Consider the following data for the reaction 
$$\text{X}_2(g) + \text{Y}_2(g) \rightleftharpoons 2\text{XY}(g)$$
at 600 K.  The $$\Delta_r G^\circ$$ (in $$kJ/mol^{-1}$$) for the reaction is :

To determine the standard Gibbs free energy change $$\Delta_r G^\circ$$ for the reaction

$$X_2(g)+Y_2(g)\rightleftharpoons2XY(g)$$

at $$600\text{ K}$$, we use the thermodynamic relation

$$\Delta_r G^\circ=\Delta_r H^\circ-T\Delta_r S^\circ.$$

The standard enthalpy change of the reaction is calculated using the standard enthalpies of formation as

$$\Delta_r H^\circ=\sum\Delta_f H^\circ(\text{products})-\sum\Delta_f H^\circ(\text{reactants}).$$

Substituting the given values,

$$\Delta_r H^\circ=[2\times\Delta_f H^\circ(XY)]-[\Delta_f H^\circ(X_2)+\Delta_f H^\circ(Y_2)],$$

$$\Delta_r H^\circ=[2\times42]-(8+80),$$

$$\Delta_r H^\circ=84-88=-4\text{ kJ mol}^{-1}.$$

The standard entropy change of the reaction is obtained from the standard molar entropies:

$$\Delta_r S^\circ=\sum S^\circ(\text{products})-\sum S^\circ(\text{reactants}).$$

Substituting the given values,

$$\Delta_r S^\circ=[2\times S^\circ(XY)]-[S^\circ(X_2)+S^\circ(Y_2)],$$

$$\Delta_r S^\circ=[2\times200]-(140+250),$$

$$\Delta_r S^\circ=400-390=10\text{ J K}^{-1}\text{ mol}^{-1}.$$

To maintain consistent units, convert entropy into $$\text{kJ K}^{-1}\text{ mol}^{-1}$$:

$$\Delta_r S^\circ=\frac{10}{1000}=0.01\text{ kJ K}^{-1}\text{ mol}^{-1}.$$

Substituting the values into the Gibbs free energy equation at $$T=600\text{ K}$$,

$$\Delta_r G^\circ=\Delta_r H^\circ-T\Delta_r S^\circ,$$

$$\Delta_r G^\circ=-4-(600\times0.01),$$

$$\Delta_r G^\circ=-4-6=-10\text{ kJ mol}^{-1}.$$

Therefore, the standard Gibbs free energy change for the reaction is

$$\Delta_r G^\circ=-10\text{ kJ mol}^{-1}.$$