PHYSICAL CHEMISTRY IN BRIEF

CHAP. 9: CHEMICAL KINETICS [CONTENTS] 270

At timeτ = 0 isx= 0 and, thus we obtain
∫x

0

dx

φ(x)

=τ. (9.15)

Example

Solve the kinetic equation−

dcA

dτ

=kcAfor a first-order reaction A→products. The initial

concentration of the reactantAiscA0.

Solution

From the initial condition and from the material balance equation (9.11) we have

dx

dτ

=k(cA0−x).

This differential equation is of the type (9.13) and we solve it using the method of separation of

variables

∫x

0

dx

cA0−x

=kτ =⇒ −ln

cA0−x

cA0

=kτ =⇒ cA=cA0e−kτ.

We derived the time dependence of the reactant concentration regardless of the number of the

reaction products and their stoichiometric coefficients. If, however, we also want to find out
the time dependence of the products concentrations, we have to specify the right side of the
reaction.
The solution of kinetic equations often leads to the integrals of rational functions of the
type ∫
dx
(x−a)n(x−b)m...

,

wheren,m,... are natural numbers. We solve these integrals by the integrand decomposition
into the sum of fractions, see basic course of mathematics.
Some kinetic equations cannot be solved using the method of separation of variables. Equa-
tion (9.130) may serve as a typical example. It is solved using the constant variation method,
see basic course of mathematics. In the case of yet more complex kinetic equations we may
fail to find any analytical solution at all, and we have to solve these equations numerically.