Question

Show that in a first order reaction, time required for completion of 99.9% is 10 times of half-life ($t_{1/2}$) of the reaction.

Answer 1

For first order reaction,

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

Given that:-

$x=99.9a$ for $t_{99.9\%}$

$x=0.5a$ for $t_{50\%}$

$\therefore \frac{t_{99.9\%}}{t_{50\%}}=\frac{\frac{2.303}{k}\log \frac{a}{a-0.999a}}{\frac{2.303}{k}\log \frac{a}{a-0.5a}}$

$\Rightarrow \frac{t_{99.9\%}}{t_{50\%}}=\frac{\log \frac{1}{0.001}}{\log \frac{1}{0.5}}$

$\Rightarrow \frac{t_{99.9\%}}{t_{50\%}}=\frac{-\log 0.001}{\log 2}$

$\Rightarrow \frac{t_{99.9\%}}{t_{50\%}}=\frac{-(-3)}{0.3010}$

$\Rightarrow t_{99.9\%}=10\times t_{50\%}$

Hence proved.