.2 19 59.2 47.4 117.6 ^; 17 1. 20 62.4 53 1 1 • / N K. 1 23.1 62.4 57 127.2 N 23.2 26 37.« (.1 116 N 16 i 10.4 I0X.H ,,i 6 m ■ II 32 »1 HO /I 124 11 I u 1 31.2 61.6 74 IM n fi 44H M 6 109 6 764 III ÍM M | >•) 1 103.'