Thermochemistry and Energetics
What this domain tests
Energetics questions test enthalpy bookkeeping: Hess's law cycles, standard enthalpies of formation and combustion, calorimetry, bond enthalpies and the Gibbs free-energy criterion for spontaneity.
Core rules
- Enthalpy is a state function, so the route does not matter - only the initial and final states. This is the basis of Hess's law.
- ΔH°rxn = ΣΔH°f(products) − ΣΔH°f(reactants). The standard enthalpy of formation of an element in its standard state is zero.
- Calorimetry: q = mcΔT for a solution (c ≈ 4.18 J g⁻¹ K⁻¹ for dilute aqueous), q = CΔT for a bomb calorimeter. If the calorimeter is the surroundings, ΔH = −q/n.
- Bond breaking is endothermic; bond forming is exothermic. Bond enthalpies apply strictly to gaseous species, so they give only an estimate for reactions involving liquids or solids.
- Gibbs energy: ΔG = ΔH − TΔS. Spontaneous when ΔG < 0; at equilibrium ΔG = 0 and T = ΔH/ΔS.
Worked example
Find ΔH° for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) using ΔH°f: CH₄ −74.8, CO₂ −393.5, H₂O(l) −285.8 kJ mol⁻¹. Products: −393.5 + 2(−285.8) = −965.1. Reactants: −74.8 + 0 = −74.8. ΔH° = −965.1 − (−74.8) = −890.3 kJ mol⁻¹. The large negative value is why methane is a fuel; note that using H₂O(g) (−241.8) would give −802.3 instead - the phase matters.
Traps that cost marks
- Sign errors when subtracting reactant enthalpies.
- Using the wrong phase for water - condensation releases a further 44 kJ mol⁻¹.
- Multiplying a formation enthalpy by its coefficient when there are several moles of that product.
- Forgetting that a reaction can be spontaneous even when endothermic, provided TΔS is large and positive.