6.7 Bond Enthalpies

Bond Enthalpy

Choose a reaction and see every bond drawn on the reactant and product molecules. A bar chart tallies energy in vs. out to show why the net ΔH is positive or negative.

ΔH from BondsBroken vs Formed5 ReactionsExo/Endothermic
Topic 6.7

Bond Enthalpies

Calculate the enthalpy change of a reaction based on the average bond energies of bonds broken and formed in the reaction.

During a chemical reaction, bonds are broken and formed, and these events change the potential energy of the system.

The two rules that govern every calculation here:

  • Breaking bonds requires energy — endothermic, positive.
  • Forming bonds releases energy — exothermic, negative.

The net enthalpy change is the balance:

ΔH ≈ Σ(bond energies of bonds broken) − Σ(bond energies of bonds formed)

Note the direction of the subtraction: broken minus formed. If the bonds in the products are stronger than those in the reactants, more energy is released than was invested and ΔH is negative.

Why this is an approximation. Tabulated bond energies are averages across many different molecules. The C–H bond in methane is not exactly the same strength as the C–H bond in ethanol. So bond-energy calculations give an estimate; enthalpies of formation (6.8) give the accurate value.

Practical procedure. Draw the Lewis structures. Count every bond of each type on each side. Only bonds that actually change need to be counted — a spectator bond appearing unchanged on both sides contributes zero to the sum. Bond energies are always positive numbers in tables; the signs come from your bookkeeping, not from the table.

Connection back to Unit 2. Bond energy increases with bond order (C–C < C=C < C≡C) and decreases with atomic size (H–F > H–Cl > H–Br > H–I). Those trends let you predict the sign of ΔH qualitatively even without numbers.

Key points

  • ΔH ≈ Σ(broken) − Σ(formed). Get the direction right and the sign follows.
  • Breaking is endothermic; forming is exothermic. Always.
  • Bond energies are averages, so this route gives an estimate.
  • Stronger bonds in the products → exothermic reaction.

Equations

  • not on the sheetNot on the equation sheet; the reasoning is expected.
    • average bond energy (always tabulated as a positive value)

Common mistakes

  • Reversing the subtraction. Formed minus broken gives exactly the wrong sign.
  • Making bond energies negative in the table. They are positive; the formula supplies the signs.
  • Miscounting bonds in a double or triple bond. C=O is one bond with one bond energy, not two C–O bonds.
  • Using bond energies for reactions involving liquids or solids. The method assumes gas-phase species, since it ignores intermolecular forces.

Worked example

Estimate ΔH for CH₄(g) + 2 Cl₂(g) → CH₂Cl₂(g) + 2 HCl(g) using these average bond energies (kJ/mol): C–H 413, Cl–Cl 243, C–Cl 328, H–Cl 431.

Bonds broken (identify only what changes):
2 × C–H = 2(413) = 826 kJ
2 × Cl–Cl = 2(243) = 486 kJ
Total broken = 1312 kJ

(The other two C–H bonds in methane survive into CH₂Cl₂, so they cancel and can be ignored.)

Bonds formed:
2 × C–Cl = 2(328) = 656 kJ
2 × H–Cl = 2(431) = 862 kJ
Total formed = 1518 kJ

ΔH ≈ 1312 − 1518 = −206 kJ

Negative, so the reaction is exothermic: the C–Cl and H–Cl bonds formed are collectively stronger than the C–H and Cl–Cl bonds broken.

Full notes for topic 6.7 →