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. Assume each system is at equilibrium before the change, and that only the stated condition changes. The system in Question 5 is invented for practice.
Question 1 (multiple choice · foundation)
Phosphorus pentachloride decomposes in a sealed container:
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) ΔH > 0
Which change increases the amount of Cl₂ present at equilibrium?
- (A) Adding more PCl₃(g)
- (B) Decreasing the volume of the container
- (C) Increasing the temperature
- (D) Adding a catalyst
Answer and explanation
Answer: (C). The forward reaction is endothermic, so heat acts as a reactant. Raising the temperature shifts the system right, making more PCl₃ and Cl₂.
- (A) adds a product, so the system shifts left and uses up Cl₂.
- (B) favours the side with fewer gas molecules: 1 on the left, 2 on the right, so the shift is left.
- (D) speeds up both directions equally; equilibrium is reached sooner but the amount of Cl₂ is the same.
Question 2 (multiple choice · core)
A pale yellow solution of iron(III) ions is mixed with thiocyanate ions. A blood-red complex forms:
Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq)
A few drops of one of the following are added to separate samples of the red equilibrium mixture. Which one makes the red colour paler?
- (A) Concentrated KSCN solution
- (B) Concentrated Fe(NO₃)₃ solution
- (C) Sodium hydroxide solution, which precipitates Fe³⁺ as Fe(OH)₃(s)
- (D) A few crystals of KNO₃, whose ions take no part in the reaction
Answer and explanation
Answer: (C). Precipitating Fe³⁺ removes a reactant. The system shifts left to replace it, so FeSCN²⁺ is used up and the red colour fades.
- (A) adds SCN⁻, a reactant: the shift is right and the red deepens.
- (B) adds Fe³⁺, a reactant: the shift is right and the red deepens.
- (D) adds spectator ions that are not in the equilibrium expression, so there is no shift.
Question 3 (multiple choice · core)
Aqueous ammonia is a weak base:
NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)
Solid ammonium nitrate, NH₄NO₃, is dissolved in a sample of aqueous ammonia at constant temperature. What happens to the pH, and why?
- (A) The pH increases, because adding NH₄⁺ shifts the system right and makes more OH⁻
- (B) The pH decreases, because adding NH₄⁺ shifts the system left and uses up OH⁻
- (C) The pH stays the same, because ammonium nitrate is a salt
- (D) The pH decreases, because adding NH₄⁺ makes K smaller
Answer and explanation
Answer: (B). NH₄⁺ is a product. Adding it shifts the system left, which uses up OH⁻ ions. A lower [OH⁻] means a lower pH.
- (A) gets the direction wrong: adding a product cannot shift the system towards that product.
- (C) ignores the fact that NH₄⁺ appears in the equilibrium.
- (D) has the right pH change for the wrong reason: at constant temperature K does not change.
Question 4 (multiple choice · core)
A sealed, heated glass vessel contains an equilibrium mixture of three gases. Of these, only iodine vapour is coloured (violet):
H₂(g) + I₂(g) ⇌ 2HI(g)
The volume of the vessel is halved at constant temperature. Which statement describes what is observed and why?
- (A) The violet colour becomes darker, and the equilibrium does not shift
- (B) The violet colour becomes paler, because the equilibrium shifts right
- (C) The colour does not change, because the equilibrium does not shift
- (D) The violet colour becomes darker, because the equilibrium shifts left
Answer and explanation
Answer: (A). There are 2 gas molecules on each side, so a change in volume causes no shift. But halving the volume doubles the concentration of every gas, including I₂, so the violet colour is darker.
- (B) assumes a shift towards fewer molecules, but the numbers are equal.
- (C) forgets that compressing the gas raises [I₂] even without any shift.
- (D) gives a shift that the equal molecule count rules out.
Question 5 (constructed response · core)
A student studies an invented metal-ion equilibrium in water. The metal ion X²⁺ is yellow; the complex XL₂ is green; L⁻ is colourless.
X²⁺(aq) + 2L⁻(aq) ⇌ XL₂(aq)
Her observations:
| Test | Change made to a sample of the equilibrium mixture | Observation |
|---|---|---|
| 1 | Warmed in a water bath | Turns more yellow |
| 2 | Solid NaL added and dissolved | Turns more green |
| 3 | Diluted with an equal volume of water | Turns more yellow (more than dilution alone explains) |
(a) Is the forward reaction exothermic or endothermic? Justify your answer using Test 1. (b) Explain the observation in Test 3 in terms of the particles in solution. (c) In Test 2, the student also recorded the temperature in an insulated cup. Predict whether the temperature rose or fell because of the shift, and explain. (Ignore any heat change from dissolving the solid.)
Worked solution and suggested Marlbridge rubric
(a) Warming made the mixture more yellow, so the system shifted left (towards X²⁺). Heating favours the endothermic direction, so the reverse reaction is endothermic and the forward reaction is exothermic.
(b) Dilution lowers the concentration of every dissolved species. The left side has 1 + 2 = 3 dissolved particles; the right side has 1. Dilution favours the side with more dissolved particles, so the system shifts left, forming more yellow X²⁺.
(c) Adding L⁻ shifts the system right. The forward reaction is exothermic (from part (a)), so heat is released and the temperature rises.
| Point | What earns it |
|---|---|
| 1 | Test 1 shows a shift left on heating |
| 1 | Concludes forward is exothermic, linking heating to the endothermic direction |
| 1 | Counts 3 dissolved particles on the left and 1 on the right (water not counted) |
| 1 | Dilution favours the side with more particles, so the shift is left |
| 1 | Temperature rises, because the shift is in the exothermic (forward) direction |
Accept a part (c) answer carried forward from an incorrect part (a) if it is consistent with it.
Question 6 (constructed response · core)
Ammonia is made from its elements in a closed reactor:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH < 0
For each change, state the direction of shift (left, right or none) and give a reason.
(a) The volume of the reactor is decreased at constant temperature. (b) The temperature is increased at constant volume. (c) Argon is pumped in at constant volume and temperature. (d) Ammonia is removed by cooling part of the gas until it liquefies.
Worked solution and suggested Marlbridge rubric
(a) Right. Left side: 1 + 3 = 4 gas molecules; right side: 2. A smaller volume favours the side with fewer gas molecules.
(b) Left. The forward reaction is exothermic, so heat is a product. Adding heat shifts the system in the endothermic direction, which is the reverse.
(c) None. Argon does not react, and at constant volume the partial pressures (concentrations) of N₂, H₂ and NH₃ do not change, so the forward and reverse rates stay equal.
(d) Right. Removing a product shifts the system to replace it.
| Point | What earns it |
|---|---|
| 1 | (a) right, with 4 → 2 gas molecules |
| 1 | (b) left, linked to the exothermic forward reaction |
| 1 | (c) no shift, because partial pressures of reacting gases are unchanged |
| 1 | (d) right, because a product is removed |
A direction without a reason earns no point.
Question 7 (explanation · stretch)
Solid ammonium hydrogen sulfide decomposes in a sealed flask at constant temperature:
NH₄HS(s) ⇌ NH₃(g) + H₂S(g)
(a) A student claims: “Adding more solid NH₄HS will increase the pressure of NH₃ at equilibrium, because more reactant is present.” Evaluate the claim. (b) Some H₂S is removed from the flask without removing anything else. Predict what happens to the amount of solid and to the pressure of NH₃ when equilibrium is re-established.
Model answer and suggested Marlbridge rubric
(a) The claim is incorrect. NH₄HS is a solid, so it does not appear in the equilibrium expression, which contains only the two gases. Adding more solid does not change the balance between the forward and reverse rates, so there is no shift and the NH₃ pressure is unchanged.
(b) Removing H₂S, a product, shifts the system right to replace it. Some solid decomposes, so the amount of solid decreases. The decomposition also makes NH₃, so the NH₃ pressure increases.
| Point | What earns it |
|---|---|
| 1 | Claim is wrong because the solid is not in the equilibrium expression |
| 1 | Removing H₂S shifts the system right |
| 1 | Amount of solid decreases and NH₃ pressure increases |
Do not award the first point for “the claim is wrong” without a reason.
How did you do?
- Q1, Q2 or Q6 wrong: re-read the four stresses in the study guide.
- Q3, Q4 or Q5(c) wrong: practise linking each shift to what you would measure (the table “What you would actually see”).
- Q5(b), Q7 incomplete: check your particle counts and leave out solids, pure liquids and solvent water.
Then tick off the topic checklist.