AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/7/7-10)
Unit 7 · Topic 7.10
7.10 Reaction Quotient and Le Châtelier's Principle
Every Le Châtelier shift can be explained by comparing Q with K. A disturbance pushes Q away from K, and the reaction runs in whichever direction brings Q back to K. Adding or removing a substance (or changing volume) changes Q only, while changing temperature changes K itself.
Key terms
- Q vs. K
- equilibrium shift
- temperature dependence of K
- new equilibrium
Disturbances change Q or K
At equilibrium, Q = K. A disturbance breaks that equality. If Q < K afterward, the reaction runs forward; if Q > K, it runs in reverse. It continues until Q equals K again. This is the same direction Le Châtelier's principle predicts, but now you can prove it with numbers.
Concentration changes move Q
Adding a reactant makes the denominator of Q bigger, so Q drops below K and the reaction goes forward. Adding a product makes the numerator bigger, so Q rises above K and the reaction goes in reverse. K is unchanged.
The new equilibrium doesn't return to the old concentrations. If you add H₂ to an H₂/I₂/HI mixture, at the new equilibrium [H₂] is higher than it was originally (but lower than just after you added it), [I₂] is lower and [HI] is higher.
Volume changes move Q (for gases)
Halving the volume of a gas mixture doubles every concentration and partial pressure at the instant it happens. How Q changes depends on Δn, the moles of gas on the product side minus the moles on the reactant side: Q is multiplied by 2 raised to the Δn power.
If Δn > 0 (more gas moles in products), compressing makes Q > K, so the reaction shifts toward reactants. If Δn < 0, compressing makes Q < K, so it shifts toward products. If Δn = 0, Q doesn't change and there's no shift.
Temperature changes K
Changing the temperature changes K, not Q (Q hasn't changed at the instant you change the temperature). For an endothermic reaction (ΔH > 0), K increases as temperature increases. For an exothermic reaction (ΔH < 0), K decreases as temperature increases.
After heating an endothermic reaction, the new K is larger than the current Q, so Q < K and the reaction goes forward. That matches 'heating favors the endothermic direction'.
| Disturbance | What changes | Result | Shift |
|---|---|---|---|
| Add reactant | Q decreases | Q < K | Forward |
| Add product | Q increases | Q > K | Reverse |
| Compress, Δn > 0 | Q increases | Q > K | Reverse |
| Compress, Δn < 0 | Q decreases | Q < K | Forward |
| Heat, ΔH > 0 | K increases | Q < K | Forward |
| Heat, ΔH < 0 | K decreases | Q > K | Reverse |
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Compressing a gas equilibrium
N₂O₄(g) ⇌ 2NO₂(g) is at equilibrium. The volume is suddenly halved at constant temperature. Use Q and K to predict the direction of the shift.
Show the solutionHide the solution
- Step 1: Kp = (P_NO₂)² / P_N₂O₄. Call the original equilibrium pressures P(NO₂) and P(N₂O₄); then Q = K.
- Step 2: Halving the volume doubles each partial pressure. New Q = (2P_NO₂)² / (2P_N₂O₄) = 4(P_NO₂)² / (2P_N₂O₄) = 2 × K.
- Step 3: Now Q > K, so the reaction runs in reverse, converting NO₂ into N₂O₄ until Q falls back to K.
- Step 4: This matches Le Châtelier: the system shifts toward fewer moles of gas.
Answer: Q becomes 2K (Q > K), so the net reaction goes toward N₂O₄.
- Example 2Calculator allowed
Adding a reactant and finding the new equilibrium
For H₂(g) + I₂(g) ⇌ 2HI(g), Kc = 50.0. An equilibrium mixture has [H₂] = [I₂] = 0.0220 M and [HI] = 0.156 M. Extra H₂ is added so that [H₂] is suddenly 0.0440 M. Show the direction of shift and find the new equilibrium concentrations.
Show the solutionHide the solution
- Step 1: New Q = (0.156)² ÷ (0.0440 × 0.0220) = 0.0243 ÷ 0.000968 = 25.1.
- Step 2: Q = 25.1 < K = 50.0, so the reaction proceeds forward.
- Step 3: ICE from the disturbed mixture: [H₂] = 0.0440 − y, [I₂] = 0.0220 − y, [HI] = 0.156 + 2y.
- Step 4: Solve 50.0 = (0.156 + 2y)² / ((0.0440 − y)(0.0220 − y)). This needs a calculator's solver; the physically sensible root is y = 0.0067.
- Step 5: New equilibrium: [H₂] = 0.0373 M, [I₂] = 0.0153 M, [HI] = 0.169 M. Check: (0.169)² ÷ (0.0373 × 0.0153) ≈ 50.
- Step 6: [H₂] ends higher than the original 0.0220 M but lower than 0.0440 M: the shift only partly offsets the change.
Answer: Q = 25.1 < K, so the shift is forward; new [H₂] ≈ 0.0373 M, [I₂] ≈ 0.0153 M, [HI] ≈ 0.169 M.
- Example 3
Trap: temperature changes K, not Q
For N₂O₄(g) ⇌ 2NO₂(g), ΔH° = +57.2 kJ/mol. An equilibrium mixture is heated. A student writes: 'Heating increases Q, so the reaction shifts left.' Correct the reasoning.
Show the solutionHide the solution
- Step 1: At the instant of heating, the partial pressures haven't changed yet, so Q hasn't changed.
- Step 2: The reaction is endothermic, so raising the temperature increases K.
- Step 3: Now Q (unchanged) is less than the new, larger K, so the reaction proceeds forward, making more NO₂. The gas turns darker brown.
Answer: Heating increases K (not Q); then Q < K, so the reaction shifts right toward NO₂.
Common mistakes
- Saying adding a reactant changes K. Only temperature changes K.
- Saying a temperature change changes Q. It changes K, and Q is then compared with the new K.
- Expecting concentrations to return to their original values after a shift.
- Forgetting that compressing affects Q only when the moles of gas differ on the two sides.
On the exam
- Free-response justifications that earn full credit name what changed (Q or K), state the comparison (Q < K or Q > K), and give the direction.
- Expect 'which stress changes K?' questions. The answer is always temperature.
Connected topics
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Check yourself
4 questions on 7.10 Reaction Quotient and Le Châtelier's Principle. Pick an answer to see if you got it, and why.
For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH° = −92 kJ/mol. An equilibrium mixture is heated to a higher temperature in a rigid container. Which of the following is true immediately after heating, before any shift occurs?
H₂(g) + I₂(g) ⇌ 2HI(g) is at equilibrium in a cylinder with a movable piston. At constant temperature, the volume is suddenly halved. Which of the following is true?
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?
N₂O₄(g) ⇌ 2NO₂(g) is at equilibrium in a cylinder with a movable piston. At constant temperature, the volume is suddenly halved. Immediately afterward, how has Q changed, and which way will the reaction go?
0 of 4 answered