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
k f [A]
(11.2-2)
where the first-order rate constant k f has 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 by dt, divide by [A], and recognize that
d[A]
dt
dt d[A]:
1
[A]
d[A]
dt
dt
1
[A]
d[A] −k f dt
(11.2-3)
The variables have been separated. If the temperature is constant, k f is constant and
we can carry out a definite integration from t 0 to t t :
[A] t
[A] 0
1
[A]
d[A] −
t
0
k f dt
ln([A] t ) − ln([A] 0 )
ln([A] t )
ln([A] 0 )
−k f t (first order,
no reverse reaction)
(11.2-4)
where the subscript on [A] indicates the time at which it is measured. Taking antilogarithms of Eq. (11.2-4),
[A] t [A] 0 e −k f t (first order, no reverse reaction)
(11.2-5)
where we have written t instead of t 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.
E X A M P L E 11.1
The gas-phase reaction
N 2 O 5 − → 2NO 2 +
1
2
O 2