Physical Chemistry Third Edition

11.2 Forward Reactions with One Reactant 491

Solution

k

ln(2)

t 1 / 2



ln(2)

7. 1 × 108 yr

 9. 8 × 10 −^10 yr−^1

τ

1

k



1

9. 8 × 10 −^10 yr−^1

 1. 02 × 109 yr

Exercise 11.3

Find the time required for a sample of^235 U 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

rrf−

d[A]

dt

kf[A]^2 (11.2-8)

where the rate constantkfhas units of concentration−^1 time−^1 (L mol−^1 s−^1 ,m^3 mol−^1

min−^1 , and so on) or pressure−^1 time−^1 (atm−^1 s−^1 , bar−^1 min−^1 , and so on). With the

variables separated,

1

[A]^2

d[A]

dt

dt

d[A]

[A]^2

−kfdt (11.2-9)

If the temperature is constant,kfis constant, and we can carry out a definite integration

from timet0 to timett′.

∫[A]t′

[A] 0

d[A]

[A]^2

−kf

∫t′

0

dt (11.2-10)

The result is

1

[A]t′

−

1

[A] 0

kft′ (second order, no reverse reaction) (11.2-11)

where[A]t′denotes the concentration of substance A at timet.

Exercise 11.4

Carry out an indefinite integration and evaluate the constant of integration to obtain Eq. (11.2-11)

in an alternative way.