Chemical and Ionic Equilibrium 1 Question 12
13. The standard Gibbs energy change at $300 \mathrm{~K}$ for the reaction, $2 A \rightleftharpoons B+C$ is 2494 . $2 \mathrm{~J}$. At a given time, the composition of the reaction mixture is $[A]=\frac{1}{2},[B]=2$ and $[C]=\frac{1}{2}$. The reaction proceeds in the
$(R=8.314 \mathrm{JK} / \mathrm{mol}, e=2.718)$
(2015, Main)
(a) forward direction because $Q>K_{c}$
(b) reverse direction because $Q>K_{c}$
(c) forward direction because $Q<K_{c}$
(d) reverse direction because $Q<K_{c}$
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Solution:
- Given, $\Delta G^{\circ}=2494.2 \mathrm{~J}$
$$ Q=\frac{[B][C]}{[A]^{2}}=\frac{2 \times \frac{1}{2}}{\left(\frac{1}{2}\right)^{2}}=4 $$
$\therefore$ We know, $\Delta G=\Delta G^{\circ}+R T \ln Q$
$$ \begin{aligned} & =2494.2+8.314 \times 300 \ln 4 \ & =28747.27 \mathrm{~J}(+ \text { ve value }) \end{aligned} $$
Also, we have $\quad \Delta G=R T \ln \frac{Q}{K}$
If $\Delta G$ is positive, $Q>K$
Therefore, reaction shifts in reverse direction.
