Thermodynamics and Thermochemistry
Thermodynamics questions ask you to track energy and entropy, then combine them into a spontaneity judgement. Expect Hess's law, formation and bond enthalpies, calorimetry, entropy changes, and Gibbs energy linked to K and E°.
Core rules
- First law: ΔU = q + w. Enthalpy ΔH = q_p, the heat exchanged at constant pressure.
- Hess's law: ΔH is path-independent. ΔH°_rxn = sum of ΔH_f°(products) - sum of ΔH_f°(reactants), with ΔH_f° = 0 for an element in its standard state.
- Bond enthalpies give estimates only: ΔH ≈ bonds broken - bonds formed, using average gaseous values.
- Calorimetry: q = mcΔT, then divide by moles and by 1000 for kJ mol-1.
- ΔS is positive for gas formation, for an increase in gas moles, and for melting or vaporisation.
- ΔG = ΔH - TΔS; spontaneous when ΔG < 0. Also ΔG° = -RT ln K and ΔG° = -nFE°_cell.
Worked example
For N2(g) + 3H2(g) → 2NH3(g), ΔH° = -92.2 kJ mol-1 and ΔS° = -198.7 J K-1 mol-1. At 298 K: ΔG° = -92.2 - 298(-0.1987) = -92.2 + 59.2 = -33.0 kJ mol-1, so the reaction is spontaneous at room temperature. The entropy term scales with T, and the sign changes when T = ΔH/ΔS = 92 200/198.7 ≈ 464 K. Above roughly 464 K the standard reaction is no longer spontaneous, which is why industrial ammonia synthesis runs hot for rate but requires high pressure to keep the equilibrium favourable.
Traps
- Mixing joules and kilojoules inside ΔG = ΔH - TΔS; convert ΔS first.
- Sign errors in q = mcΔT: a warming solution means the reaction released heat, so ΔH is negative.
- Quoting ΔH_f° = 0 for Br2(g) or C(diamond); only the standard-state form is zero.
- Expecting bond-enthalpy estimates to match formation-enthalpy values exactly.
- Reading ΔG as a rate statement; a spontaneous reaction can be kinetically invisible, as with diamond to graphite.