HomeLearning HubA Level ChemistryA2 23: Chemical energetics
A2 23

Chemical energetics

A Level · Physical chemistry · Paper 4 · builds on AS 5

🎯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.

\[ \Delta H_\mathrm{f} = \Delta H_\mathrm{at}(\text{metal}) + \text{IE(s)} + \Delta H_\mathrm{at}(\text{non-metal}) + \text{EA(s)} + \Delta H_\mathrm{latt} \]

Build it in this order, drawn as an energy-level diagram with endothermic steps going up and exothermic steps going down:

  1. elements in standard states (zero);
  2. atomise the metal (up);
  3. ionise the metal — first IE, and for a 2+ ion the second IE too (up);
  4. atomise the non-metal — as many moles of atoms as the formula needs (up);
  5. add electrons to the non-metal — first EA (down), and for a 2− ion the second EA (up);
  6. the gaseous ions come together: lattice energy (a long way down) to the solid;
  7. ΔHf connects the elements directly to the solid.

For MgCl2 remember the : 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:

\[ \Delta H_\mathrm{sol} = -\Delta H_\mathrm{latt} + \sum \Delta H_\mathrm{hyd}(\text{ions}) \]

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.
\[ \Delta S^{\ominus} = \sum S^{\ominus}(\text{products}) - \sum S^{\ominus}(\text{reactants}) \]

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

\[ \Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus} \]

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:

ΔH −, ΔS + — feasible at all temperatures
ΔH +, ΔS − — never feasible
ΔH −, ΔS − — feasible below a certain temperature
ΔH +, ΔS + — feasible above a certain temperature

The changeover temperature is where ΔG = 0:

\[ T = \frac{\Delta H^{\ominus}}{\Delta S^{\ominus}} \]

✏️Worked example

(a) Use the data to calculate the lattice energy of sodium chloride. [ΔHf(NaCl) −411; ΔHat(Na) +107; first IE(Na) +494; Cl–Cl bond energy +242; first EA(Cl) −349; all kJ mol−1] (b) For CaCO3(s) → CaO(s) + CO2(g), ΔH = +178 kJ mol−1. Standard entropies (J K−1 mol−1): CaCO3 93, CaO 40, CO2 214. Calculate ΔS, show that the decomposition is not feasible at 298 K, and find the minimum temperature at which it is.

(a) ΔHat(Cl) = ½ × 242 = +121. Around the cycle:

\[ \Delta H_\mathrm{latt} = \Delta H_\mathrm{f} - [\Delta H_\mathrm{at}(\mathrm{Na}) + \mathrm{IE}_1 + \Delta H_\mathrm{at}(\mathrm{Cl}) + \mathrm{EA}_1] \] \[ = -411 - [107 + 494 + 121 + (-349)] = -411 - 373 = -784\ \mathrm{kJ\ mol^{-1}} \]

(b)

\[ \Delta S^{\ominus} = (40 + 214) - 93 = +161\ \mathrm{J\ K^{-1}\ mol^{-1}} \] \[ \Delta G^{\ominus}_{298} = 178 - 298 \times \frac{161}{1000} = 178 - 48.0 = +130\ \mathrm{kJ\ mol^{-1}} \]

ΔG is positive, so the decomposition is not feasible at 298 K. It becomes feasible when ΔG ≤ 0:

\[ T = \frac{\Delta H^{\ominus}}{\Delta S^{\ominus}} = \frac{178\,000}{161} = 1106\ \mathrm{K} \]

So calcium carbonate decomposes above about 1110 K (about 830 °C), which is why lime kilns run so hot.

Check it. Signs first: ΔS is positive because a gas is produced from a solid; ΔH is positive (thermal decomposition). So the reaction must be feasible above some temperature, and not below it — which is what the numbers show. And the lattice energy of NaCl should be large and negative; the accepted value is about −787, so −784 is right.
Forgetting to divide ΔS by 1000. Using 161 instead of 0.161 with ΔH in kJ gives ΔG = 178 − 47 978, a huge negative number, and the false conclusion that limestone decomposes at room temperature. In Born–Haber cycles, the matching trap is using the whole Cl–Cl bond energy (242) for ΔHat instead of half of it.

📝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]
The metal needs both ionisation energies; the chlorine needs two atomisations and two electron affinities. Sum of the steps from elements to gaseous ions: 148 + 736 + 1450 + 2(121) + 2(−349) = 148 + 2186 + 242 − 698 = +1878. ΔHlatt = ΔHf − 1878 = −641 − 1878 = −2519 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.
Lattice energy comes from the electrostatic attraction between oppositely charged ions. Mg2+ and O2− each carry twice the charge of Na+ and F. With ions of similar size (and so similar separation), doubling both charges makes the attraction about four times as strong, so much more energy is released when the lattice forms.
3. Explain why the second electron affinity of oxygen is endothermic.
The second electron affinity is O(g) + e → O2−(g). The electron is being added to an ion that is already negatively charged, so it is repelled. Energy must be supplied to overcome this repulsion, so the process is endothermic. (It is still worth doing in forming an oxide, because the very large lattice energy of a lattice containing 2− ions more than pays for it.)
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.
ΔHsol = −ΔHlatt + ΣΔHhyd = +787 + (−406) + (−378) = +3 kJ mol−1. Dissolving is very slightly endothermic: the energy needed to break the lattice is almost exactly repaid by hydrating the ions. NaCl still dissolves readily, because dissolving increases entropy — a reminder that ΔH alone does not decide feasibility.
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.
(a) Positive — liquid to gas; gas particles have many more possible arrangements. (b) Negative — three moles of gas become two, so fewer possible arrangements. (c) Positive — a solid becomes two moles of gas, a large increase in disorder.
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.
ΔH and ΔS are both negative, so the reaction is feasible below T = ΔH/ΔS = −92 000 / −199 = 462 K (about 190 °C). The Haber process runs at about 450 °C (720 K), above this temperature — where ΔG is positive and K is less than 1 — because at lower temperatures the rate is too slow. High pressure and removal of ammonia push the conversion up despite the unfavourable equilibrium. Feasibility and rate are separate questions.

🔗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