Question

In a first order reaction, $10$% of the reactant is consumed in $25$ minutes. Calculate :
(1) The half-life period of the reaction.
(2) The time required for completing $87.5$% of the reaction.

Answer 1

$t=10$% reactant $\to 25$ minute
(1) $t_{1/2}=\frac{0.693}{K}$

$K=\frac{2.303}{t}\log \frac{100}{90}$

$K=0.004215$
$t_{1/2}=\frac{0.693}{K}$
$t_{1/2}=164.405$ minute.

(2) $t=\frac{2.303}{K}\log \frac{100}{12.5}$
$t=493.43$ minute.