Show that the time required for 99.9 % completion in a first order reaction is 10 times of half-life ( t_1 / 2) of the reaction [ log 2=0.3010, log…

\[\mathrm{k}=\frac{2 \cdot 303}{\mathrm{t}} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]} \] Let \(\displaystyle [\mathrm{R}]_{0}=100,[\mathrm{R}]=100-50=50\) \[\mathrm{t}_{50 \%}=\frac{2 \cdot 303}{\mathrm{k}} \log \frac{100}{50} \] \(\displaystyle =\frac{2 \cdot 303}{\mathrm{k}} \log 2\) \[=\frac{2 \cdot 303}{\mathrm{k}} \times 0 \cdot 3010 \] (ii) Divide (i) by (ii) \[\begin{gathered} \frac{\mathrm{t}_{99 \cdot 9 \%}}{\mathrm{t}_{50 \%}}=\frac{\dfrac{2 \cdot 303}{\mathrm{k}} \times 3}{\dfrac{2 \cdot 303}{\mathrm{k}} \times 0 \cdot 3010} \\ \frac{\mathrm{t}_{99 \cdot 9 \%}}{\mathrm{t}_{50 \%}}=10 \end{gathered} \] or \(\displaystyle \mathrm{t}_{99 \cdot 9 \%}=10 \mathrm{t}_{50 \%}\)