Le Chatelier's Principle: How Equilibria Respond to Change

What Is Le Chatelier's Principle?

Le Chatelier's Principle states: If a system at equilibrium is subjected to a change in conditions, the system will adjust to partially counteract that change and establish a new equilibrium.

In simpler terms: when you disturb an equilibrium, it "fights back" to reduce the effect of the disturbance.

This principle applies to all reversible reactions at equilibrium and is one of the most powerful predictive tools in chemistry.

Learning Goals: By the end of this guide, you should be able to:

State Le Chatelier's Principle precisely.

Predict the direction of shift for changes in concentration, pressure, and temperature.

Distinguish between shifts that change the position and those that change $K$ .

Apply the principle to industrial processes (Haber, Contact).

Solve multi-step exam problems using systematic reasoning.

Effect of Changing Concentration

Rule

Increase the concentration of a substance → equilibrium shifts to the opposite side to use it up.
Decrease the concentration of a substance → equilibrium shifts to the same side to replace it.

Example

N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)

Change	Shift	Effect on $[NH_3]$
Add more $N_2$	→ Right (forward)	Increases
Remove $NH_3$	→ Right (forward)	Temporarily decreases, then partially recovers
Add more $NH_3$	← Left (reverse)	Temporarily increases, then partially decreases

Key point: Changing concentration shifts the position of equilibrium but does NOT change the value of $K$ (the equilibrium constant). The system adjusts concentrations until they satisfy $K$ again.

Effect of Changing Pressure

Pressure changes only affect equilibria involving gases with different total moles on each side.

Rule

Increase pressure → equilibrium shifts to the side with fewer moles of gas.
Decrease pressure → equilibrium shifts to the side with more moles of gas.

Example

N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)

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

Change	Shift	Reason
Increase pressure	→ Right	Fewer gas moles reduces pressure
Decrease pressure	← Left	More gas moles increases pressure

Special Case: Equal Moles

H_2(g) + I_2(g) \rightleftharpoons 2HI(g)

Both sides have 2 moles of gas → no shift with pressure change.

Note: Adding an inert gas at constant volume does NOT shift equilibrium because the partial pressures of the reacting gases don't change.

Effect of Changing Temperature

Temperature is special — it is the only factor that changes the value of $K$ .

Rule

Think of heat as a "reactant" or "product":

Exothermic forward reaction ( $\Delta H < 0$ ): heat is on the product side.
Endothermic forward reaction ( $\Delta H > 0$ ): heat is on the reactant side.

Change	Exothermic ( $\Delta H < 0$ )	Endothermic ( $\Delta H > 0$ )
Increase $T$	Shift ← Left (reverse)	Shift → Right (forward)
Decrease $T$	Shift → Right (forward)	Shift ← Left (reverse)

Effect on K

	Exothermic	Endothermic
Increase $T$	$K$ decreases	$K$ increases
Decrease $T$	$K$ increases	$K$ decreases

Effect of a Catalyst

A catalyst speeds up both the forward and reverse reactions equally. It does NOT change:

The position of equilibrium
The value of $K$
The relative amounts of products and reactants

A catalyst only helps the system reach equilibrium faster.

Summary Table

Factor	Shift Direction	Changes $K$ ?
Concentration	Away from added substance	No
Pressure	Toward fewer gas moles	No
Temperature	Toward endothermic direction when heated	Yes
Catalyst	No shift	No

Le Chatelier's Principle Simulator

Adjust concentration, pressure, and temperature in real time and watch the equilibrium shift. See the position of equilibrium and K value change dynamically.

Launch Equilibrium Simulator

Industrial Application: The Haber Process

The synthesis of ammonia is the perfect case study for Le Chatelier's Principle:

N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \quad \Delta H = -92\ kJ\ mol^{-1}

What Le Chatelier Predicts

Condition	Le Chatelier Prediction	Industrial Practice
High pressure	Shift → right (fewer gas moles) → more $NH_3$	200 atm (compromise: higher is expensive)
Low temperature	Shift → right (exothermic) → more $NH_3$	450°C (compromise: too low = too slow)
Remove $NH_3$	Shift → right → produces more $NH_3$	$NH_3$ is condensed and removed continuously
Iron catalyst	No effect on position — just speeds up	Used to make 450°C viable

The actual conditions are a compromise between thermodynamic yield (Le Chatelier) and kinetic rate.

Worked Examples

Example 1: Predicting the Shift

Question: For $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ , $\Delta H = -196\ kJ\ mol^{-1}$ . Predict the effect of (a) increasing temperature and (b) increasing pressure.

Solution: (a) The forward reaction is exothermic. Increasing temperature shifts equilibrium ← left (endothermic direction). Less $SO_3$ is produced, and $K$ decreases.

(b) Left side: $2 + 1 = 3$ moles gas. Right side: $2$ moles gas. Increasing pressure shifts → right (fewer moles). More $SO_3$ produced. $K$ is unchanged.

Example 2: Effect on K

Question: The equilibrium $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$ , $\Delta H = +124\ kJ\ mol^{-1}$ . What happens to $K$ when the temperature is increased from 200°C to 300°C?

Solution: The forward reaction is endothermic ( $\Delta H > 0$ ). Increasing temperature favours the endothermic direction (forward). Therefore $K$ increases — more products at equilibrium.

Example 3: Multi-Step Reasoning

Question: In the contact process, explain why $SO_3$ yield is maximised at 450°C and 2 atm rather than at higher pressure and lower temperature.

Solution:

Lower temperature would shift equilibrium → right (exothermic forward), giving higher yield. But the rate would be too slow, even with a catalyst. 450°C is a compromise for acceptable rate.
Higher pressure would shift equilibrium → right (3 → 2 moles gas). But pressurising is expensive and the yield improvement at 2 atm is already ~99% with excess $O_2$ . The economic benefit doesn't justify higher pressure.

Common Mistakes

"A catalyst shifts the equilibrium" — No. A catalyst speeds up both forward and reverse reactions equally. It has no effect on position or $K$ .
"Adding an inert gas shifts the equilibrium" — Only if volume changes. At constant volume, adding inert gas doesn't change partial pressures of reactants/products, so no shift occurs.
Confusing $K$ changes with position changes — Only temperature changes $K$ . Concentration and pressure change the position but $K$ remains constant.
Miscounting gas moles — When predicting pressure effects, count only gaseous species. Solids and liquids are excluded.
Thinking equilibrium shifts mean complete conversion — Le Chatelier's principle says the system partially counteracts the change. It doesn't go 100% to one side.

Exam Tips (A-Level / AP / IB)

Always state the direction of shift (left/right or forward/reverse) AND explain why (e.g., "to use up the added substance" or "to counteract the increase in pressure").
For temperature questions, always mention whether the forward reaction is exothermic or endothermic first.
If asked about $K$ : "Only temperature changes $K$ ." This is a guaranteed mark in most marking schemes.
For industrial process questions, discuss the compromise between yield (Le Chatelier) and rate (kinetics/catalysts).

Frequently Asked Questions

What is the difference between equilibrium position and K?

The position of equilibrium describes the relative amounts of products and reactants at a given moment. K is the mathematical ratio of product and reactant concentrations at equilibrium. Concentration and pressure changes shift the position but not K; temperature changes both.

Does Le Chatelier's Principle apply to physical equilibria?

Yes. It applies to any dynamic equilibrium, including phase transitions (e.g., ice-water) and dissolution (e.g., solid dissolving in solution).

Why doesn't a catalyst change the equilibrium position?

A catalyst provides an alternative reaction pathway with lower activation energy for both the forward and reverse reactions equally. Since both rates increase by the same factor, the ratio of forward to reverse remains unchanged.

What happens if you increase concentration and temperature simultaneously?

Analyse each change separately. The net effect depends on which factor has a greater influence. In exam answers, discuss each change independently and then state the combined effect.

Gibbs Free Energy — Understand the thermodynamic basis for why equilibria shift with temperature.
Initial Rate Method — The kinetics side: how fast reactions reach equilibrium.
Buffer Solutions — A practical application of Le Chatelier's principle in acid-base chemistry.

What Is Le Chatelier's Principle?

Effect of Changing Concentration

Rule

Example

Effect of Changing Pressure

Rule

Example

Special Case: Equal Moles

Effect of Changing Temperature

Rule

Effect on K

Effect of a Catalyst

Summary Table

Le Chatelier's Principle Simulator

Industrial Application: The Haber Process

What Le Chatelier Predicts

Worked Examples

Example 1: Predicting the Shift

Example 2: Effect on K

Example 3: Multi-Step Reasoning

Common Mistakes

Exam Tips (A-Level / AP / IB)

Frequently Asked Questions

What is the difference between equilibrium position and K?

Does Le Chatelier's Principle apply to physical equilibria?

Why doesn't a catalyst change the equilibrium position?

What happens if you increase concentration and temperature simultaneously?

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