Question

Half life of a reaction becomes half when the initial concentration of the reactants are made double. The order of the reaction will be:

• A. 1
• B. 2
• C. 0
• D. 3

Answer 1

The correct option is B 2

We know that,
$t_{\frac{1}{2}}\propto \frac{1}{[a{]}^{n-1}}$
Where,
$\left[a\right]=\text{initial concentration of the reactant}$
$n=\text{order of the reaction}$
So,
$t_{\frac{1}{2}}=\frac{k}{[a{]}^{n-1}}$ ...(1)
$k=\text{constant}$
Again,
$\frac{t_{\frac{1}{2}}}{2}=\frac{k}{[2a{]}^{n-1}}$ ...(2)
Dividing equation (1) by equation (2), we get
$2=2^{(n-1)}$
$\Rightarrow n=2$
Thus the order of the reaction is 2.