Order of reaction 2

For a reaction $A \rightarrow B$ , let the order be $n$ . Therefore, we can write, $\displaystyle -\frac{\text{d}a}{\text{dt}} = k a^n$ . Rearranging and integrating, we obtain, $\displaystyle \int_{a_0}^a -\frac{\text{d}a}{a^n} = \int_0^t k \text{ d}t \Rightarrow \frac{1}{n-1} \left( \frac{1}{a^{n-1}} - \frac{1}{a_0^{n-1}} \right) = kt$ .

At half life time, $t = T_{1/2}$ and $a = \dfrac{a_0}{2}$ . Substituting it in the above equation, we get, $\displaystyle \frac{1}{(n-1 )a_0^{n-1}} \left( \frac{1}{2^{n-1}} - 1 \right) = kT_{1/2}$ .

Therefore, $\displaystyle T_{1/2} \propto \frac{1}{a_0^{n-1}}$ . Here we realize, $T_{1/2} \propto \frac{1}{a_0}$ . On comparing the two, we can write, $n-1 = 1 \Rightarrow \boxed{n=2}$ .

Note:

$a$ denotes the concentration of $A$ at any time $t$ while $a_0$ is its initial concentration. $T_{1/2}$ is the half life time of the reaction

General Expression for half life time for $n^{th}$ order reaction is given by $t_{1/2}=\frac{2^{n-1}-1}{(n-1)kC^{n-1}}$ See proof here: Half life time

That is $t_{1/2} \propto {C_0}^{1-n}$ When initial Concentration $C_0$ is doubled, half life time $t_{1/2}$ is reduced to half. Taking ratio, $\frac{1}{1/2}=(\frac{1}{2})^{1-n}$ That is $2=2^{n-1}$ which gives
$\boxed{n=2}$