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. (b)
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} \quad \quad …… (i) \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} \quad \quad …… (ii) \\ \dfrac{\text { Rate }}{8 \times \text { Rate }} & =\dfrac{k[M]^{n}}{k[2 M]^{n}} \Rightarrow \dfrac{1}{8}=\dfrac{1}{[2]^{n}} \\ \Rightarrow \quad[2]^{n} & =8=[2]^{3} \Rightarrow n=3 \end{aligned} $
BETA
