Question

The half life period (${\text{t}}_{1/2}$) of a reaction is reduced to $\frac{1}{4}$ times as the initial concentration of the reactant is doubled. What is the order of the reaction ?
Assume constant temperature conditions

• A. 1
• B. 2
• C. 3
• D. 0

Answer 1

The correct option is C 3

For a general reaction,
$A\to \text{Product(s)}$

$t_{1/2}=\frac{1}{k(n-1)}\left[\frac{2^{n-1}-1}{[A{]}_0^{n-1}}\right]$
Since the temperature remains constant, only thing that changes is the initial concentration.

$t_{1/2}\propto \frac{1}{[A{]}_0^{n-1}}.....eqn\left(1\right)$

From the given condition:


$\frac{1}{4}{t}_{1/2}=\frac{1}{[2A{]}_0^{n-1}}.....eqn\left(2\right)$
Dividing equation (1) by (2), we get,

$\frac{t_{1/2}}{\frac{1}{4}{t}_{1/2}}=\frac{1/[A{]}_0^{n-1}}{1/[2A{]}_0^{n-1}}$

$4=2^{n-1}$
$\Rightarrow 2^2=2^{n-1}$
so that gives,
$n-1=2n=3$
The reaction is of third order.