Bond Energy Formula
Calculate reaction enthalpy from the energy required to break and form bonds
Main Formula
ΔHrxn = Σ(BEbroken) - Σ(BEformed)
ΔHrxn = enthalpy change of reaction (kJ/mol)
BE = bond energy (kJ/mol)
Σ = sum of all bonds
Bonds Broken (Reactants)
- • Requires energy input
- • Endothermic process
- • Positive contribution to ΔH
- • Count all bonds in reactants
Bonds Formed (Products)
- • Releases energy
- • Exothermic process
- • Negative contribution to ΔH
- • Count all bonds in products
Common Bond Energies
Single Bonds
| H-H | 436 kJ/mol |
| C-H | 413 kJ/mol |
| C-C | 347 kJ/mol |
| C-O | 358 kJ/mol |
| O-H | 463 kJ/mol |
| N-H | 391 kJ/mol |
| Cl-Cl | 243 kJ/mol |
Double Bonds
| C=C | 614 kJ/mol |
| C=O | 799 kJ/mol |
| O=O | 498 kJ/mol |
| N=N | 418 kJ/mol |
| C=N | 615 kJ/mol |
Triple Bonds
| C≡C | 839 kJ/mol |
| N≡N | 945 kJ/mol |
| C≡N | 891 kJ/mol |
| C≡O | 1072 kJ/mol |
Note: These are average values. Actual bond energies vary with molecular environment.
Interpreting ΔHrxn
ΔH > 0 (Positive)
Endothermic Reaction
- More energy to break bonds than released forming bonds
- Absorbs heat from surroundings
- Feels cold
- Example: N₂ + O₂ → 2NO
ΔH < 0 (Negative)
Exothermic Reaction
- More energy released forming bonds than required to break bonds
- Releases heat to surroundings
- Feels hot
- Example: CH₄ + 2O₂ → CO₂ + 2H₂O
Worked Examples
Example 1: H₂ + Cl₂ → 2HCl
Solution:
Bonds Broken (Reactants):
1 × H-H = 1 × 436 = 436 kJ
1 × Cl-Cl = 1 × 243 = 243 kJ
Total broken = 679 kJ
Bonds Formed (Products):
2 × H-Cl = 2 × 431 = 862 kJ
Total formed = 862 kJ
Calculate ΔH:
ΔH = 679 - 862 = -183 kJ/mol
ΔH = -183 kJ/mol (Exothermic)
Example 2: CH₄ + 2O₂ → CO₂ + 2H₂O
Solution:
Bonds Broken:
CH₄: 4 × C-H = 4 × 413 = 1652 kJ
2O₂: 2 × O=O = 2 × 498 = 996 kJ
Total = 2648 kJ
Bonds Formed:
CO₂: 2 × C=O = 2 × 799 = 1598 kJ
2H₂O: 4 × O-H = 4 × 463 = 1852 kJ
Total = 3450 kJ
Calculate ΔH:
ΔH = 2648 - 3450 = -802 kJ/mol
ΔH = -802 kJ/mol (Highly Exothermic - Combustion!)
Example 3: N₂ + O₂ → 2NO
Solution:
Bonds Broken:
1 × N≡N = 945 kJ
1 × O=O = 498 kJ
Total = 1443 kJ
Bonds Formed:
2 × N=O = 2 × 607 = 1214 kJ
Total = 1214 kJ
Calculate ΔH:
ΔH = 1443 - 1214 = +229 kJ/mol
ΔH = +229 kJ/mol (Endothermic)
Breaking strong N≡N triple bond requires high energy!
Common Mistakes
Reversing the Formula
It's BROKEN - FORMED, not FORMED - BROKEN! Breaking requires energy (+), forming releases energy (-).
Counting Bonds Incorrectly
CH₄ has FOUR C-H bonds, not one! Draw Lewis structures to count accurately.
Using Single Bond Energy for Double/Triple Bonds
C=C (614) ≠ 2×C-C (694). Multiple bonds have different energies than multiples of single bonds.
Bond Energy vs. Standard Enthalpy
Bond energies give estimates. For accurate ΔH°, use ΔH°f values (formation enthalpies).