Eduqas A-level Chemistry (A410QS) · PI4 — Energy changes
Mini-Lesson
PI4 · Energy changes
Eduqas PI4 asks the deepest question in chemistry: why do reactions go? You will define the enthalpy terms precisely, build Born-Haber cycles to find lattice enthalpies, compare them with the perfect ionic model, calculate enthalpies of solution, and use entropy and ΔG = ΔH − TΔS to decide feasibility.
Work through each screen, answer every question as you go — the multiple-choice checks, the calculations and the sorting games — and collect ⭐ stars. Press Start when you are ready.
PI4 · definitions
The enthalpy definitions — say them exactly
Marks are lost here for sloppiness. Each definition needs: one mole, the state symbols, and standard conditions.
Enthalpy of atomisation — the enthalpy when one mole of gaseous atoms is formed from the element in its standard state. Always endothermic. Na(s) → Na(g)
First ionisation energy — one mole of gaseous 1+ ions from one mole of gaseous atoms. Always endothermic.
First electron affinity — one mole of gaseous 1− ions from one mole of gaseous atoms. Exothermic (the electron is attracted to the nucleus).
Second electron affinity — endothermic, because you must force a second electron onto an ion that is already negative.
Lattice enthalpy of formation — one mole of an ionic solid from its gaseous ions. Strongly exothermic. Na⁺(g) + Cl⁻(g) → NaCl(s)
Enthalpy of hydration — one mole of gaseous ions dissolved in an excess of water. Exothermic.
Enthalpy of solution — one mole of solute dissolved to infinite dilution. Can be either sign.
Match it
Match the definitions
Tap a definition on the left, then the term it defines.
Definition
Term
PI4 · Born-Haber
The Born-Haber cycle
Lattice enthalpy cannot be measured directly, so we get it by Hess's law — via a route through gaseous atoms and ions.
Sum of the "up and over" route: 107 + 496 + 122 + (−349) = +376
ΔHlatt = −411 − 376 = −787 kJ mol⁻¹
Two multiplication traps: for MgCl₂ you need IE₁ + IE₂ for the magnesium, and 2 × the atomisation enthalpy and 2 × the electron affinity for the chlorine. For an oxide you need the endothermic second electron affinity.
Calculate
Your turn — lattice enthalpy of NaCl
1Using ΔHf(NaCl) = −411, ΔHat(Na) = +107, IE₁(Na) = +496, ΔHat(Cl) = +122, EA₁(Cl) = −349 (kJ mol⁻¹), calculate the lattice enthalpy of formation of NaCl. Include the sign.
2Calculate the lattice enthalpy of formation of MgCl₂ from: ΔHf = −641, ΔHat(Mg) = +148, IE₁(Mg) = +738, IE₂(Mg) = +1451, ΔHat(Cl) = +122, EA₁(Cl) = −349 (kJ mol⁻¹). Include the sign. Remember there are two chlorines.
A theoretical lattice enthalpy can be calculated by assuming perfectly spherical ions with evenly distributed charge. Lattice enthalpy becomes more exothermic when the ions have a higher charge and a smaller radius — that is, a higher charge density.
Comparing the theoretical value with the experimental Born-Haber value is a test of the model:
Close agreement (e.g. NaCl) → the compound is close to purely ionic.
Experimental value much more exothermic than theoretical (e.g. AgI, ZnS) → there is significant covalent character. The small, highly charged cation polarises the large, easily-polarised anion, distorting its electron cloud and pulling charge into the space between the ions — extra attraction the ionic model never accounted for.
This is Fajans' reasoning: polarising power rises with small, highly charged cations; polarisability rises with large anions.
Quick check
Why the discrepancy?
?For silver iodide, the experimental (Born-Haber) lattice enthalpy is much more exothermic than the value from the perfect ionic model. What does this tell you?
PI4 · solution
Enthalpy of solution
Dissolving an ionic solid is a two-step Hess cycle: break the lattice apart into gaseous ions (endothermic — the reverse of lattice formation), then hydrate those gaseous ions (exothermic).
ΔHsol = ΣΔHhyd − ΔHlatt(formation)the minus sign is because you must REVERSE the lattice formation step
The two terms are large and nearly cancel, so ΔHsol is small — and can be either sign. That is why some salts dissolve with warming and others (ammonium nitrate in a cold pack) with cooling.
Hydration enthalpy becomes more exothermic for smaller, more highly charged ions, because the δ+ or δ− ends of the water molecules are attracted more strongly. Mg²⁺ hydrates far more exothermically than Na⁺.
Calculate
Your turn — enthalpy of solution
3For NaCl: ΔHlatt(formation) = −787, ΔHhyd(Na⁺) = −406, ΔHhyd(Cl⁻) = −363 (kJ mol⁻¹). Calculate ΔHsol. Include the sign.
Entropy, S, measures the number of ways energy and particles can be arranged — loosely, the disorder. Units: J K⁻¹ mol⁻¹ (note: joules, not kilojoules).
Entropy rises: solid < liquid << gas. The biggest jumps come with boiling and with any reaction that increases the number of gas molecules.
A perfect crystal at 0 K has S = 0.
ΔS = ΣS(products) − ΣS(reactants)multiply each standard entropy by its coefficient
Quick prediction: count the moles of gas on each side. More gas out than in → ΔS is positive. That single check answers most exam questions.
Calculate
Your turn — entropy change
4Calculate ΔS for CaCO₃(s) → CaO(s) + CO₂(g), using standard entropies in J K⁻¹ mol⁻¹: CaCO₃ = 92.9, CaO = 39.7, CO₂ = 213.8. Include the sign.
J K⁻¹ mol⁻¹
Hint: ΔS = (39.7 + 213.8) − 92.9 = 253.5 − 92.9. A gas is produced, so it must be positive.
PI4 · free energy
Free energy and feasibility
ΔG = ΔH − TΔSa reaction is FEASIBLE when ΔG ≤ 0 · T in kelvin
The units trap that costs the most marks: ΔH is in kJ mol⁻¹ but ΔS is in J K⁻¹ mol⁻¹. Divide ΔS by 1000 before you subtract.
Four combinations, and only four:
ΔH negative, ΔS positive → ΔG negative at all temperatures. Always feasible.
ΔH positive, ΔS negative → ΔG positive at all temperatures. Never feasible.
ΔH positive, ΔS positive → feasible only at high T (TΔS must outweigh ΔH).
ΔH negative, ΔS negative → feasible only at low T.
At the borderline ΔG = 0, so the minimum (or maximum) feasible temperature is T = ΔH ÷ ΔS.
Calculate
Your turn — calculate ΔG
5For CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ mol⁻¹ and ΔS = +160.6 J K⁻¹ mol⁻¹. Calculate ΔG at 298 K, in kJ mol⁻¹. Include the sign.
6Calculate the minimum temperature, in K, at which the decomposition of CaCO₃ becomes feasible. (ΔH = +178 kJ mol⁻¹, ΔS = +160.6 J K⁻¹ mol⁻¹)
K
Hint: at the threshold ΔG = 0, so T = ΔH ÷ ΔS = 178 ÷ 0.1606.
Quick check
Feasible at every temperature
?A reaction has ΔH = −150 kJ mol⁻¹ and ΔS = +80 J K⁻¹ mol⁻¹. At which temperatures is it feasible?
Sort it
When is it feasible?
Tap a pair of ΔH and ΔS values, then the box saying when ΔG becomes negative. Remember ΔG = ΔH − TΔS.
♾️ Feasible at ALL temperatures
🔥 Feasible only at HIGH T
❄️ Feasible only at LOW T
Quick check
Never feasible
?Which combination makes a reaction thermodynamically impossible at every temperature?
PI4 · the limits of ΔG
What ΔG does and does not tell you
ΔG is a thermodynamic quantity. A negative ΔG means the reaction can happen — it says nothing at all about how fast.
Diamond → graphite has a negative ΔG at room temperature. Diamonds do not visibly turn to pencil lead, because the activation energy for rearranging the covalent lattice is enormous. The reaction is feasible but kinetically inert.
Also note: ΔG = 0 is exactly the condition for equilibrium. And ΔG relates to the cell potential by ΔG = −nFE°cell, which is why a positive E°cell means a feasible reaction — the same idea in two languages.
Recap
The big ideas to know
Definitions: one mole, correct state symbols; 2nd electron affinity is ENDOthermic