Show that in case of a first order reaction, the time taken for completion of $\displaystyle 99$% reaction is twice the time required for $\displaystyle 90$% completion of the reaction. $\displaystyle (\log 10=1)$

\[t=\frac{2.303}{k} \log \frac{[A]_{0}}{[A]} \] Time required for the completion of $\displaystyle 99$% reaction \[\begin{aligned} t_{99 \%} & =\frac{2.303}{k} \log \frac{100}{1} \\ t_{99 \%} & =\frac{2.303}{k} \times 2 \end{aligned} \] Time required for the completion of $\displaystyle 90$% reaction \[\begin{aligned} t_{90 \%} & =\frac{2303}{k} \log \frac{100}{10} \\ t_{90 \%} & =\frac{2.303}{k} \log 10 \\ t_{90 \%} & =\frac{2.303}{k} \\ \frac{t_{99 \%}}{t_{90 \%}} & =\frac{\left(\dfrac{2.303}{k}\right) \times 2}{\dfrac{2.303}{k}} \\ \frac{t_{99} \%}{t_{90 \%}} & =2 \\ 99 \% & =2 \times t_{90 \%} \end{aligned} \]