Question

Show that, in a first order reaction, time required for completion of $99.9\%$ is $10$ times of half life. $(log10=1)$

Answer 1

$t_{1/2}=\frac{2.303}{K}log2$ ......$\left(1\right)$
$t_{99.9\%}=\frac{2.303}{K}log\frac{100}{0.1}$
$=\frac{2.303}{K}log{10}^3$
$=\frac{2.303}{K}3log10$ ......$\left(2\right)$
From equation $\left(1\right)$ & $\left(2\right)$
$t_{99.9\%}=10\times t_{1/2}$