IB DP Chemistry - R2.3.4 Le Châtelier’s principle - Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry – R2.3.4 Le Châtelier’s principle – Study Notes – New Syllabus
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Reactivity 2.3.4 – Le Châtelier’s Principle: Predicting Effects of Changes to Equilibrium Systems
Reactivity 2.3.4 – Le Châtelier’s Principle: Predicting Effects of Changes to Equilibrium Systems
Le Châtelier’s Principle states that when a change is applied to a system at equilibrium, the system will adjust to oppose the change and restore equilibrium. This principle can be used to predict how changes in concentration, temperature, and pressure affect the equilibrium position and equilibrium composition.
1. Effect of Concentration Changes
Changing the concentration of a species involved in a chemical equilibrium disturbs the balance of the system. According to Le Châtelier’s Principle, the system will respond by shifting the equilibrium to reduce the effect of the change.
- Increasing the concentration of a reactant: The system shifts to the right to consume some of the added reactant and form more product.
- Decreasing the concentration of a reactant: The system shifts to the left to produce more of the removed reactant.
- Increasing the concentration of a product: The system shifts to the left to convert product back into reactant.
- Decreasing the concentration of a product: The system shifts to the right to produce more of the removed product.
Effect on:
- Value of \( K \): Remains unchanged. Concentration changes do not alter the equilibrium constant; only temperature can change \( K \).
- Equilibrium Composition: The concentrations of species adjust until the ratio of products to reactants matches the value of \( K \) again.
Example
In the equilibrium system:
\( \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \)
Additional hydrogen gas is injected into the system at constant temperature. Predict the effect on the system.
▶️Answer/Explanation
The increase in [H2] disturbs equilibrium. To oppose this, the system shifts to the right — forming more HI and using up some of the added H2 and I2. The concentration of HI increases.
Value of \( K \): Remains the same because temperature has not changed.
Example
For the reaction:
\( \text{Fe}^{3+}(aq) + \text{SCN}^-(aq) \rightleftharpoons \text{FeSCN}^{2+}(aq) \)
Some FeSCN2+ is removed from the mixture by precipitation. Predict the effect on the system.
▶️Answer/Explanation
Removing FeSCN2+ (a product) disturbs equilibrium. The system shifts to the right to replace some of the removed product, consuming more Fe3+ and SCN– in the process.
Equilibrium composition: [Fe3+] and [SCN–] decrease, [FeSCN2+] increases.
Value of \( K \): No change, as temperature is constant.
2. Effect of Pressure Changes (Gaseous Systems Only)
Changes in pressure only affect equilibrium systems that involve gaseous reactants and/or products. According to Le Châtelier’s Principle, the system will shift to minimize the change in pressure by favoring the side with fewer or more gas particles (moles).
- Increasing the pressure: The system shifts toward the side with fewer moles of gas, reducing the total pressure.
- Decreasing the pressure: The system shifts toward the side with more moles of gas, increasing total pressure.
- If both sides have the same number of moles of gas: No shift in equilibrium occurs.
Effect on:
- Value of \( K \): No change — pressure changes do not alter \( K \) (unless temperature changes).
- Equilibrium Composition: Partial pressures of gaseous species adjust to restore equilibrium, affecting the amounts of reactants and products.
Note: Solids and liquids do not contribute to pressure effects — only gases count when evaluating pressure-induced shifts.
Example
In the Haber process:
\( \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \)
The total pressure is suddenly increased by decreasing the volume of the container.Predict the effect on the system.
▶️Answer/Explanation
On the left: 1 mol N2 + 3 mol H2 = 4 mol of gas.
On the right: 2 mol NH3 = 2 mol of gas.
Increasing pressure favors the side with fewer gas moles → the system shifts to the right, forming more ammonia.
Equilibrium composition: [NH3] increases; [N2] and [H2] decrease.
Value of \( K \): No change — only temperature affects \( K \).
Example
Consider the reaction:
\( \text{H}_2(g) + \text{CO}_2(g) \rightleftharpoons \text{H}_2\text{O}(g) + \text{CO}(g) \)
Predict the effect of increasing pressure on this system.
▶️Answer/Explanation
Total moles of gas on each side = 2 (1 mol H2 + 1 mol CO2 vs. 1 mol H2O + 1 mol CO).
Since both sides have equal gas moles, changing pressure has no effect on the equilibrium position or composition.
Equilibrium composition: Remains unchanged.
Value of \( K \): Remains constant.
3. Effect of Temperature Changes
Temperature directly affects both the equilibrium position and the value of the equilibrium constant \( K \). According to Le Châtelier’s Principle, the system treats heat as a “reactant” in an endothermic reaction and as a “product” in an exothermic reaction.
For Endothermic Reactions (ΔH > 0):
- Increasing temperature: Equilibrium shifts to the right (forward) to absorb the added heat.
- Decreasing temperature: Equilibrium shifts to the left (reverse) to produce heat.
For Exothermic Reactions (ΔH < 0):
- Increasing temperature: Equilibrium shifts to the left (reverse) to remove excess heat.
- Decreasing temperature: Equilibrium shifts to the right (forward) to generate heat.
Effect on:
- Value of \( K \):
- For endothermic reactions: \( K \) increases with temperature.
- For exothermic reactions: \( K \) decreases with temperature.
- Equilibrium Composition: Concentrations of reactants and products adjust depending on the direction of the shift caused by heat.
Example
The equilibrium below is endothermic:
\( \text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g) \quad \Delta H = +58\,\text{kJ/mol} \)
Predict the effect of increasing temperature on the equilibrium position and on the value of \( K \).
▶️Answer/Explanation
This is an endothermic reaction. Increasing temperature adds heat, which the system will absorb by shifting the equilibrium right (toward more NO2).
Equilibrium composition: [NO2] increases, [N2O4] decreases.
Value of \( K \): Increases, since the forward (endothermic) reaction is favored.
Example
Consider the combustion equilibrium:
\( \text{C}(s) + \text{O}_2(g) \rightleftharpoons \text{CO}_2(g) \quad \Delta H = -393\,\text{kJ/mol} \)
Predict the effect of decreasing temperature on the equilibrium and the value of \( K \).
▶️Answer/Explanation
As this reaction is exothermic, lowering the temperature removes heat. The system compensates by favoring the forward (exothermic) direction, producing more CO2.
Equilibrium composition: [CO2] increases, [O2] decreases.
Value of \( K \): Increases, as the forward reaction is favored by the temperature decrease.
4. Effect of a Catalyst
A catalyst increases the rate at which equilibrium is reached by lowering the activation energy for both the forward and reverse reactions equally. However, it does not affect the position of equilibrium or the value of the equilibrium constant \( K \), because it does not change the relative energies of the products and reactants.
- Equilibrium position: No shift.
- Equilibrium composition: Remains the same.
- Value of \( K \): Unaffected.
Example
In the reaction:
\( \text{SO}_2(g) + \frac{1}{2} \text{O}_2(g) \rightleftharpoons \text{SO}_3(g) \),
a vanadium(V) oxide catalyst is added to the system.Predict the effect on the system.
▶️Answer/Explanation
The catalyst lowers the activation energy for both the forward and reverse reactions, allowing equilibrium to be established faster.
Equilibrium composition: Remains the same.
Value of \( K \): No change.
Important: The rate of approach to equilibrium increases, but the final ratio of products to reactants stays constant.
Example
For the equilibrium:
\( \text{CH}_3\text{COOH}(aq) + \text{C}_2\text{H}_5\text{OH}(aq) \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5(aq) + \text{H}_2\text{O}(l) \)
A small amount of H2SO4 is added as a catalyst.Predict the effect on the system.
▶️Answer/Explanation
The sulfuric acid increases the rate of both the forward (ester formation) and reverse (hydrolysis) reactions, without affecting equilibrium composition.
Equilibrium position: No shift.
Value of \( K \): Remains unchanged.
Key point: The system simply reaches equilibrium faster.
5. Le Châtelier’s Principle and Heterogeneous Equilibria
A heterogeneous equilibrium involves substances in more than one phase (e.g., gas, liquid, solid, aqueous). Le Châtelier’s Principle still applies, but only species whose concentrations or pressures can change will affect the position of equilibrium.
Key points:
- Solids and pure liquids: Their concentration is constant and do not appear in the equilibrium expression — changes in their amount do not affect equilibrium position.
- Gases and aqueous species: Their concentration or pressure can change, and thus they influence equilibrium shifts.
Equilibrium constant \( K \): Still only affected by temperature, not by phase changes or pressure unless temperature is involved.
Example
The equilibrium between gaseous and aqueous carbon dioxide is:
\( \text{CO}_2(g) \rightleftharpoons \text{CO}_2(aq) \)
Predict the effect of increasing pressure of CO2 gas on this system.
▶️Answer/Explanation
Increasing the partial pressure of CO2(g) increases its concentration. The system shifts right to dissolve more CO2 into solution.
Equilibrium composition: [CO2(aq)] increases.
Value of \( K \): Remains unchanged (unless temperature changes).
Example
Consider:
\( \text{CaCO}_3(s) \rightleftharpoons \text{Ca}^{2+}(aq) + \text{CO}_3^{2-}(aq) \)
Predict the effect of adding more solid CaCO3 to this system at equilibrium.
▶️Answer/Explanation
Adding more solid does not change the concentration of CaCO3, which is constant in equilibrium expressions. Therefore, there is no shift in equilibrium.
Equilibrium composition: Unchanged.
Value of \( K \): Remains constant.