If 60% of a first order reaction was complete | vedclass

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(A) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.Given that $60\%$ is completed in $60

Vedclass · Accepted Answer

$45$

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(A) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.Given that $60\%$ is completed in $60 \ min$,$[A]_t = 100 - 60 = 40$.$k = \frac{2.303}{60} \log \frac{100}{40} = \frac{2.303}{60} \log 2.5$.Using $\log 2.5 = \log(10/4) = \log 10 - \log 4 = 1 - 0.60 = 0.40$.$k = \frac{2.303 	imes 0.40}{60} = \frac{0.9212}{60} \ min^{-1}$.For $50\%$ completion $(t_{1/2})$,$t_{1/2} = \frac{0.693}{k}$.$t_{1/2} = \frac{0.693 	imes 60}{0.9212} \approx 45.12 \ min$.Thus,the reaction is completed in approximately $45 \ min$.

If $60\%$ of a first order reaction was completed in $60 \ min$,$50\%$ of the same reaction would be completed in approximately $.......... \ min.$
$(\log \, 4 = 0.60, \, \log \, 5 = 0.69)$

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If $60\%$ of a first order reaction was completed in $60 \ min$,$50\%$ of the same reaction would be completed in approximately $.......... \ min.$$(\log \, 4 = 0.60, \, \log \, 5 = 0.69)$