Time required for 90 % completion of a first | vedclass

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(C) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$.For $90 \%$ completion,$[A]_t = [

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$3x \ 	ext{minute}$

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(C) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$.For $90 \%$ completion,$[A]_t = [A]_0 - 0.90[A]_0 = 0.10[A]_0$. Thus,$x = \frac{2.303}{k} \log_{10} \frac{[A]_0}{0.10[A]_0} = \frac{2.303}{k} \log_{10} 10 = \frac{2.303}{k} 	imes 1$.For $99.9 \%$ completion,$[A]_t = [A]_0 - 0.999[A]_0 = 0.001[A]_0$. Thus,$t_{99.9 \%} = \frac{2.303}{k} \log_{10} \frac{[A]_0}{0.001[A]_0} = \frac{2.303}{k} \log_{10} 1000 = \frac{2.303}{k} 	imes 3$.Comparing the two expressions,$t_{99.9 \%} = 3 	imes \left( \frac{2.303}{k} ight) = 3x \ 	ext{minute}$.

Time required for $90 \%$ completion of a first order reaction is '$x$' minute. Calculate the time required to complete $99.9 \%$ of the reaction at the same temperature.

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