Le Chatelier’s principle

Key idea: An equilibrium shifts to oppose a change in conditions. Compression favours fewer gas molecules, heating favours the endothermic direction, and only temperature changes the equilibrium constant.

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GCE A-Level H2 Chemistry 2027

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Industrial chemists change the concentration, pressure and temperature of an equilibrium mixture to get more of the product they want. Le Chatelier’s principle predicts which way an equilibrium moves when the conditions change. This lesson applies it to each kind of change, and shows which changes also alter the equilibrium constant.

Le Chatelier’s principle

Le Chatelier’s principle: if a change is made to a system in dynamic equilibrium, the position of equilibrium moves to oppose the change.

When the position of equilibrium moves to the right, the forward reaction goes faster than the reverse reaction for a while, until a new equilibrium is reached. When it moves to the left, the reverse reaction goes faster for a while.

“Oppose” does not mean “cancel”. The system reduces the effect of the change, but does not undo it completely.

Check your understanding 1

For A ⇌ B at equilibrium, extra A is added at constant temperature and volume. Predict the immediate rate imbalance and the later change in B before reading the graph.

Immediately after adding A

Reverse conversion exceeds forward conversionThe rates remain equal throughoutForward conversion exceeds reverse conversion

Show answer

The forward rate initially exceeds the reverse rate. B then increases as some of the added A reacts. Equal rates return at the new equilibrium.

Changing a concentration

At constant temperature:

adding a reactant, or removing a product, moves the position of equilibrium to the right;
adding a product, or removing a reactant, moves the position of equilibrium to the left.

Adding a reactant increases its forward consumption rate initially; adding a product increases its reverse consumption rate. Removing a product lowers the reverse rate initially, so forward production exceeds reverse consumption. Removing a reactant lowers the forward rate, so the reverse reaction predominates. At the new equilibrium, the two rates are equal again.

This applies only to substances whose concentrations can change. Adding more of a pure solid has no effect.

A starts at 0.600 and B at 0.400 mol dm⁻³. At 20 s, A jumps to 0.900 while B stays 0.400. A then falls towards 0.780 and B rises towards 0.520 mol dm⁻³; the total is 1.300 after addition. — Only A jumps at 20 s. The response consumes some added A and forms B, approaching a new composition with the same equilibrium ratio B/A = 2/3.

Figure data: Adding A: jump, then redistribution
Time (s)	A (reactant)	B (product)
0	0.6000	0.4000
1	0.6000	0.4000
2	0.6000	0.4000
3	0.6000	0.4000
4	0.6000	0.4000
5	0.6000	0.4000
6	0.6000	0.4000
7	0.6000	0.4000
8	0.6000	0.4000
9	0.6000	0.4000
10	0.6000	0.4000
11	0.6000	0.4000
12	0.6000	0.4000
13	0.6000	0.4000
14	0.6000	0.4000
15	0.6000	0.4000
16	0.6000	0.4000
17	0.6000	0.4000
18	0.6000	0.4000
19	0.6000	0.4000
20	0.6000	0.4000
20	0.9000	0.4000
21	0.8689	0.4311
22	0.8459	0.4541
23	0.8288	0.4712
24	0.8161	0.4839
25	0.8068	0.4932
26	0.7998	0.5002
27	0.7947	0.5053
28	0.7909	0.5091
29	0.7881	0.5119
30	0.7860	0.5140
31	0.7844	0.5156
32	0.7833	0.5167
33	0.7824	0.5176
34	0.7818	0.5182
35	0.7813	0.5187
36	0.7810	0.5190
37	0.7807	0.5193
38	0.7805	0.5195
39	0.7804	0.5196
40	0.7803	0.5197
41	0.7802	0.5198
42	0.7802	0.5198
43	0.7801	0.5199
44	0.7801	0.5199
45	0.7801	0.5199
46	0.7800	0.5200
47	0.7800	0.5200
48	0.7800	0.5200
49	0.7800	0.5200
50	0.7800	0.5200

Figure data: Adding A: jump, then redistribution
Time (s)	A (reactant)	B (product)
0	0.6000	0.4000
1	0.6000	0.4000
2	0.6000	0.4000
3	0.6000	0.4000
4	0.6000	0.4000
5	0.6000	0.4000
6	0.6000	0.4000
7	0.6000	0.4000
8	0.6000	0.4000
9	0.6000	0.4000
10	0.6000	0.4000
11	0.6000	0.4000
12	0.6000	0.4000
13	0.6000	0.4000
14	0.6000	0.4000
15	0.6000	0.4000
16	0.6000	0.4000
17	0.6000	0.4000
18	0.6000	0.4000
19	0.6000	0.4000
20	0.6000	0.4000
20	0.9000	0.4000
21	0.8689	0.4311
22	0.8459	0.4541
23	0.8288	0.4712
24	0.8161	0.4839
25	0.8068	0.4932
26	0.7998	0.5002
27	0.7947	0.5053
28	0.7909	0.5091
29	0.7881	0.5119
30	0.7860	0.5140
31	0.7844	0.5156
32	0.7833	0.5167
33	0.7824	0.5176
34	0.7818	0.5182
35	0.7813	0.5187
36	0.7810	0.5190
37	0.7807	0.5193
38	0.7805	0.5195
39	0.7804	0.5196
40	0.7803	0.5197
41	0.7802	0.5198
42	0.7802	0.5198
43	0.7801	0.5199
44	0.7801	0.5199
45	0.7801	0.5199
46	0.7800	0.5200
47	0.7800	0.5200
48	0.7800	0.5200
49	0.7800	0.5200
50	0.7800	0.5200

In this model of A ⇌ B, extra A is added at 20 s, so [A] jumps from 0.600 to 0.900 mol dm⁻³. The forward reaction then uses up some of the extra A: [A] falls to 0.780 mol dm⁻³ and [B] rises to 0.520 mol dm⁻³. The ratio [B]/[A] is 2/3 at both equilibria.

Check your understanding 2

In the model above, why does [A] at the new equilibrium (0.780 mol dm⁻³) not fall back to its original value of 0.600 mol dm⁻³?

Your explanation using total concentration and the ratio

Show answer

Adding A increased the total of A and B from 1.000 to 1.300 mol dm⁻³. The new equilibrium must still have [B]/[A] = 2/3, so with a larger total, both [A] and [B] are larger than before. The equilibrium moves to oppose the addition of A, but cannot remove all of it.

Problem

For the equilibrium

H₂(g) + I₂(g) ⇌ 2HI(g), more hydrogen is added at constant temperature and volume. State and explain the effect on the position of equilibrium and on the amount of hydrogen iodide, and state whether the equilibrium constant changes. [3 marks]

Try this before viewing the solution

Your answer

Show solution step by step

Position of equilibrium
1 mark
Method
Decide which way the equilibrium moves.
Reason
The forward reaction uses up some of the added hydrogen, opposing the change.
Working
Answer
The position of equilibrium moves to the right.
Hydrogen iodide
1 mark
Reason
Moving to the right forms more product.
Working
Answer
The amount of HI at the new equilibrium increases.
Equilibrium constant
1 mark
Working
Answer
The temperature has not changed, so the equilibrium constant is unchanged.

Self-mark with the mark scheme

Tick each mark point that your answer includes.

Mark the direction with its reason, the effect on HI and the equilibrium constant separately.

Mark points

1 mark: Moves to the right to use up the added hydrogen.
1 mark: More hydrogen iodide forms.
1 mark: Equilibrium constant unchanged because the temperature is constant.

Check your understanding 3 Adding oxygen

For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), oxygen is added at constant temperature and volume. Predict the change in the position of equilibrium. Is the new equilibrium concentration of sulfur dioxide higher or lower than before the oxygen was added? Does the equilibrium constant change?

Your shift, concentration change and constant

Show answer

The position of equilibrium moves to the right, to use up some of the added oxygen. Sulfur dioxide is used up as well, so its concentration at the new equilibrium is lower than before, and more sulfur trioxide forms. The temperature is unchanged, so the equilibrium constant is unchanged.

Changing the pressure of a gas mixture

Compressing a gas mixture at constant temperature increases its pressure. The position of equilibrium moves to reduce the pressure, by favouring the side with fewer moles of gas:

increasing the pressure favours the side with fewer moles of gas;
decreasing the pressure favours the side with more moles of gas;
if both sides have the same number of moles of gas, changing the pressure has no effect on the position of equilibrium.

Count only gases. Solids, liquids and aqueous species do not count.

Adding a gas that does not take part in the reaction, such as argon, at constant volume increases the total pressure but does not change the concentrations of the reacting gases, so the position of equilibrium does not move.

Problem

For

N₂(g) + 3H₂(g) ⇌ 2NH₃(g), predict the effect of increasing the pressure at constant temperature on the equilibrium yield of ammonia.

Show full solution

Count moles of gas
Method
Add the coefficients of the gases on each side.
Reason
Increasing the pressure favours the side with fewer moles of gas.
Working
Answer
Left: 1 + 3 = 4 mol of gas. Right: 2 mol of gas.
Direction
Method
Choose the side with fewer moles of gas.
Reason
Fewer gas molecules exert a lower pressure, which opposes the increase.
Working
Answer
The position of equilibrium moves to the right.
Yield
Working
Answer
The equilibrium yield of ammonia increases.

Quick check

This is why the Haber process is run at high pressure.

Learner claim

For

H₂(g) + Cl₂(g) ⇌ 2HCl(g), a learner says: “Increasing the pressure always moves a gas equilibrium to the right.” Explain what is wrong, and state the effect of increasing the pressure at constant temperature.

Compare the moles of gas

Effect of increasing the pressure

Position moves leftPosition moves rightPosition stays unchanged

Show solution step by step

Count moles of gas
Method
Compare the two sides.
Reason
Pressure favours the side with fewer moles of gas, so it matters only if the numbers differ.
Working
Answer
Left: 1 + 1 = 2 mol of gas. Right: 2 mol of gas.
Correct the prediction
Working
Answer
Moving either way would not change the number of gas molecules, so the position of equilibrium does not change.

Common mistake

Increasing the pressure favours the side with fewer moles of gas, which is not always the right-hand side.

Check your understanding 4

Explain why compressing H₂(g) + Cl₂(g) ⇌ 2HCl(g) at constant temperature does not favour either direction.

Your explanation

Show answer

Compare gas counts and their consequence: both sides contain 2 mol of gas per reaction as written. A shift would not reduce the number of gas molecules, so compression favours neither side. Both concentrations and reaction rates can rise without changing the equilibrium proportions.

Check your understanding 5

For N₂O₄(g) ⇌ 2NO₂(g), Δ H = +58 kJ mol⁻¹. Nitrogen dioxide is brown. Predict the colour change when the sealed mixture is cooled.

After re-equilibration on cooling

PalerUnchangedDarker

Show answer

Cooling favours the exothermic reverse reaction. The mixture becomes paler as nitrogen dioxide is converted to dinitrogen tetroxide.

Changing the temperature

Raising the temperature favours the endothermic direction, which absorbs heat. Lowering the temperature favours the exothermic direction, which releases heat. Use the sign of Δ H for the forward reaction:

if the forward reaction is exothermic (Δ H negative), raising the temperature moves the position of equilibrium to the left and lowers the yield of products;
if the forward reaction is endothermic (Δ H positive), raising the temperature moves the position of equilibrium to the right and raises the yield of products.

Raising the temperature makes both reactions faster, so equilibrium is reached sooner, even when the yield falls. The rate and the yield are separate questions.

Problem

Colourless dinitrogen tetroxide is in equilibrium with brown nitrogen dioxide:

N₂O₄(g) ⇌ 2NO₂(g),

Δ H = +58 kJ mol⁻¹. The sealed mixture is cooled. Predict the change in colour and in the equilibrium constant.

Try this before viewing the solution

Your answer

Hints

Hint 1

the other direction

The forward reaction absorbs heat. What does the reverse reaction do?

Hint 2

the rule

Lowering the temperature favours the direction that releases heat.

Show solution step by step

The exothermic direction
Method
Identify which direction releases heat.
Reason
Reversing a reaction changes the sign of Δ H.
Working
Answer
The forward reaction is endothermic, so the reverse reaction, forming N₂O₄, is exothermic.
Colour and constant
Working
Answer
Cooling favours the exothermic reverse reaction, so the position of equilibrium moves to the left: less NO₂ and the mixture becomes paler. Because the temperature changed, the equilibrium constant changes; for an endothermic forward reaction it decreases on cooling.

Quick check

Both reactions become slower on cooling, but the equilibrium mixture still contains less brown nitrogen dioxide.

Adding a catalyst

A catalyst provides a faster route in both directions. In an equilibrium mixture, the forward and reverse rates increase together and remain equal, so it has no effect on the position of equilibrium. Equilibrium is simply reached sooner.

Check your understanding 6

For A ⇌ B at equilibrium, both reaction rates are 5.0 mmol min⁻¹. After a catalyst is added at the same temperature, the forward rate becomes 15.0 mmol min⁻¹. Predict the reverse rate and the net rate of formation of B. What would differ if the catalyst were present from the start?

Reverse rate after adding catalyst

Unit: mmol min⁻¹

Net formation rate of B

Unit: mmol min⁻¹

Show answer

The reverse rate is also 15.0 mmol min⁻¹. The net formation rate of B is 15.0-15.0 = 0, so the composition stays constant. Present from the start, the catalyst would let the mixture reach this same equilibrium sooner, without changing its equilibrium composition.

Check your understanding 7

Explain why adding this catalyst leaves the composition unchanged, and what would differ if it were present from the start.

Your explanation

Show answer

Compare equal rates, zero net change and approach time: forward and reverse rates both become 15.0 mmol min⁻¹ at the same composition, so net formation is zero. A fresh mixture reaches that same equilibrium sooner with the catalyst.

Which changes alter the equilibrium constant?

For a given equation, the equilibrium constant depends only on the temperature.

Change	Position of equilibrium	Equilibrium constant
Concentration	may move	unchanged
Pressure	may move	unchanged
Catalyst	unchanged	unchanged
Temperature	moves	changes

Raising the temperature decreases the equilibrium constant for an exothermic forward reaction and increases it for an endothermic forward reaction.

Problem

For the exothermic equilibrium

2SO₂(g) + O₂(g) ⇌ 2SO₃(g), the pressure and the temperature are both increased. Predict the effect of each change on the yield of

SO₃. Can you predict the overall effect? How does the equilibrium constant change?

Try this before viewing the solution

Your answer

Hints

Hint 1

one change at a time

Count the moles of gas for the pressure change; find the endothermic direction for the temperature change.

Hint 2

opposite effects

If the two changes favour opposite directions, can you tell which is bigger?

Show solution step by step

Pressure
Method
Compare the moles of gas.
Reason
Higher pressure favours the side with fewer moles of gas.
Working
Answer
Left: 3 mol; right: 2 mol. Higher pressure favours the right, increasing the yield of SO₃.
Temperature
Method
Identify the endothermic direction.
Reason
Higher temperature favours the endothermic direction.
Working
Answer
The reverse reaction is endothermic, so higher temperature favours the left, decreasing the yield of SO₃.
Overall
Method
Combine the two effects.
Reason
Le Chatelier’s principle gives directions, not sizes.
Working
Answer
The effects are in opposite directions, so the overall effect on the yield cannot be predicted from this information.
Equilibrium constant
Working
Answer
Only the temperature change affects the equilibrium constant. The forward reaction is exothermic, so the equilibrium constant decreases.

Quick check

When two changes oppose each other, state each effect and say that the overall result cannot be predicted.

Testing one change at a time

[N₂]: 1.00 mol/dm³
[H₂]: 3.00 mol/dm³
[NH₃]: 0 mol/dm³
[NO₂]: 1.00 mol/dm³
[N₂O₄]: 3.00 mol/dm³
Q_c: 0
K_c: 0.185
Pressure: 237 atm
NH₃ at equilibrium: 25.9 %
Relative rate: 0.13

Concentration against time

Forward and backward rates against time

Try this

0 of 5 done

Let the closed mixture reach equilibrium and compare concentrations with the two reaction rates. (not done yet)
Let nitrogen and hydrogen reach equilibrium, then add nitrogen and watch equilibrium re-establish. (not done yet)
Squeeze the NO₂ and N₂O₄ syringe at equilibrium and watch the colour. (not done yet)
Heat an equilibrium mixture and compare K_c before and after. (not done yet)
In Haber conditions, compare the ammonia content and the rate at 350 °C and at 450 °C at the same pressure. (not done yet)

Keep Reaction set to Ammonia. If Play is shown, press it to start. Let the mixture settle, then record the ammonia concentration and the volume of the vessel. Work out the amount of ammonia using n = cV: when the volume changes, a higher concentration does not always mean more ammonia.

Check your understanding 9 One change at a time

Use Add H₂, reduce Vessel volume, raise Temperature, or switch on Catalyst to test how each change affects the equilibrium amount of ammonia: adding hydrogen, compressing the vessel, heating, and adding a catalyst. Record your prediction before each run. Make one change at a time, and reset between changes. Compare settled values of n = cV, not concentration alone; also compare the forward and reverse rate curves. The formation of ammonia is exothermic.

Your predictions, observations and n = cV comparison

Show answer

Compare your recorded predictions with the settled amounts and the rate curves. Adding hydrogen and compressing the vessel both increase the amount of ammonia. Heating decreases it, because the reverse reaction is endothermic. A catalyst makes the mixture reach equilibrium sooner but does not change the amount of ammonia at equilibrium.

Common mistakes

Saying that the equilibrium “cancels” a change. It only partly opposes it.
Counting solids, liquids or aqueous species when deciding the effect of pressure.
Saying that pressure always moves the equilibrium. If both sides have the same number of moles of gas, it has no effect.
Saying that adding an unreactive gas at constant volume moves the equilibrium.
Mixing up rate and yield. Raising the temperature means equilibrium is reached sooner, even when the yield falls.
Saying that a catalyst increases the yield.
Saying that a change in concentration or pressure changes the equilibrium constant. Only temperature does.

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Syllabus and review details

GCE A-Level H2 Chemistry 2027 · 9476-2027
9476 (2027), complete syllabus

Last reviewed: 2026-08-29

Le Chatelier’s principle

Check your understanding 1

Changing a concentration

Check your understanding 2

Check your understanding 3 Adding oxygen

Changing the pressure of a gas mixture

Check your understanding 4

Check your understanding 5

Changing the temperature

Adding a catalyst

Check your understanding 6

Check your understanding 7

Which changes alter the equilibrium constant?

Check your understanding 8

Testing one change at a time

Try this

Check your understanding 9 One change at a time

Common mistakes