Show that time required to complete $99.9 %$ completion of a first order reaction is $1.5$ times to $90 %$ completion.

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Solution

For $99.9 %$ completion $[A]_{0} = 100 [A] = 0.1$

$t_{99.9 %} = \frac{2.303}{K} log \frac{100}{01} = \frac{2.303}{K} \times 3$

Similarly $t_{90 %} = \frac{2.303}{K} log \frac{100}{1} = \frac{2.303}{K} \times 2$

$∴ t_{99.9 %} = 1.5 \times T_{90 %}$ .

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