Bond Breaking & Bond Forming
Section: 5. Chemical Energetics | Syllabus: Cambridge IGCSE Chemistry (0620)
Bond Breaking is Endothermic
During a reaction, bonds in the reactants are broken and new bonds in the products are formed. Breaking bonds requires energy - it is an endothermic process.
- Bonds hold atoms together with attractive forces
- Energy must be put IN to overcome these forces and separate the atoms
- The stronger the bond, the more energy is required to break it
H-H → H + H requires +436 kJ/mol
Bond Forming is Exothermic
Making new bonds releases energy - it is an exothermic process.
- When atoms bond together, they reach a more stable, lower-energy state
- The excess energy is released OUT to the surroundings
- The stronger the bond formed, the more energy is released
H + H → H-H releases −436 kJ/mol
| Process | Energy change | Type |
|---|---|---|
| Bond breaking | energy IN (absorbed) | endothermic (+) |
| Bond forming | energy OUT (released) | exothermic (−) |
Bond Energies
Bond energy is the energy required to break one mole of a particular bond in gaseous molecules, measured in kJ/mol. Bond energies are always given as positive values, since breaking a bond always requires energy.
| Bond | Bond energy (kJ/mol) |
|---|---|
| H-H | 436 |
| O=O | 498 |
| O-H | 460 |
| O-O | 150 |
| C-H | 413 |
| C-C | 347 |
| C=O | 805 |
A stronger bond (e.g. a double bond such as O=O) has a higher bond energy than a weaker single bond between the same atoms.
Calculating Enthalpy Change from Bond Energies
ΔH = (energy to break bonds in reactants) − (energy released forming bonds in products)
Method
- Draw or list every bond in the reactants and every bond in the products
- Add up the energy needed to break ALL bonds in the reactants (energy IN)
- Add up the energy released forming ALL bonds in the products (energy OUT)
- ΔH = energy IN − energy OUT
- Negative ΔH = exothermic; positive ΔH = endothermic
Worked Example: Combustion of Hydrogen
- Question: use the bond energies H-H = 436, O=O = 498, O-H = 464 kJ/mol to calculate ΔH for 2H₂ + O₂ → 2H₂O. [3]
- Answer: bonds broken: 2 × H-H + 1 × O=O = (2 × 436) + 498 = 1370 kJ [1]
- bonds formed: each H₂O has 2 O-H bonds, and there are 2 H₂O molecules, so 4 × O-H = 4 × 464 = 1856 kJ [1]
- ΔH = 1370 − 1856 = −486 kJ/mol (exothermic) [1]
Worked Example: Decomposition of Hydrogen Peroxide
- Question: hydrogen peroxide, H-O-O-H, decomposes: 2H₂O₂(l) → 2H₂O(l) + O₂(g). Given O-H = 460, O-O = 150, O=O = 496 kJ/mol, calculate ΔH. [4]
- Answer: each H₂O₂ molecule has 2 O-H bonds and 1 O-O bond; for 2 molecules, that's 4 O-H + 2 O-O
- bonds broken: (4 × 460) + (2 × 150) = 1840 + 300 = 2140 kJ [1]
- bonds formed: 2 H₂O has 4 O-H bonds; O₂ has 1 O=O bond: (4 × 460) + 496 = 1840 + 496 = 2336 kJ [1]
- ΔH = 2140 − 2336 = −196 kJ/mol (exothermic) [1]
Even though O-H bonds appear in both the reactant and the product, every one must still be counted separately - they do not cancel out
Why Some Reactions are Exothermic or Endothermic
| Condition | Result |
|---|---|
| Energy released forming bonds > energy needed breaking bonds | Exothermic (ΔH negative) - net energy released to surroundings |
| Energy needed breaking bonds > energy released forming bonds | Endothermic (ΔH positive) - net energy absorbed from surroundings |
Common Mistakes
- ❌ Forgetting that bond breaking is endothermic
- ✓ breaking bonds is always energy IN (+), regardless of whether the overall reaction is exothermic or endothermic
- ❌ Cancelling out a bond type that appears on both sides of the equation
- ✓ count every bond broken in the reactants and every bond formed in the products separately - as in the H₂O₂ example, O-H bonds must be counted on both sides
- ❌ Not counting all the bonds in a molecule
- ✓ work out how many of each bond type are in one molecule, then multiply by the number of molecules in the balanced equation
- ❌ Getting the subtraction the wrong way round
- ✓ ΔH = energy to break bonds − energy released forming bonds (bonds broken first, bonds formed second)
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