Bond Energy
Estimate a reaction’s enthalpy from the bonds broken and the bonds formed.
Bonds broken
Every bond in the reactants. Breaking them costs energy.
Bonds formed
Every bond in the products. Forming them releases energy.
2642 kJ put in to break bonds − 3450 kJ released as bonds form.
These are average bond enthalpies, so the answer is an estimate — typically within about 10 kJ/mol of the measured value, and only for gas-phase species.
Breaking a bond costs energy and forming one releases it, so a reaction’s enthalpy is the difference: ΔH ≈ Σ(bonds broken) − Σ(bonds formed). Burning methane gives −808 kJ this way, against a measured −802 kJ.
Why breaking always costs and forming always pays
A chemical bond exists because the bonded arrangement is lower in energy than the separated atoms. Pulling those atoms apart therefore requires energy in, always — there is no such thing as an exothermic bond break. Forming a bond is the same event run backwards and releases exactly the same amount.
That makes the sign convention automatic. Bonds broken enter the sum positive and bonds formed negative, and a reaction is exothermic when the products’ bonds are collectively stronger than the reactants’. Combustion is exothermic because the C=O and O–H bonds it creates are unusually strong, not because oxygen is somehow energetic on its own.
Where the number comes from
A tabulated bond energy is an average across many molecules. A C–H bond is not identical in methane, ethanol and benzene, so 413 kJ/mol is a compromise that fits them all approximately and none exactly. That averaging is the source of nearly all the error in this method, and why the answer should be quoted as an estimate.
positive ΔH is endothermic; negative is exothermic
- 1 Draw both sides and count every bond. For CH₄ + 2O₂ → CO₂ + 2H₂O the reactants hold 4 C–H and 2 O=O.
- 2 Total the energy needed to break the reactant bonds. 4 × 413 + 2 × 495 = 1652 + 990 = 2642 kJ.
- 3 Total the energy released forming the product bonds. CO₂ has 2 C=O and two waters have 4 O–H: 2 × 799 + 4 × 463 = 1598 + 1852 = 3450 kJ.
- 4 Subtract. 2642 − 3450 = −808 kJ. Negative, so the reaction is exothermic.
- 5 Read it as an estimate. The measured value is −802.3 kJ. Six kJ out of 800 is typical for this method.
Common bond energies
Average bond enthalpies in kJ/mol. The tool holds all 52; these are the ones that recur.
| Bond | kJ/mol | Bond | kJ/mol |
|---|---|---|---|
| H–H | 436 | C=C | 614 |
| C–H | 413 | C≡C | 839 |
| N–H | 391 | C=O | 799 |
| O–H | 463 | C≡N | 891 |
| C–C | 348 | N=N | 418 |
| C–O | 358 | N≡N | 941 |
| O–O | 146 | O=O | 495 |
| Cl–Cl | 242 | H–Cl | 431 |
Three limits worth knowing before you trust the answer
The first is the averaging already described, which puts the typical error around 10 kJ/mol. That is fine for deciding whether a reaction is exothermic and roughly how strongly, and useless for anything needing better than about 1% accuracy.
The second is phase. Bond energies describe gas-phase species, so the method silently ignores the energy of vaporising a liquid or dissolving a solid. Burning methane to liquid water rather than steam releases about 890 kJ, not 802 — the extra 88 kJ is condensation, which involves no bonds breaking at all. If your equation has an (l) or (aq) in it, this method is answering a different question.
The third is that it only sees covalent bonds. Ionic lattices, metallic bonding and intermolecular forces are all invisible to it, so a precipitation or neutralisation reaction cannot be estimated this way. When accuracy matters, standard enthalpies of formation give the exact value instead, and Hess’s law assembles it from known steps.