Physical Chemistry Third Edition

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 .aIf

the partial pressure of N 2 O 5 is 0.500 atm at timet0, find the partial pressure of N 2 O 5

att 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 )tP(N 2 O 5 ) 0 e−kft

(0.500 atm)exp

[

−(5. 12 × 10 −^3 s−^1 )(60.0s)

]

 0 .368 atm

aH. S. Johnston and Y. Tao,J. Am. Chem. Soc., 73 , 2948 (1951).

Thehalf-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)

kft 1 / 2 −ln

([

A]t 1 / 2

[A] 0

)

−ln

(

1

2

)

ln(2)

or

t 1 / 2 

ln(2)

kf

≈

0. 69315

kf

(first order,

no reverse reaction)

(11.2-6)

Therelaxation timeτis 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

kf



t 1 / 2

ln(2)

≈ 1. 4427 t 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.

EXAMPLE11.2

The half-life of^235 U is equal to 7. 1 × 108 years. Find the first-order rate constant and the

relaxation time.