Lesson 11: Chemical Equilibrium

Many reactions do not go to completion. Instead they reach chemical equilibrium: a state where the forward and reverse reactions continue at equal rates, so concentrations stop changing even though molecules keep reacting. It is a dynamic balance, not a standstill. This lesson quantifies equilibrium with the equilibrium constant, predicts how it responds to disturbance, and solves the table method that organises every equilibrium calculation.

The equilibrium constant and the ICE table

For aA + bB ⇌ cC + dD, the equilibrium constant is Kc = [C]^c[D]^d ÷ ([A]^a[B]^b), with concentrations at equilibrium, in mol/L. Kc is fixed at a given temperature: large Kc (≫1) means products dominate; small Kc means reactants dominate. Comparing the reaction quotient Q (same formula, current concentrations) with Kc predicts the direction of change: Q < Kc, the reaction moves forward; Q > Kc, it moves in reverse. Calculations are organised in an ICE table — Initial, Change, Equilibrium — one row per species, with changes written in terms of the reaction's stoichiometry and a single unknown x.

Worked example 1: computing Kc from equilibrium data

At equilibrium in the system N₂O₄(g) ⇌ 2NO₂(g), [N₂O₄] = 0.040 mol/L and [NO₂] = 0.200 mol/L. Kc = [NO₂]² ÷ [N₂O₄] = (0.200)² ÷ 0.040 = 0.0400 ÷ 0.040 = 1.00. Note the squared term: coefficients become powers, and forgetting them is the most common Kc error. The brown colour of NO₂ also makes this system visible — warm the tube and it darkens as the equilibrium shifts right (the forward reaction is endothermic), a colour change you can watch Le Châtelier's principle explain in the next section.

Worked example 2: an ICE table with a quadratic shortcut

Hydrogen and iodine react: H₂(g) + I₂(g) ⇌ 2HI(g), Kc = 54.3 at the temperature used. Starting with [H₂] = [I₂] = 1.00 mol/L and no HI, find the equilibrium concentration of HI. ICE: H₂ falls by x, I₂ falls by x, HI rises by 2x. At equilibrium: [H₂] = [I₂] = 1.00 − x, [HI] = 2x. Kc = (2x)² ÷ (1.00 − x)² = 54.3. Because the expression is a perfect square, take square roots instead of expanding: 2x ÷ (1.00 − x) = √54.3 = 7.37. Solve: 2x = 7.37 − 7.37x → 9.37x = 7.37 → x = 0.787. So [HI] = 2 × 0.787 = 1.57 mol/L, with [H₂] = [I₂] = 0.213 mol/L. Check by substitution: (1.57)² ÷ (0.213)² = 2.46 ÷ 0.0454 = 54.3 ✓. Always substitute back — it costs ten seconds and catches most slips.

Le Châtelier's principle

If a system at equilibrium is disturbed, it shifts to partially counteract the change. Adding a reactant shifts right; removing a product shifts right (which is why industrial processes continuously extract products). For gas equilibria, increased pressure shifts toward the side with fewer gas moles. Temperature is the exception that changes Kc itself: heating favours the endothermic direction. A catalyst speeds both directions equally — equilibrium arrives faster but does not move. The Haber process (Lesson 18) is the full orchestra of these effects played at once.

In the lab: the iron–thiocyanate equilibrium

Mixing pale yellow Fe³⁺ solution with colourless thiocyanate gives blood-red FeSCN²⁺: Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq). Adding Fe³⁺ deepens the red (shift right); adding a phosphate solution, which binds Fe³⁺ into a colourless complex, fades it (shift left) — Le Châtelier made visible. Quantitative versions measure absorbance to find Kc (a typical school value is around 100–200 L/mol at room temperature). The uncertainty discussion is rich: the equilibrium is temperature-sensitive, so a tube warmed in a student's hand measurably changes colour between readings — tubes belong in a water bath at fixed temperature, and readings taken at consistent times. The spectrophotometer contributes ±0.005 absorbance; pipetting ±0.05 mL per addition. But the deepest trap is an assumption: calculating Kc from absorbance assumes the *only* absorbing species is FeSCN²⁺ and that initial Fe³⁺ not converted stays as Fe³⁺ — if solutions are not made with Fe³⁺ in large excess, a second equilibrium (Fe(SCN)₂⁺ formation) quietly breaks the 1 : 1 stoichiometry and the computed Kc drifts with concentration. Spotting that the "constant" is not constant across dilutions is exactly how real analysts discover hidden chemistry.

Chemistry in the real world

Your blood runs on equilibrium: haemoglobin loads O₂ in the lungs (high O₂ pressure shifts Hb + 4O₂ ⇌ HbO₈ right) and unloads it in tissues (low pressure shifts left). Carbonated drinks hold CO₂(aq) ⇌ CO₂(g) under pressure in the bottle; open it and Q collapses, so the drink exhales its gas until flat.

Common mistakes

• Solids and pure liquids in Kc expressions. They are omitted — their "concentration" is constant — so CaCO₃(s) ⇌ CaO(s) + CO₂(g) has Kc = [CO₂], full stop.
• Claiming a catalyst shifts equilibrium. It changes the time to reach it, never the position.
• Coefficients forgotten as powers in Kc and Q (Worked example 1).
• Treating equilibrium as static. Both reactions continue; concentrations are constant because the rates match, a point examiners probe directly.

Exam Experience

1. For the equilibrium PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), an increase in total pressure (by compression) shifts the position: A) right B) left C) not at all D) toward more PCl₅ and more Cl₂ Answer: B. The left side has 1 mole of gas, the right side 2; compression favours the side with fewer moles. D is internally inconsistent.

2. Which change alters the value of Kc for an equilibrium? A) Adding a catalyst B) Increasing pressure C) Increasing temperature D) Adding more reactant Answer: C. Only temperature changes K; the others change concentrations or rates, and the system responds until the same K is restored.

3. At equilibrium in a 2.00 L vessel there are 0.40 mol NH₃, 0.20 mol N₂ and 0.60 mol H₂ for N₂ + 3H₂ ⇌ 2NH₃. Kc is: A) 3.70 B) 1.33 C) 14.8 D) 0.068 Answer: C. First convert to concentrations (divide by 2.00 L): [NH₃] = 0.20, [N₂] = 0.10, [H₂] = 0.30 mol/L. Kc = (0.20)² ÷ (0.10 × 0.30³) = 0.040 ÷ 0.0027 = 14.8. Distractor A uses the mole amounts without converting to concentrations; B forgets to cube [H₂]; D inverts the expression.

Constructed response (5 marks). Ethanoic acid ionises: CH₃COOH ⇌ H⁺ + CH₃COO⁻, Kc = 1.8 × 10⁻⁵. A solution is prepared with [CH₃COOH] = 0.500 mol/L and no products. (a) Set up an ICE table and write the equilibrium expression in terms of x. (b) Using the approximation that x is negligible beside 0.500, calculate [H⁺] and the pH. (c) Justify the approximation numerically. Scoring guide: (a) 1 mark: equilibrium values 0.500 − x, x, x; 1 mark: expression x² ÷ (0.500 − x) = 1.8 × 10⁻⁵. (b) 1 mark: x = √(1.8 × 10⁻⁵ × 0.500) = 3.0 × 10⁻³ mol/L; 1 mark: pH = −log(3.0 × 10⁻³) = 2.52. (c) 1 mark: x ÷ 0.500 = 0.6%, well under the usual 5% validity threshold, so neglecting x changes nothing at 3 s.f. Award full credit for the exact quadratic route giving the same result.

What the size of Kc tells you — and Kp

Kc's magnitude reads like a scoreline. Kc > 10³: the reaction effectively goes to completion (equilibrium mixture is nearly all products). Kc < 10⁻³: effectively no reaction. Between those poles lies the interesting chemistry of this lesson. Note that Kc carries no predictive power about *speed* — the equilibrium for 2H₂ + O₂ → 2H₂O has an astronomically large Kc, yet a hydrogen–oxygen mixture sits unreacted for years without a spark. Thermodynamics proposes; kinetics disposes (Lesson 12). For gas-phase work, engineers often use Kp, built from partial pressures rather than concentrations; the two are related by Kp = Kc(RT)^Δn, where Δn is the change in moles of gas across the equation. For N₂O₄ ⇌ 2NO₂, Δn = +1, so Kp and Kc differ by a factor of RT; for H₂ + I₂ ⇌ 2HI, Δn = 0 and Kp = Kc exactly — which is why Worked example 2's system is the textbook favourite.

Worked example 3: starting from products

Equilibria are reachable from both directions — a fact ICE tables handle by allowing negative changes. The esterification CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O has Kc = 4.0. A mixture starts with 1.00 mol/L of each reactant and 1.00 mol/L of each product in the same vessel. Q = (1.00 × 1.00) ÷ (1.00 × 1.00) = 1.0 < 4.0, so the system moves forward despite products being present. Let the forward change be x: equilibrium gives (1 + x)² ÷ (1 − x)² = 4.0; square-root both sides: (1 + x) ÷ (1 − x) = 2.0 → 1 + x = 2 − 2x → x = 0.333. Equilibrium: reactants 0.667 mol/L, products 1.33 mol/L. The twin morals: initial products do not mean the reverse reaction wins — Q decides — and the perfect-square structure (equal coefficients throughout) is a gift whenever it appears.

Temperature moves K: the van 't Hoff pattern in words

Only temperature changes K, and the direction is fixed by ΔH: for an exothermic forward reaction, raising temperature *decreases* K (the system treats heat as a product and retreats from it — the N₂O₄/NO₂ tube darkening in reverse: cooling a NO₂ tube pales it as N₂O₄ forms). For an endothermic forward reaction, raising temperature *increases* K. This is why the Haber converter's 450 °C is a genuine sacrifice (Lesson 18): at 450 °C ammonia's K is far smaller than at 200 °C, and the operators buy rate with equilibrium yield knowingly, then recover the loss with pressure and recycling. A quantitative habit worth forming: when a question changes temperature, answer in two clauses — what happens to K, and separately what happens to the rate. Conflating them is the commonest equilibrium error after missing powers in Kc expressions.