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(B) For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.Step $1$: Calculate $k$ using the give

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$25$

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(B) For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.Step $1$: Calculate $k$ using the given data ($75\%$ completion in $15 \ min$).$[A]_t = [A]_0 - 0.75[A]_0 = 0.25[A]_0$.$k = \frac{2.303}{15} \log \frac{[A]_0}{0.25[A]_0} = \frac{2.303}{15} \log 4 = \frac{2.303 	imes 0.602}{15} \approx 0.0924 \ min^{-1}$.Step $2$: Calculate time $t$ for $90\%$ completion.$[A]_t = [A]_0 - 0.90[A]_0 = 0.10[A]_0$.$t = \frac{2.303}{k} \log \frac{[A]_0}{0.10[A]_0} = \frac{2.303}{k} \log 10 = \frac{2.303}{k}$.Substituting $k = \frac{2.303 \log 4}{15}$:$t = \frac{2.303 	imes 15}{2.303 \log 4} = \frac{15}{0.602} \approx 24.92 \ min \approx 25 \ min$.

For a first-order reaction,if $75\%$ of the reaction is completed in $15 \ min$,then the time required for $90\%$ completion of the reaction will be ........... $min$.

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