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.