Le Châtelier’s principle — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: This chapter covers the definition of Le Châtelier’s principle, effects of concentration, pressure/volume, and temperature changes on equilibrium, the role of catalysts, and prediction of equilibrium shift direction for AP exam questions.

You should already know: The difference between reaction quotient Q and equilibrium constant K, how to write equilibrium constant expressions, and how to interpret the sign of reaction enthalpy ΔH.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Le Châtelier’s principle?

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: Equilibrium, which accounts for 7-9% of the total AP exam score per the official College Board Course and Exam Description (CED). This topic appears in both multiple-choice (MCQ) and free-response (FRQ) sections, and is often combined with calculations of K or Q, or connected to acid-base equilibria, solubility, and thermodynamics in multi-concept problems. A common synonym for the principle is the "equilibrium shift rule", since we describe the system shifting left (toward reactants) or right (toward products) to counteract the applied stress. The core statement of the principle is: 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. While it is not a fundamental physical law, it is 100% reliable for all systems tested on the AP Chemistry exam.

2. Effect of Concentration Changes

When the concentration of any reacting species (reactant or product) in an equilibrium system is changed, this creates a stress that disrupts the original Q = K balance. For a general reaction $aA+bB\rightleftharpoons cC+dD$, the reaction quotient is: $$Q=\frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}$$ If you increase the concentration of a reactant, the denominator of Q increases, so Q immediately becomes less than K. To counteract the stress, the system shifts right (toward products), consuming the added reactant to make more product until Q = K is restored. If you decrease the concentration of a reactant, Q becomes greater than K, so the system shifts left to make more reactant. The same logic applies to products: increasing product concentration shifts left, decreasing product concentration shifts right. Critically, changing concentration does not change the value of K, because K only depends on temperature. The new equilibrium will have a different set of absolute concentrations, but the ratio defined by K remains the same.

Worked Example

The reaction $N_2(g)+3H_2(g)\rightleftharpoons 2N{H}_3(g)$ 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. Identify the stress: addition of reactant $N_2$ increases $[N_2]$ immediately after injection. Original equilibrium had Q = K.
2. Compare the new Q to K: $Q=\frac{[N{H}_3{]}^2}{[N_2][H_2{]}^3}$, so the increased denominator makes Q < K.
3. By Le Châtelier’s principle, the system shifts right to consume the added $N_2$ and counteract the stress.
4. 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 concentration.

Exam tip: A common MCQ trick asks for the final concentration of the added reactant. Even after shifting to consume some of the added reactant, the final concentration of the added species is still higher than the original equilibrium concentration.

3. Effect of Pressure and Volume Changes

Pressure changes only affect gaseous equilibrium systems, because solids and liquids are nearly incompressible, so their concentrations do not change with pressure. The most common pressure change comes from changing the volume of the reaction container at constant temperature: decreasing volume increases total pressure, and increasing volume decreases total pressure. When total pressure changes, the system shifts to counteract the pressure change: it shifts toward the side with fewer moles of gas (to reduce total pressure, if pressure was increased) or toward the side with more moles of gas (to increase total pressure, if pressure was decreased). If the number of moles of gas is equal on both sides of the reaction, a pressure/volume change causes no shift. An important edge case: if pressure is increased by adding an inert (non-reacting) gas at constant volume, the partial pressures (and concentrations) of reactants and products do not change, so Q remains equal to K, and no shift occurs. Like concentration changes, pressure/volume changes do not change the value of K, only the position of equilibrium.

Worked Example

Predict the direction of equilibrium shift for the reaction $2S{O}_2(g)+O_2(g)\rightleftharpoons 2S{O}_3(g)$ when the volume of the container is decreased by half at constant temperature. Justify your answer.

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. Decreasing container volume increases the total pressure of the system, which is the applied stress.
3. 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. The product side has fewer moles of gas, so the equilibrium shifts right.

Exam tip: Always ignore solids and liquids when counting moles for pressure/volume shift problems. Their concentrations do not change with volume, so they do not affect the shift direction.

4. Effect of Temperature Changes and Catalysts

Unlike concentration and pressure changes, changing temperature always alters the value of the equilibrium constant K. The direction of shift and the direction of K change depend on whether the reaction is endothermic or exothermic. We can model heat as a reactant for endothermic reactions ($\Delta H>0$, absorbs heat) and as a product for exothermic reactions ($\Delta H<0$, releases heat). Applying Le Châtelier’s 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 the lost heat. For example, increasing temperature for an exothermic reaction (heat = product) shifts left, so product concentration decreases and K decreases. Increasing temperature for an endothermic reaction shifts right, so product concentration increases and K increases. Catalysts never cause an equilibrium shift: they lower activation energy for both the forward and reverse reactions equally, so they only reduce the time required to reach equilibrium, and do not change K or the equilibrium position.

Worked Example

The reaction $CO(g)+2H_2(g)\rightleftharpoons C{H}_{3}OH(g)\Delta H=-90.7\text{ kJ/mol}$ 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. $\Delta H$ is negative, so the reaction is exothermic, and heat is a product: $CO(g)+2H_2(g)\rightleftharpoons C{H}_{3}OH(g)+\text{heat}$.
2. The applied stress is a decrease in temperature, which corresponds to removing heat (a product) from the system.
3. Le Châtelier’s principle predicts the system shifts right toward the product side to replace the removed heat.
4. Shifting right increases the equilibrium concentration of the product $C{H}_{3}OH$, so the numerator of the K expression increases, meaning K increases.

Exam tip: Any question that asks whether K changes after a stress can immediately be answered "no change" unless the stress is a change in temperature. This is a quick check that saves time on MCQs.

5. Common Pitfalls (and how to avoid them)

• 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.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

For the endothermic decomposition reaction $CaC{O}_3(s)\rightleftharpoons CaO(s)+C{O}_2(g)$, 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 $CaC{O}_3(s)$ D) Increase the reaction temperature

Worked Solution: 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: Option A: For an endothermic reaction, decreasing temperature shifts left, so K decreases and $P_{C{O}_2}$ decreases, so A is wrong. 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, so B is wrong. Option C: Adding solid $CaC{O}_3$ does not change Q, so no shift and $P_{C{O}_2}$ stays the same, so C is wrong. Option D: Increasing temperature for an endothermic reaction shifts right, K increases, so $P_{C{O}_2}$ increases. The correct answer is D.

Question 2 (Free Response)

The reaction of hydrogen and iodine to form hydrogen iodide is: $$H_2(g)+I_2(g)\rightleftharpoons 2HI(g)\Delta H=-26.5\text{ kJ/mol}$$ (a) Predict the direction of equilibrium shift if $HI(g)$ 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.

Worked Solution: (a) Removing $HI(g)$ decreases $[HI]$ immediately after the change. For this reaction, $Q=\frac{[HI{]}^2}{[H_2][I_2]}$. Original Q = K, so decreasing $[HI]$ makes Q < K. The system shifts right (toward products) to restore Q = K. (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: $\frac{(0.5[HI]{)}^2}{(0.5[H_2])(0.5[I_2])}=\frac{0.25[HI{]}^2}{0.25[H_2][I_2]}=Q=K$. No shift occurs. (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.

Question 3 (Application / Real-World Style)

The Haber process for industrial ammonia production follows the reaction: $$N_2(g)+3H_2(g)\rightleftharpoons 2N{H}_3(g)\Delta H=-92\text{ kJ/mol}$$ 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.

Worked Solution: Yes, lowering the temperature will increase equilibrium yield of ammonia. The reaction is exothermic ($\Delta H<0$), 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. 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. In context, producers choose a moderate high temperature to balance equilibrium yield and reaction rate for maximum production output.

7. Quick Reference Cheatsheet

8. 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.

• Solubility Equilibrium
• Common Ion Effect
• Acid-Base Buffers
• Equilibrium and Thermodynamics