Q.Show that in case of first order reaction the time required for the completion of 99% is twice the time required for the completion of 90% of the reaction.

Applying the first-order integrated rate law $t=\frac{2.303}{k}\log\frac{a}{a-x}$ at 90% and 99% completion gives $t_{90\%}=\frac{2.303}{k}\log10$ and $t_{99\%}=\frac{2.303}{k}\log100$; since $\log100=2\log10$, $t_{99\%}$ is exactly $2\times t_{90\%}$.

The integrated rate law for a first-order reaction, in terms of initial concentration $a$ and the amount reacted $x$ at time $t$, is: $t=\frac{2.303}{k}\log\frac{a}{a-x}$

At 90% completion: $x=0.90a$, so $a-x=0.10a$, and $t_{90\%}=\frac{2.303}{k}\log\frac{a}{0.10a}=\frac{2.303}{k}\log10=\frac{2.303}{k}\times1=\frac{2.303}{k}$ (since $\log_{10}10=1$).

At 99% completion: $x=0.99a$, so $a-x=0.01a$, and $t_{99\%}=\frac{2.303}{k}\log\frac{a}{0.01a}=\frac{2.303}{k}\log100=\frac{2.303}{k}\times2=\frac{2\times2.303}{k}$ (since $\log_{10}100=\log_{10}10^2=2$).

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