Question

For the elementary reaction
$M\to N$
the rate of disappearence of M increases by a factor of 8 after doubling the concentration of M. The order of the reaction with respect to M is

Answer 1

Given information:
The rate of the reaction increases by a factor of 8 after the doubling the concentration of M.
Hence $\text{rate of the reaction after doubling}{r}_2$ = $8\times \text{rate of the reaction before doubling}{r}_1$

Hence $\frac{r_2}{r_1}=8$
The rate laws for the reaction can be written as,
$r_1=k[M{]}^n{r}_2=k[2M{]}^n$
Where n is the order of the reaction.
, $\frac{r_1}{r_2}=\frac{1}{8}\frac{r_1}{r_2}=\frac{1}{8}=\frac{[M{]}^n}{[2M{]}^n}onsolving\Rightarrow n=3$
Hence the order of the reaction with respect to M is 3.