The half-life of a first order reaction is 30 | 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 $t_{1/2} = 30 \, min$,so $k = \frac{0.693}{30} \, m

Vedclass · Accepted Answer

$60$

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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 $t_{1/2} = 30 \, min$,so $k = \frac{0.693}{30} \, min^{-1}$.The time $t$ required for $75 \, \%$ completion is given by the formula $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.For $75 \, \%$ completion,$[A]_t = [A]_0 - 0.75[A]_0 = 0.25[A]_0$.Thus,$t = \frac{2.303}{0.693/30} \log \frac{[A]_0}{0.25[A]_0} = \frac{2.303 	imes 30}{0.693} \log 4$.Since $\frac{2.303}{0.693} \approx \frac{1}{0.301} \approx 3.32$,we have $t = 30 	imes 2 = 60 \, min$.Alternatively,$75 \, \%$ completion corresponds to $2 	imes t_{1/2} = 2 	imes 30 = 60 \, min$.

The half-life of a first order reaction is $30 \, min$. The time required for $75 \, \%$ completion of the same reaction will be $..... \, min$

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