Physical Chemistry Third Edition

508 11 The Rates of Chemical Reactions

We can express [B] in terms of [A]. Assume that initially only substance A is present

so that[A] 0 0 and[B] 0 0:

[B][A] 0 −[A] and [B]eq[A] 0 −[A]eq

so that

[B]−[B]eq[A] 0 −[A]−([A] 0 −[A]eq)−[A]+[A]eq (11.4-9)

When this relation is substituted into Eq. (11.4-8), we obtain

−

d[A]

dt

(kf+kr)([A]−[A]eq) (11.4-10a)

Since[A]eqis a constant for any particular initial condition, we can replaced[A]/dt

byd([A]−[A]eq)/dt.

−

d

(

[A]−[A]eq

)

dt

(kf+kr)([A]−[A]eq) (11.4-10b)

Equation (11.4-10b) is the same as Eq. (11.2-2) except for the symbols used, and the

solution is obtained by transcribing Eq. (11.2-5) with appropriate changes in symbols:

[A]t′−[A]eq([A] 0 −[A]eq)e−(kf+kr)t

′

(11.4-11)

The difference[A]−[A]eqdecays exponentially, as did[A]in the case of Figure 11.2.

Exercise 11.17

Carry out the separation of variables to obtain Eq. (11.4-11).

Figure 11.5 shows the concentration of a hypothetical reactant as a function of time.

0 t

(a)

[A] [A]

eq

0 t

(b)

[A]

2

[A]

eq

Figure 11.5 Concentration of the

Reactant in a Hypothetical Reaction

with Forward and Reverse Reac-

tions.(a)[A]asafunctionoftime.

(b) [A]−[A]eqas a function of time.

We define the half-life of the reversible reaction to be the time required for

[A]−[A]eqto drop to half of its initial value. We find that

t 1 / 2 

ln(2)

kf+kr

(11.4-12)

Exercise 11.18

Verify Eq. (11.4-12).

We define therelaxation timeτas the time required for[A]−[A]eqto drop to 1/e

of its original value:

τ

1

kf+kr

(11.4-13)

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.