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Chemistry Y12 · Module 5 · Lesson 6
IQ2 — Le Chatelier's Principle

Le Chatelier's Principle — Pressure, Volume & Catalysts

Engineers designing the Haber process wanted to run it at thousands of atmospheres of pressure to maximise ammonia yield — Le Chatelier said it would work, but the steel vessels needed to contain that pressure would cost more to build than the ammonia was worth.

Understand Predict Evaluate

Misconceptions to Fix

Wrong: Le Chatelier's Principle says the system opposes the disturbance by returning to the original concentrations.

Right: Le Chatelier's Principle states the system shifts to minimise the disturbance. After adding a reactant, the system consumes some of it — but the final concentration is still higher than before. The system minimises, not eliminates, the disturbance. Original concentrations are not restored.

Learning Intentions

Predict the effect of pressure and volume changes on gas equilibria by counting moles of gas on each side
Explain why some pressure changes have no effect on equilibrium position
Describe the NO₂/N₂O₄ system as a demonstration of pressure and temperature effects simultaneously
Explain why catalysts do not shift equilibrium position or change Keq
Use the complete summary table to identify which factors change Keq and which only change equilibrium position
Printable worksheet

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🤔 Think First — A Misconception to Diagnose

A student reads the following in a textbook: "Adding a platinum catalyst to the equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) shifts the equilibrium to the right, producing more SO₃." The student underlines this as a key fact to remember. Before reading on — is the student correct? If not, what has the textbook apparently gotten wrong, and what is the correct statement? Write your analysis now.

Key Rules for This Lesson

Pressure/volume LCP rule:

  • Increase pressure (decrease volume) → shift toward side with FEWER moles of gas
  • Decrease pressure (increase volume) → shift toward side with MORE moles of gas
  • No effect if: equal moles of gas on both sides, OR no gases in the equilibrium

Important: Count ONLY gaseous species — ignore solids and aqueous species

Catalyst rule: lowers Eₐ equally for both directions → no shift in equilibrium position; no change in Keq

⚠ Keq depends ONLY on temperature — not concentration, pressure, or catalyst

01
Understand

1. Pressure and Volume Changes — The Gas Mole Count Rule

Pressure changes only affect equilibria that involve gases — and when they do, the system shifts toward whichever side has fewer gas molecules, because that reduces the pressure and partially counteracts the disturbance.

To apply this rule: count the moles of GAS on each side of the balanced equation (ignore solids, liquids, and aqueous species — only gases contribute to pressure).

Rule: Increase pressure → shift toward fewer moles of gas. Decrease pressure → shift toward more moles of gas.

Example 1: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Example 2: H₂(g) + I₂(g) ⇌ 2HI(g)

Example 3: CaCO₃(s) ⇌ CaO(s) + CO₂(g)

Must know: Only count GASEOUS species when determining the effect of pressure changes. Solids, pure liquids, and aqueous species do not contribute to gas pressure. Students frequently count all species including solids — this gives wrong predictions.
Common error: "Increasing pressure always shifts equilibrium to the right." Wrong for two reasons: (1) if there are equal moles of gas on both sides, pressure has no effect; (2) if the right side has MORE moles of gas, increasing pressure shifts LEFT. Always count moles of gas on both sides before predicting.
PRESSURE CHANGE — DECISION TREE (GAS PHASE) PRESSURE INCREASES Count gas moles left vs right Left > Right Shift → RIGHT eg. 4 mol → 2 mol Haber: 4→2 Left = Right → NO EFFECT Right > Left Shift ← LEFT eg. 1 mol → 3 mol CaCO₃: 0→1

Pressure change decision tree — always count gas moles on each side first

02
Understand

2. Pressure Changes — Collision Theory Explanation

Le Chatelier gives you the direction of shift for pressure changes — collision theory explains the mechanism at the particle level.

When the volume of a gas-phase equilibrium system is decreased (pressure increased), the concentration of all gaseous species increases simultaneously. Both forward and reverse collision frequencies increase. However, the side with MORE moles of gas experiences a proportionally larger increase in collision frequency.

For the Haber process (4 moles gas left → 2 moles gas right): when pressure is doubled by halving the volume, the forward collision rate (involving 4 moles of gas per unit reaction) increases more than the reverse collision rate (involving 2 moles per unit reaction). Forward rate now > reverse rate → net forward reaction → equilibrium shifts right → more NH₃ produced.

Must know: Pressure changes do NOT change Keq. The system shifts position to accommodate the pressure change, but the equilibrium constant remains the same — only temperature changes Keq. A common HSC question: "Describe the effect of increasing pressure on Keq." Answer: no effect.
Insight — inert gas distinction: Adding an inert gas (e.g. argon) at constant volume does NOT shift equilibrium. The partial pressures of the reacting gases are unchanged. However, adding an inert gas at constant pressure (by expanding the volume) effectively decreases the partial pressures of all reacting gases — equivalent to decreasing pressure — and shifts equilibrium toward more gas moles.
03
Apply

3. The NO₂/N₂O₄ System — Pressure and Temperature in One Experiment

The nitrogen dioxide/dinitrogen tetroxide equilibrium is a single experiment that simultaneously demonstrates pressure effects and temperature effects on equilibrium.

System Properties

Equilibrium: 2NO₂(g) ⇌ N₂O₄(g)

NO₂: brown/reddish-brown | N₂O₄: colourless

ΔH = −57 kJ/mol (forward reaction exothermic)

Gas moles: Left = 2 mol NO₂; Right = 1 mol N₂O₄

Pressure effect — compressing the gas (decrease volume, increase pressure):

Temperature effect:

Two-stage observation: In the NO₂/N₂O₄ system, immediately after compression there is a brief darker colour before the equilibrium shift makes it paler. This is because compression increases the concentration of both species initially — the paling only occurs as the forward equilibrium shift consumes NO₂. HSC questions sometimes ask about this two-stage observation.
Common error: Students predict that compression only makes the mixture darker. While the immediate observation is correct (momentarily darker), the equilibrium shift then converts NO₂ to colourless N₂O₄ — the net result after re-equilibration is a paler colour than expected for the degree of compression.
04
Understand

4. Catalysts — The Full Explanation

The catalyst misconception is the single most common error in IQ2, and it requires understanding both what catalysts do AND what they don't do.

What a catalyst does:

What a catalyst does NOT do:

For a system already at equilibrium: adding a catalyst increases both rates equally → they remain equal → no shift → concentrations unchanged.

For a system not yet at equilibrium: adding a catalyst allows the system to reach the same equilibrium position faster — but the destination is identical to without a catalyst.

The iron catalyst in the Haber process allows profitable ammonia production at 400–500°C — without it, the rate at this temperature would be too slow. The catalyst does not improve yield — only temperature and pressure can do that.

Three required statements for HSC: (1) the catalyst does not shift the equilibrium position; (2) the catalyst does not change Keq; (3) the catalyst allows equilibrium to be reached more quickly. All three are commonly tested.
Common error: "A catalyst increases the yield of products in an equilibrium reaction." Wrong. A catalyst increases the rate of reaching equilibrium, not the yield at equilibrium. Yield is determined by Keq (and therefore temperature). To increase yield, you must change temperature (or, for gases, pressure).
05
Evaluate

5. Summary — What Changes Keq and What Doesn't

One of the most reliably tested HSC questions in Module 5 is distinguishing between factors that change Keq and factors that only change the equilibrium position.

The only factor that changes Keq is TEMPERATURE. Everything else either shifts the position (concentration, pressure) or does nothing (catalyst).
Factor ChangedEffect on Equilibrium PositionEffect on Keq
Add reactantShifts right (toward products)No change
Remove reactantShifts leftNo change
Add productShifts leftNo change
Remove productShifts rightNo change
Increase pressure (gas, unequal moles)Shifts toward fewer gas molesNo change
Decrease pressure (gas, unequal moles)Shifts toward more gas molesNo change
Add catalystNo shiftNo change
Increase temperature (exothermic forward)Shifts leftDecreases
Decrease temperature (exothermic forward)Shifts rightIncreases
Increase temperature (endothermic forward)Shifts rightIncreases
Decrease temperature (endothermic forward)Shifts leftDecreases
Memorise this table. It is tested every year. Scan every row and confirm for yourself which rows have "No change" for Keq — that is every row except the temperature rows.
Common error: "Increasing pressure increases Keq." Wrong — pressure changes shift the equilibrium position but leave Keq unchanged. The system finds a new equilibrium at the same Keq value with different concentrations.
WHAT CHANGES Keq? — ONLY TEMPERATURE Does NOT change Keq • Adding/removing reactant or product • Increasing or decreasing pressure • Adding a catalyst (These change POSITION, not Keq) DOES change Keq • TEMPERATURE (only factor) Increase T (exothermic fwd) → Keq ↓ Increase T (endothermic fwd) → Keq ↑

The most commonly tested Module 5 fact: temperature is the ONLY factor that changes the value of Keq

Interactive — Gas Equilibrium Shift Predictor
Revisit Your Thinking

You diagnosed the misconception that a catalyst shifts equilibrium. A catalyst does not shift the position of equilibrium — it speeds up both the forward and reverse reactions equally, so the system reaches equilibrium faster but the final concentrations are unchanged. For pressure/volume: increasing pressure (or decreasing volume) shifts equilibrium toward the side with fewer moles of gas. Adding an inert gas at constant volume has no effect on equilibrium position.

Key Terms — scan these before reading
Catalyst ruleA substance that increases reaction rate by providing an alternative pathway with lower activation energy.
Dynamic equilibriumA state where forward and reverse reaction rates are equal.
Equilibrium constant (Keq)The ratio of product to reactant concentrations at equilibrium.
Le Chatelier's PrincipleA system at equilibrium shifts to minimise applied disturbances.
Reaction quotient (Q)The ratio of product to reactant concentrations at any instant.
Solubility product (Ksp)The equilibrium constant for dissolution of a sparingly soluble salt.
Revisit Your Initial Thinking

Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?

Choose how you work — type your answers below or write in your book.

Worked Examples

Apply

Example 1 — Predicting Pressure Effects and Identifying No-Effect Cases

Problem: For each equilibrium, predict the effect of increasing pressure by halving the container volume. State direction of shift (or no shift) and explain using mole counts. (a) N₂(g) + 3H₂(g) ⇌ 2NH₃(g). (b) H₂(g) + Cl₂(g) ⇌ 2HCl(g). (c) Fe₃O₄(s) + 4H₂(g) ⇌ 3Fe(s) + 4H₂O(g).

Step 1 (a): Count gas moles — left: 1 + 3 = 4 moles; right: 2 moles. Fewer gas moles on the right. Increase pressure → shift RIGHT. More NH₃ produced; N₂ and H₂ concentrations decrease (partially). Keq unchanged.
Step 2 (b): Count gas moles — left: 1 + 1 = 2 moles; right: 2 moles. Equal gas moles on both sides. Increase pressure → NO EFFECT on equilibrium position. Both concentrations increase (volume halved) but the ratio [HCl]²/([H₂][Cl₂]) remains equal to Keq — no shift required.
Step 3 (c): Count gas moles — left: 4 moles H₂(g) (Fe₃O₄ is solid, excluded); right: 4 moles H₂O(g) (Fe is solid, excluded). Equal gas moles on both sides (4 = 4). Increase pressure → NO EFFECT. Although the equation looks asymmetric, only the gaseous species matter — and they are equal on both sides.
Answer: (a) Shift right — 4 mol gas left, 2 mol gas right. (b) No shift — 2 mol gas on each side. (c) No shift — 4 mol gas on each side (solids excluded).
Band 5–6

Example 2 — Multi-Variable LCP: Temperature and Pressure Simultaneously

Problem: The Haber process N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ/mol, is at equilibrium at 450°C and 200 atm. An engineer simultaneously increases the temperature to 550°C and increases the pressure to 300 atm. (a) Predict the direction of shift from temperature change alone. (b) Predict the direction of shift from pressure change alone. (c) Predict the overall direction when both are applied simultaneously — do the effects reinforce or oppose?

Step 1 (a — temperature): Forward reaction is exothermic (ΔH = −92 kJ/mol). Increasing temperature favours the endothermic reverse reaction. Equilibrium shifts LEFT. Less NH₃ at new equilibrium. Keq decreases.
Step 2 (b — pressure): Left side has 4 moles of gas (1 N₂ + 3 H₂); right side has 2 moles (2 NH₃). Increasing pressure favours the side with fewer moles of gas. Equilibrium shifts RIGHT. More NH₃ at new equilibrium. Keq unchanged.
Step 3 (c — combined): The two effects OPPOSE each other. Temperature increase shifts LEFT (less NH₃); pressure increase shifts RIGHT (more NH₃). Without Keq values at both temperatures, we cannot determine the net direction quantitatively. Qualitatively: the temperature effect on yield is typically larger than the pressure effect — the increase to 550°C likely reduces yield more than the pressure increase from 200 to 300 atm restores it. Net prediction: slight shift LEFT, but with higher rate due to both factors. Classic Haber process trade-off.
Answer: (a) Shift left — exothermic forward; increase T favours reverse. (b) Shift right — more gas moles on left; increase P favours fewer gas moles side. (c) Effects oppose; temperature likely dominates → net slight left shift; yield decreases, rate improves.

Checkpoint Questions

1 mark

Q1: The equilibrium 2NO(g) + O₂(g) ⇌ 2NO₂(g) has 3 moles of gas on the left and 2 moles of gas on the right. Which correctly predicts the effect of decreasing pressure by increasing the volume?

A Shift right — pressure decrease favours the side with fewer moles of gas
B Shift left — pressure decrease favours the side with more moles of gas
C No shift — pressure changes never affect equilibrium position
D Shift right — decreasing pressure always shifts equilibrium to the right

1 mark

Q2: Which statement about the effect of a catalyst on a chemical equilibrium is correct?

A A catalyst shifts equilibrium to the right by lowering the forward activation energy more than the reverse
B A catalyst increases Keq because more products are formed per unit time
C A catalyst has no effect on equilibrium position or Keq — it only allows equilibrium to be reached more quickly
D A catalyst shifts equilibrium in the direction of the exothermic reaction because it releases additional energy

1 mark

Q3: For PCl₃(g) + Cl₂(g) ⇌ PCl₅(g), increasing pressure shifts equilibrium right. Which correctly explains why Keq is unchanged by this pressure increase?

A Keq is unchanged because the pressure increase also increases temperature, which compensates
B Keq is unchanged because pressure changes only affect the equilibrium position — the ratio of equilibrium concentrations at the new position still equals the same Keq
C Keq is unchanged because PCl₅ is a gas and its concentration increases with pressure
D Keq is unchanged because the catalyst used in this reaction prevents Keq from changing

Short Answer Practice

3 marks

Q4: A syringe containing a mixture of NO₂(g) and N₂O₄(g) at equilibrium is suddenly compressed to half its volume. Describe and explain the two-stage colour change observed, including both the immediate observation and the observation after the system re-establishes equilibrium.

3 marks

Q5: A student claims: "If I add a catalyst to a flask containing SO₂, O₂, and SO₃ at equilibrium, the equilibrium will shift right and more SO₃ will be produced." Identify what is incorrect in this statement and write the correct version.

4 marks

Q6: The reaction N₂(g) + O₂(g) ⇌ 2NO(g), ΔH = +180 kJ/mol, has equal moles of gas on both sides. An industrial chemist proposes: (a) increasing temperature; (b) increasing pressure. For each proposal, predict whether it would shift the equilibrium, which direction, and whether Keq would change. Then evaluate which variable is more useful for maximising NO yield.

Science Jump

Le Chatelier's Principle — Concentration & Pressure

Climb platforms, hit checkpoints, and answer questions on Le Chatelier's Principle — Concentration & Pressure. Quick recall from lessons 1–6.