ChemistryISC Class 11 · Chemical Equilibrium

Le Chatelier's Principle: Shifting Equilibria and Industrial Conditions

Updated 17 October 2026 · Fact-checked

Le Chatelier's principle states that a system at equilibrium that is disturbed by a change in concentration, pressure or temperature shifts in the direction that opposes the change. A catalyst speeds up both directions equally and does not change the position of equilibrium or K.

Notice that 4 gas moles turn into 2, so high pressure, low temperature and removing NH₃ push the equilibrium forward, while a catalyst only saves time.

Illustration drawn for QuestionBuddy.

Understand Le Chatelier's Principle and Industrial Equilibria

The principle. If a system at equilibrium is subjected to a change of concentration, pressure (volume) or temperature, the system adjusts in the direction that partly cancels the change.

Concentration. Adding a reactant (or removing a product) shifts the equilibrium forward; adding a product (or removing a reactant) shifts it backward. K is unchanged; Q is compared with K to confirm the direction.

Pressure and volume. For gaseous systems, an increase of pressure (decrease of volume) favours the side with fewer gas moles. If Δn = 0, pressure has no effect. Adding an inert gas at constant volume has no effect on the equilibrium.

Temperature. Raising the temperature favours the endothermic direction; lowering it favours the exothermic direction. This is the only change that alters K: K increases with T for an endothermic reaction and decreases for an exothermic one.

Catalyst. It lowers the activation energy for both forward and backward reactions equally, so equilibrium is reached faster but the composition and K are unchanged.

Industrial applications. *Haber process*, N₂ + 3H₂ ⇌ 2NH₃ (ΔH = −92.4 kJ mol⁻¹): the forward reaction is exothermic with fewer gas moles, so high pressure (about 200 atm) and low temperature favour the yield; because a low temperature makes the rate too slow, a compromise temperature of about 700 K with an iron catalyst (with promoter) is used, and the ammonia is removed as a liquid. *Contact process*, 2SO₂ + O₂ ⇌ 2SO₃ (ΔH = −198 kJ mol⁻¹): the same reasoning gives a temperature of around 700 to 720 K, V₂O₅ as a catalyst, and a slight excess of oxygen; because the equilibrium already lies far to the right, a pressure of only about 1 to 2 atm is used.

Le Chatelier's Principle and Industrial Equilibria: formulas and key facts

Concentration: Add reactant: forward | Add product: backward; K unchanged.
Pressure: Higher p shifts toward fewer gas moles; No effect if Δn = 0.
Temperature: Higher T shifts toward the endothermic direction; K changes with T.
Reaction quotient: $Q < K : f or w a r d ∣ Q > K : ba c k w a r d$; Use after a concentration change.

Le Chatelier's Principle and Industrial Equilibria in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Why ammonia plants use high pressure

Suppose a plant manager asks why N₂ + 3H₂ ⇌ 2NH₃ is run at about 200 atm.

Explain using Le Chatelier's principle.

Try it first, then show the solutionSolution

Left side has 4 moles of gas; right side has 2 moles.
Raising the pressure shifts the equilibrium towards the side with fewer gas moles, that is towards NH₃.
High pressure therefore raises the yield; the cost of strong vessels limits how high the pressure is set.

Answer: Higher pressure increases the yield of ammonia.

Memory hook: Pressure affects only the position of equilibrium, not K.

Tricks to remember Le Chatelier's Principle and Industrial Equilibria

Short memory aids. Say each one aloud once, then use it on a question the same day.

Opposes the change
Whatever you do, the system does the opposite as far as possible.
Use it when: Predicting the shift in any question.
Only T changes K
Concentration, pressure and catalysts leave K unchanged.
Use it when: Statements about K.

Le Chatelier's Principle and Industrial Equilibria: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Role of a catalyst · 1 mark · MCQBoard level

A catalyst added to a system at equilibrium:
(a) shifts the equilibrium to the right
(b) shifts the equilibrium to the left
(c) changes the value of K
(d) increases the rates of forward and reverse reactions equally

Try it first, then show the solutionSolution

It lowers the activation energy of both reactions equally.

Answer: (d) increases the rates of forward and reverse reactions equally

Example 2: Effect of conditions on ammonia equilibrium · 3 marks · Short answerBoard level

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92.4 kJ, state the effect on the equilibrium of (i) increasing the pressure, (ii) raising the temperature, (iii) adding N₂, (iv) adding a catalyst.

Try it first, then show the solutionSolution

(i) Forward: 4 mol of gas becomes 2 mol. (0.75 mark)
(ii) Backward: the forward reaction is exothermic, so heating favours the reverse. (0.75 mark)
(iii) Forward: the extra N₂ is partly used up. (0.75 mark)
(iv) No change in position or K; equilibrium is reached faster. (0.75 mark)

Answer: (i) forward; (ii) backward; (iii) forward; (iv) no shift.

Example 3: Adding a reactant · 5 marks · Short answerHarder board level

For PCl₅ ⇌ PCl₃ + Cl₂ at equilibrium, [PCl₅] = 0.15, [PCl₃] = [Cl₂] = 0.10 mol L⁻¹ (Kc = 0.0667). An extra 0.20 mol L⁻¹ of Cl₂ is added at constant volume. Find Q, say which way the reaction goes, and find the new equilibrium concentrations.

Try it first, then show the solutionSolution

Immediately after: [Cl₂] = 0.30, so Q = (0.10 × 0.30)/0.15 = 0.20 > Kc, so the reaction goes backward. (1 mark)
Let x mol L⁻¹ of PCl₃ and Cl₂ recombine: (0.10 − x)(0.30 − x)/(0.15 + x) = 0.0667. (1 mark)
Expanding: x² − 0.4667x + 0.02 = 0. (1 mark)
x = [0.4667 − √(0.2178 − 0.08)]/2 = (0.4667 − 0.3712)/2 = 0.0478. (1 mark)
[PCl₃] = 0.052, [Cl₂] = 0.252, [PCl₅] = 0.198 mol L⁻¹. Check: 0.0522 × 0.2522/0.1978 = 0.0666. (1 mark)

Answer: Q = 0.20 > K: backward; new [PCl₃] = 0.052, [Cl₂] = 0.252, [PCl₅] = 0.198 mol L⁻¹.

Example 4: Contact process · 4 marks · Short answerHarder board level

2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −198 kJ. Explain the conditions used in the manufacture of SO₃: temperature, pressure, catalyst and excess of oxygen.

Try it first, then show the solutionSolution

The forward reaction is exothermic, so a low temperature favours SO₃; but the rate is too slow, so a compromise of about 700 K is used. (1 mark)
A vanadium(V) oxide catalyst speeds up the attainment of equilibrium without changing the yield. (1 mark)
High pressure would favour SO₃ (3 mol → 2 mol), but the equilibrium already lies far to the right, so only about 1 to 2 atm is used. (1 mark)
Excess oxygen shifts the equilibrium forward and uses up SO₂. (1 mark)

Answer: About 700 K, V₂O₅ catalyst, 1 to 2 atm, excess O₂.

Example 5: Effect of volume change · 5 marks · Long answerHOTS / competency

For N₂O₄ ⇌ 2NO₂, Kc = 0.005 mol L⁻¹. 0.100 mol of N₂O₄ in a 1.0 L flask reaches equilibrium. Find the equilibrium concentrations. The volume is then suddenly halved: find Q and state the direction of shift.

Try it first, then show the solutionSolution

(2x)²/(0.100 − x) = 0.005 → 4x² + 0.005x − 0.0005 = 0. (1 mark)
x = [−0.005 + √(0.000025 + 0.008)]/8 = (−0.005 + 0.0896)/8 = 0.01057. (1 mark)
[NO₂] = 0.0211 and [N₂O₄] = 0.0894 mol L⁻¹ (α = 10.6%). (1 mark)
Halving the volume doubles the concentrations: [N₂O₄] = 0.1789 and [NO₂] = 0.0423, so Q = 0.0423²/0.1789 = 0.0100. (1 mark)
Q = 0.010 > Kc = 0.005, so the shift is backward, towards N₂O₄ (fewer moles), as the principle predicts. (1 mark)

Answer: [NO₂] = 0.0211, [N₂O₄] = 0.0894 mol L⁻¹; after compression Q = 0.010 > K: shift to N₂O₄.

Where students lose marks in Le Chatelier's Principle and Industrial Equilibria

The slips examiners see most often, and what to write instead.

Mistake: Saying a catalyst shifts the equilibrium
Writing that a catalyst increases the yield.
Do this: It speeds up the approach to equilibrium but does not change the composition.
Usually costs: 2 marks.
Mistake: Wrong effect of temperature
Saying heating favours the exothermic direction.
Do this: Heating favours the endothermic direction.
Usually costs: 1 to 2 marks.
Mistake: Applying pressure when Δn = 0
Predicting a shift for N₂ + O₂ ⇌ 2NO.
Do this: Equal gas moles on both sides: no effect.
Usually costs: 1 mark.

Exam tips

Name the change, then name the direction of shift and the reason.
Count gas moles on both sides for pressure questions.
Quote ΔH when discussing temperature.
Mention K is unchanged except for temperature changes.

Practice questions on Le Chatelier's Principle and Industrial Equilibria

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
For N₂(g) + O₂(g) ⇌ 2NO(g), an increase in pressure at constant temperature:
(a) shifts the equilibrium to the right
(b) shifts it to the left
(c) has no effect
(d) decreases K
Hint
Δn = 0.
Answer
(c) has no effect
Q2 · 2 marks · Short answerBoard level
In the Haber process NH₃ is removed as it is formed. Explain why.
Hint
Think about the effect of removing a product.
Answer
Removing the product lowers Q below K, so the equilibrium keeps shifting forward and the yield rises.
Q3 · 3 marks · Short answerHarder board level
State the conditions that favour the formation of PCl₅ in PCl₃ + Cl₂ ⇌ PCl₅ (ΔH = −88 kJ), with reasons.
Hint
Count gas moles and use the sign of ΔH.
Answer
High pressure (2 mol → 1 mol) and low temperature (the forward reaction is exothermic); also excess Cl₂ or removal of PCl₅.
Q4 · 3 marks · Short answerHarder board level
For 2NO₂(g) ⇌ N₂O₄(g), ΔH = −57 kJ. NO₂ is brown and N₂O₄ is colourless. Predict the colour change when the flask is (i) compressed and (ii) heated.
Hint
Count gas moles and use the sign of ΔH.
Answer
(i) Compression favours the side with fewer moles, N₂O₄, so the brown colour fades after the first darkening from the higher concentration. (ii) The forward reaction is exothermic, so heating favours NO₂ and the colour deepens.
Q5 · 3 marks · Short answerHarder board level
For A + B ⇌ C + D, Kc = 4. A mixture has [A] = [B] = 0.5, [C] = 1.0, [D] = 1.5 mol L⁻¹. Find Q and state the direction of the reaction.
Hint
Q = [C][D]/([A][B]).
Answer
Q = (1.0 × 1.5)/(0.5 × 0.5) = 6 > 4, so the reaction proceeds backward.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a sulphuric acid plant uses 2SO₂ + O₂ ⇌ 2SO₃ (ΔH = −198 kJ).
(i) What is the effect on the equilibrium of increasing the pressure? (1 mark)
(ii) What is the effect of raising the temperature on the yield and on K? (1 mark)
(iii) What is the effect of adding O₂? (1 mark)
(iv) What is the effect of removing SO₃? (1 mark)
Hint
Forward direction is exothermic and reduces gas moles.
Answer
(i) It shifts forward (3 mol → 2 mol), raising the yield. (ii) The yield falls and K decreases, since the forward reaction is exothermic. (iii) It shifts forward. (iv) It shifts forward.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State Le Chatelier's principle.Show answer
A system at equilibrium opposes any change in conditions.
2.Which change alters K?Show answer
Temperature.
3.Catalyst used in the Contact process?Show answer
V₂O₅.
4.Catalyst used in the Haber process?Show answer
Iron.

Le Chatelier's Principle and Industrial Equilibria: frequently asked questions

Why not use a very low temperature for ammonia synthesis?

The yield would be high at equilibrium but the rate would be too slow to be useful; a moderate temperature with a catalyst is a compromise.

What happens when an inert gas is added at constant volume?

Concentrations (and partial pressures) of the reacting gases do not change, so the equilibrium is unchanged.

Does Le Chatelier's principle give the new value of K?

No. It predicts only the direction of shift; calculations with K or Q give the extent.