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The time required for $60 \%$ completion of a first order reaction is $50 \mathrm{~min}$.
The time required for $93.6 \%$ completion of the same reaction will be
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The time required for $93.6 \%$ completion of the same reaction will be
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$150 \mathrm{~min}$
Time for $60 \%$ completion $=50 \mathrm{~min}$ for first order reaction.
Time required for $93.6 \%$ completion $=$ ? $k=\frac{2.303}{50} \log \frac{a}{a-0.6 a} \Rightarrow k=\frac{2.303}{50} \log \frac{a}{0.4 a}$ $k=\frac{2.303}{50} \log 2.5$
Also, $k=\frac{2.303}{x} \log \frac{a}{a-.936 a}$
$\therefore \frac{2.303 \log 2.5}{50}=\frac{2.303}{x} \log \frac{a}{0.064 a}$
$x=50 \times \frac{\log \frac{1000}{64}}{\log 2.5}$
$=50 \times \frac{\log 15.625}{\log 2.5}=\frac{1.19 \times 50}{0.398}=\frac{59.5}{0.398}$
$=150 \mathrm{~min}$
Time required for $93.6 \%$ completion $=$ ? $k=\frac{2.303}{50} \log \frac{a}{a-0.6 a} \Rightarrow k=\frac{2.303}{50} \log \frac{a}{0.4 a}$ $k=\frac{2.303}{50} \log 2.5$
Also, $k=\frac{2.303}{x} \log \frac{a}{a-.936 a}$
$\therefore \frac{2.303 \log 2.5}{50}=\frac{2.303}{x} \log \frac{a}{0.064 a}$
$x=50 \times \frac{\log \frac{1000}{64}}{\log 2.5}$
$=50 \times \frac{\log 15.625}{\log 2.5}=\frac{1.19 \times 50}{0.398}=\frac{59.5}{0.398}$
$=150 \mathrm{~min}$
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