Question

For a first order reaction, time required for $99.0$% completion is $x$ times for the first time required for the completion of $90$% of the reaction $x$ is:

Answer 1

For the first order reaction,

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

x=0.999(given)

$K=\frac{2.303}{k}log\frac{a}{a-0.99}$

$K=\frac{2.303}{k}log\frac{a}{0.001a}$

$K=\frac{2.303}{k}log1000$

$K=\frac{2.303}{k}3$....(i)

the half life is given by,

$t_{1/2}=\frac{2.303}{k}log\frac{a}{(a-0.5a)}=\frac{2.303}{k}log2=\frac{2.303}{k}\times \mathrm{0.3010.....}\left(ii\right)$

divide(i) by (ii)

$\frac{t_{0.999}}{t_{0.5}}=\frac{2.303}{k}3\times \frac{k}{2.303\times 0.3010}=10$

$t_{0.999}=10\times t_{0.5}$.