A first-order reaction takes 69.3 min to com | vedclass

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(C) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 69.3 \ min$,so $k = \frac{0.693}{69.3} =

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$160.97$

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(C) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 69.3 \ min$,so $k = \frac{0.693}{69.3} = 0.01 \ min^{-1}$.For $80\%$ completion,the remaining concentration is $100 - 80 = 20\%$.The time $t$ is calculated using the formula: $t = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]_t} ight)$.Substituting the values: $t = \frac{2.303}{0.01} \log \left( \frac{100}{20} ight)$.$t = 230.3 	imes \log(5) = 230.3 	imes 0.69897 \approx 160.97 \ min$.

$A$ first-order reaction takes $69.3 \ min$ to complete $50\%$ of the reaction. How much time (in $min$) will it take to complete $80\%$ of the reaction?

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