Le Châtelier’s Principle (VCE SSCE Chemistry): Revision Notes

Le Châtelier's Principle

Introduction to Le Chatelier's principle

When a chemical system reaches equilibrium, the rates of the forward and reverse reactions are equal. However, equilibrium systems can respond to changes in their conditions. Understanding how equilibria respond to changes is essential for optimising chemical reactions in industry and nature.

As a result, the position of equilibrium changes. This means the relative amounts of reactants and products at equilibrium shift. There may be an increase in either products or reactants, depending on the type of change applied.

Changes that affect equilibrium systems

The position of equilibrium can be changed by:

• Adding or removing a reactant or product
• Changing the pressure by changing the volume (for gas equilibria)
• Dilution (for equilibria in solution)
• Changing the temperature

Chemists use careful control of reaction conditions to maximise the yield of desired products by shifting the equilibrium position "to the right" (increasing the amount of products formed).

Adding or removing reactants and products

Effect of adding a reactant

Consider the equilibrium between nitrogen, hydrogen and ammonia:

$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$

When extra nitrogen gas is added to the container (without changing volume or temperature), the mixture temporarily moves out of equilibrium. The following events occur:

1. Initially, the system is at equilibrium with equal rates of forward and reverse reactions
2. Adding nitrogen increases its concentration, causing more frequent collisions between N₂ and H₂ molecules
3. The rate of the forward reaction instantly increases, producing more NH₃
4. As ammonia concentration increases, the rate of the reverse reaction also increases
5. Eventually, the forward and reverse rates become equal again at a new equilibrium position

The equilibrium position shifts "to the right" (toward products). There is a net forward reaction with an increase in ammonia concentration at the new equilibrium.

The value of $K$ remains unchanged because temperature has not changed.

Effect of adding a product

Adding more ammonia (product) to the system causes the opposite effect. The equilibrium position shifts "to the left" (toward reactants), resulting in a net reverse reaction. The overall ammonia concentration decreases compared to immediately after addition.

General effects of changes

Using the equilibrium law to predict changes

The effect of adding reactants or products can also be predicted using the equilibrium law and the reaction quotient Q.

Changing pressure by changing volume

The pressure of a gas is inversely proportional to its volume. This means pressure can be changed by increasing or decreasing the container volume at constant temperature.

Example: sulfur dioxide oxidation

Consider this equilibrium:

$2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$

On the left side: 3 gas particles (2 SO₂ + 1 O₂) On the right side: 2 gas particles (2 SO₃)

The forward reaction reduces the number of gas particles from 3 to 2, which would decrease the overall pressure. The reverse reaction increases particles from 2 to 3, which would increase the overall pressure.

Applying Le Chatelier's principle to pressure changes

Le Chatelier's principle tells us that an equilibrium system responds to an increase in pressure by adjusting to reduce the pressure. The position of equilibrium moves toward the side with fewer gas particles.

In the sulfur dioxide example:

• Increasing pressure causes a net forward reaction
• Three gaseous reactant particles become two gaseous product particles
• The amount of SO₃ increases at equilibrium

The effect can also be shown graphically:

When pressure is initially increased (by decreasing volume), all concentrations increase simultaneously. As the system adjusts, concentrations gradually change until a new equilibrium is established. The equilibrium constant $K$ remains unchanged.

Using the equilibrium law to explain pressure changes

For the equilibrium: $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$

The equilibrium constant is:

$K = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]}$

Special case: equal numbers of particles

When there are equal numbers of reactant and product gas particles, pressure changes do not shift the equilibrium position.

Example: $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$

Left side: 2 gas particles Right side: 2 gas particles

Pressure changes in liquids and solids

Pressure changes do not affect equilibrium positions of systems in liquid or solid phases. Particles in these phases are too tightly packed for pressure changes to noticeably affect volume or concentration.

Changing pressure by adding an inert gas

Adding a non-reacting (inert) gas such as helium, neon or argon to an equilibrium mixture increases the total pressure but does not change the concentrations of reactants or products.

Dilution of equilibria in solution

For equilibria occurring in solution, adding water (dilution) reduces the number of particles per unit volume. The equilibrium position shifts toward the side that produces the greater number of dissolved particles.

Example: iron(III) thiocyanate equilibrium

$\text{Fe}^{3+}(aq) + \text{SCN}^-(aq) \rightleftharpoons \text{FeSCN}^{2+}(aq)$

Left side: 2 particles in solution Right side: 1 particle in solution

When water is added to double the volume, all concentrations are momentarily halved. According to Le Chatelier's principle, a net reverse reaction occurs to increase the total concentration of particles in solution.

Temperature changes and effects on the equilibrium constant

The equilibrium constant $K$ for a reaction depends only on temperature. It is not affected by:

• Addition of reactants or products
• Changes in pressure
• Use of catalysts

How temperature affects K

The effect of temperature on $K$ depends on whether the reaction is exothermic or endothermic.

Example: nitrogen dioxide and dinitrogen tetroxide

The conversion of brown nitrogen dioxide gas to colourless dinitrogen tetroxide is exothermic:

$2\text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g) + \text{energy}$

When the temperature increases:

• The system opposes the increase in energy by absorbing energy
• The reverse reaction is endothermic (absorbs energy)
• A net reverse reaction occurs
• Less N₂O₄ and more NO₂ are present at the new equilibrium
• The mixture appears darker brown

Applying Le Chatelier's principle to temperature

For exothermic reactions (release energy):

• Increasing temperature: equilibrium shifts left (toward reactants), $K$ decreases, less product formed
• Decreasing temperature: equilibrium shifts right (toward products), $K$ increases, more product formed

For endothermic reactions (absorb energy):

• Increasing temperature: equilibrium shifts right (toward products), $K$ increases, more product formed
• Decreasing temperature: equilibrium shifts left (toward reactants), $K$ decreases, less product formed

Effect of a catalyst on equilibrium

A catalyst lowers the activation energy of both forward and reverse reactions by providing an alternative pathway.

Lower activation energy increases the number of effective collisions, which increases the rate of both forward and reverse reactions equally.

Case study: ocean acidification

Carbon dioxide dissolves in seawater through a series of equilibria:

$\text{CO}_2(g) \rightleftharpoons \text{CO}_2(aq)$

$\text{CO}_2(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_2\text{CO}_3(aq)$

$\text{H}_2\text{CO}_3(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{HCO}_3^-(aq)$

$\text{HCO}_3^-(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{CO}_3^{2-}(aq)$

Using Le Chatelier's principle, increasing atmospheric CO₂ causes:

• More CO₂ to dissolve in oceans
• Increased H₃O⁺ concentration
• Decreased pH (increased acidity)
• This process is called ocean acidification

Marine organisms use carbonate ions to build shells:

$\text{Ca}^{2+}(aq) + \text{CO}_3^{2-}(aq) \rightleftharpoons \text{CaCO}_3(s)$

Higher H₃O⁺ concentrations react with CO₃²⁻ ions, causing this equilibrium to shift left. This can dissolve calcium carbonate shells, threatening marine ecosystems.

Case study: swimming pool chemistry

Swimming pools are chlorinated to prevent bacterial and algal growth. Pool chlorine (calcium hypochlorite) dissolves to form hypochlorous acid (HOCl), an effective antibacterial agent:

$\text{Ca(OCl)}_2(s) \rightarrow \text{Ca}^{2+}(aq) + 2\text{OCl}^-(aq)$

$\text{OCl}^-(aq) + \text{H}_3\text{O}^+(aq) \rightleftharpoons \text{HOCl}(aq) + \text{H}_2\text{O}(l)$

The pool pH must be maintained between 7.2 and 7.8:

• If pH rises above 7.8: H₃O⁺ concentration decreases, equilibrium shifts left, HOCl concentration becomes too low to control bacteria and algae
• If pH falls below 7.2: H₃O⁺ concentration increases, equilibrium shifts right, too much HOCl forms, water becomes irritating to eyes and skin

Maintaining proper pH ensures adequate HOCl for hygiene while keeping water comfortable for swimmers.

Summary of Le Chatelier's principle effects

Change	Effect on equilibrium position	Effect on K
Add reactant	Shifts right (more products)	No change
Add product	Shifts left (more reactants)	No change
Remove product	Shifts right (more products)	No change
Increase pressure (decrease volume)	Shifts toward fewer gas particles	No change
Decrease pressure (increase volume)	Shifts toward more gas particles	No change
Add inert gas (constant volume)	No shift	No change
Dilution (solution equilibria)	Shifts toward more dissolved particles	No change
Increase temperature (exothermic)	Shifts left (more reactants)	Decreases
Increase temperature (endothermic)	Shifts right (more products)	Increases
Add catalyst	No shift (faster to reach equilibrium)	No change