Question

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

Answer 1

6ΔG°(f) H₂O = -229 Kj/mol

Step-by-step 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