PHYSICAL CHEMISTRY IN BRIEF

CHAP. 9: CHEMICAL KINETICS [CONTENTS] 269

Example

Consider the chemical reaction

A+ 2B→ 3 C.

The initial concentrations of substancesAandBare cA0andcB0. The initial concentration

of substanceC is zero. Write the material balance equations for the concentrations of these

substances at timeτ. Write the kinetic equation of the reaction if it is half-order with respect to

componentAand third-order with respect toB.

Solution

The material balance equations are:cA=cA0−x,cB=cB0− 2 x,cC=cC0− 3 x. The kinetic

equation is

−

dcA

dτ

=kAc^1 A/^2 c^3 B.

Using the quantityxwe may rewrite this kinetic equation into the form

dx

dτ

=kA(cA0−x)^1 /^2 (cB0− 2 x)^3.

9.1.6 Methods of solving kinetic equations

From the mathematical point of view, the kinetic equation (9.8) is a first-order ordinary differ-
ential equation. Time is an independent variable, the concentrations of the reactants, or the
quantityxin equation (9.12), are dependent variables.
In all simple reactions and in a number of other cases, the kinetic equation has the form

dx

dτ

=φ(x). (9.13)

This differential equation is solved using the method of separation of variables, see basic course
of mathematics.
dx
φ(x)

= dτ. (9.14)