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(A) For a first-order reaction, the rate constant $k$ is given by $k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]_t} \right)$.For $90\%$ completion,

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$3t$

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(A) For a first-order reaction, the rate constant $k$ is given by $k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]_t} ight)$.For $90\%$ completion, $[A]_t = [A]_0 - 0.9[A]_0 = 0.1[A]_0$. Thus, $k = \frac{2.303}{t} \log \left( \frac{[A]_0}{0.1[A]_0} ight) = \frac{2.303}{t} \log(10) = \frac{2.303}{t}$.For $99.9\%$ completion, $[A]_t = [A]_0 - 0.999[A]_0 = 0.001[A]_0$. Let the time be $t'$.$k = \frac{2.303}{t'} \log \left( \frac{[A]_0}{0.001[A]_0} ight) = \frac{2.303}{t'} \log(10^3) = \frac{2.303 	imes 3}{t'}$.Equating the two expressions for $k$: $\frac{2.303}{t} = \frac{2.303 	imes 3}{t'}$.Therefore, $t' = 3t$.

If the time required for $90\%$ completion of a first-order reaction is '$t$', what is the time required for $99.9\%$ completion of the reaction at the same temperature?

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