Question

A first order reaction has a half life of 1 min. The time required for 99.9% completion of the reaction is _____min. (Round off to the nearest integer) [ Use ln2 = 0.69, ln10 = 2.3]

Answer 1

Answer: 10
Derivation:
By first order rate equation,

$t_{99.9\text{\%}}=\frac{1}{k}ln\left(\frac{100}{0.1}\right)=\frac{1}{k}ln\left(1000\right)=\frac{3}{k}ln\left(10\right)....eqn\left(1\right)$

$t_{50\text{\%}}=\frac{1}{k}ln\left(\frac{100}{50}\right)=\frac{1}{k}ln\left(2\right).......eqn\left(2\right)$

Equation (1) divided by equation (2), we get

$\frac{t_{99.9\text{\%}}}{t_{50\text{\%}}}=\frac{3ln\left(10\right)}{ln\left(2\right)}$

$t_{99.9\text{\%}}=3\times \frac{t_{1/2}}{ln\left(2\right)}\times ln\left(10\right)$

$=3\times \left(1min\right)\times \frac{2.3}{0.69}$

$=\frac{6.9}{0.69}$

$=10min$