For a first order reaction $t_{99.9 \%}=x t_{1 / 2} x$ is\

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Professor PC Thomas Classes · Answer

$1^{	ext {st }}$ order $\mathrm{K}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}$\ $\mathrm{t} 99.9 \%=\frac{2.303}{\mathrm{~K}} \log \frac{100}{0.1}$\ $\mathrm{t} 50 \%=\frac{2.303}{\mathrm{~K}} \log \frac{100}{50}$\ $\frac{\mathrm{t} 99.9 \%}{\mathrm{t} 50 \%}=\frac{\log 1000}{\log 2}=\frac{3}{0.3}=10$\