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. All measurements are invented for practice, and every system is at constant temperature. K is written without units.
Question 1 (multiple choice · foundation)
For N₂O₄(g) ⇌ 2NO₂(g), an equilibrium mixture contains [N₂O₄] = 0.0400 mol L⁻¹ and [NO₂] = 0.0120 mol L⁻¹. What is Kc?
- (A) 3.60 × 10⁻³
- (B) 1.44 × 10⁻²
- (C) 0.300
- (D) 278
Answer and explanation
Answer: (A). Kc = [NO₂]² / [N₂O₄] = (0.0120)² / 0.0400 = 1.44 × 10⁻⁴ / 0.0400 = 3.60 × 10⁻³.
- (B) doubles [NO₂] before squaring, (2 × 0.0120)² ÷ 0.0400, treating the coefficient as a multiplier as well as a power.
- (C) forgets to square [NO₂].
- (D) is the expression upside down.
Question 2 (multiple choice · core)
For FeO(s) + CO(g) ⇌ Fe(s) + CO₂(g), the equilibrium partial pressures are P_CO = 0.80 atm and P_CO₂ = 0.32 atm. What is Kp?
- (A) 2.5
- (B) 1.12
- (C) 0.40
- (D) 0.26
Answer and explanation
Answer: (C). The solids are left out, so Kp = P_CO₂ / P_CO = 0.32 / 0.80 = 0.40.
- (A) is upside down (0.80 ÷ 0.32).
- (B) adds the two pressures.
- (D) multiplies the two pressures (0.32 × 0.80 = 0.256) instead of dividing, as if CO were a product.
Question 3 (multiple choice · core)
A vessel starts with [A] = 0.100 mol L⁻¹ and [B] = 0.100 mol L⁻¹ and no C. The reaction is A(g) + B(g) ⇌ 2C(g). At equilibrium [C] = 0.040 mol L⁻¹. What is Kc?
- (A) 0.16
- (B) 0.25
- (C) 0.44
- (D) 6.25
Answer and explanation
Answer: (B). Making 0.040 mol L⁻¹ of C uses 0.020 mol L⁻¹ each of A and B (1 : 1 : 2). So [A] = [B] = 0.080 mol L⁻¹, and Kc = (0.040)² / (0.080 × 0.080) = 0.0016 / 0.0064 = 0.25.
- (A) uses the starting values 0.100 for A and B.
- (C) subtracts 0.040 from A and B, ignoring the 1 : 2 ratio, so [A] = [B] = 0.060.
- (D) forgets to square [C].
Question 4 (multiple choice · core)
For 2A(g) ⇌ B(g), two equilibrium mixtures are made at the same temperature. In mixture 2, [A] is twice as large as in mixture 1. How does [B] in mixture 2 compare with [B] in mixture 1?
- (A) half as large
- (B) the same
- (C) twice as large
- (D) four times as large
Answer and explanation
Answer: (D). Kc = [B] / [A]² is the same in both mixtures. Doubling [A] makes [A]² four times larger, so [B] must also be four times larger to keep the ratio equal to Kc.
- (A) has the relationship upside down.
- (B) wrongly assumes [B] is fixed; K is fixed, not the individual concentrations.
- (C) ignores the power of 2 on [A].
Question 5 (constructed response · core)
A student mixes 10.0 mL of 0.00500 mol L⁻¹ Fe(NO₃)₃ with 10.0 mL of 0.00300 mol L⁻¹ KSCN. A red complex ion forms:
Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq)
The absorbance of the mixture is measured every minute. After it stops changing, a calibration graph shows [FeSCN²⁺] = 3.00 × 10⁻⁴ mol L⁻¹.
(a) Write the expression for Kc. (b) Calculate [Fe³⁺] and [SCN⁻] immediately after mixing, before any reaction. (c) Calculate the equilibrium concentrations of Fe³⁺ and SCN⁻. (d) Calculate Kc. (e) Explain why the student waited for the absorbance to stop changing.
Worked solution and suggested Marlbridge rubric
(a) Kc = [FeSCN²⁺] / ([Fe³⁺][SCN⁻]).
(b) The total volume is 20.0 mL, so each solution is diluted by a factor of 2. [Fe³⁺]₀ = 0.00500 × 10.0 / 20.0 = 2.50 × 10⁻³ mol L⁻¹; [SCN⁻]₀ = 0.00300 × 10.0 / 20.0 = 1.50 × 10⁻³ mol L⁻¹.
(c) Forming 3.00 × 10⁻⁴ mol L⁻¹ of FeSCN²⁺ uses the same amount of each reactant (1 : 1 : 1).
| Fe³⁺ | SCN⁻ | FeSCN²⁺ | |
|---|---|---|---|
| Initial | 2.50 × 10⁻³ | 1.50 × 10⁻³ | 0 |
| Change | −3.00 × 10⁻⁴ | −3.00 × 10⁻⁴ | +3.00 × 10⁻⁴ |
| Equilibrium | 2.20 × 10⁻³ | 1.20 × 10⁻³ | 3.00 × 10⁻⁴ |
(d) Kc = 3.00 × 10⁻⁴ / (2.20 × 10⁻³ × 1.20 × 10⁻³) = 3.00 × 10⁻⁴ / 2.64 × 10⁻⁶ = 114.
(e) K must be calculated from equilibrium concentrations. A constant absorbance shows that [FeSCN²⁺] has stopped changing, so the system has reached equilibrium.
| Point | What earns it |
|---|---|
| 1 | Correct Kc expression |
| 1 | Both starting concentrations halved for dilution |
| 1 | Equilibrium concentrations found by subtracting 3.00 × 10⁻⁴ from each reactant |
| 1 | Kc = 114 (accept 113–114) |
| 1 | Constant absorbance means constant concentration, so equilibrium |
Common errors: forgetting the dilution gives 23.6; using the starting concentrations without subtracting gives 80.0. Carry forward an error from (b) into (c) and (d) if the method is otherwise correct.
Question 6 (constructed response · stretch)
Pure hydrogen sulfide is sealed in a rigid container and heated to a fixed high temperature. At that temperature, before any reaction, its pressure would be 1.00 atm. It partly decomposes:
2H₂S(g) ⇌ 2H₂(g) + S₂(g)
At equilibrium the total pressure is 1.15 atm.
(a) Explain why the total pressure rises as the reaction proceeds. (b) Calculate the equilibrium partial pressure of each gas. (c) Calculate Kp.
Worked solution and suggested Marlbridge rubric
(a) Every 2 mol of gas that react form 3 mol of gas. At constant volume and temperature, pressure is proportional to the amount of gas, so the pressure rises.
(b) Let the partial pressure of S₂ formed be y atm.
| H₂S | H₂ | S₂ | |
|---|---|---|---|
| Initial | 1.00 | 0 | 0 |
| Change | −2y | +2y | +y |
| Equilibrium | 1.00 − 2y | 2y | y |
Total pressure = (1.00 − 2y) + 2y + y = 1.00 + y = 1.15, so y = 0.15 atm. P_S₂ = 0.15 atm, P_H₂ = 0.30 atm, P_H₂S = 1.00 − 0.30 = 0.70 atm.
(c) Kp = (P_H₂)² × P_S₂ / (P_H₂S)² = (0.30)² × 0.15 / (0.70)² = 0.0135 / 0.49 = 0.028 (0.02755 before rounding; two significant figures, because y = 0.15 has two).
| Point | What earns it |
|---|---|
| 1 | More moles of gas on the product side, so pressure rises at constant volume and temperature |
| 1 | Change row in the ratio −2y : +2y : +y |
| 1 | Total pressure expression 1.00 + y = 1.15, giving y = 0.15 |
| 1 | Three correct partial pressures |
| 1 | Kp ≈ 0.028 from a correct expression |
Question 7 (constructed response · core)
Three experiments study CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) at the same temperature. The equilibrium concentrations (mol L⁻¹) are:
| Experiment | [CO] | [H₂O] | [CO₂] | [H₂] |
|---|---|---|---|---|
| 1 | 0.200 | 0.300 | 0.240 | 0.250 |
| 2 | 0.100 | 0.400 | 0.200 | 0.200 |
| 3 | 0.150 | 0.150 | 0.300 | ? |
(a) Calculate Kc from experiment 1 and from experiment 2. (b) A student claims: “Experiment 2 has more water, so its K should be larger.” Evaluate the claim using your answers to (a). (c) Calculate [H₂] in experiment 3.
Worked solution and suggested Marlbridge rubric
(a) Kc = [CO₂][H₂] / ([CO][H₂O]). Experiment 1: (0.240 × 0.250) / (0.200 × 0.300) = 0.0600 / 0.0600 = 1.00. Experiment 2: (0.200 × 0.200) / (0.100 × 0.400) = 0.0400 / 0.0400 = 1.00.
(b) The claim is wrong. Both experiments give Kc = 1.00. The equilibrium concentrations are different, but K depends only on the reaction and the temperature, which are the same.
(c) Kc = 1.00 = (0.300 × [H₂]) / (0.150 × 0.150), so [H₂] = 1.00 × 0.0225 / 0.300 = 0.0750 mol L⁻¹.
| Point | What earns it |
|---|---|
| 1 | Correct expression and Kc = 1.00 for experiment 1 |
| 1 | Kc = 1.00 for experiment 2 |
| 1 | Claim wrong: same K, because K depends only on temperature for a given reaction |
| 1 | [H₂] = 0.0750 mol L⁻¹ using Kc = 1.00 |
How did you do?
- Q1 or Q2 wrong: write the expression first, then check every power and leave out solids. See Worked example 1 in the study guide.
- Q3, Q5 or Q6 wrong: practise the ICE table in Worked example 2; the change row must follow the coefficients.
- Q4 or Q7(b) wrong: re-read “Reading K as a relationship between variables”.
Then tick off the topic checklist.