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Question

(D) For a first-order reaction,the rate constant $K$ is given by $K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.For $99\%$ completion,$[A]_t = 100 - 9

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(D) For a first-order reaction,the rate constant $K$ is given by $K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.For $99\%$ completion,$[A]_t = 100 - 99 = 1$,so $K = \frac{2.303}{32} \log \frac{100}{1} = \frac{2.303}{32} 	imes 2$  ........... $(1)$For $99.9\%$ completion,$[A]_t = 100 - 99.9 = 0.1$,so $K = \frac{2.303}{t} \log \frac{100}{0.1} = \frac{2.303}{t} 	imes 3$  ............ $(2)$Equating $(1)$ and $(2)$:$\frac{2.303 	imes 2}{32} = \frac{2.303 	imes 3}{t}$$\frac{2}{32} = \frac{3}{t}$$t = \frac{3 	imes 32}{2} = 48 \ 	ext{minutes}$.

If a $99\%$ first-order reaction is completed in $32$ minutes,how many minutes will it take for $99.9\%$ of the reaction to be completed?

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