Chemical Kinetics - Result Question 19
####22. For the elementary reaction, $M \longrightarrow N$, 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
(2014 Adv.)
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Answer:
Correct Answer: 22. (a)
Solution:
- For the elementary reaction, $M \longrightarrow N$
Rate law can be written as
$$ \begin{aligned} & \text { Rate } \propto[M]^{n} \ & \text { Rate }=k[M]^{n} \end{aligned} $$
When we double the concentration of $[M]$, rate becomes 8 times, hence new rate law can be written as
$$ \begin{aligned} 8 \times \text { Rate } & =k[2 M]^{n} \ \frac{\text { Rate }}{8 \times \text { Rate }} & =\frac{k[M]^{n}}{k[2 M]^{n}} \Rightarrow \frac{1}{8}=\frac{1}{[2]^{n}} \ \Rightarrow \quad[2]^{n} & =8=[2]^{3} \Rightarrow n=3 \end{aligned} $$
