A first order reaction requires 30 min for 5 | vedclass

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(C) For a first order reaction,the half-life $(t_{1/2})$ is given as $30 \ min$. Since $50\%$ completion corresponds to one half-life,$t_{1/2} = 30 \

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) For a first order reaction,the half-life $(t_{1/2})$ is given as $30 \ min$. Since $50\%$ completion corresponds to one half-life,$t_{1/2} = 30 \ min$. For $75\%$ completion,the reaction proceeds for two half-lives $(50\% + 25\%)$. Total time $t = 2 	imes t_{1/2} = 2 	imes 30 \ min = 60 \ min$. Alternatively,using the formula $k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]_t} ight)$,where $k = \frac{0.693}{30} \ min^{-1}$,we get $t = \frac{2.303}{k} \log \left( \frac{100}{100 - 75} ight) = \frac{2.303}{(0.693/30)} \log(4) = 60 \ min$.

$A$ first order reaction requires $30 \ min$ for $50\%$ completion. The time required to complete the reaction by $75\%$ will be .......... $\min.$

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