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(D) For a first order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 10 \ minute$,so $k = \frac{0.693}{10 \ m

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$33 \ minute$

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(D) For a first order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 10 \ minute$,so $k = \frac{0.693}{10 \ min} = 0.0693 \ min^{-1}$.The integrated rate equation is $t = \frac{2.303}{k} \log_{10} \frac{[A]_0}{[A]_t}$.Here,$[A]_0 = 100$ and $[A]_t = 10$ (since concentration reduces to $10 \%$).Substituting the values: $t = \frac{2.303}{0.0693 \ min^{-1}} \log_{10} \frac{100}{10} = \frac{2.303}{0.0693} 	imes \log_{10} 10 = \frac{2.303}{0.0693} 	imes 1 \approx 33.23 \ minute$.Rounding to the nearest integer,$t = 33 \ minute$.

What is the time needed to reduce the initial concentration of reactant to $10 \%$ in a first order reaction if its half life time is $10 \ minute$?

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