HookThe reaction that feeds half the world runs at 15% conversion
In July 1909, in a Karlsruhe laboratory, Fritz Haber's bench-top apparatus produced its first steady trickle of liquid ammonia from nitrogen and hydrogen — the demonstration that persuaded BASF to buy the process and set Carl Bosch the job of scaling it a millionfold. Today the Haber–Bosch process fixes on the order of 180 million tonnes of ammonia a year, consumes between 1 and 2% of the world's energy supply, and supplies the fertiliser behind roughly half the food humanity eats: about half the nitrogen atoms in your own body's protein passed through one of these reactors. And yet, on any single pass through the converter, only around 15% of the gas actually reacts.
That 15% is not a failure of engineering — it is the chemistry itself pushing back. \(\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}\), \(\Delta H = -92\ \text{kJ mol}^{-1}\), is reversible: ammonia falls apart in the same conditions that form it, and the mixture settles into an equilibrium where both reactions run at equal speed. Section 3.1.6 gives you two instruments for dealing with such systems: Le Chatelier's principle, the qualitative lever that predicts which way an equilibrium shifts when you prod it, and Kc, the quantitative dial that states exactly where the balance point sits — and refuses to move for anything except temperature.
ModelDynamic equilibrium — constant is not equal, and still is not stopped
Seal a reversible reaction in a closed system — nothing added, nothing escaping, temperature steady — and it will find a state where the concentrations of every species stop changing. Nothing has stopped happening. The forward and reverse reactions are both still running, at equal rates: ammonia molecules are being assembled exactly as fast as others are being torn apart. That is what dynamic means, and it is the first marking point of nearly every equilibrium definition question — forward rate equals reverse rate, so macroscopic concentrations are constant.
The second marking point is the one students garble: constant does not mean equal. An equilibrium mixture can be almost all product, almost all reactant, or anywhere between; the balance point depends on the reaction and the conditions. What the equilibrium state fixes is the ratio of concentrations, not any resemblance between them.
Why a closed system? Because an open one leaks. Heat calcium carbonate in a sealed vessel and \(\text{CaCO}_3\text{(s)} \rightleftharpoons \text{CaO(s)} + \text{CO}_2\text{(g)}\) reaches equilibrium; heat it in an open kiln and the CO₂ drifts away, the reverse reaction is starved of its reactant, and the decomposition simply runs to completion — which is precisely why lime kilns work. Remove a product and there is no equilibrium to talk about.
MechanismLe Chatelier's principle — the system pushes back
Le Chatelier's principle: when a condition of a system at equilibrium is changed, the position of equilibrium shifts in the direction that opposes the change. It is a prediction machine with three standard inputs.
Concentration. Add more of a reactant and the system counteracts the excess by consuming it — the position shifts right, making more product. Remove a product as it forms and the equilibrium chases it, shifting right continuously; industrial processes exploit this by condensing ammonia out of the gas stream and recycling the rest.
Pressure. Compressing a gaseous equilibrium favours the side with fewer moles of gas, because fewer molecules exert less pressure — the opposing response. For \(\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3\), four moles become two, so high pressure drives the mixture towards ammonia. If both sides hold equal gas moles, pressure changes the rate (via concentration) but not the position.
Temperature. Heating an equilibrium favours the endothermic direction, the one that absorbs the added energy; cooling favours the exothermic direction. Ammonia synthesis is exothermic going forward, so heat actually pushes the equilibrium backwards, towards less ammonia — the seed of the compromise below. A catalyst, finally, is the trick input: it speeds forward and reverse reactions equally, so the position and the yield are untouched. It only shortens the wait — the mixture reaches the same equilibrium sooner.
CaseCompromise conditions — Haber's bargain, and ethanol's
Read the Haber equation with Le Chatelier's eyes and the ideal conditions write themselves: exothermic forward reaction, so run it cold for maximum yield; four gas moles becoming two, so run it at crushing pressure. Now read it with kinetic eyes: cold means the nitrogen triple bond barely reacts at all — near room temperature the equilibrium yield would be superb and you would wait years for it. And crushing pressure means compressors, thick-walled steel and an explosion risk that scales with every atmosphere. Real plants therefore strike a bargain: around 400–450 °C — hot enough for a workable rate, at the cost of equilibrium yield — roughly 200 atmospheres, an affordable fraction of what Le Chatelier would like, and an iron catalyst, which adds no yield but reaches the (compromised) equilibrium fast enough to be economic. The ~15% that converts per pass is then handled by engineering: cool the mixture, condense out the ammonia liquid, recycle the unreacted gas round again, and the overall conversion climbs above 95%.
The same three-way trade — yield versus rate versus cost — runs through the other industrial example AQA names: hydrating ethene to ethanol, \(\text{C}_2\text{H}_4\text{(g)} + \text{H}_2\text{O(g)} \rightleftharpoons \text{C}_2\text{H}_5\text{OH(g)}\), \(\Delta H = -45\ \text{kJ mol}^{-1}\). Exothermic and mole-reducing again, so the ideal is cold and compressed; the practice is about 300 °C and 60–70 atm over a phosphoric(V) acid catalyst, converting only ~5% per pass — recycled, as ever, to around 95% overall. When an exam question hands you any unfamiliar industrial equilibrium, run this exact scan: what does yield want, what does rate want, what does cost permit, and how does recycling rescue the leftovers.
ModelKc — writing the constant that refuses to move
Le Chatelier says which way; the equilibrium constant says how far. For a general homogeneous equilibrium — every species in the same phase — \(a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}\), the constant is \[K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}\] products over reactants, each equilibrium concentration raised to the power of its coefficient. Its size is a map of the mixture: \(K_c \gg 1\) means the equilibrium sits far to the right; \(K_c \ll 1\), far to the left. Units are built by cancellation, fresh for each reaction — insert mol dm⁻³ for every bracket and cancel. When the powers balance top and bottom, the units vanish entirely and Kc is a pure number.
Now the fact examiners test relentlessly: at a fixed temperature, Kc is constant. Add extra reactant and the concentrations are momentarily out of ratio — the system responds exactly as Le Chatelier predicts, shifting right until the ratio returns to the same Kc. The position moved; the constant did not. A catalyst changes neither, since it accelerates both directions equally. The only lever that changes Kc itself is temperature: heating an exothermic-forward reaction shifts it backwards, shrinking the product terms, so Kc falls — for ammonia synthesis it collapses by many powers of ten between room temperature and the 450 °C of a working converter. That single sentence — 'concentration and catalysts leave Kc unchanged; only temperature alters it' — settles a whole category of one-mark questions.
DataCalculating Kc — the ICE routine
Every Kc calculation is the same three-row table: Initial moles, Change, Equilibrium moles — then divide by the volume to reach concentrations, substitute, and derive units. The change row is pure stoichiometry: whatever one species loses or gains, the others follow in the ratio of the balanced equation. Exam questions usually hand you the initial amounts and one equilibrium amount, often found by titrating the mixture; the ICE table reconstructs everything else.
Work tidily and the marks fall in order — expression, concentrations, value, units. Miss the volume division and everything downstream collapses, except where mole numbers balance and the volumes cancel: spotting that cancellation is itself a skill AQA rewards, because it lets you compute Kc without ever knowing the volume.
Esterification, the classic. Mix 1.00 mol of ethanoic acid with 1.00 mol of ethanol, leave the sealed flask (with an acid catalyst) for a week, then titrate: 0.33 mol of acid remains. \(\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}\). ICE: acid went from 1.00 to 0.33, a change of −0.67; so ethanol is also 0.33, and ester and water are each 0.67 mol. With total volume V, \[K_c = \frac{(0.67/V)(0.67/V)}{(0.33/V)(0.33/V)} = \frac{0.67^2}{0.33^2} = 4.1\] — every V cancels (two concentration terms up, two down), so no volume is needed and Kc has no units. Contrast a case where units survive: 1.20 mol of PCl₅ in a 4.00 dm³ vessel dissociates, \(\text{PCl}_5\text{(g)} \rightleftharpoons \text{PCl}_3\text{(g)} + \text{Cl}_2\text{(g)}\), leaving 0.80 mol at equilibrium. Change = −0.40, so 0.40 mol of each product. Concentrations: 0.200, 0.100, 0.100 mol dm⁻³. \(K_c = (0.100 \times 0.100)/0.200 = 0.0500\ \text{mol dm}^{-3}\) — one net concentration term on top, so the units are mol dm⁻³.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Le Chatelier answers are marked as a chain, and AQA wants each link written out: state the shift direction, name the change being opposed, and supply the reason in the system's own terms — 'the equilibrium shifts right, towards fewer moles of gas, to oppose the increase in pressure' or 'shifts in the endothermic direction to absorb the added heat'. A bare direction with no opposing-the-change clause is a one-mark answer to a three-mark question. And never let a catalyst shift anything: the credited line is that it increases both rates equally, so equilibrium is reached faster with no change in position or yield.
Kc calculations climb a fixed ladder — M1 the expression with correct powers, M2 equilibrium concentrations (the ICE table, then divide by volume), M3 the evaluated answer, M4 units by cancellation. Show the units cancelling explicitly rather than memorising results; when asked for Kc of a new reaction, the units are derived, not recalled. The volume trap catches more candidates than the algebra: moles are not concentrations, and only when total mole powers balance top and bottom may you skip the division because volume cancels — say so if you use it.
The industrial 6-marker is a structure question in disguise. Organise by the three forces — what maximises equilibrium yield (Le Chatelier on temperature and pressure), what maximises rate (kinetics wants heat and a catalyst), what cost and safety permit (compression and pressure vessels are expensive) — and land on the compromise with real numbers: 400–450 °C, about 200 atm, iron catalyst, unreacted gases recycled. Quoting the per-pass conversion (~15%) and the recycled overall figure (>95%) is exactly the kind of applied detail that lifts an answer into the top level.