Have you noticed that during discussions of chemical equilibria, reaction direction, spontaneity, and other topics there is no mention of how fast a reaction proceeds? Even the chapter on thermodynamics does not investigate the rate of reaction. As a matter of fact, many texts specifically state that rate of reaction is not tied to thermodynamic considerations. The branch of chemistry that treats the rates of reactions is chemical kinetics. There are two main objectives in this chapter. The first objective is to provide a systematic approach for dealing with data relating to the dependence of the rates of reactions on controllable variables. The second objective is to show the relationship between reaction rate and the reaction’s molecular mechanism.
Homogeneous reactions are those reactions that occur within a single phase (physical state), especially a liquid or gaseous phase. Heterogeneous reactions are those reactions that take place at least in part at the interface between two phases, such as solid and liquid or liquid and gas, etc. The discussion and problems in this chapter will address homogeneous reactions unless otherwise stated. Please note that rate means that there is a time element involved which is usually expressed in seconds (s), minutes (min), or hours (h).
According to the law of mass action, the rate of chemical reaction at a constant temperature (amount of reaction per unit of time) depends only on the concentrations of the substances that influence the rate. The substances that influence the rate of reaction are usually one or more of the reactants, but can occasionally be a product. Another influence on the rate of reaction can be a catalyst that does not appear in the balanced overall chemical equation.
The dependence of the rate of reaction on the concentrations can often be expressed as a direct proportionality in which the concentrations may appear to the zero, first, or second power. The power to which the concentration of a substance appears in the rate expression (formula) is the order of the reaction with respect to that substance. Table 20-1 provides examples of rate expressions.
The order is not a result of the nature of the chemical equation; it is dependent on data collected from experiments. Those data are then used to determine a mathematical equation that fits, the rate equation. The overall order of a reaction is the sum of the orders with respect to various substances (the sum of the exponents). Further, the order of a reaction is stated with respect to a named substance in the reaction (see listing above).
The proportionality factor, k, called the rate constant or specific rate, is constant at a fixed temperature. The rate constant varies with temperature. There are dimensions to k, and the units should be expressed when the k-values are tabulated. The rate itself is defined as the change in concentration of a reactant or product per unit of time. If A is a reactant and C a product, the rate might be expressed as
where Δ[X] is the change in concentration (in math, Δ usually means a change) of X and Δt is the time interval over which the change is measured (or the experiment observed). If the rate changes rapidly, Δt will be small. The law of mass action applies only at very small values of Δt, in which case the notation is that of calculus:
This notation, then, would replace the form of ratio expression above. The minus sign is used to express the rate in terms of a reactant concentration (decreases during reaction) and a plus sign for a product concentration (increases during reaction), so that the rate is always a positive quantity.
EXAMPLE 1 For some reactions, there is no doubt in defining the rate, For example, in
the rate could be expressed as
since these three ratios are all equal when concentrations are in molarity and there are no competing or side reactions. In the following reaction, however,
the coefficients for the different substances are not the same. The concentration of H2 decreases three times as fast as that of N2 and NH3 is formed twice as fast as N2 is used up. Any one of the three ratios below could be used to specify the rate, but the choice should be specified because the numerical value of k depends on the choice. Note that the order does not depend on the choice, just the size of k.
In the particular case of a first-order reaction, where the rate is proportional to the concentration of the reactant [A], as in Table 20-1(1), the methods of integral calculus yield
The symbol e refers to the number on which natural logarithms are based, 2.71828. [A]0 is the concentration of [A] at the beginning of the experiment (initial concentration at t = 0). The logarithmic form (base 10) can be derived from the above.
The amount of time, t1/2, for the first-order reaction to go 50 percent toward completion is given by (20-2) as
The half-life, t1/2, is independent of the initial concentration of A. This independence of [A]0 is characteristic only of first-order reactions. Also, note that t1/2 and k are independent of the units in which A is expressed (although molarity is the most common unit, as indicated by [A]).
Fractional orders also occur (for example, (
)-order and (
)-order, in which the rate is proportional to [A]1/2 or [A]3/2, respectively). Some reaction rates cannot be expressed in the proportional form shown in Table 20-1 at all. An example of a complex rate expression is the following:
The rate of a heterogeneous reaction may be proportional to the interfacial area of contact between the phases, as well as to the concentrations of reactants within a particular phase, as in the case of many reactions catalyzed at surfaces.
The temperature dependence of a reaction rate may be represented by the Arrhenius equation:
The pre-exponential factor, A, is also called the frequency factor, and Ea is the energy of activation. The units of Ea (and of RT) are J/mol or cal/mol. Both A and Ea may be considered to be constant for at least a narrow temperature range. From (20-4), the rate constants at two different temperatures are related by
It has been said that a chemical equation accurately portrays the nature of the starting and final materials in a chemical reaction, but that the arrow conveniently covers our ignorance of what happens in between. Troubling questions can be raised: How many successive steps are there in the overall process? What are the spatial and energetic requirements for the interactions in each step? What is the rate of each step? A positive point is that rate measurements usually describe the overall reaction, but the measurement of rates under different conditions can often fill in the information needed for the understanding of reaction mechanisms.
The concept of molecularity specifies the number of molecules interacting in an individual reaction step. A unimolecular reaction is a step in which a single molecule spontaneously undergoes a reaction. A bimolecular reaction refers to the reaction of two molecules. In a termolecular reaction, three molecules interact in a single step. There are no known single-step reactions of higher molecularity than termolecular.
A unimolecular reaction is first-order in the species that undergoes the spontaneous rearrangement or decomposition. This is true because there is a natural probability that a molecule will undergo the reaction in a specific time interval. Also, the overall rate per unit volume is the product of this probability per unit time and the total number of molecules per unit volume (which in turn is proportional to concentration).
A bimolecular reaction rate is proportional to the frequency of collisions between the two molecules of the reacting species. It is known from kinetic theory that the frequency of collisions between two like molecules, A, is proportional to [A]2, and the frequency of collisions between an A and a B molecule is proportional to the product of the concentrations, [A][B]. If the species whose molecules collide are starting materials in limited concentrations, the reaction is second-order. This reaction follows the rate equation of either type (3) or (2), Table 20-1.
A termolecular reaction rate is proportional to the frequency of the collision of three bodies. These reactions are proportional to [A]3, [A]2[B], or [A][B][C], depending on whether the molecules taking part in the reaction are of one, two, or three species. If these species are the starting materials, the reaction is third-order; however, termolecular reactions are rare because of the low probability that three species will come together.
Although the order of a single step can be predicted from the molecularity, the molecularity of a step, or steps, cannot be predicted from the order of the overall reaction. There are a number of complications which make it impossible to automatically determine the order of the reaction. In other words, stating that a unimolecular reaction is first-order, that a bimolecular reaction is second-order, or a termolecular reaction is third-order cannot be made with ensured accuracy. In many cases, the reaction is a sequence of steps, and the overall rate is governed by the slowest step(s). Experimental conditions might interchange the relative speeds of different steps, making the order appear to change. Another complication arises because the slowest (rate-limiting) step may involve the reaction of an unstable intermediate species, and it is necessary to express the concentration of this species in terms of the reactants before the overall order of the reaction can be determined. The solved problems provide examples of some of these complexities.
Most molecular collisions do not result in reaction. Even if the appropriate number of molecules came together, only those possessing enough energy can undergo the violent distortions of bond lengths and angles necessary for the rearrangements that lead to chemical reaction. Often times, the amount of energy necessary is a great deal more than the average energy of the molecules. The energy of activation, Ea, is a measure of the energy necessary to get the molecules to react. The exponential term in the Arrhenius equation, (20-4), is of the order of magnitude of the fraction of molecules possessing the necessary amount of energy beyond the average.
The activated state has been defined as that distorted combination of reacting molecules having the minimum amount of excess energy (beyond the average energy of the molecules) so that the combination, or complex, could as easily rearrange to form the products as revert to the reactants. The energy of the activated state may be thought of as the potential energy of a mountain pass which a climber must cross in order to go from one side of the mountain to the other. For a set of reversible reactions, the same activation energy must be crossed for the reactions in the two directions. Then, for reactions carried out at constant pressure:
The factor, (1 mol), has been inserted in (20-6) so that the enthalpy change (ΔH) will, as usual, have the units of energy. The relationship between the energy of the reactants, the energy of activation required, and the energy of the products is graphically presented in Fig. 20-1. When two or more fast-moving molecules collide, their kinetic energy is converted into the high energy of the complex. When the complex breaks up, the fragments (products or reactants, depending on the direction of the reaction) fly apart, their kinetic energy coming from that energy stored in the complex.
Fig. 20-1 Energy relationships in a reaction (forward reaction is exothermic)
20.1. In a catalytic experiment involving the Haber process, N2 + 3H2 → 2NH3, the rate of reaction was measured as
If there were no side reaction, what was the rate of reaction expressed in terms of (a) N2, (b) H2?
(a) From the coefficients in the balanced equation, Δn(N2) = – Δn(NH3). Therefore
(b) Similarly,
20.2. Using concentrations expressed in molarity and time in seconds, what are the units of the rate constant, k, for (a) a zero-order reaction; (b) a first-order reaction; (c) a second-order reaction; (d) a third-order reaction; (e) a half-order reaction?
In each case, we write out the full equation and find the units of k that will satisfy the equation.
(a)
Note that the units of Δ[A], the change in concentration, are the same as the units of [A] itself; similarly for Δt.
(b)
First-order reactions are the only reactions for which k has the same numerical value, regardless of the units used for expressing the concentrations of reactants or products.
(c)
and
then,
Note that the units of k depend on the total order of the reaction, not the way the total order is composed of the orders with respect to different reactants.
(d)
(e)
20.3. Ozone is one of the indicators of polluted air. Suppose the steady-state ozone concentration is 2.0 × 10–8 mol/L, and the hourly production of O3 by all sources is estimated to be 7.2 × 10–13 mol/L. Assume the only mechanism for the destruction of O3 is the second-order reaction 2O3 → 3O2. Calculate the rate constant for the destruction reaction defined by the rate law for –Δ[O3]/Δt to maintain the steady-state concentration.
At the steady state, the rate of destruction of O3 must equal the rate of its generation, which is 7.2 × 10–13 mol · L–1 · h–1. From the second-order rate law,
20.4. A viral preparation was inactivated in a chemical bath. The inactivation process was found to be first-order in virus concentration, and at the beginning of the experiment 2.0% of the virus was found to be inactivated per minute. Evaluate k for the inactivation process in units (1/s).
From the first-order law,
It is seen that only the fractional change in concentration, –[A]/[A], is needed; namely, 0.020 when Δt = 1 min = 60 s. This form of the equation may be used for the initial rate when the value of [A] is not changing appreciably. This condition is met when only 2% is inactivated in the first minute.
20.5. For the process described in Problem 20.4, how much time would be required for the virus to become (a) 50% inactivated, (b) 75% inactivated?
The method used in Problem 20.4 cannot be used here because [A] changes appreciably over the course of the reaction. Equation (20-1), (20-2), or (20-3) is appropriate here.
(a) The time for 50% reaction is the half-life, and (20-3) gives
(b) Equation (20-2) may be used. If 75% of the virus is inactivated, the fraction remaining, [A]/[A]0, is 0.25.
An alternate solution is to apply the half-life concept twice. Since it takes 35 min for half of the virus to become inactivated, regardless of the initial concentration, the time for the virus to be reduced from 50% to 25% full strength will be another half-life. The total time for reduction to (
) strength is two half-lives (70 minutes).
Similarly, the total time required to reduce the initial activity to (
) is three half-lives; to (
), four half-lives; and so on. This method can be used only for first-order reactions.
20.6. The fermentation of sugar in an enzymatic solution is under investigation. The beginning concentration is 0.12 M. The concentration of the sugar is reduced to 0.06 M in 10 hours and 0.03 M in 20 hours. What is the order of the reaction and what is the rate constant in 1/h and 1/s?
This problem is similar to Problem 20.5. Since doubling the time reduces the amount of sugar by half, it must be a first-order reaction. Alternatively, the reduction of the sugar concentration from 0.06 M to 0.03 M may be thought of as a new experiment with an initial concentration of 0.06 M. Since the same half-life (10 hours) was observed in both experiments, the reaction must be first-order because only in a first-order reaction is the half-life independent of the initial concentration. The rate constant may be evaluated from the half-life using equation (20-3).
20.7. A reaction between substances A and B is by the reaction, A + B → C. Observations on the rate of this reaction are obtained in three separate experiments as follows:
What is the order with respect to each reactant, and what is the value of the rate constant?
Let us set up the information for the initial rate of the reaction for each experiment, noting that Δ[A] = [A]f – [A]0 in each case is small enough in size to allow the rate to be expressed in terms of changes over the entire time of the experiment.
In comparing experiments (1) and (2), we notice that [A] is the same in both, but [B] is twice as great in (2). Since the rate in (2) is 4 times that in (1), the reaction must be second-order in B.
When comparing experiments (1) and (3), we see that [B] is the same in both, but [A] is twice as great in (1). Since the rate in (1) is twice that in (3), the reaction must be first-order in A.
The rate equation may be written accordingly as
and k may be evaluated from any of the experiments. We take (1) as an example, using average values of [A] and [B]. ([B] does not change appreciably during the experiment since it is in such large excess.)
or 5.1 × 10–2 L2 · mol–2 · s–1. We suggest that you confirm that the same result will be obtained by using data from either of the other two experiments.
20.8. Acetic anhydride and ethyl alcohol react to form an ester (an organic salt) by the following:
When the reaction is carried out in dilute hexane solution, the rate may be represented by k[A][B]. When ethyl alcohol (b) is the solvent, the rate may be represented by k[A]. (The values of k are not the same in the two cases.) Explain the difference in the apparent order of the reaction.
When a solvent is also a reactant, its concentration is so large compared with the extent of reaction that it does not change. Since this is the case, the dependence of the rate on the concentration of ethyl alcohol cannot be determined unless ethyl alcohol becomes a solute in some other solvent. If another substance is the solvent, then the concentration of the alcohol can be varied, allowing the calculation. For the reaction in ethyl alcohol (ethanol), kexp = k[B] with [B] essentially constant. Such reactions are called pseudo-first-order reactions.
20.9. At elevated temperatures, gaseous chloroethane decomposes into ethene and HCl according to
A constant-volume, constant-temperature experiment was run and the following data were collected:
Test the data to determine whether or not the reaction is first-order. If the reaction is first-order, determine the rate constant.
Since the reaction is essentially complete after 100 hours, the initial concentration of chloroethane, [A]0, was 1.19 × 10–2 M and subsequent concentrations, [A], can be obtained by subtracting the ethane concentrations from this. Then, we substitute into the ratio, [A]/[A]0, and take the log. These steps are provided in the table below. (The 50-hour point, which corresponds to the least accurate value of the ratio, is omitted.)
A graph of log([A]/[A]0) vs. t is shown in Fig. 20-2. Since the points fall in nearly a straight line, the reaction is first-order. The slope of the graph, calculated on the figure, is –0.0321 h–1, which, according to equation (20-2), is equal to – k/2.303. This gives us k = 7.2 × 10–2h(2.0 × 10–5 s–1). Our numerical work can be roughly checked by inspecting t1/2 = 0.693/k = 9.6 h, and in the original table of data, in 10 h the reaction is slightly more than half completed.
Fig. 20-2 First-order test for data of Problem 20-9
20.10. The complexation of Fe2+ with the chelating agent dipyridyl (abbreviated dip) has been studied kinetically in both the forward and the reverse directions. For the complexation reaction
the rate of formation of the complex at 25°C is given by
and, for the reverse of the above reaction, the rate of disappearance of the complex is
What is Ks, the stability constant, for the complex?
Not all reactions can be studied conveniently in both directions. When it is possible, we know that the rate of formation of the complex must equal the rate of decomposition at equilibrium. This is particularly true since the concentrations of the various species stay constant (do not change).
Solving,
(In the equilibrium constant equation, as opposed to rate equations, K does not have dimensions, being the concentration relative to the standard state of 1 M.)
20.11. The decomposition of N2O into N2 and O in the presence of gaseous argon follows second-order kinetics as indicated below. What is the energy of activation of this reaction?
Comparing the equation for k in this case with (20.4), we note that the exponent of e is –Ea/RT.
20.12. The first-order rate constant for the hydrolysis of CH3Cl (methyl chloride) in H2O has a value of 3.32 × 10–10 s–1 at 25°C and 3.13 × 10–9 s–1 at 40°C. What is the value of the energy of activation?
Solve equation (20-5) for Ea.
20.13. A second-order reduction whose rate constant at 800°C was found to be 5.0 × 10–3 L · mol–1 · s–1 has an activation energy of 45 kJ · mol–1. What is the value of the rate constant at 875°C?
Equation (20-5) is solved for k2, the rate constant at the higher temperature.
20.14. The trans → cis isomerization of 1,2-dichloroethylene proceeds if supplied with the energy of activation of 55.3 kcal · mol–1. ΔH associated with the reaction is 1.0 kcal. What do you predict is the value of Ea for the reverse isomerization, cis → trans?
From equation (20-6),
20.15. A gaseous molecule, A, can undergo a unimolecular decomposition into C, if it is supplied with a critical amount of energy. An energized molecule of A, designated as A*, can be formed by a collision between two ordinary A molecules. Competing with the unimolecular decomposition of A* into C is the bimolecular deactivation of A*by collision with an ordinary A molecule.
(a) Write a balanced equation and the rate law for each of the above steps.
(b) Making the assumptions that A* disappears by all processes at the same rate at which it is formed, and that [A*] is much less than [A], what would be the rate law for the formation of C in terms of [A] and constants of individual steps only?
(c) What limiting order in A would the formation of C show at low pressures of A, and what limiting order at high pressures?
(a)
(b) Notice that A* appears in each of the three separate steps. The net change in [A*] can be evaluated by taking the summation of the three steps.
We can make a steady-state assumption by assuming that the net rate of change of [A*] is zero. Then, the right side of the equation has the value of zero.
Inserting this value into the rate law for step (3), and recognizing that – Δ[A*] = Δ[C] for this step,
Then, the formation of C follows a complex kinetic formulation and is definitely not represented by a simple order.
(c) At very low pressure (meaning small [A]), the second term in the denominator on the right side of (4) becomes insignificant when compared with the first term. After making the deletion of the term,
At very high pressures, the first term of the denominator becomes negligible in comparison with the second term. After making the deletion of the term,
Now, it appears as though the reaction is first-order with k3k1/k2 as the apparent rate constant.
This problem illustrates the concept of the rate-limiting step. The example commonly used to teach the concept is that of getting water from a source to a fire. A bucket brigade is formed for passing water from one person to the next, and so on, until it finally reaches the fire. The overall rate of transfer of water cannot be greater than the rate of the slowest person in the brigade. In the above reactions, the activation step (1) becomes the slowest step when [A] is low and the reaction step (3) becomes the slowest step when [A] is high. Step (1), depending on the square of [A], is more sensitive to pressure than step (3).
20.16. For the hydrolysis of methyl formate, HCOOCH3, in acid solutions, the reaction and rate are
[H+] is not in the equation for the reaction; propose an explanation of why [H+] appears in the rate law?
[H+] is a catalyst for the reaction. It is actually a reactant in an early intermediate stage of the reaction and is then released back into the solution at a later stage.
20.17. The conversion of the D-optical isomer of gaseous
into the L-isomer in the presence of iodine vapor follows the law rate = kP(a)P(I2)1/2, where A represents the D-isomer. (Partial pressures are a reasonable form of expression of concentrations of gases in rate laws.) Suggest a mechanism that could account for the fractional order.
I2 can undergo a slight dissociation into I atoms in a quickly established equilibrium.
The partial pressure of iodine atoms can be evaluated in terms of this equilibrium as
If an intermediate stage of the reaction involves the addition of an iodine atom to A followed by a loss of the iodine atom that was initially a part of the A molecule, and if the addition of I to A is bimolecular (rate constant k2) and the step is slow, it determines the overall rate.
The numerical value of the term in parentheses, 
is the numerical value of the apparent rate constant (experimental) for the overall 
-order reaction.
The above mechanism is plausible and is consistent with the observations. We cannot be sure from the kinetic data alone that there is not some other mechanism which is also consistent with the observations. Other types of experimentation are needed to confirm a mechanism based on rate data. However, a proposed mechanism that yields a rate law other than the observed rate law can definitely be eliminated from consideration.
20.18. For the reaction
in alkaline aqueous solution, the value of the second-order (in BrO–) rate constant at 80°C in the rate law for – Δ[BrO–]/Δt was found to be 0.056L · mol–1 · s–1. What is the rate constant when the rate law is written for (a) 
and for (b) Δ [Br–]/Δt?
Ans. (a) 0.0187 L · mol–1 · s- 1; (b) 0.037 L · mol–1 · s–1
20.19. The hydrolysis of methyl acetate in alkaline solution is
The reaction followed the rate = k[CH3COOCH3][OH–], with k = 0.137 L · mol–1 · s–1 at 25°C. A reaction mixture was prepared to have initial concentrations of methyl acetate and hydroxide ion of 0.050 M each. How long would it take for 5.0% of the methyl acetate to be hydrolyzed at 25°C?
Ans. 7.7 s
20.20. A first-order reaction in aqueous solution was too fast to be detected by a procedure that could have followed a reaction having a half-life of at least 2.0 ns. What is the minimum value of k for this reaction?
Ans. 3.5 × 108s–1
20.21. Gaseous cyclobutene isomerizes to butadiene in a first-order process which has a k-value at 153°C of 3.3 × 10–4 s–1. How many minutes would it take for the isomerization to proceed 40% to completion at this temperature?
Ans. 26 min
20.22. The approach to the following equilibrium was observed kinetically from both directions:
At 25°C, it was found that at 0.3 ionic strength:
What is the value of K′4 for the complexation of the fourth Cl– by Pt(II) (the apparent equilibrium constant for the reverse of the reaction as written above) at 0.3 ionic strength?
Ans. 54
20.23. The following reaction was studied at 25°C in benzene solution containing 0.1 M pyridine:
The following sets of data were obtained in three separate experiments:
What rate law is consistent with the above data and what is the best average value for the rate constant expressed in seconds and molar concentration units?
Ans. Rate = k[A]2[B], k = 4.6 × 10–3 L2 · mol–2 · s–1
20.24. The decomposition of substance A at a steady temperature was followed closely, and the data were recorded below. Is the reaction first-order or second-order in A? Suggestions: First plot log([A]/[A]0) against t for several points. Then calculate the ratio of (Δ[A]/Δt)/[A]2 and compare the results over several short intervals.
Ans. Second-order reaction
20.25. The rate constant for the first-order decomposition of ethylene oxide into CH4 and CO may be described by the equation
(a) What is the energy of activation for this reaction? (b) What is the value of k at 670 K?
Ans. (a) 239kJ × mol–1; (b) 5 × 10–5 s–1
20.26. The first-order gaseous decomposition of N2O4 into NO2 has a k-value of 4.5 × 103 s–1 at 1°C and an energy of activation of 58 kJ × mol–1. At what temperature would k be 1.00 × 104 s–?
Ans. 10°C
20.27. Biochemists often define Q10 for a reaction as the ratio of the rate constant at 37°C to the rate constant at 27°C. What must be the energy of activation for a reaction that has a Q10 of 2.5?
Ans. 71 kJ · mol–1
20.28. In gaseous reactions important for the understanding of the upper atmosphere, H2O and O react bimolecularly to form two OH radicals. ΔH for this reaction is 72 kJ at 500 K and Ea is 77 kJ · mol–1. Estimate Ea for the bimolecular recombination of two OH radicals to form H2O and O.
Ans. 5kJ · mol–1
20.29. H2 and I2 react bimolecularly in the gas phase to form HI. HI then decomposes bimolecularly into H2 and I2. The energies of activation for these two reactions were observed to be 163 kJ · mol–1 and 184 kJ · mol–1, respectively, at the same temperature of 100°C. What do you predict from these data for ΔH of the gaseous reaction H2 + I2 
2HI at 100°C?
Ans. –21 kJ
20.30. Predict the form of the rate law for the reaction 2A + B → products, if the first step is the reversible dimerization of A (2A A2), followed by reaction of A2 with B in a bimolecular step. Assume that the equilibrium concentration of A2 is very small compared with [A].
Ans. Rate = k[A]2[B]
20.31. The following reaction was observed in aqueous solution:
and the rate was found to be of the form k[Cu2+]2 [CN–]6. If the first step is the rapid development of the complexation equilibrium to form the relatively unstable [Cu(CN)3]–, what rate-limiting step could account for the observed kinetic data? (The ion, [Cu(CN)3]–, is unstable with respect to reversal of the complexation step.)
Ans. Bimolecular decomposition: 2[Cu(CN)3]– → 2[Cu(CN)2]– + (CN)2
20.32. The hydrolysis of (i-C3H7O)2POF was studied at different acidities. The apparent first-order rate constant, k, at a particular temperature was found to depend on pH but not on the nature or concentration of the buffer used to regulate the pH. The value of k was fairly constant from pH 4 to pH 7, but rose from this constant value with decreasing pH below 4 or with increasing pH above 7. What is the nature of the cause of this behavior?
Ans. The reaction is catalyzed by either H+ or OH–.
20.33. It has been found that the rates of reaction of a ketone in mildly basic solution are identical for the following three reactions: (a) reaction with Br2 leading to the substitution of an H on the ketone by a Br; (b) conversion of the D-isomer of the ketone into a mixture of D- and L-isomers of equal concentration; and (c) isotopic exchange of a hydrogen atom on the carbon next to the C=O group of the ketone by a deuterium atom in the solvent. The rate of each of these reactions is equal to k[ketone][OH–] and is independent of [Br2]. What can be concluded about the mechanism from these observations?
Ans. The rate-determining step for all three reactions must be the preliminary reaction of the ketone with OH–, probably leading to the conjugate base of the ketone. The conjugate base subsequently reacts very rapidly with (a) Br2; (b) some acid in the medium; or (c) the solvent with deuterium.
20.34. Many free radical reactions (polymerization and others) can be initiated by a light-sensitive substance, P. The molecules of P dissociate into two free radicals on the absorption of a photon. The following reactions are either propagation, in which one radical, R·, is produced for each radical consumed, or termination, in which two radicals combine to form products that are not radicals. The bulk of the chemical change is due to the propagation step. A small steady-state concentration of free radicals prevails throughout. Explain how such a mechanism accounts for the dependence of the rate on the square root of the light intensity, I.
Ans. Since propagation does not change the free radical concentration, the rate of initiation must equal the rate of termination. If the rate of initiation is first-order in P and proportional to the light intensity, ki [P]I, and the rate of termination is a bimolecular reaction with a rate, kt [R·]1/2, then at the steady state,
If the propagation step is first-order in [R·], i.e., proportional to [R·], then the observed rate is as shown above, proportional to I1/2.
20.35. The addition of a catalyst to a certain reaction provides an alternative pathway in which the energy of activation is less than that of the uncatalyzed reaction by 14.7 kJ/mol. By what factor is the rate constant increased by the catalyst at a temperature of 420 K? By what factor is the rate constant of the reverse reaction increased? (Assume that the frequency factor is not affected by the catalyst.)
Ans. 67, the same as for the reverse reaction
