Equilibria
Le Chatelier, Kc and Kp
Le Chatelier's principle
When an equilibrium is disturbed, the position shifts to oppose the change. Raise the temperature and it shifts in the endothermic direction; raise the pressure and it shifts to the side with fewer gas moles. A catalyst causes no shift — equilibrium is simply reached faster.
The Haber process (N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92 kJ mol⁻¹) is the classic compromise: 400–450 °C, ~200 atm and an iron catalyst. A lower temperature would favour the yield, but the reaction becomes too slow to be worth it.
Equilibrium constant
Kc = [C]c[D]d / [A]a[B]bK changes only with temperature — for an exothermic reaction, raising T decreases K. Changes in concentration or pressure move the position of equilibrium, but K itself stays put. Kp is the gas-phase version built from partial pressures, where p(X) = mole fraction × total pressure.
3.1.6.2Kc from equilibrium data — the ICE routine
- Set out the ICE table: initial 1.00 / 1.00 / 0 · change −0.80 / −0.80 / +1.60 · equilibrium 0.20 / 0.20 / 1.60 mol.
- Substitute: Kc = [HI]² ÷ ([H₂][I₂]) = 1.60² ÷ (0.20 × 0.20) = 64 (no units — the concentrations cancel).
3.1.10Kp — the same idea with pressures
Mole fraction x = moles of gas ÷ total moles, and partial pressure p = x × P(total). Kp is built exactly like Kc but with the partial pressures of gases only — solids and pure liquids never appear in Kp expressions. Like Kc, Kp changes with temperature only: pressure shifts the position, not the constant.
3.1.6.3The three industrial equilibria
| Process | Reaction | Catalyst | Conditions | Why the compromise |
|---|---|---|---|---|
| Haber | N₂ + 3H₂ ⇌ 2NH₃ (ΔH −92) | Fe | 400–450 °C, ~200 atm | low T = better yield but too slow; ~15% per pass, gases recycled |
| Contact | 2SO₂ + O₂ ⇌ 2SO₃ | V₂O₅ | ~450 °C, ~2 atm | yield already ~99.5% near atmospheric pressure — high p not worth it |
| Ethanol | C₂H₄ + H₂O ⇌ C₂H₅OH | H₃PO₄ | 300 °C, 60–70 atm | ~5% per pass; unreacted ethene recycled |
Deep dive📚 The rest of the chapter, in full
Kc from scratch: the ICE method
Kc units: build them from the expression each time. For N₂ + 3H₂ ⇌ 2NH₃, Kc = [NH₃]²/([N₂][H₂]³) has units (mol dm⁻³)² ÷ (mol dm⁻³)⁴ = mol⁻² dm⁶.
The golden rule of K
Only temperature changes K. Concentration, pressure and catalysts shift the position of equilibrium (or just the speed of reaching it) while K stays fixed. For an exothermic forward reaction, raising T decreases K; for endothermic, K increases. This one sentence — "Kc is unchanged because temperature is constant" — earns a mark in almost every equilibrium question about adding reagent or compressing the mixture.
Le Chatelier, systematically
| Change | Position shifts… | K |
|---|---|---|
| Add a reactant | Right (to remove it) | Unchanged |
| Raise pressure (gases) | To the side with fewer moles of gas | Unchanged |
| Raise temperature | In the endothermic direction | Changes |
| Catalyst | No shift — equilibrium reached faster | Unchanged |
Industrial compromises
Haber (N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92 kJ mol⁻¹): low T favours yield but is slow; high P favours yield but is expensive and hazardous. Compromise: ~450 °C, ~200 atm, iron catalyst, and ammonia is liquefied out to pull the equilibrium right. Contact (2SO₂ + O₂ ⇌ 2SO₃, exothermic): ~450 °C, ~2 atm (already 99%+ conversion, so high pressure is not worth paying for), V₂O₅ catalyst. Arguments about conditions = rate vs yield vs cost, always all three.
Kp (Year 2)
For gases use partial pressures: p(X) = mole fraction × total pressure. Kp for N₂O₄ ⇌ 2NO₂ is p(NO₂)²/p(N₂O₄), units built the same way (here: kPa). Solids and pure liquids never appear in Kp or heterogeneous Kc expressions — their "concentration" is constant. Only temperature changes Kp; compressing the mixture shifts position until the same Kp is restored.
Extended🎓 Beyond the standard course
The reaction quotient Q is Kc's expression evaluated with current concentrations. If Q < K the system shifts forward; if Q > K it shifts in reverse; if Q = K it is at equilibrium. This single comparison replaces all hand-waving about "position".
van 't Hoff: a plot of ln K against 1/T is linear with gradient −ΔH°/R. It is the quantitative version of "heating an exothermic equilibrium lowers K", and the exact mirror of Arrhenius for kinetics.
Why solids never appear in K: their concentration (density ÷ molar mass) is fixed — formally, their activity is 1. So CaCO₃(s) ⇌ CaO(s) + CO₂(g) reduces to Kp = p(CO₂): one gas controls the whole equilibrium.
Small-K approximation: when K is tiny, assuming "x is negligible against c" turns quadratics into one-liners — the same 5% rule you use for weak acids.
Mastery vault🏛 Every remaining spec point, banked
Kc when the volume refuses to cancel
Kp from first principles
Partial pressure = mole fraction × total pressure, and the partial pressures must sum to the total — use that as a built-in check. Raising the total pressure does not change Kp: the system shifts (here toward PCl₅) until the same Kp is restored with new partial pressures.
Heterogeneous equilibria omit solids and pure liquids. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp = p(CO₂) alone — at a given temperature the CO₂ pressure above limestone is fixed no matter how much solid is present.
Interpreting K values
K ≫ 1: products dominate; K ≪ 1: barely reacts; K ≈ 1: comparable amounts. K says nothing about SPEED — H₂ + O₂ has a colossal K at 298 K yet a match is needed (kinetic barrier). Comparing K at two temperatures identifies ΔH's sign: K falling as T rises → forward reaction exothermic.
Ethanol two ways — the industrial comparison
| Hydration of ethene | Fermentation | |
|---|---|---|
| Equation | C₂H₄ + H₂O ⇌ C₂H₅OH | C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ |
| Conditions | 300 °C, 60–70 atm, H₃PO₄ catalyst | ~35 °C, yeast, anaerobic |
| Rate / purity | Fast, essentially pure product | Slow, dilute — needs fractional distillation |
| Feedstock | Finite (crude oil) | Renewable (sugars) |
| Process | Continuous, low labour | Batch, higher labour |
Unreacted ethene is recycled over the catalyst — the standard fix when a compromise position leaves conversion low (same trick as the Haber loop). "Carbon-neutral" claims for fermentation ethanol fail once farming, transport and distillation energy are counted — a routine evaluation point.
Kp from Mole Fractions, Deducing K Units, and What Really Shifts K
You can build Kp even when you are given only amounts and a total pressure. The bridge is the mole fraction.
Mole fraction of a gas = (moles of that gas) ÷ (total moles of all gases). All mole fractions in a mixture sum to 1. Its partial pressure is then p(X) = mole fraction of X × total pressure.
Worked Kp: For N2(g) + 3H2(g) ⇌ 2NH3(g), an equilibrium mixture holds 1.0 mol N2, 3.0 mol H2 and 0.50 mol NH3 at a total pressure of 200 kPa.
1. Total moles = 1.0 + 3.0 + 0.50 = 4.5 mol.
| Gas | Mole fraction | Partial pressure / kPa |
| N2 | 1.0/4.5 = 0.222 | 0.222 × 200 = 44.4 |
| H2 | 3.0/4.5 = 0.667 | 0.667 × 200 = 133.3 |
| NH3 | 0.50/4.5 = 0.111 | 0.111 × 200 = 22.2 |
(Check: 44.4 + 133.3 + 22.2 = 200 kPa. ✓)
2. Substitute: Kp = p(NH3)² / [ p(N2) × p(H2)³ ]
Kp = (22.2)² / (44.4 × 133.3³) = 492.8 / (1.052 × 10⁸) = 4.68 × 10⁻⁶
3. Units: kPa² ÷ (kPa × kPa³) = kPa²⁄kPa⁴ = kPa⁻². So Kp = 4.68 × 10⁻⁶ kPa⁻².
Units are never assumed — deduce them by cancelling the units in the K expression, exactly as above.
Kc units, two cases.
Case A — N2 + 3H2 ⇌ 2NH3: Kc = [NH3]² / ([N2][H2]³). Units = (mol dm⁻³)² ÷ (mol dm⁻³)⁴ = (mol dm⁻³)⁻² = mol⁻² dm⁶.
Case B — H2 + I2 ⇌ 2HI: Kc = [HI]² / ([H2][I2]). Units = (mol dm⁻³)² ÷ (mol dm⁻³)² = they cancel, so Kc has no units.
Catalyst. A catalyst speeds the forward and reverse reactions equally, so it does not change Kc, Kp or the position of equilibrium — it only reduces the time taken to reach the same equilibrium.
| Change | Effect on position | Effect on K |
| Concentration | Shifts to oppose the change | No change |
| Pressure (volume) | Shifts toward fewer gas moles | No change |
| Temperature | Shifts in endothermic direction on heating | Changes |
| Catalyst | No change | No change |
Only temperature changes the value of K. Concentration and pressure changes move the position of equilibrium so that Q returns to the same K.
Heterogeneous equilibria. Pure solids and pure liquids have effectively constant concentration, so they are omitted from Kc and Kp. For CaCO3(s) ⇌ CaO(s) + CO2(g), only the gas remains: Kp = p(CO2), units kPa.
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