Bond Dissociation Energy

Bond Dissociation Energy

7 min read Updated Mar 26, 2026

What if you do not have ΔHf° values for the compounds in your reaction? There is another way to estimate ΔH: use the energies of the individual bonds being broken and formed. This approach is less precise than using ΔHf° (because bond energies are averages), but it gives you a fast estimate - and the MCAT tests it regularly.

The Core Principle

Every chemical bond is a store of energy. Breaking a bond requires energy input (endothermic). Forming a bond releases energy (exothermic). The overall enthalpy change of a reaction depends on the balance between these two processes.

The Bond Energy Formula

Important: This formula uses the convention that bond dissociation energies (BDEs) are always positive numbers (the energy required to break). The subtraction handles the sign:

  • If you form stronger bonds than you break → ΔH < 0 (exothermic)
  • If you break stronger bonds than you form → ΔH > 0 (endothermic)

Common Bond Energies

You do not need to memorize these - the MCAT will provide them in a table. But knowing the trends helps:

BondEnergy (kJ/mol)Trend
C-H413Single bonds
C-C348Weaker than double/triple
C=C614Stronger than single
C≡C839Strongest carbon-carbon
O-H463Strong - explains water’s stability
O=O498Must break this in combustion
N≡N941Very strong - N₂ is hard to break
H-H436
C=O799Strong bond in CO₂
N-H391

Key trend: Triple bonds > double bonds > single bonds in energy. Stronger bonds = more stable molecules = more energy released when formed.

Worked Example

Estimate ΔH for the combustion of methane: CH₄ + 2 O₂ → CO₂ + 2 H₂O

Bonds broken (reactants):

  • 4 × C-H = 4 × 413 = 1,652 kJ
  • 2 × O=O = 2 × 498 = 996 kJ
  • Total broken = 2,648 kJ

Bonds formed (products):

  • 2 × C=O (in CO₂) = 2 × 799 = 1,598 kJ
  • 4 × O-H (in 2 H₂O) = 4 × 463 = 1,852 kJ
  • Total formed = 3,450 kJ

ΔH ≈ 2,648 - 3,450 = -802 kJ

The actual value is -890 kJ. The estimate is off by about 10% because bond energies are averages across many molecules. For the MCAT, this level of accuracy is expected and acceptable.

Why Combustion Is Always Exothermic

In combustion, you break relatively weak C-H, C-C, and O=O bonds and form very strong C=O and O-H bonds. The bonds formed are stronger than the bonds broken, so energy is released. This is why all combustion reactions are exothermic - the products (CO₂ and H₂O) contain some of the strongest bonds in chemistry.

Using bond energies, is breaking a triple bond (like N≡N, 941 kJ/mol) endothermic or exothermic?
Click to reveal answer
Endothermic. Breaking ANY bond requires energy input - it is always endothermic. The 941 kJ/mol is the energy you must put IN to break one mole of N≡N bonds. This is why N₂ is so unreactive - its triple bond is extremely difficult to break.
A reaction breaks bonds worth 500 kJ total and forms bonds worth 700 kJ total. Is the reaction exothermic or endothermic, and what is the approximate ΔH?
Click to reveal answer
Exothermic, ΔH ≈ -200 kJ. ΔH = bonds broken - bonds formed = 500 - 700 = -200 kJ. More energy was released forming bonds than was consumed breaking bonds, so the reaction is exothermic (negative ΔH).