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

Le Chatelier's Principle — Pressure, Volume & Catalysts

In 1913, BASF's engineers at Oppau calculated that running N₂ + 3H₂ ⇌ 2NH₃ at 1000 atm would raise the equilibrium yield to over 70% — but the steel vessels required would cost 40 times more than running at 200 atm. Pressure shifts equilibrium, but the economics forced a compromise: 200 atm became the industrial standard.

Understand Predict Evaluate

Today's hook: In 1913, BASF's Oppau plant found that running the Haber process above 200 atm gave diminishing economic returns despite better yields. How does a system 'decide' which way to shift when pressure increases — and why does the answer have a direct cost?

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Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions

Before You Read

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.

By the end of this lesson you will:

Know

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

Understand

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

Can Do

Use the complete summary table to identify which factors change Keq and which only change equilibrium position

Vocabulary for this lesson

Pressure disturbance

Increasing pressure at constant temperature shifts equilibrium towards the side with fewer moles of gas.

Volume disturbance

Decreasing volume increases pressure, shifting equilibrium towards fewer gas moles.

Inert gas addition

Adding an inert gas at constant volume does not change partial pressures of reactants/products — no shift.

Catalyst

Increases both forward and reverse reaction rates equally; lowers Ea but does not change Keq or equilibrium position.

Mole ratio of gases

The stoichiometric coefficients of gaseous species determine how pressure changes affect the equilibrium.

Temperature vs pressure strategies

Temperature changes Keq; pressure changes only affect yield for reactions with unequal gas moles.

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

Content

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)

Left: 1 + 3 = 4 moles of gas. Right: 2 moles of gas.
Increase pressure → shift RIGHT (toward 2 moles, fewer)
Decrease pressure → shift LEFT (toward 4 moles, more)

Example 2: H₂(g) + I₂(g) ⇌ 2HI(g)

Left: 1 + 1 = 2 moles. Right: 2 moles. Equal moles on both sides.
Pressure change has NO EFFECT on equilibrium position

Example 3: CaCO₃(s) ⇌ CaO(s) + CO₂(g)

Left: 0 moles gas (CaCO₃ is solid). Right: 1 mole gas.
Increase pressure → shift LEFT (toward 0 moles gas)
Decrease pressure → shift RIGHT (toward 1 mole gas)

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 changes only affect gas-phase equilibria: increase pressure → shift toward fewer moles of gas; decrease pressure → shift toward more moles of gas; no effect if gas moles equal on both sides; count only gaseous species (ignore solids, liquids, aqueous); Keq unchanged.

Pause — copy the highlighted gas mole count rule into your book before moving on.

For H₂(g) + I₂(g) ⇌ 2HI(g), what is the effect of increasing pressure?

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

Understand

2. Pressure Changes — Collision Theory Explanation

We just saw the gas mole count rule — increase pressure, shift toward fewer gas moles. That raises a question: why does counting gas moles predict the shift — what is the collision-theory mechanism behind pressure changes? This card answers it → by showing how compression increases collision frequency unequally between the two sides.

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.

Compression increases all gas concentrations; the side with more gas moles has a proportionally larger collision frequency increase — so the rate on that side rises more, producing a net shift away from it; pressure does not change Keq; inert gas at constant volume has no effect.

Add the highlighted collision-theory mechanism for pressure to your notes before continuing.

Increasing pressure on a gas-phase equilibrium reaction always changes the value of Keq.

Apply

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

We just saw the collision-theory mechanism for pressure shifts. That raises a question: what is the NESA-specified experiment that makes pressure and temperature effects on equilibrium visible at the same time — and what are the two-stage colour observations? This card answers it → the NO₂/N₂O₄ equilibrium with its two-stage compression observation.

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):

Immediate observation: mixture becomes darker (both gas concentrations increase; more NO₂ per unit volume)
After re-equilibration: mixture becomes paler than the immediate darker colour as equilibrium shifts RIGHT (toward 1 mole N₂O₄, fewer gas moles) → NO₂ consumed → less brown

Temperature effect:

Heating: forward reaction exothermic → increase T → shift LEFT → more NO₂ produced → mixture becomes darker brown
Cooling → shift RIGHT → more N₂O₄ → mixture becomes paler

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.

2NO₂(g) ⇌ N₂O₄(g), ΔH = −57 kJ/mol: compression → immediate darker (all concentrations rise) → then paler (shift right, NO₂ consumed); heating (exothermic forward) → shift left → more NO₂ → darker; cooling → shift right → paler.

Pause — write the highlighted two-stage observation rule into your book before the check below.

Complete: After compressing the NO₂/N₂O₄ equilibrium, the immediate colour change is ________, but after re-equilibration the colour becomes ________ than the immediate post-compression colour.

Understand

4. Catalysts — The Full Explanation

We just saw the two-stage pressure and temperature effects in the NO₂/N₂O₄ system. That raises a question: what exactly does a catalyst do to equilibrium — and why doesn't it change Keq or yield? This card answers it → by separating what catalysts do (lower Eₐ equally for both directions) from what they cannot do (shift equilibrium or change Keq).

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:

Provides an alternative reaction pathway with a lower activation energy for BOTH forward and reverse reactions
Lowers Eₐ(forward) and Eₐ(reverse) by the same amount
Both forward and reverse reaction rates increase by the same factor
The transition state peak is lower on a reaction energy diagram; ΔH is unchanged

What a catalyst does NOT do:

Does not change the equilibrium position
Does not change Keq
Does not change ΔH
Does not preferentially affect either the forward or the reverse direction

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).

A catalyst lowers Eₐ equally for both forward and reverse reactions: it does NOT shift the equilibrium position, does NOT change Keq, does NOT change ΔH; it only increases the rate of reaching equilibrium — yield is unchanged.

Add the highlighted catalyst rules to your notes before the check below.

A catalyst is added to a flask containing SO₂, O₂, and SO₃ already at equilibrium. What happens?

Cross-lesson links: The 1913 BASF Oppau pressure compromise introduced in the hook is analysed quantitatively in L07 (industrial applications — rate/yield trade-off). Catalysts explored in Card 4 become the iron catalyst discussion in L07. The Keq vs position distinction in Card 5 is the most-tested concept across L09–L13 ICE table and Q calculations.

Evaluate

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

We just saw that catalysts change neither Keq nor equilibrium position. That raises a question: of all the disturbances we've studied — concentration, pressure, temperature, catalyst — which ones change Keq and which don't? This card answers it → with the complete summary table that is tested every year.

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 Changed	Effect on Equilibrium Position	Effect on Keq
Add reactant	Shifts right (toward products)	No change
Remove reactant	Shifts left	No change
Add product	Shifts left	No change
Remove product	Shifts right	No change
Increase pressure (gas, unequal moles)	Shifts toward fewer gas moles	No change
Decrease pressure (gas, unequal moles)	Shifts toward more gas moles	No change
Add catalyst	No shift	No change
Increase temperature (exothermic forward)	Shifts left	Decreases
Decrease temperature (exothermic forward)	Shifts right	Increases
Increase temperature (endothermic forward)	Shifts right	Increases
Decrease temperature (endothermic forward)	Shifts left	Decreases

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.

ONLY temperature changes Keq: concentration changes shift position (Keq unchanged); pressure changes shift position (Keq unchanged); catalyst has no effect on either; temperature increase for exothermic forward shifts left and decreases Keq.

Pause — copy the highlighted summary rule into your book before the check below.

Adding more reactant to an equilibrium mixture will shift the equilibrium position AND increase the value of Keq.

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

Activities

Worked Examples

Worked Example 1 — Apply

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).

Worked Example 2 — Band 5–6

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.

Revisit Think First

How Has Your Thinking Changed?

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.

Interactive Tool — Chemical Equilibrium Open fullscreen ↗

The Equilibrium tool shows Le Châtelier’s principle. Increasing pressure on a gaseous equilibrium shifts it toward the side with…

Predict then reveal+8 XP

1 · Predict

2 · Reveal

3 · Compare

N&sub2;(g) + 3H&sub2;(g) ⇌ 2NH&sub3;(g). The container volume is halved (pressure doubled). Predict the direction of shift and whether K_eq changes.

Confidence 50%

Complete the Learn phase to unlock practice questions.

✓

Multiple Choice

Q1 (1 mark): 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?

Q2 (1 mark): Which statement about the effect of a catalyst on a chemical equilibrium is correct?

Q3 (1 mark): For PCl₃(g) + Cl₂(g) ⇌ PCl₅(g), increasing pressure shifts equilibrium right. Which correctly explains why Keq is unchanged by this pressure increase?

Q4 (1 mark): Which of the following changes ONLY affects the rate of reaching equilibrium, without changing the equilibrium position or Keq?

Q5 (1 mark): A student performs the NO₂/N₂O₄ equilibrium experiment and compresses the syringe. The immediate colour observation is darker brown. After re-equilibration, which observation is correct?

Short Answer Questions

ApplyBand 5(3 marks) Q6: 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.

EvaluateBand 5(3 marks) Q7: 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.

EvaluateBand 6(4 marks) Q8: 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.

MODEL ANSWERS

Show Model Answers

Q6 Model Answer:

Stage 1 (immediate): The colour darkens immediately upon compression. This is because halving the volume doubles the concentration of both NO₂ (brown) and N₂O₄ (colourless) instantaneously. More NO₂ molecules per unit volume → more intense brown colour. This is a simple concentration increase, not an equilibrium shift — the shift takes time.

Stage 2 (after re-equilibration): The colour becomes paler than the immediate post-compression colour (but still darker than the original pre-compression colour). Le Chatelier's Principle: increasing pressure shifts equilibrium toward fewer moles of gas — left side has 2 mol NO₂; right side has 1 mol N₂O₄. Shift RIGHT → NO₂ (brown) is converted to N₂O₄ (colourless) → colour fades. The final colour is paler than the momentarily darkened colour because NO₂ has been partially consumed. Keq is unchanged.

Q7 Model Answer:

The student's statement is incorrect. A catalyst lowers the activation energy equally for both the forward and reverse reactions — it does not preferentially lower the forward activation energy. Both reaction rates increase by the same factor. For a system already at equilibrium (where forward rate = reverse rate), both rates increase equally → they remain equal → no net shift in equilibrium position occurs. No additional SO₃ is produced. The correct statement is: "Adding a catalyst to the flask will not shift the equilibrium or change the amount of SO₃ at equilibrium. It will allow the equilibrium to be reached more quickly if the system were disturbed, but has no effect on the equilibrium position or the value of Keq."

Q8 Model Answer:

(a) Increasing temperature: Forward reaction is endothermic (ΔH = +180 kJ/mol). Increasing temperature shifts equilibrium RIGHT (toward NO) — the endothermic direction is favoured to absorb the added heat. More NO produced. Keq increases (temperature change for endothermic forward reaction → Keq increases).

(b) Increasing pressure: Left: 1 mol N₂ + 1 mol O₂ = 2 mol gas. Right: 2 mol NO = 2 mol gas. Equal gas moles on both sides → pressure change has NO EFFECT on equilibrium position. No shift. Keq unchanged.

Evaluation: Increasing temperature is the only variable that can shift the equilibrium of this reaction to increase NO yield, because pressure has no effect (equal gas moles on both sides). However, temperature also increases the rate of reaction. There is no rate–yield trade-off for temperature in this endothermic system — both rate and yield increase with temperature. The limitation is practical (very high temperatures are costly and dangerous), not chemical.

Extended Response

Write a full Band 6 response: The Haber process N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ/mol, is operating at equilibrium. An engineer simultaneously increases the pressure to 400 atm and adds more iron catalyst. For each change, predict the effect on the equilibrium position and Keq. Include collision theory in your explanation of the pressure effect. (6 marks)

How did your thinking change?

Look back at what you wrote in the Think First section. Was the textbook statement about catalysts correct or not? Now consider the 1913 BASF Oppau decision: they chose 200 atm rather than 1000 atm for the Haber process. What does this lesson's analysis of pressure effects tell you about why higher pressure works in theory but fails economically? What is the correct statement about what a catalyst does to equilibrium position and Keq?

I have completed this lesson

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