A first order reaction is half completed in 4 | vedclass

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(B) For a first order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 45 \, 	ext{min}$,so $k = \frac{0.693}{4

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$7.5$

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(B) For a first order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 45 \, 	ext{min}$,so $k = \frac{0.693}{45} \, 	ext{min}^{-1}$.For $99.9 \%$ completion,the remaining concentration is $a - 0.999a = 0.001a$.The time taken is $t = \frac{2.303}{k} \log \left( \frac{a}{0.001a} ight) = \frac{2.303}{k} \log(10^3) = \frac{2.303 	imes 3}{k}$.Substituting $k = \frac{0.693}{45}$,we get $t = \frac{2.303 	imes 3 	imes 45}{0.693} \approx 448.5 \, 	ext{min}$.Converting to hours: $t = \frac{448.5}{60} \approx 7.475 \, 	ext{hr} \approx 7.5 \, 	ext{hr}$.

$A$ first order reaction is half completed in $45 \, \text{minutes}$. How long does it need for $99.9 \%$ of the reaction to be completed? (in $hr$)

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