Cambridge O Level Chemistry: Rates & Equilibrium

Mastering Chemical Dynamics

A comprehensive, premium study guide covering Sections 5 and 6 of the Cambridge O Level Chemistry (5070) Syllabus.

How to use this guide

This main page acts as a syllabus resume. Open the hamburger menu () in the top left to access the interactive sections, deep-dive notes, dynamic graphs, and the 150-question practice database.

5.1 Chemical Energetics

Explores the transfer of thermal energy. Covers Endothermic and Exothermic reactions, Activation Energy (E_a), and the energy mechanics of bond breaking vs. bond formation.

Energy Profile Diagrams
Bond Energy Calculations

6.2 Rate of Reaction

Analyzes the speed of chemical reactions using Collision Theory. Understand how changing conditions affects the frequency of successful collisions.

Concentration, Pressure, Surface Area
Temperature and Catalysts

6.3 Equilibrium Basics

Introduces reversible reactions and closed systems. Learn how the rate of the forward reaction equals the reverse reaction, keeping concentrations constant.

Hydrated & Anhydrous Salts
Dynamic Equilibrium defined

Industrial Applications

Applies Le Chatelier's Principle to predict equilibrium shifts. Details the exact compromise conditions used in global manufacturing.

Haber Process (Ammonia)
Contact Process (Sulfuric Acid)

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5.1 Exothermic and Endothermic Reactions

Definitions & Fundamentals

Exothermic Reaction

A reaction that transfers thermal energy to the surroundings. The temperature of the surroundings increases. The enthalpy change (ΔH) is negative.

Examples: Combustion, Neutralisation, Respiration.

Endothermic Reaction

A reaction that takes in thermal energy from the surroundings. The temperature of the surroundings decreases. The enthalpy change (ΔH) is positive.

Examples: Thermal Decomposition, Photosynthesis.

Exact Syllabus Wording (5.1.6):

"State that bond breaking is an endothermic process and bond making is an exothermic process and explain the enthalpy change of a reaction in terms of bond breaking and bond making."

Bond Dynamics: Breaking vs. Making

Chemical reactions involve two simultaneous energy steps:

Step 1: Bond Breaking (Endothermic)
Energy MUST be supplied to break the strong electrostatic forces holding reactant molecules together. It always requires an input of energy (+ve).
Step 2: Bond Making (Exothermic)
Energy is ALWAYS released when new bonds are formed to create the products (-ve).

Overall ΔH = (Energy taken in to break bonds) - (Energy released making bonds)

Reaction Pathway Diagrams (Energy Profiles)

Exothermic Profile

Endothermic Profile

Practice Questions (30)

6.2 Rate of Reaction

Collision Theory

For a reaction to occur, particles must collide. However, not all collisions are successful. A successful collision requires:

Activation Energy (E_a)

Particles must collide with kinetic energy greater than or equal to E_a to break initial bonds.

Correct Orientation

The reactive parts of the molecules must come into contact during the collision.

Interactive Kinetics Simulator

Experiment with the factors below to see how they affect the rate of gas production over time. The steeper the initial gradient, the faster the reaction.

Temperature 20°C

Increases kinetic energy AND frequency of collisions.

Concentration 1.0 mol/dm³

Increases number of particles per unit volume.

Surface Area (Solid) Chunks

ChunksChipsPowder

Add Catalyst

Provides an alternative pathway with a lower E_a.

Adjust sliders to see the initial gradient change.

Practice Questions (30)

6.3 Equilibrium

Reversible Reactions

Some chemical reactions are reversible, denoted by the symbol ⇌. This means products can react to reform the original reactants.

Classic Examples (Hydrated Salts)

Copper(II) Sulfate
CuSO₄ • 5H₂O (blue) ⇌ CuSO₄ (white) + 5H₂O Heating drives off water (forward = endothermic). Adding water reverses it (exothermic).
Cobalt(II) Chloride
CoCl₂ • 6H₂O (pink) ⇌ CoCl₂ (blue) + 6H₂O

Dynamic Equilibrium

Exact Syllabus Definition:

A reversible reaction in a closed system is at equilibrium when:
(a) the rate of the forward reaction is equal to the rate of the reverse reaction
(b) the concentrations of reactants and products are no longer changing.

Practice Questions (30)

Le Chatelier's Principle

The Principle Defined

"If a change is made to the conditions of a system at dynamic equilibrium, the position of equilibrium moves to oppose that change."

How it Works: Forward & Backward Dynamics

When you apply a change, you temporarily break the equilibrium (Rates are no longer equal). The system reacts by favouring one direction over the other until a new equilibrium is established where rates are equal again.

1. Changing Temperature

Increase Temperature:

The system opposes by absorbing excess heat. It favours the Endothermic direction.

Decrease Temperature:

The system opposes by generating heat. It favours the Exothermic direction.

2. Changing Pressure (Gases Only)

Increase Pressure:

System opposes by reducing pressure. It shifts towards the side with FEWER moles of gas.

Decrease Pressure:

System opposes by increasing pressure. It shifts towards the side with MORE moles of gas.

3. Changing Concentration

Add Reactant: System shifts right (forward) to consume it.

Remove Product: System shifts right (forward) to replace it. (Common industry trick to maximize yield).

4. Adding a Catalyst

NO EFFECT on Position of Equilibrium.

A catalyst speeds up both the forward AND backward reactions equally. It only allows the system to reach equilibrium faster.

Practice Questions (30)

Industrial Processes

In industry, time is money. We must balance thermodynamics (getting a high yield via Le Chatelier) with kinetics (getting the product quickly) and economics (cost of high pressure/heat and safety). These are called Compromise Conditions.

The Haber Process

Industrial production of Ammonia (NH₃)

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = -ve

Sources:
N₂ from air. H₂ from reacting methane with steam.
Pressure: 20,000 kPa (200 atm)
High pressure favours the right (4 moles → 2 moles) for higher yield, and increases rate. Compromise: Too high is dangerously explosive and expensive to pipe.
Temperature: 450°C
Low temp favours right (exothermic) for high yield, but reaction would be too slow. 450°C is the compromise for a decent yield at a fast rate.
Catalyst: Iron
Speeds up reaction, allowing the lower 450°C temp to be viable.

The Contact Process

Conversion of SO₂ to SO₃ (for Sulfuric Acid)

2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = -ve

Sources:
SO₂ from burning sulfur or roasting sulfide ores. O₂ from air.
Pressure: 200 kPa (2 atm)
3 moles → 2 moles. High pressure increases yield, but yield is already ~99% at 2 atm. Not worth the extreme cost of high pressure tanks.
Temperature: 450°C
Forward is exothermic. Compromise temp ensures reaction is fast enough without killing the yield.
Catalyst: Vanadium(V) oxide (V₂O₅)
Required to make the rate acceptable at 450°C.

Navigation

Mastering Chemical Dynamics

How to use this guide

5.1 Chemical Energetics

6.2 Rate of Reaction

6.3 Equilibrium Basics

Industrial Applications

5.1 Exothermic and Endothermic Reactions

Definitions & Fundamentals

Exothermic Reaction

Endothermic Reaction

Bond Dynamics: Breaking vs. Making

Reaction Pathway Diagrams (Energy Profiles)

Exothermic Profile

Endothermic Profile

Practice Questions (30)

6.2 Rate of Reaction

Collision Theory

Activation Energy (Ea)

Correct Orientation

Interactive Kinetics Simulator

Practice Questions (30)

6.3 Equilibrium

Reversible Reactions

Classic Examples (Hydrated Salts)

Dynamic Equilibrium

Practice Questions (30)

Le Chatelier's Principle

The Principle Defined

How it Works: Forward & Backward Dynamics

1. Changing Temperature

2. Changing Pressure (Gases Only)

3. Changing Concentration

4. Adding a Catalyst

Practice Questions (30)

Industrial Processes

The Haber Process

The Contact Process

Practice Questions (30)

Activation Energy (E_a)