Energetics and Thermochemistry
What this domain tests
Enthalpy changes from Hess cycles, calorimetry and bond enthalpies, including Born–Haber cycles.
Core rules
- Standard conditions are 298 K and 100 kPa; ΔH is negative for exothermic reactions.
- Hess's law: the enthalpy change is independent of route. From formation enthalpies, ΔHr = ΣΔHf(products) − ΣΔHf(reactants). From combustion enthalpies, ΔHr = ΣΔHc(reactants) − ΣΔHc(products).
- Bond enthalpy: ΔH = Σ bonds broken − Σ bonds formed, using gaseous species. Mean bond enthalpies give only approximate answers because they are averaged across different molecules.
- Calorimetry: q = mcΔT with m the mass of solution and c ≈ 4.18 J g−1 K−1; then ΔH = −q/n.
- Born–Haber cycles combine sublimation, ionisation, bond dissociation, electron affinity and lattice enthalpy.
Worked example
Calculate ΔH for CH4 + 2O2 → CO2 + 2H2O(l) given ΔHf: CH4 −74.8, CO2 −393.5, H2O(l) −285.8 kJ mol−1. ΔHr = [−393.5 + 2(−285.8)] − [−74.8] = (−965.1) − (−74.8) = −890.3 kJ mol−1. The reaction is strongly exothermic, as expected for a combustion.
Traps
Sign errors when subtracting reactant terms; using ΔHf of H2O(g) when the equation specifies liquid; mixing bond enthalpy and formation data in one cycle; forgetting to multiply by stoichiometric coefficients; ignoring heat loss in calorimetry, which makes the measured ΔH less negative than the true value.