Chemical Equilibrium and Le Chatelier's Principle

Many chemical reactions do not go to completion. Instead, they reach a state of dynamic equilibrium where the forward and reverse reactions occur at equal rates, and the concentrations of reactants and products remain constant over time. Understanding equilibrium -- and how to manipulate it -- is central to both A-level chemistry and industrial chemistry.

Dynamic Equilibrium

A system is in dynamic equilibrium when:

The rate of the forward reaction equals the rate of the reverse reaction.
The concentrations of reactants and products remain constant (but are not necessarily equal).
The reaction has not stopped -- both forward and reverse reactions continue to occur at the molecular level. It is "dynamic," not static.

For equilibrium to be established, two conditions must be met:

The system must be closed (no substances can enter or leave).
The temperature must be constant.

At the start of a reaction, only reactants are present and the forward rate is high. As products form, the reverse rate increases. Eventually, the two rates become equal and the system reaches equilibrium.

Le Chatelier's Principle

Le Chatelier's principle states that if a system at equilibrium is subjected to a change in conditions, the system will respond in a way that tends to counteract the change and restore equilibrium.

This is a qualitative prediction tool -- it tells you which direction the equilibrium shifts, but not by how much.

The following flowchart helps apply Le Chatelier's principle systematically:

flowchart TD
    A["Change applied<br/>to equilibrium"] --> B{"What type<br/>of change?"}
    B -->|"Concentration"| C{"Reactant or<br/>product?"}
    B -->|"Pressure"| D{"Compare moles<br/>of gas each side"}
    B -->|"Temperature"| E{"Exothermic or<br/>endothermic forward?"}
    B -->|"Catalyst"| F["No shift in position<br/>Reaches equilibrium faster<br/>Kc unchanged"]
    C -->|"Reactant added"| G["Shifts RIGHT<br/>(towards products)"]
    C -->|"Product removed"| G
    C -->|"Reactant removed"| H["Shifts LEFT<br/>(towards reactants)"]
    C -->|"Product added"| H
    D -->|"More moles on left"| I{"Pressure<br/>increase or<br/>decrease?"}
    D -->|"More moles on right"| J{"Pressure<br/>increase or<br/>decrease?"}
    D -->|"Equal moles"| K["No effect on position"]
    I -->|"Increase"| L["Shifts LEFT<br/>(fewer moles)"]
    I -->|"Decrease"| M["Shifts RIGHT<br/>(more moles)"]
    J -->|"Increase"| N["Shifts RIGHT<br/>(fewer moles)"]
    J -->|"Decrease"| O["Shifts LEFT<br/>(more moles)"]
    E -->|"Exothermic forward"| P{"Temperature<br/>increase or<br/>decrease?"}
    E -->|"Endothermic forward"| Q{"Temperature<br/>increase or<br/>decrease?"}
    P -->|"Increase"| R["Shifts LEFT<br/>(endothermic direction)<br/>Kc decreases"]
    P -->|"Decrease"| S["Shifts RIGHT<br/>(exothermic direction)<br/>Kc increases"]
    Q -->|"Increase"| T["Shifts RIGHT<br/>(endothermic direction)<br/>Kc increases"]
    Q -->|"Decrease"| U["Shifts LEFT<br/>(exothermic direction)<br/>Kc decreases"]

Effect of Concentration Changes

Consider the equilibrium: A + B ⇌ C + D

Increasing [A]: The system counteracts by using up some of the added A. The equilibrium shifts to the right (towards products), producing more C and D and consuming some B. The position of equilibrium shifts to the right, but the equilibrium constant Kc does not change (only temperature changes Kc).

Decreasing [C]: Removing product C disturbs the equilibrium. The system responds by producing more C -- the equilibrium shifts to the right.

Adding an inert gas at constant volume: No effect on equilibrium. The partial pressures (and concentrations) of the reacting species are unchanged.

Worked Example: Predicting Direction with the Reaction Quotient

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Kc = 0.50 mol⁻² dm⁶ at a certain temperature.

Current concentrations: [N₂] = 0.20, [H₂] = 0.30, [NH₃] = 0.10 mol dm⁻³

Qc = [NH₃]² / ([N₂][H₂]³) = (0.10)² / (0.20 × (0.30)³) = 0.010 / (0.20 × 0.027) = 0.010 / 0.0054 = 1.85

Since Qc (1.85) > Kc (0.50), there is too much product relative to equilibrium. The system shifts left (towards reactants) until Qc = Kc.

Effect of Pressure Changes (Gaseous Equilibria)

Pressure changes only affect equilibria involving gases where there is a different number of moles of gas on each side of the equation.

Consider: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Left side: 1 + 3 = 4 moles of gas. Right side: 2 moles of gas.

Increasing pressure: The system shifts towards the side with fewer moles of gas to reduce pressure. Here, it shifts to the right (2 moles < 4 moles), producing more NH₃.

Decreasing pressure: The system shifts towards more moles of gas. Here, it shifts to the left (4 moles > 2 moles).

If there are equal moles of gas on both sides (e.g., H₂ + I₂ ⇌ 2HI, which is 2 moles on each side), changing pressure has no effect on the equilibrium position.

Summary Table: Pressure Effects

Equilibrium	Moles left	Moles right	Increase P shifts...
N₂ + 3H₂ ⇌ 2NH₃	4	2	Right (→)
PCl₅ ⇌ PCl₃ + Cl₂	1	2	Left (←)
2SO₂ + O₂ ⇌ 2SO₃	3	2	Right (→)
H₂ + I₂ ⇌ 2HI	2	2	No effect
N₂O₄ ⇌ 2NO₂	1	2	Left (←)

Effect of Temperature Changes

Temperature changes are unique because they change both the position of equilibrium and the value of Kc.

Consider an exothermic forward reaction: A + B ⇌ C + D (ΔH = negative)

Increasing temperature: The system shifts in the endothermic direction (to the left, towards reactants) to absorb the extra heat. The yield of products decreases and Kc decreases.

Decreasing temperature: The system shifts in the exothermic direction (to the right, towards products) to release heat. The yield of products increases and Kc increases.

For an endothermic forward reaction (ΔH = positive):

Increasing temperature: shifts right (more products), Kc increases. Decreasing temperature: shifts left (fewer products), Kc decreases.

The key rule: increasing temperature always favours the endothermic direction.

Complete Summary Table: What Affects What?

Change	Position of equilibrium	Value of Kc
Increase [reactant]	Shifts right	Unchanged
Decrease [product]	Shifts right	Unchanged
Increase pressure	Shifts to fewer moles of gas	Unchanged
Increase temperature (exo forward)	Shifts left	Decreases
Increase temperature (endo forward)	Shifts right	Increases
Add catalyst	No shift	Unchanged
Add inert gas (constant V)	No shift	Unchanged

Effect of a Catalyst

A catalyst does not change the position of equilibrium. It speeds up both the forward and reverse reactions equally by lowering Ea for both. The equilibrium is reached faster, but the equilibrium concentrations are the same.

This is because the catalyst lowers the activation energy for the forward reaction by the same amount as for the reverse reaction (both travel over the same energy barrier from opposite directions).

The value of Kc is unchanged by a catalyst.

Industrial Applications

The Haber Process: N₂ + 3H₂ ⇌ 2NH₃ (ΔH = −92 kJ mol⁻¹)

Condition	Thermodynamic effect	Kinetic effect	Compromise
Pressure	High P shifts right (fewer moles)	Higher P = more expensive equipment	200 atm
Temperature	Low T shifts right (exothermic)	Low T = very slow rate	450 °C
Catalyst	No effect on yield	Faster equilibrium	Iron (Fe)
Remove NH₃	Shifts right (Le Chatelier)	--	Continuous removal + recycling

At 450 °C and 200 atm with an iron catalyst, the equilibrium yield is about 15% NH₃. Despite this modest yield, continuous removal of NH₃ and recycling of unreacted gases makes the process economically viable.

The Contact Process: 2SO₂ + O₂ ⇌ 2SO₃ (ΔH = −196 kJ mol⁻¹)

High pressure would favour the right (3 moles → 2 moles), but the conversion is already ~99% at 1--2 atm, so high pressure is not economically justified.
Low temperature would increase yield (exothermic forward), but the rate would be too slow.
Compromise temperature of 450 °C with V₂O₅ catalyst.

Common Mistakes to Avoid

"At equilibrium, the concentrations of reactants and products are equal." Wrong. They are constant, not equal. The ratio depends on Kc.
"Increasing pressure shifts all equilibria to the right." Wrong. It shifts towards fewer moles of gas, which could be left or right depending on the equation.
"A catalyst increases the yield." Wrong. A catalyst reaches the same equilibrium faster. It does not change the position.
"Changing concentration changes Kc." Wrong. Only temperature changes Kc. Changing concentration changes Q, and the system adjusts until Q = Kc again.
"Adding an inert gas always affects equilibrium." Wrong. At constant volume, inert gas has no effect (partial pressures unchanged). At constant pressure it can have an effect (by increasing volume).

Summary

Dynamic equilibrium is a balance between forward and reverse reaction rates. Le Chatelier's principle predicts qualitative shifts in response to changes in concentration, pressure, and temperature. Only temperature changes affect Kc. Catalysts reach equilibrium faster without changing its position. Industrial processes use compromise conditions that balance thermodynamic yield against kinetic rate.

A-Level Deep Dive: Chemical Equilibrium and Le Chatelier's Principle

Spec mapping

Edexcel 9CH0 specification, Topic 10 — Equilibrium I introduces the concept of dynamic equilibrium in reversible reactions, Le Chatelier's principle, and the qualitative effect of changes in concentration, pressure (for gas-phase systems) and temperature on the position of equilibrium and the rate at which equilibrium is reached (refer to the official specification document for exact wording). Topic 10 also covers application to industrial processes — Haber for ammonia, Contact for sulfuric acid — and the compromise between yield (favoured by certain conditions) and rate (favoured by other conditions). Examined principally on Paper 2 (Topics 11–19) and Paper 3 (synoptic with industrial chemistry). The data booklet does not list Le Chatelier's principle or industrial conditions — these must be memorised.

Worked example with full mark scheme

Question (8 marks):

For the equilibrium $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ , $\Delta H = -92$ kJ mol⁻¹.

(a) State, with reasoning, the effect on the position of equilibrium of: (i) increasing temperature; (ii) increasing pressure. (4)

(b) The Haber process operates at ~450 °C and ~200 atm. Explain why these conditions are a compromise. (4)

Solution with mark scheme:

(a) (i) M1 — increasing T favours the endothermic direction (Le Chatelier — system absorbs added heat). A1 — the forward reaction is exothermic ( $\Delta H = -92$ kJ mol⁻¹), so the reverse (endothermic) is favoured; the position shifts left (back toward N₂ + 3H₂); yield of NH₃ decreases.

(ii) M1 — counting moles of gas: 1 + 3 = 4 mol on the left; 2 mol on the right. A1 — increasing pressure favours the side with fewer gas moles to oppose the change; the position shifts right (toward NH₃); yield of NH₃ increases.

(b) M1 — yield of NH₃ is favoured by low T (exothermic forward) and high P (fewer moles on right). M1 — rate is favoured by high T (more molecules with energy ≥ $E_a$ ) and high P (higher [c] of all gases). M1 — these are conflicting requirements for T: yield wants low T, rate wants high T. Both want high P, but very high P is engineering-expensive (thicker pipes, more energy compression). A1 — 450 °C + 200 atm is the compromise: high enough T to give an acceptable rate (reach equilibrium quickly enough for industrial throughput), but not so high that yield is compromised; pressure high enough to push yield toward NH₃ but not so high as to be uneconomic.

Total: 8 marks (M5 A3, split as shown).

Specimen question modelled on the Edexcel 9CH0 Paper 2 format

Question (6 marks): For the equilibrium $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$ , $\Delta H = -197$ kJ mol⁻¹ (the rate-determining step in the Contact process for sulfuric acid).

(a) State, with reasoning, the effect of (i) removing SO₃ as it forms; (ii) lowering temperature; (iii) adding a catalyst, on the equilibrium position. (4)

(b) Industrial Contact uses ~450 °C, 1–2 atm and V₂O₅ catalyst. Justify the choice of pressure. (2)

Mark scheme decomposition by AO: