Chapter 2 The Rates of Chemical Reactions Section 2.4 Reaction Mechanisms Method Solution Comment 03 + M O + M 03 + 0->2 02 First, determine the rate law for the destruction of ozone, i.e., an expression for -d[03]/df. Then use the steady-state approximation to solve for the concentration of the intermediate, 0. Finally, substitute the 0 atom concentration into the ozone rate law and simplify. The rate law for the destruction of ozone is -d[o3; dr fc,[03][M] - t.1[02][0][M] + k2[03][0]. The steady-state equation for [O] is d[Oj it = 0 = fc,[03][M] - M02][0][M] - yo3]fO] Some algebra can be avoided by subtracting these two to obtain -d[o3; di 2/c2[03][0]. Solution of the steady-state equation gives *i[Qs][M] [O] fe[Q,] + t-i[02][M]' Substitution of this equation into the simplified ozone destruction equation gives the final answer: -d[o,; 2/c2/Cl[M][03]2 dr »j[Oj] + yo2][M]- Note that at high values of the pressure, [M] will be large enough so that the second term in the denominator will be large compared to the first. The result will then simplify to -d[03l/dr = (2k2k1/k_l) ■ [03]2/[02]. 2.4.4 Unimolecular Decomposition: The Lindemann Mechanism As an example of the use of the steady-state approximation, we consider in detail the mechanism of unimolecular decomposition. The overall reaction is A —> products, and under high-pressure conditions the rate law is -d| A]/d( = &apIAl, where 7cap is the apparent rate constant. A question that begs an answer is how the A molecules obtain enough energy to decompose. The matter was debated vigorously in the early 1900sJ It was F. A. Lindemann who first suggested in 1922k that the reac-tants obtained the necessary energy from collisions. In its simplest form, the mechanism he proposed is shown below: A + M A* + M, (2.68) In these equations P stands for the products and M represents any molecule that can energize A by collision; M might be A itself, or it might be a nonreactive molecule in which the reactant is mixed. The overall rate of the reaction is -d[A|/di, or equivalently by dfP]/dc: d[A] dr d[P] dr : k2[A* (2.69) Since A* is an intermediate in the mechanism, it will be useful to apply the steady-state approximation: d[A'] at fc,[A][M] - t_,[A*][M] - k2[A' (2.70) Here, the time dependence of A* is equal to a production term, /C|[A][M], and two destruction terms, A_1|A*][M] and k2[A']. We can then solve this last equation for the steady-state concentration of A" to obtain [A*] = fci[A][M] *_i[M] + k2 Substitution of this last equation into equation 2.69 yields the solution d[P] _ fcA,[A"[M] ~dT ~ t_,[M] (2.71) (2.72) Recall that use of the steady-state approximation assumes that the concentration of the intermediate is small compared to the concentration of the starting material. This assumption is almost always valid for the system under consideration. Rearrangement of equation 2.71 shows that [A"]/[A] is much smaller than unity when kl[M]/(k_j[M] + k2) « 1. However, even if k2 were zero, this last expression would still be satisfied since /c(/k_, is simply the equilibrium constant for the first reaction, and this equilibrium constant must be smaller than unity because A* has much more energy than A. In addition, for A* of sufficiently high energy, k2 is usually very rapid, so that the inequality kl\M]/(k_[M + k2) << 1 is ensured. Having convinced ourselves that the steady-state approximation is valid for the Lindemann mechanism, equation 2.68, it is instructive to examine the solution, equation 2.72, under two limiting conditions. Let us first consider the "high-pressure JFor an interesting discussion of rhe history of this problem, see .1. T. Steinfcld, J. S. Francisco, and William L. Hase, Chemical Kinetics and Dynamics, 2nd cd. (Prentice-Hall, Englewood Cliffs, NJ, 1999), Section 11.3, lF. A. Lindemann, Trans. Faraday Sac. 17, 598 (1922). 62 Chapter 2 The Rates of Chemical Reactions limit," for which k_1[M] » k^. In this limit, the denominator of equation 2.72 can be approximated by its first term, and division of numerator and denominator by [M] gives d[P]/d< = (kJcJk^lA). Thus, in this limit the [M] cancels and the reaction is first order. Physically, in the high-pressure limit A* is rapidly being created and destroyed, and only a small fraction goes on to form products. In the "low-pressure" limit, when fc.JM] << k2, the second term in the denominator of equation 2.72 dominates, and d[P]/d< = kt [A][M]. In this limit the reaction is second order. Physically, in this limit most of the A* that is formed lasts long enough to react to form P, and very little gets deactivated. This behavior is shown in Figure 2.13, which plots a theoretical curve on a log-log plot for the apparent first-order rate constant, defined by kap = (1/[A]) d[P]/dr, as a function of [M] for the isomerization of ci.s-but-3-cne to trans-but-2-ene. The log-log plot is necessary to show both extremes in pressure. In the high-pressure limit, we have seen that d[P]/dr = (k2k1/k_1)[A] so that log(kap) should be a constant. At high pressures, the apparent rate constant is, indeed, found to be constant, but below about 10* torr the apparent rate constant is linearly proportional to pressure. This is because, in the low-pressure limit d[P]/df = £,[A][M] so that log(£ap) should be equal to log(^,[M]); i.e., it should vary linearly with log[M], as observed in the plot. An important practical application of the Lindemann mechanism is given in Example 2.5. The high-pressure result for the Lindemann mechanism also illustrates an important point about the temperature dependence of the overall rate constant in complex reactions. In the high-pressure limit dfPI/dr = £ap[A] = (k2kllk_l)[A\. Thus, the temperature dependence of the overall rate constant /cap depends on how k,, k_{, and k^ depend on temperature. Suppose that each of the rate constants for the elementary processes can be expressed in Arrhenius form: kt = A; exp(—EjkT). Simple multiplication and division of exponentials shows that Section 2.5 Homogeneous Catalysis 63 Mi A2Ai exp[-(£2 + Ei - E-i)/kT] (2.73) ;exp(-£ap/£r), where Aap = A2AJA_] and £ap = E2 + E1 - E_v The general form of this result holds for very complicated reactions, even, as described in Problem 2.26, for reactions as complicated as those that control the rate of firefly flashing! Like most good experimental or theoretical advances, the Lindemann mechanism raised more questions than it answered. For example, how do the rate constants ftj and k2 depend on how much energy the A* molecule has? And what kind of energy is important in energizing A? We will defer these questions until Chapter 7; they form the basis for much of the current research in physical chemistry. 2.5 HOMOGENEOUS CATALYSIS A catalyst, by definition, is a substance that is neither created nor consumed in the reaction but that increases the rate of the reaction. In most catalytic mechanisms, the catalyst transforms the reactants through a series of intermediates to products, but the catalyst is regenerated in the process of making the product. Since the intermediates are usually of much lower concentration than the starting material, catalytic mechanisms can usually be analyzed using the steady-state approximation. As an example, we will study in detail below the use of the steady-state approximation to analyze enzyme reactions. We will concentrate here on forms of homogeneous catalysis, while leaving the important area of heterogeneous catalysis until Chapter 6. i -7 Figure 2.13 \ *-|[M) ((k logiu(M/torr) Theoretical dependence of the Lindemann apparent rate constant with pressure for the isomerization of m-but-2-cne. 2.5.1 Acid-Base Catalysis One prevalent form of catalysis is acid or base catalyzed hydrolysis. For example, an ester might be hydrolyzed by the following mechanism in the presence of an acid: o II R-C- t-J I + O-H H O II R—C H I O-H O H II I R—C — O + ( X* O / \ H H I I H 1 + O-R Chapter 2 The Rates of Chemical Reactions The same reaction can be base catalyzed: 0 R-C-O, R-C —O I H / Section 2.5 Homogeneous Catalysis While the reaction can be catalyzed simply by H+ ions, it proceeds much faster if the yeast enzyme is present. The name for the enzyme comes from the fact that the conversion produces an inversion in the direction of rotation of plane polarized light by the sugar solution, noted in 1832 by Persoz. Michaelis and Menten used the yeast enzyme responsible for this conversion in their classic 1914 studies that resulted in the following proposed mechanism for enzymatic action, known, not surprisingly, as the Michaelis-Menten mechanism.1 Consider the following reaction sequence: E + S ?=* X, X->P + E. (2.74) Since neither H30+ nor OH~ is consumed in these processes, the overall rate law for this mechanism would be first order in each of the two reactants, the ester and water. However, because these reactions are always carried out in aqueous solution, the concentration of water can be taken as constant and incorporated into the rate constant; i.e., the system follows pseudo-first-order kinetics. The species H30+ or OH~ is a catalyst in the reaction; it is neither created nor consumed but provided another, more favorable, mechanism for the reaction. 2.5.2 Enzyme Catalysis Enzymes are macromolecules (MW = 104 - lO6) that are remarkable in their efficiency and specificity in catalyzing reactions of biological significance. For example, the enzyme invertase (j8-fructofuranidase), a component of yeast, catalyzes the conversion of sucrose to fructose and glucose by the hydrolysis reaction: HOCH2 OH CH2OH + H20 In this mechanism, the enzyme E can reversibly bind to a substrate S to yield the intermediate X. Once bound, the enzyme can also convert the substrate to products P, which it releases while returning to its original state. The enzyme is thus available to convert more substrate. How does the rate of the reaction depend on the amount of substrate and the amount of enzyme? We analyze the sequence using the steady-state approximation for the intermediate complex X: d[X] dt ■ /c,[E][S] - *_,[X] - kJX] = 0. (2.75) Let the original concentration of enzyme be E0. Then by mass balance E0 = [E] + [X]. In a similar manner, if the initial concentration of substrate is S0, then by mass balance S0 = [S] + [X] + [P]. Substituting these into equation 2.75, we obtain d[X] dt k,(E0 - [X])(S„ - [X] - [P]) - i_,[X] - k2[X] = 0. (2.76) We now suppose, as is generally the case, that the initial concentration of enzyme is much smaller than the initial concentration of substrate, E„ << S0. Since [X] can never be larger than E0, it follows that [X] << S0, so that we may safely ignore it in the term S0 — [XJ - [P]. We now consider the initial rate of the reaction, u0 = d[P]/df as ( -> 0, where [P] ~ 0. Solving equation 2.76 for [X] gives [X] Wo £iS0 + &_i + k2 (2.77) 'Jean-Francois Persoz (1805 1868) was a professor at the Sorhonne; Leonor Michaelis (1875-1949) was a German-born physician and biochemist who did research at the Rockefeller Foundation later in life. In addition to developing with Menten his famous equation, he is responsible for finding that keratin, the chief ingredient of hair, is soluble in glycolic acid, a discovery that made possible the development of the permanent. Maude Menten received her B.A. in 1904 and her M.D. in 1911 from the University of Toronto but had to leave Canada to pursue a career as a research scientist, because in those days women were not allowed to do research in Canadian universities. After study at the Rockefeller Institute and the Western Reserve University, she went to Berlin to study with Michaelis, where the two developed the Michaelis-Menten equation. Ultimately, she received a Ph.D. in biochemistry from the University of Chicago and became a professor at the University of Pittsburgh School of Medicine. 66 Chapter 2 The Rates of Chemical Reactions Section 2.5 Homogeneous Catalysis 07 Finally, the initial rate of the reaction is d[P]| k,\X] t_i + k, (2.78) I + -f- s„ where Vm = k2BQ and Km = (ik_] + k^lkv Note that the rate is proportional to the initial concentration of enzyme and to the rate constant k2, sometimes called the turnover number of the enzyme. At very high initial substrate concentrations, the initial rate approaches d[P]/df = Vm = fc,E„, or d([P]/E0)/d( = k2. The turnover number, ft2, is thus the number of molecules of product per molecule of enzyme that can be created per unit time. Typical values are 102—103 per second, but values as large as 106 per second have been observed. A plot of equation 2.78 is shown in Figure 2.14, in which reduced units have been used. The rate at first increases linearly with S0 and then levels off to an asymptote equal to Vm. While Figure 2.14 is useful in showing the behavior of the initial rate, it is not very useful for determining the rate constants. An alternate method of analysis is to take the reciprocal of both sides of equation 2.78 to obtain the Lineweaver-Burk form Km (2.79) Slope = KJVm / ■»1A>0 l y ^ 1/V„, 1/So ■ Figure 2.15 Lineweaver-Burk plot for and enzyme reaction obeying the Michaelis-Menten mechanism. This form shows that a plot of the reciprocal of the initial rate as a function of (S0)_1 should yield a straight line whose intercept is V~' and whose slope is KJVm, as shown in Figure 2.15. An example of an enzyme whose kinetics obey the Michaelis-Menten mechanism is lactase, the enzyme responsible for catalyzing the hydrolysis of lactose to D-glucose and D-galactose. Most adults of northern European background have sufficient enzyme to digest the milk that they consume, but many of other backgrounds do not possess this enzyme, the lack of which leads to breakdown of lactose by microbial action in the large intestine rather than by the enzyme in the small intestine. The resulting fermentation in the large intestine leads to diarrhea. One strategy to make milk more digestible for those who might otherwise suffer is to add Lactobacillus acidophilus to the milk. Example 2.6 examines the enzyme kinetics of lactase in the hydrolysis of a synthetic substrate similar to lactose."1 example 2.6 Determination of the Michaelis-Menten Constant for the Lactase Catalyzed Hydrolysis of a Synthetic Substrate, o-nitrophenyl-/3-D-galactopyranoside Objective Calculate the Michaelis-Menten constant, Km, given the following data pairs for l/uu, in arbitrary units, and 1/SQ, in units of 103 M_1: (2.8,0.7), (3.2,0.9), (4.2,1.3), (6.2,2.2), and (9.0,3.3). Method Plot \lv0 as a function of 1/S0 and determine the slope (= KJVm) and the intercept (= 1/Vm). The value of Km is given by the ratio of the former to the latter. Solution A least squares fit of the data to a line gives a slope of 0.24 X 10~2 and an intercept of 1.08. The ratio is 2.2 X 10~3 M. ■ Figure 2.14 A plot of the initial rate, in units of Vm, as a function of the initial substrate concentration, in units of Km for the Michaelis-Menten mechanism. mS, F. Rnsso and L. Moothart, /. Chem. Ed- 63, 242 (1986). Chapter 2 The Rates of Chemical Reactions An important mediator in the catalytic action of enzymes is the phenomenon of enzyme inhibition. An inhibitor is a compound that decreases the enzyme-catalyzed reaction by reacting with the enzyme itself or the enzyme substrate complex. Competitive inhibition occurs when the inhibitor competes with the substrate for binding at the active site of the enzyme. The mechanism can be represented by E + S X, >P + E EI, (2.80) where it is assumed that the last reaction is always at equilibrium and that the complex EI cannot catalyze the reaction. Application of the steady-state approximation yields (Problem 2.20) V (2.81) 1 + [I] where Kl = [E] [I]/[EI] is the equilibrium constant for the reverse of the last reaction in equation 2.80. The Lineweaver-Burk equation then becomes Km (2.82) so that a plot of the inverse of the initial rate as a function of the inverse of the initial substrate concentration gives the same intercept as in the absence of inhibitor, but a different slope, as shown in Figure 2.16A. An example of a competitive inhibitor is malonic acid, CH2(COOH)2, which resembles succinic acid, (COOH)CH2CH2 (COOH) closely enough to bind to the enzyme succinic dehydrogenase and inhibit it from converting succinic acid to fumaric acid, (COOH)CH=CH(COOH). Noncompetitive inhibition occurs when the inhibitor does not bind to the active site of the enzyme but still inhibits product formation: E + S X, Section 2.5 Homogeneous Catalysis ■ Figure 2.16 Effect of (A) competitive inhibition and (B) noncompetitive inhibition on the initial reaction rate. and the Lineweaver-Burk equation is 1 VmSn 1 + ffl (2.85) Note that the presence of the inhibitor affects both the slope and intercept of the Lineweaver-Burk plot for noncompetitive inhibition, as shown in Figure 2.16B. Methemoglobin reductase is an example of an enzyme that uses NADH to maintain hemoglobin in the active oxygen-carrying form. The salicylate ion is a noncompetitive inhibitor with respect to the substrate NADH." X->P + E (2.83) E + I El X + I XI, where the last two reactions are assumed to be in equilibrium with the same equilibrium constant for their reverse: K, = [E][I]/[EI] = [X][I]/[XI], The initial rate of reaction is (Problem 2.21) K,S„ [S0 + Kj 1 + [ir (2.84) NADH (nicotinamide adenine dimicleotide) Salicylate Ion "A. O. Splittgerber, K. Mitchell, G. Dahle, M. Puffer, and K, Blomquist, /. Ckem. Ed. 52, 680 (1975). 70 Chapter 2 The Rates of Chemical Reactions 2.5.3 Autocatalysis Not all reactions obey the simple behavior illustrated in Figure 2.1, in which the reactants decrease and products increase monotonically in time. Increasing attention has recently been paid to reactions in which species concentrations oscillate. Indeed, discontinuities in the rate of oscillation for such reactions as the concentrations change have helped to open a new area of science sometimes called chaos theory." All oscillating reactions have some element of "autocatalysis," catalysis in which a product of one reaction appears as a catalyst in the same or another coupled reaction. It is interesting, and appropriate, to examine briefly such reactions here since they illustrate an interesting form of catalysis, provide another example for use of the steady-state approximation, and are fascinating in their own right. Although the "Lotka" mechanism does not, so far as anyone knows, correspond to any observed chemical system, its simple mechanism illustrates the basic principles in more complex oscillatory systems. This mechanism has three steps: A + X-H>2X, X + Y^2Y, (2.86) Suppose that reactant is continually supplied to the system so that the concentration of A does not change from its initial value |A(0)]. Then the concentrations of the intermediates X and Y will reach a steady state, while the concentration of product Z will increase with time. The steady-state equations for X and Y are d[X] it d[Y] df :0 = t,iX][A(0)]-yx][Y], 0 = *2[X][Y] - fc,[Y]. (2.87) (2.88) The steady-stale values for [X] and [Y] are thus given by i2Y„ = A,[A(0)], fc2X55 = fc, We now consider perturbing the system in some way so that the concentrations of the intermediates change. Let the new concentrations X and Y be different from their steady-state values by time-dependent differences x and y: [X] [Y] x + X„ y + y8S (2.89) We wish to determine how x and y vary with time. Substitution of equation 2.89 into equation 2.87 leads to "See, tor example, the interesting book Chaos: Making a New Science, by James Gleick (Vikj York, 1987). Section 2.5 Homogeneous Catalysis 71 J = /q[A(0)][x + Xsi ■ hi* + X„][y + Y„], or dx dt ■ = + Xss][y + Yss] - k,[y + YJ, = *[*,[A(0)] - *2YSS] + [i,X.[A(0)] - lhXmYa (2.90) /c2Xssy - k2xy, h] + feX^Yj, + k2Yssx + k2xy. dy (2.91) *,Y„ Note that equations 2.87 and 2.88 can be used to show that the terms of equation 2.91 m square brackets are zero. If x and y are small displacements, then terms like xy can be neglected with respect to terms like *YSS or yX^, so that Ax dt dy df = ^Xssy, = k,Y..x. (2.92) If we take the derivative of both sides of the first equation in equation 2.92 and then use the second equation to substitute for the derivative of y, we obtain the second-order differential equation (fx dt + k\X,-i„x = 0. (2.93) A specific solution to this equation that allows x = x0 and y = 0 at t = 0 is x — *ycos ojt, (2.94) where a? = k%L^{n = *,fc,[A(0)]. (2.95) A similar solution can be obtained for y. We thus see that, rather than decaying exponentially, the concentration displacements will oscillate indefinitely. The oscillation frequency will depend on the concentration to which A is maintained. Many actual chemical reactions have been observed to oscillate, An often-cited but quite complex example is called the Belousov-Zhabotinsky reaction; a variant called the Briggs-Rauscher reaction exhibits an oscillating color change and is often used as a class demonstration^ Z?s ^°TT' f ***** Med 145' 1958 (l95a,; A' M- Ztatottasky,Dokl. Akad. NaukSSSR 157, 392 (1964); T. S. Brigg» and W. C. Rauscher,./. Ornm. Educ. 50,496 (1973); see also "Oscillating Chemical Reactions, by E. S. Scott, R. Schreiner, L. R. Sharpe, B. Z. Shakhashiri, and O. E. Dirreen. in Chemical Demonstmlmm, A Handbook for Teachers of Chemistry, B. Z. Shakhashiri. Vol. 2, Chapter 7 (University of Wisconsin Press, Madison, 1985). Chapter 2 The Rates of Chemical Reactions 2.6 FREE RADICAL REACTIONS: CHAINS AND BRANCHED CHAINS Many gas-phase reactions take place by so-called "chain" mechanisms involving free radical carriers, molecules, or atoms with one or more unpaired electrons. These mechanisms always consist of at least three steps: an initiation step creates the radicals that cany the chain; one or more chain steps convert reactants to products using the radical(s) in the role of a catalyst; and a termination step stops the chain by consuming the chain carrying radicals. Since the radical species are usually in small concentration and are intermediates, the steady-state approximation can be used to determine the overall rate law, which sometimes takes a complex form. An example is the H2 + Br2 ->' 2 HBr reaction, which we have already noted has the overall rate law |d[HBr]/dr = A chain mechanism is consistent with the overall rate law for this reaction. The initiation step is the collisional production of Br atoms: There are two chain steps: Br2 + M->2Br + M. Br + H,->HBr + H,' H,+ Br2 —> HBr + Br, (2.96) (2.97) Note that the net result of the chain is to convert one molecule of Br2 and one of H2 into two of HBr while regenerating the radicals so that the chain can continue. The termination step in this case is simply the reverse of the initiation step: 2 Br + M-*Br, + M. (2.98) The key point to realize is that the chain steps can occur many times for every initiation or termination and that these steps are principally responsible for conversion of reactants into products. We would thus expect that ^d^ = |(fc2[Bf][H2] + yH][Br2]). (2.99) We use the steady-state approximation to solve for the concentration of Br and H radicals: ■ d[fir] „ tlí d[HJ dt = 2fc1[Br2][M] - 2£-,[Brl2ÍM] - ^[Br][H2] + fc3[H][Br2], (2.100) = 0 = + /c2[Br][H2] - fc,[H][Br2] Section 2.6 Free Radical Reactions: Chains and Branched Chains Addition of these two equations and then solution for [Br] gives ft, V [Br] [Br2 (2.101) We also note that the second of these equations implies that the two bracketed terms on the right-hand side of equation 2.99 are equal, so that 1 d[HBr] 2 dt ;(2ft2[Br][H2]) Mvy2[H2][Br2r (2.102) Note that this expression is consistent with the overall rate law provided that k = k1{kllk_])m. In actuality, this reaction is somewhat more complicated, as discussed in Problem 2.23. 2.6.2 Rice-Herzfeld Mechanism Many organic reactions also occur by free radical processes. For example, in 1934 F. O. Rice and K. F. Herzfeld showed that the decomposition of ethane to ethylene and hydrogen, while following first-order kinetics, actually has a rather more complicated mechanism. There are two initiation steps: C,Hfi ► 2CH, and CH, + C,Hň^CH4 + C,H 2n5- There are then two propagation steps: C2H5->C2H4 + H, H + C2H6ÍC2H5 + H2. Termination typically takes place by H + C2H5-^C2H6. (2.103) (2.104) (2.105) (2.106) The steady-state approximation can be made on the radicals CH3, C2H5, and H, and, after some straightforward but rather tedious algebra (Problem 2.24), we obtain °Ich6; df C,H„ (2.107) Note that this agrees with the observation that the overall reaction is first order in C2H6. A point of confusion often arises when considering radical reactions of this type. While the H2 + Br2 reaction mechanism gave no products other than the expected HBr, the ethane decomposition mechanism produces CH4 in addition to the expected H2 and C2H4. However, since the main chain is carried by the C2H5 radical in steps 3 and 4, the amount of CH4 produced per molecule of H2 or C2H4 is negligibly small. Example 2.7 provides another illustration of a Rice-Herzfeld mechanism, in this case for the decomposition of acetaldehyde. 74 Chapter 2 The Rates of Chemical Reactions Section 2.6 Free Radical Reactions: Chains and Branched Chains 75 2.6.3 Branched Chain Reactions: Explosions Explosions are simply reactions whose rates become more and more rapid as time proceeds. There are normally two causes for a rapid increase in rate, giving rise to two types of explosions. A thermal explosion takes place when the rate of heat conduction out of a vessel surrounding an exothermic series of reactions is too small to prevent a significant temperature rise in the vessel. According to the Anhenius expression, most rates increase with increasing temperature, so that as the temperature rises, more reactants are consumed per unit time, creating more heat from the exothermic process, and causing the temperature to rise further. For a spherical vessel, the rate of heat loss due to thermal conductivity is proportional to the surface area of the vessel, while the rate of heat production is proportional to the amount of reactants, i.e., to the volume of the vessel, assuming the reactant concentrations are the same in both size vessels. Since the volume increases with the cube of radius while the surface area increases as the square, heat production will overcome heat loss in a large enough vessel. Explosions can thus occur unexpectedly if an exothermic reaction that ran smoothly in a small vessel is scaled up to obtain more product without proper attention to the thermal consequences. example 2.7 The Rice-Herzfeld Mechanism for the Decomposition of Acetaldehyde Objective Show that the Rice-Herzfeld mechanism for the decomposition of acetaldehyde, listed below, is consistent with the observation that the overall rate of decomposition is (3/2)-order in acetaldehyde, and determine how the overall rate constant is related to those for the individual steps. The overall reaction is CH3CHO —> CH4 + CO, while the Rice-Herzfeld mechanism is Initiation: Chain: Termination: CH,CHO -» CH, + CHO, CH3 + CH3CHO -h> CH4 + CO + CH3, (2.108) CH, + CH, >C,Hfi. Method Determine the production rate of CH4 and then use the steady-state approximation for the CH3 intermediate. Solution The production rate of CH4 is d[CH4 dr *:2[CH3]LCH3CHO]. (2.109) The concentration of [CH3] can be found from the steady state approximation: d[CH3 = 0 = ^[CH^HO] - 2fc,[CH3 (2.110) Note that there is no net contribution to [CH3] from the second step, since for every mole consumed one mole is produced. Rearrangement of the steady-state equation yields t k, Y/2 [CH3] = Uf [CH3CHO]V2, (2.111) :\/2 and substitution of this into equation 2.109 yields d[CH4] = /_£_V0 it 2\2fc3/ Consequently, the overall rate constant is k-^kjlk^)11 mechanism is consistent with an overall order of 3/2. [CH3CHO] (2.112) and the The second type of explosion is caused by a branched chain reaction. The mechanism for the oxygen-hydrogen reaction provides a good example. The over- all reaction is, of course, 2 H2 O, > 2 H20. The mechanism is complicated, but the following steps are the most important: initiation: Chain Branching: Chain Propagation: Termination: H, o. H + 02 O + H, >H02 + H, >2 0, >2H. >OH + O, >OH + H. OH + H,^H20 + H. H + wall-s--i-H2, 2 O + wall->-02. 2 (2.113) (2.114) (2.115) (2.116) H + O, + M -> HO, M. Like other chain reactions, this mechanism has initiation, propagation, and termination steps. Initiation of the reaction might be caused, for example, by the brief exposure of the reactant mixture to rapid heating from a spark or flame, in short, any reaction that creates free oxygen or hydrogen atoms. For example, the initiation of the hydrogen-oxygen explosion on the space shuttle Challenger was the heating of the external oxygen tank by flames from the propulsion rocket that escaped through the o-ring seals. Termination is caused either by collision of radicals with the wall, where they stick and recombine, or by recombination assisted by a third body, here called M. (While technically a radical, H02 is unrcactive relative to O, H, and OH; it is treated here as a stable compound.) The propagation step is a normal 76 Chapter 2 The Rates of Chemical Reactions one, where one radical is consumed and another produced. But there are also chain branching steps, steps in which the net number of radicals increases. For example, in the first reaction of equation 2.114 an H atom is consumed but an O atom and an OH radical are produced. If reactions such as those in equation 2.114 are prevalent, the rate of the overall reaction can increase rapidly. The overall reaction rate for a stoichiometric mixture of 2 H2 + 02 is a complicated function of temperature, pressure, vessel size, and material. The rates for each step depend on temperature. The rate of termination due to diffusion of O and H to the walls depends on pressure and vessel size. The sticking probability for these radicals when they reach the wall depends on the vessel material. Finally, the rate of H + 02 recombination depends on pressure. The bottom panel of Figure 2.17 shows the dividing line in the T-p plane between the explosion and steady reaction regimes for a typical system, while the top panel schematically indicates the rate of the reaction as a function of total pressure for a temperature of 800 K. It is constructive to consider the dominant processes at 800 K as the pressure is increased. For low pressures, the mean free path of the radicals O and H is large enough so that they reach the walls of the reaction vessel with high probability. Under these conditions, the termination steps dominate and the reaction proceeds in a controlled fashion. As the pressure increases, however, the chain branching steps start to dominate, and a branched chain explosion occurs. It is interesting to note, however, that a reaction mixture at higher total pressure would produce a steady reaction, a so-called hydrogen-oxygen flame. In this region, the pressure is sufficiently high that 1 1 1 1 1 1 ...... 1 1 1 I 1 1 1 1 1 1 ii ii.........i - 1 1 I 1 1 1 Explosion 1 1 1 1 1 1 1 1 I i I 1 I iii Steady reaction — ...........I logio(p/torr) ■ Figure 2.17 Explosion limits for a stoichiometric mixture of H2 and 02. Section 2.7 Determining Mechanisms from Rate Laws 77 the termination step, H + 02 + M -> H02 + M, occurs with good probability. In this region, then, the high rate of chain termination prevents an explosion. At higher pressures still, the heat due to the exothermic reaction cannot be carried away as fasl as it is produced, so that a thermal explosion occurs. 2.7 DETERMINING MECHANISMS FROM RATE LAWS Most of the examples we have used so far indicate how to determine whether a mechanism consisting of several elementary steps is consistent with an overall rate law. A much more difficult problem is how to figure out a reasonable mechanism of elementary steps given the rate law. In general, since several mechanisms are usually consistent with an overall rate law, it is not possible to obtain a single answer to this question. Nor is it really possible to obtain any answer without what amounts to an educated guess. In this section, we examine ways to make our guesses educated. There arc two basic concepts that we can use as a guide. The first is the realization that, in many reactions, there is one step in the mechanism whose rate is much smaller than any other step. In such a case, the overall rate of the reaction is usually controlled by this rate-limiting step. The rate law then contains in its numerator the rale of this rate-limiting step. The second concept is one that we have already encountered. When the rates of steps in the mechanism are comparable, it is still usually possible to assume that the concentration of reaction intermediates is low and varies slowly. This is just the steady-state hypothesis of Section 2.4.3. Even with a few principles in hand, it is ultimately experience that provides the best guide to guessing mechanisms from overall rate laws. Fortunately, we already have some experience. Let us look back at the systems we have studied to see if we can determine how we might apply the two basic concepts to obtain clues for determining a plausible mechanism from the overall rate law. We start our excursion with the rate-law result of the Lindcmann mechanism for the overall reaction A —> products: d[P] _ *2*![A][M] df *-,[M] + k2 (2.117) One obvious clue to this reaction is that the rate law depends on something other than the reactants or products, namely, the overall pressure of the gas mixture and not just the partial pressure of the reactant. Another clue comes in the complex form of the denominator. Recall that a mechanism that leads to this rate law is A + M : A" + M, (2.118) The complex form of the denominator comes from the fact that the intermediate in this reaction can disappear in two possible ways. Thus, we have already developed two clues: (1) complex denominators likely indicate that an intermediate in steady-state can disappear in more than one way and (2) the presence of something other than a reactant or product in the numerator (or, as it also turns out, in the denominator) likely indicates that an equilibrium step appears in the mechanism.