Concise Physical Chemistry

c10 JWBS043-Rogers September 13, 2010 11:26 Printer Name: Yet to Come

150 CHEMICAL KINETICS

10.3.2 Back to Kinetics: Sequential Reactions

In this section we shall obtain the concentrationBin the a sequence of two first-order

reactions:

A

k 1

→B

k 2

→C

The differential equations are

dA

dt

=−k 1 A

dB

dt

=k 1 A−k 2 B

dC

dt

=k 2 B

From the first equation, we have

A=A 0 e−k^1 t, B 0 =C 0 = 0

where the subscripted 0 indicates the initial concentration. Substituting this result

into the equation forB(t), we obtain

dB(t)

dt

=k 1 A 0 e−k^1 t−k 2 B(t)

Now take the Laplace transform of both sides and solve forb(s), the transform of

B(t):

sb(s)−B(t=0)=

k 1 A 0

s+k 1

−k 2 b(s)

whereB(t=0)=B 0 =0, so the remaining terms are

sb(s)+k 2 b(s)=

k 1 A 0

s+k 1

b(s)(s+k 2 )=

k 1 A 0

s+k 1

b(s)=

k 1 A 0

s+k 1

1

(s+k 2 )

=k 1 A 0

1

(s+k 1 )

1

(s+k 2 )