Physical Chemistry Third Edition

11.2 Forward Reactions with One Reactant 489

First-Order Reactions

Consider a reaction at constant temperature of a single reactant:

A−→products (11.2-1)

If there is no significant reverse reaction and if the reaction is first order, the rate law is

r−

d[A]

dt

kf[A] (11.2-2)

where the first-order rate constantkfhas units of reciprocal time (s−^1 , min−^1 , and

so on). This is a differential equation that can be solved by separation of variables.

We multiply bydt, divide by [A], and recognize that

d[A]

dt

dtd[A]:

1

[A]

d[A]

dt

dt

1

[A]

d[A]−kfdt (11.2-3)

The variables have been separated. If the temperature is constant,kfis constant and

we can carry out a definite integration fromt0tott′:

∫[A]t′

[A] 0

1

[A]

d[A]−

∫t′

0

kfdt

ln([A]t′)−ln([A] 0 )

ln([A]t′)

ln([A] 0 )

−kft′

(first order,

no reverse reaction)

(11.2-4)

where the subscript on [A] indicates the time at which it is measured. Taking antiloga-

rithms of Eq. (11.2-4),

[A]t[A] 0 e−kft (first order, no reverse reaction) (11.2-5)

where we have writtentinstead oft′for the time.

Exercise 11.1

Carry out an indefinite integration of Eq. (11.2-3). Evaluate the constant of integration to obtain

Eq. (11.2-5).

For a gas-phase reaction we can use partial pressures instead of concentrations in

our rate law.

EXAMPLE11.1

The gas-phase reaction

N 2 O 5 −→2NO 2 +

1

2

O 2