Many chemical reactions are reversible — products can react to re-form reactants. When the forward and reverse reaction rates become equal, the system reaches chemical equilibrium. Understanding equilibrium and Le Chatelier's principle is essential for industrial chemistry, especially the Haber and Contact processes.
7.1 Dynamic Equilibrium and the Equilibrium Constant
Definition
Chemical equilibrium
A reaction is in chemical equilibrium when the rate of the forward reaction equals the rate of the reverse reaction.
Definition
Dynamic equilibrium
There is dynamic equilibrium in a reversible reaction when the rate of the forward reaction equals the rate of the reverse reaction. Amounts of reactants and products remain constant.
Definition
Reversible reaction
A chemical reaction that can proceed in both the forward and reverse directions.
Definition
The equilibrium constant (Kc)
The equilibrium constant Kc is the ratio of the product of the molar concentrations of the products to the product of the molar concentrations of the reactants, each raised to the power of their stoichiometric coefficients, at a given temperature. For aA + bB ⇌ cC + dD: Kc = [C]^c[D]^d / [A]^a[B]^b. If Kc >> 1, products are favoured; if Kc << 1, reactants are favoured. Kc is constant at constant temperature — only temperature changes Kc.
Formula
Equilibrium constant expression
For the reaction aA + bB ⇌ cC + dD. Square brackets denote equilibrium concentrations in mol·dm⁻³. Exponents are the stoichiometric coefficients. Kc is dimensionless.
SI unit: dimensionless
Equilibrium is dynamic, not static — both the forward and reverse reactions are still occurring, but at equal rates so that the net concentrations remain constant. The equilibrium constant Kc has a fixed value at a given temperature. It tells us the position of equilibrium: if Kc >> 1 the equilibrium lies to the right (products favoured); if Kc << 1 it lies to the left (reactants favoured). Kc only changes when temperature changes — adding reactants, changing pressure or adding a catalyst does NOT change Kc.
Worked Example
For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the equilibrium concentrations are: [N₂] = 0,50 mol·dm⁻³, [H₂] = 0,30 mol·dm⁻³, [NH₃] = 0,20 mol·dm⁻³. Calculate Kc.
Given
- [N₂] = 0,50 mol·dm⁻³
- [H₂] = 0,30 mol·dm⁻³
- [NH₃] = 0,20 mol·dm⁻³
Find
Kc
Solution
- 1Write the Kc expression: Kc = [NH₃]² / ([N₂][H₂]³)
- 2Kc = (0,20)² / (0,50 × (0,30)³)
- 3Kc = 0,04 / (0,50 × 0,027)
- 4Kc = 0,04 / 0,0135
- 5Kc ≈ 2,96
Interpreting the Value of Kc
| Property | Kc value | What it means |
|---|---|---|
| Kc >> 1 (e.g. 10⁶) | Products strongly favoured | Equilibrium lies far to the right |
| Kc ≈ 1 | Reactants and products in similar amounts | Equilibrium near the middle |
| Kc << 1 (e.g. 10⁻⁶) | Reactants strongly favoured | Equilibrium lies far to the left |
Watch Out
Pure solids and pure liquids (including water as a solvent) are NOT included in the Kc expression. Only include species whose concentrations can change — gases and aqueous species.
Practice Question
Write the Kc expression for: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
(2 marks)
Practice Question
For the reaction H₂(g) + I₂(g) ⇌ 2HI(g), equilibrium concentrations are [H₂] = 0,10, [I₂] = 0,10 and [HI] = 0,60 mol·dm⁻³. Calculate Kc.
(4 marks)
7.2 Le Chatelier's Principle
Le Chatelier's principle provides a qualitative way to predict how an equilibrium system responds when conditions are changed. The system always responds in the direction that partially counteracts the imposed change. There are three main stresses to consider: (1) changing concentration of a reactant or product; (2) changing temperature; (3) changing pressure (for gaseous equilibria). A catalyst does NOT shift the equilibrium — it only helps the system reach equilibrium faster.
Le Chatelier's Principle — Summary of Effects
- Add a reactant: equilibrium shifts to the RIGHT (uses up extra reactant, produces more products).
- Remove a product: equilibrium shifts to the RIGHT (produces more product to replace what was removed).
- Add a product: equilibrium shifts to the LEFT.
- Increase temperature: equilibrium shifts in the ENDOTHERMIC direction (absorbs extra heat).
- Decrease temperature: equilibrium shifts in the EXOTHERMIC direction.
- Increase pressure (gas): equilibrium shifts toward the side with FEWER moles of gas.
- Decrease pressure (gas): equilibrium shifts toward the side with MORE moles of gas.
- Add catalyst: no shift in equilibrium position; Kc unchanged; equilibrium reached faster.
Worked Example
Consider: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ·mol⁻¹ (exothermic). Predict and explain the effect of each change: (a) increasing pressure, (b) increasing temperature, (c) removing NH₃.
Given
- Reaction is exothermic (ΔH < 0)
- Left side: 1 + 3 = 4 mol gas; right side: 2 mol gas
Find
Direction of shift for (a), (b), (c)
Solution
- 1(a) Increasing pressure: system shifts toward fewer moles of gas → RIGHT → more NH₃ produced.
- 2(b) Increasing temperature: reaction is exothermic → forward reaction releases heat → to counteract extra heat, system shifts in the endothermic direction (reverse) → LEFT → less NH₃ at equilibrium; Kc decreases.
- 3(c) Removing NH₃: system shifts RIGHT to replace the removed product → more NH₃ produced.
Exam Tip
Exam tip: temperature is the ONLY factor that changes Kc. All other stresses (concentration, pressure, catalyst) shift the position of equilibrium but leave Kc unchanged at constant temperature.
Worked Example
For: CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g) at equilibrium. The pressure is decreased. Predict the direction of the shift and explain.
Given
- Left side: 1 + 3 = 4 mol gas
- Right side: 1 + 1 = 2 mol gas
Find
Direction of shift
Solution
- 1Decreasing pressure: system shifts toward the side with MORE moles of gas to increase pressure.
- 2Left side has 4 mol gas > right side 2 mol gas.
- 3Equilibrium shifts to the LEFT.
Practice Question
Consider: 2NO₂(g) ⇌ N₂O₄(g) ΔH = −57 kJ·mol⁻¹. Predict the direction of shift when: (a) temperature is decreased, (b) more NO₂ is added, (c) pressure is decreased.
(6 marks)
Practice Question
A student adds a catalyst to an equilibrium mixture. State TWO effects and ONE non-effect of the catalyst.
(3 marks)
7.3 Industrial Equilibria — Haber and Contact Processes
Industrial processes use Le Chatelier's principle to maximise yield while keeping costs manageable. The Haber process produces ammonia (N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92 kJ·mol⁻¹) for fertilisers. High pressure (200 atm) favours the right side (fewer moles of gas). Low temperature maximises yield but the rate would be too slow — a compromise temperature of about 450 °C and an iron catalyst are used.
Haber Process — Conditions and Reasoning
| Property | Condition | Reason (Le Chatelier) |
|---|---|---|
| High pressure (~200 atm) | Shifts right (fewer gas moles → more NH₃) | Increases yield |
| Moderate temperature (~450 °C) | Compromise: low T favours exothermic right; but rate too slow at very low T | Balances yield and rate |
| Iron catalyst | No shift in equilibrium; speeds up attainment of equilibrium | Reduces cost and time |
| Continuous removal of NH₃ | Shifts right to replace removed product | Increases overall yield |
Worked Example
In the Contact process, SO₂ is oxidised: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = −196 kJ·mol⁻¹. Explain the choice of (a) high pressure and (b) a vanadium pentoxide catalyst.
Given
- Left: 2 + 1 = 3 mol gas; Right: 2 mol gas
- Reaction is exothermic
Find
Justification for conditions
Solution
- 1(a) High pressure: reaction has 3 mol gas on left, 2 on right. Higher pressure shifts equilibrium to the right → more SO₃ produced.
- 2(b) V₂O₅ catalyst: lowers activation energy → equilibrium reached faster without changing Kc or the position of equilibrium.
Real World
Real-world connection: the Haber process produces about 150 million tonnes of ammonia per year, which feeds approximately half the world's population through fertiliser production. Without it, modern agriculture could not support current population levels.
Practice Question
Explain why a very low temperature is NOT used in the Haber process even though it would give a higher equilibrium yield of ammonia.
(4 marks)
Practice Question
Does adding a catalyst to the Haber process increase the equilibrium yield of ammonia? Explain.
(3 marks)