See the calculation on this YouTube video

Solution Using Eq. (4), we have

?H_m = ? ?H_f (products) – ? ?H_f (reactants)

       = [6 ?H_f (H₂O) + 4 ?H_f (NO)] – [4 ?H_f (NH₃) + 5 ?H_f (O₂)]

       = 6(–241.8) kJ mol^–1 + 4(90.3) kJ mol^–1 – 4(–46.1 kJ mol^–1) – 5 × 0

       = –1450.8 kJ mol^–1 + 361.2 kJ mol^–1 + 184.4 kJ mol^–1

       = –905.2 kJ mol^–1

Note that we were careful to use ?H_f [H₂O(g)] not ?H_f [H₂O(l)]. Even though water vapor is not the most stable form of water at 25°C, we can still use its ?H_f value. Also the standard enthalpy of formation of the element O₂(g) is zero by definition. Obviously it would be a waste of space to include it in the table above.