Equilibrium

Chemistry 1110 · CCEA A Level · 10 min read · Updated: 31 Aug 2026

Welcome to Equilibrium!

Welcome to one of the most exciting and fundamental topics in AS Chemistry! Have you ever wondered why some chemical reactions don't go to completion, leaving a mixture of both reactants and products? That is the magic of chemical equilibrium.

In this chapter, we will break down what dynamic equilibrium really means, explore how systems respond to change using Le Chatelier's Principle, look at how industry balances yield against cost, and master the calculations behind the equilibrium constant, \(K_c\). Don't worry if some of the maths or ideas seem unfamiliar at first — we will take it step-by-step with clear analogies, real-world examples, and handy tips to help you score top marks in your CCEA AS 2 exam.

1. Reversible Reactions and Dynamic Equilibrium

What is a Reversible Reaction?

In many reactions, reactants turn into products and stay that way (think of baking a cake or burning a match). However, many chemical reactions are reversible. This means the products can react together to reform the original reactants.

We represent reversible reactions using a special two-way equilibrium arrow: \(\rightleftharpoons\).

General equation:
\(\text{Reactants} \rightleftharpoons \text{Products}\)

What is Dynamic Equilibrium?

Imagine you are walking up a "down" escalator. If you walk upwards at the exact same speed that the escalator is moving downwards, what happens? To someone watching from the side, you stay in the exact same spot! You are still moving, and the escalator is still moving, but your overall position does not change.

This is exactly what happens in a chemical reaction at dynamic equilibrium:

• The forward reaction and the reverse reaction are happening at the exact same rate.
• The concentrations of reactants and products remain constant (unchanging), though not necessarily equal.

Essential Conditions for Dynamic Equilibrium

For dynamic equilibrium to be established and maintained, two conditions must be met:

1. A Closed System: Nothing can enter or leave the reaction container. If gases escape or heat is lost to an open environment without balance, equilibrium cannot be achieved.
2. Constant Macroscopic Properties: Observable properties like colour intensity, pressure, and temperature remain constant over time.

Did you know? The word dynamic comes from the Greek word for "power" or "motion". It reminds us that the reaction hasn't stopped — molecules are still actively reacting in both directions at equal speeds!

Key Takeaway: Dynamic equilibrium occurs in a closed system when the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant.

2. Le Chatelier's Principle

The Concept

In 1884, French chemist Henri Le Chatelier noticed that equilibrium systems behave like stubborn teenagers: whenever you try to change something, they do everything they can to oppose that change!

Le Chatelier's Principle: If a system at equilibrium is subjected to a change in conditions (such as temperature, pressure, or concentration), the position of equilibrium will shift in the direction that tends to oppose that change.

1. Effect of Concentration Changes

• Increasing the concentration of a reactant: The system opposes this by using up the extra reactant. The equilibrium shifts to the right (forward direction), producing more products.
• Decreasing the concentration of a product: The system opposes this by replacing the removed product. The equilibrium shifts to the right.
• Increasing the concentration of a product: The system opposes this by using up the product. The equilibrium shifts to the left (reverse direction), forming more reactants.

2. Effect of Pressure Changes (Gaseous Systems Only)

Pressure is caused by gas particles colliding with the walls of the container. More moles of gas mean higher pressure; fewer moles of gas mean lower pressure.

• Increasing pressure: The system opposes the increase by shifting to the side with fewer moles of gas to reduce the pressure.
• Decreasing pressure: The system opposes the decrease by shifting to the side with more moles of gas to increase the pressure.
• Equal moles of gas on both sides? A change in pressure will have no effect on the position of equilibrium.

Example:
\(\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}\)
• Left side: \(1 + 3 = 4\text{ moles of gas}\)
• Right side: \(2\text{ moles of gas}\)
If we increase pressure, the equilibrium shifts to the right (the side with fewer gas moles: 2 vs 4).

3. Effect of Temperature Changes

To predict the effect of temperature, always check the enthalpy change (\(\Delta H\)) for the forward reaction:

• Exothermic forward reaction (\(\Delta H\) is negative, \(-\)): Releases heat energy.
• Endothermic forward reaction (\(\Delta H\) is positive, \(+\)): Absorbs heat energy.

• Increasing temperature: The system opposes the added heat by absorbing it. Equilibrium shifts in the endothermic direction.
• Decreasing temperature (cooling): The system opposes the cooling by releasing heat. Equilibrium shifts in the exothermic direction.

Example:
\(2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)} \quad \Delta H = -197\text{ kJ mol}^{-1}\)
The forward reaction is exothermic (\(-\)). If we raise the temperature, the equilibrium shifts in the endothermic direction (to the left), reducing the yield of \(\text{SO}_3\).

4. Effect of a Catalyst

A catalyst increases the rate of both the forward and reverse reactions equally by providing an alternative reaction pathway with a lower activation energy (\(E_a\)).

• A catalyst does NOT alter the position of equilibrium.
• A catalyst does NOT change the yield of products.
• A catalyst simply allows the system to reach equilibrium faster.

Quick Review: Le Chatelier Summary Table
• Increase reactant concentration: Shifts Right (towards products)
• Increase pressure: Shifts to side with fewer gas molecules
• Increase temperature: Shifts in endothermic direction (\(+\Delta H\))
• Add a catalyst: No shift in position (reaches equilibrium faster)

Key Takeaway: The position of equilibrium shifts to counteract changes in concentration, pressure, and temperature. Catalysts speed up both directions equally and never alter the equilibrium position.

3. Industrial Applications and Compromise Conditions

In chemical manufacturing, chemical engineers want two things: a high yield of product and a fast rate of reaction, all at an economically viable cost. Often, equilibrium conditions and reaction rates conflict, requiring a compromise.

The Haber Process

The Haber Process synthesises ammonia from nitrogen and hydrogen:
\(\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)} \quad \Delta H = -92\text{ kJ mol}^{-1}\)

1. Temperature Analysis:
• Equilibrium consideration: Because the forward reaction is exothermic (\(-92\text{ kJ mol}^{-1}\)), a low temperature favors the forward reaction and gives a higher equilibrium yield of \(\text{NH}_3\).
• Kinetic consideration: At low temperatures, the rate of reaction is far too slow because few particles have energy \(\ge E_a\).
• The Compromise: A temperature of \(400\text{--}450^\circ\text{C}\) is chosen. This gives a reasonable yield in an acceptable time frame.

2. Pressure Analysis:
• Equilibrium consideration: There are 4 moles of gas on the left and 2 moles of gas on the right. A high pressure shifts equilibrium to the right, giving a high yield of \(\text{NH}_3\). High pressure also increases particle collision frequency, increasing the rate.
• Economic & safety consideration: Generating and containing extremely high pressures requires thick-walled steel pipes, expensive pumps, and vast amounts of electrical energy, creating significant safety hazards.
• The Compromise: A pressure of around \(200\text{ atm}\) (\(20\text{ MPa}\)) is used to balance high yield against capital and running costs.

3. The Catalyst:
An iron (Fe) catalyst is used. This allows the reaction to proceed at an acceptable rate at the moderate temperature of \(400\text{--}450^\circ\text{C}\).

Key Takeaway: Industrial conditions (like those in the Haber Process) represent a carefully chosen compromise between equilibrium yield, reaction rate, operating costs, and safety.

4. The Equilibrium Constant (\(K_c\))

Homogeneous vs. Heterogeneous Equilibria

• Homogeneous equilibrium: All reactants and products are in the same physical state (e.g., all gases or all aqueous solutions).
• Heterogeneous equilibrium: Reactants and products are present in at least two different physical states (e.g., a solid reacting with a gas).

Writing the \(K_c\) Expression

For any homogeneous equilibrium system at a constant temperature:
\(a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}\)

The equilibrium constant expression in terms of concentration, \(K_c\), is written as:
\(K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}\)

• Square brackets \([\,]\) denote equilibrium concentration in \(\text{mol dm}^{-3}\).
• Products are always in the numerator (top), reactants in the denominator (bottom).
• The balancing coefficients in the equation become the powers (indices).

How to Determine the Units of \(K_c\)

The units of \(K_c\) change depending on the number of concentration terms in the numerator and denominator. Always work them out step-by-step!

Example: Determine the units for \(K_c\) in the reaction:
\(\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}\)

Step 1: Write the expression:
\(K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}\)

Step 2: Substitute the units \(\text{mol dm}^{-3}\):
\(\text{Units} = \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^3} = \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})^4} = \frac{1}{(\text{mol dm}^{-3})^2}\)

Step 3: Invert the powers:
\(\text{Units} = (\text{mol dm}^{-3})^{-2} = \mathbf{\text{mol}^{-2}\text{ dm}^{6}}\)

Step-by-Step \(K_c\) Calculation (The RICE Method)

A classic exam calculation asks you to find \(K_c\) from initial moles and one equilibrium value. Use a RICE table:

• R - Reaction equation
• I - Initial moles
• C - Change in moles
• E - Equilibrium moles (then divide by volume in \(\text{dm}^3\) to find concentrations!)

Worked Example:
\(0.20\text{ mol}\) of \(\text{SO}_2\) and \(0.20\text{ mol}\) of \(\text{O}_2\) are placed in a sealed container of volume \(2.0\text{ dm}^3\) at a constant temperature. At equilibrium, \(0.10\text{ mol}\) of \(\text{SO}_3\) is present. Calculate \(K_c\).

\(2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)}\)

Step 1: Determine equilibrium moles
• Initial moles: \(\text{SO}_2 = 0.20\), \(\text{O}_2 = 0.20\), \(\text{SO}_3 = 0.00\)
• Change: To form \(0.10\text{ mol}\) of \(\text{SO}_3\), we need \(0.10\text{ mol}\) of \(\text{SO}_2\) (ratio 2:2) and \(0.05\text{ mol}\) of \(\text{O}_2\) (ratio 1:2).
• Equilibrium moles:
\(\text{SO}_2 = 0.20 - 0.10 = 0.10\text{ mol}\)
\(\text{O}_2 = 0.20 - 0.05 = 0.15\text{ mol}\)
\(\text{SO}_3 = 0.10\text{ mol}\)

Step 2: Calculate equilibrium concentrations (\([\text{X}] = \frac{\text{moles}}{\text{Volume}}\))
\([\text{SO}_2] = \frac{0.10}{2.0} = 0.050\text{ mol dm}^{-3}\)
\([\text{O}_2] = \frac{0.15}{2.0} = 0.075\text{ mol dm}^{-3}\)
\([\text{SO}_3] = \frac{0.10}{2.0} = 0.050\text{ mol dm}^{-3}\)

Step 3: Substitute concentrations into the \(K_c\) expression
\(K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]} = \frac{(0.050)^2}{(0.050)^2 \times 0.075} = \frac{1}{0.075} \approx \mathbf{13.3\text{ mol}^{-1}\text{ dm}^3}\)

What Factors Affect the Value of \(K_c\)?

Here is one of the most important rules to remember for your exam:

• TEMPERATURE is the ONLY factor that changes the numerical value of \(K_c\).
• If temperature change causes the equilibrium to shift to the right (more products), \(K_c\) increases.
• If temperature change causes the equilibrium to shift to the left (more reactants), \(K_c\) decreases.
• Concentration changes, pressure changes, and catalysts have NO effect on the value of \(K_c\).

Key Takeaway: \(K_c\) gives the ratio of products to reactants at equilibrium. Its units depend on the stoichiometry, and its numerical value is altered only by changing the temperature.

5. Common Mistakes to Avoid in Exams

• Mistake 1: Thinking the reaction has stopped at equilibrium.
Correction: The reaction is dynamic! Molecules are still reacting in both directions at equal rates.

• Mistake 2: Forgetting to divide moles by volume.
Correction: \(K_c\) uses concentrations in \(\text{mol dm}^{-3}\). Always divide equilibrium moles by the volume (in \(\text{dm}^3\)) before plugging them into the \(K_c\) expression (unless the total moles of reactants equal total moles of products, where volumes cancel out).

• Mistake 3: Stating that a catalyst increases the yield or alters \(K_c\).
Correction: A catalyst only increases the rate of reaching equilibrium. It has zero effect on yield and zero effect on \(K_c\).

• Mistake 4: Claiming that changing pressure or concentration changes \(K_c\).
Correction: Changing concentration or pressure causes the position of equilibrium to shift, but the system adjusts so that the ratio remains equal to the same constant value of \(K_c\). Only temperature changes \(K_c\).

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is a reversible reaction?

A reversible reaction is one where the products can react together to reform the original reactants, represented using the equilibrium arrow \(\rightleftharpoons\).

What is dynamic equilibrium and what are its two key characteristics?

Dynamic equilibrium is a state where the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant.

What are the two essential conditions required to establish dynamic equilibrium?

A closed system (nothing enters or leaves) and constant macroscopic properties (such as temperature, pressure, and colour intensity).

What does Le Chatelier's Principle state?

If a system at equilibrium is subjected to a change in conditions (temperature, pressure, or concentration), the position of equilibrium shifts to oppose that change.

How does changing reactant or product concentration affect the position of equilibrium?

Increasing reactant concentration (or removing product) shifts equilibrium to the right; increasing product concentration (or removing reactant) shifts equilibrium to the left.

How does changing the pressure affect a gaseous equilibrium system?

Increasing pressure shifts the equilibrium to the side with fewer moles of gas; decreasing pressure shifts it to the side with more moles of gas.

How does changing temperature affect the position of equilibrium?

Increasing temperature shifts the equilibrium in the endothermic direction; decreasing temperature shifts it in the exothermic direction.

What effect does adding a catalyst have on the position of equilibrium and the rate of reaching it?

A catalyst does not alter the position of equilibrium or yield; it only speeds up both forward and reverse reactions equally so equilibrium is reached faster.

Why are compromise conditions of \(400\text{--}450^\circ\text{C}\) and \(200\text{ atm}\) used in the Haber Process?

They balance a high equilibrium yield against a sufficiently fast reaction rate, economic costs, and safety considerations.

What is the difference between a homogeneous and a heterogeneous equilibrium?

In a homogeneous equilibrium, all species are in the same physical state; in a heterogeneous equilibrium, species are present in more than one physical state.

How is the equilibrium constant expression \(K_c\) written for a general reaction \(a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}\)?

\(K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}\), where square brackets denote equilibrium concentrations in \(\text{mol dm}^{-3}\).

How are the units of \(K_c\) determined?

By substituting the unit \(\text{mol dm}^{-3}\) for each concentration term into the \(K_c\) expression and simplifying the resulting powers.

What are the steps of the RICE method used to calculate \(K_c\)?

Write the Reaction (R), list Initial moles (I), calculate Change in moles (C), find Equilibrium moles (E), convert to concentrations by dividing by volume, and substitute into \(K_c\).

Which factor(s) can alter the numerical value of \(K_c\)?

Temperature is the ONLY factor that changes the value of \(K_c\); changes in concentration, pressure, and catalysts have no effect.

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