The ratio of the time taken to complete 60% a | vedclass

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(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 $20\%$ completion,$[A]_t = 100 - 2

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$4$

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(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 $20\%$ completion,$[A]_t = 100 - 20 = 80$,so $K = \frac{2.303}{t_{20}} \log \frac{100}{80} = \frac{2.303}{t_{20}} \log 1.25$.For $60\%$ completion,$[A]_t = 100 - 60 = 40$,so $K = \frac{2.303}{t_{60}} \log \frac{100}{40} = \frac{2.303}{t_{60}} \log 2.5$.Equating the two expressions for $K$: $\frac{1}{t_{20}} \log 1.25 = \frac{1}{t_{60}} \log 2.5$.Therefore,$\frac{t_{60}}{t_{20}} = \frac{\log 2.5}{\log 1.25} = \frac{\log(10/4)}{\log(10/8)} = \frac{\log 10 - \log 4}{\log 10 - \log 8} = \frac{1 - 2 \log 2}{1 - 3 \log 2}$.Given $\log 2 = 0.3$,we have $\frac{t_{60}}{t_{20}} = \frac{1 - 2(0.3)}{1 - 3(0.3)} = \frac{1 - 0.6}{1 - 0.9} = \frac{0.4}{0.1} = 4$.

The ratio of the time taken to complete $60\%$ and $20\%$ of a first-order reaction $[A \rightarrow \text{product}]$ is : $(\log 2 = 0.3)$

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