Physical Chemistry Third Edition

512 11 The Rates of Chemical Reactions

thatk 1  0 .500 s−^1 andk 2  0 .100 s−^1 .Ifk 1 <k 2 , the amount of B remains relatively

small, but ifk 1 >k 2 , the amount of B becomes fairly large before dropping eventually to

zero. Since the reverse reactions are assumed to be negligible, the final state corresponds

to complete conversion to the product F in both cases.

If steps 1 and 2 have reverse reactions that cannot be neglected, the mechanism

becomes

(1) A
B (11.5-8a)

(2) B
F (11.5-8b)

the differential equations giving the rates are

d[A]

dt

−k 1 [A]+k 1 ′[B] (11.5-9a)

d[B]

dt

k 1 [A]+k′ 1 [B]−k 2 [B]+k′ 2 [F] (11.5-9b)

d[F]

dt

k 2 [B]−k′ 2 [F] (11.5-9c)

where we use a prime (′) to label the reverse rate constants. This set of simultaneous

differential equations can be solved, but we do not present the solution.^5

Both steps are at equilibrium when the entire reaction is at equilibrium:

[B]eq

[A]eq



k 1

k′ 1

K 1 (11.5-10)

and

[F]eq

[B]eq



k 2

k′ 2

K 2 (11.5-11)

The equilibrium constantKfor the overall reaction is equal to

K

[F]eq

[A]eq



[F]eq

[B]eq

[B]eq

[A]eq

K 1 K 2 

k 1

k′ 1

k 2

k 2 ′

(11.5-12)

Relationships analogous to that shown in Eq. (11.5-12) are valid for any sequence

of steps: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 if

activity coefficients can be ignored.

Exercise 11.21

Consider the sequence of reactions:

(1) aA+bB
cC

(2) C+dD
fF+gG

Assume that the orders are equal to the stoichiometric coefficients and show that the above

assertion is correct in this case.

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