Ncert Exercises & Solutions

4.18 For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

$t_{1} = \frac{2.303}{K} \log \frac{100}{100 - 99}$
$= \frac{2.303}{K} \log 100$
$= 2 \times \frac{2.303}{K}$

For a first order reaction, the time required for 90% completion is

$t_{1} = \frac{2.303}{K} \log \frac{100}{100 - 99}$
$= \frac{2.303}{K} \log 10$
$= \frac{2.303}{K}$
$T h e r e f o r e, t_{1} = 2 t_{2}$

Hence, the time required for 99% completion of a first order reaction is twice the time

required for the completion of 90% of the reaction.