HookThe cold pack that breaks last year's rule
A physio cracks an instant cold pack, and inside a bag of ammonium nitrate bursts into water. Within seconds the pack is heading towards 0 °C against a sprained ankle. Dissolving \(\text{NH}_4\text{NO}_3\) is endothermic — about \(+26\ \text{kJ mol}^{-1}\) — so the process is absorbing heat from its surroundings while it happens, entirely of its own accord. That should bother you. The AS energetics instinct says exothermic is downhill, downhill is favourable; here is a change that runs uphill in enthalpy, spontaneously, in a £2 first-aid product.
The resolution is that enthalpy was never the only driving force. This section adds the second one — entropy — and the quantity that referees between them, the Gibbs free-energy change \(\Delta G = \Delta H - T\Delta S\). Before that, it settles an older debt: measuring the strength of an ionic lattice, a "bond" you can never isolate in a calorimeter, using the Born–Haber cycle — Hess's law with the accountancy taken seriously — and then using the numbers to test how honest the ionic model of bonding really is. This is Paper 1 heartland and the most calculation-dense of the physical topics; almost every mark in it can be booked in advance.
ModelEight definitions that gate every cycle
Born–Haber questions are won and lost on definitions, because every arrow in the cycle is a definition. All are quoted per mole under standard conditions (298 K, 100 kPa), and the word gaseous does heavy lifting throughout — drop it and the mark goes.
Enthalpy of formation: one mole of a compound formed from its elements in their standard states. Enthalpy of atomisation: one mole of gaseous atoms produced from the element — \(\text{Na(s)} \rightarrow \text{Na(g)}\) at \(+107\ \text{kJ mol}^{-1}\), and \(\tfrac{1}{2}\text{Cl}_2\text{(g)} \rightarrow \text{Cl(g)}\) at \(+122\ \text{kJ mol}^{-1}\) — note the half, because the definition counts moles of atoms, not molecules. First ionisation energy: one mole of gaseous atoms each losing one electron to form gaseous 1+ ions (\(+496\) for sodium). First electron affinity: one mole of gaseous atoms each gaining one electron to form gaseous 1− ions — \(-349\ \text{kJ mol}^{-1}\) for chlorine, exothermic because the incoming electron feels the nuclear pull.
Lattice enthalpy comes in two signs, and AQA uses both: formation (gaseous ions assemble into one mole of solid lattice; hugely exothermic) and dissociation (the lattice torn back into gaseous ions; same magnitude, positive). Read which one the question defines before you write anything. Finally the solution pair: enthalpy of hydration (gaseous ions become aqueous ions; always exothermic) and enthalpy of solution (one mole of solute dissolves to such dilution that further water changes nothing).
ModelBorn–Haber cycles — measuring what no calorimeter can see
You cannot measure lattice enthalpy directly: nobody can prepare a mole of gaseous \(\text{Na}^+\) and a mole of gaseous \(\text{Cl}^-\) and let them crash into a crystal inside a calorimeter. But Hess's law says enthalpy change is route-independent, so you build a loop out of steps that are measurable — formation, atomisation, ionisation, electron affinity — and let the one unknown fall out of the arithmetic.
AQA's preferred presentation is the energy-level diagram: endothermic steps climb, exothermic steps drop, and each level is labelled with real species and state symbols. Two rules keep the arithmetic honest. First, a reversed arrow flips the sign. Second, every value plugged in must match its definition's quantities — if your cycle needs two moles of chlorine atoms (for \(\text{MgCl}_2\)), the atomisation and electron-affinity terms are doubled, and forgetting the doubling is the classic slip in every cohort.
Sodium chloride, with the data a question would print: atomisation of sodium \(+107\), first ionisation energy of sodium \(+496\), atomisation of chlorine \(+122\), first electron affinity of chlorine \(-349\), enthalpy of formation of \(\text{NaCl(s)}\) \(-411\), all in \(\text{kJ mol}^{-1}\). The direct route from elements to lattice is formation, \(-411\). The indirect route goes elements \(\rightarrow\) gaseous atoms \((+107 + 122)\) \(\rightarrow\) gaseous ions \((+496 - 349)\) \(\rightarrow\) lattice (the unknown, \(\Delta H_{\text{latt}}\)). Hess says the routes agree: \[ \Delta H_{\text{latt}} = -411 - (107 + 496 + 122 - 349) = -411 - 376 = -787\ \text{kJ mol}^{-1} \] Sanity-check the story the numbers tell: the only exothermic input is the electron affinity; every other step costs energy, and the lattice pays it all back with interest. That \(-787\ \text{kJ mol}^{-1}\) is what holds table salt together — and why its melting point is 801 °C.
DataThe perfect ionic model — and the compounds that cheat it
There are two independent ways to a lattice enthalpy. The Born–Haber value is experimental — assembled from measured quantities. The theoretical value is calculated from the perfect ionic model: ions as undistorted spheres with their charge spread evenly, attracting each other by pure electrostatics, no electron sharing at all. Comparing the two is a truth test for the bonding.
For sodium chloride the theoretical value is about \(-770\ \text{kJ mol}^{-1}\) against the Born–Haber \(-787\) — agreement within roughly 2%, so NaCl is about as purely ionic as matter gets. For silver iodide the gap yawns: theoretical about \(-778\), Born–Haber about \(-889\) — the real lattice is roughly 14% stronger than pure electrostatics can explain. The reason is polarisation: a small or highly charged cation drags the electron cloud of a large, squashy anion towards itself, so electron density ends up shared between the nuclei. That shared density is covalent character, an extra contribution to bonding the ionic model cannot see. The exam pattern is fixed: "the Born–Haber value is more exothermic than the theoretical value — explain" earns one mark for the anion being polarised (or covalent character existing) and one for the real bonding therefore being stronger than the purely ionic prediction.
ModelDissolving — a tug-of-war the ions referee
Dissolving an ionic solid is modelled as two fictional stages. First, tear the lattice apart into gaseous ions — that costs the lattice dissociation enthalpy (\(+787\) for NaCl). Second, drop those ions into water, where polar molecules swarm them — \(\delta^-\) oxygens inward around cations, \(\delta^+\) hydrogens inward around anions — releasing the hydration enthalpies. So \[ \Delta H_{\text{sol}} = \Delta H_{\text{latt(diss)}} + \Sigma \Delta H_{\text{hyd}} \] Hydration follows charge density: the smaller the ion or the higher its charge, the closer and harder the water dipoles grip, and the more exothermic the term — compare \(\Delta H_{\text{hyd}}\) of about \(-406\ \text{kJ mol}^{-1}\) for \(\text{Na}^+\) with about \(-1{,}920\ \text{kJ mol}^{-1}\) for the smaller, doubly charged \(\text{Mg}^{2+}\).
For sodium chloride, with hydration enthalpies \(-406\) (\(\text{Na}^+\)) and \(-377\) (\(\text{Cl}^-\)): \[ \Delta H_{\text{sol}} = +787 - 406 - 377 = +4\ \text{kJ mol}^{-1} \] Slightly endothermic — dissolving salt genuinely cools the water, just barely. The discipline is in the signs: the \(+787\) must be dissociation (you are breaking the lattice), while each hydration term arrives with its own minus sign already attached. Write the cycle before the arithmetic and the signs take care of themselves; try to shortcut it and you will subtract where you should add. And notice what the answer sets up: an endothermic process that visibly happens. Salt dissolves anyway. Something other than enthalpy must be paying — that something is the next block.
ModelEntropy — the universe prefers more ways
Entropy, \(S\), measures the number of ways particles and their energy can be arranged — "disorder" is the serviceable shorthand. Its units are the section's booby trap: \(\text{J K}^{-1}\,\text{mol}^{-1}\) — joules, while every enthalpy you meet is in kilojoules. Solids sit low (particles locked in a lattice), liquids higher, gases far higher; more moles of gas means more entropy, and dissolving a crystalline solid into free-swimming aqueous ions usually raises it too.
The calculation is a straight ledger from data-book values: \(\Delta S = \Sigma S_{\text{products}} - \Sigma S_{\text{reactants}}\). Predict the sign before you compute — it catches transcription errors. Thermal decomposition of limestone, \(\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)}\), creates a mole of gas from none, so \(\Delta S\) must be positive; with standard entropies 92.9, 39.7 and 213.6 \(\text{J K}^{-1}\,\text{mol}^{-1}\), \(\Delta S = (39.7 + 213.6) - 92.9 = +160.4\ \text{J K}^{-1}\,\text{mol}^{-1}\). And the cold pack: crystalline ammonium nitrate becoming hydrated, wandering ions gives \(\Delta S \approx +108\ \text{J K}^{-1}\,\text{mol}^{-1}\). A positive \(\Delta S\) is nature's thumb on the scale — the question is how hard it presses, and that depends on temperature.
ModelΔG = ΔH − TΔS — the feasibility switch
A change is feasible (spontaneous) when \(\Delta G \le 0\), where \(\Delta G = \Delta H - T\Delta S\) and \(T\) is in kelvin. Temperature is the weighting on entropy: near absolute zero, enthalpy is almost the whole story; heat the system and the \(T\Delta S\) term buys increasing forgiveness for enthalpy's sins. Four sign combinations, two of them boring: \(\Delta H\) negative with \(\Delta S\) positive is feasible at every temperature; both reversed is feasible at none. The interesting mixed cases switch over at the temperature where \(\Delta G = 0\), i.e. \(T = \Delta H / \Delta S\) — endothermic-but-disordering reactions switch on above it, exothermic-but-ordering ones switch off.
Since \(\Delta G = -\Delta S \cdot T + \Delta H\) is linear in \(T\), AQA also likes the graph: plot \(\Delta G\) against \(T\) and the gradient is \(-\Delta S\) while the intercept is \(\Delta H\) — two thermodynamic quantities read straight off a straight line. One caveat earns evaluation marks everywhere: feasible does not mean fast. \(\Delta G\) says the destination is downhill; it says nothing about the road. Petrol and oxygen at 298 K have a hugely negative \(\Delta G\) and sit inert until a spark pays the activation energy; diamond turning to graphite is feasible and takes geological time. Thermodynamics proposes; kinetics disposes.
Limestone to quicklime, \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\): \(\Delta H = +178\ \text{kJ mol}^{-1}\), \(\Delta S = +160.4\ \text{J K}^{-1}\,\text{mol}^{-1}\). Convert units first: \(160.4\ \text{J} = 0.1604\ \text{kJ}\). At 298 K: \(\Delta G = 178 - 298 \times 0.1604 = 178 - 47.8 = +130\ \text{kJ mol}^{-1}\) — not feasible, which is why limestone cliffs survive. The crossover: \[ T = \frac{\Delta H}{\Delta S} = \frac{178}{0.1604} \approx 1{,}110\ \text{K} \] about 837 °C — which is why lime kilns have run at bright red heat since Roman engineering. Now the cold pack, full circle: \(\Delta H \approx +26\ \text{kJ mol}^{-1}\), \(\Delta S \approx +108\ \text{J K}^{-1}\,\text{mol}^{-1}\), so at 298 K \(\Delta G = 26 - 298 \times 0.108 = 26 - 32.2 = -6.2\ \text{kJ mol}^{-1}\). Negative. The endothermic pack proceeds, paid for entirely in entropy — last year's rule, properly amended.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
The single most-flagged error in examiner reports on this topic is the unit mismatch inside \(\Delta G = \Delta H - T\Delta S\): enthalpies arrive in kJ but entropies in J. Convert \(\Delta S\) to \(\text{kJ K}^{-1}\,\text{mol}^{-1}\) (divide by 1,000) before substituting, write the conversion as its own line, and keep \(T\) in kelvin — that line is usually a method mark on its own.
Definitions are strict AO1: they need "one mole of", the correct species and the word "gaseous" where it belongs (atoms for atomisation, ions for lattice and hydration). In cycle calculations, write the Hess equation from your cycle before substituting numbers — M1 is routinely the correct expression, M2 the substitution, M3 the evaluated answer with sign and units — and remember that reversing an arrow flips its sign, and that compounds like \(\text{MgCl}_2\) double the chlorine terms.
Two set-piece explanations recur. Ionic-model comparisons: Born–Haber more exothermic than theoretical → anion polarised by the cation → covalent character → real lattice stronger than the purely ionic prediction. Feasibility: state the condition \(\Delta G \le 0\), compute \(T = \Delta H/\Delta S\) for the crossover, then interpret in a sentence ("feasible above 1,110 K, so the kiln must run hot"). On graph questions, quote gradient \(= -\Delta S\) and intercept \(= \Delta H\) explicitly — AQA awards the identification, not just the numbers.