Question

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

Answer 1

Answer: 5

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