For a hypothetical gaseous reaction at 300 K : X (g) Y (g) + 2Z (g) The standard internal energy change ( U^ ) is 15.02 kJ mol ⁻¹ and the standard entropy change ( S^ ) is 80 J K ⁻¹ mol ⁻¹ . Determine the standard Gibbs free energy change ( G^ ) and predict the spontaneity of the process at this temperature. (Assume R = 8.3 J K ⁻¹ mol ⁻¹ )

Step-by-step solution

For the reaction X (g) Y (g) + 2Z (g) , the change in the number of moles of gaseous species is: n_g = (1 + 2) - 1 = 2 The standard enthalpy change is given by: H^ = U^ + n_g RT H^ = 15.02 + 2 8.3 300 10⁻³ H^ = 15.02 + 4.98 = 20.00 kJ mol ⁻¹ The standard Gibbs free energy change is given by: G^ = H^ - T S^ G^ = 20.00 - 300 80 10⁻³ G^ = 20.00 - 24.00 = -4.00 kJ mol ⁻¹ Since G^ Answer: -4.00 kJ mol ⁻¹ , spontaneous