Bond energies in kJ mol−1: N≡N 944, H–H 436, N–H 388. What is the enthalpy change for N2(g) + 3H2(g) → 2NH3(g)?
- Bonds broken: 1 N≡N and 3 H–H = 944 + (3 × 436) = 944 + 1308 = 2252 kJ.
- Bonds formed: 2 NH3 contain 6 N–H bonds = 6 × 388 = 2328 kJ.
- ΔH = bonds broken − bonds formed = 2252 − 2328 = −76 kJ mol−1.
Answer: ΔH = −76 kJ mol−1 (exothermic)