For a first order reaction,show that the time required for $99 \%$ completion is twice the time required for the completion of $90 \%$ of the reaction.

For a first order reaction,the integrated rate equation is $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.
For $99 \%$ completion,$[A]_t = [A]_0 - 0.99[A]_0 = 0.01[A]_0$. Thus,$t_{99\%} = \frac{2.303}{k} \log \frac{[A]_0}{0.01[A]_0} = \frac{2.303}{k} \log 100 = \frac{2.303}{k} \times 2$.
For $90 \%$ completion,$[A]_t = [A]_0 - 0.90[A]_0 = 0.10[A]_0$. Thus,$t_{90\%} = \frac{2.303}{k} \log \frac{[A]_0}{0.10[A]_0} = \frac{2.303}{k} \log 10 = \frac{2.303}{k} \times 1$.
Comparing the two,$t_{99\%} = 2 \times t_{90\%}$.
Therefore,the time required for $99 \%$ completion is twice the time required for $90 \%$ completion.