Le Châtelier’s principle

AP Chemistry· AP Chemistry CED — Equilibrium· 14 min read

1. Core Definition of Le Châtelier’s Principle★★☆☆☆⏱ 3 min

Le Châtelier’s principle is a qualitative predictive heuristic used to determine how a system at dynamic equilibrium responds to an external stress: a change in conditions that disrupts the equilibrium ratio of reactants and products. It is a core topic in AP Chemistry Unit 7, which accounts for 7-9% of total AP exam score, appearing in both multiple-choice and free-response questions.

📘 Definition

Le Châtelier’s principle

If a stress is applied to a system at dynamic equilibrium, the system shifts in the direction that counteracts the applied stress to restore a new equilibrium state.

Example:

Adding reactant to an equilibrium system causes a shift toward products to consume the added reactant.

2. Effect of Concentration Changes★★☆☆☆⏱ 3 min

When the concentration of any reacting species (reactant or product) changes, the original balance of $Q = K$ is disrupted. Comparing the new value of Q to K tells us the direction of shift, and the principle aligns with this comparison. For a general reaction $a A + b B ⇌ c C + d D$ , the reaction quotient is:

Q = \frac{[C]^c[D]^d}{[A]^a[B]^b}

Increasing the concentration of any species causes a shift away from that species to consume the excess
Decreasing the concentration of any species causes a shift toward that species to replace the loss
Changing concentration never changes the value of K, because K only depends on temperature

📐 Worked Example

The reaction $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ is at equilibrium. Additional $N_{2} (g)$ is injected into the rigid reaction container at constant temperature. Predict the direction of equilibrium shift, and state whether the equilibrium concentration of $H_{2}$ increases or decreases compared to the original equilibrium.

1
1. Identify the stress: addition of reactant $N_{2}$ increases $[N_{2}]$ immediately after injection. Original equilibrium had $Q = K$ .
2
1. Compare the new Q to K: for $Q = \frac{[ N H _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}}$ , the increased denominator makes $Q < K$ .
3
1. By Le Châtelier’s principle, the system shifts right to consume the added $N_{2}$ and counteract the stress.
4
1. As the shift proceeds, $H_{2}$ is consumed along with $N_{2}$ to form more $N H_{3}$ . The new equilibrium concentration of $H_{2}$ is lower than the original equilibrium.

3. Effect of Pressure and Volume Changes★★★☆☆⏱ 3 min

Pressure changes only affect gaseous equilibrium systems. Solids and liquids are nearly incompressible, so their concentrations do not change with pressure. Most pressure changes come from changing the volume of the reaction container at constant temperature: decreasing volume increases total pressure, and increasing volume decreases total pressure.

The system shifts to counteract the pressure change: it shifts toward the side with fewer moles of gas (to reduce total pressure when pressure is increased) or toward the side with more moles of gas (to increase total pressure when pressure is decreased). If moles of gas are equal on both sides, there is no shift. If pressure is increased by adding an inert (non-reacting) gas at constant volume, partial pressures of reactants/products do not change, so no shift occurs. Like concentration changes, pressure/volume changes do not change K.

📐 Worked Example

Predict the direction of equilibrium shift for the reaction $2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ when the volume of the container is decreased by half at constant temperature. Justify your answer.

1
1. Count only moles of gaseous species on each side of the reaction: reactants have $2 + 1 = 3$ moles of gas, products have 2 moles of gas.
2
1. Decreasing container volume increases the total pressure of the system, which is the applied stress.
3
1. Le Châtelier’s principle predicts the system will shift to counteract the increased pressure by reducing the total number of gas molecules, which lowers total pressure.
4
1. The product side has fewer moles of gas, so the equilibrium shifts right.

4. Effect of Temperature Changes and Catalysts★★★☆☆⏱ 3 min

Unlike concentration and pressure changes, changing temperature always alters the value of the equilibrium constant K. The direction of shift and change in K depends on whether the reaction is endothermic or exothermic. We model heat as a reactant for endothermic reactions ( $Δ H > 0$ , absorbs heat) and as a product for exothermic reactions ( $Δ H < 0$ , releases heat).

Applying the principle: increasing temperature adds heat, so the system shifts away from the side with heat to consume the added heat. Decreasing temperature removes heat, so the system shifts toward the side with heat to replace lost heat. Catalysts never cause an equilibrium shift: they lower activation energy for both forward and reverse reactions equally, only reducing the time to reach equilibrium, with no change to K or equilibrium position.

📐 Worked Example

The reaction $C O (g) + 2 H_{2} (g) ⇌ C H_{3} O H (g) Δ H = - 90.7 kJ/mol$ is at equilibrium. Predict the direction of shift when the reaction temperature is decreased, and state whether K increases, decreases, or stays the same.

1
1. $Δ H$ is negative, so the reaction is exothermic, and heat is a product: $C O (g) + 2 H_{2} (g) ⇌ C H_{3} O H (g) + heat$ .
2
1. The applied stress is a decrease in temperature, which corresponds to removing heat (a product) from the system.
3
1. Le Châtelier’s principle predicts the system shifts right toward the product side to replace the removed heat.
4
1. Shifting right increases the equilibrium concentration of the product $C H_{3} O H$ , so the numerator of the K expression increases, meaning K increases.

5. AP-Style Worked Problems★★★★☆⏱ 4 min

📐 Worked Example

For the endothermic decomposition reaction $C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$ , which of the following changes will increase the equilibrium partial pressure of $C O_{2}$ ?
A) Decrease the reaction temperature
B) Increase the volume of the container
C) Add more $C a C O_{3} (s)$
D) Increase the reaction temperature

1
First, note that the equilibrium constant expression excludes solids, so $K = P_{C O_{2}}$ , meaning the equilibrium partial pressure of $C O_{2}$ equals K. We analyze each option:
2
- Option A: For an endothermic reaction, decreasing temperature shifts left, so K decreases and $P_{C O_{2}}$ decreases. Incorrect.
3
- Option B: Increasing volume temporarily lowers $P_{C O_{2}}$ , but K does not change (no temperature change), so shifting right returns $P_{C O_{2}}$ to its original value. Incorrect.
4
- Option C: Adding solid $C a C O_{3}$ does not change Q, so no shift and $P_{C O_{2}}$ stays the same. Incorrect.
5
- Option D: Increasing temperature for an endothermic reaction shifts right, K increases, so $P_{C O_{2}}$ increases. Correct.

📐 Worked Example

The reaction of hydrogen and iodine to form hydrogen iodide is: $H_{2} (g) + I_{2} (g) ⇌ 2 H I (g) Δ H = - 26.5 kJ/mol$ (a) Predict the direction of equilibrium shift if $H I (g)$ is removed from the system at constant temperature and volume. Justify your answer in terms of Q vs K.
(b) Predict the direction of shift if the volume of the container is doubled at constant temperature. Justify your answer.
(c) A manufacturer wants to increase the equilibrium yield of HI. They add a catalyst and claim it will increase yield. Do you agree with this claim? Justify your answer.

1
(a) Removing $H I (g)$ decreases $[H I]$ immediately after the change. For this reaction, $Q = \frac{[ H I ] ^{2}}{[ H _{2} ] [ I _{2} ]}$ . Original Q = K, so decreasing $[H I]$ makes Q < K. The system shifts right (toward products) to restore Q = K.
2
(b) Doubling volume halves the concentration of all gaseous species. Counting moles of gas: 2 moles of gas on the reactant side, 2 moles on the product side. The ratio Q remains unchanged: $\frac{( 0.5 [ H I ] ) ^{2}}{( 0.5 [ H _{2} ]) ( 0.5 [ I _{2} ])} = \frac{0.25 [ H I ] ^{2}}{0.25 [ H _{2} ] [ I _{2} ]} = Q = K$ . No shift occurs.
3
(c) I do not agree. Catalysts lower activation energy for both the forward and reverse reactions equally, so they only speed up the rate at which equilibrium is reached. They do not change the position of equilibrium or the value of K, so they cannot increase equilibrium yield.

📐 Worked Example

The Haber process for industrial ammonia production follows the reaction: $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g) Δ H = - 92 kJ/mol$ An engineering intern suggests lowering the reaction temperature from 450 °C to 100 °C to increase equilibrium yield of ammonia. Would lowering the temperature increase the equilibrium yield of ammonia? Justify your answer, then explain why commercial plants use the higher temperature.

1
Yes, lowering the temperature will increase equilibrium yield of ammonia. The reaction is exothermic ( $Δ H < 0$ ), so heat is a product. Lowering temperature removes heat from the system, so Le Châtelier’s principle predicts a shift right toward products, increasing the equilibrium amount of ammonia.
2
Commercial plants use a higher temperature because the reaction rate at 100 °C is far too slow to be commercially viable, even with an iron catalyst. Higher temperatures increase reaction rate, allowing ammonia to be produced quickly enough for industrial demand, even though equilibrium yield is lower than at low temperatures. Producers choose a moderate high temperature to balance equilibrium yield and reaction rate for maximum production output.

6. Common Pitfalls

Wrong move:

Adding an inert gas to a gaseous equilibrium at constant volume causes a shift to the side with fewer moles of gas.

Why:

Students confuse pressure changes from volume change with pressure changes from adding inert gas; total pressure increases but partial pressures of reacting species do not change.

Correct move:

Only consider pressure changes that change the partial pressures (concentrations) of reacting species; inert gas at constant volume causes no shift.

Wrong move:

After increasing the concentration of a reactant, the final concentration of that reactant is lower than the original equilibrium concentration.

Why:

Students remember shifting consumes the added reactant and incorrectly assume all added material is used up.

Correct move:

Always conclude that the concentration of the added species is higher than original, and concentrations of other reactants are lower than original.

Wrong move:

Increasing temperature always increases K, because it increases reaction rate.

Why:

Students confuse the kinetic effect of temperature (faster reaction) with the thermodynamic effect on equilibrium position.

Correct move:

Check the sign of ΔH first: K increases with temperature for endothermic (ΔH > 0) reactions, and decreases for exothermic (ΔH < 0) reactions.

Wrong move:

Catalysts shift equilibrium toward products to increase reaction yield.

Why:

Students associate catalysts with faster product formation and incorrectly assume they change equilibrium.

Correct move:

Remember catalysts affect only reaction rate, not equilibrium position or K, so no shift occurs.

Wrong move:

When volume increases (total pressure decreases), shift is toward the side with fewer moles of gas.

Why:

Students memorize "pressure increase shifts to fewer moles" and incorrectly reverse the rule when pressure decreases.

Correct move:

Counteract the stress: if pressure decreases, shift to more moles of gas to increase pressure back; if pressure increases, shift to fewer moles to decrease pressure back.

Wrong move:

Adding more pure solid reactant to an equilibrium shifts the position right.

Why:

Students treat solids the same as gaseous or aqueous reactants.

Correct move:

Pure solids have constant activity (concentration) so adding them does not change Q, so no shift occurs.

7. Quick Reference Cheatsheet

Category	Rule	Notes
Core Principle	System shifts to counteract applied stress	Applies to all dynamic equilibrium systems on the AP exam
Concentration Change	Increase [X] → shift away from X; decrease [X] → shift toward X	K does not change. Pure solids/liquids are excluded
Pressure (Volume Change)	P increase → shift to fewer moles of gas; P decrease → shift to more moles of gas	Only count gaseous moles. Equal moles = no shift. K does not change
Inert Gas Addition	No shift	Only at constant volume; partial pressures of reactants/products do not change
Temperature Change (Endothermic, ΔH>0)	T increase → shift right, K increases; T decrease → shift left, K decreases	Heat is treated as a reactant for endothermic reactions
Temperature Change (Exothermic, ΔH<0)	T increase → shift left, K decreases; T decrease → shift right, K increases	Heat is treated as a product for exothermic reactions
Catalyst	No shift	Speeds up forward and reverse reactions equally. Only reduces time to reach equilibrium. K does not change

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Predict shift after temperature change
2022 · FRQ
Justify shift in terms of Q vs K

Going deeper

syllabusOfficial AP Chemistry CEDUnit 7 Equilibrium content overview

What's Next

Le Châtelier’s principle is the foundational qualitative tool for all equilibrium topics that come later in AP Chemistry, and it is required for every application of equilibrium across the rest of the course. Next, you will apply Le Châtelier’s principle to solve problems involving solubility equilibria, acid-base buffers, and the common ion effect. Without mastering the ability to correctly predict equilibrium shifts from different stresses, you will not be able to interpret the behavior of buffer solutions or predict how pH changes affect the solubility of ionic compounds. Beyond Unit 7, this principle also applies to thermodynamic equilibrium concepts in Unit 9, and is frequently tested in multi-concept FRQ problems that combine equilibrium with thermodynamics or kinetics.