Quick Simpler Start
You learn how reversible reactions settle into equilibrium and how pressure, temperature, and concentration change the balance.
When you revise this chapter, ask yourself: "Can I explain this to a friend in one easy paragraph?" If yes, you are in a good place.
Interactive Simulator
This small tool gives you a quick visual check before you move deeper into the chapter notes.
Equilibrium Shift Explorer
Choose the change and see which side the equilibrium moves toward.
1. Learning Objectives
- Explain reversible reactions and dynamic equilibrium.
- State Le Chatelier’s Principle clearly.
- Predict the effect of changing concentration, temperature and pressure.
- Apply equilibrium ideas to the Haber Process and Contact Process.
- Write equilibrium constant expressions correctly.
- HL Calculate Kc and equilibrium concentrations.
- HL Explain named equilibrium practicals and how temperature affects Kc.
2. Reversible Reactions and Dynamic Equilibrium
A reversible reaction can proceed in both the forward and reverse directions. It is shown using a double arrow.
Dynamic equilibrium is reached in a closed system when the forward reaction and reverse reaction occur at the same rate, so the concentrations of reactants and products remain constant.
3. Le Chatelier’s Principle
Le Chatelier’s Principle: If a system at equilibrium is disturbed, the system shifts in the direction that opposes the disturbance.
| Change | Equilibrium shifts to... | Simple rule |
|---|---|---|
| Increase concentration of reactant | Products | Use up added reactant. |
| Increase concentration of product | Reactants | Use up added product. |
| Increase temperature | Endothermic direction | Absorb added heat. |
| Decrease temperature | Exothermic direction | Produce heat. |
| Increase pressure | Side with fewer gas molecules | Reduce pressure. |
| Decrease pressure | Side with more gas molecules | Increase pressure. |
4. Concentration, Temperature and Pressure
Concentration
If concentration is changed, the equilibrium shifts to use up what has been added or replace what has been removed.
Temperature
Temperature change depends on whether the forward reaction is exothermic or endothermic. Temperature is the most important condition here because it can change the actual value of Kc.
Pressure
Pressure changes only matter when gases are involved. Count the number of gas molecules on each side of the equation.
Left side: 4 gas molecules. Right side: 2 gas molecules. Increasing pressure favours the right side because it has fewer gas molecules.
5. Industrial Applications
Haber Process
| Condition | Why it is used |
|---|---|
| High pressure | Favours the side with fewer gas molecules, so ammonia yield increases. |
| Moderate temperature | Lower temperature gives better yield, but reaction would be too slow; industry uses a compromise. |
| Iron catalyst | Speeds up equilibrium being reached without changing equilibrium position. |
Contact Process
This is the catalytic oxidation of sulfur dioxide to sulfur trioxide in the Contact Process. The forward reaction is exothermic and the product side has fewer gas molecules, so lower temperature and higher pressure would favour products. In practice, industry again uses compromise conditions for good yield and good rate.
| Point | Explanation |
|---|---|
| Catalyst | Vanadium(V) oxide is used. |
| Temperature | A moderate temperature is used as a compromise between rate and yield. |
| Pressure | Higher pressure favours SO3, but very high pressure is not always economical. |
6. Equilibrium Constant, Kc
Kc gives information about the position of equilibrium for a reversible reaction at a fixed temperature.
| Kc value | Meaning |
|---|---|
| Large Kc | Products are favoured. |
| Small Kc | Reactants are favoured. |
| Kc about 1 | Neither side is strongly favoured. |
7. Higher Level Kc Calculations
Writing Kc Expressions
H2(g) + I2(g) ⇌ 2HI(g)
Kc = [HI]2 / ([H2][I2])
2SO2(g) + O2(g) ⇌ 2SO3(g)
Kc = [SO3]2 / ([SO2]2[O2])
Numerical Kc Calculation
For H2 + I2 ⇌ 2HI, if [H2] = 0.10 mol L-1, [I2] = 0.20 mol L-1 and [HI] = 0.80 mol L-1, then:
Kc = (0.80)2 / (0.10 × 0.20)
Kc = 0.64 / 0.02 = 32
Equilibrium Concentration Method
For harder Higher Level questions, follow this sequence:
- Write the balanced equation.
- Write the Kc expression.
- Substitute known equilibrium concentrations.
- If one value is unknown, rearrange the equation carefully.
- Check units and whether the answer makes chemical sense.
8. What Changes Kc?
| Change applied | Does Kc change? | Why? |
|---|---|---|
| Concentration changed | No | The system shifts until the same Kc is restored. |
| Pressure changed | No | The position shifts, but Kc stays the same if temperature is constant. |
| Catalyst added | No | It speeds forward and reverse reactions equally. |
| Temperature changed | Yes | Temperature changes the equilibrium constant itself. |
9. Named Equilibrium Practicals
Cobalt Chloride Equilibrium
Cobalt chloride equilibrium is often used to show the effect of concentration and temperature changes. The pink and blue species shift depending on conditions.
- Pink side is favoured with more water.
- Blue side is favoured with more chloride ions.
- Heating can also shift the colour depending on which side is endothermic.
Chromate / Dichromate Equilibrium
- Chromate ions are yellow.
- Dichromate ions are orange.
- Adding acid favours dichromate; adding alkali favours chromate.
Iron(III) Thiocyanate Equilibrium
- The complex ion is deep red.
- Adding Fe3+ or SCN- makes the red colour deeper.
- Removing one ion or diluting the mixture can reduce the colour intensity.
10. Worked Exam Examples
N2 + 3H2 ⇌ 2NH3
Increasing pressure shifts equilibrium right because the right side has fewer gas molecules.
If the forward reaction is exothermic, increasing temperature shifts equilibrium left because the system tries to absorb the added heat.
If Kc is very large, the equilibrium mixture contains much more product than reactant.
If extra SCN- is added to Fe3+ + SCN- ⇌ FeSCN2+, the red colour deepens because the equilibrium shifts right to use up the extra thiocyanate ions.
11. Examiner Secrets, Mistakes and Traps
12. Exam Practice Questions
- Define dynamic equilibrium. [3 marks]
- State Le Chatelier’s Principle. [3 marks]
- For N2 + 3H2 ⇌ 2NH3, explain the effect of increasing pressure. [4 marks]
- Explain why a catalyst is used in the Haber Process. [3 marks]
- Write the Kc expression for 2SO2(g) + O2(g) ⇌ 2SO3(g). [3 marks]
- HL Calculate Kc for H2 + I2 ⇌ 2HI if [H2] = 0.10, [I2] = 0.20 and [HI] = 0.80 mol L-1. [4 marks]
- Explain why temperature changes Kc but a catalyst does not. [4 marks]
- Describe what happens in one named equilibrium practical when a reagent is added. [4 marks]
13. Last-Minute Revision Sheet
- Equilibrium needs a closed system.
- Dynamic equilibrium means forward rate = reverse rate.
- Le Chatelier: the system shifts to oppose a change.
- Increase pressure → favours fewer gas molecules.
- Increase temperature → favours endothermic direction.
- Kc = products over reactants.
- Use powers from the balanced equation.
- Kc changes with temperature.
- Catalysts do not change equilibrium position or Kc.
- Colour changes in named practicals must be explained by equilibrium shifts.
14. Self-Assessment Checklist
- I can define dynamic equilibrium correctly.
- I can apply Le Chatelier’s Principle to concentration, temperature and pressure.
- I can explain the Haber Process and Contact Process.
- I can write Kc expressions correctly.
- I can calculate simple Kc values.
- I can explain what changes Kc.
- I can describe cobalt chloride, chromate/dichromate, and iron(III) thiocyanate equilibrium experiments.

