HookThe reaction that feeds half the world runs at 15% conversion

Roughly 180 million tonnes of ammonia are made every year, almost all of it by one equilibrium: \(\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3\). The Haber process runs at about 450 °C and 200 atmospheres over an iron catalyst, swallows 1–2% of the world's energy supply, and the fertiliser downstream of it feeds roughly half the people alive — about half the nitrogen atoms in your own proteins have passed through a Haber reactor. Now the number nobody brags about: a single pass through the converter turns only around 15% of the gas mixture into ammonia. A century of engineering has barely moved that figure. Industry simply condenses out the ammonia and recycles the unreacted nitrogen and hydrogen, round and round.

Why tolerate 15%? Because a gaseous equilibrium answers to a constant — \(K_p\) — and constants cannot be bribed. This section is 3.1.6 rebuilt for gases: the currency changes from concentration to partial pressure, the bookkeeping runs through mole fractions, and the headline rule survives intact — of every lever a chemist can pull, only temperature changes the value of \(K_p\). Pressure moves the position of equilibrium; a catalyst shortens the wait; neither lays a finger on the constant. Keep those three effects separate and half the marks in this section are already yours.

ModelMole fraction and partial pressure — bookkeeping for gas mixtures

In a mixture of gases each component behaves, to a good approximation, as if the others were not there. The mole fraction of gas A is its share of the molecules: \(x_A = \dfrac{n_A}{n_{\text{total}}}\). Mole fractions carry no units and must sum to exactly 1 — a built-in arithmetic check examiners expect you to use.

The partial pressure of A is the pressure A would exert if it occupied the container alone, and it follows from the mole fraction directly: \(p_A = x_A \times P_{\text{total}}\). Partial pressures must add up to the total pressure — that statement is Dalton's law, and it is the second self-check. The physical picture is simple: pressure is molecules striking walls, and at a fixed temperature every molecule contributes equally on average, so each gas's share of the pressure equals its share of the molecules, regardless of which gas is heavier.

Units follow the question: AQA data usually arrive in kPa, occasionally Pa or atm. Any unit works so long as you are consistent, because \(K_p\) inherits whatever you feed it.

Worked example

An equilibrium mixture contains 0.60 mol of \(\text{N}_2\), 1.80 mol of \(\text{H}_2\) and 0.60 mol of \(\text{NH}_3\) at a total pressure of 200 kPa. Total moles \(= 3.00\). Mole fractions: \(x_{\text{N}_2} = 0.60/3.00 = 0.20\), \(x_{\text{H}_2} = 1.80/3.00 = 0.60\), \(x_{\text{NH}_3} = 0.20\) — sum 1.00, check passed. Partial pressures: \(p_{\text{N}_2} = 0.20 \times 200 = 40\ \text{kPa}\), \(p_{\text{H}_2} = 0.60 \times 200 = 120\ \text{kPa}\), \(p_{\text{NH}_3} = 40\ \text{kPa}\) — sum 200 kPa, check passed. Two self-checks, two easy marks, before any equilibrium chemistry has even started.

ModelWriting Kp — powers from the equation, units from the expression

For a homogeneous gaseous equilibrium — every species in the same phase, here all gas — \(a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}\), the equilibrium constant in partial pressures is \[ K_p = \frac{(p_{\text{C}})^c\,(p_{\text{D}})^d}{(p_{\text{A}})^a\,(p_{\text{B}})^b} \] Products on top, reactants underneath, each raised to its coefficient from the balanced equation. Note the deliberate contrast with 3.1.9: rate-equation orders come only from experiment, but equilibrium-constant powers come straight from the stoichiometry. Same-looking algebra, opposite provenance — and examiners set traps on exactly that seam.

Like the rate constant in kinetics, \(K_p\) has no fixed units: derive them fresh each time by substituting the pressure unit and cancelling. When the number of moles of gas is the same on both sides — \(\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}\) — every pressure cancels and \(K_p\) has no units at all, which you must state explicitly to collect the mark.

Worked example

For \(\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3\), using the partial pressures above: \[ K_p = \frac{(p_{\text{NH}_3})^2}{(p_{\text{N}_2})(p_{\text{H}_2})^3} = \frac{40^2}{40 \times 120^3} = \frac{1600}{6.91 \times 10^7} = 2.3 \times 10^{-5} \] Units: \(\dfrac{\text{kPa}^2}{\text{kPa} \times \text{kPa}^3} = \text{kPa}^{-2}\), so \(K_p = 2.3 \times 10^{-5}\ \text{kPa}^{-2}\). The tiny value says this equilibrium sits well to the left at this temperature — the algebraic face of that stubborn 15% conversion.

MechanismFrom starting moles to Kp — the five-step pipeline

Harder questions start before equilibrium: a stated amount of gas is sealed in, some fraction reacts, and you are handed only the total pressure. The route is always the same five steps — equilibrium moles (via the stoichiometry), total moles, mole fractions, partial pressures, \(K_p\). Lay the first step out as a small table under the equation (initial, change, equilibrium) and the arithmetic cannot ambush you.

The step that punishes carelessness is the total: when mole numbers change across the equation, the total at equilibrium is not the total you started with. Dissociating one gas into two grows the total; synthesising ammonia shrinks it. Every mole fraction downstream depends on getting that total right, so it is the first thing to re-check when an answer looks odd.

Worked example

1.00 mol of dinitrogen tetroxide is sealed in a flask and warmed; at equilibrium 0.40 mol has dissociated by \(\text{N}_2\text{O}_4 \rightleftharpoons 2\text{NO}_2\), and the total pressure is 140 kPa. Equilibrium moles: \(\text{N}_2\text{O}_4 = 1.00 - 0.40 = 0.60\); \(\text{NO}_2 = 2 \times 0.40 = 0.80\); total \(= 1.40\) mol — larger than the 1.00 mol we started with, exactly as one-into-two predicts. Mole fractions: \(0.60/1.40 = 0.429\) and \(0.80/1.40 = 0.571\). Partial pressures: \(0.429 \times 140 = 60\ \text{kPa}\) and \(0.571 \times 140 = 80\ \text{kPa}\) (sum 140, check passed). Finally \[ K_p = \frac{(p_{\text{NO}_2})^2}{p_{\text{N}_2\text{O}_4}} = \frac{80^2}{60} = 107\ \text{kPa} \] The units survive this time — \(\text{kPa}^2/\text{kPa} = \text{kPa}\) — because the mole numbers differ across the equation.

MechanismOnly temperature moves Kp — everything else is position

Compress that \(\text{N}_2\text{O}_4/\text{NO}_2\) mixture to double the total pressure and every partial pressure doubles at the instant of compression. The quotient \((p_{\text{NO}_2})^2/p_{\text{N}_2\text{O}_4}\) therefore jumps by a factor of \(2^2/2 = 2\) — it now exceeds \(K_p\). The system is out of equilibrium, and it responds by converting \(\text{NO}_2\) back into \(\text{N}_2\text{O}_4\) — towards fewer gas moles — until the quotient falls back to the unchanged \(K_p\). That is the honest mechanism behind Le Chatelier's shorthand about shifting to the side with fewer moles: the position moves precisely because the constant refuses to.

Temperature is different in kind: it changes the value of \(K_p\) itself. For an exothermic forward reaction, raising the temperature lowers \(K_p\) — less product at equilibrium; for an endothermic forward reaction it raises it. You can watch this one happen: a sealed tube of brown \(\text{NO}_2\) and colourless \(\text{N}_2\text{O}_4\) darkens in hot water and fades in iced water, because the dissociation is endothermic — hot means a bigger \(K_p\) means more brown gas.

A catalyst changes neither the constant nor the position. It accelerates the forward and reverse reactions equally, so the same equilibrium mixture simply arrives sooner. Three levers, three different targets: temperature moves \(K_p\), pressure moves the position, a catalyst moves only the clock.

CaseThe Haber compromise, run through Kp

Ammonia synthesis is exothermic (\(\Delta H = -92\ \text{kJ mol}^{-1}\)), so cooling the reactor grows \(K_p\) — at room temperature the constant is enormous and the equilibrium yield would be superb. It is also useless: at 25 °C the rate is effectively zero, and no catalyst yet discovered fixes that at industrial scale. So the industry deliberately runs hot, at 400–450 °C, accepting a much smaller \(K_p\), because a worse equilibrium reached in seconds beats a perfect one reached never.

Pressure is the recovery lever. Four moles of gas become two, so compressing to around 200 atm drives the position right and claws single-pass conversion up to that famous 15% — without touching \(K_p\) at all. Why not 1,000 atm? Compressors, pipework and safety engineering scale in cost faster than the yield improves; the equilibrium argument gives way to an economic one. The iron catalyst, meanwhile, changes no number in the \(K_p\) ledger — it exists purely so that the compromise temperature is fast enough to pay for the plant.

This is the template for every industrial-conditions question: say what each condition does to the value of \(K_p\) (temperature: changes it; pressure: no effect; catalyst: no effect), what it does to the position, and what it costs. The examiner is checking that you keep those three ledgers separate.

VocabularyKey terms the mark scheme pays for

Mole fraction

A gas's share of the molecules in a mixture: n(A) ÷ n(total). Dimensionless, and all the mole fractions in a mixture sum to exactly 1.

Partial pressure

The pressure one gas in a mixture would exert if it occupied the container alone: p(A) = x(A) × P(total).

Dalton's law

The total pressure of a gas mixture equals the sum of the partial pressures of its components — the free self-check in every Kp calculation.

The equilibrium constant written in partial pressures: products over reactants, each raised to its coefficient from the balanced equation. Constant at constant temperature.

Homogeneous equilibrium

An equilibrium in which every species is in the same phase — for Kp at A-level, all of them gases.

Units of Kp

Derived afresh each time by substituting the pressure unit and cancelling; when moles of gas are equal on both sides everything cancels and Kp has no units.

Position of equilibrium

The actual composition of the mixture. Pressure changes move the position while Kp stays fixed; temperature moves both.

Compromise conditions

Industrial temperature and pressure chosen to balance equilibrium yield against rate and cost — the Haber process's 450 °C and 200 atm.

TrapsMisconceptions that cost marks

“Increasing the pressure increases Kp — after all, you get more ammonia.”

Actually: You do get more ammonia, but Kp has not moved. Raising the pressure unbalances the quotient, and the position shifts towards fewer gas moles until the ratio equals the same old Kp. Only temperature changes the value of Kp.

“A catalyst improves the equilibrium yield.”

Actually: A catalyst speeds the forward and reverse reactions equally, so it changes neither Kp nor the position of equilibrium. The Haber iron catalyst delivers the same 15% conversion — just quickly enough to be worth building. Its effect is on time, not yield.

“The powers in Kp are found by experiment, like the orders in a rate equation.”

Actually: Opposite convention: equilibrium-constant powers come directly from the balanced equation's coefficients, while rate-equation orders come only from experimental data. Carrying the 3.1.9 habit into 3.1.10 (or vice versa) is the classic cross-topic error.

“Kp always has units of kPa or kPa².”

Actually: The units depend on the expression: kPa⁻² for ammonia synthesis, kPa for N₂O₄ dissociation, and none at all for H₂ + I₂ ⇌ 2HI, where the pressures cancel completely. Derive them every time, and write 'no units' when that is the answer.

ExamWhat examiners want

3.1.10 is examined on Paper 1 (and again in Paper 3's synoptic reach), usually as a multi-step calculation whose mark scheme runs M1–M2–M3: correct equilibrium moles, correct mole fractions or partial pressures, correct \(K_p\) with units. Run the two free self-checks every time — mole fractions sum to 1, partial pressures sum to the total — and most slips become caught slips before they cost anything.

Units are a mark, not a decoration. Derive them by substitution, cancel visibly, and when the moles of gas balance write 'no units' in so many words — leaving the line blank scores nothing. Quote the final value to the significant figures of the data (usually 2 or 3) but keep full calculator precision between steps.

The qualitative marks are pure discipline. Asked for the effect of pressure or of a catalyst on the value of \(K_p\): 'no effect' — two words, one mark, endlessly fumbled. Asked for the effect of temperature: give the direction AND tie it to the sign of \(\Delta H\) ('the forward reaction is exothermic, so raising the temperature decreases \(K_p\)'). And in industrial-context questions, which AQA marks as AO2 application, keep the three ledgers — value of \(K_p\), position of equilibrium, rate — in separate sentences, because the scheme awards them separately.