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Curriculum: Irish Leaving Certificate (ILC)
Level: Ordinary Level (OL)
Subject: Chemistry
Chapter 8: Chemical Equilibrium
Version: Complete Combined Chapter
Prepared By: ExamsLogic Academic Team
Chapter Overview
This final compiled chapter combines all four parts of Chapter 8 into one clean file and adds the missing Ordinary Level syllabus material. It covers reversible reactions, dynamic equilibrium, Le Chatelier's Principle, Kc, the Haber process, the catalytic oxidation of sulfur dioxide to sulfur trioxide, and the mandatory equilibrium experiments.
Syllabus Match: This version is designed to cover both 8.1 Chemical Equilibrium and 8.2 Le Chatelier's Principle, including Mandatory Experiment 8.1 and the named industrial applications.
Contents
Part 1: Reversible reactions, closed systems and dynamic equilibrium
Part 2: Le Chatelier's Principle and industrial equilibrium
Part 3: Equilibrium constant Kc
Mandatory Experiment 8.1: three equilibrium demonstrations
Industrial application: catalytic oxidation of sulfur dioxide to sulfur trioxide
Part 4: final revision, summary tables and exam practice
Part 1: Reversible Reactions and Dynamic Equilibrium
This section compiles the content from part 1: reversible reactions and dynamic equilibrium.
1. Learning Objectives
Define a reversible reaction.
Use the reversible reaction symbol ⇌ correctly.
Explain the meaning of dynamic equilibrium.
State the conditions needed for equilibrium.
Explain why equilibrium needs a closed system.
Interpret particle diagrams and rate graphs for equilibrium.
Answer Ordinary Level exam-style questions on reversible reactions and equilibrium.
2. Reversible Reactions
A reversible reaction is a reaction that can happen in both directions. The reactants form products, and the products can react to form the original reactants again.
Reactants ⇌ Products
The symbol ⇌ means the reaction is reversible. It shows that the forward and reverse reactions can both occur.
Direction
Meaning
Example Language
Forward reaction
Reactants change into products.
A + B → C + D
Reverse reaction
Products change back into reactants.
C + D → A + B
A reversible reaction can move forwards and backwards.
Examiner Tip: In reversible reaction questions, use the correct symbol ⇌ instead of a one-way arrow.
3. Everyday and Chemical Examples of Reversible Changes
Some reversible changes are physical, while others are chemical. Equilibrium is mainly about reversible chemical reactions, but physical examples help the idea make sense.
Example
Forward Change
Reverse Change
Water freezing/melting
Water → ice
Ice → water
Hydrated copper sulfate
Blue crystals lose water and turn white
White solid gains water and turns blue
Ammonium chloride
Solid breaks down on heating
Gases recombine on cooling
Haber process
Nitrogen and hydrogen form ammonia
Ammonia can decompose back
Examiner Trap: Reversible does not mean “easy to reverse by hand”. It means the chemical reaction can proceed in both directions under suitable conditions.
4. Closed Systems
A closed system is a system where substances cannot escape and new substances cannot enter.
Equilibrium can only be reached in a closed system.
If gases or vapours escape, the reverse reaction may not happen properly, so equilibrium cannot be maintained.
A closed system allows forward and reverse reactions to continue.
Common Mistake: Saying equilibrium can be reached in any container. For reversible reactions, the system must be closed.
5. Dynamic Equilibrium
Dynamic equilibrium is reached when the rate of the forward reaction equals the rate of the reverse reaction in a closed system.
Dynamic equilibrium: forward rate = reverse rate
The word dynamic means the reactions have not stopped. Both reactions are still happening, but at equal rates.
At Dynamic Equilibrium
What it Means
Forward reaction continues
Reactants are still forming products.
Reverse reaction continues
Products are still forming reactants.
Rates are equal
Forward rate = reverse rate.
Concentrations stay constant
Amounts do not appear to change overall.
System is closed
No substance escapes or enters.
At equilibrium, the forward and reverse rates become equal.
Examiner Trap: Equilibrium does not mean the reaction has stopped. It means both reactions continue at equal rates.
6. Concentrations at Equilibrium
At equilibrium, concentrations of reactants and products remain constant, but they are not necessarily equal.
Constant does not always mean equal.
Possible Equilibrium Mixture A: More products than reactants. This means equilibrium lies more to the product side.
Possible Equilibrium Mixture B: More reactants than products. This means equilibrium lies more to the reactant side.
Common Mistake: Saying reactants and products must have the same concentration at equilibrium. They do not. Only the forward and reverse rates are equal.
7. Particle-Level View of Equilibrium
Particles are still reacting at equilibrium. A product molecule may break down at the same rate as reactant molecules combine.
The mixture looks unchanged overall, but particles are still reacting.
8. Interactive Simulators and Virtual Labs
Simulator 1: PhET – Reversible Reactions
Use for: Seeing forward and reverse reaction rates change over time.
Student Task: Observe when the forward and reverse rates become equal.
Exam Link: Dynamic equilibrium and closed systems.
Example 1: In A + B ⇌ C + D, the forward reaction is A + B → C + D and the reverse reaction is C + D → A + B.
Example 2: At dynamic equilibrium, the forward and reverse reactions are still happening, but at equal rates.
Example 3: A gas escapes from an open flask. Equilibrium may not be maintained because the system is not closed.
10. Examiner Secrets, Mistakes and Traps
Examiner Secret: The key phrase for this topic is rate of forward reaction equals rate of reverse reaction.
Examiner Tip: Always mention closed system when defining dynamic equilibrium.
Common Mistake: Saying equilibrium means equal amounts of reactants and products. No — it means equal rates.
Common Mistake: Saying reactions stop at equilibrium. The reactions continue.
Examiner Trap: Constant concentration means the amounts are no longer changing overall, not that particles are inactive.
11. Exam Practice Questions
Q1. What is meant by a reversible reaction? [2 marks]
Q2. Write the symbol used for a reversible reaction. [1 mark]
Q3. Define dynamic equilibrium. [3 marks]
Q4. Explain why equilibrium requires a closed system. [2 marks]
Q5. At equilibrium, are the concentrations of reactants and products always equal? Explain. [2 marks]
Q6. A student says, “At equilibrium, the reaction has stopped.” Explain why this is wrong. [3 marks]
MCQs with Explanations
1. The symbol for a reversible reaction is: A. → B. ⇌ C. + D. =
Answer: B. The ⇌ symbol shows the forward and reverse reactions can occur.
2. Dynamic equilibrium is reached when: A. reactions stop B. forward rate equals reverse rate C. products disappear D. reactants equal products exactly
Answer: B. Dynamic equilibrium means both reactions continue at equal rates.
3. Equilibrium is best maintained in: A. open system B. closed system C. broken test tube D. evaporating dish only
Answer: B. A closed system prevents substances from escaping.
12. Last-Minute Revision Sheet
Reversible reactions can happen in both directions.
The reversible reaction symbol is ⇌.
Forward reaction: reactants form products.
Reverse reaction: products form reactants.
Equilibrium needs a closed system.
Dynamic equilibrium means forward rate = reverse rate.
At equilibrium, reactions continue.
Concentrations stay constant but are not necessarily equal.
Constant does not mean equal.
Equilibrium is a balance of rates, not a stopping point.
13. Self-Assessment Checklist
I can define reversible reaction.
I can use the ⇌ symbol correctly.
I can identify forward and reverse reactions.
I can define dynamic equilibrium.
I can explain why a closed system is needed.
I can explain why equilibrium does not mean reactions stop.
I can interpret simple rate graphs for equilibrium.
I can avoid saying concentrations must be equal.
14. Mark Scheme
Q1. A reaction that can proceed in both directions [1]; products can react to reform reactants [1].
Q2. ⇌ [1].
Q3. Occurs in a closed system [1]; forward and reverse reactions continue [1]; forward rate equals reverse rate [1].
Q4. Substances cannot escape or enter [1]; so both forward and reverse reactions can continue and equilibrium can be maintained [1].
Q5. No [1]; concentrations are constant but not necessarily equal [1].
Q6. Equilibrium is dynamic [1]; forward and reverse reactions still occur [1]; but at equal rates so no overall change is observed [1].
Part 2: Le Chatelier's Principle
This section compiles the content from part 2: le chatelier's principle.
1. Learning Objectives
State Le Chatelier's Principle in Ordinary Level language.
Predict the effect of changing concentration, temperature and pressure.
Apply the principle to gaseous equilibria and the Haber Process.
Explain equilibrium shifts using clear exam language.
2. Le Chatelier's Principle
Le Chatelier's Principle predicts what happens when a system at equilibrium is disturbed.
If a system at equilibrium is disturbed, the equilibrium shifts to oppose the change.
The disturbance may be a change in concentration, temperature, or pressure for gaseous reactions.
The equilibrium shifts to reduce the effect of the change.
Examiner Tip: Use the words shifts left or shifts right, then explain why.
3. Changing Concentration
If concentration is changed, the equilibrium shifts to use up the substance added or replace the substance removed.
Change
Equilibrium Response
Simple Reason
Add more reactant
Shifts right
Uses up the added reactant by making more product.
Remove reactant
Shifts left
Replaces some of the removed reactant.
Add more product
Shifts left
Uses up the added product.
Remove product
Shifts right
Replaces some of the removed product.
Add reactant → shift to products | Add product → shift to reactants
Worked Example: For A + B ⇌ C + D, if more A is added, equilibrium shifts right to use up some A and produce more C and D.
Examiner Trap: Do not say it shifts “to the bigger side”. For concentration, it shifts to reduce the concentration change.
4. Changing Temperature
Temperature changes depend on whether the forward reaction is exothermic or endothermic.
Exothermic Direction
Releases heat. Heat behaves like a product.
Endothermic Direction
Takes in heat. Heat behaves like a reactant.
Change
Equilibrium shifts toward
Why?
Increase temperature
Endothermic direction
Uses up added heat.
Decrease temperature
Exothermic direction
Replaces heat that was removed.
Temperature questions are about heat: added heat is opposed, removed heat is replaced.
Common Mistake: Higher temperature does not always give more product. It depends on whether product formation is exothermic or endothermic.
5. Changing Pressure
Pressure only matters for equilibria involving gases. The equilibrium shifts to reduce the pressure change.
Change
Equilibrium Response
Reason
Increase pressure
Shifts to side with fewer gas molecules
Fewer gas particles reduce pressure.
Decrease pressure
Shifts to side with more gas molecules
More gas particles increase pressure.
Same number of gas molecules on both sides
No major shift
Both sides have equal gas particle numbers.
Higher pressure → fewer gas molecules | Lower pressure → more gas molecules
Worked Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). Left side has 4 gas molecules; right side has 2. Increasing pressure shifts right.
Examiner Tip: Count gas molecules using balanced equation coefficients. Ignore solids and liquids for pressure questions.
6. Haber Process: OL Industrial Application
The Haber Process makes ammonia from nitrogen and hydrogen.
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) forward reaction is exothermic
Condition Change
Effect on Ammonia Yield
Reason
Increase pressure
Increases ammonia yield
Shifts to side with fewer gas molecules: 2 NH₃ instead of 4 reactant molecules.
Decrease temperature
Increases ammonia yield
Forward reaction is exothermic, so lower temperature favours product formation.
Increase temperature
Decreases ammonia yield but increases rate
Higher temperature favours the endothermic reverse reaction, but particles react faster.
Use catalyst
No change in equilibrium position
Catalyst speeds up both forward and reverse reactions equally.
High pressure favours the side with fewer gas molecules.
Examiner Trap: A catalyst does not increase equilibrium yield. It only helps equilibrium be reached faster.
7. Interactive Simulators and Virtual Labs
Simulator 1: PhET – Reversible Reactions
Use for: Seeing how equilibrium responds to changes.
Student Task: Change reactant/product amounts and observe the shift.
Example 1: A + B ⇌ C + D. If C is removed, equilibrium shifts right to replace some removed C.
Example 2: N₂ + 3H₂ ⇌ 2NH₃. Increasing pressure shifts right because the product side has fewer gas molecules.
Example 3: If the forward reaction is exothermic, decreasing temperature shifts equilibrium forward.
9. Examiner Secrets, Mistakes and Traps
Examiner Secret: The best answers say the change, the direction of shift, and the reason.
Examiner Tip: For pressure, count gas molecules. For temperature, identify exothermic/endothermic direction.
Common Mistake: Saying catalysts shift equilibrium. They do not.
Common Mistake: Forgetting pressure only applies to gases.
Examiner Trap: Higher temperature favours the endothermic direction, not automatically the product side.
10. Exam Practice Questions
Q1. State Le Chatelier's Principle. [3 marks]
Q2. For A + B ⇌ C + D, predict the effect of adding more A. [2 marks]
Q3. Explain the effect of increasing pressure on N₂(g) + 3H₂(g) ⇌ 2NH₃(g). [3 marks]
Q4. The forward reaction is exothermic. What happens to product yield when temperature is increased? Explain. [3 marks]
Q5. Explain why a catalyst does not change equilibrium yield. [2 marks]
MCQs with Explanations
1. Increasing pressure favours the side with: A. more gas molecules B. fewer gas molecules C. more solids D. higher colour
Answer: B. Fewer gas molecules reduce pressure.
2. Increasing temperature favours the: A. exothermic direction B. endothermic direction C. side with fewer gases D. catalyst
Answer: B. The system uses up added heat by favouring the endothermic direction.
3. A catalyst affects equilibrium by: A. increasing yield B. shifting right C. shifting left D. reaching equilibrium faster
Answer: D. A catalyst speeds both directions equally and does not shift equilibrium.
11. Last-Minute Revision Sheet
Le Chatelier's Principle: equilibrium shifts to oppose a change.
Add reactant → shift right.
Add product → shift left.
Remove product → shift right.
Increase temperature → favours endothermic direction.
Decrease temperature → favours exothermic direction.
Increase pressure → favours fewer gas molecules.
Decrease pressure → favours more gas molecules.
Catalysts do not change equilibrium position.
Haber Process: high pressure favours ammonia.
12. Self-Assessment Checklist
I can state Le Chatelier's Principle.
I can predict concentration changes.
I can predict temperature changes.
I can count gas molecules for pressure questions.
I can apply equilibrium ideas to the Haber Process.
I can explain why catalysts do not change yield.
13. Mark Scheme
Q1. If a system at equilibrium is disturbed [1], the equilibrium shifts [1] to oppose the change [1].
Q2. Shifts right [1] to use up added A / produce more C and D [1].
Q3. Left side has 4 gas molecules and right side has 2 [1]; increasing pressure favours fewer gas molecules [1]; shifts right, increasing ammonia yield [1].
Q4. Product yield decreases [1]; higher temperature favours endothermic direction [1]; if forward is exothermic, reverse direction is favoured [1].
Q5. Catalyst speeds up forward and reverse reactions equally [1]; so equilibrium position/yield is unchanged [1].
Part 3: Equilibrium Constant Kc
This section compiles the content from part 3: equilibrium constant kc.
1. Learning Objectives
Explain what the equilibrium constant Kc represents.
Write simple Kc expressions for reversible reactions.
Use products over reactants in Kc expressions.
Interpret whether equilibrium favours products or reactants.
Understand that Kc changes with temperature only.
Answer OL-style Kc questions without overcomplicating the maths.
2. What is Kc?
Kc is the equilibrium constant. It gives information about the position of equilibrium for a reversible reaction at a fixed temperature.
Kc shows whether the equilibrium mixture contains more products or more reactants.
Kc Value
Meaning
Exam Interpretation
Kc is large
More products than reactants at equilibrium
Equilibrium lies to the right.
Kc is small
More reactants than products at equilibrium
Equilibrium lies to the left.
Kc is about 1
Similar amounts of products and reactants
Neither side is strongly favoured.
Large Kc means products are favoured; small Kc means reactants are favoured.
Examiner Tip: For OL, interpretation matters more than heavy calculation. Large Kc = products; small Kc = reactants.
3. Writing a Kc Expression
For a reversible reaction, the Kc expression places the concentration of products on top and reactants on the bottom.
Kc = products ÷ reactants
General Reaction
A + B ⇌ C + D
Kc = [C][D] ÷ [A][B]
The square brackets mean concentration.
Symbol
Meaning
[A]
Concentration of A at equilibrium
[B]
Concentration of B at equilibrium
[C]
Concentration of C at equilibrium
[D]
Concentration of D at equilibrium
Examiner Trap: Do not use starting concentrations. Kc uses equilibrium concentrations.
4. Coefficients in Kc Expressions
If a balanced equation has a number in front of a substance, that number becomes a power in the Kc expression.
aA + bB ⇌ cC + dD
Kc = [C]c[D]d ÷ [A]a[B]b
Example: N₂ + 3H₂ ⇌ 2NH₃
Kc = [NH₃]2 ÷ [N₂][H₂]3
Common Mistake: Forgetting powers. In N₂ + 3H₂ ⇌ 2NH₃, ammonia is squared and hydrogen is cubed.
5. Kc Expression Examples
Reaction
Kc Expression
H₂ + I₂ ⇌ 2HI
Kc = [HI]2 ÷ [H₂][I₂]
N₂ + 3H₂ ⇌ 2NH₃
Kc = [NH₃]2 ÷ [N₂][H₂]3
2SO₂ + O₂ ⇌ 2SO₃
Kc = [SO₃]2 ÷ [SO₂]2[O₂]
A + 2B ⇌ C
Kc = [C] ÷ [A][B]2
A safe memory trick: products over reactants.
6. Solids and Liquids in Kc
At Ordinary Level, most Kc questions will focus on gases or aqueous solutions. Pure solids and pure liquids are usually not included in Kc expressions.
Include gases (g) and aqueous substances (aq). Usually omit pure solids (s) and pure liquids (l).
Example: CaCO₃(s) ⇌ CaO(s) + CO₂(g)
Kc expression includes only CO₂: Kc = [CO₂]
Examiner Tip: Check state symbols. If a substance is solid or pure liquid, it is usually left out of Kc.
7. What Changes Kc?
Kc is constant only at a particular temperature. If the temperature changes, Kc may change.
Change
Does Kc Change?
Reason
Temperature changed
Yes
Equilibrium position changes in a way that changes the ratio.
Concentration changed
No, if temperature stays same
System shifts until the same Kc is restored.
Pressure changed
No, if temperature stays same
System shifts until the same Kc is restored.
Catalyst added
No
Catalyst does not change equilibrium position.
Examiner Trap: A catalyst does not change Kc. It only helps equilibrium be reached faster.
8. Interactive Simulators and Virtual Labs
Simulator 1: PhET – Reversible Reactions
Use for: Seeing equilibrium concentrations change before becoming constant.
Student Task: Observe product and reactant amounts and decide if Kc is likely large or small.
Example 2: If Kc is very large, what does this tell you?
Answer: The equilibrium mixture contains more products than reactants. The equilibrium lies to the right.
Example 3: For 2SO₂ + O₂ ⇌ 2SO₃, write Kc.
Answer: Kc = [SO₃]2 ÷ [SO₂]2[O₂]
10. Examiner Secrets, Mistakes and Traps
Examiner Secret: Most Kc mistakes are not chemistry mistakes — they are bracket, power, or upside-down expression mistakes.
Examiner Tip: Write products first, then draw a line, then write reactants underneath.
Common Mistake: Putting reactants on top. Kc is products over reactants.
Common Mistake: Forgetting powers from the balanced equation.
Examiner Trap: Large Kc does not mean the reaction is fast. It means products are favoured at equilibrium.
11. Exam Practice Questions
Q1. What does Kc tell us about an equilibrium mixture? [2 marks]
Q2. Write the Kc expression for A + B ⇌ C + D. [2 marks]
Q3. Write the Kc expression for H₂ + I₂ ⇌ 2HI. [3 marks]
Q4. Write the Kc expression for N₂ + 3H₂ ⇌ 2NH₃. [3 marks]
Q5. If Kc is very small, what does this tell you about the equilibrium position? [2 marks]
Q6. State one factor that changes Kc. [1 mark]
MCQs with Explanations
1. In a Kc expression, products are written: A. on top B. at the bottom C. ignored D. only if solid
Answer: A. Kc is products over reactants.
2. A large Kc means equilibrium favours: A. reactants B. products C. catalyst D. no reaction
Answer: B. Large Kc means the equilibrium mixture contains more products.
3. Kc changes when: A. catalyst is added B. temperature changes C. flask is labelled D. colour is observed
Answer: B. Kc depends on temperature.
4. For N₂ + 3H₂ ⇌ 2NH₃, the ammonia term is: A. [NH₃] B. [NH₃]² C. [NH₃]³ D. [NH₃]/2
Answer: B. The coefficient 2 becomes the power 2.
12. Last-Minute Revision Sheet
Kc is the equilibrium constant.
Kc tells whether products or reactants are favoured.
Kc = products over reactants.
Square brackets mean concentration.
Balanced equation numbers become powers.
Large Kc means products are favoured.
Small Kc means reactants are favoured.
Kc changes with temperature.
Catalysts do not change Kc.
Use equilibrium concentrations, not starting concentrations.
13. Self-Assessment Checklist
I can explain what Kc represents.
I can write simple Kc expressions.
I can use powers from balanced equations.
I can interpret large and small Kc values.
I can explain why catalysts do not change Kc.
I can avoid putting reactants on top.
14. Mark Scheme
Q1. Kc tells the position of equilibrium [1] and whether products or reactants are favoured [1].
Q2. Kc = [C][D] ÷ [A][B] [2].
Q3. Products on top [1]; [HI]² used [1]; denominator [H₂][I₂] [1].
Q4. Kc = [NH₃]² ÷ [N₂][H₂]³ [3].
Q5. Reactants are favoured [1]; equilibrium lies to the left [1].
Q6. Temperature [1].
Added Syllabus Content: Industrial Application in the Contact Process
The Ordinary Level syllabus specifically names the catalytic oxidation of sulfur dioxide to sulfur trioxide as an industrial application of Le Chatelier's Principle. This reaction is part of the Contact Process used to make sulfuric acid.
2SO2(g) + O2(g) ⇌ 2SO3(g) forward reaction is exothermic
Condition
Effect on SO3 yield
Reason
Increase pressure
Favours sulfur trioxide slightly
3 gas molecules become 2 gas molecules, so higher pressure favours the side with fewer gas molecules.
Decrease temperature
Favours sulfur trioxide
The forward reaction is exothermic, so lower temperature favours product formation.
Use V2O5 catalyst
No change in equilibrium position
The catalyst speeds up both forward and reverse reactions equally.
Industrial compromise: Industry does not use extremely low temperature because the reaction would be too slow. Instead, a moderate temperature is used so the reaction is fast enough while still giving a good yield.
Examiner Tip: For both the Haber process and the Contact Process, the key industrial idea is compromise between yield, rate, cost and safety.
Added Syllabus Content: Mandatory Experiment 8.1
The syllabus requires simple experiments to illustrate Le Chatelier's Principle. These experiments show that when an equilibrium mixture is disturbed, the equilibrium shifts to oppose the change.
Practical theme: In each experiment, you change temperature or concentration and then observe a colour change. The colour change tells you the direction of equilibrium shift.
More FeSCN2+ has formed; equilibrium shifts right.
Blood-red colour becomes paler
Less FeSCN2+ is present; equilibrium shifts left.
Adding Fe3+ or SCN− shifts equilibrium to the right, deepening the red colour.
Removing one reactant or diluting the mixture can shift equilibrium to the left, making the red colour paler.
How to write the conclusion
Model conclusion: The equilibrium mixture changed colour when disturbed. This shows that the equilibrium shifted to oppose the change, which agrees with Le Chatelier's Principle.
Part 4: Final Revision and Practical Applications
This section compiles the content from part 4: final revision and practical applications.
1. Learning Objectives
Review reversible reactions, dynamic equilibrium and closed systems.
Apply Le Chatelier's Principle to concentration, temperature and pressure changes.
Interpret Kc values and simple Kc expressions.
Explain industrial equilibrium choices using the Haber Process.
Answer mixed OL exam questions across the full chapter.
2. Chemical Equilibrium Concept Map
This final part joins the whole chapter together. Equilibrium questions usually test the same few ideas in different ways.
Every equilibrium question comes back to reversible reactions, rates and shifts.
3. Full Chapter Summary Table
Key Idea
Meaning
Exam Phrase
Reversible reaction
Reaction can happen forwards and backwards.
Use the symbol ⇌.
Closed system
No substances enter or leave.
Needed for equilibrium.
Dynamic equilibrium
Forward and reverse reactions continue at equal rates.
Forward rate = reverse rate.
Constant concentration
Amounts do not change overall.
Constant does not mean equal.
Le Chatelier's Principle
Equilibrium shifts to oppose a change.
State direction and reason.
Kc
Equilibrium constant.
Large Kc favours products; small Kc favours reactants.
Catalyst
Speeds up both directions equally.
No change in equilibrium position or Kc.
4. Le Chatelier Quick Decision Table
Change
Shift
Memory Rule
Add reactant
Right
Use up added reactant.
Remove reactant
Left
Replace removed reactant.
Add product
Left
Use up added product.
Remove product
Right
Replace removed product.
Increase temperature
Endothermic direction
Use up added heat.
Decrease temperature
Exothermic direction
Replace removed heat.
Increase pressure
Side with fewer gas molecules
Reduce pressure.
Decrease pressure
Side with more gas molecules
Increase pressure back.
Examiner Tip: For every Le Chatelier question, use the pattern: change → shift → reason.
Example 1: For A + B ⇌ C + D, removing D shifts the equilibrium right to replace D.
Example 2: In the Haber Process, increasing pressure shifts right because 2 gas molecules are on the product side compared with 4 on the reactant side.
Example 3: If Kc is very small, reactants are favoured and equilibrium lies to the left.
Example 4: A catalyst does not change Kc because it speeds up forward and reverse reactions equally.
9. Examiner Secrets, Mistakes and Traps
Examiner Secret: Most marks come from precise language: closed system, equal rates, shifts to oppose change, fewer gas molecules.
Examiner Tip: For pressure questions, write the number of gas molecules on each side before deciding the shift.
Common Mistake: Saying equilibrium means equal amounts. It means equal rates.
Common Mistake: Saying catalysts increase equilibrium yield. They do not.
Examiner Trap: A large Kc does not mean a fast reaction. Kc describes equilibrium position, not reaction speed.
10. Final Chapter 8 Exam Practice
Q1. Define dynamic equilibrium. [3 marks]
Q2. Explain why equilibrium requires a closed system. [2 marks]
Q3. State Le Chatelier's Principle. [3 marks]
Q4. For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), explain the effect of increasing pressure. [3 marks]
Q5. The forward reaction is exothermic. Predict the effect of increasing temperature on product yield. [3 marks]
Q6. Write the Kc expression for H₂ + I₂ ⇌ 2HI. [3 marks]
Q7. Explain why a catalyst does not change equilibrium yield. [2 marks]
Q8. If Kc is large, what does this tell you about the equilibrium mixture? [2 marks]
MCQs with Explanations
1. Dynamic equilibrium means: A. reactions stop B. equal amounts C. forward rate equals reverse rate D. only products remain
Answer: C. Dynamic equilibrium is a balance of rates.
2. Increasing pressure favours the side with: A. fewer gas molecules B. more gas molecules C. more solids D. no catalyst
Answer: A. Fewer gas molecules reduce pressure.
3. Kc is changed by: A. catalyst B. temperature C. stirring D. filter paper
Answer: B. Kc changes with temperature.
4. A catalyst: A. shifts equilibrium right B. shifts equilibrium left C. changes Kc D. reaches equilibrium faster
Answer: D. It speeds both forward and reverse reactions equally.
11. Full Chapter 8 Last-Minute Revision Sheet
Reversible reactions use ⇌.
Equilibrium needs a closed system.
Dynamic equilibrium: forward rate = reverse rate.
Reactions continue at equilibrium.
Concentrations are constant, not necessarily equal.
Le Chatelier: equilibrium shifts to oppose a change.
Increase temperature → favours endothermic direction.
Decrease temperature → favours exothermic direction.
Increase pressure → favours fewer gas molecules.
Catalyst does not change yield or Kc.
Kc = products over reactants.
Large Kc favours products.
Small Kc favours reactants.
Haber Process uses compromise conditions for rate, yield, cost and safety.
12. Self-Assessment Checklist
I can define reversible reaction.
I can define dynamic equilibrium.
I can explain why a closed system is needed.
I can apply Le Chatelier's Principle to concentration changes.
I can apply Le Chatelier's Principle to temperature changes.
I can apply Le Chatelier's Principle to pressure changes.
I can explain the Haber Process equilibrium choices.
I can write simple Kc expressions.
I can interpret large and small Kc values.
I can avoid common equilibrium traps.
13. Mark Scheme
Q1. Occurs in a closed system [1]; forward and reverse reactions continue [1]; rates are equal [1].
Q2. Substances cannot escape or enter [1]; this allows forward and reverse reactions to continue and equilibrium to be maintained [1].
Q3. If a system at equilibrium is disturbed [1], the equilibrium shifts [1] to oppose the change [1].
Q4. Left side has 4 gas molecules and right side has 2 [1]; increasing pressure favours fewer gas molecules [1]; equilibrium shifts right/increases ammonia yield [1].
Q5. Product yield decreases [1]; higher temperature favours endothermic direction [1]; reverse reaction is favoured if forward is exothermic [1].
Q6. Products on top [1]; [HI]² [1]; denominator [H₂][I₂] [1].
Q7. Catalyst speeds up forward and reverse reactions equally [1]; equilibrium position/yield is unchanged [1].
Q8. Products are favoured [1]; equilibrium lies to the right / more products than reactants at equilibrium [1].