330 Chapter 11First-order differential equations
Therefore, by formula (11.53),
0 Exercises 33–40
11.6 An example of linear equations in chemical kinetics
We saw in Section 11.4 that many (in fact, most) chemical reactions proceed through
several elementary steps, and a description of the overall rate process then involves
several simultaneous first-order differential equations. Such a system of equations can
be solved only in simple cases without resorting to approximations, and numerical
methods must otherwise be used. One of the simplest solvable systems is that of two
consecutive first-order processes,
in which Ais transformed into Cthrough the intermediate species B. The process is
modelled by the two first-order equations
(11.54)
(11.55)
We assume that the initial concentrations, at timet 1 = 10 , are[A]
01 = 1 a,[B]
01 = 10 , and
[C]
01 = 10 , and that the concentrations at time tare[A] 1 = 1 a 1 − 1 xand[B] 1 = 1 y, so that
[C] 1 = 1 x 1 − 1 y. Then
(11.56)
(11.57)
The first of these equations is the separable equation (11.32), with solution
a 1 − 1 x 1 = 1 ae
−k
1t(11.58)
Substitution of this result into equation (11.57) gives
dy
dt
ak e k y
kt=−
−121dy
dt
=−−ka x ky
12()
da x
dt
ka x
()
()
−
=− −
1d
dt
kk
[]
[] []
B
=−AB
12d
dt
k
[]
[]
A
=− A
1ABC
kk
12
→→
y
xc
x
=+
422
xy x xdx c
x
c
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