Question

For a given reaction of $n^{th}$ order :
$A\to \text{Products}$
When the inital concentration i.e. $[A{]}_0$ was $0.96M$, the half life was found out to be $4min$. On changing the initial concentration (at constant temperature) to $0.64M$, half life period changes to $9min$.
The overall order of the reaction is:

Answer 1

$A\to \text{Products}$
$t_{1/2}=\frac{1}{k(n-1)}\left[\frac{2^{n-1}-1}{[A{]}_0^{n-1}}\right]$
Since the temperature remains constant, only thing that changes is the initial concentration.
$t_{1/2}\propto \frac{1}{[A{]}_0^{n-1}}$
Case 1:
$4\propto \frac{1}{(0.96{)}^{n-1}}$
Case 2:
$9\propto \frac{1}{(0.64{)}^{n-1}}$
Dividing both equations
$\frac{4}{9}={\left(\frac{0.64}{0.96}\right)}^{n-1}$
${\left(\frac{2}{3}\right)}^2={\left(\frac{2}{3}\right)}^{n-1}$
$n-1=2n=3$
hence it is a thord order reaction.