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.)
Passage Based Questions
Passage
Carbon-14 is used to determine the age of organic material. The procedure is based on the formation of ${ }^{14} Cby$ neutron capture in the upper atmosphere.
$$ { } _7^{14} N+{ } _0 n^{1} \longrightarrow{ } _6^{14} C+{ } _1 p^{1} $$
${ }^{14} C$ is absorbed by living organisms during photosynthesis. The ${ }^{14} C$ content is constant in living organism once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of ${ }^{14} C$ in the dead being, falls due to the decay which $C$ - 14 underoges
$$ { } _6^{14} C \longrightarrow{ } _7^{14} N+\beta^{-} $$
The half-life period of ${ }^{14} C$ is $5770 yr$.
The decay constant $(\lambda)$ can be calculated by using the following formula $\lambda=\frac{0.693}{t _{1 / 2}}$. The comparison of the $\beta^{-}$activity of the dead matter with that of the carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle. The method however, ceases to be accurate over periods longer than 30,000 yr. The proportion of ${ }^{14} C$ to ${ }^{12} C$ in living matter is $1: 10^{12}$.
$(2006,3 \times 4 M=12 M)$
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Answer:
Correct Answer: 47. 0.0015
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
$$ A _f=4 A _b \quad \text { or } \quad \frac{A _f}{A _b}=4 $$
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
$$ \begin{array}{ll} \text { i.e., } & k _f=k _b \text { or } \frac{k _f}{k _b}=k _{e q} \ \text { so } & k _{e q}=\frac{A _f e^{-E _a / R T}}{A _b e^{-E _{a b} / R T}}=\frac{A _f}{A _b} e^{-\left(E _{a _f}-E _{a _b}\right) / R T} \end{array} $$
After putting the given values
$$ k _{e q}=4 e^{2} \quad\left(\text { as } E _{a _b}-E _{a _f}=2 R T \text { and } \frac{A _f}{A _b}=4\right) $$
$$ \text { Now, } \quad \begin{aligned} \Delta G^{\circ} & =-R T \ln K _{eq}=-2500 \ln \left(4 e^{2}\right) \ & =-2500\left(\ln 4+\ln e^{2}\right)=-2500(1.4+2) \ & =-2500 \times 3.4=-8500 J / mol \end{aligned} $$
Absolute value $=8500 J / mol$
