ΔH from mean bond enthalpies
A-Level (AQA) · unreviewed
Combustion of methane (all gases) with mean bond enthalpies for every bond involved.
Worked example
Use the mean bond enthalpies C–H 413 kJ mol⁻¹, O=O 497 kJ mol⁻¹, C=O 801 kJ mol⁻¹ and O–H 463 kJ mol⁻¹ to calculate the enthalpy change for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g).
1. Weighted sum of 4·C–H + 2·O=O → Bonds broken
= 2646 kJ mol⁻¹
2. Weighted sum of 2·C=O + 4·O–H → Bonds made
= 3454 kJ mol⁻¹
3. Subtract Bonds broken − Bonds made → ΔH = broken − made
= −808.0 kJ mol⁻¹
Answer: ΔcH = −808 kJ mol⁻¹
Method
Bonds broken: 4 C–H and 2 O=O. Bonds made: 2 C=O and 4 O–H. ΔH = Σ broken − Σ made. The answer differs from the data-book ΔcH (−890 kJ mol⁻¹) because mean bond enthalpies are averages over many compounds and because water is a liquid under standard conditions.
Common mistakes
- ΔH = Σ(bonds broken) − Σ(bonds made). Breaking bonds takes in energy; making them releases it.
- Each H₂O has two O–H bonds and two H₂O form, so four O–H bonds are made.
More thermodynamics questions
- ΔG from ΔH and ΔS
- K from ΔG°
- ΔG° from K
- K from ΔH° and ΔS°
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- Moles of an ideal gas
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- Boiling point at reduced pressure
- Entropy change on heating
- Hess's law from enthalpies of formation
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- Enthalpy of neutralisation
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- ΔS from standard entropies
- Temperature at which a reaction becomes feasible
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- Enthalpy of solution from a thermodynamic cycle
- Methanol synthesis: K from tabulated data