For the gas-phase reaction X₂(g) + Y₂(g) 2XY(g) , the equilibrium concentrations of X₂ , Y₂ , and XY at 500 K are 0.01 mol L ⁻¹ , 0.01 mol L ⁻¹ , and 0.1 mol L ⁻¹ respectively. Calculate the standard Gibbs free energy change ( G^ ) for this reaction. (Given: R = 8.314 J K ⁻¹ mol ⁻¹ , 10 = 2.303 )

Step-by-step solution

The equilibrium constant K_c is given by: K_c = [XY]^2 [X₂][Y₂] Substituting the given equilibrium concentrations: K_c = (0.1)^2 (0.01)(0.01) = 0.01 0.0001 = 100 Since the change in the number of moles of gas 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: G^ = -RT K_p Substituting the values: G^ = -8.314 500 (100) G^ = -8.314 500 2 10 G^ = -8.314 500 2 2.303 G^ = -19147.14 J mol ⁻¹