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. Concentrations are in mol L⁻¹ and partial pressures in atm. Every K value is given in the question and was chosen for practice; Q and K are written without units.
Question 1 (multiple choice · foundation)
Which is the equilibrium expression Kc for the reaction below?
2H₂S(g) + 3O₂(g) ⇌ 2SO₂(g) + 2H₂O(g)
- (A) [H₂S]²[O₂]³ / ([SO₂]²[H₂O]²)
- (B) [SO₂]²[H₂O]² / ([H₂S]²[O₂]³)
- (C) [SO₂]² / ([H₂S]²[O₂]³)
- (D) (2[SO₂])(2[H₂O]) / ((2[H₂S])(3[O₂]))
Answer and explanation
Answer: (B). Products go on top, reactants underneath, and each concentration is raised to the power of its coefficient. All four species are gases, so all four appear.
- (A) is upside down: it puts the reactants on top.
- (C) leaves out water. Water is only left out when it is a liquid; here it is H₂O(g).
- (D) multiplies by the coefficients instead of using them as powers.
Question 2 (multiple choice · core)
When sodium hydrogencarbonate is heated in a closed container, this equilibrium is set up:
2NaHCO₃(s) ⇌ Na₂CO₃(s) + CO₂(g) + H₂O(g)
Which is the expression for Kp?
- (A) P_CO₂
- (B) P_CO₂ + P_H₂O
- (C) P_CO₂ × P_H₂O
- (D) P_CO₂ × P_H₂O / (P_NaHCO₃)²
Answer and explanation
Answer: (C). The two solids are left out, so only the two gaseous products remain. Terms are multiplied, and each has a power of 1.
- (A) wrongly leaves out water vapour, which is a gas.
- (B) adds the pressures. Terms in Q and K are always multiplied.
- (D) includes a solid. A solid has no partial pressure, and its “concentration” does not depend on its amount, so it is never part of K.
Question 3 (multiple choice · core)
Dichromate ions and chromate ions are linked by this equilibrium in water:
Cr₂O₇²⁻(aq) + H₂O(l) ⇌ 2CrO₄²⁻(aq) + 2H⁺(aq)
Which is the expression for Kc?
- (A) [CrO₄²⁻]²[H⁺]² / [Cr₂O₇²⁻]
- (B) [CrO₄²⁻]²[H⁺]² / ([Cr₂O₇²⁻][H₂O])
- (C) [CrO₄²⁻][H⁺] / [Cr₂O₇²⁻]
- (D) [Cr₂O₇²⁻] / ([CrO₄²⁻]²[H⁺]²)
Answer and explanation
Answer: (A). Water is the liquid solvent, so it is left out. The two products each have a coefficient of 2, so both are squared.
- (B) includes H₂O(l), a pure liquid.
- (C) ignores the coefficients of 2.
- (D) is upside down.
Question 4 (multiple choice · core)
For the reaction 2NO(g) + Br₂(g) ⇌ 2NOBr(g), a mixture at one moment has [NO] = 0.20 mol L⁻¹, [Br₂] = 0.10 mol L⁻¹ and [NOBr] = 0.30 mol L⁻¹. What is the value of Qc?
- (A) 0.044
- (B) 4.5
- (C) 15
- (D) 22.5
Answer and explanation
Answer: (D). Qc = [NOBr]² / ([NO]²[Br₂]) = (0.30)² / ((0.20)² × 0.10) = 0.090 / 0.0040 = 22.5.
- (A) is the expression upside down (1 ÷ 22.5 = 0.044).
- (B) squares [NOBr] but forgets to square [NO]: 0.090 ÷ (0.20 × 0.10) = 4.5.
- (C) leaves out both powers: 0.30 ÷ (0.20 × 0.10) = 15.
Question 5 (constructed response · core)
Methanol can be made from carbon monoxide and hydrogen:
CO(g) + 2H₂(g) ⇌ CH₃OH(g)
(a) Write the expression for Kc and the expression for Kp. (b) A student says: “Kc and Kp are just two names for the same number.” Explain why this is not correct. (c) At one moment a reaction vessel contains P_CO = 0.50 atm, P_H₂ = 1.00 atm and P_CH₃OH = 0.10 atm. Calculate Qp. (d) At the temperature of the vessel, Kp = 0.050. Is the mixture at equilibrium? As the system approaches equilibrium, will Qp increase or decrease? Justify your answers.
Worked solution and suggested Marlbridge rubric
(a) Kc = [CH₃OH] / ([CO][H₂]²); Kp = P_CH₃OH / (P_CO × (P_H₂)²).
(b) Kc is built from concentrations in mol L⁻¹, while Kp is built from partial pressures in atm. The same mixture gives different numbers in the two forms, so for most reactions Kc and Kp have different values. A question must be answered with the one it names.
(c) Qp = 0.10 / (0.50 × (1.00)²) = 0.10 / 0.50 = 0.20.
(d) Qp = 0.20 is not equal to Kp = 0.050, so the mixture is not at equilibrium. The reaction quotient always moves towards K, so Qp will decrease from 0.20 until it reaches 0.050 (it is four times too large now).
| Point | What earns it |
|---|---|
| 1 | Correct Kc and Kp expressions, with [H₂]² and (P_H₂)² |
| 1 | States that Kc uses concentrations and Kp uses partial pressures, so their values generally differ |
| 1 | Qp = 0.20 with substitution shown |
| 1 | Not at equilibrium, because Qp ≠ Kp |
| 1 | Qp decreases, because Q moves towards K and 0.20 > 0.050 |
Accept a wrong Qp from (c) carried forward into (d) if the reasoning is consistent. Do not award the (b) point for “they are different” with no reason.
Question 6 (constructed response · core)
A sealed 1.00 L flask at constant temperature starts with 0.500 mol of a gas X and no Y. The reaction is X(g) ⇌ 2Y(g). The table shows the measured concentrations.
| Time (s) | [X] (mol L⁻¹) | [Y] (mol L⁻¹) |
|---|---|---|
| 0 | 0.500 | 0.000 |
| 10 | 0.450 | 0.100 |
| 20 | 0.420 | 0.160 |
| 30 | 0.405 | 0.190 |
| 40 | 0.400 | 0.200 |
| 50 | 0.400 | 0.200 |
(a) Write the expression for Qc. (b) Calculate Qc at each time. (c) Describe how you would plot Qc against time, including the axes and scale, and sketch the shape of the graph. (d) State the value of Kc and the earliest time in the table at which the system is shown to be at equilibrium. (e) A second flask at the same temperature starts with pure Y. What value will Qc have once that flask reaches equilibrium? Explain.
Worked solution and suggested Marlbridge rubric
(a) Qc = [Y]² / [X].
(b)
| Time (s) | 0 | 10 | 20 | 30 | 40 | 50 |
|---|---|---|---|---|---|---|
| Qc | 0 | 0.0222 | 0.0610 | 0.0891 | 0.100 | 0.100 |
For example, at 20 s: (0.160)² ÷ 0.420 = 0.0256 ÷ 0.420 = 0.0610.
(c) Time (s) on the horizontal axis from 0 to 50 in steps of 10; Qc (no unit) on the vertical axis from 0 to 0.12 in even steps such as 0.02. Plot the six points and join them with a smooth curve. The curve rises steeply at first, then more slowly, and becomes a horizontal line at Qc = 0.100.
(d) Kc = 0.100. The concentrations, and so Qc, stop changing by 40 s. (Equilibrium may have been reached a little earlier, between 30 s and 40 s; the table cannot show exactly when.)
(e) Qc will reach 0.100. Kc depends only on the reaction and the temperature, not on the starting mixture. Starting with pure Y, there is no X at first, so Qc has no finite value. As soon as a little X forms, Qc is very large, and it then falls until it equals Kc.
| Point | What earns it |
|---|---|
| 1 | Correct expression with [Y] squared |
| 1 | At least four correct Qc values |
| 1 | Axes labelled with quantity (and unit for time), even scales, curve levelling off at 0.100 |
| 1 | Kc = 0.100 and equilibrium shown by 40 s |
| 1 | 0.100, because K depends only on temperature for a given reaction, not on starting amounts |
Question 7 (explanation · stretch)
Magnesium carbonate decomposes on heating: MgCO₃(s) ⇌ MgO(s) + CO₂(g). Two identical sealed, evacuated flasks are heated to the same temperature. Flask 1 contains 5.0 g of MgCO₃ and flask 2 contains 20.0 g. In both flasks some MgCO₃ and some MgO remain at equilibrium.
(a) Write the expression for Kp. (b) A student predicts that the equilibrium pressure of CO₂ in flask 2 will be four times that in flask 1. Evaluate this prediction. (c) Explain why solids are left out of the expression for K.
Model answer and suggested Marlbridge rubric
(a) Kp = P_CO₂.
(b) The prediction is wrong. Both flasks are at the same temperature, so Kp is the same. Since Kp = P_CO₂, the equilibrium pressure of CO₂ is the same in both flasks. Flask 2 simply has more solid left over at equilibrium.
(c) The concentration of a pure solid (moles per litre of the solid itself) is fixed by its density. A bigger lump has more moles but also more volume, so the ratio does not change. Because adding or removing solid cannot change this value, it cannot shift the equilibrium, so it is left out of Q and K. The solids must still be present for the equilibrium to exist.
| Point | What earns it |
|---|---|
| 1 | Kp = P_CO₂ (no solids) |
| 1 | Prediction wrong: same Kp at the same temperature, so the same P_CO₂ |
| 1 | The concentration of a pure solid does not depend on the amount present (fixed by density) |
| 1 | So changing the amount of solid does not change Q or the equilibrium mixture |
How did you do?
- Q1–Q3 wrong: go back to the three rules and the table of what to leave out in the study guide.
- Q4 or Q5(c) wrong: write the expression first, then substitute, and check every power.
- Q6 incomplete: compare your graph with Figure 1 in the study guide; the flat part gives K.
- Q7 incomplete: your answer needs the reason solids are left out, not only the rule.
Then tick off the topic checklist.