The time required for 60 % completion of a fi | vedclass

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(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 $60 \%$ completion,$[A]_t = [A]_0

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$150$

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(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 $60 \%$ completion,$[A]_t = [A]_0 - 0.60[A]_0 = 0.40[A]_0$ and $t = 50 \ min$.$k = \frac{2.303}{50} \log \frac{[A]_0}{0.40[A]_0} = \frac{2.303}{50} \log 2.5$.For $93.6 \%$ completion,$[A]_t = [A]_0 - 0.936[A]_0 = 0.064[A]_0$.$k = \frac{2.303}{t'} \log \frac{[A]_0}{0.064[A]_0} = \frac{2.303}{t'} \log \frac{1}{0.064} = \frac{2.303}{t'} \log 15.625$.Equating the two expressions for $k$:$\frac{2.303}{50} \log 2.5 = \frac{2.303}{t'} \log 15.625$.$t' = 50 	imes \frac{\log 15.625}{\log 2.5} = 50 	imes \frac{\log (2.5)^3}{\log 2.5} = 50 	imes 3 = 150 \ min$.

The time required for $60 \%$ completion of a first order reaction is $50 \ min$. The time required for $93.6 \%$ completion of the same reaction will be (in $min$)

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