Show that in a first order reaction, time required for completion of $75 %$ is twice of half life of the reaction. $(l o g 2 = 0.3010)$

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Solution

$t_{1 / 2} = \frac{2.303}{K} l o g \frac{a_{0}}{a_{0} - \frac{a_{0}}{2}}$
$= \frac{2.303}{K} l o g 2$ ...... $(1)$
$t_{75 %} = \frac{2.303}{K} l o g \frac{a_{0}}{a_{0} - \frac{3}{4} a_{0}}$
$= \frac{2.303}{K} l o g_{4} = \frac{2.303}{K} .2 l o g 2$ ........(2)
$∴$ from equation $(1)$ & $(2)$
$t_{75 % = 2 \times t_{1 / 2}}$

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