Question

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

Answer 1

For a first order reaction, the time required for 99% completion is

$t_1 = \frac {2.303}{k} log \ \frac {100}{100-99}$

$t_1 = \frac {2.303}{k} log \ 100$

$t_1 = 2 \times \frac {2.303}{k}$

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

$t_2 = \frac {2.303}{k} log \ \frac {100}{100-90}$

$t_2 = \frac {2.303}{k} log \ 10$

$t_2 = \frac {2.303}{k}$

$Therefore,\ t_1= 2t_2$

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.