Question

For the elementary reaction $M⟶N$, the rate of disappearance of M increases by a factor of 8 upon doubling the concentration of M. The order of the reaction with respect to M is

Answer 1

$\text{Rate}\propto {\left[M\right]}^n$

$\text{Rate}=k{\left[M\right]}^n......\left(i\right)$

When we double the concentration of [M], rate becomes 8 times, hence new rate law can be written as

$8\times \text{Rate}=k{\left[2M\right]}^n......\left(ii\right)$

$\frac{\text{Rate}}{8\times \text{Rate}}=\frac{k{\left[M\right]}^n}{k{\left[2M\right]}^n}$

$\frac{1}{8}=\frac{1}{2^n}$

$\Rightarrow 2^n=8\Rightarrow n=3$