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It takes $1 \mathrm{~h}$ for a first order reaction to go to $50 \%$ completion. The total time required for the same reaction to reach $87.5 \%$ completion will be
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$3.00 \mathrm{~h}$
$\mathrm{t}_{1 / 2}=1 \mathrm{hr}$
$\mathrm{t}_{87.5}=\frac{2.303}{\mathrm{k}} \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}$
$=\frac{2.303}{\frac{0.693}{1}} \log \frac{\mathrm{a}}{\mathrm{a}-.875}$
$=\frac{2.303}{0.693} \log 8$
$=\frac{2.303}{0.693} \times 3 \times .3010$
$=3$
$\mathrm{t}_{87.5}=\frac{2.303}{\mathrm{k}} \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}$
$=\frac{2.303}{\frac{0.693}{1}} \log \frac{\mathrm{a}}{\mathrm{a}-.875}$
$=\frac{2.303}{0.693} \log 8$
$=\frac{2.303}{0.693} \times 3 \times .3010$
$=3$
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