Le Châtelier’s principle — AP Chemistry

AP Chemistry · Unit 7: 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.

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 $aA + bB \rightleftharpoons cC + dD$, 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 $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ 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: for $Q = \frac{[NH_3]^2}{[N_2][H_2]^3}$, 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 $NH_3$. 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.

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 ($\Delta H > 0$, absorbs heat) and as a product for exothermic reactions ($\Delta H < 0$, 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 $CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(g) \quad \Delta H = -90.7 \text{ kJ/mol}$ 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: $CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(g) + \text{heat}$.
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 $CH_3OH$, 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 $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$, which of the following changes will increase the equilibrium partial pressure of $CO_2$? A) Decrease the reaction temperature B) Increase the volume of the container C) Add more $CaCO_3(s)$ D) Increase the reaction temperature

First, note that the equilibrium constant expression excludes solids, so $K = P_{CO_2}$, meaning the equilibrium partial pressure of $CO_2$ equals K. We analyze each option:
- Option A: For an endothermic reaction, decreasing temperature shifts left, so K decreases and $P_{CO_2}$ decreases. Incorrect.
- Option B: Increasing volume temporarily lowers $P_{CO_2}$, but K does not change (no temperature change), so shifting right returns $P_{CO_2}$ to its original value. Incorrect.
- Option C: Adding solid $CaCO_3$ does not change Q, so no shift and $P_{CO_2}$ stays the same. Incorrect.
- Option D: Increasing temperature for an endothermic reaction shifts right, K increases, so $P_{CO_2}$ increases. Correct.

📐 Worked Example

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

(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.

📐 Worked Example

The Haber process for industrial ammonia production follows the reaction: $$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \quad \Delta H = -92 \text{ kJ/mol}$$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.

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. Producers choose a moderate high temperature to balance equilibrium yield and reaction rate for maximum production output.

Common Pitfalls

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.

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

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

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

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

Why: Students treat solids the same as gaseous or aqueous reactants.

Le Châtelier’s principle — AP Chemistry

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

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

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

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

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

Common Pitfalls

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