Lesson 13: Dynamic Equilibrium

Seal a reacting mixture in a closed vessel and something remarkable happens: the reaction does not run to completion and stop — it settles into a dynamic equilibrium (Reactivity 2.1), where the forward and reverse reactions continue at equal rates, so concentrations hold steady while molecules keep interconverting. The macroscopic stillness hides microscopic traffic. This lesson builds equilibrium thinking: the characteristics of the equilibrium state, the equilibrium constant K_c as the quantitative fingerprint of position, Le Chatelier's principle as the qualitative steering wheel, and the industrial compromises — Haber and Contact — where equilibrium yield negotiates with rate and cost.

The equilibrium constant

For aA + bB ⇌ cC + dD at equilibrium, K_c = [C]^c [D]^d ÷ ([A]^a [B]^b), concentrations in mol dm⁻³ at that temperature. The expression follows the balanced equation exactly: products over reactants, each raised to its stoichiometric power. Pure solids and liquids are omitted — their "concentrations" are effectively constant and are absorbed into K_c. K_c has a plain-language reading: K_c ≫ 1 means the equilibrium lies far to the right (products dominate); K_c ≪ 1, far to the left; K_c near 1, a genuine mixture. Crucially, K_c is fixed only by temperature: change concentrations, pressure or catalyst and the position shifts, but the system shifts precisely so that the ratio returns to the same K_c. A catalyst therefore cannot change yield — it speeds the arrival at equilibrium from both directions equally. Writing the expression for esterification, CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O, gives K_c = [ester][H₂O] ÷ ([acid][alcohol]) — no species omitted because all are liquids dissolved in the mixture (water here is a reactant-phase component, not the solvent excess case — a subtlety examiners test with heterogeneous examples, where pure solids/liquids genuinely are left out).

Le Chatelier: the system fights back

Le Chatelier's principle: when a system at equilibrium is disturbed, it shifts in the direction that partially counteracts the change. Add reactant → shift right, consuming some of it. Remove product → shift right, replacing it. Compress a gas mixture → shift towards the side with fewer gas moles. Raise the temperature → shift in the endothermic direction (treating heat as a reactant for endothermic forward reactions), which is also the only change that alters K_c itself: for an exothermic forward reaction, heating *decreases* K_c. Dilute a solution equilibrium → shift towards the side with more dissolved particles. The principle is qualitative but never wrong when applied to the stated change — its classic misuse is applying it to catalysts (no effect on position) and to inert-gas additions at constant volume (no concentration changes, no shift).

Worked example 1: equilibrium composition from an ICE table

The esterification above reaches equilibrium in a mixture that initially contains 1.00 mol of ethanoic acid and 1.00 mol of ethanol (and no products) in a fixed volume; at equilibrium 0.667 mol of ester has formed. Find K_c. Initial: acid 1.00, ethanol 1.00, ester 0, water 0. Change: −0.667, −0.667, +0.667, +0.667 (1 : 1 : 1 : 1 stoichiometry). Equilibrium: 0.333, 0.333, 0.667, 0.667 mol. Because all species share the same volume, mole ratios substitute directly for concentration ratios here. K_c = (0.667 × 0.667) ÷ (0.333 × 0.333) = 0.445 ÷ 0.111 = 4.01. The famous value for this reaction — K_c ≈ 4 means a respectable but incomplete conversion: even with perfect catalysis, a 1 : 1 feed leaves a third of the acid unreacted, which is why industry distils the ester out as it forms, dragging the position right by Le Chatelier. The ICE table's discipline — a row per species, changes tied by the mole ratio, equilibrium by addition — is the entire technique; most lost marks come from change rows that ignore the stoichiometry.

Worked example 2: the iron(III)–thiocyanate equilibrium — Le Chatelier in colour (practical)

Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq) sets up a blood-red equilibrium ideal for the bench: the colour intensity tracks [FeSCN²⁺] directly. Add Fe³⁺ → deeper red (shift right); add SCN⁻ → deeper red; add F⁻, which binds Fe³⁺ into a colourless complex → paler (Fe³⁺ removed, shift left); warm the tube → the colour fades if the forward reaction is exothermic, as it is here. To make it quantitative, prepare a series in a colorimeter: measure absorbance of the equilibrium mixture, and calibrate against solutions where excess Fe³⁺ drives SCN⁻ essentially fully to FeSCN²⁺, converting absorbance into concentration — then K_c follows from the ICE arithmetic above. Uncertainty thinking: absorbance reads to ±0.005, but the calibration assumes complete conversion in the standards (an assumption to state, and to test by doubling the excess Fe³⁺ and checking the absorbance plateaus); temperature must be controlled because K_c is a function of T — a water bath, not the bench, and record it. This is a complete IA in embryo: measurable variable, calibratable instrument, a constant to extract, and assumptions to interrogate.

Nature of Science — and the real world

Le Chatelier's principle is an empirical generalisation (1884) that thermodynamics later explained — K_c's temperature dependence is entropy and enthalpy negotiating (Lesson 22 makes this quantitative with ΔG° = −RT ln K). The Haber process is its monument: N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92 kJ mol⁻¹. Low temperature favours yield but strangles rate; high pressure favours yield (4 gas moles → 2) but costs money and steel. The industrial compromise — ~450 °C, 200 atm, iron catalyst, continuous removal of NH₃ by liquefaction — feeds roughly half the nitrogen in every human body alive today via fertiliser, and its inventor's other legacy (chemical weapons) keeps it the standard case study in the ethics of dual-use science.

Common mistakes

• Including solids or pure liquids in the K_c expression for heterogeneous equilibria.
• Writing K_c the wrong way up or forgetting the stoichiometric powers.
• Saying a catalyst "shifts equilibrium to the right" — it changes no position, only the time taken to reach it.
• Applying Le Chatelier to a temperature change but then claiming K_c is unchanged. Temperature is the one change that rewrites K_c.

IB Exam Experience

Question 1. For the equilibrium PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), what happens to the position of equilibrium and to K_c when the temperature is increased? (The forward reaction is endothermic.)

• (A) Position shifts right; K_c increases
• (B) Position shifts right; K_c unchanged
• (C) Position shifts left; K_c increases
• (D) Position unchanged; K_c increases

Question 2. Which change will increase the yield of ammonia in N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ mol⁻¹?

• (A) Increasing the temperature
• (B) Adding a catalyst
• (C) Increasing the pressure
• (D) Adding argon gas at constant volume

Question 3. For the reaction H₂(g) + I₂(g) ⇌ 2HI(g) at equilibrium, K_c = 54. [H₂] = [I₂] = 0.050 mol dm⁻³. What is [HI]?

• (A) 0.018 mol dm⁻³
• (B) 0.18 mol dm⁻³
• (C) 0.37 mol dm⁻³
• (D) 0.018 mol dm⁻⁶

Answer key. 1. (A) — heating favours the endothermic direction (right) and K_c rises for an endothermic forward reaction. 2. (C) — compression shifts towards fewer gas moles (2 vs 4); a catalyst changes no yield, heating shifts left. 3. (C) — K_c = [HI]² ÷ (0.050 × 0.050) = 54, so [HI]² = 0.135 and [HI] = 0.37 mol dm⁻³.

Structured question (Paper 2 style). Methanol is manufactured by CO(g) + 2H₂(g) ⇌ CH₃OH(g), ΔH = −91 kJ mol⁻¹. (a) Write the equilibrium constant expression, K_c, for this reaction. [1] (b) State and explain the effect on the position of equilibrium of (i) increasing the pressure at constant temperature, (ii) increasing the temperature at constant pressure. [4] (c) A catalyst is added to the equilibrium mixture. State the effect on the position of equilibrium and on the value of K_c. [1] Marking notes. (a) K_c = [CH₃OH] ÷ ([CO][H₂]²) [1]. (b) (i) Shifts right/towards products [1] — the product side has fewer moles of gas (1 vs 3), so compression is partially opposed by forming product [1]. (ii) Shifts left/towards reactants [1] — the forward reaction is exothermic, so heating favours the endothermic (reverse) direction [1]. (c) No effect on either the position or K_c — the catalyst speeds both directions equally [1].

IA idea: Determine the equilibrium constant for the iron(III)–thiocyanate reaction colorimetrically at several temperatures and use the trend to estimate the reaction's enthalpy change.

Two refinements: heterogeneous equilibria and a full K_c workout

For heterogeneous equilibria, the K_c expression includes only species whose concentrations can vary — gases and solutes. Calcium carbonate's decomposition, CaCO₃(s) ⇌ CaO(s) + CO₂(g), has K_c = [CO₂]: the equilibrium CO₂ concentration (and pressure) at a given temperature is fixed no matter how much solid is present, provided some of each solid exists. Add more CaCO₃ and nothing changes; that invariance is the experimental signature of a heterogeneous equilibrium and the working principle of the lime kiln (Lesson 22 will price it with ΔG). A companion subtlety: pure liquids *as solvents in great excess* are likewise absorbed into K — in dilute aqueous equilibria, water's 55.5 mol dm⁻³ never appears in expressions — but in the esterification of Lesson 13's worked example water is a stoichiometric participant in a non-aqueous mixture and stays in. Context decides; the question's phases tell you which.

Full workout, ammonia synthesis at a stated temperature: a 1.00 dm³ vessel starts with 1.00 mol N₂ and 3.00 mol H₂; at equilibrium it contains 0.400 mol NH₃. ICE by the 1 : 3 : 2 ratio: NH₃ formed 0.400 → N₂ consumed 0.200, H₂ consumed 0.600. Equilibrium amounts: N₂ 0.800, H₂ 2.40, NH₃ 0.400 mol (concentrations identical numerically, volume 1.00 dm³). K_c = [NH₃]² ÷ ([N₂][H₂]³) = 0.400² ÷ (0.800 × 2.40³) = 0.160 ÷ 11.06 = 0.0145 (dm⁶ mol⁻² in school convention; the IB treats K_c as dimensionless through standard-state ratios — quote the number, and do not let unit bookkeeping derail the chemistry). A small K at this temperature... yet industry still runs here, because the mixture is continuously harvested and recycled — equilibrium position is a snapshot, process design is the film.

Direction before equilibrium: Q with numbers

One numerical pass makes Q mechanical. For N₂O₄(g) ⇌ 2NO₂(g) at a temperature where K_c = 0.200, a mixture contains [N₂O₄] = 0.100 and [NO₂] = 0.200 mol dm⁻³. Q = [NO₂]² ÷ [N₂O₄] = 0.200² ÷ 0.100 = 0.400. Since Q (0.400) > K_c (0.200), the system carries too much product: the reverse reaction runs net, NO₂ is consumed, and the mixture drifts left until Q falls to 0.200. Halve the NO₂ concentration instead ([NO₂] = 0.100) and Q = 0.100 < K_c: forward reaction, drift right. The discipline: Q is calculated from *whatever concentrations exist now*, K is the target, and the shift is always the direction that moves Q towards K. Every "predict the direction" and most "explain what happens when more X is added" items in Paper 2 are this comparison in prose form — adding a reactant drops Q below K (the denominator grows... precisely: the reactant's concentration rises, Q falls), adding a product raises Q above K. Learn to run the comparison numerically even when the question only asks qualitatively, and the qualitative answer stops being a guess.