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Dynamic Equilibrium And Le Chatelier

Learn and apply Dynamic Equilibrium And Le Chatelier in the published Chemistry course sequence.

  • GCE A-Level H1 Chemistry 8873-2027
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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:

ChangeEquilibrium positionK value
concentrationmay shiftunchanged
pressure (gases)may shiftunchanged
catalystno shiftunchanged
temperatureshifts depending on Δ Hchanges
Quick Recall (Answer Skeleton)
  1. 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)

Reaching Dynamic Equilibrium in a Closed System (Example). A (reactant), B (product) plotted as Concentration against Time.

Reaching Dynamic Equilibrium in a Closed System (Example). A (reactant), B (product) plotted as Concentration against Time.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).
Open full-size graph
View figure data
Values for Reaching Dynamic Equilibrium in a Closed System (Example)
Time (arbitrary units)A (reactant)B (product)
010
50.820.18
100.720.28
200.630.37
300.60.4
400.60.4

Disturbing Equilibrium by Adding Reactant (Example)

Disturbing Equilibrium by Adding Reactant (Example). A (reactant), B (product) plotted as Concentration against Time.

Scroll across the graph to read all labels.

Disturbing Equilibrium by Adding Reactant (Example). A (reactant), B (product) plotted as Concentration against Time.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.
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View figure data
Values and uncertainty for Disturbing Equilibrium by Adding Reactant (Example)
SeriesTime (arbitrary units)Time uncertaintyConcentration (mol dm^-3)Concentration uncertainty
A (reactant)01
A (reactant)100.72
A (reactant)200.6
A (reactant)200.9
A (reactant)300.8
A (reactant)400.75
A (reactant)500.75
B (product)00
B (product)100.28
B (product)200.4
B (product)300.5
B (product)400.55
B (product)500.55

Detailed Explanations

A. A repeatable workflow (Le Chatelier questions)

  1. Write the balanced equilibrium equation (with states).
  2. Identify the disturbance (what changed?).
  3. Decide which side opposes it (uses up what was added, replaces what was removed).
  4. 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

Modelled example 1

Model a pressure disturbance in ammonia formation

Core

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

  1. 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.

  2. 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.

  3. 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.

Guided practice 2

Guide a temperature prediction

About 6 min

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

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 K.
View solution step by step

  1. 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.

  2. 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.

  3. 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.

Common misconception 3

Correct an automatic pressure-shift claim

Find and correct the mistake

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

View solution step by step

  1. 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.

  2. 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.

Examiner practice 4

Explain a concentration disturbance

3 marks

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

View solution step by step

  1. 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.

  2. 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.

  3. 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.

Challenge 5

Separate competing equilibrium effects

Minimal support

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

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

  1. 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.

  2. 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.

  3. 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.

  4. 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.

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.