The Rates of Chemical Reactions

The Rates of Chemical Reactions

402

12 The Rates of Chemical Reactions

The Macroscopic Description of Chemically Reacting Systems In Chapter 11 we described nonequilibrium processes in fluid systems that do not undergo any chemical reactions. We now discuss uniform fluid systems in which chemical reactions can occur at constant temperature. These conditions can usually be met by carrying out the reaction in a constant-temperature bath and either stirring the system or using a sufficiently small system. It is possible to treat reactions in systems in which the temperature is not uniform, 1 but we will not discuss such reactions. Consider a chemical reaction (12.1-1)

aA + bB --+ dD + f F

where the capital letters stand for chemical formulas, and the lower-case letters are stoichiometric coefficients. We define the rate of the reaction, denoted by r: r .

1 d[A]

. a

. dt

.

1 d[B]

1 d[D]

. . b dt

where [A] is the concentration (molL -1 or concentration of substance B, etc., and where An equation similar to Eq. (12.1-2a) can be appropriate stoichiometric coefficients. In Eq. written in the form

. d

dt

1 d[F] f

dt

(12.1-2a)

m o l m -3) of substance A, [B] is the t is the time. written for any other reaction, using the (8.1-1) a general reaction equation was

0 -- ~ YiJ~i i=l

for which the rate can be written as 1 d[~il r = -- ~ Yi dt

(12.1-2b)

This equation gives the same value for the rate for any choice of substance i taking part in the reaction. We will assume that the rate of the reaction is a function of temperature, pressure, and the concentrations of the substances in the system. For reactions in nonideal solutions, it would be a better approximation to assume that reaction rates are functions of activities instead of concentrations, but we will assume that concentrations and activities are equal. In most reactions the rate will depend only on the concentrations of the substances occurring in the chemical equation. If a substance that does not occur in the chemical equation increases the rate, it is a catalyst. The macroscopic rate of a chemical reaction is determined by measuring concentrations and is actually a net rate: r = ?'net ~- Ff -

Fr

(12.1-3)

where rf is the forward rate, or the rate at which products form, and where r r is the reverse rate, or the rate at which reactants form. Chemical reactions usually proceed smoothly toward a macroscopic equilibrium state in which the forward and reverse rates

1 See for example R. G. Mortimer,

J. Phys. Chem., 67,

1938 (1963).

12.1 The Macroscopic Description of Chemically Reacting Systems

403

cancel each other. However, some oscillatory reactions do exist. 2 Figure 12.1 shows schematically how a nonoscillatory chemical reaction, A--+ B, approaches equilibrium. In most reactions the forward rate depends only on the concentrations of the reactants, so that the forward reaction can be observed in the absence of the reverse reaction if the products are absent. If A and B are the reactants,

1 d[A]

rf . . . . . . a dt

rf([A], [B]) (no reverse reaction)

(12.1-4)

If the function represented in Eq. (12.1-4) is known, it is called the rate law of the forward reaction. Similarly, the reverse reaction rate usually depends only on the concentrations of the products. If D and F are the reactants, (12.1-5) Equation (12.1-5) is the rate law of the reverse reaction. There is a large class of chemical reactions in which the forward reaction rate is proportional to the concentration of each reactant raised to some power. For example, rf =

1 d[A] a

aft

= kf[A]~[B]/~

(12 1-6)

Equation (12.1-6) is called a rate law with definite orders. The exponent ~ is called the order with respect to substance A and the exponent fl is called the order with respect to substance B. If there is more than one reactant in the rate law, these orders are sometimes called partial orders. The sum of the orders with respect to the different substances is called the overall order. If ~ and fl are both equal unity, the reaction is said to be first order with respect to substance A, first order with respect to substance B, and second order overall. Higher orders are similarly assigned. The orders are usually small positive integers, but other cases do occur. There are some reactions that are not described by Eq. (12.1-6). Such reactions are said not to have a definite order. The proportionality Constant kf in Eq. (12.1-6) is independent of the concentrations, and is called the forward rate constant. Rate constants depend on temperature and pressure, although the pressure dependence is generally negligible. 3 We will discuss the temperature dependence of reaction rates in Chapter 13, and will discuss only constanttemperature systems in this chapter. One of the objectives of the kinetic study of a reaction is to determine the rate law. For a reaction with definite orders, this means to determine the orders with respect to each reactant and the value of the rate constant. There are at least two reasons why this is useful. The first is that it allows us to predict the rates of the reaction for new values of the concentrations without doing additional experiments. The second is that the form of the rate law usually provides information about the sequence of molecular steps making up the reaction (the mechanism of the reaction).

2 R. J. Field and M. Burger, Oscillations and Traveling Waves in Chemical Systems, Wiley, New York, 1985. 3 R. E. Weston and H. A. Schwarz, Chemical Kinetics, Prentice-Hall, Englewood Cliffs, N.J., 1972, pp. 181ft.

404

12

The Rates of Chemical Reactions

Forward Reactions with One Reactant In this section we discuss reactions with definite orders, a single reactant, and with negligible reverse reaction. Inspection of Figure 12.1 shows that when a reaction that converts A to B has approached equilibrium, the reverse rate cannot be neglected. We must avoid applying the results of this section to that case. However, many reactions proceed essentially to completion and in that case we can neglect the reverse reaction for nearly the entire reaction. The "classical" method for determining the rate law for a reaction is to mix the reactants and to determine the concentration of one of the reactants or products as a function of time as the reaction proceeds at constant temperature. A variety of analytical methods have been used to determine concentrations, including measurement of the following: 1. The absorbance of radiation at some wavelength at which a given product or reactant absorbs 2. The intensity of the emission spectrum of the system at a wavelength at which a given product or reactant emits 3. The volume of a solution required to titrate an aliquot of the system 4. The pressure of the system (for a reaction at constant volume) 5. The volume of the system (for a reaction at constant pressure) 6. The electrical conductance of the system 7. The mass spectrum of the system 8. The ESR or NMR spectrum of the system 9. The dielectric constant or index of refraction of the system 10. The mass loss or gain if a gas is lost or absorbed Once we know from experiment how the concentration of a reactant or product depends on time, we must have a formula for the concentration as a function of time to compare this with. We now proceed to integrate differential rate laws for a number of cases to obtain such formulas.

First-Order Reactions Consider a first-order reaction without significant reverse reaction: A ~ products

( 12.2-1 )

The differential rate law is

r-

-

d[A] d--7-= kf[A]

(12.2-2)

To separate the variables we multiply Eq. (12.2-2) by dt and divide by [A]:

1 d[A]

1

ltrA--q- - ~ dt - t~l~ d[A] - - k f dt

(12.2-3)

405

12.2 Forward Reactions with One Reactant

We carry out a definite integration from time t = 0 to t = t', where t' is a specific value of the time:

i

t' 1 o [-~ d [ A ] -

I t' -

o kf

dt

(12.2-4)

in([A]t, ) -ln([A]o ) = - k f t ' where the subscript on a concentration indicates the time at which it is measured. Taking antilogarithms of Eq. (12.2-4), (12.2-5) where we have written t instead of t'. An indefinite integration can also be carried out, followed by evaluation of the constant of integration.

Exercise 12.1 Carry out an indefinite integration of Eq. (12.2-3). Evaluate the constant of integration to obtain Eq. (12.2-5). For a reaction that involves a change in the number of moles of gaseous substances the progress of the reaction can be monitored by measuring the pressure. In this case, it is convenient to use partial pressures instead of concentrations.

The half-life, t 1/2, is defined as the time required for half of the starting compound to react. Since [A]q/2 - [A]0/2, we can write from Eq. (12.2-4),

kftl/2

-- _

Ink, [A]o )

ln(1)

4H. S. Johnston and Y. Tao, J. Am. Chem. Soc., 73, 2948 (1951).

406

12

The Rates of Chemical Reactions

or

(12.2-6)

The relaxation time z is the time for the amount of reactant to drop to a fraction 1/e (approximately 0.3679) of its original value. Substitution of this definition into Eq. (12.2-4) gives 1 tl/2 r -- ~ -- ln(2) "~ 1"4427q/2

(12.2-7)

For a first-order reaction kf has units of time -1 (s -1, min -1, etc.).

Exercise 12.2 *a. Find the half-life and the relaxation time for the reaction of Example 12.1. b. Verify Eq. (12.2-7). Like some chemical reactions, the decay of radioactive nuclides obeys first-order kinetics with rate constants that do not appear to depend on temperature. Half-lives, not rate constants, are tabulated for radioactive nuclides.

*Exercise 12.3 The half-life of 235U is equal to 7.1 x 108years. a. Find the first-order rate constant. b. Find the time required to a sample of 235U to decay to 10.0% of its original amount.

Second-Order Reactions The rate law for a second-order reaction with a single reactant and negligible reverse reaction is r--

d[A] _ kf[A]2 d--~-

(12.2-8)

With the variables separated,

1 d[A]dt_d[A]__kfd t dt [A] 2 -

(12.2-9)

[A] 2

We carry out a definite integration from time t = 0 to time t = t'. d[A__j --kf at f.tA~,,_

(12.2-10)

31110 [A] 2

The result of the integration is (12.2-11)

12.2 Forward Reactions with One Reactant

407

For a second-order reaction, kf has units of concentration -1 time -1 (Lmo1-1 s -1, m 3 mo1-1 min -1, etc.).

Exercise 12.4 Carry out an indefinite integration and evaluate the constant of integration to obtain Eq. (12.2-11) in an alternative way.

*Exercise 12.5 a. Find the concentration of NO2 in the experiment of Example 12.2 after a total elapsed time of 145 min. b. Find the concentration after 145 min if the initial concentration is 0.100molL -1. The half-life of the reaction is the time necessary for half of the initial amount of reactant to be consumed: 1

1

[A]0/2

[A]0

2

1

[A]0

[A]0

=

= kftl/2

(12.2-12)

where we have used the fact that [A]t,/2 - [A]0/2. Equation (12.2-12) leads to (12.2-13)

*Exercise 12.6 a. Find the half-life of the reaction of Example 12.2 with the given initial concentration. b. Find the half-life of the reaction of Example 12.2 if the initial concentration is equal to 0.0200 molL -1 . c. Find the half-life of the reaction of Example 12.2 if the initial concentration is equal to 0.1000 mol L -1 . If there is considerable experimental error, it might be difficult to tell a first-order reaction from a second-order reaction by inspection of a graph of the concentration versus time if the graph extends over only one half-life. Figure 12.2 shows a

408

12 The Rates of Chemical Reactions

hypothetical case: substance A undergoes a first-order reaction and substance B undergoes a second-order reaction with the same half-life. The concentrations in the two cases differ only slightly for times up to tl/2.

*Exercise 12.7 a. Find expressions for the time required for the concentration of the reactant in each of the reactions of Figure 12.2 to drop to one-sixteenth of its original value, assuming that the reverse reaction is absent. Express this time in terms of tl/2. b. Express this time in terms of the two forward rate constants and the initial concentrations.

nth-Order Reactions The rate law for an nth-order reaction with a single reactant is d[A]

r-

dt

(12.2-14)

= kf[A]n

where n is not necessarily an integer but is not equal to unity or zero. The variables can be separated by division by [A]" and multiplication by dt, giving d[A] [A] n = kf dt We perform a definite integration from t -

(12.2-15)

0 to t -

t' It

[AIr' d[A____]]_- k f I o dt

(12.2-16)

J[A]o [A] n

The result is 1 n-1

[

1 [A]t, 1

'

n -

1 ]_kft, [A]~)-l

(12.2-17)

The half-life of an nth-order reaction without reverse reaction is found by substituting [A]q n = [A]o/2 into Eq. (12.2-17). The result is 2 n-1

-

1

tl/2 - (n - 1)kf[A]~)-1

This formula is not valid if n -

1 or n -

0. If n -

(12.2-18)

1, Eq. (12.2-6) applies.

Exercise 12.8 a. Verify Eq. (12.2-18). b. For a third-order reaction with a single reactant and negligible reverse reaction, find an expression for the time required for 80% of the reactant to react. e. In terms of tl/2, how long will it take for 7 of the reactant of part (b) to react?

12.2

409

Forward Reactions with One Reactant

Zero-Order Reactions In the rare case that a reaction o f a single reactant is zero order (the rate is independent of the concentration of the reactant), the rate law for the forward reaction is r=-

d[A] = kf[A]O = kf d--7-

(12.2-19)

The solution of this equation is

[Alt--

[A]0 0

kft

if 0 < t < [A]0/k f if[A]0/k f
(12.2-20)

where the first line of the solution is obtained from Eq. (12.2-19). The second line of the solution is obtained from the fact that the reaction stops when the reactant has been consumed.

*Exercise 12.9 a. A hypothetical zero-order reaction has a rate constant equal to 0.0150molL -1 s -1 at a certain temperature. If the initial concentration of the single reactant is 1.000molL -1, find the concentration after a reaction time of 5.00 s at this temperature. b. Find the time required for all of the reactant of part (a) to react at this temperature. c. Find an expression for the half-life of a zero-order reaction and the value of the half-life of the reaction of part (a).

Determination of Reaction Order by Comparison of Experimental Data with Integrated Rate Laws Since linear graphs are the easiest type to use and since concentrations are not linear functions o f time except for zero order, one does not ordinarily use a graph of the concentration as a function of time to determine the reaction order. For each order, one plots the appropriate function of the concentration that will give a linear graph. To test for zero order, one makes a graph of [A]t as a function of t. To test for first order, one makes a graph of ln([A]) as a function of t. To test for second order, one makes a graph o f 1/[A] as a function of t. To test for third order, one makes a graph of 1/2[A] 2 as a function of t, etc. To test for nonintegral orders, graphs of 1/((n - 1)[A] n-l) for various nonintegral values of n can also be made. The graph that is most nearly linear corresponds to the correct order. Figure 12.3 shows schematic graphs for zero, first, second, and third orders. Using commercially available software packages such as Excel, MathCad, Mathematica, CricketGraph and KaleidaGraph, one can construct the graphs and carry out least-squares (regression) fits to the data. The software automatically calculates the correlation coefficient (or its square), which is a measure of the accuracy of the fit of the function to the data. Some packages will print out a list o f residuals, which are the differences between the data point and the least-squares line.

The Method of Half-Lives Another way to determine the order of a reaction is to determine how the half-life of the reaction depends on the initial concentration. If the half-life of the reaction is

410

12

The Rates of Chemical Reactions

independent of concentration the reaction is first order. To test for other orders we take the logarithm of Eq. (12.2-18) to obtain:

ln(tl/2)

--

( 2 " ~ - 1) l n \ ( n - l i ~ - (n - 1)ln([A]o )

(n :fi 1, n :fi O)

(12.2-21)

To use Eq. (12.2-21), one could perform a set of different experiments at the same temperature but with different initial concentrations, determine the half-life for each, and construct a plot of ln(tl/2) versus ln([A]0 ). A straight line should result, with slope equal to - ( n - 1) and with intercept equal to the first term on the right-hand-side of Eq. (12.2-21). One can also take the data for a single experiment and regard different times during the experiment as "initial" times. The reverse reaction must be negligible for the entire experiment.

12.2 Forward Reactions with One Reactant

411

The Method of Initial Rates In this method, one compares data directly with the differential rate laws instead of with the integrated rate laws. The method has two advantages: it is not necessary to integrate the rate law, and there is almost certainly no interference by the reverse reaction. The reaction is followed for a short time At for which there is a change A[A]: A[A] = [A]At - [ A ] 0

(12.2-22)

The time At must be short enough that A[A] << [A]. The initial rate is approximated by a quotient of finite differences:

rinitial

d[A] -- ~dt~ --

A[A]

At

(12.2-23)

One could alternatively measure the concentration at several times and use a graphical or numerical procedure to obtain a better approximation to the derivative d[A]/dt at t=0.

412

12

The Rates of Chemical Reactions

If there is only one reactant, the logarithm of Eq. (12.1-6) is ln(rinitial) = ln(kf) + ~ ln([A]0 )

(12.2-24)

To determine the order and the rate constant, one carries out several experiments at the same temperature but with different initial values of [A]. Plots of the logarithm of the initial rate as a function of the logarithm of the initial concentration are constructed or a linear least-squares fit is performed. The slope of the line best fitting the data points is the order of the reaction, and the intercept is the logarithm of the rate constant. The method of initial rates has the disadvantage that several experiments must be carried out. A modification to the above method would be to determine the value of A[A]/At and the value of [A] at different times in a single experiment and to use these values in Eq. (12.2-24).

Forward Reactions with More Than One Reactant Reactions with more than one reactant are harder to deal with than reactions of a single reactant. In a few cases the differential rate law can be integrated to obtain an integrated rate law, and in other cases we can use the method of initial rates or the method of isolation.

Integration of the Differential Rate Law That Is First Order in Each of Two Reactants Consider a reaction aA + bB --+ products

(12.3-1)

12.3 Forward Reactions with More Than One Reactant

413

where a and b represent stoichiometric coefficients. The rate law is 1 d[Al

r =

a dt

= ky[A][B]

(12.3-2)

We first assume that the reactants are in the stoichiometric ratio: [A]0 a [B]0 = ~

(12.3-3)

The concentrations will remain in this ratio during the reaction.

Exercise 12.10 Show that if Eq. (12.3-3) holds then [A]t

a

[B]t = ~

(12.3-4)

for all values of t greater than 0. Equation (12.3-2) appears to have two dependent variables, [A] and [B]. However, we can express the equation in terms of a single dependent variable x(t):

x(t) --

[A]o - [ A ] t [B]o --[g]t = a b

(12.3-5)

We can now write Eq. (12.3-2) in the form

dx dt =

bx)

(12.3-6a)

-kfab([A~]b~ -X)([B-~]b~-X ) - kfab([A~]a~-X) 2

(12.3-6b)

kf([A]~ - ax)([B]~ -

where we have used the fact that [B]o/b = [A]o/a for a stoichiometric mixture. The variables can be separated in Eq. (12.3-6b) by dividing by ( [ A ] o / a - x) 2 and multiplying by dt: dx 2 = kfab

dt

(12.3-7)

_ x) We integrate both sides of Eq. (12.3-7) from t =- 0 to t 1

[A]0 =

x({)

[A]~ a

t"

kfabt

which, if we replace t' by t, is the same as 1

1

[hit

[A]o

-- krbt

(12.3-8)

414

12

The Rates of Chemical Reactions

Equation (12.3-8) is the same as Eq. (12.2-11) except for the appearance o f the factor b in the fight-hand side.

*Exercise 12.11 a. Find the expression for the half-life of the reaction of Eq. (12.3-1) for the case of a stoichiometric mixture. b. Find the half-life of the reaction in Example 12.5 with the given initial concentrations. c. Find the half-life of the reaction in Example 12.5 with initial concentrations both equal to 5.00 x 10 -5 mol L -1.

If the reactants are not mixed in the stoichiometric ratio, we separate the variables in Eq. (12.3-6a) to obtain

1 ([A]0 - ax)([B]0 - bx)

dx - kf dt

(12.3-9)

This equation can be integrated by the method of partial fractions. We write

1

=

([A]o - ax)([B]o - bx)

G [A]o - ax

+

H [B]o - bx

(12.3-10)

where G and H are guaranteed by a theorem of algebra to be constants. These constants are G -

1 [B]o - b[A]o/a

and

H --

1 [A]o - a[B]o/b

Exercise 12.12 Verify the expressions for G and H. When the expressions for G and H are substituted into Eq. (12.3-10) and the resulting expression is substituted into Eq. (12.3-9), a definite integration gives a[B] ~

, -

b[A] ~ In

(tBl,IAl0 \[A]t[B]o j

- kft

(12.3-11)

12.3 Forward Reactions with More Than One Reactant

415

Exercise 12.13 Verify Eq. (12.3-11).

The Method of Initial Rates for Two or More Reactants For a reaction with several reactants, the method of initial rates provides the most convenient way to determine the rate law and the rate constant. The method is very similar to that with a single reactant, which led to Eq. (12.2-14). Consider the reaction: aA + bB + f F --+ products

(12.3-12)

for which the initial rate can be written as

1 A[A] a

At

' ~ /"initial--

kf[A]~)[Blo~[F]or

(12.3-13)

Several experiments are carried out at the same fixed temperature and with the same values of [B]o and [F]o, but with different values of [A]o. The initial rate is determined for each experiment. We write ln(rinitial) -- ln(kf[BloB[F]o~) + ~ ln([A]o)

(12.3-14)

The first term on the right-hand side of this equation has the same value in all of these experiments. A plot of ln(rinitial) as a function of ln([A]0 ) should give a straight line with slope equal to ct. Additional experiments are carried out in which the initial concentration of each of the other substances is varied in turn, keeping all other initial concentrations fixed. Equations analogous to Eq. (12.3-14) allow the order with respect to each substance to be determined. After all of the orders are determined, everything in the fight-hand side of Eq. (12.3-13) is known but kf, so kf can be computed from any one of the experiments. Since there is always some experimental error, a reasonable policy is to calculate the rate constant separately from each experiment and then to average the values.

416

12 The Rates of Chemical Reactions

The Method of Isolation In this method, an experiment is carried out in which the initial concentration of one reactant is made much smaller than the concentrations of the other reactants. During the reaction the fractional changes in the large concentrations are negligible, and these concentrations are treated as constants. The species having the small concentration is monitored as in the case of a single reactant. For example, in the reaction of Eq.

12.4

Inclusion of a Reverse Reaction. Chemical Equilibrium

417

(12.3-12), if [A] is much smaller than [B] and [F], the relative changes in [B] and [F] will be small. We write

1 d[A] - - ~ = (kf[B]fl[F]O)[A] ~ a dt

(12.3-15)

where the quantity in parentheses is approximately constant. Data from this kind of experiment can be treated like data from reactions with a single reactant. For example, if c~ = 2, Eq. (12.2-11) can be transcribed to obtain

1

1 = ~ + [A]t [A]0

(kf[B]fl[F]~)t

(12.3-16)

Sets of experiments can also be carried out in which [B] is made much smaller than [A] and [F], and then in which [F] is made much smaller than [A] and [B], in order to determine t , ~b, and kf. If a reaction in a solution includes the solvent as a reactant, the concentration of the solvent is usually much larger than the concentrations of other reactants and is almost constant. Assume that the solvent S is involved in the reaction A + S ---->products

(12.3-17)

and that the rate law is

r=

d[A] = k[SI~[A] ~ _ kapp[A]~ dt

(12.3-18)

where a is the order with respect to the solvent. [S] is almost constant in a dilute solution. The quantity /Capp is equal to k[S] G and is almost constant. It is called an apparent rate constant. The order with respect to substance A and the apparent rate constant can be determined by any of the methods that apply to a single reactant. However, the actual rate constant k and the order with respect to the solvent cannot be determined unless the concentration of the solvent can be varied. If the reaction is first order with respect to substance A and of unknown order with respect to the solvent, the reaction is called a pseudo first-order reaction. If the reaction is second order with respect to substance A, the reaction is called pseudo second-order, and so on.

Inclusion of a Reverse Reaction. Chemical Equilibrium So far, we have treated only reactions for which the reverse reaction could be neglected. We now include a reverse reaction for the simplest case, that the reaction is first order in both directions and that there is one product and one reactant: kf A ~,-~-B kr

(12.4-1)

where kf is the rate constant for the forward reaction and kr is the rate constant for the reverse reaction. Other cases, such as second order in one direction and first order in the other direction and reactions with more, than two substances, are much more difficult to treat and we will not discuss them explicitly.

418

12 The Rates of Chemical Reactions

The net (observable) rate of the reaction is given by the difference between the forward rate and the reverse rate: rnet=--

d[A] dt

=kr[Al-kr[B]

(12.4-2)

At equilibrium, rnet(eq ) --" 0 = k f [ A ] e q -- k r [ B l e q

(12.4-3)

Equation (12.4-3) is the same as

(12.4-4)

where Keq is the equilibrium constant for the reaction. A large value for the equilibrium constant means that the rate constant for the forward reaction is large compared with the rate constant for the reverse reaction. A small value means that the rate constant for the forward reaction is small compared with the rate constant for the reverse reaction. Equation (12.4-4) can apply to a more general case as shown in the following exercise.

Exercise 12.14 Assume that for the reaction kf

aA + bB ~ dD + f F k~

(12.4-5)

the forward and reverse rates of the reaction are given by rf = kf[A]a[B] b

and

r r = kr[D]d[F]f

(12.4-6)

That is, assume that the order with respect to each substance is equal to its stoichiometric coefficient (which might or might not be the case in a real reaction). Show that

kf

Keq = ~r

(12.4-7)

We subtract Eq. (12.4-3) from Eq. (12.4-2) to obtain d[A] dt = kf([A] -[A]eq) - kr([B ] -[B]eq)

(12.4-8)

Equation (12.4-8) appears to contain two dependent variables. However, we can express [B] in terms of [A]. Assume that initially only substance A is present so that [B]0 = 0: [B] = [A]0 - [ A ]

and

[B]eq = [A]0 -[A]e q

so that [B] - [ B ] e q = [A]0 - [ A ] - ([A]o - [A]eq) = - [ A ] -+- [A]eq

(12.4-9)

When the relation of Eq. (12.4-9) is substituted into Eq. (12.4-8), we get d[A] dt = (kf + kr)([A ] -[A]eq)

(12.4-10)

Since [A]e q is a constant for any particular initial condition, we can replace d[A]/dt by d ( [ A ] - [A]eq)/dt. Equation (12.4-10) is identical to Eq. (12.2-2) except for the

12.4

Inclusion of a Reverse Reaction. Chemical Equilibrium

419

symbols used, and the solution is obtained by transcribing Eq. (12.2-5) with appropriate changes in symbols: [ A ] r - [A]e q -- ([A]0 -[A]eq)e-(kf+kr)

t'

(12.4-11)

Exercise 12.15 Carry out the separation of variables to obtain Eq. (12.4-11). Figure 12.5 shows the concentration of a hypothetical reactant as a function of time. [ A ] - [A]e q decays exponentially, as did [A] in the case of Figure 12.2. We define the half-life of the reversible reaction to be the time required for [A] - [A]e q to drop to half of its initial value. We find that

tl/2

ln(2) = kf _i_ kr

(12.4-12)

Exercise 12.16 Verify Eq. (12.4-12).

The relaxation time z is the time required for [ A ] - [A]e q to drop to 1/e of its original value:

z = ~

1

kf+kr

(12.4-13)

A large value of the reverse rate constant is as effective in giving a rapid relaxation to equilibrium as is a large value of the forward rate constant, even if there is no product initially present.

*Exercise 12.17 For a hypothetical is0merization with kf = 17.7 min -1 and kr --32.2 min -1, find the final composition if the initial concentration of A is equal to 0.175 molL -1 and the initial concentration of B is equal to zero. Find the half-life and the relaxation time. Find the composition at time t - 0.100 min.

Exercise 12.18 The treatment of the reaction kf A ~,-~-B kr

can be carried out for [B]0 -r O. Carry out this analysis and compare your results with those corresponding to the assumption that [B]0 = O.

420

12 The Rates of Chemical Reactions

Consecutive Reactions and Competing Reactions We now consider two cases in which two reactions are linked together. The first case is that of consecutive reactions in which the product of the first reaction is the reactant of a succeeding reaction. The second case is that of competition between two reactions with the same reactant. We consider only the cases in which both reactions are first order with a single reactant.

Consecutive Reactions Consider the case that a single reactant forms a second substance in a first-order reaction with a negligible reverse reaction, and that this substance forms a product in another first-order reaction with a negligible reverse reaction: kl

A

>B

k2

>F

(12.5-1a)

where kl is the rate constant for the first step and k2 is the rate constant for the second step. We number the reactions as follows: (1)

A ~

B

(12.5-1b)

(2)

B ~

F

(12.5-1c)

Equation (12.5-1) constitutes a reaction mechanism. Many chemical reactions consist of two or more sequential steps analogous to Eqs. (12.5-1b) and (12.5-1c). When the steps in a sequence of reactions are numbered, the rate constants for the steps are labeled with the number of the step. If both forward and reverse reactions are included in step number i, the rate constant for the forward reaction will be called k i and the rate constant for the reverse reaction will be called k~. The reaction shown in Eq (12.5-1b) has the same rate law as Eq. (12.2-2), alIA] dt

=-k~

dt

(12.5-2)

The rate law for the reaction of Eq. (12.5-1c) is d[B] dt

-- k l [ A ] - k2[B]

(12.5-3)

Equations (12.5-2) and (12.5-3) are a set of simultaneous differential equations. However, the first equation does not contain [B] and can be solved separately. Its solution has already been obtained: [A], = [A]0e-kit

(12.5-4)

This solution can be substituted into Eq. (12.5-3) to obtain a single differential equation relating [B] and t: d[B] dt

= kl[A]~

- k2[B]

(12.5-5)

The solution of this equation is carried out in Appendix B for the case that no B or F is present at time t = 0. The solution is kl [A]0 (e_k,t _ e_k2t) [Bit = k2 _ kl

(12.5-6)

12.5 Consecutive Reactions and Competing Reactions

421

Exercise 12.19 Substitute the function of Eq. (12.5-6) into the original differential equation of Eq. (12.5-5)and show that it satisfies this equation. The concentration of the final product F is obtained from IF] = [A]0 - [ A ] -

[B]

(12.5-7)

Figure 12.6a shows the concentrations of all three substances for the case that k 1 = 0.100 s -1 and k 2 - - 0 . 5 0 0 s -1, and Figure 12.6b shows the concentrations for the case that k 1 = 0 . 5 0 0 s - ! and k 2 = 0.100 s -1. Since the reverse reactions are assumed to be negligible, the final state is complete conversion to the product F in both cases. If k 1 < k2, the amount of B remains relatively small, but if k 1 > k 2, the amount of B becomes fairly large before dropping eventually to zero. If k 1 = k2, the solution of (12.5-6) cannot be used since it is not permissible to divide by zero. See Problem 12.44 for the solution in this case. If steps 1 and 2 have reverse reactions, (1)

A ~--- B

(12.5-8a)

(2)

B ~ F

(12.5-8b)

422

12

The Rates of Chemical Reactions

the differential equations giving the rates are d[A] = -kl[A] + UI[B ] dt

(12.5-9a)

d[B] dt

(12.5-9b)

-

'

-

k 1[A] + k'1[B] - k2[B ] + k~[F]

d[F] = k2[B ] - k;[F] dt

(12.5-9c)

This set of simultaneous differential equations presents a more difficult mathematical problem than if the reverse reactions are negligible. It can be solved but we will not present the solution. 5 In this case, both steps are at equilibrium when the entire reaction is at equilibrium: [B]eq

[A]eq

__ k_.[ ~_

K~

(12.5-10)

k~l

and

[F]eq __ k2 [B]eq -- ~22-- K2

(12.5-11)

The equilibrium constant K for the overall reaction is equal to

K-[F]eq[A]e q

[F]eq [B]eq-" K 1 K 2 - kl k2 [B]e q [A]e q kll k~

(12.5-12)

The relationships shown in Eq. (12.5-12) are valid for any stepwise reaction: If the orders in all steps are equal to the stoichiometric coefficients, the equilibrium constant is equal to the product of all of the rate constants for the forward reactions divided by the product of all of the rate constants for the reverse reactions.

Competing (Parallel) Reactions In many syntheses a side reaction occurs that consumes part of the reactants but gives a product other than the one desired. We consider the case of two such competing reactions, both of which are assumed to be first order with negligible reverse reaction. (1)

A --+ F

(12.5-13a)

(2)

A --+ G

(12.5-13b)

The rates of the two reactions combine to give d[A] at = (k~ + k2)[A ]

(12.5-14)

This equation is the same as Eq. (12.2-2) except that kf is replaced by k 1 + k2, and the solution is [A]t = [A]oe-(k, +k2)t'

5 T. M. Lowry and W. T. John, J. Chem. Soc., 97, 2634 (1910).

(12.5-15)

12.5 ConsecutiveReactions and Competing Reactions

423

The half-life for the disappearance of A is ln(2) tl/2 -- kl + k2

(12.5-16)

In other words, [A] decreases as in a single first-order reaction except that the sum of the rate constants appears instead of a single rate constant. Consider the concentrations of the two products. We have d[F] = kl [A] - k 1[m]0 e-(kl +k2)t

(12.5-17)

dt

The fight-hand side of this equation does not contain [F], so we can multiply by dt and integrate to obtain -kl[A]~ (e -(kl+k2)t' - 1) [ F ] t , - [F]0 = [F]t, -- kl q- k 2

(12.5-18)

where we assume that [F]o - 0. A similar treatment for [G] gives the same result except that k 1 is replaced by k2 in the numerator: -k2 [A]o (e-(kl +k2)t' [ G l t , - [G]o = [Glt, -- kl + ~ - 1)

(12.5-19)

The ratio [F]/[G] is thus the same at any time:

[F]

[~1

k1 = --

k2

(12.5-20)

Exercise 12.20 For the reactions shown in Eq. (12.5-13) assume that [A]0 - 0.500 mol L -1, that k 1 -- 0.100 S-1 and that k2 = 0.0100 s-1. Construct a graph showing [A], IF], and [G] for t ranging from 0 to 20s. If the reverse reactions cannot be neglected, the situation can be different. If the system comes to equilibrium, [F]eq - kl - Ka [A]eq ka

[G]eq k2 [A]eq -- k22 -- K2

(12.5-21 a)

(12.5-21 b)

so that

[F]eq _. k__.l.1~2 ._. k_..l_lk~ = g___l_l [G]e q k~ k 2 k 2 k 1 K 2

(12.5-22)

Depending on the values of the four rate constants, this ratio might differ significantly from the ratio in Eq. (12.5-20). If F is a desired product and G is an undesired product of a side reaction, the ratio of [F] to [G] might be optimized by allowing the system to come to equilibrium (using "thermodynamic control") or by stopping the reaction before it comes to equilibrium (using "kinetic control").

424

12 The Rates of Chemical Reactions

The Experimental Study of Fast Reactions The "classical" method of studying reaction rates is to mix the reactants and then to determine the concentration of some reactant or product as a function of time. This method is clearly inadequate if the reaction time is comparable to or shorter than the time required to mix the reactants.

Flow Techniques There are two common flow methods that can be used to speed up the mixing of liquids or gases. In the continuous-flow method, two fluids are forced into a mixing chamber. The newly mixed fluid passes into a transparent tube of uniform diameter. The flow rates into the mixing chamber are kept constant so that the distance along the tube is proportional to the elapsed time after mixing. The concentration of a reactant or product is determined spectrophotometrically as a function of position along the tube, using the tube as a spectrophotometer cell. The position dependence of the concentration is translated into time dependence from knowledge of the flow rate. In the stopped-flow method, two fluids are forced into a mixing chamber as in the continuous flow method. After a steady state is attained, the flow of solutions into the chamber is suddenly stopped and the concentration of a product or reactant is determined spectrophotometrically as a function of time, using the mixing chamber as a spectrophotometer cell. Figure 12.7 schematically shows a stopped-flow apparatus. Flow systems have been designed that can mix two liquids in a tenth of a millisecond, so that reactions with half-lives of from 1 ms to 1 s can be studied by either of the two flow methods.

12.6 The Experimental Study of Fast Reactions

425

Relaxation Techniques These techniques do not rely on mixing but use the fact that equilibrium compositions depend on temperature and sometimes on pressure. The experiment begins with a system at equilibrium and the temperature or pressure is suddenly changed so that the system is no longer at equilibrium. The relaxation of the system to its new equilibrium state is then monitored. Figure 12.8 shows the range of reaction half-lives for which each of several techniques can be used. In the shock-tube method a reaction vessel is constructed with two chambers separated by a diaphragm that can be ruptured suddenly. On one side is a mixture of gaseous reactants and products at equilibrium at a fairly low pressure. On the other side is a "driver" gas at a much higher pressure. When the diaphragm is ruptured the driver gas moves quickly into the low-pressure chamber. Collisions of the driver gas molecules with the other molecules produce a shock wave that propagates through the low-pressure gas and heats it. The reacting system will then relax to the equilibrium state for the new temperature and the concentration of a reactant or product is monitored spectrophotometrically. This method is applied to reactions that have halflives in the range from 1 ms to 1 ps, but it is limited to gas-phase reactions. In the flash photolysis method 6 a brief burst of light irradiates the system. If this light is absorbed it can quickly change the composition of the system and its temperature. The concentration of a reactant or product is then measured spectroscopically as a function of time as the system relaxes to its new equilibrium. Figure 12.9 shows schematically an apparatus for flash photolysis. Flash photolysis differs from the shock-wave technique in that the irradiation ordinarily does more than change the temperature of the system. Photochemical processes can produce new species so that the system is far from equilibrium immediately after the irradiation. In the temperature-jump ("T-jump") and the pressure-jump ("P-jump") methods a gaseous or liquid system is subjected to a rapid heating or a rapid change in pressure. A heating pulse can be delivered by a burst of microwave radiation or by the passage of a brief pulse of electric current if the system is electrically conductive. A rapid change in pressure can be achieved by rupturing a diaphragm. The T-jump technique usually

6See G. Porter, J Chem. Soc., Faraday Trans. 2, 82, 2445 (1986) for a historical survey.

426

12 The Rates of Chemical Reactions

produces a larger effect and is more commonly used than the P-jump technique, which will produce a significant change in equilibrium composition only for a gas-phase reaction. After the temperature or pressure change, the system relaxes to its new equilibrium state. The concentration of a reactant or product is usually monitored spectroscopically, although the reaction of hydrogen ions and hydroxide ions was monitored by measurement of the electrical conductivity. Consider a system in which the reaction can occur: (1)

A+B~C

(12.6-1)

Assume that this reaction is second order overall in the forward direction and first order in the reverse direction. Suppose that a temperature or pressure jump is suddenly imposed on the system at time t - 0. The system then relaxes to its new equilibrium state under conditions of constant temperature and pressure. The time to accomplish the T-jump or P-jump should not be greater than 10% of the half-life of the reaction. Figure 12.10 shows schematically the concentrations of A, B, and C before and after a T-jump. The initial concentration [A]0 was the equilibrium concentration at the temperature prior to the temperature jump, but since the equilibrium constant of the reaction depends on temperature, [A]o is not equal to the new equilibrium concentration, denoted by [A]e q. The same is true of [B] and [C]. We now let A[A] = [A]t - [A]e q

(12.6-2a)

A[B] = [B]t- [B]e q

(12.6-2b)

A[C] = [C]t- [C]eq

(12.6-2c)

We assume that ]A[A]I << [A], IA[B]I << [B], and IA[C]] << [C], since it is not possible to change the equilibrium composition very much with a temperature jump or a pressure jump. A temperature jump is usually limited to about 20~ and a pressure jump is limited to 2 or 3 atm. From the stoichiometry of the reaction shown in Eq. (12.6-1), A[A] = A[B] = -A[C]

(12.6-3)

12.6 The Experimental Study of Fast Reactions

427

so that we can express the concentrations in terms of A[C]" [C] - [ C ] e q + A[C]

(12.6-4a)

[A] - [ A ] e q - A[C]

(12.6-4b)

[B] -- [B]e q - A[C]

(12.6-4c)

The differential equation for the net rate is dA[C] -_ kl[A][B]_ k,l[C] R a t e - - d---7-

(12.6-5)

To solve this differential equation, we must write it in terms of one dependent variable, which we do using Eq. (12.6-4): dA[C] dt

= kl ([A]eq - A[C])([B]eq - A[r

- k~l([C]eq + A[C])

= k 1[A]eq[g]eq - k'1[Cle q -- k 1([A]eq + [g]eq)A[C] -- k'IA[C ] + kl(A[C]) 2

(12.6-6)

The first two terms on the right-hand side of the final version of Eq. (12.6-6) cancel because the first is the forward rate at equilibrium and the second is the reverse rate at equilibrium. The final term on the right-hand side is much smaller than the others because the deviation from equilibrium is small so that (A[C]) 2 << ]A[C]]. We neglect this term, which linearizes the equation: dA[C____~]= _(kl([A]eq "nt- [B]eq) + k,1) A[C]

(12.6-7)

dt

Equation (12.6-7) is exactly like Eq. (12.2-2) except for the symbols used, so we can write the solution: A [ C ] - A[C]0e-t/~

(12.6-8)

428

12

The Rates of Chemical Reactions

where 1

- = k 1([A]e q 4- [B]eq) 4- k' 1

r

(12.6-9)

The quantity z is the relaxation time for A[C]. The relaxation is exponential because we linearized the equation.

Exercise

12.21

a. Verify the steps of algebra leading to Eq. (12.6-7). b. Verify that Eq. (12.6-8), with Eq. (12.6-9), is a solution to Eq. (12.6-7). e. Write the expressions for A[A] and A[B]. The reaction of hydrogen ions and hydroxide ions in water can be written as H3 O+ 4 - O H - ~ 2H20

(12.6-10)

which has the general form A+B

~ 2C

(12.6-11)

Assume that the reaction is second order in both directions. We write [C] = [C]eq 4- A[C]

(12.6-12a)

[A] = [A]eq - I A[C]

(12.6-12b)

[B] - [B]eq - 21-A[C]

(12.6-12c)

When Eqs. (12.6-12) are substituted into the differential equation for the rate of the reaction and the necessary steps of algebra are carried out with neglect of terms proportional to (A[C]) 2, we obtain A[C] = A[C]0 e-'/~

(12.6-13)

where 1 _

-

kl [A]eq 4- [B]eq 2 + 2k'1[C]eq

(12.6-14)

12.6 The Experimental Study of Fast Reactions

429

Similar expressions for the relaxation time can be derived for other rate laws. 7

Exercise 12.22 a. Verify Eq. (12.6-14). b. For the reaction equation A+B ~--X+Y assumed second order in both directions, show that the relaxation is exponential, with relaxation time given by 1

- = kl([A]eq + [B]eq) + k'l ([X]eq nt-[Y]eq) T

(12.6-15)

Summary of the Chapter A rate law of the form R a t e - k[A]~[B] ~ is said to have definite order, with order c~ with respect to A and with order/~ with respect to B. The proportionality constant k is called the rate constant although it depends on temperature. We solved several such differential rate laws to obtain the integrated rate laws. Some techniques for experimental determination of the rate law involve comparison of the integrated rate equation with concentration data. The method of initial rates allows direct comparison of the differential rate law with the experimental data. In the method of isolation, the concentration of one reactant is made much smaller than the concentrations of the other reactants. During the reaction, the fractional changes in the larger concentrations are negligible, and the small concentration behaves like the concentration in a reaction with one reactant. For a reversible reaction, the difference between the concentration of the reactant and its equilibrium value relaxes exponentially, and it was found that the relaxation time and the half-life are both inversely proportional to the sum of the two rate constants. The case of two consecutive first-order reactions without reverse reactions was considered. This is a simple example of a reaction mechanism. It was found that the concentration of the reactive intermediate rose and then fell as the reaction proceeded. Some techniques were presented for studying fast reactions that cannot be studied by classical experimental techniques. These techniques included continuous-flow and

7K. J. Laidler, Chemical Kinetics, 3d ed., Harper and Row, New York, 1987, p. 38.

430

12

The Rates of Chemical Reactions

stopped-flow techniques, which are rapid mixing methods, as well as relaxation techniques. The relaxation techniques included shock-tube methods, flash photolysis, and T-jump and P-jump methods. Equations were derived for the relaxation of a reaction after a small perturbation, giving an exponential relaxation for a variety of rate laws.

Problems

431

432

12

The Rates of Chemical Reactions

Problems

433

434

i. In the case of two consecutive reactions with reverse reactions, the concentration of the intermediate always remains small compared with the initial concentration of the reactant. ]. The linearization of the rate equation that is done in the study of the temperature-jump method is usually a good

12

The Rates of Chemical Reactions

approximation, because the state immediately after the temperature jump does not deviate very much from the final equilibrium state. k. First-order processes occur only in chemical processes. 1. All rate laws can be written in a form with definite orders.