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. Data: ΔG°f at 298 K: CaCO₃(s) −1129.1, CaO(s) −603.3, CO₂(g) −394.39 kJ mol⁻¹. T(K) = T(°C) + 273.15. Reactions X, Z and the dimerisation in Question 7 use invented data, made up for practice.
Question 1 (multiple choice · foundation)
A reaction is endothermic, and the number of moles of gas decreases from reactants to products. Under standard conditions, at which temperatures is the reaction thermodynamically favoured?
- (A) At all temperatures
- (B) Only at high temperatures
- (C) Only at low temperatures
- (D) At no temperature
Answer and explanation
Answer: (D). Endothermic means ΔH° > 0. Fewer moles of gas means matter is less dispersed, so ΔS° < 0 and −TΔS° > 0. Both terms in ΔG° = ΔH° − TΔS° are positive, so ΔG° > 0 at every temperature.
- (A) is the case ΔH° < 0, ΔS° > 0.
- (B) is the case where both are positive; here ΔS° is negative.
- (C) is the case where both are negative; here ΔH° is positive.
Question 2 (multiple choice · core)
For a reaction, ΔH° = −120 kJ mol⁻¹ and ΔS° = −150 J K⁻¹ mol⁻¹. What is ΔG° at 400 K?
- (A) −180 kJ mol⁻¹
- (B) −60.0 kJ mol⁻¹
- (C) +60.0 kJ mol⁻¹
- (D) +59 880 kJ mol⁻¹
Answer and explanation
Answer: (B). Convert ΔS° to −0.150 kJ K⁻¹ mol⁻¹. Then ΔG° = −120 − (400)(−0.150) = −120 + 60.0 = −60.0 kJ mol⁻¹. The reaction is favoured at 400 K, although the entropy term has cancelled half of ΔH°.
- (A) adds TΔS° instead of subtracting it: −120 + (400)(−0.150).
- (C) calculates TΔS° − ΔH°, the right numbers in the wrong order.
- (D) leaves ΔS° in J: −120 − (400)(−150). Mixing kJ and J gives a huge, meaningless number.
Question 3 (multiple choice · core)
Quicklime absorbs carbon dioxide: CaO(s) + CO₂(g) → CaCO₃(s). Using the data at the top of the page, what is ΔG° at 298 K?
- (A) +131.4 kJ mol⁻¹
- (B) −131.4 kJ mol⁻¹
- (C) −1129.1 kJ mol⁻¹
- (D) −2126.8 kJ mol⁻¹
Answer and explanation
Answer: (B). ΔG° = ΔG°f(CaCO₃) − [ΔG°f(CaO) + ΔG°f(CO₂)] = −1129.1 − (−603.3 − 394.39) = −1129.1 + 997.69 = −131.4 kJ mol⁻¹. ΔG° < 0, so the reaction is favoured at 298 K.
- (A) is reactants minus products, which reverses the sign.
- (C) uses only the product value and ignores the reactants.
- (D) adds all three values instead of subtracting the reactants.
Question 4 (multiple choice · core)
Solid sodium nitrate, NaNO₃, dissolves readily in water at 25 °C, and the temperature of the solution falls. Which statement best explains why the dissolving is thermodynamically favoured?
- (A) The process is exothermic, so ΔH° < 0 makes ΔG° negative.
- (B) The process is endothermic, but ΔS° > 0 as ions spread through the solution, and TΔS° is larger than ΔH°.
- (C) The process absorbs energy from the surroundings, and any process that absorbs energy is favoured.
- (D) ΔG° is positive, but the solid dissolves anyway because dissolving is fast.
Answer and explanation
Answer: (B). The falling temperature shows the process takes in energy, so ΔH° > 0. Ions leave fixed lattice positions and spread out among the water molecules, so ΔS° > 0. Since the solid does dissolve, ΔG° < 0, which means TΔS° > ΔH° at this temperature.
- (A) misreads the temperature drop: an exothermic process would warm the solution.
- (C) is false in general: many endothermic processes (such as melting ice at −10 °C) are not favoured.
- (D) confuses speed with favourability. Rate decides how quickly a process happens, not which direction is favoured; a fast rate cannot make an unfavoured process go.
Question 5 (calculation · core)
Reaction X has ΔH° = +64.8 kJ mol⁻¹ and ΔS° = +172 J K⁻¹ mol⁻¹. Assume both are constant with temperature.
(a) Calculate ΔG° at 298 K and state whether X is favoured. (b) Calculate the temperature above which X is favoured. (c) Calculate ΔG° at 450 K.
Worked solution
(a) ΔS° = 0.172 kJ K⁻¹ mol⁻¹. TΔS° = 298 × 0.172 = 51.26 kJ mol⁻¹. ΔG° = 64.8 − 51.26 = +13.5 kJ mol⁻¹. Not favoured at 298 K.
(b) ΔG° = 0 when T = ΔH° ÷ ΔS° = 64 800 ÷ 172 = 377 K (about 104 °C). Above this, −TΔS° outweighs ΔH°.
(c) ΔG° = 64.8 − 450 × 0.172 = 64.8 − 77.4 = −12.6 kJ mol⁻¹. Favoured, as (b) predicts.
Suggested mark points (4): 1 for converting ΔS° to kJ (or ΔH° to J); 1 for +13.5 kJ mol⁻¹ with “not favoured”; 1 for 377 K; 1 for −12.6 kJ mol⁻¹. Accept 376.7 K.
Question 6 (constructed response · core)
Salt Z dissolves in water: Z(s) → Z(aq). For this process, ΔH° = +26.0 kJ mol⁻¹ and ΔS° = +105 J K⁻¹ mol⁻¹.
(a) Explain, in terms of particles, why ΔS° is positive. (b) Calculate ΔG° at 298 K. Is the dissolving favoured? (c) A student cools the water to 285 K. Calculate ΔG° and state whether the conclusion in (b) changes. (d) Identify whether enthalpy or entropy drives the dissolving, and justify your answer.
Worked solution and suggested Marlbridge rubric
(a) In the solid the ions are held in fixed positions in a lattice. In solution they move freely and are spread through a much larger volume, so matter is more dispersed and ΔS° > 0.
(b) ΔG° = 26.0 − 298(0.105) = 26.0 − 31.29 = −5.3 kJ mol⁻¹. ΔG° < 0, so dissolving is favoured.
(c) ΔG° = 26.0 − 285(0.105) = 26.0 − 29.93 = −3.9 kJ mol⁻¹. Still negative, so still favoured, but less strongly, because the entropy term is smaller at lower T.
(d) Entropy. ΔH° > 0 works against dissolving; only the −TΔS° term is negative, and it is larger in size than ΔH°.
| Point | What earns it |
|---|---|
| 1 | Ions go from fixed lattice positions to moving freely through the solution (more dispersed) |
| 1 | −5.3 kJ mol⁻¹ with ΔS° converted to kJ, and “favoured” |
| 1 | −3.9 kJ mol⁻¹ and “still favoured” |
| 1 | Entropy, because ΔH° is positive (unfavourable) and −TΔS° outweighs it |
Accept answers to (c) carried forward from a unit error in (b) only if the same method is used consistently.
Question 7 (constructed response · stretch)
A gas A dimerises: 2A(g) → A₂(g). For this reaction, ΔH° = −58.0 kJ mol⁻¹ and ΔS° = −176 J K⁻¹ mol⁻¹.
(a) Justify the sign of ΔS° at the particle level. (b) Determine the range of temperatures over which the dimerisation is thermodynamically favoured. (c) A student claims: “Because ΔH° is negative, the dimerisation is favoured at every temperature.” Evaluate the claim, using a calculation at 400 K.
Worked solution and suggested Marlbridge rubric
(a) Two moles of gas particles become one mole. Fewer independent particles moving through the container means matter is less dispersed, so ΔS° < 0.
(b) Both ΔH° and ΔS° are negative, so the reaction is favoured only at low T. The changeover is T = ΔH° ÷ ΔS° = (−58 000) ÷ (−176) = 329.5 K. The dimerisation is favoured below about 330 K (about 56 °C).
(c) At 400 K: ΔG° = −58.0 − 400(−0.176) = −58.0 + 70.4 = +12.4 kJ mol⁻¹. ΔG° > 0, so the dimerisation is not favoured at 400 K, and the claim is wrong. A negative ΔH° favours the reaction, but the −TΔS° term is positive and grows with temperature; above 330 K it is larger than the size of ΔH°.
| Point | What earns it |
|---|---|
| 1 | Fewer moles of gas, so less dispersal of matter |
| 1 | Favoured only below about 330 K, with T = ΔH° ÷ ΔS° shown |
| 1 | +12.4 kJ mol⁻¹ at 400 K |
| 1 | Claim rejected, with the reason that −TΔS° is positive and outweighs ΔH° at high T |
Do not award the last point for “the claim is wrong” without a reason linked to the entropy term.
How did you do?
- Q1 or Q7(b) wrong: re-learn the four sign combinations (Figure 1 in the study guide).
- Q2, Q5 or Q6 off by a factor of 1000: convert ΔS° to kJ K⁻¹ mol⁻¹ before you multiply by T.
- Q3 wrong: use products minus reactants, as in Worked example 1.
- Q4, Q6(d) or Q7(c) incomplete: your argument needs both terms, ΔH° and −TΔS°, and which one is larger.
Then tick off the topic checklist.