11.4 Separable equations in chemical kinetics 327
This is the same as in case (2)above except for the factor 2, which distinguishes case
A 1 + 1 Afrom caseA 1 + 1 B.
(b)The initial concentrations of Aand Bare not equal.
Separating the variables in (11.41) and integrating,
The integral over the variable xis evaluated by expressing the integrand in partial
fractions (see Sections 2.7 and 6.6, and Examples 2.28 and 6.17). Write
It is required that
11 = 1 A(b 1 − 1 x) 1 + 1 B(a 1 − 1 x)
be true for all values of x. Settingx 1 = 1 bgivesB 1 = 112 (a 1 − 1 b)and settingx 1 = 1 agives
A 1 = 112 (b 1 − 1 a). Therefore
and
Therefore
At timet 1 = 10 ,x 1 = 10 , and so that
(11.45)
1
ba
ab x
ba x
kt
−
−
−
ln =
()
()
c
ba
b
a
=
−
1
ln ,
1
ba
bx
ax
kt c
−
−
−
ln =+
=
−
−−+−
()=
−
−
−
11
ba
ax bx
ba
bx
ax
ln( ) ln( ) ln
ZZ
dx
axbx ba ax bx
dx
()()−−
=
−−
−
−
111
1111
()()axbx baax bx−−
=
−−
−
−
1
()()
()()
axbx ()(
A
ax
B
bx
Ab x Ba x
−− axb
=
−
−
=
−+ −
−−xx)
dx
axbx
kdt
dx
axbx
kt c
()()−− ()()
=,
−−