AP Chemistry · Lesson 2

How an equilibrium responds to change

Know Know it

Le Chatelier’s principle: when a change disturbs a system at equilibrium, the net reaction shifts in the direction that partly undoes the change. The shift never fully cancels the change.

Adding or removing a substance. Add a reactant (or remove a product) and Q falls below K, so the reaction shifts forward. Add a product (or remove a reactant) and it shifts in reverse. Adding more of a pure solid or liquid changes nothing, since it is not in Q.

Changing the volume of a gas mixture. Making the volume smaller raises every partial pressure, and the reaction shifts toward the side with fewer moles of gas. If both sides have the same number of moles of gas, there is no shift. Adding an inert gas at constant volume changes no partial pressure, so there is no shift either.

Changing the temperature. This is the only change that alters K. Treat heat as a reactant for an endothermic reaction and as a product for an exothermic one. Heating an endothermic reaction raises K and shifts it forward; heating an exothermic reaction lowers K and shifts it in reverse.

Adding a catalyst speeds up both directions equally. Equilibrium is reached sooner, but the amounts at equilibrium and K stay the same.

What each change does
ChangeEffect on QEffect on KNet shift
Add a reactantQ falls below Knoneforward
Add a productQ rises above Knonereverse
Smaller volume (gases)moves away from K unless moles of gas are equalnonetoward fewer moles of gas
Inert gas, constant volumenonenonenone
Heat an endothermic reactionnone at firstK risesforward
Heat an exothermic reactionnone at firstK fallsreverse
Add a catalystnonenonenone

Worked example Easy

For CO(g)+2H2(g)⇌CH3OH(g)\text{CO}(g) + 2\text{H}_{2}(g) \rightleftharpoons \text{CH}_{3}\text{OH}(g), ΔH∘=−91\Delta H^\circ = -91 kJ/mol. Which change increases the amount of CH3OH\text{CH}_{3}\text{OH} present at equilibrium?

  1. Raising the temperature
  2. Shrinking the volume of the containerAnswer
  3. Adding a catalyst
  4. Removing some H2(g)\text{H}_{2}(g)

How to solve it

  1. Count moles of gas: 1 + 2 = 3 on the left, 1 on the right.
  2. A smaller volume shifts toward the side with fewer moles of gas, which is the methanol side.
  3. The negative ΔH\Delta H means heating would shift the other way.

Why each choice is right or wrong

  • A. Incorrect. Raising the temperature favors the endothermic direction, which here is the reverse reaction, so methanol is lost.
  • B. Correct. A smaller volume shifts the reaction toward fewer moles of gas: 1 mol of CH3OH\text{CH}_3\text{OH} against 3 mol of reactant gas.
  • C. Incorrect. A catalyst speeds up both directions equally and does not change the amounts at equilibrium.
  • D. Incorrect. Removing H2\text{H}_2, a reactant, makes Q larger than K, so the reaction shifts in reverse.

Trap Temperature changes K; the rest change Q

Adding a substance, or changing the volume when the two sides have different moles of gas, moves Q away from K, and the system reacts until Q equals the same K again. Only a temperature change gives a new value of K. And a catalyst changes neither.

Explain Explain it

Task verbs: explain, predict

You can back up any Le Chatelier prediction (except a temperature change) with Q. For CO(g)+2H2(g)⇌CH3OH(g)\text{CO}(g) + 2\text{H}_{2}(g) \rightleftharpoons \text{CH}_{3}\text{OH}(g), Q=[CH3OH][CO][H2]2Q = \frac{[\text{CH}_{3}\text{OH}]}{[\text{CO}][\text{H}_{2}]^2}. Halving the volume doubles every concentration, so Q is multiplied by 2/(2⋅22)=1/42/(2 \cdot 2^2) = 1/4. Now Q is less than K, so the reaction runs forward until Q climbs back to K.

For a temperature change, use a table of K at different temperatures. If K falls as the temperature rises, the forward reaction is exothermic; if K rises, it is endothermic.

Worked example Hard

A chemist measures K for a gas-phase reaction A(g)+B(g)⇌C(g)\text{A}(g) + \text{B}(g) \rightleftharpoons \text{C}(g) at three temperatures.

K at three temperatures

Temperature (K)K
3002.4×1032.4 \times 10^{3}
40044
5003.9

Which conclusion about this reaction is supported by the data?

  1. The forward reaction is endothermic, so heating shifts the equilibrium toward C
  2. The forward reaction is exothermic, so heating shifts the equilibrium toward A and BAnswer
  3. K falls because the higher temperature speeds up only the reverse reaction
  4. The forward reaction is exothermic, so heating raises the amount of C at equilibrium

How to solve it

  1. Read the trend: K drops from 2.4×1032.4 \times 10^{3} at 300 K to 3.9 at 500 K.
  2. A K that falls on heating means heat acts as a product, so the forward reaction is exothermic.
  3. Heating then favors the reverse direction, toward A and B.

Why each choice is right or wrong

  • A. Incorrect. If the forward reaction were endothermic, K would rise with temperature; here it falls from 2.4×1032.4 \times 10^{3} to 3.9.
  • B. Correct. K falls as the temperature rises, so the forward reaction releases heat, and heating shifts the mixture back toward A and B.
  • C. Incorrect. Heating speeds up both directions; K falls because the reverse rate constant grows more than the forward one, not because only the reverse reaction speeds up.
  • D. Incorrect. An exothermic reaction loses product when heated, so the amount of C falls, not rises.

Explain Your turn

The equilibrium CO2(g)+H2(g)⇌CO(g)+H2O(g)\text{CO}_{2}(g) + \text{H}_{2}(g) \rightleftharpoons \text{CO}(g) + \text{H}_{2}\text{O}(g) is set up in a cylinder with a movable piston. At constant temperature, the piston is pushed in until the volume is half of what it was. Explain, using Q and K, why the number of moles of each gas at equilibrium does not change.

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Compare with a model answer

A full answer does this

  • Says that halving the volume doubles the concentration (or partial pressure) of every gas
  • Shows that Q is unchanged because the factors of 2 cancel, so Q still equals K and there is no net shift

Model answer

Halving the volume doubles the concentration of every gas. In Q=[CO][H2O][CO2][H2]Q = \frac{[\text{CO}][\text{H}_{2}\text{O}]}{[\text{CO}_{2}][\text{H}_{2}]}, the numerator is multiplied by 2×2=42 \times 2 = 4 and so is the denominator, so the factors cancel and Q is unchanged. Since Q still equals K, there is no net reaction, and the moles of each gas stay the same. This happens because each side has 2 mol of gas.

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