1. Given reaction

$2A \rightleftharpoons B + C$

We are given:

• Standard Gibbs energy change: $\Delta G^\circ = 2494.2\,\text{J mol}^{-1}$
• Temperature: $T = 300\,\text{K}$
• $R = 8.314\,\text{J K}^{-1}\text{mol}^{-1}$

At a given time:

• $[A] = \frac{1}{2}$
• $[B] = 2$
• $[C] = \frac{1}{2}$

We must determine the direction in which the reaction proceeds.

2. Find the equilibrium constant using

$\Delta G^\circ = -RT \ln K$

So,

$\ln K = -\frac{\Delta G^\circ}{RT}$

Substitute values:

$\ln K = -\frac{2494.2}{8.314 \times 300}$

Since,

$8.314 \times 300 = 2494.2$

therefore,

$\ln K = -1$

Hence,

$K = e^{-1} = \frac{1}{e} = \frac{1}{2.718} \approx 0.368$

3. Calculate reaction quotient

For the reaction

$2A \rightleftharpoons B + C$

$Q_c = \frac{[B][C]}{[A]^2}$

Substitute the given concentrations:

$Q_c = \frac{(2)\left(\frac{1}{2}\right)}{\left(\frac{1}{2}\right)^2}$

$Q_c = \frac{1}{\frac{1}{4}} = 4$

4. Compare $Q_c$ and $K_c$

We have:

$Q_c = 4$ $K_c \approx 0.368$

Thus,

$Q_c > K_c$

When $Q > K$, the reaction mixture contains excess products compared to equilibrium, so the reaction proceeds in the reverse direction to form more reactant $A$.

5. Check options

• A: reverse direction because $Q > K_c$ ✅
• B: forward direction because $Q < K_c$ ❌
• C: reverse direction because $Q < K_c$ ❌
• D: forward direction because $Q > K_c$ ❌

Final answer: Option A is correct.