UNIT 2 CHEMISTRY · PEARSON AAQ · D1.4.2
Le Chatelier’s Principle
FREE CHEMISTRY REVISION · TOPIC 44
Le Chatelier’s Principle
Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · D1.4.2
What you will learn
Predict how an equilibrium responds to concentration, gas pressure and temperature changes, and explain the role of catalysts. Separate equilibrium position and yield from reaction rate and the equilibrium constant. This lesson uses the meaning and temperature dependence of the constant; its expressions and calculations are reserved for D1.4.3 and D1.4.4.
1. Le Chatelier’s principle: a response to a disturbance
When a system at equilibrium is disturbed, its composition changes in the direction that tends to oppose the imposed change, until a new equilibrium is established. State the disturbance, identify the favoured direction and explain how that response opposes it. “Oppose” means reduce the effect, not completely undo the imposed change.
“Shifts right” means net forward reaction during readjustment; “shifts left” means net reverse reaction. At the new equilibrium both directions continue at equal rates again. A temporary imbalance of rates drives the change. The rates need not return to their old numerical values.
| Term | Meaning | Important distinction |
|---|---|---|
| Equilibrium position | Relative composition of the equilibrium mixture | Can change at fixed temperature when concentration or volume is changed |
| Equilibrium yield | Amount or proportion of desired product at equilibrium, on a stated starting basis | Distinguish moles from concentration; also distinguish equilibrium yield from yield collected after a short time |
| Reaction rate | Speed of conversion, related to change per unit time | Faster does not necessarily mean more equilibrium product |
| Equilibrium constant | A temperature-dependent measure of the equilibrium relationship for a defined reaction | At fixed temperature it is unchanged by concentration, volume/pressure changes or catalysis |
The equilibrium constant is not simply the amount of product or the product concentration. Changing composition can restore the same constant at a fixed temperature. Reversing the equation changes what is called “forward”, so read the written equation and sign of ΔH carefully.
2. Concentration: add or remove a reacting species
At fixed temperature and volume, adding a reactant generally favours the forward direction, consuming some of that reactant. Adding a product favours the reverse direction. Removing a product favours forward reaction to replace some of it; removing a reactant favours reverse reaction. These rules concern species whose effective concentration changes in the equilibrium relationship.
The reaction only partly offsets the change. If product is added, its eventual concentration may still be higher than before addition even though the equilibrium shifts left. The added species can show an instantaneous concentration jump, followed by gradual chemical readjustment. Species not added have no instantaneous jump if volume is unchanged.
For heterogeneous equilibria, adding more of a pure solid or liquid does not change its effective concentration while that phase is already present. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), adding extra CaCO₃ to a mixture containing both solids does not shift equilibrium at fixed temperature and gas volume. Do not apply the simple “add reactant” rule blindly to a pure solid.
3. Gas pressure: specify how it changes
For a gas equilibrium at fixed temperature, increasing pressure by decreasing volume favours the side with fewer moles of gas. Decreasing pressure by expansion favours more gas moles. Count balanced gaseous coefficients only; ignore solids and pure liquids in this count. If both sides have equal gas-mole totals, compression gives no shift in equilibrium position under the usual ideal-gas treatment.
Compression instantly increases all gas concentrations before reaction occurs. A subsequent shift can still increase the amount of one side relative to the other. A higher product concentration after compression does not, on its own, prove a shift: even an equal-gas-mole reaction can have increased concentrations with unchanged equilibrium mole fractions.
The way pressure is changed matters. Adding an inert gas at constant volume raises total pressure but leaves reacting-gas partial pressures unchanged in the ideal-gas model, so there is no shift. Adding inert gas at constant total pressure causes expansion and can favour the side with more gas moles. These qualifications prevent the false rule “any pressure increase shifts towards fewer gas molecules”.
4. Temperature: use the enthalpy sign
Heating favours the endothermic direction, which absorbs some of the added thermal energy. Cooling favours the exothermic direction, which releases thermal energy. If the forward ΔH is negative, forward is exothermic: heating shifts left and cooling shifts right. If forward ΔH is positive, heating shifts right and cooling shifts left.
Temperature changes the equilibrium constant. Heating an exothermic forward reaction decreases its equilibrium constant; heating an endothermic forward reaction increases it. Concentration and pressure changes at fixed temperature do not change this constant. Temperature usually speeds both reaction directions when raised, while their relative kinetic changes determine the new equilibrium composition.
For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH < 0 for the forward reaction. Lower temperature favours ammonia at equilibrium, but can make its formation slow. A high rate and high equilibrium yield are separate goals. For N₂O₄(g) ⇌ 2NO₂(g), forward ΔH > 0: heating favours brown NO₂; cooling favours the paler dimer.
5. Catalysts: reach the same equilibrium sooner
A catalyst offers an alternative reaction mechanism with lower activation barriers and accelerates both forward and reverse reactions. It does not change the thermodynamic difference between the starting and finishing states. Therefore at fixed temperature it changes neither equilibrium position nor the equilibrium constant, and gives the same equilibrium yield for the same overall starting conditions.
If a catalyst is added to a mixture already at equilibrium, opposing rates increase while remaining balanced, so composition does not shift. If added before equilibrium is reached, it reduces the waiting time. More product may be collected after a short fixed time, but that is a faster approach to the same equilibrium yield, not a greater equilibrium yield. A catalyst is not a substitute for changing temperature or pressure when the goal is equilibrium composition.
6. Comparison table
| Change | Equilibrium response | Constant and rate |
|---|---|---|
| Add a dissolved/gaseous reactant at fixed volume | Usually towards products, consuming some addition | K unchanged at fixed T; rates temporarily unbalanced |
| Remove product | Towards products, replacing some removed material | K unchanged at fixed T |
| Compress gaseous equilibrium at fixed T | Towards fewer gas moles; no shift for equal totals | K unchanged; collision rates can change |
| Heat an exothermic forward reaction | Left; lower forward-product equilibrium yield on the same basis | K decreases; both directions can become faster |
| Heat an endothermic forward reaction | Right; higher forward-product equilibrium yield on the same basis | K increases; both directions can become faster |
| Add catalyst at fixed T | No shift; same equilibrium yield | K unchanged; faster attainment of equilibrium |
7. Worked predictions
Example 1 — add hydrogen to the Haber equilibrium
N₂(g) + 3H₂(g) ⇌ 2NH₃(g). At fixed temperature and volume, add H₂. Its concentration rises immediately. Net forward reaction then consumes some added H₂ with N₂, making more NH₃. The equilibrium shifts right; the equilibrium constant stays unchanged. The final hydrogen concentration need not return to its original value.
Example 2 — remove ammonia
For the same mixture, remove some NH₃ at fixed temperature and volume. Net forward reaction replaces part of the removed product, consuming N₂ and H₂. The shift is right, with unchanged K. This statement concerns readjustment after removal; the instantaneous amount of ammonia first decreases, and removal need not be fully compensated.
Example 3 — compress the Haber mixture
Decrease volume at constant temperature. Left side: 1 + 3 = 4 moles of gas; right side: 2. Equilibrium shifts right towards fewer gas particles, increasing ammonia’s equilibrium fraction. Count molecules, not the total number of atoms. K is unchanged because temperature is fixed.
Example 4 — an equal-gas-mole reaction
H₂(g) + I₂(g) ⇌ 2HI(g), with all substances gaseous at the stated temperature. There are two gas moles on each side. Compression at fixed temperature causes no equilibrium shift in the ideal-gas treatment. Concentrations increase immediately and reaction rates can change; unchanged position does not mean unchanged concentrations.
Example 5 — heat the exothermic Haber equilibrium
For forward ΔH < 0, the reverse direction absorbs heat. Heating therefore shifts left and decreases ammonia equilibrium yield on the same initial-material basis. K decreases. Reactions can nevertheless proceed faster at the higher temperature: faster rate does not establish higher equilibrium yield.
Example 6 — heat an endothermic dissociation
N₂O₄(g) ⇌ 2NO₂(g), forward ΔH > 0. Heating shifts right, absorbing some heat and increasing the fraction of brown NO₂; K increases. Cooling shifts left and reduces its brown colour. Pressure must be controlled when isolating the temperature effect.
Example 7 — add a catalyst
Add a suitable catalyst to a sealed mixture approaching equilibrium at fixed temperature. Both directions accelerate; equilibrium is reached sooner. The final composition, equilibrium yield and K are unchanged compared with an uncatalysed experiment allowed to reach equilibrium from the same starting mixture. At a short fixed time the amounts may differ.
Example 8 — decompose calcium carbonate
CaCO₃(s) ⇌ CaO(s) + CO₂(g), forward ΔH > 0. Heating favours right. Compressing the gas at fixed temperature favours left (one gas mole versus zero). Adding more CaCO₃ when both solids are already present does not shift equilibrium. These predictions assume the phases remain available.
Example 9 — distinguish total and reacting-gas pressure
Add helium to a gas equilibrium at fixed volume and temperature. Total pressure rises, but each reacting gas still has the same amount, volume and temperature, so its partial pressure is unchanged. There is no shift in the ideal-gas model. This differs from compressing the same reacting mixture.
Example 10 — read the addition graph
In the A ⇌ B graph, B is added at 40 s: [B] jumps 0.25 → 0.65 mol dm⁻³ while [A] initially remains 0.75. Net reverse reaction then raises [A] towards 1.05 and lowers [B] towards 0.35. The system shifts left, yet final [B] is higher than its original value. The same kinetic model gives final opposing rates of 0.021 mol dm⁻³ s⁻¹, higher than the old 0.015 although equal to each other.
Common misconceptions
- The system cancels the disturbance completely: readjustment only partly opposes it.
- More concentrated product always means a right shift: an addition or volume change can alter concentration instantly.
- Count all coefficients for pressure: count gaseous species only.
- A pressure increase always causes a shift: specify compression and compare gas-mole totals.
- Higher temperature always gives more product: use the enthalpy sign of the written forward direction.
- Catalyst increases equilibrium yield or K: it changes how quickly equilibrium is attained.
- A change in position requires a change in K: at fixed T, composition can readjust with the same K.
Exam tips
- Write the balanced equilibrium with states and the forward ΔH sign.
- State the exact imposed change, including fixed temperature or volume.
- Use “favours the direction which…” and explain the opposition.
- Separate instantaneous physical changes from slower chemical readjustment.
- State whether K changes; only a temperature change changes it for a fixed reaction.
- Use equilibrium yield only when equilibrium has actually been reached.
- For catalyst questions, mention both directions and unchanged final composition.
Check your understanding
Choose one answer, select Check Answer, then read the explanation. Try Again resets the question. Use Tab between controls, arrow keys within a radio group, and Enter or Space on buttons. Answers are not saved.