Chemical Energetics: Lattice Energy, Entropy and Gibbs Free Energy

This guide covers subtopics 23.1, Lattice energy and Born-Haber cycles, 23.2, Enthalpies of solution and hydration, 23.3, Entropy change, and 23.4, Gibbs free energy change, from Topic 23, Chemical energetics, of Cambridge International AS & A Level Chemistry 9701, 2025–2027 series. This is A Level content, building directly on AS Level enthalpy work.

Before studying this

This resource assumes Chemical Energetics: Hess’s Law and Enthalpy Cycles — the enthalpy-cycle method (Hess’s law, indirect enthalpy determination) is reused here without re-explanation, now applied to ionic-lattice and dissolving processes. It also assumes ionic bonding and ionic radius trends from Chemical Bonding: Electronegativity, Ionic and Metallic Bonds and Atomic Structure: Particles, Radius and Isotopes.

Syllabus coverage

CAMBRIDGE INTERNATIONAL AS & A LEVEL CHEMISTRY 9701 — A Level, Topic 23

23.1 Lattice energy and Born-Haber cycles — defining enthalpy change of atomisation and lattice energy; defining first electron affinity and explaining the factors affecting it, including the trends across Group 16 and Group 17; constructing and using Born-Haber cycles for ionic solids (+1/+2 cations, −1/−2 anions); explaining qualitatively the effect of ionic charge and radius on the magnitude of a lattice energy.

23.2 Enthalpies of solution and hydration — defining enthalpy change of hydration and of solution; constructing and using an energy cycle linking enthalpy change of solution, lattice energy and enthalpy change of hydration; explaining qualitatively the effect of ionic charge and radius on the magnitude of an enthalpy change of hydration.

23.3 Entropy change — defining entropy as the number of possible arrangements of particles and their energy in a system; predicting and explaining the sign of entropy changes for changes of state, temperature changes, and reactions with a change in the number of gaseous molecules; calculating entropy change from standard entropies of reactants and products.

23.4 Gibbs free energy change — stating and using ΔG = ΔH − TΔS; performing calculations with this equation; deciding feasibility from the sign of ΔG; predicting the effect of a temperature change on feasibility.

Lattice energy and Born-Haber cycles

Enthalpy change of atomisation, ΔH_at, is the enthalpy change when one mole of gaseous atoms forms from an element in its standard state — always endothermic, since it involves breaking bonds or overcoming intermolecular/metallic forces to produce isolated gaseous atoms.

Lattice energy, ΔH_latt, is the enthalpy change when one mole of a solid ionic lattice forms from its constituent gaseous ions — always exothermic (strongly so), since it’s dominated by the electrostatic attraction between oppositely charged ions coming together.

First electron affinity, EA₁, is the enthalpy change when one mole of gaseous 1− ions forms from one mole of gaseous atoms — usually exothermic (the atom’s nucleus attracts the incoming electron), though second electron affinities (relevant to O²⁻ or S²⁻) are endothermic, since adding an electron to an already-negative ion involves overcoming repulsion. Across Group 16 and Group 17, electron affinity becomes less exothermic going down each group, because the added electron enters a shell further from the nucleus, on a larger atom with more shielding — the same nuclear-attraction logic used for ionisation energy trends, in reverse.

A Born-Haber cycle is an enthalpy cycle (Hess’s law applied to ionic compound formation) linking the enthalpy change of formation of an ionic solid to atomisation, ionisation, electron affinity and lattice energy steps.

Worked example. Construct a Born-Haber cycle for NaCl and use it to find the lattice energy, given: ΔH_f(NaCl) = −411 kJ mol⁻¹, ΔH_at(Na) = +107 kJ mol⁻¹, first ionisation energy of Na = +496 kJ mol⁻¹, ΔH_at(Cl) = +122 kJ mol⁻¹, EA₁(Cl) = −349 kJ mol⁻¹.

By Hess’s law, the direct route (formation) equals the sum of the indirect route (atomise both elements, ionise Na, add an electron to Cl, then form the lattice):

ΔH_f = ΔH_at(Na) + IE₁(Na) + ΔH_at(Cl) + EA₁(Cl) + ΔH_latt

−411 = (+107) + (+496) + (+122) + (−349) + ΔH_latt

−411 = +376 + ΔH_latt

ΔH_latt(NaCl) = −411 − 376 = −787 kJ mol⁻¹

The large negative value confirms lattice formation is strongly exothermic, as expected for the electrostatic attraction between Na⁺ and Cl⁻ ions coming together into a solid.

Charge and radius effects on lattice energy. A lattice energy becomes more exothermic (numerically larger) as ionic charge increases (stronger electrostatic attraction, e.g. MgO’s lattice energy is far more exothermic than NaCl’s) and as ionic radius decreases (ions can approach more closely, increasing the attractive force, which depends on 1/distance²).

Enthalpies of solution and hydration

Enthalpy change of hydration, ΔH_hyd, is the enthalpy change when one mole of a gaseous ion dissolves in water to form one mole of aqueous ions — always exothermic, since it’s dominated by the attraction between the ion and the polar water molecules surrounding it.

Enthalpy change of solution, ΔH_sol, is the enthalpy change when one mole of an ionic solid dissolves in enough water to form an infinitely dilute solution. It can be either sign, depending on the balance between the energy needed to break apart the lattice and the energy released hydrating the separated ions.

The energy cycle: dissolving can be thought of as two steps — first break the lattice apart into gaseous ions (the reverse of lattice energy, so +|ΔH_latt|), then hydrate those gaseous ions (ΔH_hyd of the cation plus ΔH_hyd of the anion):

ΔH_sol = −ΔH_latt + ΔH_hyd(cation) + ΔH_hyd(anion)

Worked example. Given ΔH_latt(NaCl) = −787 kJ mol⁻¹, ΔH_hyd(Na⁺) = −406 kJ mol⁻¹ and ΔH_hyd(Cl⁻) = −364 kJ mol⁻¹, calculate ΔH_sol(NaCl).

ΔH_sol = −(−787) + (−406) + (−364) = 787 − 770 = +17 kJ mol⁻¹

A small positive (endothermic) value is typical for simple 1:1 salts like NaCl — the lattice-breaking and hydration terms are both large but nearly cancel, so the small residual can fall on either side of zero.

Charge and radius effects on hydration enthalpy. Hydration enthalpy becomes more exothermic as ionic charge increases and as ionic radius decreases — a smaller, more highly charged ion has a more concentrated charge density, attracting the surrounding water dipoles more strongly. This is the same size/charge logic as lattice energy, which is why the two often (though not always) move together and partly cancel in the solution cycle.

Entropy change, ΔS

Entropy, S, measures the number of possible arrangements of the particles and their energy in a system — loosely, how “spread out” or disordered a system is. A system with more accessible microscopic arrangements has higher entropy.

Predicting the sign of ΔS:

• Change of state: entropy increases (ΔS positive) going from solid → liquid → gas, since particles gain more freedom of movement and more ways to arrange their energy; the reverse changes (condensing, freezing) have ΔS negative.
• Temperature increase: entropy increases, since particles have access to a wider spread of energy states.
• Reactions with a change in the number of gaseous moles: entropy increases if the number of gas moles increases (e.g. a solid decomposing to give a gas), and decreases if gas moles decrease (e.g. gas-phase addition reactions, or a gas being absorbed into a solid or solution).

Calculating ΔS:

ΔS = ΣS°(products) − ΣS°(reactants)

Worked example. Given standard entropies S°(CaCO₃, s) = 92.9 J K⁻¹ mol⁻¹, S°(CaO, s) = 40.0 J K⁻¹ mol⁻¹ and S°(CO₂, g) = 213.6 J K⁻¹ mol⁻¹, calculate ΔS for CaCO₃(s) → CaO(s) + CO₂(g).

ΔS = [S°(CaO) + S°(CO₂)] − S°(CaCO₃) = (40.0 + 213.6) − 92.9 = 253.6 − 92.9 = +160.7 J K⁻¹ mol⁻¹

Positive, as expected — one mole of gas appears where there was none before.

Gibbs free energy change, ΔG

A reaction’s overall feasibility depends on both enthalpy and entropy together, combined in the Gibbs equation:

ΔG = ΔH − TΔS

where T is the absolute temperature in kelvin. A reaction is thermodynamically feasible when ΔG is negative or zero. (Feasible does not mean fast — this says nothing about the rate, only about whether the reaction is energetically favourable.)

Worked example. For the thermal decomposition CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ mol⁻¹ and ΔS = +165 J K⁻¹ mol⁻¹. Estimate the temperature above which the decomposition becomes feasible.

The reaction becomes feasible once ΔG ≤ 0, so find the temperature where ΔG = 0:

0 = ΔH − TΔS, so T = ΔH / ΔS

Converting ΔH to J mol⁻¹ to match the units of ΔS: T = 178 000 / 165 = 1079 K (about 806 °C)

Below this temperature the positive ΔH term dominates and ΔG is positive (not feasible); above it, the TΔS term is large enough to make ΔG negative. This matches the real-world observation that limestone is thermally stable at room temperature but decomposes in a hot lime kiln.

Predicting the effect of temperature on feasibility, given the signs of ΔH and ΔS:

ΔH	ΔS	Effect of temperature
−	+	Feasible at all temperatures (ΔG always negative)
+	−	Never feasible (ΔG always positive)
+	+	Feasible only above a certain temperature (as above)
−	−	Feasible only below a certain temperature

Common mistakes

Forgetting that lattice energy is defined from gaseous ions to solid (exothermic), not the reverse. Some questions ask for the reverse process (breaking the lattice apart) — that value is +|ΔH_latt|, the same size but opposite sign.

Mixing up electron affinity and electronegativity. They’re related in concept but not interchangeable: electron affinity is an enthalpy change in kJ mol⁻¹ measured for an isolated gaseous atom gaining an electron; electronegativity is a relative, dimensionless measure of an atom’s pull on bonding electrons within a covalent bond.

Ignoring the sign of ΔS when a reaction produces fewer gas moles than it starts with. Not every reaction has ΔS positive — always check the change in gas moles specifically, not just “does it feel more chaotic.”

Forgetting to convert kJ to J (or vice versa) before combining ΔH and TΔS in the Gibbs equation. ΔH is usually quoted in kJ mol⁻¹ and ΔS in J K⁻¹ mol⁻¹ — mixing the units without converting is one of the most common numerical errors in this topic.

Quick revision checklist

• Born-Haber cycle: atomisation (both elements) → ionisation → electron affinity → lattice energy = enthalpy of formation
• Lattice energy more exothermic with higher charge, smaller radius
• Solution cycle: ΔH_sol = −ΔH_latt + ΣΔH_hyd
• Hydration enthalpy more exothermic with higher charge, smaller radius
• Entropy increases: solid → liquid → gas; heating; more gas moles produced
• ΔS = ΣS°(products) − ΣS°(reactants)
• ΔG = ΔH − TΔS; feasible when ΔG ≤ 0
• Feasibility temperature: T = ΔH / ΔS, when ΔH and ΔS have the same sign

Related resources

• Chemical Energetics: Hess’s Law and Enthalpy Cycles — the enthalpy-cycle method this topic builds on
• Chemical Bonding: Electronegativity, Ionic and Metallic Bonds — ionic bonding and radius trends
• Group 2: Quantitative Trends in Thermal Stability and Solubility — applies lattice energy and hydration enthalpy reasoning to a real trend
• Cambridge AS & A Level Chemistry hub

Written against Cambridge International AS & A Level Chemistry 9701, 2025–2027 series. Always check the current syllabus for your examination year.