Page 81 - Bruno Linder Elementary Physical Chemistry

August 18, 2010 11:36 9in x 6in b985-ch07 Elementary Physical Chemistry

66 Elementary Physical Chemistry

7.6. Half-Lives

The half-life of a reaction, t 1/2 , is the time taken for the concentration of
the substance to reduce to half its value. In a first-order reaction, from
Eq. (7.18b),

k A t 1/2 =ln{[A 0 ]/[A]} =ln{[A 0 ]/{(1/2)[A 0 ]} (7.20a)
and thus,

t 1/2 =ln 2/k A (7.20b)
Example 7.3. The half-life of an enzyme-catalyzed reaction is t 1/2 = 138 s.
The reaction is first order. If the initial concentration [A] 0 =1.28 mmol L −1 .
How long will it take for the concentration to fall to 0.040 mmol L −1 .

Solution

k A =ln 2/138 s = 5.022 × 10 −3 −1
s
Since

k At =ln 1.28 mmolL −1 /0.040 mmolL −1
=3.4687

we obtain
t = 690 s.

Note: In a first-order reaction, the half-life is independent of concen-
tration. But in a second-order reaction, the half-life depends on the
initial concentration. Exercise: Show that t 1/2 =1/{k A[A] 0 }.

Example 7.4. The reaction rate for the reaction

2N 2 O 5 → 4NO 2 +O 2
is of the form r = −1/2d[N 2O 5 ]/dt = k[N 2O 5 ]
a) Write an expression in terms of k[N 2 O 5 ] for the rate of production of
NO 2 , i.e. d[NO 2 ]/dt.
s
b) If k =1.73 × 10 −5 −1 what is the half-life of N 2 O 5 ?