If the half-life of a first-order reaction is | vedclass

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(A) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 4 \ min$,so $k = \frac{0.693}{4} \ min^{

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$40$

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(A) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 4 \ min$,so $k = \frac{0.693}{4} \ min^{-1}$.The time $t$ required for a reaction to complete is given by $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.For $99.9 \%$ completion,$[A]_t = [A]_0 - 0.999[A]_0 = 0.001[A]_0$.Thus,$\frac{[A]_0}{[A]_t} = \frac{1}{0.001} = 1000 = 10^3$.Substituting the values: $t = \frac{2.303 	imes 4}{0.693} \log(10^3) = \frac{2.303 	imes 4}{0.693} 	imes 3$.Since $\frac{2.303}{0.693} \approx \frac{1}{0.301} \approx 3.32$,we have $t \approx 3.32 	imes 4 	imes 3 \approx 39.84 \approx 40 \ min$.

If the half-life of a first-order reaction is $4 \ min$,then after how many minutes will the reaction be $99.9 \%$ complete?

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