11.4 Separable equations in chemical kinetics 325
If the initial concentration of Ais[A]
0, at timet 1 = 10 , thenc 1 = 1 ln[A]
0and the solution
of the differential rate equation is the integrated rate equation
ln[A] 1 = 1 −kt 1 + 1 ln[A]
0(11.34)
A plot ofln[A]against tis a straight line (Figure 11.1)
whose slope is −kand whose intercept with theln[A]axis
isln[A]
0.
The more conventional expression for the integrated rate
equation is obtained from (11.34) by taking the exponential
of each side,
[A] 1 = 1 [A]
01 e
−kt(11.35)
and shows that the first-order process is an example of
exponential decay.
The half-life
The half-lifeτ
122of an exponential decay is the interval in which the amount of
reactant is halved:
or, from (11.35),
Therefore and, taking logs,
0 Exercise 29
(2) The second-order process: 2A 1 → 1 products
The rate equation is
(11.36)
or,
(11.37)
dx
dt
=− 2 kx
2v=− =
1
2
2d
dt
k
[]
[]
A
A
τ
122
=
ln
.
k
e
−k=
τ1212
[] []
()AA
00121
2
ee
ktkt−+−=
τxt()()+=τ xt
1212..............................................................................................................................................................................0
ln[A]
0−k
t
ln[A]
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Figure 11.1