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, G, H, J, M, R, S, T, U, V, W, Z) are fictional, and every temperature is constant. Concentrations are in mol L⁻¹ and partial pressures in atm; K has no units.
Question 1 (multiple choice · foundation)
For the reaction 2 D(g) ⇌ D₂(g), Kc = 4.0. At one moment a flask contains [D] = 0.50 mol L⁻¹ and [D₂] = 2.0 mol L⁻¹. Which statement is correct?
- (A) Q < K, so there is a net forward reaction and [D₂] increases.
- (B) Q > K, so there is a net reverse reaction and [D] increases.
- (C) Q = K, so the mixture is at equilibrium and nothing changes.
- (D) Q > K, so there is a net reverse reaction and [D] decreases.
Answer and explanation
Answer: (B). Q = [D₂] / [D]² = 2.0 / (0.50)² = 2.0 / 0.25 = 8.0. Since 8.0 > 4.0, there is too much product. The reaction runs backwards: D₂ is used up and D forms.
- (A) uses the upside-down expression, [D]² / [D₂] = 0.125, which makes Q look smaller than K.
- (C) forgets to square [D]: 2.0 / 0.50 = 4.0, which happens to equal K.
- (D) gets the direction right but the consequence wrong. A reverse reaction makes reactant, so [D] rises.
Question 2 (multiple choice · core)
For the reaction 2 G(g) + H(g) ⇌ 3 J(g), a flask starts with [G] = 0.40 mol L⁻¹, some H and no J. At equilibrium [J] = 0.12 mol L⁻¹. What is [G] at equilibrium?
- (A) 0.16 mol L⁻¹
- (B) 0.28 mol L⁻¹
- (C) 0.32 mol L⁻¹
- (D) 0.36 mol L⁻¹
Answer and explanation
Answer: (C). J rises by 3x = 0.12, so x = 0.040. G falls by 2x = 0.080. [G] = 0.40 − 0.080 = 0.32 mol L⁻¹.
- (A) takes G’s change as twice J’s change (0.24) instead of two-thirds of it.
- (B) assumes every species changes by the same amount (1 : 1).
- (D) uses H’s change (x = 0.040) for G, ignoring G’s coefficient of 2.
Question 3 (multiple choice · core)
For M₂(g) ⇌ 2 M(g), Kc = 1.6 × 10⁻⁷. A flask starts with [M₂] = 0.25 mol L⁻¹ and no M. What is [M] at equilibrium?
- (A) 4.0 × 10⁻⁸ mol L⁻¹
- (B) 1.0 × 10⁻⁴ mol L⁻¹
- (C) 2.0 × 10⁻⁴ mol L⁻¹
- (D) 4.0 × 10⁻⁴ mol L⁻¹
Answer and explanation
Answer: (C). Kc = (2x)² / (0.25 − x). K is tiny, so 0.25 − x ≈ 0.25. Then 4x² = 1.6 × 10⁻⁷ × 0.25 = 4.0 × 10⁻⁸, x² = 1.0 × 10⁻⁸ and x = 1.0 × 10⁻⁴. [M] = 2x = 2.0 × 10⁻⁴ mol L⁻¹. Check: x is 0.04% of 0.25, so the shortcut is valid.
- (A) forgets to square [M], writing 2x = K × 0.25.
- (B) is x, not 2x. It is the amount of M₂ that split, not the concentration of M.
- (D) is √K. It leaves out [M₂] altogether.
Question 4 (multiple choice · foundation)
A reversible reaction in a sealed flask reaches the point where Q = K. Which statement describes the flask from then on, at constant temperature?
- (A) Both the forward and reverse reactions have stopped.
- (B) The forward and reverse reactions continue at equal rates.
- (C) The concentrations of reactants and products are equal.
- (D) The forward reaction continues faster than the reverse reaction, but too slowly to measure.
Answer and explanation
Answer: (B). Equilibrium is dynamic. Each species is formed as fast as it is used up, so the concentrations, and the ratio of products to reactants, stay constant.
- (A) confuses “no net change” with “no reaction”.
- (C) confuses constant concentrations with equal concentrations. Their ratio is fixed by K, which is rarely 1.
- (D) would mean products keep building up slowly, so Q would keep rising and would no longer equal K.
Question 5 (calculation · core)
For the reaction U₂(g) + V₂(g) ⇌ 2 UV(g), Kp = 49.0. A container is filled with UV only, at a partial pressure of 1.80 atm. Calculate the equilibrium partial pressures of U₂, V₂ and UV.
Worked solution
- Direction. There are no reactants, so Q is undefined (treated as infinitely large) and Q > K. The reaction goes backwards.
- ICE table.
| U₂ | V₂ | 2 UV | |
|---|---|---|---|
| I (atm) | 0 | 0 | 1.80 |
| C (atm) | +x | +x | −2x |
| E (atm) | x | x | 1.80 − 2x |
- Solve. (1.80 − 2x)² / x² = 49.0. Square root: (1.80 − 2x) / x = 7.00, so 1.80 = 9x and x = 0.200 atm.
- Answer. P(U₂) = P(V₂) = 0.200 atm; P(UV) = 1.80 − 0.400 = 1.40 atm.
- Check. (1.40)² / (0.200 × 0.200) = 1.96 / 0.0400 = 49.0. ✓
Suggested mark points (3): 1 for recognising the reverse direction with a correct Change row (+x, +x, −2x); 1 for the perfect-square step giving x = 0.200; 1 for all three partial pressures.
Common error: writing −x for UV gives (1.80 − x)/x = 7, so x = 0.225 atm and P(UV) = 1.575 atm. Substituting these back into K still gives 49.0, so the K check cannot catch this mistake. An atom check can: two UV molecules must break to form one U₂ and one V₂, so 1.80 = P(UV) + 2 × P(U₂). The correct answer gives 1.40 + 0.400 = 1.80 ✓; the wrong one gives 1.575 + 0.450 = 2.025 ✗.
Question 6 (constructed response · core)
0.80 mol of the fictional gas R is placed in an empty 2.00 L flask. It decomposes: R(g) ⇌ S(g) + T(g), Kc = 0.50.
(a) Explain how you know the reaction will go forwards. (b) A student tries the small-x shortcut, writing x² / 0.40 = 0.50. Show that the shortcut fails here. (c) Calculate the equilibrium concentrations of R, S and T. (d) Sketch a concentration–time graph for R and S from the start until after equilibrium is reached.
Worked solution and suggested Marlbridge rubric
(a) There is no S or T at the start, so Q = 0. Since Q < K, there is a net forward reaction.
(b) The shortcut gives x² = 0.50 × 0.40 = 0.20, so x = 0.447 mol L⁻¹. That is more than the starting [R] of 0.40 mol L⁻¹, which is impossible: you cannot use up more R than you had. K is not small compared with 0.40, so x cannot be ignored.
(c) [R]₀ = 0.80 mol ÷ 2.00 L = 0.40 mol L⁻¹.
| R | S | T | |
|---|---|---|---|
| I (mol L⁻¹) | 0.40 | 0 | 0 |
| C (mol L⁻¹) | −x | +x | +x |
| E (mol L⁻¹) | 0.40 − x | x | x |
x² / (0.40 − x) = 0.50, so x² = 0.20 − 0.50x and x² + 0.50x − 0.20 = 0. x = [−0.50 + √(0.25 + 0.80)] / 2 = (−0.50 + 1.0247) / 2 = 0.262 mol L⁻¹. (The other root, −0.762, would make [S] negative.) [S] = [T] = 0.26 mol L⁻¹; [R] = 0.40 − 0.262 = 0.14 mol L⁻¹. Check: (0.2623)² / 0.1377 = 0.50. ✓
(d) Features of a good sketch: vertical axis “Concentration (mol L⁻¹)” with an even scale from 0 to at least 0.40; horizontal axis “Time”; the R curve starts at 0.40 and falls, levelling off at 0.14; the S curve starts at 0 and rises, levelling off at 0.26 (T is the same curve as S); both curves become flat at the same time; the fall in R (0.26) equals the rise in S.
| Point | What earns it |
|---|---|
| 1 | (a) Q = 0 (no products), so Q < K and the reaction goes forwards |
| 1 | (b) Shows x ≈ 0.447 is larger than 0.40, so the assumption is invalid |
| 1 | (c) Correct quadratic, x² + 0.50x − 0.20 = 0 (or equivalent) from a correct ICE table |
| 1 | (c) [S] = [T] = 0.26 and [R] = 0.14 mol L⁻¹, rejecting the negative root |
| 1 | (d) Labelled concentration axis with units and an even scale |
| 1 | (d) Curves start and end at the right values, change by equal amounts and flatten at the same time |
Accept answers to 2 or 3 significant figures. Carry forward an error from (c) into (d) if the graph matches the student’s values.
Question 7 (calculation and explanation · stretch)
Two fictional solutes combine in water: W(aq) + Z(aq) ⇌ WZ(aq), Kc = 2.0 × 10⁶. Solutions are mixed so that, before any reaction, [W] = 0.300 mol L⁻¹, [Z] = 0.100 mol L⁻¹ and [WZ] = 0.
(a) Explain why it is reasonable to assume that almost all of the Z reacts. (b) Calculate [WZ], [W] and [Z] at equilibrium.
Worked solution and suggested Marlbridge rubric
(a) K is very large, so at equilibrium the products are strongly favoured and the reaction goes almost to completion. Z is the limiting reactant (0.100 mol L⁻¹ against 0.300 mol L⁻¹ of W, reacting 1 : 1), so nearly all of it is used up.
(b) Two stages make the algebra easy.
- Go to completion on paper. All 0.100 mol L⁻¹ of Z reacts: [WZ] = 0.100, [W] = 0.300 − 0.100 = 0.200, [Z] = 0.
- Let a small amount, y, come back. [WZ] = 0.100 − y ≈ 0.100; [W] = 0.200 + y ≈ 0.200; [Z] = y.
- Substitute. 2.0 × 10⁶ = 0.100 / (0.200 × y), so y = 0.100 / (0.200 × 2.0 × 10⁶) = 2.5 × 10⁻⁷ mol L⁻¹.
Answer. [WZ] = 0.100 mol L⁻¹, [W] = 0.200 mol L⁻¹, [Z] = 2.5 × 10⁻⁷ mol L⁻¹. y is far less than 5% of 0.100, so the approximations hold.
| Point | What earns it |
|---|---|
| 1 | (a) Links the very large K to the reaction going almost to completion |
| 1 | (a) Identifies Z as the limiting reactant |
| 1 | (b) [WZ] ≈ 0.100 and [W] ≈ 0.200 mol L⁻¹ |
| 1 | (b) [Z] = 2.5 × 10⁻⁷ mol L⁻¹ from a correct K expression |
A full ICE table with −x for W and Z and +x for WZ, solved exactly, gives the same values and earns full credit. Do not award (b) if the student reports [Z] = 0: some Z must remain, or K would be infinite.
How did you do?
- Q1 or Q4 wrong: re-read “Step 1: compare Q with K” in the study guide.
- Q2 or Q5 wrong: check the Change row follows the coefficients, as in Worked example 1.
- Q3 or Q6(b) wrong: go back to Worked example 2 and the 5% check.
- Q6(d) incomplete: compare your sketch with Figure 1 and its four-point checklist.
- Q7 incomplete: use the “Choosing a method” table, large-K row.
Then tick off the topic checklist.