If half life of a first order reaction involving reactant A is 5 min. How long will it take [A] to reach 25% of its initial concentration?

For the first order reaction the Rate constant

$\mathrm{k}=\frac{0.693}{{\mathrm{t}}_{1/2}}$

${\mathrm{t}}_{1/2}$ is given 5 min, hence      

$\mathrm{k}=\frac{0.693}{5}=0.1386 {\min }^{-1}$

Let initial concentration ‘a’ of A be 100 mol L^–1. To reach 25% of initial concentration means, (a – x) = 25 mol L^–1.

$\mathrm{k} = \frac{2.303}{\mathrm{t}}\log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}$

Or

 $\mathrm{t}=\frac{2.303}{0.1386}\log \frac{100}{25}$


or                         

$$\begin{gathered} \mathrm{t}=\frac{2.303}{0.1386}\log 4 \\ \mathrm{t} = \frac{2.303}{0.1386}\times 0.6021 \mathrm{or} \mathrm{t} = 10.00 \min . \end{gathered}$$


Ostwald Isolation Method: In this method, the concentration of all the reactants are taken in large excess except that of one. The concentration change only for this reactant is significant as other are so much in excess that practically there is no change in their concentrations. The constant terms may be combined with the rate constant and we may write

$\mathrm{Rate} = \mathrm{k}{\left[\mathrm{A}\right]}^{\mathrm{\alpha }} {\left[\mathrm{B}\right]}^{\mathrm{\beta }} {\left[\mathrm{C}\right]}^{\mathrm{\gamma }} = {\mathrm{k}}_0{\left[\mathrm{A}\right]}^{\mathrm{\alpha }}$

The value of ‘a’, i.e., the order of reaction with respect to A can be determined by the methods given above.