All question types · Thermodynamics
ΔfH from enthalpies of combustion
A-Level (AQA) · unreviewed
Enthalpy of formation of a hydrocarbon from the combustion enthalpies of it and its elements.
Worked example
Calculate ΔfH for propane from these enthalpies of combustion: carbon −393.5 kJ mol⁻¹, hydrogen −285.8 kJ mol⁻¹, propane −2218.7 kJ mol⁻¹.
1. Weighted sum of 3·ΔcH(C) + 4·ΔcH(H₂) − 1·ΔcH(C₃H₈) → ΔfH(C₃H₈)
= −105.0 kJ mol⁻¹
Answer: ΔfH(C₃H₈) = −105 kJ mol⁻¹
Method
Both the elements and propane burn to the same products (CO₂ and H₂O), so ΔfH = 3ΔcH(C) + 4ΔcH(H₂) − ΔcH(C₃H₈). Draw the cycle with the combustion products at the bottom if in doubt.
Common mistakes
- From combustion data, ΔfH = Σ ΔcH(reactants) − Σ ΔcH(products): the elements minus the compound, not the other way round.
- Forming one C₃H₈ needs 3 C and 4 H₂, so multiply their combustion enthalpies by 3 and 4.
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