Chemical Kinetics - Result Question 47
47. Consider the following reversible reaction,
$ A(g)+B(g) \rightleftharpoons A B(g) $
The activation energy of the backward reaction exceeds that of the forward reaction by $2 R T$ (in $J$ $mol^{-1}$ ). If the pre-exponential factor of the forward reaction is 4 times that of the reverse reaction, the absolute value of $\Delta G^{\ominus}$ (in $J$ $mol^{-1}$ ) for the reaction at $300 K$ is
(Given; $\ln (2)=0.7 R T=2500 J mol^{-1}$ at $300 K$ and $G$ is the Gibbs energy)
(2018 Adv.)
$(2006,3 \times 4 M=12 M)$
Show Answer
Answer:
Correct Answer: 47. $(+8500 \mathrm{~J} / \mathrm{mol})$
Solution:
- For the reaction,
$ A(g)+B(g) \rightleftharpoons A B(g) $
Given $\quad E _{a _b}=E _{a _f}+2 R T$ or $E _{a _b}-E _{a _f}=2 R T$
Further
$ A _f=4 A _b \quad \text { or } \quad \dfrac{A _f}{A _b}=4 $
Now, rate constant for forward reaction,
$ k _f=A _f e^{-E _{a _f} / R T} $
Likewise, rate constant for backward reaction,
$ k _b=A _b e^{-E _{a _b} / R T} $
At equilibrium,
Rate of forward reaction $=$ Rate of backward reaction
$ \begin{array}{ll} \text { i.e., } & k _f=k _b \text { or } \dfrac{k _f}{k _b}=k _{e q} \\ \text { so } & k _{e q}=\dfrac{A _f e^{-E _a / R T}}{A _b e^{-E _{a b} / R T}}=\dfrac{A _f}{A _b} e^{-\left(E _{a _f}-E _{a _b}\right) / R T} \end{array} $
After putting the given values
$ k _{e q}=4 e^{2} \quad\left(\text { as } E _{a _b}-E _{a _f}=2 R T \text { and } \dfrac{A _f}{A _b}=4\right) $
$ \text { Now, } \quad \begin{aligned} \Delta G^{\circ} & =-R T \ln K _{eq}=-2500 \ln \left(4 e^{2}\right) \\ & =-2500\left(\ln 4+\ln e^{2}\right)=-2500(1.4+2) \\ & =-2500 \times 3.4=-8500 J / mol \end{aligned} $
Absolute value $=8500 J / mol$
BETA
