Question

The $t_{1/2}$ of a reaction is doubled as the initial concentration of the reactant is doubled. What is the order of the reaction?

• A. Zero order
• B. First order
• C. Second order
• D. Third order

Answer 1

The correct option is A Zero order

We know,
$t_{\frac{1}{2}}\propto \frac{1}{a_0^{n-1}}$
Where,
$t_{\frac{1}{2}}=\text{half-life of the reaction}{a}_0=\text{initial concentration}n=\text{order of reaction}$
According to the question,
$\frac{t_{1/2}}{2t_{1/2}}=\frac{1/\left(a_0^{n-1}\right)}{1/(2a_0{)}^{n-1}}$
$\Rightarrow \frac{1}{2}=2^{n-1}$
$\Rightarrow 2^{-1}=2^{n-1}$
$\Rightarrow n-1=-1$
$\Rightarrow n=0$
Hence, the reaction is of the zero order.