The standard Gibbs energy change at 300 K for the reaction $$2A \rightleftharpoons B + C$$ is 2494.2 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 J/K-mol, e = 2.718] {Given antilog (-0.44) = 0.36}

For a chemical reaction, the equilibrium constant $$K_C$$ is linked to the standard Gibbs energy change $$\Delta G^\circ$$ by the formula

$$\Delta G^\circ = -RT\ln K_C.$$

We have $$\Delta G^\circ = 2494.2\ \text{J},\; R = 8.314\ \text{J\,K}^{-1}\text{mol}^{-1},\; T = 300\ \text{K}.$$ Substituting,

$$\ln K_C = -\frac{\Delta G^\circ}{RT} = -\frac{2494.2}{8.314 \times 300}.$$

The denominator is $$8.314 \times 300 = 2494.2,$$ so

$$\ln K_C = -\frac{2494.2}{2494.2} = -1.$$

Now we take the antilogarithm (base $$e$$):

$$K_C = e^{-1} \approx \frac{1}{2.718} \approx 0.37.$$

Next, we evaluate the reaction quotient $$Q$$ for the current concentrations. For the reaction $$2A \rightleftharpoons B + C,$$ the expression is

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

With $$[A] = \tfrac12,\; [B] = 2,\; [C] = \tfrac12,$$ we get

$$Q = \frac{2 \times \tfrac12}{\left(\tfrac12\right)^2} = \frac{1}{\tfrac14} = 4.$$

So we have $$Q = 4$$ and $$K_C \approx 0.37.$$ Clearly,

$$Q > K_C.$$

When $$Q$$ is larger than $$K_C,$$ the system contains more products than it would at equilibrium, so the reaction shifts in the reverse direction to form more reactant and reduce $$Q$$ to $$K_C.$$

Hence, the correct answer is Option C.