ΔG° = ΔH° - TΔS°

Question

For the reaction

H₂S(g) + 2H₂O(l)3H₂(g) + SO₂(g)

H° = 295.4 kJ and S° = 294.7 J/K

The standard free energy change for the reaction of 2.28 moles of H₂S(g) at 325 K, 1 atm would be _____ kJ.

This reaction is (reactant, product) _____ favored under standard conditions at 325 K.

Assume that H° and S° are independent of temperature.

Answer 1

When using this equation, remember to change units to match for ∆H and ∆S. I'll change ∆S to kJ instead of J

∆Gº = 295.4 kJ/mol + (325K)(0.2947 kJ/mol-K)= 295.4 + 95.78

∆Gº = 391.2 kJ/mol

For 2.28 moles H2S, ∆Gº = 2.28 mol x 391.2 kJ/mol = 891.9 kJ

This reaction reactant favored (because ∆Gº is positive)