A first order reaction is 75 % completed in 6 | vedclass

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(D) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$.Given that $75 \%$ is completed i

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$30$

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(D) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$.Given that $75 \%$ is completed in $60 \ min$,the remaining concentration $[A]_t$ is $100 - 75 = 25 \%$ of $[A]_0$.$k = \frac{2.303}{60} \log_{10} \frac{100}{25} = \frac{2.303}{60} \log_{10} 4 = \frac{2.303 	imes 0.6020}{60} \approx 0.0231 \ min^{-1}$.The time required for $50 \%$ completion is the half-life $t_{1/2}$.$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.0231} = 30 \ min$.

$A$ first order reaction is $75 \%$ completed in $60 \ min$. The time required for its $50 \%$ completion is: (in $min$)

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