A first order reaction is half completed in 4 | vedclass

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

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$7.5$

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(A) For a first order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 45 \ min$,so $k = \frac{0.693}{45} \ min^{-1}$.For $99.9 \%$ completion,the amount remaining $(a-x)$ is $100 - 99.9 = 0.1$.The time $t$ required is $t = \frac{2.303}{k} \log \frac{a}{a-x}$.Substituting the values: $t = \frac{2.303}{0.693 / 45} \log \frac{100}{0.1} = \frac{2.303 	imes 45}{0.693} \log 1000$.Since $\log 1000 = 3$,we have $t = \frac{2.303 	imes 45 	imes 3}{0.693} \approx 448.5 \ min$.Converting to hours: $t = \frac{448.5}{60} \approx 7.475 \ hrs \approx 7.5 \ hours$.

$A$ first order reaction is half completed in $45 \ min$. How long does it need $99.9 \%$ of the reaction to be completed (in $Hours$)?

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