Enthalpy of formation

Question

A scientist measures the standard enthalpy change for the following reaction to be -716.4 kJ:

2NH₃(g) + 2 O₂(g)N₂O(g) + 3 H₂O(l)

Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of H₂O(l) is  kJ/mol.

Answer 1

2NH₃(g) + 2O₂(g) ==> N₂O(g) + 3H₂O(l) ... ∆H = -716.4 kJ

Looking up standard enthalpy of formation for the other substances, I find the following:

NH₃(g) = -46.2 kJ/mol

O₂(g) = 0

N₂O(g) = +82.05 kJ/mol

To find ∆H_f for H₂O, we take the difference between the products and reactants and set that equal to ∆H_rxn:

-716.4 = (82.05 + 3x∆H_fH₂O) - (2x-46.2 + 0)

-716.4 = 82.05 + 92.4 + 3x∆H_fH₂O

3x∆H_fH₂O = -890.85

∆_fH₂O = 296.95 kJ/mol