nth Order Reaction - Practice Questions & MCQ

n^th order kinetics

The rates of the reaction is proportional to nth power of reactant

      $\begin{aligned} & \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=-\mathrm{k}[\mathrm{A}]^{\mathrm{n}} \\ \Rightarrow & \frac{\mathrm{d}[\mathrm{A}]}{[\mathrm{A}]^{\mathrm{n}}}=-\mathrm{kdt} \\ \Rightarrow & \int_{\mathrm{A}_0}^{[\mathrm{A}]_{\mathrm{t}}} \frac{\mathrm{d}[\mathrm{A}]}{[\mathrm{A}]^{\mathrm{n}}}=-\mathrm{k} \int_0^{\mathrm{t}} \mathrm{dt} \\ \Rightarrow & {\left[\frac{[\mathrm{A}]^{1-\mathrm{n}}}{1-\mathrm{n}}\right]_{[\mathrm{A}]_0}^{[\mathrm{A}]_{\mathrm{t}}}=-\mathrm{k}[\mathrm{t}]_0^{\mathrm{t}} }\end{aligned}$

$\Rightarrow \frac{1}{(\mathrm{n}-1)}\left[\frac{1}{[\mathrm{~A}]_{\mathrm{t}}^{(\mathrm{n}-1)}}-\frac{1}{[\mathrm{~A}]_0^{(\mathrm{n}-1)}}\right]=\mathrm{k}(\mathrm{t})$

Half life for any n^th order reaction

$\mathrm{t}_{\frac{1}{2}}=\frac{1}{(\mathrm{k})(\mathrm{n}-1)\left([\mathrm{A}]_0^{\mathrm{n}-1}\right)}\left[2^{\mathrm{n}-1}-1\right]$

Thus for any general nth order reaction it is evident that,

$\mathrm{t}_{\frac{1}{2}} \propto[\mathrm{~A}]_0^{1-\mathrm{n}}$

It is to be noted that the above formula is applicable for any general nth order reaction except n=1.

Can you think of the reason why this is not applicable for a first order reaction?