These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. All substances named with letters (D, E, F, G, H, R, U, V, W) are fictional. Every box in a question has the same volume, and the temperature is constant. Where a question asks you to draw a box, draw it on paper (or list the counts), then compare with the solution.
Question 1 (multiple choice · foundation)
A sequence of particle diagrams shows a sealed container. The counts of every species are the same in the fifth, sixth and seventh snapshots, but the particles are in different positions each time. Which statement best explains this?
- (A) The reactions stopped at the fifth snapshot, and the particles are only moving around.
- (B) The forward and reverse reactions are happening at equal rates.
- (C) The reactant and product particles are present in equal numbers.
- (D) The forward reaction is complete, so only the reverse reaction is happening.
Answer and explanation
Answer: (B). Constant counts mean Q = K. Equilibrium is dynamic: molecules keep reacting both ways at the same rate, so the counts stay fixed while individual particles change.
- (A) confuses “no net change” with “no reaction”. Products still break apart and re-form.
- (C) is not required. The counts are constant, not necessarily equal.
- (D) If only the reverse reaction were happening, reactant counts would rise and product counts would fall.
Question 2 (multiple choice · core)
The box shows an equilibrium mixture for D₂(g) + E₂(g) ⇌ 2 DE(g).
What is the value of K?
- (A) 0.083
- (B) 1.5
- (C) 2.0
- (D) 12
Answer and explanation
Answer: (D). Count: 1 D₂, 3 E₂, 6 DE. Both sides have two particles, so counts can be used directly. K = (DE)² ÷ (D₂ × E₂) = 6² ÷ (1 × 3) = 36 ÷ 3 = 12.
- (A) is the expression upside down (3 ÷ 36).
- (B) adds the reactants instead of multiplying them and forgets the square: 6 ÷ (1 + 3).
- (C) forgets to square the DE count: 6 ÷ 3.
Question 3 (multiple choice · core)
A box starts with 6 G atoms and 4 H atoms. They react: G(g) + H(g) ⇌ GH(g), and K is very large. Which box best represents the mixture at equilibrium?
| Box | G | H | GH |
|---|---|---|---|
| (A) | 2 | 0 | 4 |
| (B) | 4 | 2 | 2 |
| (C) | 6 | 4 | 0 |
| (D) | 0 | 0 | 6 |
- (A) Box A
- (B) Box B
- (C) Box C
- (D) Box D
Answer and explanation
Answer: (A). A very large K means the reaction goes almost to completion. H is the limiting reactant (4 against 6 G), so nearly all of it forms GH: 4 GH, 2 G left over and essentially no H. Atoms check: G = 2 + 4 = 6 and H = 0 + 4 = 4. The tiny amount of H that remains is far less than one particle, so it cannot be drawn.
- (B) conserves atoms but shows plenty of both reactants, which fits a moderate or small K.
- (C) shows no reaction at all, which would need K close to zero.
- (D) has 6 H atoms in its 6 GH, but only 4 H atoms were in the box. It breaks atom conservation.
Question 4 (multiple choice · core)
For 2 R(g) ⇌ R₂(g), Kc = 0.20. A 1.0 L box holds 6 R and 1 R₂, and each particle represents 0.10 mol. What happens next?
- (A) Q < K, so more R₂ forms.
- (B) Q > K, so more R forms.
- (C) Q > K, so more R₂ forms.
- (D) Q = K, so the counts do not change.
Answer and explanation
Answer: (B). The sides have different numbers of particles, so convert first. [R] = 6 × 0.10 = 0.60 mol L⁻¹; [R₂] = 0.10 mol L⁻¹. Q = 0.10 ÷ (0.60)² = 0.28. Since 0.28 > 0.20, the reverse reaction wins and R₂ splits into R.
- (A) puts raw counts into Q: 1 ÷ 6² = 0.028, which looks smaller than K. The unconverted factor makes Q ten times too small here.
- (C) has the right comparison but the wrong consequence. Q > K means the reaction goes backwards, making reactant.
- (D) Q is 0.28, not 0.20, so the box is not at equilibrium.
Question 5 (drawing · core)
For U₂(g) + V₂(g) ⇌ 2 UV(g), K = 9.0. A box starts with 5 U₂ and 5 V₂ molecules and no UV.
(a) Explain which way the reaction goes. (b) Determine the number of each kind of molecule in the box at equilibrium. (c) Describe or draw the equilibrium box, with a key.
Worked solution and suggested Marlbridge rubric
(a) No UV, so Q = 0 < K. There is a net forward reaction.
(b) Let x molecules of U₂ react (with x of V₂), making 2x UV. Both sides have two particles, so counts can go into K: (2x)² ÷ (5 − x)² = 9.0. Square root: 2x ÷ (5 − x) = 3, so 2x = 15 − 3x and x = 3. Equilibrium: 2 U₂, 2 V₂, 6 UV. Check: 6² ÷ (2 × 2) = 9.0. ✓
(c) A box the same size as the starting box, with a key (for example filled circle = U, open circle = V), containing 2 filled pairs, 2 open pairs and 6 mixed pairs. U atoms: 2 × 2 + 6 = 10, the same as at the start (5 × 2). ✓
| Point | What earns it |
|---|---|
| 1 | (a) Q = 0 < K, so forward |
| 1 | (b) Correct expression in counts and x = 3 |
| 1 | (b) 2 U₂, 2 V₂, 6 UV |
| 1 | (c) Box with a key, the right counts and conserved atoms |
Question 6 (constructed response · core)
A container holds the fictional gas W₂, which splits: W₂(g) ⇌ 2 W(g). Four snapshots are taken. Each box represents 2.0 L, and each particle represents 0.20 mol.
| Snapshot | W₂ | W |
|---|---|---|
| t₀ | 6 | 0 |
| t₁ | 4 | 4 |
| t₂ | 3 | 6 |
| t₃ | 3 | 6 |
(a) Show that the snapshots conserve atoms. (b) State the earliest snapshot at which the system is at equilibrium, and justify your answer. (c) Calculate Kc. (d) A student puts the counts straight into K and gets 12. Explain why this is wrong. (e) Compare the forward and reverse rates at t₁ and at t₃.
Worked solution and suggested Marlbridge rubric
(a) W atoms = 2 × (W₂) + (W): t₀ 12; t₁ 8 + 4 = 12; t₂ 6 + 6 = 12; t₃ 12. The same in every box.
(b) t₂. The counts at t₂ and t₃ are the same, so there is no net change from t₂ onwards. (Strictly, the data show equilibrium was reached at or before t₂ and after t₁.)
(c) Each particle gives 0.20 mol ÷ 2.0 L = 0.10 mol L⁻¹. At equilibrium, [W] = 6 × 0.10 = 0.60 mol L⁻¹ and [W₂] = 3 × 0.10 = 0.30 mol L⁻¹. Kc = [W]² ÷ [W₂] = (0.60)² ÷ 0.30 = 0.36 ÷ 0.30 = 1.2.
(d) The equation has one particle on the left and two on the right, so the conversion factor (0.10 mol L⁻¹ per particle) does not cancel. Raw counts give 6² ÷ 3 = 12, which is ten times too big.
(e) At t₁, Q = (0.40)² ÷ 0.40 = 0.40, which is less than K, so the forward rate is greater than the reverse rate. At t₃ the forward and reverse rates are equal (and not zero).
| Point | What earns it |
|---|---|
| 1 | (a) Shows 12 W atoms in every snapshot |
| 1 | (b) t₂, because the counts no longer change |
| 1 | (c) Converts counts to concentrations correctly |
| 1 | (c) Kc = 1.2 |
| 1 | (d) Links the error to the different numbers of particles on each side |
| 1 | (e) Forward faster at t₁; rates equal at t₃ |
Question 7 (evaluation · stretch)
For F₂(g) + G₂(g) ⇌ 2 FG(g), Kc = 1.0. A box starts with 3 F₂ and 3 G₂. A student draws the equilibrium box with 2 F₂, 2 G₂ and 3 FG.
(a) Identify two errors in the student’s box. (b) Determine the correct equilibrium box.
Model answer and suggested Marlbridge rubric
(a) Error 1: atoms are not conserved. The start has 6 F atoms; the student’s box has 2 × 2 + 3 = 7. Error 2: the box does not match K. Both sides have two particles, so counts can be used: Q = 3² ÷ (2 × 2) = 2.25, not 1.0.
(b) Let x F₂ react: (2x)² ÷ (3 − x)² = 1.0, so 2x = 3 − x and x = 1. The correct box has 2 F₂, 2 G₂ and 2 FG. Check: Q = 2² ÷ (2 × 2) = 1.0 ✓; F atoms = 4 + 2 = 6 ✓.
| Point | What earns it |
|---|---|
| 1 | (a) Atom conservation error, with a count |
| 1 | (a) Q ≠ K error, with Q calculated |
| 1 | (b) Correct method from the starting box |
| 1 | (b) 2 F₂, 2 G₂, 2 FG, with a check |
Equal counts of all three species here are a feature of this particular K, not a general rule for equilibrium.
How did you do?
- Q1 wrong: re-read “Before and at equilibrium” and Figure 1 in the study guide.
- Q2, Q4 or Q6(c)–(d) wrong: go back to “From particles to Q and K” and Worked example 1.
- Q3 wrong: use the table in “Linking the picture to the size of K”, and always count atoms.
- Q5 or Q7 incomplete: follow the “Drawing your own particle diagram” checklist and Worked example 2.
Then tick off the topic checklist.