The Chemistry Maths Book, Second Edition

326 Chapter 11First-order differential equations

Separation of the variables and integration gives

If x

0

is the initial concentration of reactant thenc 1 = 112 x

0

and

(11.38)

or

(11.39)

Aplot of 12 [A]against tis a straight line (Figure 11.2)

whose slope is 2 kand whose intercept with the vertical axis

is 12 [A]

0

.

(3) The second-order process: A 1 + 1 B 1 → 1 products

The rate equation is

(11.40)

Let the initial concentrations of Aand Bbe[A]

0

1 = 1 aand[B]

0

1 = 1 b, respectively, and

let[A] 1 = 1 a 1 −xand[B] 1 = 1 b 1 −xat time t. The rate equation is then

(11.41)

Two cases need to be distinguished.

(a)The initial concentrations of Aand Bare equal;[A]

0

1 = 1 [B]

0

,a 1 = 1 b.

The rate equation (11.41) is

(11.42)

and, separating the variables and integrating,

(11.43)

Thenc 1 = 112 a 1 = 112 [A]

0

and the integrated rate equation is

(11.44)

11

0

[] []AA

−=kt

dx

ax

kdt

ax

kt c

()

[]

−

=,

−

==+

2

11

A

dx

dt

=−ka x()

2

−

−

== − −

da x

dt

dx

dt

ka x b x

()

()()

v=− =− =

d

dt

d

dt

k

[] []

[][]

AB

AB

11

2

12

0

0

0

[] []

[]

[]

AA []

A

A

A

−=, =

• 
  

kt

kt

11

2

0

xx

−=kt

−= , = +

dx

x

kdt

x

kt c

2

2

1

2

0

1 /[A]

0

• 
  

2 k

t

1 /[A]

...

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Figure 11.2