For the gas-phase reaction X_2(g) + Y_2(g) leftharpoons 2XY(g) ,...

The equilibrium constant $K_c$ is given by:

$K_c = \dfrac{[XY]^2}{[X_2][Y_2]}$

Substituting the given equilibrium concentrations:

$K_c = \dfrac{(0.1)^2}{(0.01)(0.01)} = \dfrac{0.01}{0.0001} = 100$

Since the change in the number of moles of gas $\Delta n_g = 2 - (1+1) = 0$, the equilibrium constant in terms of pressure is $K_p = K_c = 100$.

The standard Gibbs free energy change is related to the equilibrium constant by the equation:

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

Substituting the values:

$\Delta G^\circ = -8.314 \times 500 \times \ln(100)$

$\Delta G^\circ = -8.314 \times 500 \times 2 \ln 10$

$\Delta G^\circ = -8.314 \times 500 \times 2 \times 2.303$

$\Delta G^\circ = -19147.14 \text{ J mol}^{-1}$

$\Delta G^\circ = -19.15 \text{ kJ mol}^{-1}$

Answer: $-19.15\text{ kJ mol}^{-1}$