Time required for \(90 \%\) completion of a first order reaction is ' \(x\) ' minute. Calculate the time required to complete \(99.9 \%\) of the reaction at same temperature.

\(\mathrm{t}_{90 \%} =\frac{2.303}{\mathrm{k}} \log _{10} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_t}=\frac{2.303}{\mathrm{k}} \log _{10} \frac{100}{10} \)
\( =\frac{2.303}{\mathrm{k}} \log _{10} 10 \)
\( \mathrm{t}_{99.9 \%} =\frac{2.303}{\mathrm{k}} \log _{10} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}}=\frac{2.303}{\mathrm{k}} \log _{10} \frac{100}{0.1} \)
\( =\frac{2.303}{\mathrm{k}} \log _{10} 1000 \)
\( \frac{\mathrm{t}_{99.9 \%}}{\mathrm{t}_{90 \%}} =\frac{\frac{2.303}{\mathrm{k}} \log _{10} 1000}{\frac{2.303}{\mathrm{k}} \log _{10} 10}=\frac{\log _{10} 1000}{\log _{10} 10}=\frac{3}{1} \)
\( \therefore \mathrm{t}_{99.9 \%} =3 \times \mathrm{t}_{90 \%}=3 x \text { minute}\)\(\text { (since, } \mathrm{t}_{90 \%}=x \text { minute)}\)