AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/7/7-9)
Unit 7 · Topic 7.9
7.9 Introduction to Le Châtelier's Principle
Le Châtelier's principle says that if you disturb a system at equilibrium, it shifts in the direction that partly undoes the disturbance. Adding or removing a substance, changing the temperature, squeezing or expanding a gas mixture, or diluting a solution can all cause shifts you can see as changes in color, temperature or pH.
Key terms
- Le Châtelier's principle
- stress
- equilibrium shift
- endothermic
- exothermic
The principle
A system at equilibrium responds to a change (often called a stress) by running the reaction in whichever direction counteracts it. It never fully cancels the change; it partly offsets it and settles into a new equilibrium.
'Shifts right' means a net forward reaction (more products form). 'Shifts left' means a net reverse reaction.
Adding or removing a substance
Add a reactant, or remove a product: the system shifts right to use up some of the added reactant or replace some of the removed product. Add a product, or remove a reactant: the system shifts left.
Adding more of a pure solid or pure liquid causes no shift, because it doesn't appear in Q.
You can remove a species chemically. For example, adding Ag⁺ to a solution removes Cl⁻ by precipitating AgCl, and adding acid removes OH⁻.
Changing volume or pressure of gases
Decreasing the volume (which increases pressure) shifts the equilibrium toward the side with fewer moles of gas, because that lowers the pressure. Increasing the volume shifts it toward the side with more moles of gas. If both sides have the same number of moles of gas, changing the volume causes no shift.
Adding an unreactive gas such as argon to a rigid container raises the total pressure but doesn't change any partial pressures, so there is no shift.
Changing temperature
Treat heat as if it were a reactant in an endothermic reaction and a product in an exothermic one. Heating the system shifts it in the endothermic direction, which absorbs some of the added energy. Cooling shifts it in the exothermic direction.
Temperature is the only one of these stresses that changes the value of K (topic 7.10).
Dilution and catalysts
Adding water to an aqueous equilibrium lowers all the dissolved concentrations. The system shifts toward the side with more dissolved particles, to partly make up for the dilution.
A catalyst speeds up the forward and reverse reactions equally. It helps the system reach equilibrium faster, but it doesn't shift the equilibrium or change K.
| Stress on N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH° = −92 kJ | Shift | Effect on K |
|---|---|---|
| Add N₂ | Right | None |
| Remove NH₃ | Right | None |
| Decrease volume | Right (4 mol gas → 2 mol gas) | None |
| Add argon at constant volume | None | None |
| Raise temperature | Left (endothermic direction) | K decreases |
| Add a catalyst | None | None |
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Predicting color changes
In solution, pink Co(H₂O)₆²⁺(aq) + 4Cl⁻(aq) ⇌ blue CoCl₄²⁻(aq) + 6H₂O(l), and the forward reaction is endothermic. Predict the color change when you (a) add concentrated HCl, (b) put the tube in an ice bath, (c) add AgNO₃ solution, (d) add water.
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- Step 1: (a) HCl adds Cl⁻, a reactant. The system shifts right to use some up, so the solution turns more blue.
- Step 2: (b) Cooling favors the exothermic direction, which is the reverse reaction here. The system shifts left, so it turns more pink.
- Step 3: (c) Ag⁺ removes Cl⁻ by forming solid AgCl. Removing a reactant shifts the system left, so it turns more pink (and a white precipitate forms).
- Step 4: (d) Dilution lowers every dissolved concentration. The left side has 5 dissolved particles (1 complex + 4 Cl⁻) and the right side has 1, so the system shifts left toward more dissolved particles: more pink.
Answer: (a) bluer; (b) pinker; (c) pinker, with a white precipitate; (d) pinker
- Example 2
Trap: adding an inert gas
A rigid flask contains N₂O₄(g) ⇌ 2NO₂(g) at equilibrium. Argon is added, and the total pressure rises. A student predicts a shift toward N₂O₄ because the pressure went up. Is the student right? What would happen if, instead, the flask's volume were halved?
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- Step 1: Argon doesn't react, and the volume didn't change, so the amount of N₂O₄ and NO₂ per liter is unchanged. Their partial pressures are unchanged, so Q still equals K. No shift.
- Step 2: The total pressure went up only because argon contributes its own partial pressure, which isn't in the K expression.
- Step 3: Halving the volume is different: it raises the partial pressures of both gases, and the system shifts toward the side with fewer moles of gas, N₂O₄ (1 mol versus 2 mol).
Answer: No shift when argon is added at constant volume; halving the volume shifts toward N₂O₄.
Common mistakes
- Thinking a catalyst shifts an equilibrium. It only gets the system there faster.
- Treating any rise in total pressure as a stress. Only changes in the partial pressures of reacting gases matter.
- Predicting a shift when a pure solid or liquid is added.
- Getting temperature backward: heating shifts toward the endothermic direction, not toward 'the side with heat removed'.
On the exam
- Explain shifts with the change and the response: 'Adding Cl⁻ increases a reactant concentration, so the system shifts toward products to consume some of it.' For full credit, many questions want you to justify it with Q and K (topic 7.10).
- Expect lab-style questions where the evidence is a color change, a temperature change or a pH change, and you must connect it to a shift.
Connected topics
Videos
Check yourself
4 questions on 7.9 Introduction to Le Châtelier's Principle. Pick an answer to see if you got it, and why.
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is at equilibrium in a cylinder with a movable piston. At constant temperature, the piston is pushed in to halve the volume. Which of the following describes the result?
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is at equilibrium in a rigid container. Argon gas is added at constant temperature, raising the total pressure. Which of the following describes the result?
Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq) is at equilibrium, and the solution is deep red because of FeSCN²⁺. A few drops of AgNO₃ are added, and a white solid, AgSCN, forms. Which of the following is observed, and why?
2NO₂(g) ⇌ N₂O₄(g), ΔH° = −57 kJ/mol. NO₂ is brown and N₂O₄ is colorless. A sealed tube containing an equilibrium mixture is moved from room temperature into an ice bath. Which of the following correctly describes the result?
0 of 4 answered