Le Chatelier's Principle

Overview

Le Chatelier's Principle states that when a system at equilibrium is disturbed, it shifts to counteract the disturbance and establish a new equilibrium. This principle helps predict how equilibrium responds to changes.

Statement

"If a stress is applied to a system at equilibrium, the system will shift to relieve the stress and restore equilibrium."

Types of Stress

1. Concentration Changes

2. Pressure/Volume Changes (gases)

3. Temperature Changes

4. Addition of Catalyst

Concentration Changes

Adding a Substance

Equilibrium shifts AWAY from added substance.

Removing a Substance

Equilibrium shifts TOWARD removed substance.

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

Change	Shift Direction	Effect
Add N₂	Right (→)	More NH₃
Add H₂	Right (→)	More NH₃
Add NH₃	Left (←)	More N₂, H₂
Remove NH₃	Right (→)	More NH₃
Remove N₂	Left (←)	Less NH₃

Pressure/Volume Changes

Only affects gaseous equilibria.

Increasing Pressure (Decreasing Volume)

Shifts toward side with FEWER moles of gas.

Decreasing Pressure (Increasing Volume)

Shifts toward side with MORE moles of gas.

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

• Left side: $1 + 3 = 4$ moles of gas
• Right side: 2 moles of gas

Change	Shift	Result
Increase P	Right (→)	More NH₃
Decrease P	Left (←)	More N₂, H₂

When Moles Are Equal

If moles of gas are equal on both sides, pressure change has NO EFFECT.

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

2 moles = 2 moles → No shift with pressure change

Adding Inert Gas

Condition	Effect
Constant volume	No effect (partial pressures unchanged)
Constant pressure	Volume increases → shifts to more moles

Temperature Changes

Temperature changes affect K value itself.

For Exothermic Reactions ($\Delta H < 0$)

Think of heat as a product:

$$\text{Reactants} \rightleftharpoons \text{Products} + \text{Heat}$$

Change	Shift	K Value
Increase T	Left (←)	Decreases
Decrease T	Right (→)	Increases

For Endothermic Reactions ($\Delta H > 0$)

Think of heat as a reactant:

$$\text{Heat} + \text{Reactants} \rightleftharpoons \text{Products}$$

Change	Shift	K Value
Increase T	Right (→)	Increases
Decrease T	Left (←)	Decreases

Example: Haber Process

$$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \quad \Delta H = -92 \text{ kJ (exothermic)}$$

• High T: Faster rate but lower yield (K smaller)
• Low T: Higher yield but slower rate
• Compromise: ~450°C used industrially

Effect of Catalyst

A catalyst:

• Does NOT shift equilibrium position
• Does NOT change K
• Speeds up BOTH forward and reverse reactions equally
• Helps reach equilibrium FASTER

Summary Table

Stress	Shift Direction	Effect on K
Add reactant	Right →	Unchanged
Add product	Left ←	Unchanged
Remove reactant	Left ←	Unchanged
Remove product	Right →	Unchanged
Increase P	Fewer moles side	Unchanged
Decrease P	More moles side	Unchanged
Increase T (exo)	Left ←	Decreases
Decrease T (exo)	Right →	Increases
Increase T (endo)	Right →	Increases
Decrease T (endo)	Left ←	Decreases
Add catalyst	No shift	Unchanged

Industrial Applications

Haber Process: $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ ($\Delta H = -92$ kJ)

Condition	Le Chatelier Prediction	Industrial Choice
Pressure	High P favors NH₃	~200 atm
Temperature	Low T favors NH₃	~450°C (compromise)
Remove NH₃	Shifts right	Continuous removal
Catalyst	Faster equilibrium	Iron catalyst

Contact Process: $2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3$ ($\Delta H = -198$ kJ)

Condition	Le Chatelier Prediction	Industrial Choice
Pressure	High P favors SO₃	1-2 atm (economical)
Temperature	Low T favors SO₃	~450°C (compromise)
Excess O₂	Shifts right	Used
Catalyst	Faster equilibrium	V₂O₅

Problem-Solving Strategy

1. Write the balanced equation
2. Identify the stress applied
3. Determine which direction relieves stress
4. Predict the shift and its effects
5. Remember: Only T changes affect K