The half-life period of a first order reaction is $1$ hr. What is the time in hour taken for $87.5$ % completion of the reaction?

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Solution

The correct option is A $3$
$t_{1 / 2} = \frac{I n 2}{k} = 1 h r$ => k=In2$
For 87.5% completion, $a_{t} = \frac{100 - 87.5}{100} \times a_{0} = \frac{125}{1000} a - 0 = \frac{a_{0}}{8}$
$k t = I n \frac{a_{0}}{a_{t}}$
$(I n 2) . t = I n \frac{a_{0}}{a_{0} / 8} = 3 I n 2$
t=3 hr

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