Equilibria

What you'll learn

This revision guide covers everything you need to know about equilibria for WJEC GCSE Chemistry. You'll learn about reversible reactions, dynamic equilibrium, and how to predict the effects of changing conditions using Le Chatelier's principle. Understanding equilibria is essential for industrial chemistry applications and will form the basis of questions worth 6-10 marks in your exam.

Key terms and definitions

Reversible reaction — a chemical reaction where the products can react to reform the original reactants, indicated by the symbol ⇌

Dynamic equilibrium — the state in a reversible reaction where the forward and backward reactions occur at the same rate, and the concentrations of reactants and products remain constant

Le Chatelier's principle — when a system at equilibrium is subjected to a change in conditions (temperature, pressure, or concentration), the position of equilibrium shifts to counteract that change

Position of equilibrium — the relative amounts of reactants and products in an equilibrium mixture; "to the right" means more products, "to the left" means more reactants

Closed system — a system where no substances can enter or leave, allowing equilibrium to be established

Catalyst — a substance that increases the rate of both forward and backward reactions equally without affecting the position of equilibrium

Exothermic reaction — a reaction that transfers energy to the surroundings, increasing the temperature of the surroundings

Endothermic reaction — a reaction that takes in energy from the surroundings, decreasing the temperature of the surroundings

Core concepts

Reversible reactions and the equilibrium symbol

Many chemical reactions are reversible, meaning they can proceed in both directions. When writing equations for reversible reactions, we use the ⇌ symbol instead of →.

Examples of reversible reactions include:

• Hydrated copper(II) sulfate ⇌ anhydrous copper(II) sulfate + water
  • CuSO₄·5H₂O(s) ⇌ CuSO₄(s) + 5H₂O(l)
• Ammonium chloride ⇌ ammonia + hydrogen chloride
  • NH₄Cl(s) ⇌ NH₃(g) + HCl(g)

The forward reaction proceeds from left to right (reactants → products). The backward reaction proceeds from right to left (products → reactants).

In an open system, products may escape (like steam evaporating or gases escaping to the atmosphere), preventing equilibrium from being reached. Only in a closed system can dynamic equilibrium be established.

Understanding dynamic equilibrium

When a reversible reaction is carried out in a closed system, it eventually reaches dynamic equilibrium. At this point:

• The forward and backward reactions continue to occur
• Both reactions proceed at exactly the same rate
• The concentrations of all reactants and products remain constant
• There is no overall change in the amounts of substances present
• The system appears static, but reactions are still happening at the molecular level

It's crucial to understand that equilibrium is dynamic, not static. Molecules are constantly reacting in both directions, but because the rates are equal, there's no net change in concentrations.

The position of equilibrium can lie anywhere between mostly reactants and mostly products, depending on the specific reaction and conditions.

Le Chatelier's principle: changes in concentration

Le Chatelier's principle allows us to predict how the position of equilibrium will shift when conditions change.

When concentration of a reactant is increased:

• The system responds by shifting the position of equilibrium to the right
• More product is formed to use up the added reactant
• This partially counteracts the increase in reactant concentration

When concentration of a product is increased:

• The position of equilibrium shifts to the left
• More reactant is formed to use up the added product

When concentration of a reactant is decreased:

• The position of equilibrium shifts to the left
• More reactant is formed to replace what was removed

When concentration of a product is decreased:

• The position of equilibrium shifts to the right
• More product is formed to replace what was removed

Le Chatelier's principle: changes in temperature

The effect of temperature changes depends on whether the forward reaction is exothermic or endothermic.

For an exothermic forward reaction (releases energy):

• Increasing temperature shifts equilibrium to the left (favours the backward reaction)
• The system counteracts the temperature rise by favouring the endothermic direction
• Decreasing temperature shifts equilibrium to the right (favours the forward reaction)

For an endothermic forward reaction (absorbs energy):

• Increasing temperature shifts equilibrium to the right (favours the forward reaction)
• The system counteracts the temperature rise by favouring the direction that absorbs energy
• Decreasing temperature shifts equilibrium to the left (favours the backward reaction)

Consider the reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = -92 kJ/mol

The forward reaction is exothermic (releases 92 kJ per mole of nitrogen reacted). Increasing temperature would shift the equilibrium to the left, producing less ammonia.

Le Chatelier's principle: changes in pressure

Pressure changes only affect equilibria involving gases. The key is counting the number of gas molecules on each side of the equation.

When pressure is increased:

• The position of equilibrium shifts towards the side with fewer gas molecules
• This partially counteracts the pressure increase by reducing the total number of molecules

When pressure is decreased:

• The position of equilibrium shifts towards the side with more gas molecules
• This partially counteracts the pressure decrease by increasing the total number of molecules

Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

• Left side: 1 + 3 = 4 gas molecules
• Right side: 2 gas molecules
• Increasing pressure shifts equilibrium to the right (towards fewer molecules)
• Decreasing pressure shifts equilibrium to the left (towards more molecules)

If there are equal numbers of gas molecules on both sides, changing pressure has no effect on the position of equilibrium.

Example: H₂(g) + Cl₂(g) ⇌ 2HCl(g)

• Both sides have 2 gas molecules
• Pressure changes have no effect on equilibrium position

The effect of catalysts on equilibrium

Catalysts are important in industry but don't affect the position of equilibrium:

• Catalysts increase the rate of both forward and backward reactions equally
• Equilibrium is reached more quickly
• The position of equilibrium (amounts of products and reactants at equilibrium) remains unchanged
• The same equilibrium mixture is achieved, just faster

This is economically important in industrial processes because reaching equilibrium faster means higher productivity without needing to compromise on yield by changing temperature or pressure.

Industrial applications: The Haber Process

The Haber Process synthesizes ammonia from nitrogen and hydrogen. It demonstrates the practical application of equilibrium principles:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = -92 kJ/mol

Conditions used industrially:

• Temperature: 450°C
• Pressure: 200 atmospheres
• Catalyst: iron

Equilibrium considerations:

Temperature: The forward reaction is exothermic, so lower temperatures favour ammonia production. However, at low temperatures the reaction rate is too slow. 450°C is a compromise — high enough for a reasonable reaction rate but not so high that yield becomes very poor.

Pressure: There are 4 gas molecules on the left and 2 on the right. Higher pressure favours ammonia production. 200 atmospheres is used as a compromise — high enough to give good yield but not so high that equipment costs and safety concerns become prohibitive.

Catalyst: Iron catalyst speeds up both forward and backward reactions equally, allowing equilibrium to be reached faster without changing the yield. This improves productivity.

In practice, unreacted nitrogen and hydrogen are recycled, which improves the overall efficiency and yield of the process.

Worked examples

Example 1: Predicting equilibrium shifts

Question: Sulfur dioxide reacts with oxygen to form sulfur trioxide in a reversible reaction:

2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = -197 kJ/mol

(a) State what happens to the position of equilibrium when: (i) the pressure is increased [2 marks] (ii) the temperature is decreased [2 marks]

(b) Explain why adding a vanadium(V) oxide catalyst does not affect the position of equilibrium. [2 marks]

Answer:

(a)(i)

• The equilibrium shifts to the right / towards the products [1]
• Because there are fewer molecules on the right side (3 molecules on left, 2 on right) [1]

(a)(ii)

• The equilibrium shifts to the right / towards the products [1]
• Because the forward reaction is exothermic, and lowering temperature favours the exothermic direction [1]

(b)

• The catalyst increases the rate of both forward and backward reactions equally [1]
• So the position of equilibrium remains the same / equilibrium is reached faster but with the same proportions [1]

Example 2: Applying Le Chatelier's principle

Question: Nitrogen dioxide exists in equilibrium with dinitrogen tetroxide:

2NO₂(g) ⇌ N₂O₄(g) ΔH = -58 kJ/mol

Nitrogen dioxide is a brown gas. Dinitrogen tetroxide is colourless.

A sealed gas syringe contains the equilibrium mixture and appears light brown.

(a) Explain what would be observed if the plunger is pushed in to increase the pressure. [3 marks]

(b) Explain what would be observed if the gas syringe is placed in a beaker of hot water. [3 marks]

Answer:

(a)

• The mixture would become lighter / less brown [1]
• Because increasing pressure shifts equilibrium to the right / towards N₂O₄ [1]
• There are fewer molecules on the right (2 molecules on left, 1 on right), so more colourless N₂O₄ is formed [1]

(b)

• The mixture would become darker / more brown [1]
• Because increasing temperature shifts equilibrium to the left / towards NO₂ [1]
• The forward reaction is exothermic, so increasing temperature favours the endothermic backward direction, producing more brown NO₂ [1]

Example 3: Industrial equilibrium

Question: Methanol can be produced from carbon monoxide and hydrogen:

CO(g) + 2H₂(g) ⇌ CH₃OH(g) ΔH = -91 kJ/mol

Industrial conditions: 250°C, 50-100 atmospheres, copper catalyst

Explain why these conditions are used. [6 marks]

Answer:

Temperature:

• The forward reaction is exothermic [1]
• So lower temperatures would increase the yield of methanol [1]
• But 250°C is used because at lower temperatures the reaction would be too slow / it's a compromise between yield and rate [1]

Pressure:

• There are more molecules on the left (3 total) than the right (1) [1]
• Higher pressure shifts equilibrium to the right / increases methanol yield [1]
• 50-100 atmospheres is high enough for good yield without excessive equipment costs [1]

(Note: Catalyst could also be mentioned — it increases the rate without affecting yield, allowing equilibrium to be reached faster)

Common mistakes and how to avoid them

• Confusing "equilibrium shifts right" with "reaction goes to completion" — At equilibrium, both reactions continue; shifting right just means the ratio changes to favour products more, not that all reactants are used up.

• Thinking catalysts increase yield — Catalysts speed up the rate of reaching equilibrium but don't change the position of equilibrium or the amounts of products formed. They affect rate, not position.

• Forgetting to count all molecules when considering pressure effects — You must count the total number of gaseous molecules on each side. Don't forget coefficients: N₂ + 3H₂ = 4 molecules total, not 2.

• Stating that "concentration increases" at equilibrium — At equilibrium, concentrations remain constant. Say "the equilibrium shifts to produce more..." or "the concentration was increased by adding..."

• Mixing up exothermic and endothermic effects — If forward is exothermic, increasing temperature shifts equilibrium left (favours the endothermic backward reaction). Write out "forward is exothermic" first to help you work it out.

• Applying pressure changes to reactions with solids or liquids — Pressure changes only significantly affect equilibria involving gases. Solids and liquids are essentially incompressible at normal pressures.

Exam technique for "Equilibria"

• "Explain" questions require reasoning — Don't just state the direction of shift. You must explain why it shifts that way (e.g., "to counteract the change" or "because there are fewer molecules on the right"). These are typically worth 2-3 marks.

• Link direction to observations — Questions often describe colour changes or other observations. Link the equilibrium shift to which substance increases/decreases to explain the observation (1 mark for direction, 1 mark for linking to observation).

• Show your working for molecule counting — When explaining pressure effects, write "3 molecules on left, 2 on right" to show you've counted correctly. This can earn method marks even if you make an error elsewhere.

• Industrial process questions combine multiple factors — You may need to explain temperature AND pressure AND catalyst in one question (6 marks). Structure your answer clearly with subheadings or separate paragraphs for each factor.

Quick revision summary

Reversible reactions reach dynamic equilibrium in closed systems when forward and backward rates are equal. Le Chatelier's principle predicts equilibrium shifts: increasing reactant concentration, decreasing temperature (for exothermic forward), or increasing pressure (if fewer product molecules) all shift equilibrium right. Catalysts speed both reactions equally without affecting position. Industrial processes like the Haber Process use compromise conditions balancing yield and rate while considering economics and safety.