Bond energies in kJ/mol: H–H 436, Br–Br 193, H–Br 366. What is the enthalpy change for H2 + Br2 → 2HBr?
- Bonds broken: one H–H and one Br–Br = 436 + 193 = 629 kJ taken in.
- Bonds made: two H–Br = 2 × 366 = 732 kJ given out.
- ΔH = broken − made = 629 − 732 = −103 kJ/mol.
Answer: ΔH = −103 kJ/mol. The sign is negative, so the reaction is exothermic.