Chemical Kinetics (1102-346) PDF

Chemical Kinetics (1102-346) Department of Chemistry Faculty of Science Northern Border University Arar Dr. Ali El-Rayyes  chapter 15...

Chemical Kinetics (1102-346) Department of Chemistry Faculty of Science Northern Border University Arar Dr. Ali El-Rayyes  chapter 15 Chemical Kinetics The rate of chemical reactions is a very complicated subject. This statement is to be interpreted as a challenge to enthusiastic and vigorous chemists; it is not to be interpreted as a sad sigh of defeat. —Harold S. Johnston* The aims of studying chemical kinetics are to determine experimentally the rate of a reaction and its dependence on parameters such as concentration, temperature, and catalysts, and to understand the mechanism of a reaction, that is, the number of steps involved and the nature of intermediates formed. The subject of chemical kinetics is conceptually easier to understand than some other topics in physical chemistry, such as thermodynamics and quantum mechan- ics, although rigorous theoretical treatment of the energetics involved is possible only for very simple systems in the gas phase. Nevertheless, the macroscopic, empirical approach to the subject can provide much useful information. In this chapter, we discuss general topics in chemical kinetics and consider some important examples, including fast reactions and enzyme kinetics. 15.1 Reaction Rate The rate of a reaction is expressed as the change in reactant concentration with time. Consider the stoichiometrically simple reaction R→P Let the concentrations ^in mol L−1h of R at times t1 and t2 ^t2. t1h be 6R@ 1 and 6R@ 2. The rate of the reaction over the time interval ^t2 2 t1h is given by 6R@ 2 2 6R@ 1 Δ6R@ 5 t 2 2 t1 Δt Because 6R@ 2 , 6R@ 1, we introduce a minus sign so that the rate will be a positive quantity: Δ6R@ rate 5 2 Δt * Johnston, H. S., Gas Phase Reaction Rate Theory, The Ronald Press, New York, 1966. Used by permission. 671 1 672 Chapter 15: Chemical Kinetics The rate can be expressed also in terms of the appearance of a product 6P@ 2 2 6P@ 1 Δ6P@ rate 5 5 t 2 2 t1 Δt In this case, we have 6P@ 2. 6P@ 1. In practice, we find that the quantity of interest is not the rate over a certain time interval ^because this is only an average quantity whose value depends on the particular value of Δth; rather, we are interested in the instantaneous rate. In the language of calculus, as Δt becomes smaller and eventually approaches zero, the rate of the foregoing reaction at a specific time t is given by d6R@ d 6P@ rate 5 2 dt 5 dt The units of reaction rates are usually M s−1 or M min−1. For stoichiometrically more complicated reactions, the rate must be expressed in an unambiguous manner. Suppose that the reaction of interest is 2R → P The ratios 2d6R@ / dt and d6P@ / dt still express the rate of change of the reactant and the product, respectively, but they are no longer equal to each other because the reactant is disappearing twice as fast as the product is appearing. For this reason, we write the rate of this reaction as 1 d6R@ d6P@ rate 5 2 5 2 dt dt In general, for the reaction aA 1 bB → cC 1 d D the rate is given by 1 d6A@ 1 d6B@ 1 d6C@ 1 d6D@ rate 5 2 52 5 5 ^15.1h a dt b dt c dt d dt where the expressions in brackets refer to the concentrations of the reactants and prod- ucts at time t after the start of the reaction. 15.2 Reaction Order The relationship between the rate of a chemical reaction and the concentrations of the reactants is a complex one that must be determined experimentally. Referring to the general equation above, however, we find that usually ^but by no means alwaysh the reaction rate can be expressed as rate ∝ 6A@ x 6B@ y 5 k 6A@ x 6B@ y ^15.2h 2  15.2 Reaction Order 673 This equation, known as the rate law, tells us that the rate of a reaction is not con- stant; its value at any time, t, is proportional to the concentrations of A and B raised to some powers. The proportionality constant, k, is called the rate constant. The rate law is defined in terms of the reactant concentrations, but the rate constant for a given reaction does not depend on the concentrations of the reactants. The rate constant is affected only by temperature, as we shall see later. Expressing the rate of a reaction as shown in Equation 15.2 enables us to define the order of a reaction. We say that the reaction is x order with respect to A and y order with respect to B. Thus, the reaction has an overall order of ^x 1 yh. It is important to This must be the case understand that, in general, there is no connection between the order of a reactant in because a chemical equation the rate expression and its stoichiometric coefficient in the balanced chemical equa- can be balanced in many tion. For example, the rate of the reaction different ways. 2N2O5^gh → 4NO2^gh 1 O2^gh is given by rate 5 k 6N2O5@ The reaction is first order in N2O5—not second order as we might have inferred from the balanced equation. The order of a reaction specifies the empirical dependence of the rate on concen- trations. It may be zero, an integer, a negative integer, or even a noninteger. We can use the rate law to determine the concentrations of reactants at any time during the course of a reaction. To do so, we need to integrate the rate law expressions. For sim- plicity, we shall focus only on reactions that have positive integral orders. Zero-Order Reactions The rate law for a zero-order reaction of the type A → product Rate is given by d6A@ rate 5 2 5 k 6A@ 0 5 k ^15.3h dt [A] The quantity k ^M s−1h is the zero-order rate constant. As you can see, the rate of Figure 15.1 Zero-order reaction. The the reaction is independent of the reactant concentration ^Figure 15.1h. Rearranging rate is independent of Equation 15.3, we obtain concentration. d6A@ 5 2kdt Integration between t 5 0 and t 5 t at concentrations 6A@ 0 and 6A@ gives 6A@ t < d6A@ 5 6A@ 2 6A@ 0 5 2< kdt 5 2kt 6A@ 0 0 or 6A@ 5 6A@ 0 2 kt ^15.4h 3 674 Chapter 15: Chemical Kinetics Note that Equation 15.4 gives the time dependence of 6A@, but cannot be the full description of the factors affecting the rate. To illustrate this point, consider the decomposition of gaseous ammonia on a tungsten surface: NH3^gh → 1–2 N2^gh 1 3–2 H2^gh Under certain conditions, this reaction obeys zero-order rate law. Such a zero-order reaction can occur if the rate is limited, for example, by the concentration of a catalyst. The rate of the reaction is given by rate 5 k′θA where k′ is a constant, θ is the fraction of metal surface covered by the adsorbed ammonia molecules, and A is the total catalyst surface area. If the pressure of ammo- nia is large enough, θ 5 1, and the reaction is zero order in ammonia. At sufficiently low pressures, however, θ is proportional to 6NH3@ in the gas phase and the reaction becomes first order in ammonia. Note that the rate will also depend on the amount of catalyst, that is, on the area A. First-Order Reactions A first-order reaction is one in which the rate of the reaction depends only on the con- centration of the reactant raised to the first power, d6A@ rate 5 2 5 k 6A@ ^15.5h dt where k ^s−1h is the first-order rate constant. Rearranging Equation 15.5, we get d6A@ 6A@ 2 5 kdt Integrating between t 5 0 and t 5 t at concentrations 6A@ 0 and 6A@, we obtain 6A@ t < d6A@ 6A@ 0 6A@ 5 2< kdt 0 6A@ ^15.6h 6A@ 0 ln 5 2kt or 6A@ 5 6A@ 0 e−kt ^15.7h A plot of ln^6A@/ 6A@ 0h versus t gives a straight line whose slope, which is negative, is given by 2k ^Figure 15.2ah. Equation 15.7 shows that in first-order reactions, the decrease in reactant concentration with time is exponential ^Figure 15.2bh. Radioactive decays fit first-order kinetics. An example is 222Rn → 218 86 84Po 1 α 4  15.2 Reaction Order 675 Slope k [A] ln [A] [A] 0 t t (a) (b) Figure 15.2 First-order reaction. ^ah Plot based on Equation 15.6 with a slope of 2k. ^bh Exponential decay of 6A@ with time according to Equation 15.7. where α represents the helium nucleus ^He2+h. The thermal decomposition of N2O5 mentioned earlier is first order in N2O5. Another example is the rearrangement of methyl isonitrile to acetonitrile: CH3NC^gh → CH3CN^gh Half-Life of a Reaction. A measure of considerable practical importance in kinetic studies is the half-life ^t1/2h of a reaction. The half-life of a reaction is defined as the time it takes for the concentration of the reactant to decrease by half of its original value. For example, in a first-order reaction, as 6A@ 5 6A@ 0 / 2, t 5 t1/2 and Equation 15.6 becomes 6A@ 0 / 2 6A@ 0 ln 5 2kt1/2 or ^15.8h ln 2 0.693 t1/2 5 5 k k Thus, the half-life of a first-order reaction is independent of the initial concentration ^Figure 15.3h. For A to decrease from 1 M to 0.5 M takes just as much time as it does 1.0 0.8 0.6 Figure 15.3 [A] /M The half-lives of a first-order reaction ^A → producth. 0.4 The initial concentration is arbitrarily set at 1 M, and A reacts with a constant half-life of 50 s. 0.2 0.0 0 50 100 150 200 250 300 t /s 5 676 Chapter 15: Chemical Kinetics Table 15.1 Half-lives of Some Common Radioisotopes Isotope Decay process t1/2 3 3 1H 1H → 32He 1 −10β 12.3 yr 14 14 6C 6C → 147N 1 −10β 5.73 × 103 yr 24 24 11 Na 11 Na → 24 0 12Mg 1 −1 β 15 h 32 32 32 15P 15P → 16 S 1 −10β 14.3 d 35 35 35 16S 16S → 17Cl 1 −10β 88 d 60 27Co Emission of γ rays 5.26 yr 99m a 43Tc Emission of γ rays 6.0 h 131 131 53I 53I → 131 0 54Xe 1 −1 β 8.05 d a The superscript m denotes the metastable excited nuclear energy state. for A to decrease from 0.1 M to 0.05 M. Table 15.1 lists the half-lives of radioactive isotopes that are used extensively in biochemical research and medicine. In contrast to first-order reactions, the half-lives of other types of reaction all depend on the initial concentration. In general, we can show that ^see Appendix 15.1 on p. 724h ^15.9h 1 t1/2 ∝ 6A@ 0n−1 where n is the order of the reaction. 6 678 Chapter 15: Chemical Kinetics Second-Order Reactions We consider two types of second-order reactions here. In one type, there is just one reactant. The second type involves two different reactants. The first type is repre- sented by the general reaction A → products and this rate is d6A@ rate 5 2 5 k6A@ 2 ^15.10h dt That is, the rate is proportional to the concentration of A raised to the second power, and k ^M −1 s−1h is the second-order rate constant. Separating the variables and inte- grating, we obtain 6A@ t < d6A@ 5 2< kdt 6A@ 0 6A@ 2 0 1 1 5 kt 6A@ 6A@ 0 2 or ^15.11h 1 1 5 kt 1 6A@ 6A@ 0 7  15.2 Reaction Order 679 Slope k ([B] 0 [A] 0) Slope k 1 [B][A] 0 ln [A] [A][B] 0 t t (a) (b) Figure 15.5 Second-order reactions. ^ah Plot based on Equation 15.11. ^bh Plot based on Equation 15.14. where 6A@ 0 is the initial concentration. Thus, a plot of 1/ 6A@ versus t gives a straight line with a slope equal to k ^Figure 15.5ah. To derive the half-life of this type of second- order reaction, we set 6A@ 5 6A@ 0 / 2 in Equation 15.11 and write 1 1 5 kt1/2 1 6A@ 0 / 2 6A@ 0 or ^15.12h 1 t1/2 5 k6A@ 0 As mentioned earlier, except for first-order reactions, the half-lives of all other reac- tions are concentration dependent. The second type of second-order reaction is represented by A 1 B → products and d6A@ d6B@ rate 5 2 52 5 k6A@ 6B@ ^15.13h dt dt This reaction is first order in A, first order in B, and second order overall. Let 6A@ 5 6A@ 0 2 x 6B@ 5 6B@ 0 2 x where x ^in mol L−1h is the amount of A and B consumed in time t. From Equation 15.13, d6A@ d^6A@ 0 2 xh dx 2 52 5 5 k6A@ 6B@ dt dt dt 5 k^6A@ 0 2 xh^6B@ 0 2 xh 8 680 Chapter 15: Chemical Kinetics Rearranging, we obtain dx 5 kdt ^6A@ 0 2 xh^6B@ 0 2 xh By the somewhat tedious but straightforward method of integration by partial func- tions, we can obtain the final result: 1 ^6B@ 0 2 xh6A@ 0 ln 5 kt 6B@ 0 2 6A@ 0 ^6A@ 0 2 xh6B@ 0 or 6B@ 6A@ 0 ^15.14h 1 ln 5 kt 6B@ 0 2 6A@ 0 6A@ 6B@ 0 Equation 15.14 was derived by assuming that 6A@ 0 , 6B@ 0. If 6A@ 0 5 6B@ 0, the solu- tion is the same as that in Equation 15.11. Note that Equation 15.11 cannot be obtained from Equation 15.14 by setting 6A@ 0 5 6B@ 0. A plot of Equation 15.14 is shown in Figure 15.5b. Below are a few examples of second-order reactions: CH3CHO^gh → CH4^gh 1 CO^gh 2NO2^gh → 2NO^gh 1 O2^gh C2H5Br^aqh 1 OH−^aqh → C2H5OH^aqh 1 Br−^aqh Pseudo-First-Order Reactions. An interesting special case of second-order reac- tions occurs when one of the reactants is present in great excess. An example is the hydrolysis of acetyl chloride: CH3COCl^aqh 1 H2O^lh → CH3COOH^aqh 1 HCl^aqh Because the concentration of water in the acetyl chloride solution is quite high ^about 55.5 M, the concentration of pure waterh and the concentration of acetyl chloride is of the order of 1 M or less, the amount of water consumed is negligible compared with the amount of water originally present. Thus, we can express the rate as d6CH3COCl@ 5 k′ 6CH3COCl@ 6H2O@ dt 5 k6CH3COCl@ ^15.15h where k 5 k′6H2O@. The reaction therefore appears to follow first-order kinetics and is called a pseudo-first-order reaction. ^To measure k′, the second-order rate constant, we need to measure k for many different initial concentrations of H2O, then a plot of k versus 6H2O@ will yield a straight line with a slope equal to k′.h 9  15.2 Reaction Order 681 Table 15.2 Summary of Rate Equations for A → Products Units of the Order Differential form Integrated form Half-life rate constant d6A@ 6A@ 0 0 2 dt 5k 6A@ 0 2 6A@ 5 kt 2k M s−1 d6A@ 6A@ 5 6A@ 0 e−kt ln 2 1 2 5 k6A@ s−1 dt k d6A@ 1 1 1 5 kt 6A@ 6A@ 0 6A@ 0k 2 2 5 k6A@ 2 2 M −1 s−1 dt d6A@ 6B@ 6A@ 0 5 k6A@ 6B@ 1 2a M −1 s−1 6B@ 0 2 6A@ 0 6A@ 6B@ 0 2 ln 5 kt — dt a For A 1 B → products. Table 15.2 summarizes the rate laws and half-life expressions for zero-, first-, and second-order reactions. Third-order reactions are known but are uncommon, and so we shall not discuss them. Determination of Reaction Order In the study of chemical kinetics, one of the first tasks is to determine the order of the reaction. Several methods are available for determining the order of a reaction, and we shall briefly discuss four common approaches. 1. Integration Method. An obvious procedure is to measure the concentration of the reactant^sh at various time intervals of a reaction and to substitute the data into the equations listed in Table 15.2. The equation giving the most constant value of the rate constant for a series of time intervals is the one that corresponds best to the correct order of the reaction. In practice, this method is not precise enough to do more than to distinguish between, say, first- and second-order reactions. 2. Differential Method. This method was developed by van’t Hoff in 1884. Because the rate υ of an nth-order reaction is proportional to the nth power of the concentration of the reactant, we write υ 5 k6A@ n ^15.16h Taking common logarithms of both sides, we obtain log υ 5 n log6A@ 1 log k ^15.17h Thus, by measuring υ at several different concentrations of A, we can obtain the value of n from a plot of log υ versus log6A@. A satisfactory procedure is to measure the 10 682 Chapter 15: Chemical Kinetics [A] Figure 15.6 log v 0 ^ah Measurement of the initial rates, υ 0, Slope n of a reaction at different concentrations. ^bh Plot of log υ 0 versus log 6A@. t log [A] 0 (a) (b) initial rates ^υ 0h of the reaction for several different starting concentrations of A, as shown in Figure 15.6. The advantages of using initial rates are ^1h that it avoids pos- sible complications due to the presence of products that might affect the order of the reaction and ^2h that the reactant concentrations are known most accurately at this time. 3. Half-Life Method. Another simple method of determining reaction order is to find the dependence of the half-life of a reaction on the initial concentration, again using the equations in Table 15.2 or Equation 15.9. Thus, measuring the half-life of a reaction will help us determine the order of the reaction. This procedure is particularly useful for first-order reactions because their half-lives are independent of concentration. 4. Isolation Method. If a reaction involves more than one type of reactant, we can keep the concentrations of all but one reactant constant and measure the rate as a func- tion of its concentration. Any change in rate must be due to that reactant alone. Once we have determined the order with respect to this reactant, we repeat the procedure for the second reactant, and so on. In this way, we can obtain the overall order of the reaction. The four methods described above apply only in ideal cases. In practice, deter- mining reaction order can be very difficult because of uncertainty in concentration measurements ^e.g., when there are small concentration changes in initial rate deter- minationsh as well as the complexity of the reactions ^e.g., when reactions are revers- ible and reactions occur between reactants and productsh. To a certain extent, the procedure is one of trial and error. The use of computers has significantly facilitated the analysis of kinetic data. Once the order of the reaction has been determined, the rate constant at a particu- lar temperature can be calculated from the ratio of the rate and the concentrations of the reactants, each raised to the power of its order. Knowing the order and rate con- stant then enables us to write the rate law for the reaction. 11 12 13 14 15 16 17  15.3 Molecularity of a Reaction 683 15.3 Molecularity of a Reaction Knowing the order of a reaction is but one step toward a detailed understanding of how a reaction occurs. A reaction seldom takes place in the manner suggested by a balanced chemical equation. Typically, the overall reaction is the sum of several steps; the sequence of steps by which a reaction occurs is called the mechanism of the reac- tion. To know the mechanism of a reaction is to know how molecules approach one another during a collision, how chemical bonds are broken and formed, how charges are transferred, and so on when the reactant molecules are in close proximity. The mechanism proposed for a given reaction must account for the overall stoichiometry, the rate law, and other known facts. Consider the decomposition of hydrogen peroxide: 2H2O2^aqh → 2H2O^lh 1 O2^gh When this reaction is catalyzed by iodide ions, the rate law is found to be rate 5 k6H2O2@ 6I−@ Thus, the reaction is first order with respect to both H2O2 and I−. Whereas the word order reflects the overall change in going from the reactants to products, the molecularity of a reaction refers to a single, definite kinetic process that may be only one step in the overall reaction. For example, evidence suggests that the decomposition of hydrogen peroxide takes place in two steps, as follows: Step ^1h k H2O2 1 I− $1 H2O 1 IO− Step ^2h k H2O2 1 IO− $ H2O 1 O2 1 I− 2 Each of these so-called elementary steps describes what actually happens at the molec- ular level. How do we account for the observed rate dependence in terms of these two steps? We simply assume that the rate for the first step is much slower than that for the second step, that is, k1 ,, k 2. The overall rate of decomposition, then, is completely controlled by the rate of the first step, which is aptly called the rate-determining step, and we have: rate 5 k16H2O2@ 6I−@, where k1 5 k. Note that the sum of steps 1 and 2 gives us the overall reaction, because the species IO− cancels out. Such a species is called an intermediate because it appears in the mechanism of the reaction but is not in the overall balanced chemical equation. An intermediate is always formed in an early elementary step and consumed in a later elementary step. On the other hand, a catalyst ^I− in this caseh appears as a reactant in the initial elementary step. It invariably forms an intermediate and is regenerated at the end of the reaction. The preceding discussion shows that our insight into a reaction comes from an understanding of molecularity, not of order. Once we know the mechanism and the rate-determining step, we can write the rate law for the reaction, which must agree with the experimentally determined rate law. Although most reactions are kinetically As in a court of law, we complex, the mechanisms for some of them are sufficiently well understood to be dis- ask only for proof beyond a cussed in molecular terms. In general, however, proving the uniqueness of a mecha- reasonable doubt. nism is very difficult or impossible, especially for a complex reaction. We shall now examine the three different types of molecularity. Unlike the order of a reaction, molecularity cannot be zero or a noninteger. 18 684 Chapter 15: Chemical Kinetics Unimolecular Reactions Reactions such as cis-trans isomerization, thermal decomposition, ring opening, and racemization are usually unimolecular, that is, they involve only one reactant mol- ecule in the elementary step. For example, the following gas-phase elementary steps are unimolecular: Unimolecular reactions often follow a first-order rate law. Because these reac- tions presumably occur as the result of a binary collision through which the reactant molecules acquire the necessary energy to change forms, we would expect them to be bimolecular processes and hence second-order reactions. How do we account for the discrepancy between the predicted and observed rate laws? To answer this question, let us consider the treatment put forward by the British chemist Frederick Alexander Lindemann ^1886–1957h in 1922.† Every now and then a reactant molecule, A, collides with another A molecule, and one becomes energetically excited at the expense of the other: k A 1 A* 1 A1A $ where the asterisk denotes the activated molecule. The activated molecule can form the desired product according to the elementary step k A* $2 product Another process that may also be going on is the deactivation of the A* molecule: k A* 1 A $ A 1 A −1 The rate of product formation is given by d6product@ 5 k 2 6A*@ ^15.18h dt All that remains for us to do is to derive an expression for 6A*@. Because A* is an ener- getically excited species, it has little stability and a short lifetime. Its concentration in the gas phase is not only low but probably fairly constant as well. Using this assump- tion, we can apply the steady-state approximation as follows.‡ The rate of change of 6A*@ is given by the steps leading to the formation of A* minus the steps leading to † A similar treatment was proposed independently and almost simultaneously by the Danish chemist Jens Anton Christiansen ^1888–1969h. ‡ Note that the steady-state approximation does not always apply to intermediates. Its use must be justified by either experimental evidence or theoretical considerations. 19  15.3 Molecularity of a Reaction 685 the removal of A*. According to the steady-state approximation, however, this rate of change must be zero. Mathematically, we have d6A*@ 5 0 5 k16A@ 2 2 k−16A@ 6A*@ 2k 2 6A*@ ^15.19h dt Solving for 6A*@, we obtain k16A@ 2 6A*@ 5 ^15.20h k 2 1 k−16A@ The rate of product formation is now given by d6product@ k k 6A@ 2 5 k 2 6A*@ 5 1 2 ^15.21h dt k 2 1 k−16A@ Two important limiting cases may be applied to this equation. At higher pressures ^$ 1 atmh, most A* molecules will be deactivated instead of forming product, and we have k−16A@ 6A*@.. k 2 6A*@ or k−16A@.. k 2 The rate in this case is given by d6product@ k1k 2 5 6A@ ^15.22h dt k−1 and the reaction is first order in A. On the other hand, if the reaction is run at low pressures ^, 0.01 atmh so that most A* molecules form the product instead of being deactivated, the following inequality will hold: k−16A@ 6A*@ ,, k 2 6A*@ or k−16A@ ,, k 2 The rate now becomes d6product@ 5 k16A@ 2 ^15.23h dt which is second order in A. Lindemann’s theory has been tested for a number of systems and is found to be essentially correct. The analysis for the intermediate case ^i.e., k−16A@ 6A*@ < k 2 6A*@h is more complex and will not be discussed here. 20 686 Chapter 15: Chemical Kinetics Bimolecular Reactions Any elementary step that involves two reactant molecules is a bimolecular reaction. Some of the examples in the gas phase are H 1 H2 → H2 1 H NO2 1 CO → NO 1 CO2 2NOCl → 2NO 1 Cl2 In the solution phase we have 2CH3COOH → ^CH3COOHh2 ^in nonpolar solventsh Fe2+ 1 Fe3+ → Fe3+ 1 Fe2+ Termolecular Reactions Finally, an elementary step that involves the simultaneous encounter of three reactant molecules is called a termolecular reaction. The probability of a three-body collision is usually quite small and only a few such reactions are known. Interestingly, they all involve nitric oxide as one of the reactants: 2NO^gh 1 X2^gh → 2NOX^gh where X 5 Cl, Br, or I. Another type of termolecular “reaction” involves atomic recombinations in the gas phase; for example, H 1 H 1 M → H2 1 M I 1 I 1 M → I2 1 M where M is usually some inert gas such as N2 or Ar. When atoms combine to form diatomic molecules, they possess an excess of kinetic energy, which is converted to vibrational motion, resulting in bond dissociation. Through three-body collisions, the M species can take away some of this excess energy to prevent the break-up of the diatomic molecules. No elementary steps with a molecularity greater than three are known. 15.4 More Complex Reactions All the reactions discussed so far are simple in the sense that only one reaction is taking place in each case. Unfortunately, this condition is often not satisfied in actual practice. Three examples of more complex reactions will now be discussed. Reversible Reactions Most reactions are reversible to a certain extent, and we must consider both the forward and reverse rates. For the reversible reaction that proceeds by two elementary steps, k1 A k−1 B 21  15.4 More Complex Reactions 687 we represent the net rate of change in 6A@ as d6A@ 5 2 k16A@ 1 k−16B@ ^15.24h dt At equilibrium, there is no net change in the concentration of A with time, that is, d6A@ /dt 5 0, so that k16A@ 5 k−16B@ ^15.25h This expression leads to 6B@ k 5 1 5K ^15.26h 6A@ k−1 where K is the equilibrium constant. The discussion of the relationship between reaction rates and equilibria is rooted in a principle of great importance in chemical kinetics. The principle of microscopic reversibility states that at equilibrium, the rates of the forward and reverse processes are equal for every elementary reaction occurring.* It means that the process A → B is exactly balanced by B → A so that equilibrium cannot be maintained by a cyclic process in which the forward reaction is A → B and the reverse reaction is B → C → A: Instead, for every elementary reaction we must write a reverse reaction as follows: such that k16A@ 5 k−16B@ k 2 6B@ 5 k−2 6C@ k 3 6C@ 5 k−3 6A@ These rate constants are not all independent. By simple algebraic manipulation, we can show that k1k2k 3 5 k−1k−2k−3 ^see Problem 15.65h. The usefulness of the principle of microscopic reversibility is that it tells us that the reaction pathway for the reverse of * The principle of microscopic reversibility is a consequence of the fact that the fundamen- tal equations for the microscopic dynamics of a system ^i.e., Newton’s laws or the Schrödinger equationh have the same form when time t is replaced by 2t and when the signs of all velocities are also reversed. See B. H. Mahan, J. Chem. Educ. 52, 299 ^1975h. 22 688 Chapter 15: Chemical Kinetics a reaction at equilibrium is the exact opposite of the pathway for the forward reaction. Therefore, the transition states* for the forward and reverse reactions are identical. Consider the base-catalyzed esterification between acetic acid and ethanol, where B is a base ^e.g., OH−h. The species formed in the first step is a tetrahedral intermediate. Now, according to the principle of microscopic reversibility, the reverse reaction, that is, the hydrolysis of ethyl acetate, must involve the acid-catalyzed expul- sion of ethoxide ion from the same tetrahedral intermediate: Thus, when the likelihood of a certain mechanism is being considered, we can always turn to the principle for guidance. If the reverse mechanism looks implausible, then chances are that the proposed mechanism is wrong and we must search for another mechanism. Consecutive Reactions A consecutive reaction is one in which the product from the first step becomes the reactant for the second step, and so on. The thermal decomposition of acetone in the gas phase is an example: CH3COCH3 → CH25CO 1 CH4 CH25CO → CO 1 1–2 C2H4 Many nuclear decays are also consecutive reactions. For example, upon the capture of a neutron, a uranium-238 isotope is converted to a uranium-239 isotope, which then decays as follows: 239 U → 239 0 92 93Np 1 −1β 239 Np → 239 0 93 94 Pu 1 −1β * The transition state of a reaction is the complex formed between the reactants and prod- ucts along the reaction coordinate ^discussed further in Section 15.7h. 23  15.4 More Complex Reactions 689 For a two-step consecutive reaction, we have 1 k 2 k A $ B$C Because each step is first order, the rate law equations are d6A@ 5 2k16A@ ^15.27h dt d6B@ 5 k16A@ 2 k 2 6B@ ^15.28h dt d6C@ 5 k 2 6B@ ^15.29h dt We assume that initially only A is present and its concentration is 6A@ 0 so that 6A@ 5 6A@ 0 e−k1t ^15.30h The rate equation for the intermediate B is quite complex and will not be fully discussed here. The treatment can be simplified, however, by applying the steady-state approximation to B, that is, by assuming that the concentration of B remains constant over a certain time period so that we can write d6B@ 5 0 5 k16A@ 2 k 2 6B@ ^15.31h dt or k1 k 6B@ 5 6A@ 5 1 6A@ 0 e−k1t ^15.32h k2 k2 Equation 15.32 holds if k 2.. k1. Under this condition, B molecules are converted to C as soon as they are formed, so 6B@ is kept constant and low compared to 6A@. To get an expression for 6C@, we note that at any instant we have 6A@ 0 5 6A@ 1 6B@ 1 6C@. Therefore, from Equations 15.30 and 15.32 we obtain 6C@ 5 6A@ 0 2 6A@ 2 6B@ k1 −k t 5 6A@ 0d1 2 e−k1t 2 e 1n k2 5 6A@ 0^1 2 e−k1th ^15.33h The ^k1/k 2hexp^2k1th term is eliminated because it is much smaller than 1. Figure 15.7 shows plots of 6A@, 6B@, and 6C@ with time for different rate constant ratios. In all cases, 6A@ falls steadily from 6A@ 0 to zero while 6C@ rises from zero to 6A@ 0. The concentration of B rises from zero to a maximum and then falls back to zero. Note that as k 2 becomes larger than k1, the steady-state approximation becomes valid over the time period when 6B@ remains constant ^Figure 15.7ch. 24 690 Chapter 15: Chemical Kinetics Figure 15.7 Variation in the concentrations of A, B, and C with time for a consecutive reaction A $ B $ C. ^ah k1 5 k2; ^bh k1 5 2k2; ^ch k1 5 0.5k2. k1 k2 A more complicated but common consecutive reaction is k1 k 2 A1B k−1 C$ P where P denotes product. This scheme involves a pre-equilibrium, in which an inter- mediate is in equilibrium with the reactants. A pre-equilibrium arises when the rates of formation of the intermediate and of its decay back into reactants are much faster than its rate of formation of products, that is, when k−1.. k 2. Because A, B, and C are assumed to be in equilibrium, we can write 6C@ k 5 1 6A@ 6B@ k−1 K5 and the rate of formation of P is given by d6P@ 5k 2 6C@ 5 k 2K 6A@ 6B@ dt k1k 2 5 6A@ 6B@ ^15.34h k−1 Chain Reactions One of the best-known gas-phase chain reactions involves the formation of hydrogen bromide from molecular hydrogen and bromine between 230°C and 300°C: H2^gh 1 Br2^gh → 2HBr^gh The complexity of this reaction is indicated by the rate equation, d6HBr@ α 6H2@ 6Br2@ 1/2 ^15.35h 1 1 β 6HBr@ / 6Br2@ 5 dt 25  15.5 The Effect of Temperature on Reaction Rate 691 where α and β are some constants. Thus, the reaction does not have an integral reac- tion order. It has taken many experiments and considerable chemical intuition to come up with Equation 15.35. We assume that a chain of reactions proceeds as follows: k1 Br2 $ 2Br chain initiation 2 k Br 1 H2 $ HBr 1 H chain propagation 3 k H 1 Br2 $ HBr 1 Br chain propagation 4 k H 1 HBr $ H2 1 Br chain inhibition 5 k Br 1 Br $ Br2 chain termination The following reactions play only a minor role in determining the rate: H2 → 2H chain initiation Br 1 HBr → Br2 1 H chain inhibition H 1 H → H2 chain termination H 1 Br → HBr chain termination For this reason, they are not included in the kinetic analysis. By applying the steady- state approximation to the intermediates H and Br, we can derive Equation 15.35 using the first five elementary steps ^see Problem 15.20h. 15.5 The Effect of Temperature on Reaction Rate Figure 15.8 shows four types of temperature dependence for reaction rate constants. Type ^ah represents normal reactions whose rates increase with increasing tempera- ture. Type ^bh shows a rate that initially increases with temperature, reaches a maxi- mum, and finally decreases with further temperature rise. Type ^ch shows a steady decrease of rate with temperature. The behavior outlined in ^bh and ^ch may be sur- prising, because we might expect the rate of a reaction to depend on two quantities: the number of collisions per second and the fraction of collisions that activate molecules k k k k T T T T (a) (b) (c) (d) Figure 15.8 Four types of temperature dependence for rate constants. See text. 26 692 Chapter 15: Chemical Kinetics for the reaction. Both quantities should increase with increasing temperature. The complex nature of the reaction mechanism accounts for this ostensibly unusual behav- ior. For example, in an enzyme-catalyzed reaction, the enzyme molecule must be in a specific conformation to react with the substrate molecule. When the enzyme is in the native state, the reaction rate does increase with temperature. At higher temperatures, the molecule may undergo denaturation, thereby losing its effectiveness as a catalyst. Consequently, the rate will decrease with increasing temperature. The behavior shown in Figure 15.8c is known for only a few systems. Consider the formation of nitrogen dioxide from nitric oxide and oxygen: 2NO^gh 1 O2^gh ⇋ 2NO2^gh The rate law is rate 5 k6NO@ 2 6O2@ The mechanism is believed to involve two bimolecular steps: 6^NOh2@ Rapid: 2NO ⇋ ^NOh2 6NO@ 2 This is an example K5 of pre-equilibrium, Slow, rate determining: ^NOh2 1 O2 $ 2NO2 discussed on p. 690. k′ Thus, the overall rate is rate 5 k′6^NOh2@ 6O2@ 5 k′K 6NO@ 2 6O2@ 5 k6NO@ 2 6O2@ where k 5 k′K. Furthermore, the equilibrium between 2NO and ^NOh2 is exothermic from left to right. Because the decrease in K with temperature outweighs the increase in k′ with temperature, the overall rate decreases with increasing temperature over a certain range of temperature. Finally, the behavior shown in Figure 15.8d corresponds to a chain reaction. At first, the rate rises gradually with temperature. At a particular temperature, the chain propagation reactions become significant, and the reaction is literally explosive. The Arrhenius Equation In 1889, Arrhenius discovered that the temperature dependence of many reactions could be described by the equation, k 5 Ae2Ea /RT ^15.36h where k is the rate constant, A is called the frequency factor or pre-exponential factor, Ea is the activation energy ^kJ mol−1h, R is the gas constant, and T is the absolute tem- perature. The activation energy is the minimum amount of energy required to initiate a chemical reaction. The frequency factor, A, represents the frequency of collisions between reactant molecules. The factor exp^2Ea /RTh resembles the Boltzmann distri- bution law ^see Equation 2.33h; it represents the fraction of molecular collisions that 27  15.5 The Effect of Temperature on Reaction Rate 693 Activated complex E a for forward Energy reaction E a for reverse reaction Figure 15.9 Reactants Schematic diagram of activation energy for an exothermic reaction. Products have energy equal to or greater than the activation energy, Ea ^Figure 15.9h. Because the exponential term is a number, the units of A are the same as the units of the rate constant ^s−1 for first-order rate constants; M −1 s−1 for second-order rate constants, and so onh. As we shall see later, because the frequency factor A is related to molecular colli- sions, it is temperature dependent. For a limited temperature range ^# 50 Kh, however, the predominant temperature variation is contained in the exponential term. Taking the natural logarithm of Equation 15.36, we obtain Ea ln k 5 ln A 2 ^15.37h RT Thus, a plot of ln k versus 1/T yields a straight line whose slope, which is negative, is equal to 2Ea /R ^Figure 15.10h. Note that in Equation 15.37 k and A are treated as dimensionless quantities. Alternatively, if we know the rate constants k1 and k 2 at T1 and T2, we have, from Equation 15.37 Ea ln k1 5 ln A 2 RT1 Ea ln k 2 5 ln A 2 RT2 Taking the difference between these two equations, we obtain k2 E 1 1 The formulation is valid for ln 5 2 ad 2 n ^15.38h any order rate constant. k1 R T2 T1 Ea Slope R ln k Figure 15.10 An Arrhenius plot. The slope of the straight line is equal to 2Ea / R. 1 T 28 694 Chapter 15: Chemical Kinetics Equation 15.38 enables us to calculate the rate constant at a different temperature if Ea is known. From the standpoint of Arrhenius’s rate equation, a complete understanding of the factors determining the rate constant of a reaction requires that we be able to calculate the values of both A and Ea. Considerable effort has been devoted to this problem, as we shall see below. 15.6 Potential-Energy Surfaces To discuss activation energy in more detail, we need to learn something about the energetics of a reaction. One of the simplest reactions is the combination of two atoms to form a diatomic molecule, such as H 1 H → H2. Basically, we would like to describe more complex reactions in terms of a potential-energy curve such as that shown in Figure 3.11. However, potential-energy diagrams are prohibitively complex for all but the simplest systems. One of the simplest and most studied systems is the exchange reaction between the hydrogen atom and the hydrogen molecule: H 1 H2 → H2 1 H Even for a three-atom system such as this, we need a four-dimensional plot, describ- ing three bond lengths, or two bond lengths and a bond angle, versus energy. The problem is greatly simplified by assuming that the minimum energy configuration is linear so that only two bond lengths need to be specified. Consequently, only a Potential energy curve for H B H C Potential energy Reaction path Activated complex HA H B H C HA HB HC Activated complex r AB r BC r AB Potential energy HA HB HC HA HB HC curve for HA H B Completely dissociated state HA H B H C r BC (a) (b) Figure 15.11 The H 1 H2 → H2 1 H reaction. ^ah Potential-energy surface. ^bh Contour diagram of the potential-energy surface. 29  15.7 Theories of Reaction Rates 695 Potential energy Potential energy Potential energy Reactants Products Reactants Products Products Reactants Reaction coordinate Reaction coordinate Reaction coordinate (a) (b) (c) Figure 15.12 Potential-energy profile along the minimum energy path for ^ah the H 1 H2 reaction, ^bh an exothermic reaction, and ^ch an endothermic reaction. three-dimensional plot is required ^Figure 15.11h. Labeling the atoms A, B, and C, we can represent the reaction as HA 1 HB2HC → 6HA · · · HB · · · HC@ → HA2HB 1 HC activated complex The plot, called the potential-energy surface, is a map of potential energies corre- sponding to different values of rAB and rBC, which are separations between the atoms. Although the reaction can proceed along any path, the one that requires the mini- mum amount of energy is shown by the red curve. The system travels along this path through the first valley and over the saddle point, which is the location of the activated complex, and then moves down the second valley. We represent this path in a plot of the potential energy versus the reaction coordinate, which describes the positions of the atoms during the course of a reaction. Figure 15.12a shows the plot for the H 1 H2 reaction. The plots shown in Figures 15.12b and 15.12c are customarily employed for reactions in general, where the products differ from the reactants. You need to understand that these plots provide only a qualitative description for the reaction path because of the complexities involved for large molecules. Much effort has gone into the calculation of the activation energy for the H 1 H 2 reaction. The correspondence between the calculated and measured values of Ea ^36.8 kJ mol−1h lends support to the validity of the model ^i.e., linear activated com- plexh.* An interesting side note is that if the reaction took the path involving the dissociation of the H2 molecule followed by recombination, an activation energy of 432 kJ mol−1 would be required. 15.7 Theories of Reaction Rates At this point, we are ready to consider two important theories of reaction rates: colli- sion theory and transition-state theory. These theories provide us with greater insight into the energetic and mechanistic aspects of reactions. * A theoretical analysis shows that the activation energy for this reaction is 40.2 kJ mol−1. The calculation for this simple reaction required 80 days of computer time! See D. D. Diedrich and J. B. Anderson, Science 258, 786 ^1992h. 30 31 32 756 20 PHYSICAL CHEMISTRY COLLISION THEORY OF REACTION RATES According to this theory, a chemical reaction takes place only by collisions between the reacting molecules. But not all collisions are effective. Only a small fraction of the collisions produce a reaction. The two main conditions for a collision between the reacting molecules to be productive are : (1) The colliding molecules must posses sufficient kinetic energy to cause a reaction. (2) The reacting molecules must collide with proper orientation. Now let us have a closer look at these two postulates of the collision theory. (1) The molecules must collide with sufficient kinetic energy Let us consider a reaction A – A + B – B ⎯⎯ → 2A – B A chemical reaction occurs by breaking bonds between the atoms of the reacting molecules and forming new bonds in the product molecules. The energy for the breaking of bonds comes from the kinetic energy possessed by the reacting molecules before the collision. Fig. 20.9 shows the energy of molecules A2 and B2 as the reaction A2 + B2 → 2AB progresses. Energy barrier Ea Energy A–A B–B Reactants 2A – B Product Reaction progress Figure 20.9 The energy of the colliding molecules as the reaction A2 + B2 2AB proceeds. The activation energy Ea provides the energy barrier. The Fig. 20.9 also shows the activation energy, Ea, that is the minimum energy necessary to cause a reaction between the colliding molecules. Only the molecules that collide with a kinetic energy greater than Ea, are able to get over the barrier and react. The molecules colliding with kinetic energies less that Ea fail to surmount the barrier. The collisions between them are unproductive and the molecules simply bounce off one another. (2) The molecules must collide with correct orientation The reactant molecules must collide with favourable orientation (relative position). The correct orientation is that which ensure direct contact between the atoms involved in the breaking and forming of bonds. (Fig. 20.10) From the above discussion it is clear that : Only the molecules colliding with kinetic energy greater that Ea and with correct orientation can cause reaction. 33  CHEMICAL KINETICS 757 Effective collision Ineffective collision Figure 20.10 Orientations of reacting molecules A2 and B2 which lead to an effective and ineffective collision. Collision Theory and Reaction Rate Expression Taking into account the two postulates of the collision theory, the reaction rate for the elementary process. A + B ⎯⎯ → C+D is given by the expression rate = f × p × z where f = fraction of molecules which possess sufficient energy to react; p = probable fraction of collisions with effective orientations, and z = collision frequency. b Limitations of the Collision Theory The collision theory of reaction rates is logical and correct. However, it has been oversimplified and suffers from the following weaknesses. (1) The theory applies to simple gaseous reactions only. It is also valid for solutions in which the reacting species exist as simple molecules. 34  CHEMICAL KINETICS 765 (2) The values of rate constant calculated from the collision theory expression (Arrhenius equation) are in agreement with the experimental values only for simple bimolecular reactions. For reactions involving complex molecules, the experimental rate constants are quite different from the calculated values. (3) There is no method for determining the steric effect (p) for a reaction whose rate constant has not been determined experimentally. (4) In the collision theory it is supposed that only the kinetic energy of the colliding molecules contributes to the energy required for surmounting the energy barrier. There is no reason why the rotational and vibrational energies of molecules should be ignored. (5) The collision theory is silent on the cleavage and formation of bonds involved in the reaction. The various drawbacks in the simple collision theory do not appear in the modern transition-state theory. TRANSITION STATE THEORY The transition state or activated complex theory was developed by Henry Erying (1935). This theory is also called the absolute rate theory because with its help it is possible to get the absolute value of the rate constant. The transition state theory assume that simply a collision between the reactant molecules does not really causes a reaction. During the collision, the reactant molecules form a transition state or activated complex which decomposes to give the products. Thus, A + B—C ABC A—B + C Reactants Activated Products complex The double dagger superscript ( ) is used to identify the activated complex. The transition state theory may be summarised as follows : (1) In a collision, the fast approaching reactant molecules (A and BC) slow down due to gradual repulsion between their electron clouds. In the process the kinetic energy of the two molecules is converted into potential energy. (2) As the molecules come close, the interpenetration of their electron clouds occurs which allows the rearrangement of valence electrons. (3) A partial bond is formed between the atoms A and B with corresponding weakening of B – C bond. This leads to formation of an activated complex or transition state. The activated complex is momentary and decomposes to give the products (A–B + C) Partial Bond bond weakens A B C A B C Bond Activated complex breaks A B C A B + C Products A + B—C A B C A—B + C A B C A B C A B C The activated complex theory may be illustrated by the reaction energy diagram (Fig. 20.13). 35 766 20 PHYSICAL CHEMISTRY Activated complex A B C A B C Ea Potential energy A + B—C ΔE Reactants A B C A—B + C Products A B C Reaction coordinate Figure 20.13 Change of potential energy during a collision between the reactant molecules for an exothermic reaction. Here the potential energy of the system undergoing reaction is plotted against the reaction coordinate (the progress of the reaction). The difference in the potential energy between the reactants and the activated complex is the activation energy, Ea.The reactants must have this minimum energy to undergo the reaction through the transition state. As evident from the energy diagram, energy is required by the reactants to reach the transition state. Also, the energy is obtained in passing from the transition state to the products. If the potential energy of the products is less than that of the reactants (Fig. 20.14) the energy obtained in going from the activated complex to products will be more than the activation energy (Ea). Thus such a reaction will be exothermic. A B C A—B + C Potential energy Ea E A + B—C Reaction coordinate Figure 20.14 A potential energy diagram for an endothermic reaction. 36  CHEMICAL KINETICS 767 On the other hand, if the potential energy of the products is greater than that of the reactants, the energy released in going from the activated complex to products will be less than the activation energy and the reaction will be endothermic. ACTIVATION ENERGY AND CATALYSIS We know that for each reaction a certain energy barrier must be surmounted. As shown in the energy diagram shown in Fig. 20.15, the reactant molecules must possess the activation energy, Ea, for the reaction to occur. Activated complex Activated complex involving catalyst Ea Potential energy Ecat Reactants E Products Reaction coordinate Figure 20.15 Energy diagram for a catalysed and uncatalysed reaction showing a lowering of activation energy by a catalyst. The catalyst functions by providing another pathway with lower activation energy, Ecat. Thus a much large number of collisions becomes effective at a given temperature. Since the rate of reaction is proportional to effective collisions, the presence of a catalyst makes the reaction go faster, other conditions remaining the same. It may be noted from the above diagram that although a catalyst lowers the activation energy, the energy difference, ΔE, between products and reactants remains the same. 37  15.8 Isotope Effects in Chemical Reactions 703 15.8 Isotope Effects in Chemical Reactions When an atom in a reactant molecule is replaced by one of its isotopes, both the equi- librium constant of the reaction and the rate constants may change. The term equilib- rium isotope effects refers to changes in equilibrium constants that result from isotope substitution. Rate variations caused by this exchange are known as kinetic isotope effects. The study of isotope effects provides information on reaction mechanisms that has applications in many branches of chemistry. The underlying theory is complex, requiring both quantum mechanics and statistical mechanics; therefore, only a quali- tative description will be given here. Isotopic replacement in a molecule does not result in a change in the electronic structure of the molecule or in the potential-energy surface for any reaction the mol- ecule might undergo, yet the rate of the reaction can be profoundly affected by the substitution. To see why, let us consider the H 2, HD, and D2 molecules, whose zero- point energies, that is, the ground-state vibrational energies, are 26.5 kJ mol−1, 21.6 kJ mol−1, and 17.9 kJ mol−1, respectively.* Because D2 has the lowest zero-point energy ^due to the fact that it has the largest reduced massh, more energy is required to dis- sociate this molecule than to break apart H2 or HD ^Figure 15.14ah. Consequently, the reaction rate for D2 → 2D will be the slowest compared with the other two corre- sponding dissociations. As a rough estimate, we can calculate the ratio of the rate con- stants for the dissociation of H2 and D2, k H /k D, as follows. According to Figure 15.14b, the activation energies for these two processes, EH and ED, are given by EH 5 Estretch 2 EH0 ED 5 Estretch 2 ED0 1 * The zero-point energy is given by Evib 5 –2 hν, where ν is the fundamental frequency of vibration. This frequency is given by v 5 ^1/2πhÏ·k /µ, where k is the force constant of the bond and µ the reduced mass, given by m1m2 /^m1 1 m2h. Because D2 has the largest value of µ, it possesses the smallest frequency of vibration and hence of Evib. The reverse holds true for H2. Activated complex Figure 15.14 Schematic energy-level diagrams. ^ah Ground vibrational energy levels for H2, HD, and D2. ^bh Potential energy Potential energy Activation energies for r bond rupture in H2 and D2. 0 E H0 EH and E D0 are the ground E stretch vibrational energies of H2 and D2 ^see texth. E D0 H2 Reactant HD D2 Product (a) (b) 38  704 Chapter 15: Chemical Kinetics where EH0 and ED0 are the zero-point energies and Estretch is the difference between the lowest possible potential energy and the potential energy of the activated complex. Using the Arrhenius expression ^Equation 15.36h, we write k H Ae−^Estretch 2 EH0 h/RT 5 k D Ae−^Estretch 2 ED0 h/RT 5 e^EH0 2ED0 h/RT 5 e^26.5 – 17.9h × 1000 J mol−1/^8.314 J K−1 mol−1h^300 Kh < 31 which is quite a large number. The example above dramatizes the difference between the rate constants of D2 and H2. In practice, however, we are concerned more with the breaking of a bond between hydrogen and some other atom, such as carbon. Consider, for example, the reactions, C H B C H B C D B C D B where B is some group that can take up a hydrogen atom. Again, we would predict a kinetic isotope effect, because the fundamental frequencies of vibration are differ- ent for the C—H and C—D bonds. But the difference is not as great as that between H2 and D2 because the reduced masses are closer to each other in this case ^see Problem 15.39h. Still, the ratio kC2H /kC2D might be appreciable, on the order of 5 or so. We assume that this bond- Isotope effects that reflect isotope substitution for an atom involved in a bond- breaking process is also the breaking process are called primary kinetic isotope effects. Such effects are most rate-determining step. pronounced for light elements such as H, D, and T. Reactions involving isotopes of mercury ^199Hg and 201Hgh, for example, would show hardly any detectable difference in rates. A secondary kinetic isotope effect occurs when the isotope is not directly involved in the bond rupture. We would expect a small change in the reaction rate in this case. Indeed, this prediction has been confirmed experimentally. How does the isotope effect change an equilibrium process? Although the forward and reverse directions of an equilibrium process must trace the same reaction path- way, the isotope effect on the two rate constants need not be the same. Consequently, there can be an isotope effect on the equilibrium constant. As a simple example, let us consider the dissociation of a monoprotic acid, such as acetic acid, in H2O and D2O. The dissociations are 6H+@ 6CH3COO−@ 6CH3COOH@ CH3COOH ⇋ CH3COO− 1 H+ KH 5 6D+@ 6CH3COO−@ 6CH3COOD@ CH3COOD ⇋ CH3COO− 1 D+ KD 5 39  15.9 Reactions in Solution 705 ^In D2O, all the ionizable protons are replaced by deuterons.h Experimentally, we find that KH /KD 5 3.3. The greater acid strength of CH3COOH over CH3COOD can be explained by noting that the undeuterated molecule has a higher zero-point vibrational level ^for the O2H bondh, and less energy is required to dissociate the hydrogen than to dissociate the deuterium in CH3COOD. A useful general rule with regard to the isotope effect on equilibria is that substitution with a heavier isotope will favor the formation of a stronger bond. As for acetic acid, when we replace H with D, the O2D bond becomes stronger, and the resulting molecule has less of a tendency to dissociate. 15.9 Reactions in Solution The major difference between gas-phase reactions and reactions in solution lies in the role of the solvent. In many cases, the solvent plays a minor role, and rates do not differ much in the two phases. In terms of simple kinetic theory, the frequency of collisions between reacting molecules depends only on the concentrations of the reactants; it is not affected by solvent molecules. There is a difference, however, in the outcome of an encounter between reactant molecules in solution compared with the collision of molecules in the gas phase. If two molecules collide in the gas phase and do not react, they will normally move away from each other. There is very little likelihood that this same pair will collide again. In contrast, when two solute molecules diffuse together in a solution, they cannot move apart again quickly after the initial encounter because they are surrounded closely by solvent molecules. In this case, the reactants are temporarily trapped in a “cage” of solvent ^Figure 15.15h. To be sure, the cage is not rigid, as the solvent molecules are constantly in motion and changing positions. Nevertheless, the cage effect causes the reactant molecules to remain together for a longer time than they would in the gas phase, and they may collide with each other hundreds of times before they drift apart.* For reactions that have relatively low acti- vation energies, the cage effect virtually ensures reaction during each encounter; the steric factor no longer plays an important role, because the reacting molecules would sooner or later become properly oriented for reaction during the collisions. Under * We speak of molecular collisions in the gas phase and molecular encounters in solution. The difference is that after each encounter in solution, the molecules might collide many times before they move away from each other. Figure 15.15 Solute molecules ^red spheresh diffuse into a solvent “cage” and encounter each other. Hundreds of collisions between solute molecules occur before the cage is destroyed. 40 706 Chapter 15: Chemical Kinetics 0.6 4 0.4 2 Figure 15.16