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
–228.6 kJ/mol
–206.4 kJ/mol
46.7 kJ/mol
90.7 kJ/mol

Answer 1

Use ΔG°rxn = ΣνΔGf°(products) − ΣνΔGf°(reactants).

−957.9 = 4(86.71) + 6 ΔGf°(H2O) − 4(−16.66)

−957.9 = 346.84 + 6 ΔGf°(H2O) + 66.64 = 413.48 + 6 ΔGf°(H2O)

6 ΔGf°(H2O) = −1371.38 ⇒ ΔGf°(H2O) = −1371.38/6 = −228.6 kJ·mol⁻¹.