Question:
A reaction has a half life of $1 \mathrm{~min}$. The time required for $99.9 \%$ completion of the reaction is____________ min. (Round off to the Nearest integer)
$[$ Use $: \ln 2=0.69, \ln 10=2.3]$
Solution:
(10)
$\frac{\mathrm{t}_{99.9 \%}}{\mathrm{t}_{50 \%}}=\frac{\frac{1}{\mathrm{~K}} \ln \frac{100}{0.1}}{\frac{1}{\mathrm{~K}} \ln 2}$
$=\frac{\ln 1000}{\ln 2} \times \mathrm{t}_{50 \%}$
$=\frac{3 \ln 10}{\ln 2} \times 1$
$=\frac{3 \times 2.3}{0.69}=10$
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