Physical Chemistry Third Edition

B Some Useful Mathematics 1245

B.4 Solution of a Differential Equation from

the Two-Step Mechanism of Chapter 11

Equation (11.5-5) is

d[B]

dt

k 1 [A] 0 e−k^1 t−k 2 [B] (B-51)

We will see that it can be put into the form

dzMd[B]+Ndt 0 (B-52)

whereMandNare functions oftand [B], the concentration of substance B. The

equation shown in Eq. (B-52) represents a class of equations calledPfaffian differential

equations. An equation of this type is called anexact differential equationifdzis an

exact differential. Ifdzis exact,MandNmust be derivatives of the functionzand

must obey the Euler reciprocity relation shown in Eq. (B-13):

∂M

∂t



∂^2 z

∂t∂[B]



∂^2 z

∂[B]∂t



∂N

∂[B]

(B-53)

We multiply Eq. (B-51) bydtand recognize that (d[B]/dt)dtd[B] to obtain an

equation in Pfaffian form:

d[B]+

(

k 2 [B]−k 1 [A] 0 e−k^1 t

)

dt 0 (B-54)

This equation corresponds toM1 andN(k 2 [B]−k 1 [A] 0 e−k^1 t). The differential

is not exact since the derivative ofMwith respect totequals zero and the derivative

ofNwith respect to [B] equalsk 2. However, if the equation is multiplied by the factor

ek^2 t, we get the exact differential equation

ek^2 td[B]+

(

k 2 [B]ek^2 t−k 1 [A] 0 e(k^2 −k^1 )t

)

dt 0 (B-55)

as can be checked by differentiation. A factor that converts an inexact Pfaffian

differential equation into an exact differential equation is called anintegrating

factor. Multiplication of any equation on both sides by the same nonzero factor yields

a valid equation with the same solution as the original equation.

We consider the special case that no B orFis present at timet0. We denote the

differential in Eq. (B-55) bydzand perform a line integral ofdzon the path shown in

Figure B.7. The result of the integration must equal zero, since the differential equals

zero if it satisfies the differential equation. That is, the functionzmust have the same

value at the two ends of the path.

0 t

[B]

(t 9 , [B] 9 )

(t 9 ,0)

Figure B.7 The Path of Integration
to Solve Eq. (11.5-5). Thed[B] term indzgives no contribution on the first leg of the path. On the second

legtis equal tot′so that

∫[B]

t′

0

ek^2 t

′

d[B]ek^2 t

′

[B]t′ (B-56)

On the first leg of the path we replace [B] by zero, and the result is

−

∫t′

0

k 1 [A] 0 e(k^2 −k^1 )tdt−

k 1 [A] 0

k 2 −k 1

(e(k^2 −k^1 )t

′

−1) (B-57)