Question

The time required for a first-order reaction for $99$% completion is $x$ times for the time required for $90$% completion. $x$ is:

Answer 1

Time required of a first-order reaction for 99% completion is x times for the time required for 90% completion. x is 2.

$t=\frac{2.303}{k}log\frac{a}{a-x}$

For 90% completion,

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

For 99% completion,

$t_{99}=\frac{2.303}{k}log\frac{100}{100-99}=\frac{4.605}{k}$

Hence, $\frac{t_{99}}{t_{90}}=\frac{\frac{4.605}{k}}{\frac{2.303}{k}}=2=x$

$\therefore x=2$