4 RELATIONSHIP BETWEEN RATES AND MECHANISMS Mil cis I In objective of this chapter is to examine the relationship between rates and mechanisms. Nil I.ii in this book we have treated rate equations as empirical relationships that arc lit III i .iii- data. However, in fact, rate equations have a more fundamental basis. The form III the rate equation is determined by the elementary chemical reactions that occur as a Ii i1 lion proceeds. The rate constants are related to the rate of the elementary reactions. II ie objective of this chapter is to explore the relationship between rates and .....lunisms. We will introduce the concept of a mechanism, and show how one I in use the mechanism and the pseudo-steady-state approximation to calculate a rate I quation for a given reaction. Most of the students who take our course already know lllill reactions obey distinct mechanisms, and many have seen the pseudo-steady-statc Approximation previously. However, most students do not know where the pseudo-steady-i 11. approximation comes from, when it works, and when it fails. The objective of this lupin is to extend students' knowledge so that they will know how the pseudo-steady-Kllltc approximation arises, and where it fails. We will also define some more key terms ili.il we will use later in the book. •I 1 INTRODUCTION Nludies of the relationship between rates and mechanisms have had a long history. Work uliirtcd with Dobereiner (1829) and Wilhelmy (1850), who supposed that rates of reaction would be simply related to the stoichiometry of the reaction. However, in 1878, Van't I loll showed that the rate of reaction had little correlation to stoichiometry. For example, I'lible 2.5 (page 14) shows the rate equation for several reactions discussed by Van't Hoff. Notice that the rate of phosphine oxidation is first-order in the phosphine concentration iiul half-order in the oxygen concentration. Yet during the reaction, each phosphine reacts Willi two oxygens to yield products. Further, one needs two phosphines to yield each P2O5. 157 4 RELATIONSHIP BETWEEN RATES AND MECHANISMS I'll! CIS i hi objective of this chapter is to examine the relationship between rates and mechanisms 'in I.ii in this book we have treated rate equations as empirical relationships thai arc In In rule data. However, in fact, rate equations have a more fundamental basis. The form III the rate equation is determined by the elementary chemical reactions that occur as a i< ii i ichi proceeds. The rate constants are related to the rate of the elementary reactions. I ho objective of this chapter is to explore the relationship between rates and mechanisms. We will introduce the concept of a mechanism, and show how one I .....sc (he mechanism and the pseudo-steady-state approximation to calculate a rate in.iiion for a given reaction. Most of the students who take our course already know Ihul reactions obey distinct mechanisms, and many have seen the pseudo-steady-state approximation previously. However, most students do not know where the pseudo-stcady- i itc approximation comes from, when it works, and when it fails. The objective of this . luipter is to extend students' knowledge so that they will know how the pseudo-stcady-illltc approximation arises, and where it fails. We will also define some more key terms Ihul we will use later in the book. I I INTRODUCTION Studies of the relationship between rates and mechanisms have had a long history. Work darted with Dobereiner (1829) and Wilhelmy (1850), who supposed that rates of reaction would be simply related to the stoichiometry of the reaction. However, in 1878, Van't I lull' showed that the rate of reaction had little correlation to stoichiometry. For example, liible 2.5 (page 14) shows the rate equation for several reactions discussed by Van't Hoff. Notice that the rate of phosphine oxidation is first-order in the phosphine concentration ukI half-order in the oxygen concentration. Yet during the reaction, each phosphine reacts M ill) two oxygens to yield products. Further, one needs two phosphines to yield each P2O5. 157 MIMI...... AI I lvi IIVII W I lir. rxiliuplr '.Imu ; ili.it Mícu' is nu i > 1. 111< >i i 1111 > between llir i nic equation lni n rem hun .nul the slon lunimliv ul Ihc ic.u Hun The purpose of this chapter is to try to understand the moleculai basis ol the rate equation, anil in particular why rales do nol correlate with stoichiometry. Section 4.2 will discuss the idea that reactions actually occur by a complex series of chemical reactions. Sections 4.3 and 4.4 will review the rates of elementary reactions and how the rates arc related to the sloichiometry of the reactions. Next, we will examine rate equations. In Section 4.5 we will show that if one knows the mechanism one will be able to derive a set of differential equations for the change in concentration of all of the species in the reactor. The differential equations are the fundamental rate equations for a chemical reaction, and they fully determine the behavior. We will briefly review the behavior to give the reader a qualitative picture of the results. Next, we describe the pseudo-steady-state approximation and see how it can be used to determine a rate equation. We will also discuss the rate-determining-step approximation. Finally, we will close by discussing the failure of the methods, and what one does in such a case. The discussion will assume that the reader has had an introduction to these topics before. If you have not had an introduction, please look back over the kinetics section of your physical chemistry text before proceeding with this chapter. 4.2 HISTORICAL OVERVIEW To start, it is useful to review some history. In the 1896 edition of his book, I'Etudes Dynamique Chemie, Van't Hoff speculated why the kinetics of a reaction did not correspond to the stoichiometry of the reaction. Van't Hoff supposed that when only one molecule participated in a critical step of reaction, the reaction would be first-order. In contrast, if two molecules participated in some critical step of a reaction, the reaction would be second-order. Consequently, Van't Hoff proposed that the kinetics of a reaction were related to what Van't Hoff termed the molecularity of a reaction, where the molecularity was loosely defined as the number of molecules participating in some critical step in the reaction. For further reference, we will need to know that there are unimolecular reactions, where only one reactive molecule participates; bimolecular reactions, where two molecules participate; and termolecular reactions, where three molecules participate. An example of a unimolecular reaction is Cyclopropane-> propylene (4.1) where one cyclopropane molecule rearranges to yield propylene. An example of a bimolecular reaction is OH + C2H„ => H20 + C2H5 (4.2) where a hydroxyl grabs a hydrogen from an ethane molecule. An example of a termolecular reaction is i in . i ample is more complex; two ineihvl radicals ut reacting to form ethane. However, ...........hi occurs unless ■ nitrogen collides with the Iwo methyls. The nitrogen lucililiiles ii.. n in lion Viiii'I Hull proposed thai all lusl ordei reactions are unimolecular. all second ordci ......... ure bimolecular, and all third ordei reactions are termolecular, II one looks at the stoichiometry of the reaction, though, one finds that complicated ......., huvc simple rale equations. For example, according to the informatlo..... i .ii ' I (page 14) the reaction 4PH, P4 + 6H2 ,1 . ||| i order in the phosphine pressure. Yet, it is hard to imagine how reaction (4.4) could i i unimolecular reaction. After all, one needs four phosphines to produce a single I', hi i Hull incd to get around this difficulty by assuming that only one molecule WM i.....i Ipating in some critical step in reaction (4.4). However, when Van't Hull's bunk i wiiticn (1896), it was not obvious how it was possible to have only a single phosphine .....|i, ill, participate in the reaction when four phosphines are needed to produce a P< In the time between 1890 and 1919, a number of papers were written thai looked n the details of many different reactions. An attempt was made to try to understand how the kinetics of a reaction was related to the stoichiometry of the reaction. David i hapman (1913) Muriel Chapman, and Max Bodenstein (1907) examined the kinetics of in I formation. They showed that the reaction H2 + Cl2 2HCI (4.5) .i>.I i„,i occur via a direct reaction between molecular chlorine and molecular hydrogen in ir.nl. the chlorine needed to dissociate into atoms before the reaction proceeded. Slowly ......sensus emerged that the reactants are not directly converted into products during i , hemical reaction. Rather, the reactants are converted into species called reactive Intermediates, and then the reactive intermediates are converted into products. For • numple, during the acid-catalyzed isomerization of 1-butene to 2-butene in solution CH,CH2HC=CH2 I he 111 reacts with the 1-butenc to produce a CH3HC=CHCH3 H / \ CH3CH2HC — CH2 i I in intermediate where I he hydrogen is held in a three-center two-electron bond. The intermediate then loses a 3-proton to form 2-butene. The sequence of reactions is H CH,CH2HC—CH: CH3 + CH3 + N2 C2H6 + N2 (4.3) CH3CH2HC=CH2+ H+ H CH3CH2HC—CH2 H CH3CH2HC—CH2 CH3HC=CHCH3 + H+ (4.7a) (4.7b) Ill'lll IIIIOAI (IVI IIVII W Kit Hun- .in- (wo ii-ih lions in met li.iiiisin (I /) I hiiU'iii' is the leiicliint II < ■!!,( line -en, is the inlcrmcdiatc, and 2-bulcnc is the product, Fach ol the- steps in mechanism 4.7 are called elementary reactions, while reaction (4.6) is called the overall reaction, or the stoichiometric reaction. More precisely, an elementary reaction is defined as a chemical reaction going From reactants to products without going through any stable intermediates. In this context, a species is said to be stable if it has a lifetime longer than -iO"" seconds (i.e., much longer than vibrational or collisional times). There are some important notations. We will use a single arrow -> to designate an elementary reaction but a double arrow => to designate an overall reaction. In this notation, the reaction H2 + Br2 => 2HBr (4.8) will be an overall reaction, while the reaction H + Br2 -> HBr + Br will be an elementary reaction. In the gas phase, most elementary reactions are of the type A + B C + D (4.9) (4.10) where at least two reactants come together to produce at least two products. Examples include H + HBr-»H2 + Br (4.11) and X + H2 2H + X (4.12) where X is a something called a collision partner. The collision partner is a species that collides with the H2 and initiates the reaction. Next, we want to define a concept called the mechanism of a reaction. The mechanism of a reaction is defined as the sequence of elementary reactions that occur at appreciable rates when the reactants come together and react to form products. For example, if one actually runs reaction (4.6) in solution, one finds that four elementary reactions can occur: CH3CH2HC=CH2 + H+ H CH3CH2HC— CH2 H CH3CH2HC—NCH2 CH3HC=CHCH3 + H+ (4.13a) (4.13b) II 1 II 1 II.IK (II ('ll,(-H2HC—CH2 - t 11,1 II.IK (II. 1 II' H,C=('IK IU'11, 1 M' (I I lo) l I I (,ll II.. lust two steps in (4.13) are the steps listed in reaction (4.7). The last two intermediate loses a proton u> 1. |. me processes where the H CH,CH2HC — CH2 ,.....I,,.. either l-butene or the equivalent 3-butene. Note that the proton that leaves the intermediate in steps (4.13c) and (4.13d) may be different from H / \ 1 II.CIl.HC—CH2 1i1. proton which reacts in step (4.13a).1 I In- sequence of elementary reactions listed in (4.13) is called the mechanism of acid ■ ululyzed isomerization of l-butene and 2-butene. I here are two subtle points in the definition of a mechanism that students often nuss I 11 it, notice that there are four elementary reactions in equation (4.13). However, only Ihi lust two of the four lead from reactants to products. The other two steps lead bat k In ilir icaetants. Therefore, it is incorrect to think of the mechanism as consisting of just I hi i< ps leading from reactants and products. Rather, the mechanism includes the steps ih.it lead from reactions to products, plus other steps that either lead to new products or .....veil the intermediates back to the reactants. \ second subtlety is that when someone reports a mechanism in the literature, they till 11.it necessarily list all of the reactions that would occur. They list only reactions that ■ mi it appreciable rates. For example, during reaction (4.6), one also gets the reaction H CH3CH2HC—CH2 [CH3]+ + CH2=CH—CH3 (4.14) Reaction (4.14) is much slower than reactions (4.13a)-(4.13d). Consequently, people ..11,11 ignore reaction (4.14) even though it occurs at a nonnegligible rate. The mechanism ..1 .1 reaction, then, does not necessarily include all reactions that occur. Rather, they in. lude only reactions that occur at nonnegligible rates. An important point is that livery overall chemical reaction can be divided into a sequence of elementary reaction. I.very reaction has a mechanism. Sometimes reactions have more than one mechanism. This occurs when there arc, for . numple, competing reaction pathways. ' II reaction (4.13) would occur in the gas phase, one would need collision partners. For the purposes here, wc will ignore the collision partners, since in solution, the solvent can act as a collision partner. '•"*< luniN...... i I Mi In iHiillvcK ■ Im..!,. n..... ruction ''"' '""'""i'li'n Mh- ii. i it, 'III'.! follows the following mechanism: Br2 —2Br Br+ H2 —HBr + H H + Br2 —HBr + Br 2Br Br2 H + HBr —5-». H2 + Br (4.16) The reaction 2H2 + 02 • 2H20 (4.17) follows a complex mechanism with over 30 steps. However, one can get the structure of the solution by assuming that the reaction obeys the following mechantm H2 +02 OH. + H2 20H. H20 + H. H. + Q2 —i-> OH. + O. 0: + H2 II. OH. + H. wall (4.18) H. + Oz + X H02. + X HO:. + X —'—t wall 2HO + X —5-». H202 + X Notice that even through the overall reactions in (4.15) and (4.17) appear to be simple, the mechanisms are not simple. This is typical. Physically, what is happening is that reactive intermediates are, by definition, reactive. They undergo many reactions. As a result, there are usually many reactions in an overall mechanism. I i I '. Icpeill lll.lt ,1^1111 Ih'IIUIM' ll I*i ll'llllv llll|>Otllllll Reactive Intermediates are by definition reactive. They undergo many reactions. We will describe the mechanisms of many different reactions in Chapters 5 and 12. However, the thing to remember for now is that a typical chemical reaction occurs via . i......bei (.I elementary steps. One needs to understand the kinetics of those individual |l |.'. to determine the mechanism of a reaction. i i KINETICS OF ELEMENTARY REACTIONS I we want to discuss how the mechanism of a reaction is related to the kinetics III iln reaction. We will assume for the moment that you already know the mechanism III ii reaction, and ask how one can determine the rate equation of reaction from die in. i hanism. We will start with some material that we will eventually derive in Chapter 9. Considd in elementary reaction between an atom or molecule, A, and another atom or molecule, ii in form two product atoms or molecules, P and Q: A + B P + Q In i hapter 9, we will show that r2, the rate of reaction (4.19), is given by r2 = k2[A][B] ilarly, the rate of formation of A is given by rA = -k2[A][B] (4.19) (4.20) (4.21 where |A] and [B] are the concentrations of A and B, respectively, and k2 is a constant, liquation (4.20) is a key result. It says that if we know the stoichiometry of an elementary ii it) lion, we also know the rate equation for the reaction. Notice that the rate equation for mi elementary reaction is proportional to the product of the concentrations of the reactanls ol ihe elementary reaction. However, the rate does not depend on the concentrations of ilie products of the elementary reactions. ()ne can also consider elementary reactions of the form -> P + Q 2A — In ihis case the rate of reaction, r4, obeys r4 = k4[A]2 '.....ilarly, the rate of formation of A, rA is given by rA = -2k4[A]2 where lc* is the rate constant for reaction (4.22). (4.22) (4.23) (4.24) lili' lili till III ' III ft 11 111 I it >l ■ i I Ml IS WIS llll|llllllllll I III' IlUlOI III .' .DIM'S III! .IIISI' Iwo molecules ol A are destroyed cveiy mm n-action (I .'.'» cms once. Students often forget ihe factors of 2 when they derive rule e(|tialions. The readci should In- wis careful to avoid thai error. One way to avoid this error is lo use the definitions from Chapter 2 to quantify the rate of formation of any species. For example, consider species A participating in a group of reactions numbered 1,2.....i. One can show that the net rate of formation of the species, rA, is given by rA =ßA,in +ßA,2r2H----+ pA,ifi (4.25) where r(, r2,...,rj are the rates at reactions 1,2____, i, respectively, and p"A-i, p\,2, • • •. p\.i are the stoichiometric coefficients of species A in reaction 1,2.....i. Again, we remind you that in our definition, the stoichiometric coefficient is negative for a reac-tant and positive for a product. In reaction (4.22), pV4.26's ~2, which is why there is a —2 in equation (4.24). When I am deriving rate equations, I find it easier to remember that the coefficient in the rate equation is ALWAYS the stoichiometric coefficient instead of having to work out the coefficient every time. Equation (4.20) applies only to elementary reaction of the form in (4.19). However, people often generalize equations (4.20) to first- and third-order reactions. For example, with an elementary reaction of the form ■> P it is often incorrectly assumed that the reaction obeys rA =k,[A] (4.26) (4.27) where [A] is the concentration of A and k] is a constant. Similarly, with an elementary reaction of the form A + B + C-► products (4.28) it is often assumed that the reaction obeys: rA = -k3[A][B|[C] (4.29) where [A], [B] and [C] are the concentration of A, B, and C, respectively, and k3 is a constant. In actual practice, elementary reactions of the form in equation (4.26) arc never seen experimentally and reactions of the form in equation (4.28) are rarely seen. In Chapter 8, we will show that the rate constant for any elementary reaction of the form in (4.26) (i.e., where one reactant is converted into one product without colliding with any other molecule in the system) is ZERO, which means that elementary reactions of the form in equation (4.26) are impossible. Similarly, we will show that the rate constant for reactions of the form in equation (4.28) is very small. Therefore, in practice, one rarely sees reactions of the form in equations (4.26) and (4.28). One can, however, see reactions of the form ..... \ In .ms olhei molecule in I hi s sli in I tint ..in collide with A We cull \ Ihr hIIimhii iiittlnri Keuclion (■! III) obeys MAIIM (I II I In n |A| and |X| are the concentrations of A and X, and k. is a constant I i|ii;il..... it t| i is very similar to equation (4.27) in cases where |X| does not change duini)' tin .....i As a result, people often say that the kinetics of a reaction like that in iculiun i I tin obey (4.27) approximately, even though they do not lit exactly. I In ii' is an important point I want to repeat here that lor emphasis. One c;.....evei havi III I. in. nl.iiy reaction with only one reactant or only one product, ll looks as llitmj'li smi i.i, ii.isr one reactant or one product in some elementary reaction. There will alwayi I" 11......Mi.'i species, X, contributing to the reaction. Therefore, the elementary reaction! IT -> B +C + X -> P + X mul NOT -» B + C C A + X A + B + X A A + B 4 -1 THE RELATIONSHIP BETWEEN KINETICS AND EQUILIBRIUM (MICROSCOPIC REVERSIBILITY) Ii. inn- we proceed, it is useful to point out that the results in Section 4.3 imply thai then I. .1 simple relationship between the kinetics of a reaction and the equilibrium constant im the reaction. Consider the simple reversible elementary reaction: A + B ^ C + D 2 i in equilibrium constant for the reaction, K,i2, is given by [C][D] Ki 2 — [AHB] (4.32) (4 O.) • hue | A|, [B], |C|, and |DJ are the concentrations of A, B, C, and D. There is a principle . tilled microscopic reversibility, which states At equilibrium, the rate of any forward chemical reaction (elementary or not) must equal the rate of the reverse chemical reaction. (4.34) A + X ^ P + X (4.30) Principle (4.34) implies that at equilibrium, the rate of all processes must equal the rate nl the backward processes. II wr apply principle (•! 14) In mil ci.....pic, we rum hide 11 ml ill ri|inlill where k| and k2 are the rate constants lor reactions I and 2. Rearranging equal ion (4.35) yields [C][D] k, = — (4.36) [A][B] k2 Substituting equation (4.36) into equation (4.33) yields K|.2 = — (4.37) Consequently, the equilibrium constant for a reversible elementary reaction is equal to the ratio of the forward and reverse rate constants for the reaction. One can generalize this result to a more complex reaction: A + B <=^=» C 4- D E + F «==± G + H <-^-» I + J ?==± L + M (4.38) At equilibrium [E][F] [A][B] [L][M] [A]fB] [L][M] [G][H] k|k2 k_,k„2 k|k2k3k4k5 k_lk_2k_3k_4k_5 k4k5 k_4k_ (4.39) (4.40) (4.41) where [A], [B], [E], [F], [G], [H], [L], and [M] are the concentrations of A, B, E, F, G, H, L, and M, respectively, and k|, k_i, and so on are the rate constants for reactions 1,-1, and so forth. I find that a good mnemonic to remember these results is that if I want to calculate the equilibrium concentration of two species G and H and two other species C and D, I say that the product of the concentration of G and H divided by the concentrations of C and D is equal to the product of all the rate constants leading from C and D to G and H divided by the rate constants for the reverse reactions. There are two rate constants leading from C and D to G and H, k2 and k3. Therefore [G][H] = k2k3 [C][D] k_2k_3 (4.42) Similarly, I say that the product of the concentration of C and D divided by the concentrations of G and H is equal to the product of all of the rate constants leading from G and H to C and D divided by the rate constants for the reverse reactions. There are two rate constants leading from G and H to C and D, k_2 and k_3. Therefore [C][D] [G][H] k-2k_; k2k3 (4.43) IIAII '......VI MAI I III AC linn. 11 -i we will change topics slightly mid use die results in Setiion 4.2 In derive a rule i......mi I'm in overall reaction. Oui approach will he In cnnsidci the met hams...... • >|n.iiinii (I I 11 and derive an equahnn. We will then generalize In a mechanism. i . mi . n lii tin- mechanism in equation (4.13). During I he mechanism in read inn (I I I), i huti in reacts with a pinion to yield an intermediate. I he intermediate then decomposed .....ili' i I<>■ in .' bulené or decomposes back to form l-bulene. If we call I bulené. "A". i.......r T ", anil (he intermediate "I", then one can view the reaction as follows: A + H P + H+ III ii wc have ignored the collision partners since wc are running the reaction in solution In equation (4.44), we call the reaction that converts A into I "reaction I". The reaction ||........veils I back to A is called "reaction 2". The reaction that converts I to I' is called i. ii linn 3". Next, we will to derive a general equation for the change in the concentration nl all il 11 it- species in the system. Our approach will be to use equation (4.25) to derive i ■ Mi' rcntial equation for the production rate of all of the species. We will then integrate iix differential equation to compute all of the rates. i '• t Derivation of a Differential Equation for the Concentration i'ii we derive a general differential equation for the rate of production of each of the 1» i us. The general approach will be to plug into equation (4.25) to obtain a differential ■ i|uulion for each species. < 'onsider the intermediate I. The intermediate I is formed in reaction 1 and destroyed in reactions 2 and 3. Therefore, according to equation (4.25), one can write the change in the concentration of the intermediate I as dm dt (4.45) ii here 11. r2, and r3 are the rates of reactions 1, 2, and 3 respectively; [I] is the concentration nl the intermediate I; and t is time. Similarly, [A], the concentration of A, obeys: d[A] dt ri where n, and r2 are the rates of reactions 1 and 2, respectively. \i cording to equation (4.21), r, is given by r, =k,[A]lH+] (4.46) (4.47) where ki is the rate constant for reaction 1, [A] is the concentration of A, and [H+| is the concentration of protons. Similarly, r2 and r3 are given by r2 = k2[I] r3 = k3[I] (4.48) (4.49) /i mill i i IN) lulu < 1111.11...... i ll.i m, |,|. ,l| \| ell •v-MI hlAHII'l Similarly, substituting equations (4.47)-(4.49) inlo equation (4.45) yields dm dt :k,[A][H+]-(k2 + k3)[I] (4 .id (4.51 Equations (4.50) and (4.51) are the fundamental differential equations for the behavior of the system. They are the key results in this section. 4.5.2 Integration of the Rate Equation Next, we want to integrate the equations to calculate the overall rate of reaction. There are three approaches: . Analytic integration of the differential equations . Numerical integration of the differential equations • Approximate integration of the rate equation In this section we will concentrate on the analytic treatment. Numerical integration is discussed in the supplementary material. An approximation scheme is discussed in Section 4.7. In order to simplify the algebra, we will assume that [H+] is constant [equation (4.50)] and that we start with pure A (no intermediate or product) so that [A] = |AJ° [I| = 0 [P] = 0 att = 0 (4.52) One can then solve equations (4.50) and (4.51) simultaneously to calculate the concentration of A and I as a function of time. The algebra is complex, but the answer is [A| = [A]H / — - k«)exP(-k5t) ~ ~ kft)exp(-k4t) "1 I (k4 - k5) J = [A]°{ (k4-k6)(k6-k5) k2(k4-k5) where [A]0 is the initial concentration of A and (exp(-k5t) - exp(-k4t)J k4 = ifc + k2 + k, + 1/(k6 + k2 + k3)2-4k6k3) k3 = i(k« + k2 + k, - v/(k6 + k2 + k3)2-4k6k3) k6 = k,[H+] (4.53) (4.54) (4.55) (4.56) (4.57) Note that every mole of A that is used up must be converted into P or I. The number of moles of A used up is the initial concentration of A minus the current concentration of A. Therefore, the sum of the number of moles of I and P is given by [P] + [I] = [Aj° - [A] Time, minutes ii,i,im,4.1 [A], [i], and [P] as a function of time calculated from equations (4.53), (4.54), and (4.58), lively, with k^h*] = 0.2/minute, k2 = 5.7 x 106/minute, and k3 = 3.8 x 107/minute. Will rc |A|, [I], and [P] are the concentrations of A, I, and P, respectively, at any time ......i;' ihe reaction, and [A]° is the initial concentration of A. Equation (4.58) can be used III i ulculate [P]. One should verify equations (4.53) and (4.54) before proceeding willi ihr. chapter. Figure 4.1 shows a plot of [A], [I], and [P] as a function of time calculated from dilutions (4.53), (4.54), and (4.58), respectively, for some typical values of the parameters typically k,[H+], the rate constant for the formation of the reactive intermediate, will be inn, h less than k2 and k3, the rate constants for the destruction of the reactive intermediate, ii one substitutes values into equation (4.54), one finds that the concentration of the Intermediates is small. I S3 An Approximation i oi iuiurc reference, Figure 4.1 also provides a plot of another quantity, [I]s, where |l|s ■lined by [I]S = [I] k2 + k; k,[H+] (4.59) Nniice that [I]s and [A] are almost equal, except near t = 0. Next, we will compute a new quantity, where [1JX is an approximation to [I], by mibstituting [A] for [I]s in equation (4.59), substituting [I]x for [I] and rearranging. The irsiilt is (4-60) [I]x k2 + k3 (4.58) i igure 4.2 compares [I] calculated from equation (4.54) to [I[x computed from equation i i 60). The left curve shows the behavior from t = 0 to t < 10-7 minutes; the right curve shows the behavior from 2 x 10"2 minutes on. Notice that there are some deviations for i lir7 minutes. However, the two curves are virtually identical when t > 10"7 minutes. o O 5E-08 1E-07 1.5E-07 2E-07 0 2 3 4 Time, minutes Time, minutes Figure 4.2 A comparison of [I] calculated from equation (4.54) and [l]x computed from equation (4.60). We rarely care about the first 10"7 minutes in a process that takes several minutes. Consequently, the approximation in (4.60) works quite well in most practical cases. 4.6 GENERALIZATION TO OTHER REACTIONS Next, we want to generalize these results to other reactions. First, it is important to note that there is nothing special about the example we considered in Section 4.4. If we start with any reaction, we can always set up a series of differential equations similar to equations (4.50) and (4.51), and then solve the equations to predict the behavior of the system. In a general case, the differential equations can be complex. However, one can always numerically integrate the differential equations to calculate all of the concentrations. There are some details that one needs to worry about when one does the numerical integrations because initially the intermediate concentration changes much more quickly than does the rcactant concentration. Still, the equations can be solved numerically, using standard commercial software as shown in the solved examples at the end of this chapter. Consequently, if one knows the rate equations for a mechanism, one can always solve the rale equations to calculate the behavior of the reacting system. Algorithms are given in Example 4.B. The algorithms are very general, and can be used for any rate equation. Generally, the numerical solution is the most accurate way to understand the behavior of a reaction. 4.7 THE PSEUDO-STEADY-STATE APPROXIMATION Often however, one needs an analytical expression for the rate equation that one then uses in, for example, a reactor design. In order to get the approximation, we will generalize the approximation in equation (4.60) and use that approximation to estimate the concentrations ' ii il.....hiiiieiliiiles In p.ulii ill.ii .hi h ill deilve wluil is i ailed llic />wlulu \/>Wv \lulr .(/ocxytooi 5E-08 1E-07 1.5E-07 2E-07 Time, minutes 1 2 3 Time, minutes Figure 4.3 The size of various terms in equation (4.51). ll'Sllll, l'l|ll!!llOII I I V|), ,'Uipl HI I II I lll'lt'ltilf. I >lll- ( ill I l I > 1111 >l 11 > I ISC t III piopcltlCS hy iiNMimniK llinl ilu" ik-iivillivc- in t-i|iiiiiion i I M ) is /fin Physically, the derivative li the net rate ol formation ol the intermediate, n we sei th< derivalive equal lo zero, we are in effect assuming that the intermediates react as soon as they form. Physically, even though the intermediates form relatively quickly, they read very quickly, too, so that the net rate of formation of the intermediates is small. One can easily generalize this result using something called the pseudo-stoadv-slalc approximation, discussed above. Most of our readers have already seen the pseudo steady-state approximation. However, I want to state it a little differently than you have seen before According to the pseudo-steady-state approximation, one can compute accurate values of the concentrations of all of the intermediates in a reaction by assuming that the net rate of formation of the intermediates is negligible (i.e., the derivatives with respect to time of the concentrations of all intermediates are negligible). Note that there is a subtle point here. One is not actually assuming that the derivatives are exactly zero. Rather, one is assuming that the derivatives are much smaller than the other terms in the rate equations, so that the intermediates are consumed as quickly as the intermediates are formed. Consequently, the net rate of formation of the intermediates (i.e., the difference between the rate that the intermediates are formed and destroyed) can be neglected when one uses the rate equation to compute the concentration of all of the intermediates. Physically, what is happening is that the concentrations of the intermediates are usually tiny, so the derivatives of the intermediate concentrations are tiny, too. Consequently, the derivatives of the intermediate concentration can be ignored in equation (4.51). One needs to be careful because one cannot just ignore all of the terms in the rate equation containing the intermediate concentration; one can ignore only the derivatives. For example, in equation (4.51), there is a term: r3 = k3[I] (4.63) If one proceeded naively, one might guess that according to equation (4.62), the intermediate concentration in equation (4.63) is tiny, so r3 is tiny, too. However, note that we plotted r3 in Figure 4.3, and r3 is not tiny. The reason is that even though the intermediate concentration is tiny (6 x 1(T9), k3 is huge (2.8 x 10+7). Consequently, r3 is not negligible. This is a general result. While the concentrations of all of the reactive intermediates are generally small, the intermediates are by definition very reactive; the rate constants for reactions involving the intermediates are generally huge. As a result, one cannot simply eliminate the terms involving the intermediate concentrations in the rate equation. Rather, one does have to consider the intermediate concentrations explicitly. One can, however, often eliminate terms that have to do with the derivatives of the intermediate concentration since the derivatives are usually small, except near time = 0. There is another subtle point—the pseudo-steady-state approximation works only for reactive intermediates. If you consider a stable species such as a reactant, a product, or even a side product, you cannot assume that the species will be consumed as soon as it is formed. In most, but not all cases, radicals, hot molecules, highly strained species, or •|ii-i les wllh Nome unusual bonding 1«' It. the il"" i enlei bonded inleitnediitle in leiiclntn ,i i 111 will oIm-y die pseudo steady stale approximation There me exi options, ol course i , mnipie, in the uppei itmotphert, <>n radical! and n<> ipecies are stable. Mora ,.....11 , rlicria foi when the pseudo steady state approximation works will be given in ...... i'» i he key point, however, is that the pseudo-steady-state approximation works .......i highly reactive species. Consequently, this approximation is quite useful. i | APPLICATIONS OF THE PSEUDO-STEADY-STATE APPROXIMATION lo ill i l.MMINE THE KINETICS OF A REACTION we will use the pseudo-steady-state approximation to derive a rate law for a mimhei .....pic reactions. I realize that most of the readers have seen these types of derivations However, we needed some of the results for the discussion later in this chapter. Alio, the material is a useful review. w, will start with the simple example that was discussed in Section 3.14. Considei ili, leiielion ->• I (4.64) Hiding to the analysis in Section 3.9, [I], the concentration of the intermediate, can be mputcd exactly from equation (3.84). In this section, we will use the pseudo-steady-state imalion to calculate an approximation for [I], according to equation (4.31), r,, the net rate of formation of the intermediate, is plstii by r, = k,[A]-k2[I] (4.65) |A| and [I] are the concentrations of A and I, respectively, and ki and k2 are .....me constants for reactions 1 and 2, respectively. According to the pseudo-steady- i 111 approximation, n will be much smaller than the other terms in equation (4.65). quently, r, can be ignored in equation (4.65). Setting r, to zero in equation (4.65) mil mlving for [I] yields [I]=~IA| (4.66) k2 n , useful to compare equation (4.66) to equation (3.87). The notation is different in Ihi two equations, but otherwise the two equations are identical. Figure 3.28 shows that .pi.iiion (3.87) is an excellent approximation whenever k2/ki is greater than about 10. w nil a typical intermediate, k2/k, is in the order of 106. Clearly, the pseudo-steady-statc ipproximation works in such a case. Next, let's consider a more complex example, the reaction between hydrogen and bromine to yield HBr: H2 + Br2 ==> 2HBr (4.67) I he mechanism of the reaction is X + Br2 Br + Hj 2Br + X HBr + H ,» ri ii /Mil IINM c II llll I-.I I II H i '.II Al I, '.I All AIM'III i" IMA III IN 1 7B II I III. Illli I III il 68) X + 2Br ——* Br2 + X ll i lllti 11 i \u where X is a collision partner (i.e., any other species that can collide wilh Br., as described in Section 4.2). Next, let's derive an expression lor the rate that Br, is used up during the reaction. Note that Br2 is formed in reaction 4 and destroyed in reactions I and j Therefore fBr, = - r3 + r4 (4.69) where rj, r3, and r4 are the rates of reactions 1, 3, and 4, respectively. One can compute and r4 from equation (4.28): ri, r3, r, =k,[XJ[Br2l r3 = k3[H][Br2] r4 = k,[X][Br]2 (4.70) (4.71) (4.72) where [Br2], [Br], [H], and [X] are respectively the concentrations of bromine molecules, bromine atoms, hydrogen atoms, and collision partners. Substituting equations (4.70)-(4.72) into equation (4.69) yields rBr2 = -kriX][Br2l - k3[H][Br2] + k4[X][Br]2 (4.73) Equation (4.73) is an exact expression for the rate of consumption of bromide during the reaction. In principle, therefore, equation (4.73) can be used to calculate the concentration of all of the species in the system. In practice, however, the concentration's of bromine atoms and hydrogen atoms are tiny. Therefore, equation (4.73) is not very useful in practice. Now, one might think that one could just ignore the second and third terms on the right side of (4.73). After all, if one runs the reaction at 1 atm and 800 K, one finds that the concentrations of hydrogen atoms and bromine atoms are less than 10"8 mol/liter. By comparison, the Br2 concentration is 0.01 mol/liter under the same conditions. However, it is incorrect to ignore the second and third terms in equation (4.73) because the rate constants for the second the third terms in that equation (4.73) are large. According to the data in Benson (1976), k3 is about 1 x I010 A3/(molecule-second) at 800 K, while ki is only 1 x 102 A /(molecule-second). Plugging in numbers shows that the second term on the right side of equation (4.73) is about 100 times bigger than the first term in the equation. A similar argument also applies to the third term on the right side of equation (4.73). As a result, it would be certainly incorrect to ignore the second and third terms on the right side of equation (4.73). Instead, one must include these two terms in the analysis. With three terms, we have a very complicated equation. Fortunately, the pseudo-steady-state approximation can be used to get simple expressions for the bromine and hydrogen atom concentrations in equation (4.73). 111. inline alums tin* louiicd in mictions I. I. and V mid losl in u-.u turns .' ami •! I .i.... >i, r. I i, Jr., I is i I / li d|Hl I dl I In I .ii litis, ol 2 iii equation (4.74) arise because two bromine atoms are formed in reaction i in.i iwo .limns of bromine are lost in reaction 4. Substituting expressions similai to 11j11111h mi', (4.70) (4.72) into equation (4.74) yields d| Bi | dt rB, = 2k,|X||Br,| - k2|H2||Br] + k3[H||Bi,| -2k4[Br]2[X] + k5[H][HBr] \ mill.ii analysis shows d|H| dt = rH = k2[H2][Br] - k3[H][HBr2] - k5(H][HBr] i I /•>. (4.76) i \ note for students: Notice that I first wrote out the equations in terms of the rales ol ii individual reactions, and then substituted in the rate laws. In my experience, students i.ii lewer errors if they first write down the rates in terms of ri,r2,r3,... and then ■ i mule the rate in terms of laws into the subsequent equations.) Next, we will assume that the concentrations of the bromine atoms and hydrogen atoms follow the pseudo-steady-state approximation, that is, that the derivatives in equations i ,md (4.76) are negligible. Physically, we can make that assumption because the .....i. Mirations of the reactive intermediates are so small that the derivatives of (he .......nediate concentrations will be small, too. We note again, however, that k2, k,, k4, mil I , ate huge, so one cannot assume that the other terms in equations (4.75) and (4.76) .....egligible. II one assumes that the derivatives in equations (4.75) and (4.76) are negligible, one Minis 0 = 2k, [X][Br2] - k2[H2][Br] + k3[H][Br2] -2k4[Br]2[X] + k5[Hl[HBrl 0 = +k2[H2][Br] - k3[H][Br2] - k5[H][HBr] (4.77) (4.78) Wi now have two equations — equations (4.77) and (4.78) — and two unknowns — [H| in.I | Br] — so we can solve the equations simultaneously to calculate [H] and [Br]. \dding equations (4.77) and (4.78) yields 0 = 2k,[X][Br2]-2k4[Br]2[X] '..'King equation (4.79) for [Br] yields [Br] = k4 1/2 [Br2 11/2 (4.79) (4.80) Equation (4.80) is the pseudo-steady-state expression for the concentration of bromine .itiims produced during reaction (4.67). .........-in nut riiAll AIM'Ml IXIMAIII IN 177 Nril. we will iihlillll nil r h pi cnsii >n Ini I lit' nun euliiiliiin ul IiviIiul-iii alums Sulistiliiline i'i|ii.iliuii i I Hill inlii i -1111.11 ii id (I /Kl vti'liK (£) [Br2]1/2 = (4.81 ] Solving equation (4.81) for [H] yields k2 [HI k4 1/2 (H,][Br2 11/2 k3[Br2| + k5[HBr] (4.82) Equation (4.82) is the pseudo-steady-state approximation for the concentration of hydrogen atoms produced during reaction (4.67). At this point, we have expressions for all of the unknown quantities in equation (4.73), so we can calculate an expression for the rate of destruction of bromine atoms. Substituting equation (4.82) and (4.80) into equation (4.73) and rearranging yields k., [H2|[Br2] 1/2 1 + k, [HBr] k, |Bn| (4.83) Equation (4.83) is the pseudo-steady-slate approximation to the rate of reaction (4.67). The analysis above was for reaction (4.67). However, one can use an analysis similar to that above to calculate a rate equation for any reaction when one knows the mechanism of the reaction. For example, we can use the steady approximation to calculate a rate equation for the reaction between hydrogen and oxygen by assuming that the reaction follows the mechanism in equation (4.17). Generally, one has to do some algebra to solve all of the equations. However, the process is straightforward. For the work that follows, we will need to consider one other example — the reaction between hydrogen and oxygen: 2H2 + Q2 2H2Q (4.84) Reaction (4.84) follows a complex mechanism with over 40 important steps. Still, one can get a reasonable approximation to the rate if one assumes that the reaction obeys the following approximate mechanism: H2 +02—^20H. OH. + H2—^H20 + H. H. + 02—-+OH. + 0: O: + H2->OH. + H. (4.85) H.—i-»wall H. + 02 + X—^->HO... I X IK).. + X- • wall III ■ uuation (4.85), the term "wall" in reactions 5 and 7 refers to processes where radii ill lh ...i. onto the walls of the vessel. We want to emphasize thai reaction (4.85) is no) the ......ii i mechanism of the reaction, but one gets a reasonable rale equation anywaj In the ....ik in the next section, we will need an expression for |H.|. the concentration ol die i.....cn atoms during this reaction. We will use the pscudo-steady-siale approximation Ul i .mil 6. Consequently, rH, the rate of formation of hydrogen atoms is given b) Hi r3 + r4 - r5 - r6 (4.86) • here i ■, rt, r4, r5, and r6 are the rate of reactions 2-6. Substituting the rate reactions fol ■ Ii in his 2-6 into equation (4.86) yields d[H. = rH- = k2[OH.][H2] -k,[H.J|0:] + k4[0.][H2] - ks[H.] - k6[H.][02][X] (I 8/1 • line |H2|, [H.], [OH.], [02] [0:1, and [X] are the concentration of hydrogen molecules, hydrogen atoms, hydroxyl radicals, oxygen molecules, oxygen atoms, and collision l>.niiiors, respectively. Similarly, the rate of formation of hydroxyl radicals and oxygen Htoms, rl)H and r0, is given by d[OH dt d[0: = rOH = 2k,[H2][02] - k2[OH.][H2] + k3[H.][02] + k4[0:][H2] = r0 = k3[H.l[02] - k4[Q:][H2] (4.88 i (4.89) II we assume that rOH, rH and r0 are negligible, we can solve equations (4.87), (4.88), and i I 89) simultaneously to calculate expressions for [H], [OH], and [O]. Adding equation i 1.87), (4.88) and 2 times equation (4.89), and neglecting r0h, rH, and r0 yields 0 = 2k,[H2||02] + 2k3[H.][02J - k5|H.| - kr,[H][02][X| (4.90) Solving equation (4.90) for [H.] yields 2k,[H2][02] LH.] = k5 + (k6[X]-2k3)02 (4.911 Equation (4.91) is the pseudo-steady-state approximation for the hydrogen atom coneen Iralion. MAM III M MMININi i Ml I"• 178 We |ivi seu-i.il niliii examples ol llit- um- ni 11 it- pseuilu Ulomiv slulo appioxmialion in the problems ^11 the end ol iins chaptei i ho pteudo iteady state approximation is vory important, so ii is important thai loader's solve some ol those examples themselves lo verily thai ihey know how to use the pseudo-steady-state approximation to calculate I rate equation. 4.8.1 Tricks to Simplify the Rate Equation The one difficulty with the pseudo-steady-state approximation is that the algebra is complicated. I find that students often get lost in the algebra and never get lo an answer. Just to keep track of everything, in the example at the end of Section 4.7, we are trying to get an expression for the hydrogen atom concentration in terms of the concentration of all of the other stable species. The general approach we take is to 1. Set up the differential equation for the species of interest in terms of rate of all of the elementary reactions using equation (4.25) to keep track of the coefficients. 2. Substitute the expression for the rate of each of the elementary reactions using equations from Section 4.3. 3. Set the derivatives of the intermediate concentrations to zero. 4. Eliminate terms in the expression in (1) that contain the concentrations of unstable intermediates other than the species of interest. 5. Solve the resultant expression for the concentration of the species of interest. For example, if one wanted to calculate the concentration of hydrogen atoms during reaction (4.85), one would start with the expression for change in the concentration of hydrogen atoms: d[H.] - =r2 - r3 + r4 - r5 - r5 (4.92) dl One substitutes in the appropriate rate terms, and sets the derivative of the intermediate concentration to zero. 0 d[H. ~d7 = k2[.OH][H2] - k3[H.][02] + k4[.0][H2] - k5[H.] - k6[H.][02][X] (4.93) Next, one would have to determine what terms in this expression need to be eliminated. Notice that the k3, k5, and k^ terms contain [H»]. We are trying to find [H«], so those terms are okay in the expression. In contrast, the k2 and k4 terms contain [0»] or [OH.]. These terms need to be eliminated. The k2 term contains the hydroxyl concentration, so one can eliminate it by looking at the steady-state approximation for hydroxyls. Setting the derivative in equation (4.88) to zero and solving for k2[OH][H2] yields k2[OH.][H2] = 2k,[H2][02] + k3 - [H.][02] + k4[0:][H2] (4.94) Substituting equation (4.94) into equation (4.93) yields 0 = 2k,[H2][02] + 2k4[0:][H2] - k5[H.] - ke[H.][02][X] (4.95) i |iiul..... (I1'*!) has only olio U-1 in In elnmiiiilc Iho I loim i he I loini ioniums ii.......< t-nl i nl i« >n ol oxygon moms One cum eliminate the lenn limn i ho steud\ sinio ii|i|iiu<......lion on oxygon alums in equation (4.89) by soiling Iho derivative in equation i i h'ii equal to zero; then, solving I'm k, |o||ll<| yields k.,|():||ll2| = kt|H.||()..| i I Hi' iiinliii)' equalion It.90) into equation (4.95) yields 0 = 2k,[H2]|02| + 2k,[H.][02] - k,|H.| - k„|H||0.||X| (4.97) i in)' equalion (4.97) for |H»| yields 2k,[H2|[02| |H. ks + (k6[X]-2k3)02 (4.98) Mm i>t>1111 of this exercise is that you need to focus on where you are going when you nullifying the rate equation. Look at the terms that you need to eliminate, and luiil a i h i eliminate them. It usually works out that other terms cancel as well. The algebra is ii n.illv simple unless you have a reaction where two other reactive intermediates collide in I.....i products. In that case, only brute-force algebra works. I lie key thing to remember when simplifying rate expressions are . Write down the differential equations in terms of the rales and then substitute in the i.tie equations. . Keep track of what terms you want to eliminate, and eliminate them. I lie solutions are normally straightforward and the methods work when you keep track ■ ii where you are going. •I il RATE-DETERMINING STEPS I he pseudo-steady-statc approximation is the key approximation that people use to derive i ii. equations. Consequently, this approximation is very important. There are a series of niliei approximations that are occasionally used. I do not think they are important. Slill, I wanted to mention them so that you know what people are doing when they use another tpproximation. i >nc key simplification occurs if one can assume that one step, called the rate-ilrli i mining step or rate-limiting step, is much slower than all of the other steps in I he mechanism. The objective of this section is to work out the details. One can skip llns •' lion without loss of continuity. I el us start by considering a simple reaction mechanism where the idea of a rate determining step might be important. Consider a reaction where molecule A collides with ii ollision partner to yield an unstable excited species, A*. Then the excited species reads in form a product B: A + X ^J=^ A* + X 2 a* __L^ r (4.99) MAII HI II IIMIIIIMi . Ml I". Hll I lir i,ill' ul In......lion nl II. iM is pis en I > v i„ k,|A'| ( I 100) where [A!| is the eoneeniration of excited A molecules and k< is the rate constant for reaction 3. One can derive an equation for [A*] using ihe pseudo-steady state approximation: r* = k, [A][X] - k2[A*]LX] - k3[A*] «0 (4.101) where r^ is the rate of formation of excited A molecules; [AJ, [A*], and |X| are respectively the concentrations of A molecules, excited A molecules, and collision partners; and, k,, k2, and k3 are the rate constants for reactions 1, 2, and 3, respectively. Solving equation (4.101) for [A*] and then substituting that into equation (4.100) yields rB = k.,k,[AKX] k2[X] + k3 (4.102) Now consider the case where k2[X] » k3 so that reaction 3 is much slower than the rest. In that limit, equation (4.102) simplifies to i'll k,k, [A] (4.103) This is a key result because it shows that the reaction looks first order when k2[X] ^> k3. Next, we will show that one could have gotten to equation (4.103) more simply. When k2[Xl 3> k3, reactions 1 and 2 are just about in equilibrium. At equilibrium, the ratio of the concentration of excited A molecules to total A molecules is given by the equilibrium constant K| 2: 7ZT = K>-2 = r (4-104) [A] k2 In equation (4.104), we noted that the equilibrium constant for a reaction is given by the ratio of the equilibrium constants for a reaction, as noted in Section 4.3. Substituting equation (4.104) into equation (4.100) yields rB = -r— [A] k2 (4.105) Notice that equations (4.103) and (4.105) are identical. Therefore, in the limit that k2[X] » k3, one can calculate an accurate rate equation by assuming that reactions 1 and 2 are in equilibrium. One can generalize this process to any reaction. The idea is to assume that one step, called the rate-determining step or rate-limiting step, is much slower than the rest of the steps in the mechanism. One can show that if one step is extremely slow, then all of the steps before the slow step can be assumed to be in equilibrium with the reactants, and all of the steps after the rate-determining step can be considered to be in equilibrium with the products. If so, one can often derive a suitable rate equation for the reaction using somewhat less algebra than in the case where none of the steps are slow. I mi I'xiunple, considei a reversible iciictioii A I H C + D I i I • (I Ml ......, write, i \. the rate formation of the reactant A by rA = r i - r, + J?==fcL+M (4.10ft) i I in i 11, ii r, and i , are the rates of reactions 1 and -I, respectively. The Stead) SUN ipprmlmation on C shows did dt = ri - r._] - r2 + r_ 0 (4.108) i,. ,, r|, r i, r2, and r_2 are respectively the rales of reaction 1, -1, 2, and -2. Similarl) ii, teady-state approximation on E shows d|H| dt = r2 - r_2 - r3 + r_ 0 (4.109) Where ... r 2. r3, and r_3 are the rates of reaction 2, -2, 3 and -3, respectively. Adding luuations (4.107)-(4.109) yields rA = r_3-r3 (4.110) iibsiituiing the expressions for r3 and r_3 into equation (4.110) yields rA = k_3lG][H]-k3[EUF] (4.111) I quation (4.110) is a key result. It says that if one has a reaction consisting of several Heps, ihen at steady state, the net rates of each of the steps are equal. Consequently, une can calculate the rate of reaction by considering the net rate of reaction at any point .lining the reaction sequence. Now, let's assume that reaction 3 and -3 are much slower than the rest. Then it will he permissible to assume that the concentrations of E and F are in equilibrium with the ,. uctants. The analysis in Section 4.4 shows; I Ell F| - k,k? [A]|B| k_ik_2 Similarly, the concentration of G and H will be in equilibrium with the products: [L][M] _ Inks [G)[H] ~ k_4k-5 Substituting equation (4.112) and (4.113) into equation (4.111) yields rA = k-3k-4k.5[L][M]_k1l^[A|[B| (4.112) k4k5 k_,k (4.113) (4.114) -lK-2 I «111.111• >ii i I llhr. ilit- i-t1111\nlcnl ul llic |im'iiiIii sternly slult- ,i|i|iio\iiii.iliiiii when k, and k | IN slllllll. lii 111v experience, iherc is nol thai much simplification in assuming thul one ol ihc steps in a mechanism is al equilibrium, and one often misses some key information. Therefore, (here is some uncertainty in using the idea ol a rale determining Step to derive a rate equation. Still, the idea of a rate-determining step is a useful construct to understand reaction*! Typically, there is one step in a reaction mechanism, called the raw-ilcn,niiiiiiiif> step, which is much slower than the other reactions in the mechanism. This slow step is the bottleneck. Anything one can do to eliminate the bottleneck will greatly enhance the rate of reaction. 4.10 TESTING THE ACCURACY OF THE PSEUDO-STEADY-STATE APPROXIMATION Next, I want to change topics and consider when the pseudo-steady-state approximation is accurate. In practice, this approximation is used to calculate rate equations for a wide range of reactions. It usually works but unfortunately, there are a few exceptions. The exceptions come in systems that can explode, or in systems with other unusual behavior, such as oscillations. These are rare examples, but there are still a few cases where the pseudo-steady-state approximation docs not work. The fact that the pseudo-steady-state approximation may not work is pretty significant. For example, let's assume that you are trying to determine the kinetics for a reaction that has never been studied before. In principle, you can derive a mechanism and then use the pseudo-steady-state approximation to derive a rate equation. The difficulty, though, is that at that point you will not know whether the pseudo-steady-state approximation is accurate. As a result, you will not know whether to believe the results derived using the pseudo-steady-state approximation. In this section, we will discuss how one can determine whether the pseudo-steady-state approximation is accurate for a given reaction network. Let's start with the case discussed in Section 4.4: A + H+ <=!=>: I P + H2 (4.115) Figure 4.4 compares the intermediate concentration calculated via the pseudo-steady-state approximation [equation (4.62)] to the concentration calculated via the exact result [equation (4.54)] for several values of the rate constants. Notice that in the exact case (k2 = 5.7 x 106/minute), the pseudo-steady-state approximation works extremely well, and in fact, this approximation is still accurate when k2 is reduced by a factor of 105. Significant deviations are seen only when k2 is reduced to about 1. Notice that at the point where the pseudo-steady-state approximation fails, the intermediate concentration is already a few percent of the reactant concentration. This is a general finding. The pseudo-steady-state approximation is essentially exact when the intermediate concentration is much smaller than the reactant and product concentrations. One can, however, start to see deviations when the intermediate concentration begins to be significant compared to the reactant (or product) concentration. Later in this chapter, we will note that there are some cases where the pseudo-steady-state approximation works even though the intermediate concentrations are significant. However, the key 1 2 3 Time, minutes 1 2 Time, minutes 0.04 1 2 3 Time, minutes 1 2 3 Time, minutes i m.....4.4 A comparison of the intermediate concentration calculated via the steady-state approximation i ii......ncentration calculated via the exact result: equations (4.54) to (4.57) with l^ = 0.2, = 5kj, and viiiimin values of k2. I'.'ini to remember now is that the pseudo-steady-state approximation always winks when the intermediate concentrations are much less than the reactant or product . i mi nitrations. Physically, when we set the net rate of formation of the intermediate to zero, wc an laying that the amount of the intermediate that we are forming is negligible. Thai is | (•ihhI approximation if the intermediate concentration is tiny so that the amount we are forming is negligible. However, it is a bad approximation if we form a nonnegligible amount of impurity. One can use the idea that the concentration of the intermediates is negligible to check whether the pseudo-steady-state approximation works for a specific example. The idea ll i" use the pseudo-steady-state approximation to calculate the concentrations of all ol the Intermediates and see whether the concentration of any of the intermediates is signifii ani compared to the reactants or products. For example, there are two intermediates in reaction (4.67): [H] and [Br]. One can .sinuate the intermediate concentrations from equation (4.80) and (4.82). Rate constant! arc given in Westley (1980). When we plug numbers into equations (4.80) and (4.82), we find that at 300 K the intermediate concentrations are a factor of 10~8 lower than the reactant concentrations. The intermediate concentrations are low, which implies thai ihe sieady-state approximation will work. ■Ill I All 111(1 Ol 11II I'M IJI)<> Ml AI)Y SI All AI'I'I l( )XIMA 11( )N 4.11.1 Explosions The reaction between hydrogen and oxygen |reaction <4.X4)| is dil'l'erenl. In Section I. /. we derived an expression for the pseudo-steady-state approximation to |ll|, the hydrogen atom concentration. Rearranging the expression yields [H] = 2k,[H2][02] k5 + k6[X|[02] -2k3[02] (4.1 16) Figure 4.5 is a plot of the radical concentration as a function of the oxygen pressure calculated from equation (4.116) for some typical values of the parameters. Notice that the radical concentration goes first to infinity and then to negative infinity and then comes back again. This weird behavior occurs because there is a negative term in the denominator of equation (4.116). That term increases as the concentration of molecular oxygen in the reactor increases. When the denominator of equation (4.116) decreases, the calculated value of the hydrogen atom concentration increases. If one plugs in the rate constants, one finds that at temperatures above 450 K, the denominator of equation (4.116) goes to zero at moderate oxygen concentrations. Consequently, according to the pseudo-steady-state approximation, under some conditions, the radical concentration should approach inlinity. One can even get to the situation where the denominator is negative, which means that according to the pseudo-steady-state approximation, the radical concentration is negative. What is going on is that the pseudo-steady-state approximation fails in cases where the hydrogen atom concentration is comparable to or larger than the H2 concentration, or when the calculated concentrations are negative. Physically, it is impossible for the hydrogen atom concentration to be negative. Large hydrogen atom concentrations are impossible as well since the only hydrogen source is the H2 initially fed into the reactor. E o ra c CD en o 3E-05 2E-05 1E-05 0 ■o -1E-05 T3 in -2E-05 3 -3E-05 i i " Stable i i i Explosion Stable ________________i J i i I, , J 1E-05 3E-05 1E-04 3E-04 1E-03 3E-03 1E-02 3E-02 Oxygen cone, mol/liter Figure 4.5 A plot of the hydrogen atom concentration calculated from equation (4.116) as a function of the oxygen concentration. i ohm quenllv. the hydrogen atom i om enlinlion can nevei exceed twice the initial ii. . nil. i llll.lllOII. Iln pxcudo steady stale appioxiinalion lads whenever the Intermediate i om enlialioii It i oi11|i.iiable lo die ii. concentration, which one can test by plugging numbers into ill ,ii.m i I I 16). The weird behavior in figure 4.5 is associated with the failure ol the i. i.l\ stale approximation. iii.H ,i,ies mu imply that the pseudo-steady-state approximation was uselen In .....umple. If one runs the hydrogen/oxygen reaction under conditions wheie die ............n..i m equation (4.116) is small or negative, one finds thai the concentration ol radii iK does increase rapidly when one loads the rcaetants into the reactor. The radical mi.m..n does not actually increase to inlinity; instead the radical concentre!......I ....... However, there is a rapid production of radicals, which means that the reaction i in be very large. One calls a rapid increase in the radical concentration .m . > illusion You already know that a hydrogen/oxygen mixture can explode. Now you | .m iiuilhcmatically how the explosion happens. II one actually runs the hydrogen/oxygen reaction experimentally, one finds thai there .....ic complex behavior because the rate of the reactions with the walls [i.e., reactions -. umí / m equation (4.85)1 depends on the rate at which molecules diffuse through the ......Mine. One can also find circumstances where the heat generated during a re n n.m .i up the reaction mixture leading to an explosion. However, the thing to reincinhei ll ..... m explosion happens when the concentration of reactive intermediates grows quickly .luiiii). the course of the reaction. Once you start to build up radicals, the radicals aie so n n live that the reaction rate quickly grows toward a large value. In the H2 + Oj ease. H.....I.serves an upper and lower explosion limit. The lower explosion occurs because, m smi laise the oxygen pressure, the -k3[02] term in the denominator of equation (II 16) ,'lows until the denominator becomes zero. The upper explosion limit occurs because ii i... h pressure, the lc6[Xl[02] term becomes important, and so the denominator becomes punitive again. One can also get a thermal explosion, where the heat generated by the . .. non raises the temperature, which, in turn, raises the radical concentration. there are many other examples where the pseudo-steady-state approximation fails POl imple, OH and H02 radicals are stable in the upper atmosphere, so one cannot appl) I he pseudo-steady-state approximation to them. The pseudo-steady-state approximation winks only when the intermediate concentrations are much smaller than the ie.ui.un nul product concentrations. The pseudo-steady-statc approximation does not work well m u ics where the intermediate concentrations are comparable to the reactant concentration! In ihose cases, one can solve the rate equations numerically as described in Section 4.5. I he pseudo-steady-state approximation is quite useful, but it does not work all the time 4.11.2 Oscillations I here is one major class of reactions where the pseudo-steady-state approximation fails dismally: systems where oscillations in rate are observed. We briefly mentioned oscillating reactions in Chapter 2. The classic example is a modified Bclousov-Zhabotinskii (BZ) reaction: HOOCCH2COOH + HBrO, products (4.117) II you run reaction (4.117) in a beaker with phcnanthrolinc to complex the iron, the color of the mixture oscillates from red to blue. This makes a great demonstration during class. In a batch reactor, the oscillations eventually stop because one runs out of reactanis However, iiilť i (in also mih llic n-iii lion in .1 How n.i, loi il one > milium.ii,l\ Iccds ic.u l.nils Into iii>' Novt rati tor, the oiclllatloni etn ba itiiuincd forevci (see Figure 4.<>). Clearly, il i In- reliction rule oscillates, the pseudo-steady state approximation is not going tO work. Unfortunately, the 11/ reaction is very complicated, so it does not make a good example for the discussion here. Therefore, I would like to instead consider a simpler example the oxidation of a molecule called nicotinamide adenine dineuclotide hydride (NAI)II). shown in Figure 4.7. The oxidation reaction can be written NADH + H+ + iQ, NAD+ + H20 (4.1 IS) The oxidation is catalyzed by an enzyme called peroxidase. Reaction (4.118) is one of the key steps in the energy cycle of cells. It is also used to produce cell walls in plains. Benson and Scheeline (1996) examined the chemistry in detail using the mechanism in Figure 4.6. They find that the key reactions are NAD+ + H,0, PerJ+ + H202 + 2H+ —Í-*. Per5+ + 2H20 (4.119) (4.120) 100 200 300 400 Time, seconds 500 600 Figure 4.6 Rate oscillations during reaction (4.117). (Unpublished data of G. E. Poisson, D. A Tuchmann and R. I. Masel). O-CH. I OP-' CONH, O-CH. i Of-q OH OH 0=P-O° H_^ 0=P-O° O •ch2 n' O-CH 0=P-0, OH OH ~hrJ °~c^r OH OH NAD^ OH OH OH OH NADH NAD* Figure 4.7 NAD+, NADH and NAD". IVi" I NAHM • Per4' I NAD. I II' Per" I NADU ' • Per" I NAD. I II1 NAD. I O. NAD 1 + 02 • Per3-1 I ()2 . + 4H1 —!U Per6' + 2H2C) 7 JO. . + 2H i H202 + 02 Per1" + NAD.-► Per' + NAD NADHR -> NADH NADH ——+ NADHr 2(g) () 2(aq) -> o2(aq) -> o2(g) 2NAD.-► (NAD)2 i I I 'I i i i I ' ' i d 123) n imi (4.125) (4.126) (4.127) (4.128) (4.129) (4.1 10) íl 131) . In it- NAD. is the NAD radical shown in Figure 4.7; Per,+ , Per4'. Per'" . and Per'" aie .ii different oxidation states of the peroxidase enzyme; and NADHK is a different isomei ..I NADU. I ij'iirc 4.8 shows a pictorial representation of the mechanism. The mechanism contains i i lassie catalytic cycle with five important isomers of horseradish peroxidase, and two I......s of 02 (i.e., 02 and 02~ with different properties. I he NAD radicals are classic unstable intermediates. As a result, one would cxpcii ili. ii concentrations to be low. However, the radicals can form dimers. The dimers are liable and can accumulate in the system. Consequently, one cannot use the steady sLiir ipproximation to calculate the dimer concentration. What one needs to do, instead, is to follow the procedures in Section 4.4 and sci up tin- differential equations for all of the species in the system. For example, one would s.is ili.a the concentration of NAD radicals follows d[NAD'] = k,[NADH][Per5+] + k4[NAD.][Per4+] - k?[NAD.][02| dt - ks|NAD+][Per6+l - k,,[NAD.]2 (4.132) Benson and Scheeline (1996) list the other differential equations to describe the behaviot nl the species produced during NADH oxidation. One can numerically integrate these differential equations to calculate the behavior Of I he system using a modification of the computer program at the end of Example 4.B.6.( Itenson and Scheeline did so, and the key result is shown in Figure 4.9. Notice thai the concentrations of many species oscillate with time. Therefore, the system never reaches steady state. nahm ii' i nai)» -« Peroxidase4* NADH H++NAD» H202 + 2H Peroxidase3 Dimers Figure 4.8 Peroxidase2* The mechanism „, NADH oxidat|Qn ^ ^ ^ ^ ^ 3000 Time, seconds Figure 4.9 Solution of the differential 4000 equations of Bensen and Scheeline (1996). There are many other systems that oscillate. Generally, one can determine whether a given reaction shows oscillations by numerically integrating the rate equations for all of the species in the system. If the system can oscillate, then one will observe oscillations in the numerical solutions. One can also do analysis on the differential equations for each of the species to see when oscillations can occur. The details are beyond the scope of this book. See Gray and Scott (1990) for more information. The key point to remember is that the steady-state approximation fails when the concentrations of the reactive intermediates become significant. One must integrate the rate equations for the system in cases where the steady-state approximation fails. In such a case, one can get unusual behavior, including, possibly, oscillations or explosions. liniNn tin i"ii iiiki tin aiiy niAii aithi iximaihin kiini i m mi < mami\m\ lint < i iIMINQ THE PSEUDO-STEADY-STATE APPROXIMAI ION TO INFER - -i i iiani'.MS ' ......•. 4.6 and 4.Id. we discussed eases where the mechanism ol a reaction wits n .mil showed how one can use the pseudo-steady-state approximation lo derive .i title i......... lot the reaction. I have lo tell you that in practice I have rarely used the pseudo ' id i.Mr approximation in that way. It is usually hard to determine the mechanism ol lion Iioiii experiments. Kinetic measurements are much easier. Consequently, il .ill Ii ii.......mill do wiili the pseudo-steady-statc approximation were lo calculate something i \ in measure, namely, a rale equation, from something that is haul lo uie.isuir n mo lianism, the pseudo-sleady-state approximation would not be very uselul in i h.ipier 5, we will show that one can often predict mechanisms of reactions from III ' piuii iples. In that case, the pseudo-steady-statc approximation is useful. Howei..... general case, one will not know the mechanism very well. In that case, one I Ul i i iIh pscudo-steady-state approximation in reverse and use kinetic measurement! to .......elhing about the mechanism of a reaction. 1 in i.illy. I find that it is best to try to infer the rate law and the mechai..... i.....Iluneously. The general procedure is to first do some kinetic measurements. You in know if the reaction is closer to first-order or second-order in each reaelanl 01 iliui)' in between. Once you have some kinetic information, you I i mess at a mechanism. i i the pseudo-steady-state approximation to derive a rate equation Ibi lli.it mechanism. fake rale data. i See if the rate data fits the mechanism. Generally, once you have a mechanism, you need to take more rate data to verify that the rate equation fits the kinetic data i Moilily the proposed mechanism and iterate. i nfortunately, there is a lot of black art in trying to find a mechanism. For yean i'm'pIr mainly guessed, using their best judgment, and then iterated. Such a procedure (Hows you to fit kinetic data. However, often more than one mechanism will fit the data. • mi cquently, it is incorrect to assume that just because a mechanism fits the rate data, Mm mechanism is correct. On the other hand, if the rate equation does not fit the data, the .....lianism is incorrect. Consequently, one can use the pseudo-steady-state approximation In i iniiinale proposed mechanisms. I el's solve an equation to show how one can use these ideas to fit rate data. Assume tli.it you were trying to determine the mechanism of the following reaction: H2 + Br2 2HBr (4.133) \ i first guess, let's assume that you could write the rate equation for the reaction as lnllows: rHBr = kHBr|Br2]nfH2]m (4.134) here imsr is the rate of HBr formation; |H2| and [Br2] are the concentrations of hydrogen nut bromine in the reactor; and, kHBr. ni and n are constants. Now the question is how In hi Ihc constants kHBr. m, and n. Well, the procedures in Chapter 3 give you a clue. For example, one could use the \\ imping method. You could load Br2 and a large excess of hydrogen into a glass vessel.