To calculate the Gibbs free energy change G for the formation of 2 moles of water molecules from hydrogen gas and oxygen gas at standard conditions, we need to use the following equation:G = H - TSwhere H is the change in enthalpy, T is the temperature in Kelvin, and S is the change in entropy.The balanced chemical equation for the formation of water is:2H g + O g 2HO l The standard enthalpy change H for this reaction is -285.8 kJ/mol for the formation of 1 mole of water. Since we are considering the formation of 2 moles of water, the H for this reaction will be:H = -285.8 kJ/mol 2 = -571.6 kJThe standard entropy change S for this reaction can be calculated using the standard molar entropies of the reactants and products:S = S products - S reactants The standard molar entropies for the substances involved in the reaction are:S H = 130.6 J/molKS O = 205.1 J/molKS HO = 69.9 J/molKNow, we can calculate the S for the reaction:S = 2 S HO - 2 S H + S O S = 2 69.9 J/molK - 2 130.6 J/molK + 205.1 J/molK S = 139.8 J/molK - 261.2 J/molK + 205.1 J/molK S = -326.5 J/molKNow, we can calculate the Gibbs free energy change G using the equation:G = H - TSAt standard conditions, the temperature T is 298 K. So, we have:G = -571.6 kJ - 298 K -326.5 J/molK 1 kJ/1000 J G = -571.6 kJ + 97.3 kJG = -474.3 kJSince we are considering the formation of 2 moles of water, the Gibbs free energy change per mole of water is:G per mole = -474.3 kJ / 2 = -237.15 kJ/molTherefore, the Gibbs free energy change for the formation of 2 moles of water molecules from hydrogen gas and oxygen gas at standard conditions 298 K and 1 atm pressure is -237.15 kJ/mol.