The big idea: Every chemical reaction is just breaking some bonds and making new ones.
- Breaking a bond always costs energy → it is endothermic (energy in). - Making a bond always releases energy → it is exothermic (energy out).
The overall enthalpy change, ΔH, is simply the difference between the energy put in to break bonds and the energy given out when new bonds form.
Bonds broken (reactants → top) absorb energy; bonds made (top → products) release energy. If more is released than absorbed, the reaction is exothermic (ΔH < 0).
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Define: bond enthalpy: The (average) bond enthalpy is the energy needed to break one mole of a particular bond in the gaseous state.
It is always a positive value, because breaking a bond always requires energy. The bigger the bond enthalpy, the stronger the bond.
To estimate ΔH for a reaction, add up the bond enthalpies of all the bonds broken in the reactants, then subtract the total of all the bonds made in the products.
- enthalpy change of reaction (kJ mol⁻¹)
- total bond enthalpy of all bonds broken in the reactants (endothermic, +)
- total bond enthalpy of all bonds formed in the products (exothermic, −)
Mind the sign: If the answer is negative, more energy was released making bonds than was used breaking them → exothermic.
If the answer is positive, breaking the bonds cost more than was released → endothermic.
Always draw out the full structural formulas first so you count every bond — this is where most marks are lost.
You read the bond enthalpy values straight from the data booklet. A sample set (used in the worked examples below):
| Bond | Average bond enthalpy / kJ mol⁻¹ |
|---|---|
| C–H | 414 |
| C–C | 346 |
| C=C | 614 |
| C–O | 358 |
| O–H | 463 |
| H–H | 436 |
| Cl–Cl | 242 |
| H–Cl | 431 |
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Worked example — combustion of hydrogen
Estimate the enthalpy change for 2H2(g) + O2(g) → 2H2O(g). (Bond enthalpies / kJ mol⁻¹: H–H = 436, O=O = 498, O–H = 463.)
Solution
- Bonds broken (reactants) — two H–H bonds and one O=O bond:
- Bonds made (products) — each H2O has two O–H bonds, and there are two molecules → four O–H bonds:
- Formula first — bonds broken minus bonds made:
- Work it out — the sign tells you it is exothermic:
Final answer
ΔH = −482 kJ mol⁻¹ (exothermic — more energy is released forming O–H bonds than is used breaking H–H and O=O).
Worked example — hydrogenation of ethene
Estimate ΔH for the hydrogenation of ethene: C2H4(g) + H2(g) → C2H6(g). Use C=C = 614, C–H = 414, H–H = 436 and C–C = 346 kJ mol⁻¹.
Solution
- Only the bonds that change matter. Bonds broken: the C=C and the H–H:
- Bonds made: a new C–C (the double becomes single) and two new C–H:
- Formula first — subtract:
- Work it out:
Final answer
ΔH = −124 kJ mol⁻¹ (exothermic). Tip: you only need the bonds that actually break or form.
How this is tested: Bond enthalpies appear as a Paper 2 determine/calculate question (3 marks): given a table of bond enthalpies, estimate ΔH for a reaction — often an organic one (hydration of an alkene, cracking, combustion).
On Paper 1A you may be asked which reactions can be done with bond enthalpies alone — the answer is those with all species gaseous, because bond enthalpies are defined for the gaseous state.
Three marks, three steps: (1) Total the bonds broken. (2) Total the bonds made. (3) ΔH = broken − made (watch the sign). Draw the structures so you count every bond — that is the step examiners reward.
IB-style question — hydration of ethene to ethanol
Ethene reacts with steam to form ethanol: C2H4(g) + H2O(g) → C2H5OH(g). Using the bond enthalpies C=C = 614, C–C = 346, C–H = 414, O–H = 463 and C–O = 358 kJ mol⁻¹, determine the enthalpy change for this reaction. [3]
Solution
- Bonds broken — the C=C of ethene and one O–H of water (the rest are unchanged):
- Bonds made — a new C–C, a new C–H and a new C–O (the H and the OH add across the double bond):
- Formula first — bonds broken minus bonds made:
- Work it out:
Final answer
ΔH = −41 kJ mol⁻¹ (exothermic). Only the bonds that break or form across the C=C are counted.
Why are the values 'average'?: A bond like C–H exists in thousands of different molecules, and its exact strength varies a little with the rest of the molecule.
The data booklet quotes an average taken over many compounds. That is why a bond-enthalpy ΔH is only an estimate — and why it disagrees slightly with the experimental value.