Time required for completion of 90 % of a fir | vedclass

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

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$3 t$

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(C) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.For $90 \%$ completion,$[A]_t = [A]_0 - 0.90[A]_0 = 0.10[A]_0$. So,$k = \frac{2.303}{t} \log \frac{[A]_0}{0.10[A]_0} = \frac{2.303}{t} \log(10) = \frac{2.303}{t}$.Thus,$t = \frac{2.303}{k}$.For $99.9 \%$ completion,$[A]_t = [A]_0 - 0.999[A]_0 = 0.001[A]_0$. Let the time be $t'$.$t' = \frac{2.303}{k} \log \frac{[A]_0}{0.001[A]_0} = \frac{2.303}{k} \log(1000) = \frac{2.303}{k} 	imes 3$.Substituting $t = \frac{2.303}{k}$,we get $t' = 3t$.

Time required for completion of $90 \%$ of a first order reaction is '$t$'. What is the time required for completion of $99.9 \%$ of the reaction?

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