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(C) For a first order reaction,the time $t$ is given by $t = \frac{2.303}{k} \log_{10} \frac{a}{a-x}$.Let the initial concentration $a = 100$.For $75

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$4.8$

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(C) For a first order reaction,the time $t$ is given by $t = \frac{2.303}{k} \log_{10} \frac{a}{a-x}$.Let the initial concentration $a = 100$.For $75 \%$ completion,$x = 75$,so the remaining concentration is $100 - 75 = 25$. Thus,$t_{75\%} = \frac{2.303}{k} \log \frac{100}{25} = \frac{2.303}{k} \log 4$.For $25 \%$ completion,$x = 25$,so the remaining concentration is $100 - 25 = 75$. Thus,$t_{25\%} = \frac{2.303}{k} \log \frac{100}{75} = \frac{2.303}{k} \log \frac{4}{3}$.The ratio is $\frac{t_{75\%}}{t_{25\%}} = \frac{\log 4}{\log (4/3)} = \frac{\log 4}{\log 4 - \log 3}$.Using $\log 4 \approx 0.6020$ and $\log 3 \approx 0.4771$,we get $\frac{0.6020}{0.6020 - 0.4771} = \frac{0.6020}{0.1249} \approx 4.82$.

For a first order reaction at $27^{\circ} C$,the ratio of time required for $75 \%$ completion to $25 \%$ completion of reaction is

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