Everything from the 4 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.
The lattice energy is the enthalpy change when 1 mol of an ionic compound is formed from its gaseous ions under standard conditions, e.g. Na+(g) + Cl−(g) → NaCl(s); it is always exothermic. A Born-Haber cycle is a Hess cycle linking it to measurable steps: enthalpy change of atomisation of each element, ionisation energies of the metal, electron affinities of the non-metal and the enthalpy change of formation. For NaCl: LE = −411 − (+107 + 496 + 122 − 349) = −787 kJ mol−1. Lattice energy becomes more exothermic when the ions have higher charge and smaller radius, because the electrostatic attraction between them is stronger.
Lattice energy goes from gaseous ions to the solid, so it is negative; do not reverse it. Use ½ the bond energy for the atomisation of Cl or O, and remember MgO needs both IE1 + IE2 and EA1 + EA2, with EA2 positive. For MgCl2, the chlorine steps are doubled.
The enthalpy change of hydration is the enthalpy change when 1 mol of gaseous ions dissolves in water, e.g. Na+(g) → Na+(aq); it is always exothermic because water dipoles are attracted to the ion. The enthalpy change of solution is the enthalpy change when 1 mol of an ionic solid dissolves in enough water to give an infinitely dilute solution, e.g. NaCl(s) → Na+(aq) + Cl−(aq). A Hess cycle through the gaseous ions gives LE + ΔHsol = ΣΔHhyd, so for NaCl ΔHsol = (−406 − 377) − (−787) = +4 kJ mol−1. Hydration is more exothermic for ions of higher charge and smaller radius (higher charge density).
Hydration starts from gaseous ions, not the solid. In the cycle, the solid goes to gaseous ions by the reverse of lattice energy, so ΔHsol = ΣΔHhyd − LE; adding LE instead gives a hugely negative answer. Multiply each hydration enthalpy by the number of ions in the formula.
Entropy, S, is a measure of the dispersal of energy in a system: the more ways the particles and their energy can be arranged, the greater the entropy. Its units are J K−1 mol−1, not kJ. Entropy increases from solid to liquid to gas (ice 48, water 70, steam 189), when temperature rises, when a solid dissolves, and when a reaction produces more moles of gas. A positive ΔS means energy becomes more dispersed. The standard entropy change of a reaction is ΔS° = ΣS°(products) − ΣS°(reactants), using the stoichiometric coefficients; for CaCO3 → CaO + CO2 it is 40 + 214 − 93 = +161 J K−1 mol−1.
Entropy values are in J K−1 mol−1, not kJ. Elements do not have zero entropy, so O2, H2 and N2 must be included. Multiply by the coefficients, and always do products minus reactants. The state symbol matters: H2O(l) is 70 but H2O(g) is 189.
The Gibbs free energy change combines enthalpy and entropy: ΔG° = ΔH° − TΔS°, with T in kelvin. ΔH is in kJ mol−1 but ΔS is in J K−1 mol−1, so divide ΔS by 1000 first. A reaction is feasible when ΔG ≤ 0. When ΔH and ΔS have the same sign, temperature decides: the switch-over happens where ΔG = 0, at T = ΔH / ΔS. For CaCO3 → CaO + CO2 (ΔH +178 kJ mol−1, ΔS +161 J K−1 mol−1), T = 178 / 0.161 = 1106 K. A feasible reaction may still be very slow if its activation energy is high.
The classic slip is forgetting to divide ΔS by 1000: 178 − 298 × 161 gives a nonsense −47 800 kJ mol−1. Temperature must be in kelvin. ΔG ≤ 0 means feasible, not fast. When ΔH and ΔS have the same sign, always state whether the reaction is feasible above or below the calculated temperature.