75% of the first order reaction was completed | vedclass

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(B) 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 $75\%$ of the reaction is c

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(B) 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 $75\%$ of the reaction is completed in $32$ minutes,the remaining concentration $[A]_t = 100\% - 75\% = 25\% = 0.25 [A]_0$.Substituting these values: $k = \frac{2.303}{32} \log \frac{[A]_0}{0.25 [A]_0} = \frac{2.303}{32} \log(4) = \frac{2.303 	imes 0.602}{32}$.The half-life $t_{1/2}$ is given by $t_{1/2} = \frac{0.693}{k}$.Substituting $k$: $t_{1/2} = \frac{0.693 	imes 32}{2.303 	imes 0.602} \approx 16 	ext{ minutes}$.Alternatively,$75\%$ completion corresponds to $2$ half-lives $(2 	imes t_{1/2} = 32 	ext{ min})$,so $t_{1/2} = 16 	ext{ minutes}$.

$75\%$ of the first order reaction was completed in $32$ minutes. When was $50\%$ of the reaction completed? (in $\text{minutes}$)

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