Edexcel A-level Chemistry (9CH0) · Topic 13: Energetics II
Mini-Lesson
Energetics II
This mini-lesson covers the whole of Edexcel Topic 13: lattice enthalpy and Born–Haber cycles, enthalpies of solution and hydration, entropy (ΔS), and Gibbs free energy (ΔG) — the quantity that decides whether a reaction is feasible, and at what temperature.
Work through each screen, answer the questions as you go (several are full Born–Haber and Gibbs calculations) and collect ⭐ stars. Every calculation is worked through for you first. Press Start when you're ready.
Definitions
The enthalpy changes you must be able to define
Born–Haber questions are usually lost on definitions, not arithmetic. Learn these exactly — the state symbols and the words "one mole" earn the marks.
Lattice enthalpy of formation, ΔLEH — the enthalpy change when 1 mol of an ionic solid is formed from its gaseous ions. Always exothermic (negative): e.g. Na⁺(g) + Cl⁻(g) → NaCl(s).
Enthalpy of atomisation, ΔatH — the enthalpy change when 1 mol of gaseous atoms is formed from the element in its standard state: ½Cl₂(g) → Cl(g). Always endothermic.
First ionisation energy — X(g) → X⁺(g) + e⁻ (endothermic).
First electron affinity — the enthalpy change when 1 mol of gaseous atoms each gain an electron: X(g) + e⁻ → X⁻(g). The first EA is exothermic; the second EA (e.g. O⁻(g) + e⁻ → O²⁻(g)) is endothermic because you push an electron onto an already-negative ion.
Enthalpy of formation, ΔfH — 1 mol of a compound from its elements in their standard states.
Watch the sign convention. Edexcel uses lattice enthalpy of formation (gaseous ions → solid, negative). Some textbooks quote lattice enthalpy of dissociation (solid → gaseous ions), which is the same magnitude but positive. Read the question.
Born–Haber
Building a Born–Haber cycle
A lattice enthalpy cannot be measured directly, so we get it indirectly using Hess's Law. A Born–Haber cycle is just Hess's Law drawn on an energy-level diagram: the route from elements → gaseous ions → ionic solid must have the same total enthalpy change as the direct route (ΔfH).
ΔfH = ΔatH(metal) + IE + ΔatH(non-metal) + EA + ΔLEHrearranged: ΔLEH = ΔfH − (all the steps that make the gaseous ions)
Sum of the steps making the gaseous ions = 89 + 419 + 122 − 349 = +281 kJ mol⁻¹
ΔLEH = −437 − (+281) = −718 kJ mol⁻¹ (exothermic, as it must be)
Method that never fails: add up every step from the elements to the separate gaseous ions, then subtract that total from ΔfH.
Calculate
Your turn — Born–Haber for NaCl
1Use this data (kJ mol⁻¹) to calculate the lattice enthalpy of formation of NaCl: ΔfH(NaCl) = −411; ΔatH(Na) = +107; IE₁(Na) = +496; ΔatH(Cl) = +122; EA₁(Cl) = −349.
kJ mol⁻¹
Hint: steps to the gaseous ions = 107 + 496 + 122 − 349 = +376. Then ΔLEH = −411 − 376. Remember the minus sign.
Born–Haber
When the ions are not 1+ and 1−
For a compound such as MgCl₂ the cycle grows, but the method is identical. You now need:
Both the 1st and 2nd ionisation energies of magnesium (Mg → Mg²⁺ needs two electrons removed).
Two chlorine atoms: 2 × ΔatH(Cl), i.e. Cl₂(g) → 2Cl(g).
Two electron affinities: 2 × EA₁(Cl).
Exam trap: multiply both the atomisation enthalpy and the electron affinity by 2 — candidates routinely double one and forget the other.
For an oxide such as MgO you also need the second electron affinity of oxygen (+798 kJ mol⁻¹, endothermic). MgO is still strongly exothermic overall because its lattice enthalpy (≈ −3791 kJ mol⁻¹) is enormous: 2+ and 2− ions, both small.
Calculate
Your turn — Born–Haber for MgCl₂
2Data (kJ mol⁻¹): ΔfH(MgCl₂) = −641; ΔatH(Mg) = +148; IE₁(Mg) = +738; IE₂(Mg) = +1451; ΔatH(Cl) = +121; EA₁(Cl) = −349. Calculate the lattice enthalpy of formation of MgCl₂.
3Which equation represents the lattice enthalpy of formation of magnesium oxide?
Ionic model
Theoretical vs experimental lattice enthalpies
A theoretical lattice enthalpy is calculated from the perfect ionic model: spherical ions, point charges, purely electrostatic attraction. It gets bigger (more exothermic) when the ions have higher charge and smaller radius — i.e. higher charge density.
The experimental value comes from a Born–Haber cycle. For NaCl the two agree closely (≈ −766 theoretical vs −787 experimental), so the perfect ionic model is a good description.
For AgI, AgCl or MgI₂ the experimental value is much more exothermic than the theoretical one. The difference is covalent character: a small, highly charged cation polarises a large, easily distorted anion, pulling electron density into the space between the ions. The bonding is stronger than the pure ionic model predicts.
Say it properly: "the cation has a high charge density and polarises the large anion, so the compound has some covalent character and the experimental lattice enthalpy is more exothermic than the theoretical value."
Quick check
Why do the two values disagree?
4For AgI the Born–Haber (experimental) lattice enthalpy is −889 kJ mol⁻¹ but the value calculated from the perfect ionic model is −778 kJ mol⁻¹. The best explanation is that…
Dissolving
Enthalpy of solution and enthalpy of hydration
When an ionic solid dissolves, two things happen and they pull in opposite directions:
The lattice must be broken up into gaseous ions — endothermic, equal to −ΔLEH.
Those gaseous ions are hydrated by water molecules — exothermic. The enthalpy of hydration, ΔhydH, is the enthalpy change when 1 mol of gaseous ions dissolves in enough water to give an infinitely dilute solution: X⁺(g) + aq → X⁺(aq).
ΔsolH = −ΔLEH + ΣΔhydHi.e. ΔsolH = (energy in to break the lattice) + (energy out from hydrating both ions)
Hydration is exothermic because the δ⁻ oxygen of water is attracted to cations and the δ⁺ hydrogens to anions — ion–dipole attractions form. ΔhydH is more exothermic for ions of higher charge density (small ion, big charge): ΔhydH(Mg²⁺) = −1920 but ΔhydH(Na⁺) = only −406 kJ mol⁻¹.
6Which of these ions has the most exothermic enthalpy of hydration?
Entropy
Entropy: the number of ways of arranging things
Entropy, S, measures the disorder of a system — more precisely, the number of ways the energy and particles can be arranged. Units: J K⁻¹ mol⁻¹ (note: joules, not kilojoules — this catches people out constantly).
Entropy increases: solid < liquid < gas. Gases are hugely more disordered.
Entropy increases when the number of moles of gas increases.
Entropy increases when a solid dissolves (ordered lattice → free-moving aqueous ions).
A perfect crystal at 0 K has S = 0 — so every substance has a positive standard entropy (unlike ΔfH).
ΔS⦵system = ΣS⦵(products) − ΣS⦵(reactants)same "products minus reactants" pattern as Hess's Law — but do not reverse any signs
ΔS = 188.7 − 69.9 = +118.8 J K⁻¹ mol⁻¹ — strongly positive, as expected for liquid → gas.
Calculate
Your turn — entropy change of the system
7For CaCO₃(s) → CaO(s) + CO₂(g), the standard entropies (J K⁻¹ mol⁻¹) are: CaCO₃ = 92.9, CaO = 39.7, CO₂ = 213.6. Calculate ΔS⦵system.
J K⁻¹ mol⁻¹
Hint: products − reactants = (39.7 + 213.6) − 92.9. Positive, because a gas is made from a solid.
Gibbs
Gibbs free energy decides feasibility
A reaction is thermodynamically feasible (spontaneous) when the Gibbs free energy change is negative:
ΔG = ΔH − TΔSΔG < 0 → feasible · ΔG = 0 → at the point of becoming feasible · ΔG > 0 → not feasible
The unit trap. ΔH is in kJ mol⁻¹ but ΔS is in J K⁻¹ mol⁻¹. Divide ΔS by 1000 (or multiply ΔH by 1000) before you use the equation. Miss this and your answer is out by a factor of a thousand.
The four cases are worth memorising:
ΔH negative, ΔS positive → ΔG negative at all temperatures — always feasible.
ΔH positive, ΔS negative → never feasible.
ΔH positive, ΔS positive → feasible only at high T (the TΔS term must outweigh ΔH).
ΔH negative, ΔS negative → feasible only at low T.
Worked example — is CaCO₃ decomposition feasible at 298 K?
The temperature at which a reaction becomes feasible
A reaction with ΔH positive and ΔS positive is not feasible at low temperature, but as T rises the TΔS term grows until it overtakes ΔH. The changeover happens exactly when ΔG = 0:
0 = ΔH − TΔS ⟹ T = ΔH ÷ ΔSΔH in J mol⁻¹ and ΔS in J K⁻¹ mol⁻¹ ⟹ T comes out in kelvin
Above about 1110 K the decomposition becomes feasible — and real industrial lime kilns run at roughly 1100–1200 K. The thermodynamics is not an abstraction: it sets the furnace temperature.
Both signs negative? T = ΔH ÷ ΔS still gives the changeover temperature, but now the reaction is feasible below it, not above it.
Calculate
Your turn — find the changeover temperature
9For the decomposition MgCO₃(s) → MgO(s) + CO₂(g): ΔH = +117.0 kJ mol⁻¹ and ΔS = +175.0 J K⁻¹ mol⁻¹. Calculate the minimum temperature (in K) at which the reaction becomes feasible.
K
Hint: set ΔG = 0, so T = ΔH ÷ ΔS = 117 000 J mol⁻¹ ÷ 175.0 J K⁻¹ mol⁻¹.
Calculate
Your turn — feasible at low temperature
10For N₂(g) + 3H₂(g) → 2NH₃(g): ΔH = −92.0 kJ mol⁻¹ and ΔS = −199.0 J K⁻¹ mol⁻¹. Above what temperature (in K) does the reaction stop being feasible?
K
Hint: the changeover is still at ΔG = 0, so T = ΔH ÷ ΔS = (−92 000) ÷ (−199.0). The two minus signs cancel.
Quick check
Reading the signs
11A reaction has ΔH = −50 kJ mol⁻¹ and ΔS = −120 J K⁻¹ mol⁻¹. Which statement is correct?
Quick check
The limitation of ΔG
12The combustion of petrol has a large negative ΔG at 298 K, yet a bucket of petrol sitting in air does not burst into flames. Why not?
Quick check
Why do endothermic salts dissolve?
13Ammonium nitrate dissolves in water even though ΔsolH is positive (+26 kJ mol⁻¹) — the solution gets cold. Why is dissolving still feasible at 298 K?
Sort it
What happens to the entropy of the system?
Tap a change, then tap the bin for the sign of ΔSsystem. Count the moles of gas on each side first.
🟦 ΔS positive
🟩 ΔS negative
🟪 ΔS ≈ zero
Match it
Match the term to its definition
Tap a term on the left, then its definition on the right.
Term
Definition
Recap
The big ideas to know
Lattice enthalpy of formation: gaseous ions → 1 mol ionic solid; always exothermic; bigger for high charge and small radius
Born–Haber: ΔLEH = ΔfH − (ΔatH + IEs + ΔatH + EAs) — double the halogen terms for MX₂
Experimental more exothermic than theoretical: the cation polarises the anion → covalent character (AgI, MgI₂)
Dissolving: ΔsolH = −ΔLEH + ΣΔhydH; ΔhydH is more exothermic for high charge density