Question

The half life period (${\text{t}}_{1/2}$) of a reaction is halved as the initial concentration of the reactant is doubled. What is the order of the reaction ?

• A. Zero
• B. First
• C. Second
• D. Pseudo first

Answer 1

The correct option is C Second

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.

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

When the initial concentration of reactant is doubled, the half life period is halved.
$\therefore$

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

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

$2=2^{n-1}$
$\Rightarrow 2^1=2^{n-1}$
so that,
$n-1=1n=2$
The reaction is of the second order.