Show that for a first order reaction, the time required for 99.9% completion of the reaction is 10 times that required for 50% completion.

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Solution

$t = \frac{2.303}{k} l o g \frac{a}{a - x}$
$t_{99.9} = \frac{2.303}{k} l o g \frac{100}{100 - 99.9} = \frac{2.303}{k} l o g_{10} \frac{100}{1}$
$= \frac{2.303}{k} l o g_{10} 1000 . . . . . (i)$
$t_{50} = \frac{2.303}{k} l o g \frac{100}{100 - 50}$
$= \frac{2.303}{k} l o g_{10} \frac{100}{50}$
$= \frac{2.303}{k} l o g_{10}^{2} . . . . . (i i)$
$\frac{i}{i i} = \frac{99.9}{50} = \frac{\frac{2.303}{K} l o g_{10} 1000}{\frac{2.303}{K} l o g_{10}^{2}}$
$= \frac{3.000}{0.3010} = 10$

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