Question

For the elementary reaction, M $\to$ 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

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

The correct option is B 3

For the elementary reaction, M $\to$ N, the rate law can be written as

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

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

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

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

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

$\Rightarrow [2{]}^n=8=[2{]}^3\Rightarrow n=3$