Question

The half-life period of a reaction becomes 16 times when reactant concentration is halved. The order of reaction is ______.

Answer 1

For nth order reaction, $t_{1/2}\propto a^{(1-n)}$

When the reactant concentration is halved, the half-life period becomes 16 times.

$\frac{t_{1/2}}{t_{1/2}^{\prime }}=\frac{a^{(1-n)}}{a^{\prime (1-n)}}$

Substitute values in the above expression.

$\frac{1}{16}={\left(\frac{2}{1}\right)}^{1-n}$

Hence, $n=5$