Unit 7 · Topic 7.9 Beta

Introduction to Le Châtelier's Principle

Le Châtelier’s principle: a system at equilibrium that is disturbed shifts to partly counteract the change.

Practice 4: Model AnalysisPractice 6: Argumentation

Question set for this topic

Part 1 · Hook

Why this matters

A tube of pale purple solution turns deep blue when you drop it into hot water and pink when you plunge it into ice. Pull it out and it slowly turns purple again. Nothing was added or removed. The solution is an equilibrium, and it responds to every push in a way you can predict.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Adding a reactant to a system at equilibrium first changes which rate?

  1. The forward rate
  2. The reverse rate
  3. Both rates equally
  4. Neither rate
Show the answer

The forward reaction uses the reactant, so its rate rises first.

  • Correct: The forward rate:
  • The reverse rate:
  • Both rates equally:
  • Neither rate:

2. For a given reaction equation, what changes the value of K?

  1. Temperature
  2. Adding a reactant
  3. Changing the volume
  4. Adding a catalyst
Show the answer

K depends only on temperature.

  • Correct: Temperature:
  • Adding a reactant:
  • Changing the volume:
  • Adding a catalyst:

3. An endothermic reaction

  1. absorbs heat from the surroundings
  2. releases heat to the surroundings
  3. has a negative ΔH
  4. needs no activation energy
Show the answer

Endothermic: heat flows into the system, ΔH > 0.

  • Correct: absorbs heat from the surroundings:
  • releases heat to the surroundings:
  • has a negative ΔH:
  • needs no activation energy:

Part 4 · See it

See it first

A table of changes made to a system at equilibrium, the direction of shift, and whether K changes. Add a reactant or remove a product: shift toward products, K unchanged. Add a product or remove a reactant: toward reactants, K unchanged. Decrease the volume of a gas mixture: toward fewer gas moles, K unchanged. Add an inert gas at constant volume: no shift. Add a catalyst: no shift. Raise the temperature of an endothermic reaction: toward products, and K rises. Raise the temperature of an exothermic reaction: toward reactants, and K falls.
Concentration and volume changes shift the equilibrium without changing K. A temperature change shifts it by changing K. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. A system at equilibrium is disturbedit is no longer at equilibrium, and a net reaction starts
  2. A reactant is added or a product removedthe system shifts toward products, partly undoing the change
  3. The volume of a gas mixture is decreasedthe system shifts toward the side with fewer gas moles
  4. The temperature is raisedK changes: up for an endothermic reaction, down for an exothermic one, and the system shifts to match
  5. A catalyst or an inert gas at constant volume is addedno shift: the equilibrium mixture stays the same

Part 6 · Key ideas

Key ideas

  • Le Châtelier’s principle: a disturbed equilibrium shifts in the direction that partly counteracts the change.
  • Add a species: shift away from it. Remove a species: shift toward it. Solids and pure liquids do not count.
  • Smaller volume: shift toward fewer gas moles. No shift if gas moles are equal, for a catalyst, or for an inert gas at constant volume.
  • Temperature changes K: heating favors the endothermic direction.

Part 7 · Misconception

A common mistake

The wrong idea: Adding a catalyst or an inert gas shifts an equilibrium toward more product.

What actually happens: A catalyst speeds both directions equally and an inert gas at constant volume changes no reacting concentration, so neither shifts the equilibrium. Only concentration, volume (for unequal gas moles) and temperature changes do.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

Disturbing a PCl₅ equilibrium

A sealed 1.00 L flask holds PCl5, PCl3 and Cl2 at equilibrium at constant temperature: PCl5(g) ⇌ PCl3(g) + Cl2(g), Kc = 0.040. At 50 s, one change is made to the flask. The graph shows the three concentrations.

00.050.10.150.20.250255075100125150Time (s)Concentration (M)

PCl₅PCl₃Cl₂

Data table
Time (s)PCl₅PCl₃Cl₂
00.1280.0720.072
250.1280.0720.072
500.1280.0720.072
600.1490.0510.179
700.1570.0430.171
800.160.040.169
1000.1610.0390.167
1250.1610.0390.167
1500.1610.0390.167

1. What change was made at 50 s?

  1. Cl2 was added.
  2. PCl5 was added.
  3. The temperature was raised.
  4. The volume was halved.
Show the answer

A sudden vertical jump in one curve means that substance was added (or removed, for a drop). Gradual changes afterward are the system’s response.

  • Correct: Cl2 was added.: Right: [Cl₂] jumps straight up at 50 s while the other two do not jump.
  • PCl5 was added.: Added PCl₅ would make the PCl₅ curve jump. It rises gradually, as a result of the shift.
  • The temperature was raised.: A temperature change does not make one concentration jump; the curves would bend smoothly to new values.
  • The volume was halved.: Halving the volume would make all three concentrations jump at once.

2. Which statement describes the response after 50 s?

  1. The net reaction runs in reverse, using some of the added Cl2.
  2. The net reaction runs forward, making more Cl2.
  3. There is no net reaction, because K has not changed.
  4. The net reaction runs in reverse until [Cl2] is back to its old value.
Show the answer

Adding a product makes the system shift toward reactants, partly using up what was added. It does not undo the change completely.

  • Correct: The net reaction runs in reverse, using some of the added Cl2.: Right: [Cl₂] and [PCl₃] fall while [PCl₅] rises, the reverse direction, which partly uses up the added Cl₂.
  • The net reaction runs forward, making more Cl2.: More Cl₂ forming would push [Cl₂] even higher. It falls after the jump.
  • There is no net reaction, because K has not changed.: K is unchanged, but Q has changed, so the system must react until Q = K again.
  • The net reaction runs in reverse until [Cl2] is back to its old value.: The shift only partly counteracts the change: [Cl₂] ends above its old value.

3. How does Kc after 100 s compare with Kc before 50 s?

  1. It is the same, 0.040.
  2. It is larger, because more Cl2 is present.
  3. It is smaller, because the reaction shifted in reverse.
  4. It is undefined until the added Cl2 is used up.
Show the answer

K depends only on temperature. A concentration change shifts the equilibrium position, not the constant.

  • Correct: It is the same, 0.040.: Right: the temperature did not change, so K did not change. The concentrations rearranged to fit the same K.
  • It is larger, because more Cl2 is present.: Adding Cl₂ changed Q, not K. The system shifted until Q equaled the original K.
  • It is smaller, because the reaction shifted in reverse.: A shift changes concentrations, not K.
  • It is undefined until the added Cl2 is used up.: The added Cl₂ is never all used up, and K is defined at any equilibrium.

Data table

The pink and blue cobalt equilibrium

In water, cobalt(II) ions take part in this equilibrium:

Co(H2O)62+(aq) + 4 Cl−(aq) ⇌ CoCl42−(aq) + 6 H2O(l)
pink                       blue

A student starts with a purple solution (both ions present) in five test tubes and makes one change to each.

Observations
TubeChangeColor afterward
1none (control)purple
2placed in a hot-water bathblue
3placed in an ice bathpink
4a few drops of concentrated HCl(aq), a source of Cl−, addedblue
5AgNO3(aq) added; a white solid formspink

4. What do tubes 2 and 3 show about the forward reaction?

  1. It is endothermic.
  2. It is exothermic.
  3. It is faster when cold.
  4. It does not depend on temperature.
Show the answer

For an endothermic reaction, heat acts like a reactant: adding heat shifts the equilibrium forward (blue), removing heat shifts it back (pink).

  • Correct: It is endothermic.: Right: heating shifts the equilibrium toward the blue product, so the forward reaction absorbs heat.
  • It is exothermic.: If it were exothermic, heating would shift toward the reactants and turn the tube pink. It turned blue.
  • It is faster when cold.: The colors show where equilibrium lies, not how fast it is reached.
  • It does not depend on temperature.: The color changed with temperature, so the equilibrium position did.

5. Why does tube 5 turn pink?

  1. Ag+ removes Cl− as AgCl(s).
  2. Ag+ adds to the blue ion, turning it pink.
  3. Adding a solution dilutes the tube, which turns it pink.
  4. NO3− reacts with the pink ion.
Show the answer

Ag⁺(aq) + Cl⁻(aq) → AgCl(s) removes a reactant. The equilibrium shifts toward the side that makes Cl⁻: the pink side.

  • Correct: Ag+ removes Cl− as AgCl(s).: Right: lowering [Cl⁻], a reactant, makes the system shift in reverse to replace some of it.
  • Ag+ adds to the blue ion, turning it pink.: The white solid is AgCl. The color change comes from the shift, not from silver joining a cobalt ion.
  • Adding a solution dilutes the tube, which turns it pink.: A few drops of solution barely dilute it. The precipitate shows Cl⁻ is being removed.
  • NO3− reacts with the pink ion.: Nitrate is a spectator ion here; Ag⁺ is the active ion.

6. How does Kc in tube 2 compare with Kc in tube 1?

  1. Greater: higher T, endothermic forward reaction.
  2. The same, because no substance was added.
  3. Smaller, because the reaction shifted toward the products.
  4. The same, because K is a constant.
Show the answer

Temperature is the one change that alters K. For an endothermic forward reaction, a higher temperature means a larger K.

  • Correct: Greater: higher T, endothermic forward reaction.: Right: raising the temperature of an endothermic reaction increases K, which is why the mixture shifts toward the blue product.
  • The same, because no substance was added.: K stays the same for concentration changes, but a temperature change alters K itself.
  • Smaller, because the reaction shifted toward the products.: A shift toward products at higher temperature means K got larger, not smaller.
  • The same, because K is a constant.: K is constant only at a fixed temperature.

7. For N2(g) + 3 H2(g) ⇌ 2 NH3(g), ΔH° = −92 kJ/molrxn, at equilibrium in a sealed container. Predict how the amount of NH3 at the new equilibrium compares with before each change.

VariableChange
The volume of the container is halved—
The temperature is raised—
A catalyst is added—
Some H2 is removed—
Show the answer

Le Châtelier’s principle: a change in concentration, volume or temperature makes the equilibrium shift to partly counteract it. A catalyst causes no shift.

  • The volume of the container is halved: increases. Smaller volume favors the side with fewer gas molecules: 2 NH₃ versus 4 reactant molecules.
  • The temperature is raised: decreases. The forward reaction is exothermic, so heat acts as a product; adding heat shifts the equilibrium in reverse.
  • A catalyst is added: no change. A catalyst speeds both directions equally; the equilibrium amounts do not change.
  • Some H2 is removed: decreases. Removing a reactant shifts the equilibrium in reverse to replace some of it.

Part 9 · Summary

Summary

Le Châtelier’s principle: a system at equilibrium that is disturbed shifts to partly counteract the change. Adding a species shifts the equilibrium away from it and removing one shifts toward it; decreasing the volume of a gas mixture shifts toward fewer gas moles. These changes leave K unchanged. A temperature change alters K, favoring the endothermic direction on heating. Catalysts, inert gases at constant volume and extra solid cause no shift.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections