Given that Delta. G for the reaction below is - 957.9 kJ, what is Delta. Gf of H2O?

4NH3 (g) + 5O2 (g) Right arrow. 4NO (g) + 6H2O (g)

Delta. Gf, NH3 = - 16.66 kJ/mol

Delta. Gf, NO = 86.71 kJ/mol

6ΔG° (f) H₂O = - 229 Kj/mol

Explanation:

4NH₃ (g) + 5O₂ (g) = > 4NO (g) + 6H₂O (g)

ΔG° (f) 4mol (-16.66Kj/mol) | 5mol (0Kj/mol) || 4mol (+86.71Kj/mol) | 6ΔG° (f) H₂O

Hess's Law

ΔG° (Rxn) = ∑ΔG° (f) Products - ∑ΔG° (f) Reactants

-957.9 Kj = [ (4mol (+86.71Kj/mol)) + 6ΔG° (f) H₂O (g) ] - [4mol (-16.66Kj/mol) + 5mol (0Kj/mol) ]

-957.9 Kj = [4 (86.7) Kj + 6ΔG° (f) H₂O] - [4 (-16.66) Kj] = 346.84Kj + 6ΔG° (f) H₂O + 66.64Kj

ΔG° (f) H₂O = ((-957.9 - 346.84 - 66.64) / 6) Kj = - 228.56 Kj ≅ - 228.6 Kj*

*Verified with Standard Heat of Formation Table