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. No data are needed to answer the questions. Where an answer quotes a value to confirm a prediction, it is a standard entropy change at 298 K calculated from published standard molar entropies, in J K⁻¹ mol⁻¹.
Question 1 (multiple choice · foundation)
For which process does the entropy of the system decrease?
- (A) Solid carbon dioxide (dry ice) turns directly into CO₂ gas.
- (B) Water vapour condenses into droplets on a cold window.
- (C) Helium in a sealed, rigid container is warmed from 20 °C to 60 °C.
- (D) N₂O₄(g) → 2 NO₂(g)
Answer and explanation
Answer: (B). Gas becomes liquid. The water molecules go from moving freely through a large volume to being held close together in droplets, so matter is much less dispersed: ΔS < 0.
- (A) is sublimation, solid → gas. Matter becomes far more dispersed: ΔS > 0.
- (C) has no change in volume, but the temperature rises. The kinetic-energy distribution widens, so energy is more dispersed: ΔS > 0.
- (D) goes from 1 mol of gas to 2 mol of gas: ΔS > 0.
Question 2 (multiple choice · core)
Which process has the largest positive entropy change per mole of reaction as written?
- (A) NH₄Cl(s) → NH₃(g) + HCl(g)
- (B) CaCO₃(s) → CaO(s) + CO₂(g)
- (C) H₂O(s) → H₂O(l)
- (D) 2 SO₃(g) → 2 SO₂(g) + O₂(g)
Answer and explanation
Answer: (A). A solid produces 2 mol of gas from no gas at all. That is the largest increase in moles of gas among the options, so it gives the largest increase in matter dispersal.
- (B) produces only 1 mol of gas from solids. Positive, but smaller than (A).
- (C) is melting. The particles become freer to move but stay close together, so ΔS is positive and small. Melting steps are much smaller than steps that create gas.
- (D) goes from 2 mol to 3 mol of gas, an increase of 1 mol, and the reactant was already a gas. Positive, but smaller than (A).
Question 3 (multiple choice · core)
A sample of argon is heated from 300 K to 450 K at constant volume. Which statement best explains why the entropy of the argon increases?
- (A) The kinetic energies of the atoms are spread over a wider range of values.
- (B) The atoms move into a larger volume, so matter is more dispersed.
- (C) Heating increases the number of moles of gas in the container.
- (D) All of the atoms now have the same, higher kinetic energy.
Answer and explanation
Answer: (A). At the higher temperature the kinetic-energy distribution is lower and wider: energy is more dispersed among the atoms. That is why S rises with temperature.
- (B) is wrong because the volume is constant. Matter dispersal does not change here.
- (C) is wrong: heating does not create atoms. The amount of argon is fixed.
- (D) describes the opposite of what happens. Atoms never all share one energy; at a higher temperature the spread of energies becomes wider, not narrower.
Question 4 (multiple choice · core)
Which statement about the reaction N₂(g) + O₂(g) → 2 NO(g) is best supported by counting moles of gas?
- (A) ΔS is large and positive, because the reaction makes more molecules.
- (B) ΔS is large and negative, because two different gases combine.
- (C) ΔS is small, and its sign cannot be predicted reliably by counting moles of gas.
- (D) ΔS is exactly zero, because the moles of gas are equal on both sides.
Answer and explanation
Answer: (C). There are 2 mol of gas on each side, so there is no large change in matter dispersal. The entropy change must be small, but its sign depends on details of the particular molecules, which counting cannot reveal. (Using data, as you will in Topic 9.2, the value turns out to be small and positive.)
- (A) miscounts: 1 + 1 = 2 mol of reactant gas and 2 mol of product gas.
- (B) has no basis: “two gases combine” is not a reason for a large decrease when the moles of gas do not change.
- (D) over-reaches. Equal moles of gas make ΔS small, not exactly zero.
Question 5 (constructed response · core)
Three identical samples of the same gas are described below. Each holds the same number of molecules.
- Sample P: in a 1.0 L container at 300 K.
- Sample Q: in a 3.0 L container at 300 K.
- Sample R: in a 1.0 L container at 500 K.
(a) Which has the higher entropy, P or Q? Justify your answer at the particle level. (b) Which has the higher entropy, P or R? Justify your answer at the particle level. (c) A student says: “Q and R must have the same entropy, because each has been changed once from P.” Explain whether you can decide which of Q and R has the higher entropy from the information given.
Model answer and suggested Marlbridge rubric
(a) Q. Both are at 300 K, so the energy distribution is the same. In Q the molecules can move through three times the volume, so matter is more dispersed.
(b) R. The volume is the same, so matter dispersal is the same. R is hotter, so its kinetic-energy distribution is lower and wider: energy is more dispersed.
(c) You cannot decide. Q and R are each higher than P, but for different reasons (matter dispersal for Q, energy dispersal for R). The rules in this topic tell you the direction of each effect, not which one is bigger, so the comparison needs data. “Changed once” is not a valid reason for equal entropy.
| Point | What earns it |
|---|---|
| 1 | Q, with the reason: same temperature, larger volume, so molecules (matter) more dispersed |
| 1 | R, with the reason: same volume, higher temperature, so a wider spread of kinetic energies (energy more dispersed) |
| 1 | States that Q and R cannot be ranked from the information given, because the two changes involve different effects whose sizes are not known |
Do not award (a) or (b) for the letter alone, or for “more disorder” with no particle detail.
Question 6 (constructed response · core)
Ethane burns in oxygen:
2 C₂H₆(g) + 7 O₂(g) → 4 CO₂(g) + 6 H₂O(l)
(a) Predict the sign of ΔS for this reaction. Justify your prediction. (b) The same reaction is run at a high temperature, where the water is produced as H₂O(g). Predict how the entropy change differs from part (a), with a reason. (c) A student says: “Combustion releases a lot of energy, so ΔS must be negative.” Evaluate this claim.
Model answer and suggested Marlbridge rubric
(a) Negative. Moles of gas: 2 + 7 = 9 on the left; 4 on the right (the water is liquid). The products contain 5 fewer moles of gas, so matter is much less dispersed.
(b) Now the products contain 4 + 6 = 10 mol of gas, one more than the 9 mol of reactant gas. So ΔS is much more positive than in (a), and counting suggests it is small and positive. The difference comes from the 6 mol of water being gas instead of liquid, and vaporisation has a large positive ΔS.
For reference, standard molar entropies give about −620 J K⁻¹ mol⁻¹ for (a) and about +93 J K⁻¹ mol⁻¹ for (b), which agrees with both predictions.
(c) The claim is not valid. Releasing energy is about ΔH (enthalpy), not ΔS. The sign of ΔH does not fix the sign of ΔS. In (a) ΔS is negative because of the drop in moles of gas, not because the reaction is exothermic. Part (b) shows the same exothermic combustion with a positive ΔS.
| Point | What earns it |
|---|---|
| 1 | Negative, with correct moles of gas (9 → 4) or a clear statement that products have fewer moles of gas |
| 1 | (b): ΔS more positive (or less negative) than in (a), because 6 mol of water are now gas (10 mol vs 9 mol of gas) |
| 1 | (c): claim rejected, with the reason that ΔH and ΔS are independent; the sign of ΔS comes from dispersal, not from heat released |
Accept “small positive” or “close to zero but positive” in (b) if the reasoning compares 10 mol with 9 mol of gas.
Question 7 (explanation · stretch)
A bottle of sparkling water is sealed under pressure, so extra carbon dioxide is dissolved in it:
CO₂(g) → CO₂(aq)
(a) Predict the sign of ΔS for this dissolving process. Justify your prediction at the particle level. (b) When the bottle is opened, bubbles form and CO₂ escapes into the room. Explain, in terms of entropy, why this change happens readily, and identify one other factor that would also need to be considered to give a full explanation.
Model answer and suggested Marlbridge rubric
(a) Negative. As a gas, the CO₂ molecules move freely through a large volume. Once dissolved, they are confined to the small volume of the liquid and surrounded by water molecules, so they are much less free to move. Matter is less dispersed, so ΔS < 0.
(b) Escaping is the reverse change, CO₂(aq) → CO₂(g), so for the CO₂ ΔS > 0: the molecules become free to spread through the whole room, a large increase in matter dispersal. A full explanation must also consider the energy change (enthalpy) of the process, and how ΔH and ΔS combine at the temperature of the room; that is the idea of thermodynamic favourability in Topic 9.3. Accept also: the drop in pressure above the liquid when the bottle is opened.
| Point | What earns it |
|---|---|
| 1 | (a): negative, because gas molecules become confined in the liquid (less volume, less freedom to move) |
| 1 | (b): escaping gas gains entropy because the molecules spread through a much larger volume |
| 1 | (b): names a second factor, such as the enthalpy change or the pressure drop, as needed for a full explanation |
Do not accept “dissolving always increases entropy”. It often does for solids, but not for a gas dissolving in a liquid.
How did you do?
- Q1 or Q3 wrong: re-read the two kinds of dispersal and Figure 2 in the study guide.
- Q2, Q4 or Q6(a) wrong: practise counting moles of gas only, as in Worked example 1 of the study guide.
- Q5 or Q7 incomplete: every claim needs a particle-level reason: say what spreads out, matter or energy.
- Q6(c) wrong: keep ΔH and ΔS apart; they are separate quantities.
Then tick off the topic checklist.