Chemical Kinetics - Result Question 12
13. For an elementary chemical reaction, $A _2 \underset{k _{-1}}{\stackrel{k _1}{\rightleftharpoons}} 2 A$, the expression for $\dfrac{d[A]}{d t}$ is
(a) $2 k _1\left[A _2\right]-k _{-1}[A]^{2}$
(b) $k _1\left[A _2\right]-k _{-1}[A]^{2}$
(c) $2 k _1\left[A _2\right]-2 k _{-1}[A]^{2}$
(d) $k _1\left[A _2\right]+k _{-1}[A]^{2}$
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Answer:
Correct Answer: 13. (c)
Solution:
- The elementary reaction, $A _2 \underset{k _{-1}}{\stackrel{k _1}{\rightleftharpoons}} 2 A$
follows opposing or reversible kinetics,
(i) Rate of the reaction,
$ \begin{aligned} r & =r _{\text {forward }}-r _{\text {backward }} \\ & =k _1\left[A _2\right]-k _{-1}[A]^{2} \quad …….. (i) \end{aligned} $
(ii) Again, rate of the reaction can be expressed as,
$ r=-\dfrac{d\left[A _2\right]}{d t}=+\dfrac{1}{2} \dfrac{d[A]}{d t} $
So, the rate of appearance of $A$, i.e.
$\dfrac{d[A]}{d t}=2 r=2 k _1\left[A _2\right]-2 k _{-1}[A]^{2}\quad $ [from Eq. (i)]
BETA
