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(B) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}$.Given that the concentration is reduced

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$20 \ min$

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(B) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}$.Given that the concentration is reduced to $12.5 \%$ in $1 \ hr$ $(60 \ min)$,we have $[A]_0 = 100$ and $[A] = 12.5$.$k = \frac{2.303}{60} \log \frac{100}{12.5} = \frac{2.303}{60} \log 8 = \frac{2.303}{60} 	imes 0.903 = 0.0346 \ min^{-1}$.The half-life $t_{1/2}$ is calculated as $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.0346} \approx 20 \ min$.Alternatively,$12.5 \% = (1/2)^3$,which means $3$ half-lives have passed in $60 \ min$. Thus,$t_{1/2} = 60/3 = 20 \ min$.

Time $(s)$	$0$	$30$	$60$	$90$
Concentration $(mol \ L^{-1})$	$0.551$	$0.312$	$0.173$	$0.085$

In a first order reaction,the concentration of the reactant is reduced to $12.5 \%$ in one hour. When was it half completed?

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