Introduction to Le Châtelier's Principle: Study Guide (Chemistry 7.9)

What Le Châtelier’s principle says

At equilibrium, the forward and reverse reactions run at the same rate, so the amounts of reactants and products stay constant. A stress is any change that upsets this balance. The four stresses you need are:

1. adding or removing a reactant or product;
2. changing the temperature;
3. changing the volume (and so the pressure) of a gas-phase system;
4. diluting a reaction mixture in solution.

Le Châtelier’s principle: when a system at equilibrium is disturbed, it shifts in the direction that partly counteracts the disturbance, until a new equilibrium is reached.

“Shifts to the right” (or “forward”) means there is a net change from reactants to products. “Shifts to the left” (or “in reverse”) means a net change from products to reactants. After the shift, the forward and reverse rates are equal again, but the concentrations are different from before.

Why it works: rates

The principle is not magic. It follows from the rates you met in Topic 7.2. Suppose you add more of a reactant. Reactant particles now collide more often, so the forward rate jumps above the reverse rate. There is a net forward reaction. As it runs, reactants are used up (so the forward rate falls) and products build up (so the reverse rate rises). When the two rates are equal again, the system is at a new equilibrium.

Stress 1: adding or removing a species

• Add a reactant or product → the system shifts to use it up.
• Remove a reactant or product → the system shifts to replace it.

You can remove a species without taking it out of the flask yourself. Adding a reagent that turns it into a precipitate, or that reacts with it, removes it from the equilibrium just as well. For example, adding silver nitrate removes chloride ions as solid AgCl.

What does not count as a stress. A solid or a pure liquid does not appear in the equilibrium expression (Topic 7.3). Its “concentration” does not change when you add more of it, so adding or removing some (as long as some remains) causes no shift. The same applies to water when it is the solvent.

Stress 2: changing the temperature

Temperature is different from the other stresses because it changes the value of K. For predicting the direction, the simplest method is to write heat into the equation:

• Exothermic forward reaction (ΔH < 0): heat is a product. A + B ⇌ C + heat.
• Endothermic forward reaction (ΔH > 0): heat is a reactant. heat + A + B ⇌ C.

Then treat heating as “adding heat” and cooling as “removing heat”. Heating always favours the endothermic direction; cooling favours the exothermic direction.

You can also run this backwards. If heating a mixture increases the amount of product, the forward reaction must be endothermic.

Stress 3: changing the volume of a gas mixture

For a system of gases, count the gas molecules on each side of the balanced equation, using the coefficients.

• Decrease the volume (which increases the total pressure) → shift towards the side with fewer gas molecules. Fewer molecules partly reduce the pressure.
• Increase the volume → shift towards the side with more gas molecules.
• Equal numbers of gas molecules on both sides → no shift.

Only count gases. In CaCO₃(s) ⇌ CaO(s) + CO₂(g) there is 0 mol of gas on the left and 1 mol on the right.

One special case: adding an unreactive gas such as argon at constant volume raises the total pressure but does not change the partial pressure or concentration of any reacting gas. There is no shift. A pressure change matters only when it changes the concentrations of the reacting species.

Stress 4: diluting a solution

Diluting a solution with water lowers the concentration of every dissolved species by the same factor. This is the solution version of increasing the volume of a gas:

• Dilution favours the side with more dissolved particles (count ions and molecules in solution, using coefficients).
• Equal numbers of dissolved particles on both sides → dilution causes no shift.

Water as the solvent is not counted, because it is a pure liquid.

What you would actually see

Exam questions often ask what an observer would measure, not just which way the system shifts. Link the shift to the property:

Property	How the shift shows up
Colour	If one species is coloured, a shift towards it deepens that colour; a shift away from it makes the colour fade or change.
pH	If H⁺ or OH⁻ is in the equation, a shift that makes or uses it changes the pH.
Temperature	In an insulated vessel, a shift in the exothermic direction releases heat and the temperature rises; a shift in the endothermic direction absorbs heat and the temperature falls.

For example, in aqueous ammonia, NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq). Dissolving some ammonium chloride adds NH₄⁺, so the system shifts left. Hydroxide ions are used up, [OH⁻] falls, and the pH decreases. A pH meter would show this directly.

Worked example 1: one coloured equilibrium, four stresses

Question. In a solution containing cobalt(II) ions and chloride ions, this equilibrium is set up. The forward reaction is endothermic.

[Co(H₂O)₆]²⁺(aq) + 4Cl⁻(aq) ⇌ [CoCl₄]²⁻(aq) + 6H₂O(l) ΔH > 0

The ion [Co(H₂O)₆]²⁺ is pink and the ion [CoCl₄]²⁻ is blue. The mixture is purple (both colours present). Predict the colour change for each stress: (a) adding concentrated hydrochloric acid; (b) adding silver nitrate solution; (c) placing the tube in hot water; (d) adding a large volume of water.

1. (a) Add HCl. This adds Cl⁻, a reactant. The system shifts right to use it up. More blue [CoCl₄]²⁻ forms: the solution turns bluer.
2. (b) Add AgNO₃. Ag⁺ removes Cl⁻ as a white precipitate of AgCl(s). Removing a reactant shifts the system left to replace it. More pink [Co(H₂O)₆]²⁺ forms: the solution turns pinker (and goes cloudy from the precipitate).
3. (c) Heat. The forward reaction is endothermic, so heat is a reactant. Adding heat shifts the system right: the solution turns bluer. (Cooling in ice would turn it pinker.)
4. (d) Dilute. Water is the solvent, so it is not counted. Count dissolved particles: left side 1 + 4 = 5; right side 1. Dilution favours the side with more dissolved particles, so the system shifts left: the solution turns pinker, beyond the simple fading that dilution causes.

Check. In each case, ask: “Does the shift partly undo what I did?” Adding Cl⁻ → Cl⁻ is used up. Removing Cl⁻ → Cl⁻ is released. Adding heat → heat is absorbed. Diluting → the number of dissolved particles increases. All four agree with the principle.

Worked example 2: squeezing a gas syringe

Question. A sealed glass syringe holds an equilibrium mixture of dinitrogen tetroxide (colourless) and nitrogen dioxide (brown) at room temperature:

N₂O₄(g) ⇌ 2NO₂(g) ΔH > 0

(a) The plunger is pushed in quickly so the volume halves, and the temperature is kept constant. Describe what you see immediately and over the next few seconds. (b) The syringe is then put in an ice bath. Predict the colour change.

1. (a) The instant of compression. Nothing has reacted yet, but every gas is squeezed into half the space, so [NO₂] doubles. The mixture looks darker brown straight away.
2. The shift. Count gas molecules: 1 on the left, 2 on the right. A smaller volume favours the side with fewer gas molecules, so the system shifts left. Some NO₂ combines to form N₂O₄, and the brown colour fades a little.
3. The new equilibrium. The shift only partly undoes the stress. The final colour is lighter than just after compression but still darker than the original mixture, because [NO₂] at the new equilibrium is higher than before.
4. (b) Ice bath. The forward reaction is endothermic, so heat is a reactant. Cooling removes heat, so the system shifts left (the exothermic direction). More colourless N₂O₄ forms and the gas turns paler.

Why step 1 matters. If an exam question asks what you observe, the darkening caused by compression is part of the answer. The shift explains the partial fading afterwards, not the darkening.

Changes that cause no shift

Change	Why there is no shift
Adding a catalyst	It speeds up the forward and reverse reactions by the same factor. Equilibrium is reached sooner, but its position is the same.
Adding more solid or pure liquid	Not in the equilibrium expression, so the balance is not upset.
Adding an unreactive gas at constant volume	The partial pressures of the reacting gases do not change.
Changing the volume when gas molecules are equal on both sides	Both sides are affected equally.

Common misconceptions

• “The system returns to the old concentrations.” No. A new equilibrium is reached with different concentrations. If you add a reactant, its final concentration is still higher than before (Figure 1).
• “Adding a reactant increases K.” K changes only with temperature. Adding a species changes the position of equilibrium, not K.
• “Adding more solid shifts the equilibrium.” Solids and pure liquids are not in the equilibrium expression, so adding more has no effect on the position.
• “Higher pressure always shifts the equilibrium.” Only if it changes the concentrations of reacting gases, and only if the numbers of gas molecules on the two sides differ.
• “Heating always shifts to the right.” Heating favours the endothermic direction, which may be the reverse reaction.
• “A catalyst increases the yield.” It gets you to equilibrium faster but does not change how much product is present at equilibrium.
• “Darker colour after compression proves the system shifted towards the coloured gas.” The concentration jump from compression can darken the colour even when the shift is the other way (Worked example 2).

Where this leads

Le Châtelier’s principle tells you the direction of a shift but not why it must go that way, or by how much. In Topic 7.10, Reaction Quotient and Le Châtelier’s Principle, you will compare the reaction quotient Q with K to prove each prediction with numbers. Previous topic: Representations of Equilibrium. Now try the practice questions, then use the revision notes and the checklist.