Using average bond enthalpies (linked above), estimate the enthalpy change for the following reaction: 4NH3(g) + 5O2(g)4NO(g) + 6H2O(g)?

Refer to the values of average bond enthalpies.
The values of average bond enthalpies from difference sources may be slight different. This causes the result to vary slightly.

Equation for the reaction :
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g) …… ΔH
Each NH₃ molecule contains 3 N-H bonds.
Each O₂ molecule contains 1 O=O bond.
Each NO molecule contains 1 N=O bond.
Each H₂O molecule contains 2 O-H bonds.

Energy needed for bond breaking
= (391×3×4 + 495×5) kJ
= 7167 kJ

Energy released in bond formation
= (607×4 + 463×2×6) kJ
= 7984 kJ

ΔH
= (Energy needed for bond breaking) - (Energy released in bond formation)
= (7167 - 7984) kJ
= -817 kJ