Chemical energetics
🎯What you need to be able to do
- Define and use enthalpy change of atomisation and lattice energy.
- Define first electron affinity, explain what affects it, and describe the trends in Groups 16 and 17.
- Construct Born–Haber cycles for ionic solids (1+ and 2+ cations, 1− and 2− anions) and calculate with them.
- Explain how ionic charge and radius affect the size of a lattice energy and of an enthalpy change of hydration.
- Define enthalpy changes of hydration and solution, and use the cycle linking them to lattice energy.
- Define entropy, predict the sign of ΔS for changes of state, temperature and number of gas molecules, and calculate ΔS⦵ from standard entropies.
- Use ΔG⦵ = ΔH⦵ − TΔS⦵ to decide feasibility and how temperature affects it.
📚The chemistry
AS energetics (topic 5) measured enthalpy changes and used Hess’s law. This topic applies Hess’s law to ionic compounds, which cannot be formed in one step from gaseous ions, and then asks a deeper question: why do some reactions happen and others not? The answer needs entropy as well as enthalpy.
23.1 Lattice energy and Born–Haber cycles
The new terms
- Standard enthalpy change of atomisation, ΔHat⦵ — the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state. Na(s) → Na(g); ½Cl2(g) → Cl(g). Always endothermic. For a diatomic gas it is half the bond energy: ΔHat(Cl) = ½ × 242 = +121 kJ mol−1.
- First electron affinity, EA — the enthalpy change when one electron is added to each atom in one mole of gaseous atoms, forming one mole of gaseous 1− ions: Cl(g) + e− → Cl−(g). Usually exothermic.
- Lattice energy, ΔHlatt⦵ — the enthalpy change when one mole of an ionic compound is formed from its gaseous ions: Na+(g) + Cl−(g) → NaCl(s). This syllabus defines it in the direction gas ions → solid, so lattice energies are always negative (exothermic).
Electron affinity: factors and trends
An electron is attracted by the nucleus, so adding one releases energy. The same factors as ionisation energy decide how much: nuclear charge, atomic radius (distance of the incoming electron from the nucleus) and shielding.
- Group 17: the first EA becomes less exothermic down the group from chlorine to iodine, because the incoming electron enters a shell further from the nucleus and more shielded. Fluorine is the exception: its EA is less exothermic than chlorine’s, because its atom is so small that the incoming electron is repelled strongly by the electrons already crowded in the second shell.
- Group 16: the same pattern — sulfur’s first EA is more exothermic than oxygen’s, for the same reason as Cl and F.
- Second electron affinities are endothermic. O−(g) + e− → O2−(g) means forcing an electron onto an ion that is already negative, against repulsion. Energy must be put in.
Born–Haber cycles
Lattice energy cannot be measured directly. A Born–Haber cycle finds it by Hess’s law: forming the solid from its elements directly (ΔHf) must equal forming it through gaseous atoms and then gaseous ions.
Build it in this order, drawn as an energy-level diagram with endothermic steps going up and exothermic steps going down:
- elements in standard states (zero);
- atomise the metal (up);
- ionise the metal — first IE, and for a 2+ ion the second IE too (up);
- atomise the non-metal — as many moles of atoms as the formula needs (up);
- add electrons to the non-metal — first EA (down), and for a 2− ion the second EA (up);
- the gaseous ions come together: lattice energy (a long way down) to the solid;
- ΔHf connects the elements directly to the solid.
For MgCl2 remember the 2×: two chlorine atoms to atomise, two electron affinities. For MgO, the first and second EA of oxygen.
What controls the size of a lattice energy
Lattice energy comes from the electrostatic attraction between the ions, so it is more exothermic when:
- the ionic charges are larger — MgO (2+, 2−) has a lattice energy about four times that of NaCl (1+, 1−);
- the ions are smaller — they can get closer together. NaF is more exothermic than NaCl, and LiCl than KCl.
The two ideas are often combined as charge density: high charge on a small ion.
23.2 Enthalpies of solution and hydration
- Enthalpy change of hydration, ΔHhyd⦵ — the enthalpy change when one mole of gaseous ions dissolves in water to form an infinitely dilute solution: Na+(g) → Na+(aq). Always exothermic: water molecules are attracted to the ion (the δ− O to cations, the δ+ H to anions).
- Enthalpy change of solution, ΔHsol⦵ — the enthalpy change when one mole of an ionic solid dissolves in water to form an infinitely dilute solution: NaCl(s) → Na+(aq) + Cl−(aq). May be exothermic or endothermic.
The cycle: dissolving a solid is the same as breaking the lattice into gaseous ions (the reverse of lattice energy) and then hydrating the ions:
Hydration enthalpy follows the same rule as lattice energy: it is more exothermic for smaller, more highly charged ions, which attract water molecules more strongly. Mg2+ is hydrated far more exothermically than Na+, and Li+ more than K+.
23.3 Entropy
Entropy, S, is a measure of the number of possible arrangements of the particles and their energy in a system. The more ways the particles and their energy can be arranged, the higher the entropy. Standard entropies, S⦵, are measured in J K−1 mol−1 (note: joules, not kilojoules) and are always positive.
Predicting the sign of ΔS:
- Change of state: solid → liquid → gas increases entropy (melting, boiling, sublimation: ΔS positive); the reverse decreases it. Dissolving a solid usually increases entropy, as the ions or molecules spread through the solvent.
- Temperature: heating a substance increases its entropy, because the particles have more energy and more ways of distributing it.
- Number of gas molecules: a reaction that increases the number of gas molecules has a positive ΔS. Gases dominate, because gas particles have far more possible arrangements than solids or liquids. CaCO3(s) → CaO(s) + CO2(g): ΔS positive. N2(g) + 3H2(g) → 2NH3(g): four gas molecules become two, ΔS negative.
Unlike ΔHf, the standard entropy of an element is not zero. Every substance in the equation counts, multiplied by its coefficient.
23.4 Gibbs free energy
A reaction is feasible (can happen spontaneously) when ΔG is negative or zero. Feasible does not mean fast: a feasible reaction may still have a high activation energy and not happen at a measurable rate.
Units need care: ΔH is in kJ mol−1 and ΔS in J K−1 mol−1, so divide ΔS by 1000 before substituting. T is in kelvin.
How temperature affects feasibility depends on the signs:
The changeover temperature is where ΔG = 0:
✏️Worked example
(a) ΔHat(Cl) = ½ × 242 = +121. Around the cycle:
(b)
ΔG is positive, so the decomposition is not feasible at 298 K. It becomes feasible when ΔG ≤ 0:
So calcium carbonate decomposes above about 1110 K (about 830 °C), which is why lime kilns run so hot.
📝Practise
Work through these, then reveal the answer. Each question targets a different objective from the list above.
1. Calculate the lattice energy of MgCl2. [ΔHf −641; ΔHat(Mg) +148; IE1(Mg) +736; IE2(Mg) +1450; ΔHat(Cl) +121; EA(Cl) −349; kJ mol−1]
2. Explain why the lattice energy of MgO is much more exothermic than that of NaF, although the ions are about the same size.
3. Explain why the second electron affinity of oxygen is endothermic.
4. The lattice energy of NaCl is −787 kJ mol−1 and the hydration enthalpies are Na+ −406 and Cl− −378 kJ mol−1. Calculate ΔHsol of NaCl and comment on the result.
5. Predict the sign of ΔS for: (a) H2O(l) → H2O(g); (b) 2SO2(g) + O2(g) → 2SO3(g); (c) NH4Cl(s) → NH3(g) + HCl(g). Explain.
6. For N2(g) + 3H2(g) → 2NH3(g), ΔH⦵ = −92 kJ mol−1 and ΔS⦵ = −199 J K−1 mol−1. Above what temperature is the reaction no longer feasible? Comment on the Haber process.
🔗Go deeper — other people’s work
These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.
- Chemguide (Jim Clark) — the Born–Haber cycles and entropy pages, with the cycles drawn as energy-level diagrams
- Royal Society of Chemistry — resources on entropy and Gibbs free energy