Dynamic Equilibrium and Le Chatelier’s Principle

Learn and apply Dynamic Equilibrium and Le Chatelier’s Principle for H2 Chemistry 9476.

GCE A-Level H2 Chemistry 9476-2027

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Dynamic Equilibrium and Le Chatelier’s Principle: Orientation

Equilibrium questions are mostly “language + logic”: define equilibrium correctly, then apply Le Chatelier with one clean chain (disturbance → shift → effect on named species → whether K changes).

Use the course selector and topic navigator on this page to move between this equilibrium concept lesson, the equilibrium-constant lesson for your course, and the Chemical Equilibria hub.

Definitions (Must Know)

A. Reversible reaction

A reversible reaction can proceed in both the forward and reverse directions under the same conditions.

B. Dynamic equilibrium

A system is at dynamic equilibrium when the forward and reverse reaction rates are equal, so macroscopic quantities (e.g. concentration, pressure) remain constant.

C. Le Chatelier’s principle

Le Chatelier’s principle: when a system at equilibrium is disturbed, the position of equilibrium shifts to oppose the disturbance.

D. Position of equilibrium

The position of equilibrium describes which side has a higher proportion of products/reactants at equilibrium (not “equal amounts”).

Key Ideas (What Earns Marks)

Equilibrium must be in a closed system (no loss of reactants/products).
At equilibrium, rates are equal, but concentrations are not necessarily equal.
A catalyst does not change the position of equilibrium; it increases both forward and reverse rates.
Concentration/pressure changes shift equilibrium to reduce the change; temperature changes shift based on endothermic/exothermic direction.
Changing conditions does not change K unless temperature changes.

Fast decision table:

Change	Equilibrium position	K value
concentration	may shift	unchanged
pressure (gases)	may shift	unchanged
catalyst	no shift	unchanged
temperature	shifts depending on Δ H	changes

State the disturbance. 2) State the shift direction. 3) State the effect on a named species. 4) State whether K changes.

Quick visuals (equilibrium reached + disturbance):

Reaching Dynamic Equilibrium in a Closed System (Example). A (reactant), B (product) plotted as Concentration against Time. — Example concentrations for A ⇌ B: A falls and B rises until both become constant at equilibrium (forward rate = reverse rate).

Values for Reaching Dynamic Equilibrium in a Closed System (Example)
Time (arbitrary units)	A (reactant)	B (product)
0	1	0
5	0.82	0.18
10	0.72	0.28
20	0.63	0.37
30	0.6	0.4
40	0.6	0.4

Values for Reaching Dynamic Equilibrium in a Closed System (Example)
Time (arbitrary units)	A (reactant)	B (product)
0	1	0
5	0.82	0.18
10	0.72	0.28
20	0.63	0.37
30	0.6	0.4
40	0.6	0.4

Disturbing Equilibrium by Adding Reactant (Example). A (reactant), B (product) plotted as Concentration against Time. — At time 20, A is added (instant jump). The system shifts to oppose the change: A falls and B rises to a new equilibrium. K is unchanged if temperature is constant.

Values and uncertainty for Disturbing Equilibrium by Adding Reactant (Example)
Series	Time (arbitrary units)	Concentration (mol dm^-3)
A (reactant)	0	1
A (reactant)	10	0.72
A (reactant)	20	0.6
A (reactant)	20	0.9
A (reactant)	30	0.8
A (reactant)	40	0.75
A (reactant)	50	0.75
B (product)	0	0
B (product)	10	0.28
B (product)	20	0.4
B (product)	30	0.5
B (product)	40	0.55
B (product)	50	0.55

Values and uncertainty for Disturbing Equilibrium by Adding Reactant (Example)
Series	Time (arbitrary units)	Concentration (mol dm^-3)
A (reactant)	0	1
A (reactant)	10	0.72
A (reactant)	20	0.6
A (reactant)	20	0.9
A (reactant)	30	0.8
A (reactant)	40	0.75
A (reactant)	50	0.75
B (product)	0	0
B (product)	10	0.28
B (product)	20	0.4
B (product)	30	0.5
B (product)	40	0.55
B (product)	50	0.55

Detailed Explanations

A. A repeatable workflow (Le Chatelier questions)

Write the balanced equilibrium equation (with states).
Identify the disturbance (what changed?).
Decide which side opposes it (uses up what was added, replaces what was removed).
Conclude: “shifts left/right”, then state “more/less of X”.

Mini example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) If H₂ is added, the system removes added reactant by forming more NH₃ → shifts right.

B. Concentration changes

For: aA + bB ⇌ cC + dD

Increasing a reactant concentration shifts equilibrium right (to use it up).
Removing a product shifts equilibrium right (to replace it).

Because concentration changes do not change the reaction’s temperature, they do not change K; therefore the system shifts position until the new equilibrium concentrations satisfy the same K value.

C. Pressure changes (gases only)

Pressure effects matter when there are different total moles of gas on each side.

Increasing pressure shifts to the side with fewer moles of gas.
Decreasing pressure shifts to the side with more moles of gas.

Mini example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) Left: 4 mol gas, right: 2 mol gas, so increasing pressure shifts right (more NH₃).

D. Temperature changes

Treat heat as a reagent:

If forward reaction is exothermic, heat is like a product. Increasing temperature shifts left.
If forward reaction is endothermic, heat is like a reactant. Increasing temperature shifts right.

Because temperature changes change the relative favourability of products vs reactants, temperature is the only factor here that changes K.

E. Catalysts

A catalyst speeds up reaching equilibrium but does not change K or the final equilibrium composition.

Worked Examples

Problem

For

N₂(g) + 3H₂(g) ⇌ 2NH₃(g), predict the shift when pressure is increased and state the effect on ammonia yield.

Study the worked solution

Count gaseous amounts
Method
Add the stoichiometric coefficients of gaseous species on each side.
Reason
At constant temperature, increasing pressure favours the side occupying fewer gaseous moles.
Working
Left: 1 + 3 = 4 mol gas; right: 2 mol gas.
Oppose the disturbance
Method
Select the side with fewer gaseous moles.
Reason
A shift to that side reduces the pressure increase.
Working
The equilibrium shifts right.
Name the composition effect
Method
State which product becomes more abundant at the new equilibrium.
Reason
A rightward shift consumes nitrogen and hydrogen and forms ammonia.
Working
The equilibrium yield of NH₃ increases.

Problem

For an equilibrium whose forward reaction is exothermic, temperature is increased. Predict the change in the equilibrium yield of the forward products and explain whether

K changes.

Try this before viewing the solution

Heat placement, shift, yield and K

Hints

Hint 1: place heat

For an exothermic forward reaction, treat heat as a product.

Hint 2: oppose added heat

Increasing temperature adds heat, so choose the direction that consumes it; then recall which disturbance can change

View solution step by step

Represent the thermal change
Method
Place heat on the product side of the exothermic forward reaction.
Reason
That makes the reverse direction endothermic and able to absorb added heat.
Working
reactants ⇌ products + heat.
Predict the shift and yield
Method
Shift left to oppose the temperature increase.
Reason
The endothermic reverse reaction consumes some of the added heat.
Working
The equilibrium yield of forward products decreases.
State the constant effect
Method
State that the equilibrium constant changes.
Reason
Temperature changes the relative favourability of the forward and reverse reactions.
Working
For this exothermic forward reaction, increasing temperature decreases K.

Learner claim

For

H₂(g) + Cl₂(g) ⇌ 2HCl(g), a learner says, “Increasing pressure always shifts a gaseous equilibrium to the right.” Diagnose the error and state what happens at constant temperature.

Compare gaseous counts

Effect of increased pressure

No equilibrium shiftShift rightShift left

View solution step by step

Test the pressure condition
Method
Count gaseous coefficients on both sides.
Reason
Pressure favours a side only when the total gaseous mole counts differ.
Working
Left: 1 + 1 = 2 mol gas; right: 2 mol gas.
Correct the prediction
Method
State that neither side is favoured.
Reason
Shifting either way would not reduce the total gaseous mole count.
Working
There is no shift in equilibrium position, and K is unchanged because temperature is constant.

Problem

For

H₂(g) + I₂(g) ⇌ 2HI(g) at constant temperature, more

H₂ is added. Predict the shift, state the effect on

HI and state whether

K changes. [3 marks]

Try this before viewing the solution

Shift, named composition effect and K

View solution step by step

Predict the response
1 mark
Method
State that equilibrium shifts right.
Reason
The forward reaction consumes some of the added hydrogen.
Working
Added reactant H₂ → rightward shift.
Name the composition effect
1 mark
Method
State that more hydrogen iodide is formed.
Reason
A rightward shift favours the product side.
Working
The amount of HI at the new equilibrium increases.
Treat the constant
1 mark
Method
State that K is unchanged.
Reason
The temperature has not changed.
Working
Concentration changes the equilibrium position, not K at fixed temperature.

Self-mark with the mark scheme

Compare your response with each mark point. Select a point only when your response contains that evidence.

Credit the direction, the named product effect and the fixed-temperature K statement separately.

Mark points

1 mark: Predicts a rightward shift to consume added hydrogen.
1 mark: States that more hydrogen iodide forms.
1 mark: States that K is unchanged because temperature is constant.

Problem

For the exothermic equilibrium

2SO₂(g) + O₂(g) ⇌ 2SO₃(g), pressure and temperature are both increased. Predict the direction favoured by each change, decide whether the overall change in equilibrium

SO₃ yield can be determined from Le Chatelier’s principle alone, and state how

K changes.

Try this before viewing the solution

Separate pressure, temperature, net yield and K

Hints

Hint 1: analyse separately

Compare three gaseous moles with two for pressure; place heat on the product side for temperature.

Hint 2: do not force a net direction

If the two disturbances favour opposite directions, qualitative rules alone do not show which effect is larger.

View solution step by step

Analyse pressure
Method
Compare the gaseous coefficients.
Reason
Higher pressure favours the side with fewer gaseous moles.
Working
Left: 3 mol gas; right: 2 mol gas, so pressure favours the right.
Analyse temperature
Method
Treat heat as a product of the exothermic forward reaction.
Reason
Higher temperature favours the endothermic reverse direction.
Working
Temperature favours the left.
Judge the combined outcome
Method
Keep the two qualitative effects separate.
Reason
They favour opposite directions, and Le Chatelier’s principle does not quantify their relative sizes.
Working
The net change in equilibrium SO₃ yield cannot be determined from this information alone.
State the K effect
Method
Apply the temperature-only rule for the equilibrium constant.
Reason
Increasing temperature disfavors products for an exothermic forward reaction.
Working
K decreases; the pressure change itself does not alter K.

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Hydrogen and iodine only are sealed in a flask at 700 K and react: H₂(g) + I₂(g) ⇌ 2HI(g). Which statement describes the rates as the mixture approaches equilibrium?

The forward rate falls and the reverse rate rises from zero until the two rates are equal; both reactions then continue.
The forward rate falls to zero at equilibrium, and the reverse rate stays at zero.
Both rates rise until the concentrations of H₂, I₂ and HI are equal.
The reverse reaction begins only after all the H₂ and I₂ has reacted.

N₂O₄(g) ⇌ 2NO₂(g), ΔH = +58 kJ mol⁻¹. The equilibrium mixture is compressed to a smaller volume at constant temperature. How does the position of equilibrium change?

It shifts to the left, towards the side with fewer gas molecules, which partly opposes the pressure increase.
It shifts to the right, because the higher pressure makes collisions more frequent.
It does not change, because the temperature is constant.
It shifts to the right, because the forward reaction is endothermic.

2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −197 kJ mol⁻¹. Which change alters the value of Kc?

Raising the temperature, which decreases Kc.
Raising the pressure, which increases Kc.
Adding a catalyst, which increases Kc.
Removing SO₃ as it forms, which decreases Kc.

2NO(g) + O₂(g) ⇌ 2NO₂(g). At equilibrium a 1.00 dm³ vessel contains 0.20 mol NO, 0.10 mol O₂ and 0.40 mol NO₂. What is the value of Kc?

40 mol⁻¹ dm³
20 mol⁻¹ dm³
0.025 mol dm⁻³
40 mol dm⁻³

1.00 mol CO and 3.00 mol H₂ are sealed in a vessel: CO(g) + 2H₂(g) ⇌ CH₃OH(g). At equilibrium the vessel contains 0.40 mol CH₃OH. What amounts of CO and H₂ are present at equilibrium?

0.60 mol CO and 2.20 mol H₂
0.60 mol CO and 2.60 mol H₂
0.60 mol CO and 2.80 mol H₂
0.40 mol CO and 0.80 mol H₂

N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ mol⁻¹. Why is the Haber process operated at about 450 °C rather than a much lower temperature?

A lower temperature would give a higher equilibrium yield, but equilibrium would be reached too slowly even with the iron catalyst; 450 °C is a compromise between yield and rate.
A higher temperature gives a higher equilibrium yield of ammonia.
The iron catalyst increases the equilibrium yield only when it is hot.
Below 450 °C the forward reaction becomes endothermic.

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A little radioactive iodine is added to an equilibrium mixture of H₂(g), I₂(g) and HI(g) at constant temperature. Later, radioactivity is found in both I₂ and HI, but the concentrations of all three substances are unchanged. What does this show?

The forward and reverse reactions are both still happening, at equal rates.
The position of equilibrium has shifted to the right.
The mixture had not yet reached equilibrium.
The radioactive iodine acts as a catalyst for the reaction.

CO(g) + 2H₂(g) ⇌ CH₃OH(g), ΔH = −91 kJ mol⁻¹. The temperature of the equilibrium mixture is raised at constant pressure. What happens to the equilibrium amount of CH₃OH?

It decreases, because the position shifts in the endothermic (reverse) direction to absorb the added heat.
It increases, because both reactions become faster.
It is unchanged, because temperature affects only the rate.
It increases, because the position shifts to the side with fewer gas molecules.

N₂O₄(g) ⇌ 2NO₂(g), ΔH = +58 kJ mol⁻¹. Which change increases the value of Kc?

Raising the temperature.
Lowering the pressure at constant temperature.
Removing NO₂ as it forms at constant temperature.
Adding a catalyst.

2SO₂(g) + O₂(g) ⇌ 2SO₃(g). The equilibrium concentrations are [SO₂] = 0.20, [O₂] = 0.10 and [SO₃] = 0.60 mol dm⁻³. What is the value of Kc?

90 mol⁻¹ dm³
30 mol⁻¹ dm³
90 mol dm⁻³
0.011 mol dm⁻³

0.80 mol H₂S is heated in a sealed vessel: 2H₂S(g) ⇌ 2H₂(g) + S₂(g). At equilibrium 25% of the H₂S has dissociated. What amount of S₂ is present at equilibrium?

0.10 mol
0.20 mol
0.40 mol
0.60 mol

In the Haber process a pressure of about 200 atm is used. A higher pressure would give a higher equilibrium yield of ammonia. Why is a much higher pressure not used?

The extra yield does not justify the higher cost of stronger plant and of the energy for compression, and the greater safety risk.
Higher pressure would decrease Kc, lowering the yield.
Higher pressure would shift the position to the left, towards more gas molecules.
Higher pressure would poison the iron catalyst.

CaCO₃(s) ⇌ CaO(s) + CO₂(g). At 1100 K the equilibrium partial pressure of carbon dioxide is 0.45 atm. What is Kp at this temperature?

0.45 atm
2.2 atm⁻¹
It cannot be found without the amounts of CaCO₃ and CaO.
0.45, with no units

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The forward and reverse reactions are both still happening, at equal rates.
The position of equilibrium has shifted to the right.
The mixture had not yet reached equilibrium.
The radioactive iodine acts as a catalyst for the reaction.

CO(g) + 2H₂(g) ⇌ CH₃OH(g), ΔH = −91 kJ mol⁻¹. The temperature of the equilibrium mixture is raised at constant pressure. What happens to the equilibrium amount of CH₃OH?

It decreases, because the position shifts in the endothermic (reverse) direction to absorb the added heat.
It increases, because both reactions become faster.
It is unchanged, because temperature affects only the rate.
It increases, because the position shifts to the side with fewer gas molecules.

N₂O₄(g) ⇌ 2NO₂(g), ΔH = +58 kJ mol⁻¹. Which change increases the value of Kc?

Raising the temperature.
Lowering the pressure at constant temperature.
Removing NO₂ as it forms at constant temperature.
Adding a catalyst.

2SO₂(g) + O₂(g) ⇌ 2SO₃(g). The equilibrium concentrations are [SO₂] = 0.20, [O₂] = 0.10 and [SO₃] = 0.60 mol dm⁻³. What is the value of Kc?

90 mol⁻¹ dm³
30 mol⁻¹ dm³
90 mol dm⁻³
0.011 mol dm⁻³

0.80 mol H₂S is heated in a sealed vessel: 2H₂S(g) ⇌ 2H₂(g) + S₂(g). At equilibrium 25% of the H₂S has dissociated. What amount of S₂ is present at equilibrium?

0.10 mol
0.20 mol
0.40 mol
0.60 mol

In the Haber process a pressure of about 200 atm is used. A higher pressure would give a higher equilibrium yield of ammonia. Why is a much higher pressure not used?

The extra yield does not justify the higher cost of stronger plant and of the energy for compression, and the greater safety risk.
Higher pressure would decrease Kc, lowering the yield.
Higher pressure would shift the position to the left, towards more gas molecules.
Higher pressure would poison the iron catalyst.

CaCO₃(s) ⇌ CaO(s) + CO₂(g). At 1100 K the equilibrium partial pressure of carbon dioxide is 0.45 atm. What is Kp at this temperature?

0.45 atm
2.2 atm⁻¹
It cannot be found without the amounts of CaCO₃ and CaO.
0.45, with no units

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Mind Stretchers

Mind stretcher 1Extension

Explain why adding a catalyst does not change the equilibrium composition.

Show Hint

Separate the immediate disturbance from the subsequent shift, then decide independently whether temperature changed.

Show Answer

Mark scheme:

A catalyst lowers activation energy for both forward and reverse reactions.
Both rates increase, so equilibrium is reached faster.
At equilibrium, rates are equal; the ratio of forward/reverse rate constants at a given temperature (related to K) is unchanged.

Mind stretcher 2: Disturbance, response and KExtension

Question. For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), oxygen is added at constant temperature and volume. Predict the shift, the eventual change in [SO₂], and whether K_c changes.

Show Hint

The system consumes some added oxygen by favouring the forward reaction; concentration changes alone do not change Kc.

Show Answer

The equilibrium shifts right, consuming SO₂ and producing more SO₃, so the eventual SO₂ concentration is lower than immediately before the disturbance. K_c is unchanged because temperature is unchanged.

Continue where you stopped

Dynamic Equilibrium and Le Chatelier’s Principle: Orientation

Definitions (Must Know)

A. Reversible reaction

B. Dynamic equilibrium

C. Le Chatelier’s principle

D. Position of equilibrium

Key Ideas (What Earns Marks)

Detailed Explanations

A. A repeatable workflow (Le Chatelier questions)

B. Concentration changes

C. Pressure changes (gases only)

D. Temperature changes

E. Catalysts

Worked Examples

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Mind Stretchers

Mind stretcher 1Extension

Mind stretcher 2: Disturbance, response and KExtension