The Chemistry Maths Book, Second Edition

11.6 An example of linear equations in chemical kinetics 331

or

(11.59)

This is a linear equation withp(t) 1 = 1 k

2

,r(t) 1 = 1 ak

1

e

−k

1

t

, and integrating factor

The solution of equation (11.59) is therefore

By the initial conditions,y 1 = 10 whent 1 = 10. We have

and the required particular solution is

(11.60)

The concentration of each species present at time tis therefore

[A] 1 = 1 [A]

0

e

−k

1

t

(11.61)

[C] 1 = 1 [A]

0

1 − 1 [A] 1 − 1 [B]

Figure 11.3 illustrates the three types of behaviour exhibited by the system. In each

case[A]

0

1 = 11 andk

1

1 = 11 (in some appropriate units). Figure (a), withk

2

2 k

1

1 = 110 , shows

how the system behaves whenk

2

1 >> 1 k

1

; the rate-determining step isA 1 → 1 B, and the

[]

[]

[]

B

A

if

A

=

−

−

()

≠

−−

01

21

12

01

12

k

kk

ee kk

kte

kt kt

−−

=















kt

kk

1

12

if

y

ak

kk

ee kk

ak te

kt kt

kt

=

−

−

()

≠

−−

−

1

21

12

1

12

1

if

ifkk

12

=















c

ak

kk

kk

kk

=

−

−

≠

=















1

21

12

12

0

if

if

eyake dtc

ak

kk

ec

kt k k t

kkt

221

21

1

1

21

=+=

−

• 
  

−

Z

()

(–)

iif

if

kk

ak t c k k

12

1

12

≠

+=



















Fty Ftrtdt c()=+Z ()()

Ft e e

pt dt kt

()

()

=

∫

=

2

dy

dt

ky ake

kt

+=

−

21

1