How many times of the half-life will require to complete $75$ % of a reaction of first order?

1

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2

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3

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Solution

The correct option is B $2$
75% completion of a reaction corresponds to product concentration becoming ${\frac{3}{4}}^{t h}$ of initial concentration i.e. concentration of reactant at that time, $a_{t} = \frac{a_{0}}{4}$
$k t_{3 / 4} = I n \frac{a_{0}}{a_{0} / 4} = 2 I n 2$
$t_{3 / 4} = 2 (\frac{I n 2}{k}) = 2 t_{1 / 2}$

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75 %

(l o g 2 = 0.3010)

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The half-life period of a first order chemical reaction is 6.93 minutes.
The time required for 99% completion of reaction will be (log 2 = 0.301)

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