For a first-order reaction,the ratio of t_{3/ | vedclass

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(C) For a first-order reaction,the time required for a fraction of reaction to complete is given by $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.For

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$2 : 1$

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(C) For a first-order reaction,the time required for a fraction of reaction to complete is given by $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.For $t_{1/2}$ (half-life),$[A]_t = \frac{[A]_0}{2}$,so $t_{1/2} = \frac{0.693}{k}$.For $t_{3/4}$,the amount remaining is $[A]_t = [A]_0 - \frac{3}{4}[A]_0 = \frac{1}{4}[A]_0$.Thus,$t_{3/4} = \frac{2.303}{k} \log \frac{[A]_0}{[A]_0 / 4} = \frac{2.303}{k} \log 4 = \frac{2.303}{k} 	imes 2 \log 2$.Since $\log 2 \approx 0.3010$,$t_{3/4} = 2 	imes t_{1/2}$.Therefore,the ratio $\frac{t_{3/4}}{t_{1/2}} = 2 : 1$.

For a first-order reaction,the ratio of $t_{3/4}$ to $t_{1/2}$ is ...

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