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(B) For a first-order reaction,the rate constant $K$ is given by: $K = \frac{2.303}{t} \log \frac{a}{a-x}$.For $75\%$ completion,$x = 0.75a$,so $a-x =

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$16$

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(B) For a first-order reaction,the rate constant $K$ is given by: $K = \frac{2.303}{t} \log \frac{a}{a-x}$.For $75\%$ completion,$x = 0.75a$,so $a-x = 0.25a$ and $t = 32 	ext{ min}$.$K = \frac{2.303}{32} \log \frac{100}{25} = \frac{2.303}{32} \log 4 = \frac{2.303 	imes 0.602}{32}$.Now,for $50\%$ completion (half-life,$t_{1/2}$),the formula is $t_{1/2} = \frac{0.693}{K}$.Substituting $K = \frac{2.303 	imes 0.602}{32} \approx \frac{0.693}{32} 	imes 2 = \frac{0.693}{16}$.Therefore,$t_{1/2} = \frac{0.693}{0.693 / 16} = 16 	ext{ minutes}$.

If a $75\%$ first-order reaction is completed in $32 \text{ minutes}$,how many minutes will it take for the same reaction to be $50\%$ completed?

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