If the half-life (t_{1/2}) for a first order | vedclass

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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 $t_{1/2} = 1 	ext{ minute}$,so $k = 0.693 	ext{ m

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$10 	ext{ minutes}$

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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 $t_{1/2} = 1 	ext{ minute}$,so $k = 0.693 	ext{ min}^{-1}$.The time $t$ required for a reaction to reach a certain percentage completion is given by $t = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]_t} ight)$.For $99.9\%$ completion,$[A]_t = [A]_0 - 0.999[A]_0 = 0.001[A]_0$.Thus,$t = \frac{2.303}{0.693} \log \left( \frac{[A]_0}{0.001[A]_0} ight) = \frac{2.303}{0.693} \log(1000) = \frac{2.303}{0.693} 	imes 3$.Since $\frac{2.303}{0.693} \approx 3.32$,we have $t \approx 3.32 	imes 3 = 9.96 	ext{ minutes}$.This is closest to $10 	ext{ minutes}$.

If the half-life $(t_{1/2})$ for a first order reaction is $1 \text{ minute}$,then the time required for $99.9\%$ completion of the reaction is closest to

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