Question

The initial concentration of both the reactants of a second order reaction are equal and $60 \%$ of the reaction gets completed in $30\, s$. How much time will be taken in $20 \%$ completion of the reaction?

Answer 1

For second order:
$\left.k_2=\frac{1}{t} \cdot \frac{x}{a(a-x)} \text { [Let } a=1\right] $
$=\frac{1}{30 s } \times \frac{0.6}{1(1-0.6)}=\frac{1}{30} \times \frac{0.6}{0.4}$
Now for $20 \%$ completion
$k_2=\frac{1}{t} \cdot \frac{x}{a(1-x)}$
(Since $k_2$ is constant)
$\frac{1}{30} \times \frac{0.6}{0.4}=\frac{1}{t} \times \frac{1}{4} $
$t=\frac{30}{0.6} \times \frac{0.4}{4}=5 \,s$