For a first order reaction,the half-life peri | vedclass

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(D) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 5 \ min$,so $k = \frac{0.693}{5} \ min^{

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$50$

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(D) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 5 \ min$,so $k = \frac{0.693}{5} \ min^{-1}$.The time $t$ required for $99.9 \%$ completion is given by the formula $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.For $99.9 \%$ completion,$[A]_t = [A]_0 - 0.999[A]_0 = 0.001[A]_0$.Substituting the values: $t = \frac{2.303}{0.693/5} \log \frac{[A]_0}{0.001[A]_0}$.$t = \frac{2.303 	imes 5}{0.693} \log(1000)$.Since $\log(1000) = 3$ and $\frac{2.303}{0.693} \approx 3.32$,we have $t \approx 3.32 	imes 5 	imes 3 = 49.8 \approx 50 \ min$.Thus,the correct option is $D$.

$S. No.$	$Time$ $(s)$	$Total$ $pressure$ $(atm)$
$1$	$0$	$0.5$
$2$	$100$	$0.6$

For a first order reaction,the half-life period is $5 \ min$. What time will it take for $99.9 \%$ completion of the reaction (in $min$)?

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