How to use this review
Use this page after the Unit 7 diagnostic and the topics it flagged. Answer all six questions on paper before opening any answer.
These are original Marlbridge practice questions, not past exam questions. Every K value, rate constant and lettered substance is invented. Each rubric is a suggested Marlbridge rubric, not official scoring, and your total gives no predicted score.
Big ideas of the unit
- A closed system settles into a dynamic equilibrium. Amounts stop changing, but both processes continue (Topic 7.1).
- Rates decide the direction. Net change follows the faster process; equal rates mean equilibrium (Topic 7.2).
- Q describes any mixture; K is the value Q reaches at equilibrium. Solids and pure liquids are left out (Topic 7.3).
- K comes from measurements, and K gives back amounts. An ICE table works in both directions (Topic 7.4, Topic 7.7).
- The size of K tells you where equilibrium lies, not how fast it is reached (Topic 7.5).
- K follows the equation you write, and so does Q (Topic 7.6).
- Particle diagrams are the same mathematics in pictures (Topic 7.8).
- A stress moves Q away from K, and the system shifts to restore Q = K. Only temperature changes K itself (Topic 7.9, Topic 7.10).
- Dissolving a salt is one more equilibrium. A common ion lowers the solubility without changing K_sp (Topic 7.11, Topic 7.12).
Key relationships and methods
| Idea | Relationship or method | Topic |
|---|---|---|
| Direction from rates | forward > reverse: net to the right; equal rates: equilibrium | 7.1, 7.2 |
| Expressions | Kc = [C]^c[D]^d / ([A]^a[B]^b); Kp uses pressures; leave out (s) and (l) | 7.3 |
| K from data | ICE table, then substitute equilibrium values | 7.4 |
| Size of K | K ≫ 1: mostly products; K ≪ 1: mostly reactants; never exactly zero | 7.5 |
| Changing the equation | reverse: 1/K; × c: K^c; add: K₁ × K₂ | 7.6 |
| Direction from Q | Q < K: forward; Q > K: reverse; Q = K: equilibrium | 7.7 |
| Particle boxes | counts → concentrations → Q or K | 7.8 |
| Stresses | species, volume and dilution change Q; temperature changes K | 7.9, 7.10 |
| Solubility | MX: K_sp = s²; MX₂: 4s³; MX₃: 27s⁴; compare s, not K_sp | 7.11 |
| Common ion | put its concentration into K_sp, solve for s, check s is small | 7.12 |
Question 1 (multiple choice · mixed)
Two particle diagrams show equilibrium mixtures for A₂(g) ⇌ 2A(g) in boxes of the same volume. At 300 K the box holds 4 A₂ and 2 A. At 500 K it holds 2 A₂ and 6 A. Which conclusion is supported?
- (A) K is larger at 500 K, so the forward reaction is endothermic.
- (B) K is larger at 500 K, so the forward reaction is exothermic.
- (C) K is the same at both temperatures, because both boxes are at equilibrium.
- (D) K is smaller at 500 K, because there are fewer A₂ molecules.
Answer and explanation
Answer: (A). Both boxes hold 10 A atoms, so they show the same system (Topic 7.8). With counts in equal volumes, K = 2² ÷ 4 = 1 at 300 K and 6² ÷ 2 = 18 at 500 K. Heating raised K, and heating favours the endothermic direction (Topics 7.9 and 7.10), so the forward reaction is endothermic.
- (B) Heating favours the endothermic direction, not the exothermic one.
- (C) Temperature is the one stress that changes K.
- (D) Fewer A₂ and more A means a larger K, not a smaller one.
Question 2 (multiple choice · mixed)
The fictional salt MY₂ has K_sp = 4.0 × 10⁻¹². Equal volumes of an M(NO₃)₂ solution and a NaY solution are mixed. Which pair of starting concentrations gives a precipitate?
- (A) 2.0 × 10⁻³ M M(NO₃)₂ and 2.0 × 10⁻⁴ M NaY
- (B) 2.0 × 10⁻⁴ M M(NO₃)₂ and 2.0 × 10⁻⁴ M NaY
- (C) 4.0 × 10⁻⁴ M M(NO₃)₂ and 1.0 × 10⁻⁴ M NaY
- (D) 1.0 × 10⁻³ M M(NO₃)₂ and 1.4 × 10⁻⁴ M NaY
Answer and explanation
Answer: (A). Mixing equal volumes halves each concentration. Then compare Q = [M²⁺][Y⁻]² with K_sp (Topics 7.10 and 7.11). For (A): Q = (1.0 × 10⁻³)(1.0 × 10⁻⁴)² = 1.0 × 10⁻¹¹ > K_sp, so a precipitate forms.
- (B) Q = 1.0 × 10⁻¹², below K_sp.
- (C) Q = 5.0 × 10⁻¹³. Not squaring [Y⁻] gives 1.0 × 10⁻⁸, a false “yes”.
- (D) Q = 2.5 × 10⁻¹². Forgetting the dilution gives 2.0 × 10⁻¹¹, a false “yes”.
Question 3 (constructed response · mixed)
A fictional compound M dissolves in water and slowly splits into two identical fragments: M(aq) ⇌ 2N(aq). A solution starts with [M] = 0.150 M and no N.
| Time (min) | 0 | 2 | 4 | 6 | 8 | 10 |
|---|---|---|---|---|---|---|
| [N] (M) | 0.000 | 0.050 | 0.080 | 0.094 | 0.100 | 0.100 |
(a) Give the earliest time in the table at which the system is shown to be at equilibrium. Justify your answer. (b) Calculate [M] at equilibrium, and Kc. (c) A second solution at the same temperature starts with [M] = 0.200 M and [N] = 0.200 M. Calculate Qc and predict the direction of the net reaction. (d) Calculate [M] and [N] when the second solution reaches equilibrium. (e) Describe how [M] and [N] change with time in the second solution, and explain the shape in terms of rates.
Worked solution and suggested Marlbridge rubric
(a) 8 min. [N] is the same at 8 and 10 min, so it has stopped changing. At 6 min it was still rising.
(b) Each M gives 2N: [M] = 0.150 − 0.100 ÷ 2 = 0.100 M. Kc = [N]² ÷ [M] = 0.100² ÷ 0.100 = 0.100.
(c) Qc = 0.200² ÷ 0.200 = 0.200. Q > K, so the net reaction goes in reverse: N forms M.
(d) Let x M of M form: [M] = 0.200 + x, [N] = 0.200 − 2x.
- (0.200 − 2x)² = 0.100(0.200 + x), so 4x² − 0.900x + 0.0200 = 0.
- x = (0.900 − 0.700) ÷ 8 = 0.0250. (The other root, 0.200, makes [N] negative.)
- [M] = 0.225 M, [N] = 0.150 M. Check: 0.150² ÷ 0.225 = 0.100 ✓.
(e) [N] falls from 0.200 M to 0.150 M; [M] rises from 0.200 M to 0.225 M, half as much. Both change quickly at first, then level off at the same time. The reverse rate falls as [N] falls and the forward rate rises as [M] rises, until the rates are equal.
| Point | What earns it |
|---|---|
| 1 | (a) 8 min, with constant concentration as the reason |
| 1 | (b) [M] = 0.100 M using the 1 : 2 ratio, and Kc = 0.100 |
| 1 | (c) Q = 0.200 > K, so net reverse reaction |
| 1 | (d) ICE set up in the reverse direction, with 2x for N |
| 1 | (d) [M] = 0.225 M and [N] = 0.150 M |
| 1 | (e) Correct start and end values, the 2 : 1 change, flat from the same time |
| 1 | (e) Shape explained by the rates becoming equal |
Total: 7 points. Accept carried-forward values.
Question 4 (constructed response · mixed)
The fictional gases D and E combine: D(g) + E(g) ⇌ DE(g), Kp = 2.0. A cylinder with a piston holds an equilibrium mixture with P_D = 0.50 atm, P_E = 0.90 atm and P_DE = 0.90 atm.
(a) Show that the mixture is at equilibrium. (b) The volume is halved at constant temperature. Calculate Qp just after the change and predict the direction of the net reaction. Show that Le Châtelier’s principle agrees. (c) Calculate the partial pressures at the new equilibrium. (d) The forward reaction is exothermic. The temperature is now raised at constant volume. State what happens to Kp and to the amount of DE, and explain using Q and K. (e) Calculate Kp at the original temperature for 2DE(g) ⇌ 2D(g) + 2E(g).
Worked solution and suggested Marlbridge rubric
(a) Qp = 0.90 ÷ (0.50 × 0.90) = 2.0 = Kp ✓.
(b) Every partial pressure doubles: 1.00, 1.80 and 1.80 atm. Qp = 1.80 ÷ (1.00 × 1.80) = 1.0 = ½Kp. Q < K, so the net reaction goes forward. Le Châtelier agrees: a smaller volume favours the side with fewer gas molecules (1 rather than 2) ✓.
(c) Let x atm of D react: 2.0 = (1.80 + x) ÷ ((1.00 − x)(1.80 − x)).
- 2x² − 6.60x + 1.80 = 0, or x² − 3.30x + 0.90 = 0.
- x = (3.30 − 2.70) ÷ 2 = 0.30. (The other root, 3.0, is larger than P_D.)
- P_D = 0.70 atm, P_E = 1.50 atm, P_DE = 2.10 atm. Check: 2.10 ÷ (0.70 × 1.50) = 2.0 ✓.
(d) Kp decreases and the amount of DE decreases. Heating favours the endothermic (reverse) direction, so K falls. Heating at constant volume does not change the concentrations, so Qc is unchanged and now greater than the smaller Kc. The net reaction runs in reverse until Q falls to K.
(e) Reversed (invert) and doubled (square): Kp = (1 ÷ 2.0)² = 0.25.
| Point | What earns it |
|---|---|
| 1 | (a) Q = 2.0 = K |
| 1 | (b) Pressures doubled and Q = 1.0 |
| 1 | (b) Q < K so forward, and the gas-molecule argument |
| 1 | (c) Correct expression and quadratic |
| 1 | (c) 0.70, 1.50 and 2.10 atm |
| 1 | (d) K decreases because the forward reaction is exothermic |
| 1 | (d) Q > new K, so net reverse and less DE |
| 1 | (e) 0.25 |
Total: 8 points.
Question 5 (constructed response · mixed)
The fictional hydroxide Z(OH)₂ is only slightly soluble: Z(OH)₂(s) ⇌ Z²⁺(aq) + 2OH⁻(aq). A student stirs excess solid with water at 25 °C, filters, and titrates 25.00 mL of the saturated solution with 0.0100 M HCl. The end point is at 12.00 mL. Assume all the OH⁻ comes from the salt.
(a) Calculate [OH⁻] and the molar solubility of Z(OH)₂. (b) Calculate K_sp. (c) Predict the molar solubility of Z(OH)₂ in 0.050 M NaOH, and show that your approximation is valid. (d) Explain your answer to (c) in terms of the equilibrium. (e) The student forgot to filter, and a few particles of solid entered the titration flask. State and explain the effect on her K_sp. (f) A fictional hydroxide WOH (1:1) has K_sp = 4.0 × 10⁻⁶. A classmate says that WOH must be more soluble than Z(OH)₂. Evaluate the claim.
Worked solution and suggested Marlbridge rubric
(a) n(OH⁻) = n(H⁺) = 0.01200 L × 0.0100 M = 1.20 × 10⁻⁴ mol. [OH⁻] = 1.20 × 10⁻⁴ ÷ 0.02500 = 4.80 × 10⁻³ M, so s = 2.40 × 10⁻³ M.
(b) K_sp = (2.40 × 10⁻³)(4.80 × 10⁻³)² = 5.53 × 10⁻⁸.
(c) s = K_sp ÷ (0.050)² = 2.2 × 10⁻⁵ M. The salt adds 2s = 4.4 × 10⁻⁵ M OH⁻, under 0.1% of 0.050 M, so the approximation holds.
(d) The common ion OH⁻ shifts the dissolving equilibrium left, so Q reaches K_sp after about 100 times less solid has dissolved. K_sp does not change.
(e) Too high. The solid also reacts with HCl, so the titre, [OH⁻], s and K_sp all come out too large.
(f) Wrong. For WOH, s = √(4.0 × 10⁻⁶) = 2.0 × 10⁻³ M, less than 2.40 × 10⁻³ M for Z(OH)₂. K_sp expressions differ in form, so compare molar solubilities, not K_sp values.
| Point | What earns it |
|---|---|
| 1 | (a) [OH⁻] = 4.80 × 10⁻³ M from the titration |
| 1 | (a) s = 2.40 × 10⁻³ M (half of [OH⁻]) |
| 1 | (b) K_sp = 5.53 × 10⁻⁸, with [OH⁻] squared |
| 1 | (c) 2.2 × 10⁻⁵ M with a validity check |
| 1 | (d) Common ion shifts the equilibrium left; K_sp unchanged |
| 1 | (e) Too high, because undissolved solid adds to the titre |
| 1 | (f) s for WOH compared, with the reason |
Total: 7 points. Accept carried-forward values.
Question 6 (constructed response · mixed)
A fictional reaction X₂(g) + 2Y(g) ⇌ 2XY(g) happens in two elementary steps:
- Step 1: X₂(g) ⇌ 2X(g); forward rate = k₁[X₂], reverse rate = k₋₁[X]², with k₁ = 4.0 × 10⁻² s⁻¹ and k₋₁ = 8.0 × 10³ M⁻¹ s⁻¹.
- Step 2: X(g) + Y(g) ⇌ XY(g), K₂ = 2.0 × 10³.
(a) Use equal rates at equilibrium to show that K₁ = k₁ ÷ k₋₁, and calculate K₁. (b) A flask containing only X₂ and X has [X₂] = 0.10 M and [X] = 1.0 × 10⁻³ M. Calculate both rates for step 1 and state the direction of net change. Show that comparing Q with K₁ agrees. (c) Show how the steps combine to give the overall equation, and calculate the overall K. (d) Calculate K for XY(g) ⇌ ½X₂(g) + Y(g). (e) A student says: “K₁ is tiny, so the overall reaction barely happens.” Evaluate the claim.
Worked solution and suggested Marlbridge rubric
(a) At equilibrium k₁[X₂] = k₋₁[X]², so [X]² ÷ [X₂] = k₁ ÷ k₋₁, and the left side is K₁. K₁ = 4.0 × 10⁻² ÷ 8.0 × 10³ = 5.0 × 10⁻⁶.
(b) Forward: 4.0 × 10⁻² × 0.10 = 4.0 × 10⁻³ M s⁻¹. Reverse: 8.0 × 10³ × (1.0 × 10⁻³)² = 8.0 × 10⁻³ M s⁻¹. Reverse is faster, so there is net formation of X₂ (Topic 7.2). Q = (1.0 × 10⁻³)² ÷ 0.10 = 1.0 × 10⁻⁵ > K₁: reverse ✓ (Topic 7.7).
(c) Double step 2 (K₂²) and add step 1: X₂ + 2X + 2Y ⇌ 2X + 2XY; cancel 2X. K = K₁ × K₂² = 5.0 × 10⁻⁶ × 4.0 × 10⁶ = 20.
(d) Reverse and halve: K = 1 ÷ √20 = 0.22.
(e) Not justified. K₁ only says that little X is present at any time. The overall K = 20 is greater than 1, so products are favoured at equilibrium: as step 2 uses up X, more X₂ splits. (K says nothing about speed.)
| Point | What earns it |
|---|---|
| 1 | (a) Rates set equal and rearranged |
| 1 | (a) K₁ = 5.0 × 10⁻⁶ |
| 1 | (b) Both rates, and net formation of X₂ |
| 1 | (b) Q > K₁, the same conclusion |
| 1 | (c) Step 2 doubled, steps added, X cancelled |
| 1 | (c) K = 20 |
| 1 | (d) 0.22 |
| 1 | (e) Rejects the claim using the overall K |
Total: 8 points. K₁ × K₂ = 1.0 × 10⁻² in (c) means step 2 was not doubled.
How did you do?
Total: 32 points (2 multiple choice, 30 constructed response). Where you lost points matters more.
- Q, K and direction (2, 3(c), 4(b), 6(b)): revisit Topics 7.2, 7.3 and 7.7.
- Calculating K or equilibrium amounts (3, 4(c), 5(b)): revisit Topics 7.4 and 7.7.
- Changing the equation (4(e), 6(c)–(d)): revisit Topic 7.6.
- Stresses and temperature (1, 4(b), 4(d)): revisit Topics 7.9 and 7.10.
- Particle pictures or the size of K (1, 6(e)): revisit Topics 7.5 and 7.8.
- Solubility and common ions (2, 5): revisit Topics 7.11 and 7.12.
- Explanations marked down: say whether Q or K changed, how they compare and which way the system moves.
If several topics were weak, return to the Unit 7 diagnostic. Then tick off each topic checklist: