Chapter 15: Principles of Chemical Dynamics

Note that the method of finite differences, like the method based on integrated rate laws, uses the full range of concentration-time data, thus avoiding the major problem with the method of initial rates. For the method to be successful, however, it is important that the time intervals over which average rates are calculated be short relative to the total reaction time. In the example above, the 10-second time intervals used in the early stages of the reaction, when concentration changes most rapidly, are small compared with the overall reaction time of 240 seconds. As a useful guideline, time intervals should not exceed 5% of the total reaction time; otherwise inaccurate rate constants will result from the finite difference approach.

Electronic spreadsheets enable the calculations required for the finite difference method to be performed rapidly and conveniently using a computer. Further, the various data plots can be made quickly and professionally using the spreadsheet. With this modern tool, calculation by finite difference methods is routine. We will consider one additional example of the approach.

Example 15-9. The decomposition of N₂O occurs according to the following equation:

In a kinetics run, the concentration of N₂O was found to change with time as indicated in the first 2 columns of Table 15-1. The numbers in the remaining columns, calculated using an electronic spreadsheet, represent the average [N₂O] over each time interval, squares of the N₂O concentrations, and the average rates of N₂O concentration change over the successive time intervals. Figure 15-5a and b show spreadsheet-generated plots of Rate versus [N₂O] and Rate versus [N₂O]², respectively. Plot a shows slight but distinct curvature and in addition has a non-zero intercept. The linearity of the plot and lack of intercept in b confirms that the reaction follows the rate law in 15-3-13:

The rate constant, obtained from the slope of the plot, is 6.0 * 10^-3 L/mole-min.

Pseudo-Order Methods. Many chemical reactions obey a mixed second-order rate law of the type shown in 15-3-14:

The reaction is first order in each of two reactants. An example of a reaction governed by this rate law is given in 15-3-15.

Suppose that a kinetics experiment is carried out by preparing a solution containing similar initial concentrations of the two reactants, NH₄⁺ and NO₂^-, and monitoring the appearance of N₂(g) as a function of time in order to measure the rate. Since the two reactants are present in similar concentrations, both substances undergo significant decreases in concentration over the course of the reaction. The rate of reaction reflects this simultaneous decrease in concentration of both reactants. Consequently, a plot of Rate versus [NH₄⁺] (NH₄⁺ = A) will be non-linear, even though the reaction is first order in NH₄⁺, because the Rate also responds to the change in concentration of NO₂^- (B) significantly during reaction. Although it is possible to integrate 15-3-14 to obtain a rather complex equation that may then be used to plot the data, a simpler approach to the kinetics study is often possible. In this approach, a kinetics run is performed on a solution that initially contains a substantial excess of one of the reactants, say B. If the initial concentration of B is more than 10 times larger than the initial concentration of A, the concentration of B will be nearly the same at the end of the reaction as at the beginning; consequently it may be considered to be constant during reaction. To illustrate this, consider a solution initially containing 0.002 M NH₄⁺ and 0.10 M NO₂-. If reaction is allowed to proceed until all NH₄⁺ is consumed, the concentration of NO₂- will have decreased by only 0.002 M to 0.098 M:

Thus the concentration of NO₂- remains essentially constant while NH₄⁺ is consumed, and the rate law simplifies as shown in 15-3-16:

Here the constancy of [NO₂-] has been recognized, and this concentration has been lumped into the rate constant to produce a new constant k'. The rate law now has a simple first-order form, even though the reaction is actually second-order. This simplified rate law, which may be integrated or subjected to finite difference analysis, is called a pseudo-first order rate law. The prefix, pseudo-, reminds us that the reaction is not truly first-order, but has only been made to appear that way by imposition of a large concentration of NO₂-.

If one carries out kinetic runs using several (relatively large) concentrations of [B], a pseudo-first order rate constant, k', will be obtained for each run. Of course, these will have different values depending on [B]. A plot of k' versus [B] is then made to obtain the true rate constant, k, from the slope.

Example 15-10. Pseudo-First Order Kinetics. For the reaction of NH₄⁺ with NO₂-, 15-3-15, several kinetics runs were carried out using the initial reactant concentrations in the first two columns of the table below. Plots of Rate versus [NH₄⁺] for each run were found to be linear, indicating that the reaction is first-order in NH₄⁺. The first-order rate constants obtained from these plots varied with [NO₂^-] as indicated in the third column of the table. What is the order of the reaction in NO₂-? What is the true rate constant for the reaction?

Solution. The order of reaction with respect to NH₄⁺ is known to be 1, based on linear Rate vs. [NH₄⁺] plots. We thus focus on the relationship between the observed pseudo-first-order rate constants and the concentration of [NO₂-]. Note that for all three runs, the initial NO₂- concentration is much larger than the initial [NH₄⁺], guaranteeing that [NO₂-] remains essentially constant during the entire course of reaction. A linear plot of k' versus [NO₂-] raised to a power, n, indicates that the reaction is n-order in [NO₂-]. We can determine n by trying several plots, first with n = 1, then, if necessary, with n = 2, 3, and so on, until we obtain linearity.

Plots of k' versus [NO₂-] and [NO₂-]² are shown in Figure 15-6. The top plot is linear, so the reaction is first order in [NO₂-]. The overall second-order rate constant, k = 3.6 * 10^-4 L/mole-s, is obtained from the slope.

The Experimental Measurement of Concentration. In order to carry out a kinetics study, it is necessary to have an experimental "handle" on at least one of the reactants or products; that is, the change in concentration of a reactant or product must be accompanied by a change in some measurable property of the system. For reactions involving gases, the total pressure of the reaction mixture changes as reaction proceeds, provided that there is a net change in the amount of gas in the system. Thus, the rate of reaction 15-3-17 could be studied by monitoring the total pressure with time. Total pressure decreases as reaction proceeds because only 1 mole of gas forms for each 2 moles consumed.

On the other hand, no change in total pressure accompanies 15-3-18 because there is no change in moles of gas; reaction rate can not be studied by monitoring total pressure in this case.

Rather than measuring system pressure at constant volume, one could measure system volume at constant pressure. In either case, the measured quantity is proportional to the number of moles of gas in the system at a particular time during reaction. Pressure (or volume) measurements are particularly convenient when only a single gaseous reactant or product is involved. In this case, the pressure (or volume) provides a direct measure of the amount of that reactant or product. An alternative to pressure or volume measurements is to use gas chromatography to monitor the amounts of gaseous reactants and products as a function of time. The gaseous reaction mixture is periodically sampled with a syringe and injected into a gas chromatograph. In the ideal situation, it may be possible to simultaneously determine the amounts of all reactants and products at each sampling time.

For reactions carried out in solution, a number of methods are used to monitor concentration. For 15-3-19, the acidity of the solution changes during the course of reaction. Provided the reaction is slow enough, the kineticist could periodically "sample" an aliquot of reaction solution and carry out an acid-base titration using a suitable indicator to determine the [H₃O⁺] present at the known sampling time.

Alternately, since ions are produced but not consumed in the reaction, the kinetics could be monitored by measuring the electrical conductivity of the solution as a function of time. The measured conductivity values can then be related to the total concentration of ions present at that time. Perhaps the most common method for following the time course of reactions in solution is to monitor the absorption of light by the reacting solution. If one of the reactants (or products) has an absorption band in the UV-visible, infrared, or NMR region of the electromagnetic spectrum, the change with time of the amount of light absorbed is a measure of the rate of appearance or disappearance of the absorbing substance. The kinetics of 15-3-20 can easily be studied by this method.

The product, Cr(EDTA)^-, has an absorption maximum at a wavelength of 542 nm in the visible region of the spectrum. As reaction proceeds, absorbance at this wavelength increases in direct proportion to the concentration of Cr(EDTA)^-, providing a "handle" on the reaction kinetics.

15-4 Elementary Processes and Reaction Mechanism. A balanced chemical equation is intended to represent the net transformation of reactants to products that occurs when reaction takes place. It does not, and is not intended to, tell us anything about the details of the reaction pathway. In fact, most reactions occur via a series of relatively simple steps, each involving reaction of only a few (one, two, and very occasionally, three) molecules. Each of these simple steps is called an elementary process. The equation for an elementary process represents an actual collision event at the molecular level. The sequence of elementary processes by which a reaction is proposed to occur is called the mechanism for the reaction. For example, the reaction in 15-4-1 is thought to occur by the 2-step mechanism in 15-4-2a and 15-4-2b.

with the first step proposed to be slow, the second one fast. Each of the two steps in the mechanism is understood to be an elementary process. We can make several important generalizations about the mechanism in 15-4-2. First, each elementary process involves reaction (collision) of only two species, even though the overall reaction involves 3 moles of reactants. Second, F atoms, produced in the first step and consumed in the second, do not appear in the overall equation for the reaction. Such a species--first produced, then consumed--is called a reaction intermediate. Third, the elementary processes add to give the overall equation for the reaction. This must always be true.

It is important to realize that a mechanism for a reaction is a hypothesis about what goes on at the molecular level, made by the chemist based on her or his experimental observations. The overall equation for a reaction is NOT a hypothesis; it is an experimental fact, observable in the laboratory. The mechanism in all of its detail is seldom observable; it must be deduced from experiment, must be consistent with all known experimental facts about the reaction, and is subject to change if in conflict with a new experimental fact.

Types of Elementary Processes. There are three types of elementary process, named for their molecularity--the number of reactant molecules that they involve. A unimolecular process involves only one reactant molecule. Two examples of unimolecular processes are given in 15-4-3 and 15-4-4.

A bimolecular process involves two reactant molecules, which may be the same or different. Examples are given in 15-4-5 and 15-4-6. Bimolecular elementary processes are by far the most common type.

A termolecular process requires the simultaneous collision of three reactant molecules. 15-4-7 is an example.

Such three-body collisions are extremely improbable. Consequently, termolecular processes are rare. Elementary processes with molecularity greater than three do not occur, due to the overwhelming improbability of collisions involving four or more molecules.

In contrast to the situation for the overall (net) reaction, the rate law for an elementary process may be written directly from its stoichiometry, because an elementary process is an actual, not a net, process. Thus for the elementary process,

the rate law is -d[A]/dt = k[A][B]. Unimolecular processes have first-order rate laws. For 15-4-3, the rate law is

Bimolecular processes have rate laws that are second-order overall; and termolecular processes have overall third order rate laws. For 15-4-5 and 15-4-7

In summary, an elementary process of molecularity n has a rate law of order n. The converse is NOT true: a first-order reaction is not necessarily either an elementary process or unimolecular.

The difference between order and molecularity is subtle but important. The concept of molecularity is appropriate only in the context of an elementary process, and indicates the number of reactant molecules involved in (undergoing collision) in the elementary step. The order of a reaction is the number to which the concentration of a reactant must be raised in the experimental rate law for the overall (net) reaction. This rate law, and the orders in it, do not necessarily correspond to a particular elementary step in the mechanism of the net reaction. We will say more about this shortly.

Rate Determining Step. A particular overall reaction consists of a sequence of elementary steps, which usually occur with different rates. Very often one of the elementary processes occurs much more slowly than do the others and limits the rate at which the overall reaction occurs. This slow step is then referred to as the rate-determining step (rds). The overall rate is equal to the rate of the rds. The fact that the overall reaction rate is limited by the rate of a particular elementary step is the reason why rate laws frequently do not correspond with the stoichiometry of the overall reaction.

A familiar example of an overall process occuring at a rate limited by a slow step is illustrated in Figure 15-7. Pictured is a theater filled to capacity, connected by a set of wide double doors to a smaller lobby, which is linked to the outside via a revolving door. When the movie is over, people begin to move from the theater to the lobby, and from there to the outside. We suppose that the double doors between theater and lobby are large enough so that 10 people per second can pass through; but that only 1 person per second can exit the revolving door. We then represent the overall process of leaving the theater in terms of 2 steps:

Initially, people pass from theater to lobby at the rate of 10 per second, and begin to exit the revolving door one by one. However, the lobby quickly fills up, because it cannot be emptied nearly as rapidly as it was filled. The process "piles up" behind the slow step, which is passage through the revolving door. The maximum rate at which people can leave the theater is 1 per second, which is the rate of the slowest step in the process.

Rate Law and Reaction Mechanism. Chemists are very interested in the details of how a reaction occurs; i.e., in the reaction mechanism. The approach to the elucidation of the mechanism of a reaction of known stoichiometry involves at least the following steps:

1. Do experiments to determine the experimental rate law for the overall reaction;
2. Propose a mechanism for the overall reaction, based on the observed rate law;
3. Derive the theoretical rate law implied by the proposed mechanism;
4. Compare the rate laws in 1 and 3. If they have the same form, it is safe to conclude that the mechanism is consistent with (not in conflict with) the experimental rate law. If they do not have the same form, go back to 2.

Step 2 makes use of the fact that the observed rate law indicates what atoms are involved as reactants in the rate-determining step of the mechanism. This will be made clear in specific examples to follow. It is the most difficult step of the above procedure, requiring extensive experience and well-developed "chemical intuition." At this level of study, we will not attempt to deduce mechanism from the rate law. Rather, our focus will be this: given the form of the experimental rate law, and one or more proposed mechanisms, decide whether each mechanism is or is not consistent with experiment.

Can either of these mechanisms be ruled out as being in conflict with experiment?

Solution. We examine mechanism 1 first. To obtain the theoretical rate law (implied rate law) for this mechanism, we proceed as follows:

1) Write the rate law for the rds. The rate of this step is the same as the rate of the overall reaction.

The rds for mechanism 1 is step 2.

Although this rate law is valid, it is not expressed solely in terms of concentrations of the reactants in the overall reaction. Since these concentrations are the ones we actually observe, we must somehow rewrite the rate law so as to eliminate the concentration of the intermediate, F, which is not generally observable, and replace it with concentrations of reactants in the net reaction. We can accomplish this making use of the preceding rapid steps, which we assume to occur rapidly enough to come to a pseudo equilibrium. So,

2) Use the rapid pre-equilibrium to solve for the concentration of the intermediate in terms of known concentrations. Since the first step of the mechanism is in equilibrium, the forward and reverse rates of this elementary process are the same, and we can write

Solving for [F] gives

This expression for the concentration of the intermediate can be substituted into the rate law for the rds:

This is the rate law implied by the mechanism. It is the rate law that the reaction should follow if the proposed mechanism is correct. Clearly, though, it is not in agreement with the experimental rate law, which is first order in NO₂, and shows no dependence on NO₂F. The conclusion is that Mechanism 1 can NOT be correct; it is in conflict with experiment.

We next consider Mechanism 2, using the same general approach.

1) Write the rate law for the rds, which in this case is the first step in the mechanism.

Further work is unnecessary here, because the rds rate law is already expressed in terms of concentrations of participants in the overall reaction. The implied rate law for Mechanism 2 has the same form as the experimental rate law (first order in NO₂, first order in F₂, second order overall), and is therefore not in conflict with experiment.

Can we conclude, then, that Mechanism 2 is the correct mechanism for reaction 15-4-1? The answer is an emphatic NO. If a mechanism does not agree with experiment, it must be wrong; but just because a mechanism agrees with experiment does not mean that it is correct. A mechanism is a theory--it can never be proved. To underscore this point, consider the following possible mechanism for 15-4-1:

This also is consistent with experiment, giving the implied rate law

based on the first step, which is the slow one. You might like to assess the validity of yet a fourth possible mechanism, given below:

We can now appreciate the significance of a statement made earlier: the experimental rate law tells us what atoms are contained in the reactants of the rate-determining step. The experimental rate law in Example 15-11, Rate = k [NO₂][F₂], reveals that the reactants in the slow step must contain a total of 1 N atom, 2 O atoms, and 2 F atoms. Based on this, we can immediately rule out mechanism 1, for which the rds involves only one F atom; and mechanism 4, for which the same is true. Both mechanisms 2 and 3 have rds's containing the correct numbers of atoms, so neither is in conflict with experiment.

In general, chemists try to propose mechanisms that agree with the experimental rate law and with any other experimental evidence; and that are chemically reasonable. It is the assessment of chemical reasonability that requires experience.

We consider one additional example of mechanism assessment.

Example 15-12. Several mechanisms might be considered for the overall reaction

with rate law, Rate = -d[O₂]/dt = k_obs[NO]²[O₂]. The symbol, k_obs, where "obs" stands for "observed", is commonly used for an experimentally measured rate constant.

Can any of these mechanisms be ruled out?

Solution. Mechanism 3 can be eliminated right away, because the reactants in the rds do not contain the correct numbers of atoms. The experimental rate law demands that there be two N atoms and four O atoms in the rds reactants; however, only one N and three O atoms are involved in the rds of mechanism 3. Mechanisms 1 and 2 show the correct atoms in the rds. You should confirm that the implied rate laws are

both of which are consistent with experiment. Attempts to draw Lewis structures for the intermediates, N₂O₂ and OONO, are shown in Figure 15-8. The intermediate OONO is a free radical (it has an odd electron), and should therefore react very rapidly with the free radical NO. It is thus unlikely that the second step of Mechanism 2 would be slow. Mechanism 1 is preferable, based on chemical reasonability. There is yet a fourth mechanism, involving a one-step termolecular process. It is currently believed that this is the correct one.

Let's summarize what we have just learned about reaction mechanisms:

• The steps in a mechanism are numbered from 1 to n. The subscripts on the elementary process rate constants are given the same numbers, with k_n being the forward rate constant and k_-n the reverse rate constant for step n. The steps must add to give the overall reaction.
• The mechanism often (but not always) involves at least one slow step. This is the rate-determining step.
• Any fast step preceding the slow step is treated as reversible, and assumed to come to equilibrium.
• To obtain the rate law, we
  • Write the rate law for the elementary rds;
  • Eliminate concentrations of intermediates using rapid pre-equilibrium expressions. This is necessary because the concentrations of intermediates are not usually observable.
• That the reaction proceeds in several steps having different rate constants can cause its rate law to deviate from the overall stoichiometry. However, the rate law tells us which atoms are involved in the reactants of the rate determining step, so is strongly suggestive of mechanism.
• Not all elementary step rate constants need appear in the implied rate law for the mechanism.
• Mechanisms can be disproven, but not proven.

15-5 Temperature Dependence of Rate--Kinetic Molecular Theory. The rates of chemical reactions increase with increasing temperature, often dramatically. Coal reacts with oxygen at a negligible rate at room temperature, but burns spontaneously with the liberation of heat and light at 500 ^oC. A sirloin steak "cooks" (oxidizes) very slowly at room temperature, but requires only 20 minutes at 180 ^oC. Reactions are accelerated by temperature increase because the elementary processes by which they occur take place more rapidly. We begin this section with two questions:

• Why are some elementary processes fast and others slow at a particular temperature?
• Why are reaction rates (rate constants) so dramatically temperature dependent?

We can arrive at answers to these questions by looking in more detail at the collision process.

Requirements for Collision to Lead to Reaction. As discussed earlier, rate is proportional to the number of molecular collisions occuring per unit volume per unit time. For the decomposition of hydrogen iodide to hydrogen and iodine, shown in 15-5-1, assuming that [HI] = 0.001 M and a temperature of 500 ^oC, the rate would

be 5.8 * 10⁴ M/s if all collisions led to formation of products. The observed rate, however, is only 1.2 * 10^-3 M/s, 5 * 10⁷ times slower than it would be if all collisions between HI molecules were effective. Our conclusion must be that only a small fraction of molecular collisions lead to formation of products; i.e., are effective. In general, for the elementary process

Rate = Number of effective collisions/volume-time, which should be proportional to the total number of collisions per unit volume per unit time. We call this total number of collisions Z. We can relate Z to the concentrations of A and B via equation 15-5-3:

where Z_o is the number of collisions per volume per time when A and B have 1 molar concentrations. The number of effective collisions is less than the total number of collisions for two reasons. First, two colliding molecules must be properly oriented if they are to react. Reaction requires that bonds within reactants be broken and that new bonds form between atoms in the colliding molecules. A bond cannot form between atoms in separate molecules unless they are near each other during collision. Figure 15-9 shows two of the many possible collision orientations of CO and NO₂, which are reactants in 15-5-4.

This bimolecular elementary process involves transfer of an oxygen atom from NO₂ to CO. For this to occur, it is necessary that an oxygen atom of NO₂ approach the carbon atom of CO during collision, so that a bond may form between them. This requirement is satisfied in the orientation in Figure 15-9a. The relative orientation in 15-9b brings the nitrogen atom of NO₂ and the oxygen atom of CO together, so cannot be effective in advancing the process in 15-5-4. Similarly, many other possible orientations of these two molecules are unsuitable for transfer of an oxygen atom from N to C, so will not be effective. We symbolize the fraction of collisions having the proper orientation of molecules as p, and recognize that p is a small number.

Every elementary process can be depicted as a single-humped potential energy curve, with an activated complex at the maximum. Letting the fraction of collisions with sufficient kinetic energy to react be f, then

We are now in a position to answer the first of the two questions posed at the beginning of this section. Some elementary processes are slower than others either because they have very restrictive orientation requirements, corresponding to a very small value for p; or they have high potential energy barriers (E_a), corresponding to a small value for f.

Next we consider the effect of an increase in temperature. The distribution of kinetic energies in a collection of molecules is governed by the Maxwell-Boltzmann distribution, first introduced in Chapter 5. The M-B distribution curve for temperature T₁ is shown in Figure 15-12. The value of the activation energy for elementary process 15-5-2 has been indicated on the graph. The fraction of molecules with kinetic energy greater than E_a is given by the area under the M-B curve to the right of E_a. As temperature is increased to some new value T₂, the M-B curve shifts to the right and flattens, indicating the increase in both the average molecular kinetic energy and the breadth of the distribution. The high temperature curve is also shown in the figure. Notice that the area under the curve to the right of E_a is much larger for the high-temperature curve, leading to a dramatic increase in the number of effective molecular collisions. Quantitatively, the fraction of effective collision can be expressed as in 15-5-7.

Thus

Equation 15-5-8 has the form of a rate law if we identify the terms preceding the concentrations of A and B as the rate constant, k:

where A contains information about the orientation requirements. Taking the logarithm of both sides of 15-5-9 gives 15-5-10, called the Arrhenius equation.

The value of the activation energy for a process is obtained by measuring the rate constant at each of several temperatures, and constructing a plot of lnk versus 1/T. The plot should be linear, with slope = -E_a/R and intercept = lnA.

We are now able to offer an answer to the second of the two questions posed at the outset: reaction rates are temperature dependent because of the requirement of a minimum kinetic energy for effective collisions.

Example 15-13. Aqueous hydrogen peroxide and iodide ion react according to the following equation.

The reaction is first order in peroxide and first order in iodide, second order overall. The reaction was studied at several temperatures, with the following results. What is the activation energy for the reaction?

Solution. The activation energy can be obtained from the slope of a plot of ln k_obs versus reciprocal Kelvin temperature. The plot is in Figure 15-13. The activation energy obtained from the slope is 53.6 kJ /mole.

Overall (Multistep) Reactions. Thus far we have discussed the temperature dependence of a single elementary process. Similar considerations apply to overall (net) reactions consisting of several steps. We make several general statements about reaction coordinate diagrams for net reactions before looking at a specific example.

• The number of maxima (activated complexes) in a reaction coordinate diagram is equal to the number of steps in the mechanism.
• The slowest (rate-determining) step usually has the highest E_a.
• The rate law for the overall reaction indicates the atoms present in the activated complex for the slowest elementary step.
• Intermediates occur at minima in the reaction coordinate diagram.
• The E_a measured experimentally for the multistep reaction is the difference between the potential energy at the highest point in the diagram and the potential energy of the reactants. (This is analogous to the process of rolling a ball over a succession of hills of varying heights; to make it, the ball must have kinetic energy sufficient to get over the highest hill.)

Example 15-14. For the reaction

the rate law is -d[F₂]/dt = k_obs[NO₂][F₂]. Propose a mechanism for the reaction, and draw an appropriate reaction coordinate diagram.

Solution. The rate law tells us that the activated complex of the slow step contains two oxygen atoms, two fluorine atoms, and one nitrogen atom. This suggests the following mechanism:

Because reaction involves the breaking of the F-F bond and formation of a bond between N and F, the structure shown in Figure 15-14a is a reasonable arrangement of atoms in the activated complex. Finally, Figure 15-14b shows the reaction coordinate diagram corresponding to the proposed mechanism. There are two "humps" (maxima), one corresponding to each elementary step in the mechanism; the intermediate, F, is located at the minimum between maxima; the activation energy for the overall reaction is indicated; and the overall reaction is exothermic. The activation energy for the first step is substantially larger than that for the second step, consistent with the first step being rate-determining. Thus the reaction coordinate diagram reveals by its profile which is the slow step in the overall reaction.

Example 15-15. The reaction coordinate diagram for an unspecified reaction is shown in Figure 15-15. From the details of the diagram, say as much as possible about the net reaction to which it corresponds.

Solution. Because there are three maxima in the diagram, the mechanism must consist of three steps. Step 2 must be the rds, because it has the largest E_a. There is one activated complex for each step, for a total of three. There are two minima, indicating two intermediates. The experimental rate law for the reaction should indicate the composition of the activated complex of the second step. The overall reaction is exothermic, because products are lower on the energy axis than reactants. Finally, the measured E_a is the energy difference between the highest maximum in the diagram and the reactants. It is shown in the diagram.

The temperature dependence of the observed rate law for an overall reaction consisting of several steps is usually qualitatively similar to that for a single elementary process. However, there are complex mechanisms for which this is not the case. Indeed, examples are known of reactions for which the observed rate constant is either independent of temperature or decreases with increasing temperature. We will not be concerned with such situations in this book.

15-6 Kinetics and Equilibrium. The connection between the equilibrium constant for a reaction and the rate constants of the elementary steps by which it occurs is an important one. We begin by stating the relationship. We will then demonstrate, for a specific case, that it is true.

The equilibrium constant for a net reaction is the ratio of the product of the forward rate constants for all steps in the mechanism to the product of the reverse rate constants for all steps in the mechanism:

We now show for a specific net reaction and a specific (proposed) mechanism that this statement is true. Consider again the reaction of NO₂ and F₂ to produce NO₂F, reaction 15-4-1:

We have written the reaction with the double arrow because we are now interested in the forward and reverse rates at equilibrium. At equilibrium, the overall forward rate and overall reverse rate must be equal:

This must be true regardless of the detailed pathway by which the reaction occurs. All available experimental evidence indicates that the mechanism for 15-4-1 is the two-step process in 15-4-2a and b:

When the overall reaction is in equilibrium, so must each elementary step be. Consequently, we have used double arrows in the elementary steps as well. The equality of the forward and reverse rates of the individual steps is expressed mathematically as follows.

Eliminating the concentration of the intermediate, F, gives 15-6-2.

The statement made at the outset is thus seen to be true for this specific case: the equilibrium constant is the ratio of the product of forward rate constants to the product of reverse rate constants. For the general overall reaction occuring by an n-step mechanism,

Objection is frequently made to the method of derivation that we have just used, because the mechanism for 15-4-1 is not known with certainty; it is and will remain a hypothesis, however well based in experiment. If it is found that the two step mechanism in 15-4-2 is indeed NOT correct, then the relationship in 15-6-2 is invalid. This is certainly true. However, even if the mechanism above is incorrect, 15-4-1 must occur by some mechanism. At equilibrium each step in the mechanism must be balanced in rate, enabling us to eliminate concentrations of any and all intermediates just as we did above, to arrive at a modified version of 15-6-2. The validity of the general relationship in 15-6-3 is independent of the details of mechanism.

Thus the intuitively-expected connection between k and K does indeed exist. In fact, there are many examples in which the equilibrium constant for a reaction has been obtained from kinetics studies. This process can never be reversed however; it is not possible to obtain k by performing equilibrium studies. Before leaving this matter, we make one more very important point. There is a tendency to believe that reactions with large equilibrium constants are fast, whereas those with small equilibrium constants are slow. However, neither of these beliefs is true. A large value for K_eq means that the product of forward rate constants is much larger than the product of reverse rate constants. It does not follow, however, that the forward rate constants are LARGE. It means only that they are larger than the reverse rate constants.

A large K_eq can result from the ratio of a slow forward rate to an even slower reverse rate.

15-7 Catalysis. A number of chemical reactions that ordinarily occur slowly can be induced to occur more rapidly by the addition of a suitable substance called a catalyst. We will define a catalyst as a substance that speeds up a reaction without itself being consumed or chemically changed in the overall reaction process. This definition is in practice rather restrictive, because eventually, all real catalysts become deactivated by "poisoning" or via irreversible structural changes; thus they cannot function indefinitely. For our purposes, here, however, this definition will serve. The reactant molecule that is affected by the catalyst is called the substrate. The process by which a catalyst effects its action is called catalysis. Catalysis is one of the most intensely studied areas in science because it is of tremendous biological and industrial importance. We begin our exploration of catalysts by discussing how they work.

The Mode of Catalyst Function. A catalyst functions either by entering into a slow step of the uncatalyzed reaction mechanism, or by creating an entirely new mechanism for the reaction. In either case, it is thought that the catalyzed pathway has a lower activation energy than the uncatalyzed path. We can illustrate catalytic action generically in terms of the following reaction scheme, where A, B, D, and F are reactants and products in the overall reaction, and C is a catalyst for the reaction.

Uncatalyzed mechanism:

Catalyzed mechanism:

The effect of the catalyst on the activation energy of the slow step of the reaction is shown schematically in Figure 15-16. As shown in the figure, the catalyst affects only E_a; it does not affect the energy of either the reactants or products of the reaction. The intermediate, CAB, in which both reactants are bound to the catalyst, is of particular interest. A possible (and fairly common) structural motif for this intermediate is shown in Figure 15-16. Notice that, although A and B are bound to different sites on the catalyst, C, they are in proximity and can effectively interact. Binding of A and B to C causes shifts in electron density that may facilitate bond breaking within A and/or B and bond formation between A and B or fragments of them. Thus the catalyst provides an organizing center for A and B, facilitating their interaction. In this manner the catalyst can overcome the orientation factor discussed earlier.

As soon as the molecule of catalyst is regenerated in the last step of the 4-step mechanism, it may bind another molecule of A and proceed once again through the sequence, converting A and B to D and F. This sequence is repeated a large number of times, so that generally only a small amount of catalyst is required to convert a substantial quantity of reactant to product. Chemists call the process by which the catalyst cycles through the reaction over and over again a catalytic loop, and represent it as in Figure 15-17. The unbound form of the catalyst, C, is placed at the 12 oclock position of the loop, and the various forms in which the catalyst is found are placed more or less evenly around the remainder of the loop. Arrows show the direction of reaction around the loop, with reactants brought in from outside the loop, and products ejected out of the loop. This is a very effective visual presentation of catalyst action. The number of times that a molecule of catalyst cycles through the loop per unit time is called the turnover number of the catalyst. An effective measure of turnover number is moles product produced per mole catalyst present per time.

Let's look at a specific example of the effect of a catalyst on a simple reaction, from which we can draw some general conclusions.

Example 15-16. The reaction of ethanol (ethyl alcohol) with bromide ion to produce ethyl bromide and hydroxide ion is shown in 15-7-4.

The reaction is quite slow in neutral or basic solution, but occurs quite rapidly in acidic solution because it is catalyzed by the hydronium ion, H₃O⁺. Discuss the mechanisms for the uncatalyzed and catalyzed processes.

Solution. The uncatalyzed reaction is thought to occur in a single elementary step, in which bromide ion attacks the hydroxyl carbon atom of ethanol while the hydroxide group simultaneously departs:

This mechanism is proposed based on the experimental rate law, which is overall second order. The observed rate law for the catalyzed process is

The following three-step mechanism is consistent with this rate law if k_obs is identified with k₂K₁:

The reaction coordinate diagrams for the uncatalyzed and catalyzed pathways are shown in Figure 15-18.

Several general statements about catalysis are evident from this example.

• Because the catalyst does not appear in the overall equation for the reaction, it does not affect the equilibrium; it affects only the kinetics. The catalyst speeds up not only the forward reaction, but also the reverse, by the same factor. The position of equilibrium remains the same.
• the activation barrier is lower in the catalyzed mechanism. Even though the catalyzed path involves more steps, the overall reaction occurs more rapidly.
• The concentration of the catalyst appears in the rate law, even though the catalyst does not appear as a reactant or product in the overall equation.
• Although a catalyst and an intermediate share the property of not appearing in the overall equation, they are different types of things. First, the catalyst appears in the rate law; an intermediate cannot. Second, the catalyst appears first as a reactant, then is generated later as a product. An intermediate appears first as a product, then later as a reactant.

Examples of Catalysis. We will now briefly discuss several examples of catalysis that are actually used on a mammoth scale in the chemical industry. Each of these processes, in its own way, has a major impact on the quality of our lives.

The polymerization of ethylene. Ethylene is a very simple molecule with formula C₂H₄. It is a gas, obtained as a byproduct during the catalytic cracking of petroleum. Under certain conditions, ethylene molecules can be made to join together end-to-end to form very long chain-like molecules of polyethylene, so called because it consists of many (poly) ethylenes. The process is represented in 15-7-7.

Even at high temperature and pressure of ethylene, this process occurs negligibly slowly. In the presence of a small amount of a modified form of TiCl₃ (called tickle-3 in the plastics industry), however, it occurs rapidly at only moderately high temperature and pressure. The process is referred to as Ziegler-Natta catalysis after its two coinventors, who jointly received the Nobel Prize in Chemistry in 1963. Since the discovery of this process in the 1950's, the entire plastics industry has developed and grown to huge proportions. To this day, the mechanism of Ziegler-Natta catalysis is not fully understood. Mechanistic studies are difficult for several reasons, one of which is that the process is heterogeneous; that is, the catalyst (TiCl₃) and substrate (ethylene) are in different phases. The catalytic process takes place on the surface of crystals of TiCl₃, which rapidly become covered with and blocked from view by the resulting polyethylene. Much effort is ongoing in the US chemical industry to develop more efficient and easily handled Ziegler-Natta catalysts.

The Production of Sulfuric Acid. Year after year, sulfuric acid ranks first on the list of the top ten chemical substances produced in the United States: billions of pounds are produced annually. Sulfuric acid is synthesized by the so-called Contact Process, which involves the four sequential steps below:

The reaction of SO₃ with water is very exothermic and causes extensive spattering and production of a fine mist of highly acidic water. For this reason, direct reaction of SO₃ with water in the third step is impractical. Instead, SO₃ is bubbled into pure sulfuric acid, with which it reacts smoothly to give fuming sulfuric acid, H₂S₂O₇. This can then be treated with the stoichiometrically correct amount of water to give sulfuric acid. The contact process has been so perfected that sulfuric acid is very inexpensive to produce. Consequently, it is used in any industrial process requiring acid. It finds it major uses in the production of phosphate fertilizers; in paper manufacture; in the petroleum industry; in steel production; and in the production of detergents. The importance of these products in our lives is obvious.

Catalytic Converters. For some years now, catalysts have been placed within the exhaust systems of automobiles to reduce the amount of poisonous or otherwise harmful emissions. These noble metal catalysts (based on platinum and palladium) carry out a dual function. First, they facilitate oxidation of carbon monoxide, resulting from incomplete hydrocarbon combustion, to carbon dioxide:

Second, they catalyze decomposition of nitric oxide, produced during engine operation and oxidized rapidly to toxic NO₂ by atmospheric oxygen, to N₂ and O₂. Unfortunately, catalytic converters also facilitate oxidation of SO₂ to SO₃, which is the precursor of acid rain. Low-sulfur petroleum distillates are therefore essential.

The Haber Process. Vast quantities of ammonia are synthesized each year for use as fertilizer. Currently the most efficient process for ammonia synthesis is the Haber-Bosch Process, developed during the ten-year period preceding 1913, in which nitrogen and hydrogen react directly at high temperature and pressure and in the presence of an activated iron catalyst to form ammonia.

A catalyst and high temperature are necessary to cause the reaction to go at a reasonable rate. Unfortunately, high temperature makes the exothermic reaction less favored, so very high pressure is used to favor products. Even with conditions optimized, reaction is incomplete, and unreacted hydrogen gas is recycled for maximum efficiency of ammonia production. Nitrogen fertilizers produced from ammonia are largely responsible for the incredible growth in agricultural production over the decades since the First World War, when the process was first put on line in Germany. Ironically, the original motivation for development of the process was the requirement of explosives for the war effort.

The processes discussed above have at least three features in common. First, catalysis is heterogeneous; the catalyst is in all cases a solid, interacting with the substrate in the gas phase. Second, the mechanisms for these processes are incompletely understood. Thus catalysts are used successfully on a huge scale, even though we do not understand how they work. Third, in all cases the catalyst involves a transition metal, from the d block of the periodic table. Transition metals are often versatile catalysts because they are flexible in coordination number (that is, the number of atoms, ions, or molecules to which they may bind in a Lewis acid-base interaction); in stereochemistry (that is, in the shapes that their adducts assume); and in oxidation state (that is, in the charge that they carry). Nature has chosen transition metals to serve as the centerpieces for many of its catalysts, the enzymes, most probably for these same reasons.

15-8 Complex Reactions: Parallel Processes, Consecutive Processes, and Reversible Processes. Thus far we have discussed relatively simple reactions, which occur by mechanisms that involve one step that is substantially slower than the others, and which go to completion. In this situation, rate laws may be readily written from the mechanism and compared with the laboratory-determined rate law. As you might expect, most reactions are not so simple. In this section, we briefly describe three frequently encountered complications: parallel reactions (two or more simultaneous processes leading to the same or different products); consecutive reactions having comparable rates; reversible reactions (those in which the reverse process plays an important role in determining the observed rate).

Parallel Reactions. Parallel reaction paths may be represented schematically as in 15-8-1.

To simplify discussion, we assume that each reaction occurs by a single bimolecular elementary step. If so, then we may write the rate laws directly. The rate of disappearance of A is the sum of the rates at which it is used in 15-8-1a and 15-8-1b:

If pseudo-order conditions are imposed, this simplifies to the pseudo first-order rate law in 15-8-3.

where k₁' = k₁[B] and k₂' = k₂[C]. A plot of Rate versus [A] is linear, with slope k₁' + k₂'. Thus if one follows the reaction rate by monitoring [A], only the sum of the two rate constants is obtained. If it is possible to monitor [P] or [Q], k₁' or k₂' may be separately determined, and the other one calculated by difference.

Example 15-17. The overall reaction below is thought to occur by two different pathways operating simultaneously.

The first pathway involves attack by H₂O at the platinum atom, with simultaneous loss of Cl-. Product is thus formed in a single elementary step with the same equation as the overall reaction, with rate constant k₁. In the second pathway, Cl- leaves the platinum compound in a slow first step; this is followed by a rapid second step in which H₂O forms a bond with platinum:

What rate law will be exhibited if this hypothesis is correct?

Solution. The overall rate law is the sum of the rate laws for the two parallel processes:

(Be sure that you understand why the rate laws for the two mechanisms have these forms.)

Consecutive Reactions. We have previously examined systems consisting of two consecutive reactions (e.g., Example 15-11) under what might be called limiting conditions: either the first reaction was much more rapid than the second and could be treated as an equilibrium; or the first reaction was much slower than the second and completely determined the overall reaction rate. It is instructive to examine how the concentrations of A, B, and C vary with time in these situations, as well as the one in which the two steps have comparable rate constants.

Consider the following two step mechanism:

We present three situations involving different relative sizes of k₁ and k₂. We assume for simplicity that only A is present initially; [B] = [C] = 0 at t = 0. When k₁ >> k₂, reaction 15-8-4a goes to completion before 15-8-4b gets under way. When k₁ = 20k₂, the concentrations of A, B, and C vary as shown in Figure 15-19. A rapidly disappears and is replaced by B; B then very slowly converts to C. If the reaction rate is monitored by measuring the time dependence of [C], only the rate constant k₂ is obtained. On the other hand, if [A] is monitored, its rapid disappearance is governed by k₁, which is obtainable from either a plot of ln [A] versus time, or a plot of -d[A]/dt vs [A]. When k₁ << k₂, B is formed very slowly via 15-8-4a, but is then immediately consumed by 15-8-4b. The concentration of B thus never rises much above zero, and C appears at the same rate that A disappears. This corresponds to the second set of limiting conditions mentioned above. When k₂ = 20k₁, the concentrations of the 3 species vary as shown in Figure 15-20. The rate constant determined experimentally is k₁, the smaller of the two. In this situation, it is very difficult to measure k₂ as [B] is never appreciable. Only if B can be prepared independently under conditions where it does not convert to C can the rate constant, k₂, be measured.

There are many situations in which k₁ and k₂ are approximately equal. We illustrate these by looking at the case in which k₁ = k₂. The time dependence of the concentrations of A, B, and C is presented in Figure 15-21. At the very beginning of reaction, t = 0, the rate of the second reaction is zero because [B] = 0. The initial rate is the rate of disappearance of A, from which the rate constant k₁ can be determined. Before reaction has progressed very far, however, [B] has risen sufficiently that the second process occurs simultaneously with the first. This tends to drain the concentration of B as it is converted to C. In the Figure, we see [B] rise to a maximum, then fall off because its rate of production from A slows as A is consumed. At any time, it must be true that [A] + [B] + [C] = constant = [A]_o. If [C] is monitored and [A] is calculated from the value of k₁, [A]_o, and t, [B ] can be calculated as a function of time. It is then possible to determine k₂ from appropriate plots.

Finally, the mechanism in 15-8-4a and b can be modified slightly to allow reversibility in the first (but not the second) step:

(When k_-1 = 0 this mechanism reduces to the one in 15-8-4a and b). Figure 15-22 shows the time dependence of the concentrations of A, B, and C assuming k₁, k_-1 >> k₂ . This corresponds to the first limiting situation mentioned at the beginning of this section. Figure 15-22a shows the concentration behavior for a short time early in the reaction. It is very clear from this figure that A and B rapidly come to equilibrium and maintain a constant concentration ratio thereafter. The total of their concentrations is then very slowly reduced as B converts to C. Figure 15-22b shows the course of reaction over a longer time.

The plots in Figures 15-19 to 15-22 were obtained using finite difference methods to calculate the concentrations of the three species, A, B, and C, at many times during the course of reaction. These repetitive calculations were carried out using an electronic spreadsheet. Table 15-2 contains the data used in plotting Figure 15-21. Table 15-3 outlines the method of calculation programmed into the spreadsheet. It would be instructive for you to set up your own spreadsheet according to the information in Table 15-3 to see that you can produce plots similar to those in Figures 15-19 to 15-22.

Reversible Reactions. Our last example of complexity in chemical dynamics will focus on reversible reactions: those that approach an equilibrium position in which appreciable amounts of both reactants and products are present. The simplest such process is shown in 15-8-6.

k₁ and k_-1 are the rate constants for the forward and reverse processes, respectively, which are both assumed to be first order. The rate law in 15-8-7 follows:

This equation states that the rate of change in the concentration of A is the difference between its rate of consumption via the forward process and its rate of production via the reverse process. If the initial concentration of B is 0, then at any point during the process, [B] = [A]_o - [A]. Equation 15-8-7 becomes

This can be integrated, but the resulting expression is not readily plottable. Thus, even though the reaction is first-order, a plot of ln[A] versus t will not be linear. To eliminate the troublesome k_-1[A]_o term, we recognize that 15-8-8 applies at all points during reaction, including equilibrium. Letting [A]_e represent the equilibrium concentration of A, 15-8-9 holds when the reaction reaches equilibrium:

Subtracting 15-8-9 from 15-8-8 eliminates the k_-1[A]_o term, giving

This equation is readily integrated in the variable [A] - [A]_e, which is the deviation of the concentration of A from the equilibrium value. Thus 15-8-10 says that in a reversible first-order process, the approach to equilibrium is governed by first-order kinetics. A plot of ln([A] - [A]_e) against t will be linear, with slope -k_tot, the sum of the forward and reverse rate constants. We make several observations about reversible first-order processes.

• The rate constant for approach to equilibrium is the sum of forward and reverse rate constants because equilibrium is approached by simultaneous reaction in both directions.
• When 15-8-6 is irreversible, k_-1 = 0 and [A]_e = 0; thus 15-8-10 reduces to 15-3-6 (with n = 1).
• When k₁ >> k_-1, the measured rate constant, k_tot, is essentially equal to k₁. The reverse process is unimportant until the reaction is nearly at equilibrium, because k_-1[B] remains small. At equilibrium, the forward and reverse reactions, given by the following expressions, are equal:

  Rate_f = k₁[A] = Rate_r = k_-1[B]

Both rates are small since at equilibrium [A] and k_-1 are both small. Similar considerations hold when k₁<< k_-1. In this case, the measured rate constant is essentially k_-1.

Let's examine a particular example of a reversible first order process.

Example 15-18. The kinetics of the reaction below were monitored by following the change in absorbance with time of R.

The following data were obtained.

Solution. Determine the rate constant for approach to equilibrium. The absorbance value at 60 minutes represents the equilibrium value, since enough time has elapsed for the reaction to have reached equilibrium. Thus A_e = 0.15. A value of A - A_e is calculated for each time and a plot of ln(A-A_e) constructed. Figure 15-23 shows such a plot. The linearity of the plot substantiates a first order approach to equilibrium. The slope of the plot gives the rate constant, 0.27 min^-1. This is actually the sum of the forward and reverse rate constants for the reaction.

More complicated processes can be made to conform to 15-8-10 by imposing pseudo-order conditions. For example, reaction 15-8-11 involves forward and reverse processes that are second order:

Suppose that initially [B]_o and [D]_o are both made much larger than [A]_o, and [C]_o = 0. Then the concentrations of B and D will remain constant during the reaction, and A will be converted to C in a pseudo-first-order process in which the forward rate constant, k_f, is equal to k₁[B]; and the reverse rate constant, k_r, is equal to k_-1[D]. By analogy with 15-8-10, we write the rate law for approach to equilibrium:

k_tot is studied first as a function of [B] with [D] held constant, in which case 15-8-13 applies:

k₁ can be obtained from a plot of k_tot versus [B]. Then k_tot is studied as a function of [D] with [B] held constant, when 15-8-14 applies:

k_-1 is obtained from a plot of k_tot versus [D].

We will have more to say about enzyme structure in a subsequent chapter. In this section, our focus will be on the kinetics of enzyme catalysis. Despite the vast number of enzyme structures and the equally vast number of reactions catalyzed, most enzyme-catalyzed processes exhibit essentially the same kinetics. Considering the complexity of the molecules and reactions, the kinetics is relatively simple. We approach it first from the experimental standpoint, which produces an observed rate law; we then address the question of mechanism. Equation 15-9-1 shows a typical enzyme catalyzed reaction in generic form.

Here, E stands for the enzyme, and is written over the arrow to indicate that the reaction is catalyzed. The reactant molecule on which the enzyme works is called the substrate. Although there may be more than one reactant molecule, in most cases only one of them interacts directly with the enzyme. We will condense 15-9-1 to the form in 15-9-2.

Kinetics studies of enzyme catalyzed reactions usually involve the measurement of initial rates, for reasons that need not concern us here. Throughout the discussion that follows, it will be assumed that the word, rate, stands for initial rate. The general rate law, applicable to most enzyme catalyzed processes for which studies have been done, is in 15-9-3.

Here [E]_T stands for the total concentration of enzyme present in the solution, and k_obs is the observed first-order rate constant. The observed rate constant exhibits an interesting dependence on the concentration of substrate, S. When [S] is small, k_obs increases nearly linearly with [S]; in other words, the rate is first order in S. However, as larger and larger concentrations of S are used, the increases in k_obs become smaller and smaller; k_obs eventually levels off to a constant value that is independent of further increases in [S]. At this point, the rate is zero-order in S. The relationship between k_obs and [S] is shown in Figure 15-25. To an experienced kineticist, this behavior is known as saturation kinetics, and can be explained by a rate law of the form in 15-9-4.

Here A and B are constants that are presumably related to rate constants of the elementary steps in the reaction mechanism. We interpret the observed kinetics in terms of 15-9-4 as follows. When [S] is small, B is the dominant term in the denominator. Under these conditions, 15-9-4 simplifies to

Thus the rate is first order in E, first order in S, with k_obs = A/B. When [S] is large, it is the dominant term in the denominator, and 15-9-4 simplifies to

The rate is first order in E, zero-order in S, with k_obs = A. The form of the expression in 15-9-4 is consistent with the observed kinetics. For historical reasons, enzyme-substrate systems that obey the kinetic form in 15-9-4 are said to exhibit Michaelis-Menten kinetics, after two early investigators in the field. The constant, B, is called the Michaelis constant, and is usually given the symbol, K_M. We will not use this symbol, however, because it implies that B is an equilibrium constant. Instead, we will use the symbol M. We make the switch at this point:

The plot of Rate versus substrate concentration nicely shows the saturation behavior of the kinetics; unfortunately, it is not possible to obtain accurate values for A and M from this plot. However, there are two rearrangements of 15-9-8 that allow the experimental data to be plotted in linear form. The first of these is obtained by reciprocating both sides of 15-9-8:

A plot of 1/k_obs versus 1/[S] should be linear with slope M/A and intercept 1/A. Figure 15-26 shows such a plot for the data of Figure 15-25. A reciprocal plot of this type is called a Lineweaver-Burk plot. The values of A and M obtained from Figure 15-26 are A = 0.01 s^-1, M = 0.0055 M.

The second linear form is obtained from 15-9-8 by, first, multiplication of both sides by k_obs;

second, multiplication by A;

and third, subtraction of the first term on the right from both sides.

A plot of k_obs versus k_obs/[S] should be linear with slope -M and intercept A. This plot, called an Eadie-Hofstee Plot, is shown in Figure 15-27. It produces M = 0.0055 M, A = 0.01 s^-1.

The physical interpretation of Michaelis-Menten (saturation) kinetics goes something as follows. First, the rate is first order in both enzyme and substrate when [S] is small. This implies that 1 molecule of enzyme and 1 molecule of substrate must collide and interact in order to form products. Second, the rate is first order in E and zero-order in S when S is large. The implication of this observation is that the substrate forms an adduct of some type with the enzyme, which we might symbolize as ES. This adduct then reacts further to form products. When [S] becomes large enough, Le Chatelier's Principle requires that all of the free enzyme is converted to the adduct. Once this happens, the rate no longer increases with increases in [S], because [ES] has reached its maximum possible value. This type of reasoning led biochemists to the simplest mechanism consistent with the observations:

We must now develop the rate law implied by this mechanism, to see that it matches what is observed. It is not legitimate to make assumptions at the outset about the relative magnitudes of the rate constants, because we have no information about that. The development of the rate law must be carried out as generally as possible, then examined for limiting behavior. We use the so-called steady state approximation. The essence of this approximation is that the concentration of the intermediate adduct, ES, stays essentially constant as reaction proceeds. This idea is fairly simple to implement. First, we recognize that the rate of formation of products, P, is given by

This rate law is strictly correct and always true within the context of the mechanism in 15-9-13. However, it is difficult to use because [ES] is not accessible. As usual, we need to express the rate law in terms of measurable concentrations: those of E and S. Further work is required. Now we invoke the steady state assumption, which is that

This is a mathematical way to say what we said above: the concentration of ES remains constant as reaction proceeds. We now use the elementary steps to obtain an expression for [ES]:

Solving for [ES] gives

which can be substituted for [ES] in the rate law, 15-9-14.

We are close, but not quite done. Recall that the experimental rate law is written in terms of the total enzyme concentration, [E]_T. Interpreted in terms of our mechanism,

In words, the enzyme is partitioned between the unbound and substrate bound forms. We cannot at the outset expect to know how much enzyme is in each form. However, we do know the total amount; it is the amount we put into the solution. We must therefore obtain an expression for [E] in terms of [E]_T, which can be done using 15-9-17 and 15-9-19:

Solving for [E] gives

Substituting this expression into the rate law, 15-9-18, and doing some simplifying arithmetic gives

This has a familiar look. In fact, if we identify k_cat with A, and (k_cat + k_-1)/k₁ with M, we obtain

which is identical in form with the experimental law.

Within the context of mechanism 15-9-13, we can draw the following conclusions:

• the value of A gives a direct measure of the rate constant k_cat, which governs the rate at which the enzyme substrate adduct reacts to form products.
• The Michaelis constant, M, is seen to be (k_cat + k_-1)/k₁. This gives the ratio of the rate of consumption of ES to its rate of production. Importantly, it is clear that M is not an equilibrium constant, because it involves the sum of rate constants from two different elementary processes.

15-10. Mechanisms of Substitution Reactions. A substitution reaction is represented generally in equation 1:

This reaction type, in which a group Y replaces a group X in a molecule, is widespread in chemistry. Several examples of substitution reactions are presented in equations 2-5.

Reaction (2) is a Bronsted-Lowry acid base reaction, discussed in Chapter 13. It is also a substitution reaction in which one base, F-, replaces another base, CN-, in forming a bond to the proton, H+. Reaction (15-10-3) shows the formation of an alcohol from an alkyl halide by substitution of a hydroxide anion for a chloride anion at carbon. Reaction (15-10-4) involves the replacement of one ligand by another in a transition metal complex. We will discuss these in more detail below. Finally, (15-10-5) is a double substitution (or double displacement) process in which OH- and Cl- substitute for each other. Because substitution reactions are so common, they have attracted the attention of chemists for many years. Of particular interest has been the mechanism(s) by which they take place. What are the details of the process the net result of which is replacement of X by Y? Do all substitution reactions occur in the same way, or are there different pathways that depend in some understandable way on the structures of A, X, and Y? In this section we will explore these questions and attempt to supply some answers. As we will see, kinetics studies have played the central role in the discovery of mechanism. Our discussion will parallel the historical development of the subject, beginning with the mechanisms of substitution reactions at carbon.

Substitution Reactions at Carbon. The replacement of a halide by a hydroxide anion is shown in 15-10-6:

R represents a carbon-containing group of atoms that may be simple (for example, CH₃) or more complex (for example, (CH₃)₃C-). (15-10-3) is a specific example of the general process in (15-10-6). It has been shown via many research studies over many years that (15-10-6) occurs by two different pathways, depending on the nature of the carbon-containing group R.

The second-order pathway. Consider first the kinetics of (15-10-3), in which R can be identified with the CH₃CH₂- portion of the reactant, CH₃CH₂Cl. The reaction can be studied under pseudo-order conditions in CH₃CH₂Cl. Recall that this means that the concentration of OH- is maintained at a value at least 10 times larger than the concentration of CH₃CH₂Cl, so that [OH-] changes imperceptibly during reaction. Studied in this manner, the kinetics are first-order in CH₃CH₂Cl, with a pseudo-first order rate constant, k’, that is directly proportional to the concentration of OH-. The experimental rate law thus has the form in (15-10-7).

Thus (15-10-3) is experimentally observed to follow second-order kinetics. From this experimental basis, the chemist must then speculate as to the possible mechanisms that could give rise to a rate law of this form. In other words, the chemist must propose a theory about the elementary steps involved in the reaction process. Reaction (15-10-3) seems quite simple, so it is conceivable that it could occur in a single step in which the Cl- ion leaves simultaneously with attack by the OH- ion. In other words, (15-10-3) could be an elementary process. As we know, such processes have rate laws that follow directly from their stoichiometries. Thus, IF (15-10-6) were to occur in the single step (15-10-7), it would necessarily obey the rate law (15-10-8).

This agrees with the experimental rate law, and so the single-step mechanism is consistent with experiment. However, the thorough chemist would not stop here. Rather, s/he would ask whether there are other mechanisms that are consistent with the experimentally observed second-order kinetics. In fact, there is at least one other mechanism that should be considered. It is the two-step mechanism in (15-10-9):

In the first step of this process, the OH- group forms a bond to the carbon atom that is attached to Cl-, forming an intermediate in which this carbon atom is attached to 5 other groups. In the second step, the bond to the Cl- ion breaks, generating the products. The reverse arrow in the first step is included to recognize that the just-formed bond to the OH- ion could also break, regenerating the reactants. A steady-state treatment of the mechanism, in which the concentration of the intermediate is assumed to stay constant during the course of reaction, yields the theoretical rate law in (15-10-10):

The 2-step mechanism in (15-10-9) also predicts second-order kinetics. In this case, the experimental k_obs would be identified with the expression k₁k₂/(k_-1+k₂), which involves the rate constants of all of the elementary steps in mechanism (15-10-9). The important point here is that the experimental rate law does not enable us to choose between the one-step elementary mechanism in (15-10-7) and the more elaborate two-step mechanism in (15-10-9). Indeed, there are probably other even more elaborate mechanisms that also predict second-order kinetics. ANY MECHANISM THAT PREDICTS A RATE LAW HAVING THE SAME FORM AS THE EXPERIMENTAL RATE LAW MUST BE CONSIDERED AS POSSIBLE. Experiments of other types must be designed in an attempt to narrow the field of possible mechanisms. Many such experiments have been designed and carried out. Based on the results, most chemists now believe that reaction (15-10-3) occurs by the elementary mechanism in (15-10-7).

Because reaction (15-10-3) is only a specific example of reactions of type (15-10-6), is it then reasonable to conclude that all reactions of type (15-10-6) occur by the same single-step process? As is so often the case in science, the situation is not nearly so clean and simple! Thus, although many substitution reactions at carbon show second-order kinetics and are therefore consistent with the one-step mechanism, other reactions show rate laws that are first order in RCl and zero-order in OH-. An example is shown in (15-10-11).

Kinetics studies of (15-10-11) yield the experimental rate law in (15-10-12).

Interestingly, the rate of (15-10-11) is independent of how much OH- is made available in the system. How can this be? What mechanism can be proposed to explain this behavior? First-order kinetics implies that the process occurs in at least two steps, and that the slow step does not involve the incoming group OH-. One mechanism that produces a rate law of the form of (15-10-12) is given in (15-10-13).

Because the overall rate is determined by the rate of the slowest (rate-determining) step, the theoretical rate law for mechanism (15-10-13) is

This is consistent with the experimental rate law.

Mechanism 15-10-13 is oversimplified, however, because it ignores the possibilty that the Cl- ion could reattach to the carbon species, (CH₃)₃C⁺. We can fix this by including the reverse process in (15-10-13a). The amended mechanism is in (15-10-15).

A steady-state treatment on the intermediate, (CH₃)₃C⁺, gives the rate law in (15-10-16).

Allowing for the reverse of the first step therefore greatly complicates the theoretical rate law. The most noticeable feature of (15-10-16) is that it is no longer obvious that the kinetics are simple first-order in (CH₃)₃CCl. Instead, the overall reaction order depends on the relative size of the 2 terms in the denominator. Under conditions where k₂[OH-] >> k_-1[Cl-], the k₂[OH-] term dominates and will cancel with the k₂[OH-] term in the numerator. The resulting rate law is in (15-10-17):

This is consistent with the observed first order behavior. However, under conditons where k₂[OH-] << k_-1[Cl-], the latter term will dominate the denominator, and the rate law becomes

Under conditions of small [OH-], the rate should be first order in (CH₃)₃Cl, first order in OH-, and inverse first order in Cl-. Indeed, this suggests that this mechanism could be tested by studying the kinetics using a small [OH-]. Observation of an inverse dependence on [Cl-] would provide support for mechanism (15-10-15).

In summary, to account for the varied kinetics behavior displayed by reactions of type (15-10-6), chemists have proposed two limiting mechanisms. These are shown below with their traditional names:

The S_N2 Mechanism:

The S_N1 Mechanism:

The abbreviation, S_N2, stands for substitution nucleophilic second-order. This name is a mouthful, but it is easily explained. "Substitution" and "second-order" are self-explanatory. The adjective nucleophilic (nucleus-loving) is used to describe an atom, group of atoms, or ion that is attracted to a positive or partially-positive center in a molecule. Both X and Y are nucleophilic groups that compete for the positive center, R+. In the S_N2 mechanism, Y attacks the carbon atom in R-X and forms a weak bond. Simultaneously, the bond between R and X weakens. The result is a transition state, Y--R--X, in which both the leaving group (X) and the entering group (Y) are weakly bonded to R. The reaction is completed when the bond to X is severed and replaced with a full bond to Y. The reaction thus occurs in a single elementary step and therefore must display second-order kinetics, consistent with its stoichiometry. The abbreviation, S_N1, stands for substitution nucleophilic first-order. The name suggests a substitution involving the competition of 2 nucleophiles for a positive center that displays first order kinetics.

Substitution in Transition Metal Complexes. Elements in the d block of the periodic table, sandwiched between the s block elements at the left and the p-block elements at the right, are called transition metals. These elements form a class of compounds in which the transition metal is bonded to a number of ions or molecules via covalent bonds. When attached to a transition metal ion, such ions or molecules are called ligands. Most ligands are also Bronsted-Lowry bases, so that we can recognize them by the availability of at least one lone pair of electrons. In fact, it is a lone pair of electrons on the ligand that is used to form the bond to the transition metal ion. Some examples of transition metal complexes are given below.

Fe(CN)₅(NO)^3-. The transition metal ion is Fe²⁺. It is attached to 6 ligands, five of them cyanide anions (CN-) and the sixth a nitric oxide molecule. The complex has a net charge of 3-, obtained by summing the charges on the metal ion and the ligands. The six ligands are arranged about the metal ion in an octahedron. The complex is brown in color.

Co(NH₃)₅Cl²⁺. The transition metal ion is Co³⁺. It is attached to 6 ligands, five of them ammonia molecules (NH₃) and the sixth a chloride anion. The net charge of 2+ comes from summing the metal charge (3+) and the chloride charge (1-). The six ligands are arranged about the metal ion in an octahedron. The complex is green.

CoCl₄^2-. The metal ion is Co²⁺, which is attached to four Cl- ligands arranged about it in a tetrahedron. The overall charge is the sum of the ion charge (2+) and the chloride charges (4-). The complex is intensely blue in color.

Ni(CN)₄^2-. The metal ion is Ni²⁺, attached to four cyanide ligands which form a square planar arrangement. The complex is yellow in color.

These few examples illustrate some generalities about transition metal complexes:

The great variety of colors exhibited by transition metal complexes has captured the interest of chemists for centuries. Long ago it was observed that color changes sometimes occur when a complex is put in the presence of another potential ligand different from those already contained in it. The color changes are the result of substitution of one ligand by another, as for example in equation (15-10-19):

The mechanisms of ligand substitution reactions have been intensely investigated since the 1950s. The rate at which one ligand substitutes for another depends tremendously on what the metal is, and on what the other ligands are. In complexes of Cu²⁺, one ligand can substitute for another at a rate of up to 10¹⁰ times per second, whereas in complexes of Co³⁺ and Cr³⁺, substitution can take hours or days. It is not surprising that study of complexes of the latter two ions has been very popular, and quite a bit is now known about their mechanisms of substitution.

Substitution at Co³⁺. Consider reaction (15-10-20), in which a water ligand is replaced by a chloride ion in aqueous solution:

Kinetics studies of this reaction give the rate law, Rate = k_obs[Co(NH₃)₅(H₂O)][Cl-]. This is consistent with several mechanisms, shown below:

Mechanism 1, one-step:

This mechanism implies the rate law, Rate = k₁[Co(NH₃)₅(H₂O)³⁺][Cl-], where k₁ is the rate constant for the single elementary step process shown.

Mechanism 2, two-step:

The steady-state assumption for the intermediate, Co(NH₃)₅³⁺, yields the rate law in (15-10-23).

In aqueous solution, [H₂O] is huge and constant, so the first denominator term is expected to be much larger than the second term. The rate law then simplifies to

which agrees with the observed rate law if k_obs = k₁k₂/k_-1[H₂O]. This is called the Dissociative Mechanism, because the outgoing ligand first dissociates from the metal before the incoming ligand bonds.

Mechanism 3, two-step:

The implied rate law is

which agrees with the observed rate law if k_obs = k₁k₂/(k_-1+k₂). This is called the Associative Mechanism, because the incoming ligand is proposed to first associate with the metal before the outgoing ligand leaves.

Mechanism 4, two-step:

The implied rate law for this mechanism is

where [Co]_T represents the sum of the concentrations of Co(NH₃)₅(H₂O)³⁺ and [Co(NH₃)₅(H₂O]· · · Cl]²⁺. This, too, predicts second order kinetics under conditions where K_eq[Cl-] << 1. This mechanism is called the Interchange mechanism, because the RDS involves an interchange of the incoming and outgoing ligands at the metal.

How are these mechanisms to be distinguished? Examination of the implied rate laws in (15-10-21), (15-10-23), (15-10-26), (15-10-28) provides the answer. Both the dissociative (15-10-23) and the interchange (15-10-26) mechanisms predict that at sufficiently high concentration of the incoming ligand, the rate should become zero-order in incoming ligand. This is because the denominator term involving the incoming ligand should eventually become large enough to dominate, and would then cancel the incoming ligand factor in the numerator. This behavior has been experimentally observed in the reaction of a few complexes of Co³⁺ with anions, and on this basis the one-step and associative mechanisms have been tentatively ruled out for ligand substitution at this metal ion. It is now accepted by most chemists that substitution in Co³⁺ complexes occurs by the interchange mechanism because, despite much work, an intermediate of the type proposed in the dissociative mechanism has not been experimentally observed.

Substitution at Fe²⁺. In some ligand environments, substitution reactions at the Fe²⁺ ion are slow enough to be conveniently measured at room temperature. The complex, Fe(CN)₅X^3-, where X is a variety of neutral molecules, is particularly convenient to study because the CN- ligands are very difficult to replace. Reactions in which X is replaced by some other group, Y, can be studied without complication in aqueous solution:

It has been shown experimentally that the reaction obeys a rate law of the form in (15-10-30):

where [Fe] is shorthand for the reactant iron complex. It is thought that substitutions (15-10-29) occur by the dissociative mechanism in (15-10-31).

A steady-state assumption applied to the intermediate, Fe(CN)₅^3-, produces the rate law (15-10-32), which has the same form as the observed rate law.

Because the reaction is reversible, the rate law is expressed in terms of the deviation of the concentration of Fe(CN)₅X^3- from its equilibrium value. "Rate" then represents the rate of approach of the reaction to equilibrium. By studying the manner in which k_obs varies with [X] and [Y], it is possible to assign values to the rate constants k₁, k_-1, k₂, and k_-2.

Substitution at Pt²⁺. Most complexes of Pt²⁺ have four ligands arranged in a square planar arrangement around the Pt²⁺ ion. This leaves the Pt²⁺ center open to attack from above and below the plane of the complex, and suggests that an associative substitution mechanism might be possible.

Fundamental Substitution Mechanisms. There are four mechanisms that have been proposed to account for the details of substitution reactions. These are presented below.

Concerted One-Step Substitution

Associative Substitution

Dissociative Substitution

Interchange

Supplement: Picosecond Reactions; Laser Promoted Processes.

Supplement: Kinetics, Mechanisms, and Occam's Razor.

Applications

15-1. Express the rate of the following reaction in terms of the change in concentration per unit time of each participating species:

15-2. Complete the following statements for the reaction 2N₂O₅(g) --> 4NO₂(g) + O₂(g).

15-3. For the reaction in Application 1, what is the rate of appearance of ammonia if the rate of consumption of nitrogen is 0.01 moles L^-1 s^-1? What is the reaction rate?

15-4. In the reaction 3 ClO^-(aq) ---> 2 Cl^-(aq) + ClO₃^-(aq), the rate of formation of Cl^- is 3.6 mol/L-min. What is the rate of reaction of ClO^-? What is the unique rate of reaction?

15-5. Dinitrogen pentoxide, N₂O₅, decomposes by a first-order reaction. What is the initial rate of decomposition of N₂O₅ when 2.00 g of N₂O₅ is confined in a 1.00 L container and heated to 65 ^oC? The rate constant for the reaction is 5.2 * 10^-3 s^-1 at 65 ^oC.

15-6. When the NO concentration is doubled, the rate of the reaction 2NO(g) + O₂(g) --> 2NO₂(g) increases by a factor of 4. When both the O₂ and NO concentrations are doubled, the rate increases by a factor of 8. What are the reactant orders and overall order of the reaction?

15-7. Write the rate law for the consumption of persulfate ions in the reaction S₂O₈^2-(aq) + 3I^-(aq) ---> 2 SO₄^2-(aq) + I₃^-(aq) with respect to each reactant and determine the value of k from the following data.

15-8. The following kinetics data were obtained for the reaction, 2ICl(g) + H₂(g) ---> I₂(g) + 2HCl(g).

15-9. The following data were obtained for the reaction A + B + C --> products.

15-10. Determine the rate constants for the following first-order reactions

15-11. Dinitrogen pentoxide, N₂O₅, decomposes by first-order kinetics with a rate constant of 3.7 * 10^-5 s^-1 at 298 K.

15-12. The half-life for the first-order decomposition of A is 200 s. How much time must elapse for the concentration of A to decrease to a) one-half, b) one-sixteenth, c) one-ninth of its initial concentration?

15-13. Sulfuryl chloride, SO₂Cl₂, decomposes by first-order kinetics with k = 2.81 * 10^-3 min^-1 at a certain temperature.

15-14. For the first-order reaction A ---> 3B + C, when [A]_o = 0.015 mol/L, the concentration of B increases to 0.020 mole/L in 3.0 min.

15-15. The following data were collected for the reaction 2N₂O₅(g) ---> 4NO₂(g) + O₂(g) at 25 ^oC.

15-16. The following data were collected for the reaction 2HI(g) ---> H₂(g) + I₂(g) at 580 ^oC.

15-17. The half-life for the second-order reaction of a substance A is 50.5 s when [A]_o = 0.84 mol/L. Calculate the time needed for the concentration of A to decrease to a) one-sixteenth; b) one-fourth; c) one-fifth of its original value.

15-18. Determine the rate constant for the following second-order reactions

15-19. Calculate the activation energy for the conversion of cyclopropane to propene from an Arrhenius plot of the following data:

15-20. Write the overall reaction for the mechanism below, and identify intermediates:

15-21. Write the overall reaction for the mechanism below and identify intermediates:

15-22. Develop the rate law implied by the following mechanism:

15-23. Develop the rate law implied by the following mechanism:

15-24. Develop the rate law implied by the following mechanism, proposed for the formation of phosgene from CO and Cl₂:

15-25. Three mechanisms have been proposed for the reaction

The observed rate law is Rate = k[NO₂]². Which mechanisms are feasible?

15-26. For each mechanism in application 15-25, draw a reaction coordinate diagram.

15-27. Critically analyze each of the following statements about elementary processes:

15-28. For the reversible elementary process 2A <===> B + C, the rate constant for the forward reaction is 265 L*mol^-1*min^-1; the rate constant for the reverse reaction is 392 L*mol^-1*min^-1; the activation energy for the forward reaction is 39.7 kJ/mol; and the activation energy for the reverse reaction is 25.4 kJ/mol.

15-29. For the reversible elementary process, A + B <===> C + D, the forward rate constant = 36.4 L*mol^-1*h^-1; the reverse rate constant = 24.3 L*mol^-1*h^-1; the forward activation energy is 33.8 kJ/mol; and the reverse activation energy is 45.4 kJ/mol.

15-30. Suppose that a catalyst lowers E_a for a particular reaction from 100 to 50 kJ/mol. By what factor does the reaction rate increase at 400 K, all other factors being equal?

15-31. A reaction rate increases by a factor of 10³ in the presence of a catalyst at 25 ^oC. The activation energy of the uncatalysed path is 98 kJ/mol. What is the activation energy of the catalysed path, all other factors being equal?

15-32. The reaction of nitric oxide and hydrogen,

is thought to proceed by the mechanism below.

The rate law for the reaction is Rate = k[NO]² at high pressure of H₂. At low H₂ pressure, the rate law is Rate = k[H₂][NO]². Apply the steady state approximation to the mechanism to obtain the full implied rate law, and show that it agrees with the experimental observations under the indicated conditions.

15-34. For the second-order reaction A ---> B + C, the concentration of A falls from 0.040 mol/L to 0.0050 mol/L in 12 h. What is the rate constant?

15-35. All radioactive decay processes follow first-order kinetics. The half-life of the radioactive isotope tritium, ³H or T, is 12.3 years. How much of a 1.0 mg ample of tritium would remain after 5.2 years?

15-36. The rate law for the reaction 2NO(g) + 2H₂(g) ---> N₂(g) + 2H₂O(g) is Rate = k[NO]²[H₂]. The following mechanism has been proposed:

a) Which step is likely to be rate determining?
b) Sketch a reaction profile for the reaction, which is exothermic. Label the activation energy for each step and the overall reaction enthalpy.

15-37. Use the following data to determine the activation energy for the reaction of ethyl bromide with hydroxide ions in aqueous solution.

15-38. Explain in narrative form (i.e., in words) why an enzyme catalyzed reaction approaches a maximum velocity when substrate concentration is steadily increased.

15-39. The decarboxylation of a b-keto acid catalyzed by a decarboxylation enzyme can be measured by the rate of CO₂ formation. From the initial rates given in the table, determine the Michaelis-Menten constant for the enzyme and the maximum velocity by a graphical method.

15-40. The hydration of CO₂,

is catalyzed by the enzyme, carbonic anhydrase. The kinetics of the forward (hydration) and reverse (dehydration) reactions at pH 7.1, 0.5 ^oC, and 2*10^-3 M phosphate buffer were studied using bovine carbonic anhydrase. Some results follow:

Determine the Michaelis constant and the rate constant, k₂, for the decomposition of the enzyme-substrate complex to form product for:

15-41. At pH 7 the measured Michaelis constant and maximum velocity for the enzymatic conversion of fumarate to L-malate,

are 4.0*10^-6 M and 1.3*10³[E]_o s^-1, respectively, where [E]_o is the total molarity of the enzyme. The Michaelis constant and maximum velocity for the reverse reaction are 1.0*10^-5 M and 800[E]_o s^-1, respectively. What is the equilibrium constant for the hydration reaction?

15-42. The simple Michaelis-Menton mechanism for an enzmye catalyzed reaction is

The following data were obtained:

15-43. Explain in simple terms some of the reasons why enzymes are so efficient and so specific.

15-44. A particular enzyme is present at a concentration of 1.2*10^-8 M. The rate of substrate disappearance under a particular set of concentration and temperature conditions is 0.1 M/min. What is the turnover number of the enzyme?

15-45. For the enzyme in the previous problem, an increase of pH by 1 unit slows the rate at which the enzyme converts substrate to product by a substantial factor. What tentative conclusions can be drawn about the chemical environment at the active site of the enzyme?

15-46. It is generally stated that a catalyst lowers the activation energy for a reaction by providing an alternative mechanism. Provided this is true, what general statement(s) can you make about the temperature dependence of the rate of an enzyme-catalyzed reaction versus that for the uncatalyzed reaction?

15-47.Equation 11 of Kinetics Handout 2 shows the rate law for a reaction catalyzed by an enzyme in the presence of a competitive inhibitor of the enzyme. The rate law is obtained from a steady state treatment of the preceding mechanism. Carry out the steady state treatment and show that the rate law is correct.

15-48. The basic mechanism for enzyme catalysis is usually written as if it involves only 2 steps: E + S <==> ES; ES --> E + P. The mechanism of action of chymotrypsin is discussed in Kinetics Handout 1. How many substeps are involved in the "second step" for chymotrypsin?

15-49. Suppose that the equilibrium constant for binding of a molecule of substrate to a molecule of enzyme is 1.0*10³ M^-1. What fraction of the enzyme will be in the form of the enzyme-substrate complex when substrate concentration is 0.001 M. When substrate concentration is 0.01 M. By what factor will the substrate conversion rate increase when [S] is increased from 0.001 to 0.01 M?

15-50. The equilibrium constant for binding of a molecule of substrate to a molecule of enzyme is 10⁴ M^-1. What is the minimum substrate concentration necessary to guarantee that the enzyme is operating at more than 99% of its maximum conversion rate?

15-51. You have obtained the following data for the destruction of thiamine by an enzyme, both with and without 0.20*10^-4 M inhibitor.