99 % of a first order reaction was completed | vedclass

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Question

(D) For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.For $99 \%$ completion,$[A]_t = [A]_0

Vedclass · Accepted Answer

$96$

Vedclass · Answer

(D) For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.For $99 \%$ completion,$[A]_t = [A]_0 - 0.99[A]_0 = 0.01[A]_0$. Thus,$64 = \frac{2.303}{k} \log \frac{100}{1} = \frac{2.303}{k} 	imes 2$.For $99.9 \%$ completion,$[A]_t = [A]_0 - 0.999[A]_0 = 0.001[A]_0$. Thus,$t = \frac{2.303}{k} \log \frac{1000}{1} = \frac{2.303}{k} 	imes 3$.Dividing the two equations: $\frac{t}{64} = \frac{3}{2}$.Therefore,$t = 64 	imes 1.5 = 96 	ext{ min}$.

$99 \%$ of a first order reaction was completed in $64 \text{ min}$. In how many minutes will $99.9 \%$ of the reaction be complete?

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