Concise Physical Chemistry

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

148 CHEMICAL KINETICS

If the reactants are mixed so that the initial concentrationsAandBare equal, they will

remain equal throughout, because for each molecule of one that reacts, a molecule of

the other has also reacted. As a result,AB=A^2 throughout, giving the special case

of the rate equation

−

dA

dt

=kAB=kA^2

and

dA

A^2

=−kdt

Integrating

∫

1

A^2

dA=−k

∫

dt

gives

−

1

A

=−kt+C

To evaluate the constant of integration, lett=0. The concentrationAis its initial

valueA 0 ,so−

1

A 0

=Cand

1

A

=kt+

1

A 0

which leads to

A 0 −A

A 0 A

=kt

The relation betweent 1 / 2 andkfor this restricted form of the second-order reaction

iskt 1 / 2 A 0 =1.

An example of second-order kinetics in which the rate law−dA/dt=kAB=kA^2

applies is the thermal decomposition of NO 2 :

2NO 2 (g)→2NO(g)+O 2 (g)

The rate is controlled by collisions of NO 2 molecules with other NO 2 molecules,

hence the stipulationA=Bthroughout is met. Although in this case the order of

the reaction is the same as the stoichiometric coefficient, that need not be the case as

exemplified by the thermal decomposition of N 2 O 5 :

2N 2 O 5 (g)→4NO 2 (g)+O 2 (g)

which isfirstorder.