75% of a first-order reaction was completed i | vedclass

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Question

(A) For a first-order reaction,the rate constant $k$ is given by: $k = \frac{2.303}{t} \log \left( \frac{a}{a-x} \right)$.For $75\%$ completion,$x = 0

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(A) For a first-order reaction,the rate constant $k$ is given by: $k = \frac{2.303}{t} \log \left( \frac{a}{a-x} ight)$.For $75\%$ completion,$x = 0.75a$ and $t = 32 	ext{ min}$:$k = \frac{2.303}{32} \log \left( \frac{a}{a-0.75a} ight) = \frac{2.303}{32} \log(4) = \frac{2.303}{32} 	imes 2 \log(2) \dots (i)$.For $50\%$ completion (half-life),$x = 0.5a$:$k = \frac{2.303}{t_{1/2}} \log \left( \frac{a}{a-0.5a} ight) = \frac{2.303}{t_{1/2}} \log(2) \dots (ii)$.Equating $(i)$ and $(ii)$:$\frac{2.303}{32} 	imes 2 \log(2) = \frac{2.303}{t_{1/2}} \log(2)$.$t_{1/2} = \frac{32}{2} = 16 	ext{ minutes}$.

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

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