A reaction has a rate constant k = 2.4 times | vedclass

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(C) The unit of $k$ is $s^{-1}$,which indicates that it is a first-order reaction.For a first-order reaction,the time required for a certain percentag

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$10$

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(C) The unit of $k$ is $s^{-1}$,which indicates that it is a first-order reaction.For a first-order reaction,the time required for a certain percentage of completion is given by $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.For $t_{99.9}$,the remaining concentration is $0.1 \%$ of the initial concentration,so $t_{99.9} = \frac{2.303}{k} \log \frac{100}{0.1} = \frac{2.303}{k} \log 1000 = \frac{2.303 	imes 3}{k}$.For $t_{50}$,the time is the half-life,$t_{50} = \frac{0.693}{k} = \frac{2.303 	imes 0.301}{k}$.The ratio $\frac{t_{99.9}}{t_{50}} = \frac{3 	imes 2.303 / k}{0.301 	imes 2.303 / k} = \frac{3}{0.301} \approx 10$.

$A$ reaction has a rate constant $k = 2.4 \times 10^{-4} \ s^{-1}$. Find the ratio of $t_{99.9}$ to $t_{50}$.

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