490
11 The Rates of Chemical Reactions
is found to be first order. The rate constant at 337.6 K is equal to 5.12 × 10 −3 s −1 . a If
the partial pressure of N 2 O 5 is 0.500 atm at time t 0, find the partial pressure of N 2 O 5
at t 60.0 s, neglecting any reverse reaction and assuming that the system is at constant
volume.
Solution
Assuming the gas to be ideal, the partial pressure of a gas is proportional to the concentration.
In the case of constant temperature Eq. (11.2-5) can be rewritten in terms of pressure:
P(N 2 O 5 ) t P(N 2 O 5 ) 0 e −k f t
(0.500 atm)exp
−(5.12 × 10 −3 s −1 )(60.0 s)
0.368 atm
a H. S. Johnston and Y. Tao, J. Am. Chem. Soc., 73, 2948 (1951).
The half-life, t 1/2 , is defined as the time required for half of the original amount of
the reactant to be consumed. Since [A] t 1/2 [A] 0 /2, we can write from Eq. (11.2-4)
k f t 1/2 − ln
[A] t 1/2
[A] 0
−ln
1
2
ln(2)
or
t 1/2
ln(2)
k f
≈
0.69315
k f
(first order,
no reverse reaction)
(11.2-6)
The relaxation time τ is the time for the amount of reactant to drop to a fraction
1/e (approximately 0.3679) of its original value. Substitution of this definition into
Eq. (11.2-4) gives
τ
1
k f
t 1/2
ln(2)
≈ 1.4427t 1/2
(first order,
no reverse reaction)
(11.2-7)
Exercise 11.2
a. Verify Eq. (11.2-7).
b. Find the half-life and the relaxation time for the reaction of Example 11.1.
The decay of radioactive nuclides obeys first-order kinetics with rate constants that
do not appear to depend on temperature. Half-lives, not rate constants, are tabulated
for radioactive nuclides.
E X A M P L E 11.2
The half-life of 235 U is equal to 7.1 × 10 8 years. Find the first-order rate constant and the
relaxation time.