Question

Given that G for the reaction below is –957.9 kJ, what is Gf of H2O? 4NH3(g) + 5O2(g) =4NO(g) + 6H2O(g)

Answer 1

Answer:

- 228.6 kJ/mol.

Explanation:

∵ ΔGrxn = ∑ΔGf, products - ∑ΔGf, reactants

• For the reaction:

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O,  

∴ ΔGrxn = [6(ΔGf, H₂O) + 4(ΔGf, NO)] - [4(ΔGf, NH₃) + 5(ΔGf, O₂)]

ΔGrxn = – 957.9 kJ, ΔGf, H₂O = ??? KJ/mol, ΔGf, NO = 86.71 kJ/mol, ΔGf, NH₃ = - 16.66 kJ/mol, ΔGf, O₂ = 0.0 kJ/mol.

∴ – 957.9 kJ = [6(ΔGf, H₂O) + 4(86.71 kJ/mol)] - [4(- 16.66 kJ/mol) + 5(0.0 kJ/mol)].

– 957.9 kJ = 6(ΔGf, H₂O) + 346.8 kJ/mol + 66.64 kJ/mol

– 957.9 kJ = 6(ΔGf, H₂O) + 413.8 kJ/mol

6(ΔGf, H₂O) = – 957.9 kJ - 413.8 kJ/mol = - 1372 kJ/mol.

∴ ΔGf, H₂O = (- 1372 kJ/mol)/6 = - 228.6 kJ/mol.