Question

Describe the half-life method for determining the order of a reaction.

Answer 1

The half-life (t_1/2) of a reaction is the time in which the concentration of the reactant reduces to half of its initial value.

The order of a reaction can be determined by measuring the half-life of the reaction.

The half-life (t_1/2) of a reaction is related to the initial concentration of the reactant as

\[\ce{t_{1/2} \propto \frac{1}{[A]{^{n - 1}_{0}}}}\]

Where, [A]₀ is the initial concentration of the reactant and n is the order of reaction. 

The half-lives of a reaction at different beginning concentrations are measured in order to determine the order of the reaction. Suppose t_1/2 is the half-life of a reaction when the initial concentration is [A]₀ and \[\ce{t^'_{1/2}}\] is the half-life of the same reaction when the initial concentration is \[\ce{[A^'_{0}]}\]. Thus, we have

\[\ce{t_{1/2} \propto \frac{1}{[A]{^{n - 1}_{0}}}}\]    ...(i)

and \[\ce{t^'_{1/2} \propto \frac{1}{[A]{^{' n - 1}_{0}}}}\]    ...(ii)

Dividing the first equation by the second one, we get

\[\ce{\frac{t_{1/2}}{t{^{'}_{1/2}}} = (\frac{[A]{^{'}_{0}}}{[A]_0})^{n - 1}}\]    ...(iii)

or, \[\ce{log_10 \frac{t_{1/2}}{t{^{'}_{1/2}}} = (n - 1) log_10 (\frac{[A]{^{'}_{0}}}{[A]_0})}\]

or, \[\ce{n = 1 + \frac{log_10(t_{1/2}/t{^{'}_{1/2}})}{log_10([A]{^{'}_{0}}/[A]_0)}}\]    ...(iv)

Thus, knowing the values of t_1/2, \[\ce{t{^{'}_{1/2}}}\], [A]₀, \[\ce{[A]{^{'}_{0}}}\], the order n of the reaction can be obtained with the help of equation (iv).