Question

Calculate the Gibbs free energy change for the reaction: 2H2(g) + O2(g) → 2H2O(g) at standard conditions (298 K and 1 atm pressure) given the following data:ΔH˚f[H2O(g)] = -241.8 kJ/molΔH˚f[H2(g)] = 0 kJ/molΔH˚f[O2(g)] = 0 kJ/molΔS˚[H2(g)] = 130.7 J/K·molΔS˚[O2(g)] = 205.0 J/K·molΔS˚[H2O(g)] = 188.8 J/K·mol

Answer 1

To calculate the Gibbs free energy change  G  for the reaction, we can use the equation:G = H - TSwhere H is the change in enthalpy, T is the temperature in Kelvin, and S is the change in entropy.First, we need to find the change in enthalpy  H  for the reaction. We can do this using the given standard enthalpies of formation  Hf  for the reactants and products:H = [2 * Hf[H2O g ] -  2 * Hf[H2 g ] + Hf[O2 g ] ]H = [2 *  -241.8 kJ/mol  -  2 * 0 kJ/mol + 0 kJ/mol ]H = -483.6 kJ/molNext, we need to find the change in entropy  S  for the reaction. We can do this using the given standard entropies  S  for the reactants and products:S = [2 * S[H2O g ] -  2 * S[H2 g ] + S[O2 g ] ]S = [2 * 188.8 J/Kmol -  2 * 130.7 J/Kmol + 205.0 J/Kmol ]S = 377.6 J/Kmol -  261.4 J/Kmol + 205.0 J/Kmol S = 377.6 J/Kmol - 466.4 J/KmolS = -88.8 J/KmolNow, we can calculate the Gibbs free energy change  G  for the reaction using the equation:G = H - TSG = -483.6 kJ/mol -  298 K *  -88.8 J/Kmol  Since we need to convert J to kJ, we divide by 1000:G = -483.6 kJ/mol -  298 K *  -0.0888 kJ/Kmol  G = -483.6 kJ/mol + 26.5 kJ/molG = -457.1 kJ/molThe Gibbs free energy change for the reaction 2H2 g  + O2 g   2H2O g  at standard conditions  298 K and 1 atm pressure  is -457.1 kJ/mol.