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 for every question: standard molar entropies at 298 K, in J K⁻¹ mol⁻¹.
| Substance | S° | Substance | S° |
|---|---|---|---|
| C(graphite) | 5.7 | C₆H₆(g) | 269.2 |
| CO(g) | 197.7 | H₂(g) | 130.7 |
| CO₂(g) | 213.8 | O₂(g) | 205.2 |
| CH₄(g) | 186.3 | H₂O(l) | 70.0 |
| C₂H₆(g) | 229.2 | H₂O(g) | 188.8 |
| C₂H₅OH(l) | 160.7 | HCl(g) | 186.9 |
| C₆H₆(l) | 173.4 | Cl₂(g) | 223.1 |
Molar masses: H 1.008, C 12.01, O 16.00 g mol⁻¹.
Question 1 (multiple choice · foundation)
What is ΔS° for the reaction C(graphite) + CO₂(g) → 2 CO(g)?
- (A) −21.8 J K⁻¹ mol⁻¹
- (B) −175.9 J K⁻¹ mol⁻¹
- (C) +175.9 J K⁻¹ mol⁻¹
- (D) +181.6 J K⁻¹ mol⁻¹
Answer and explanation
Answer: (C). ΔS° = 2 S°(CO) − [S°(C) + S°(CO₂)] = 2(197.7) − (5.7 + 213.8) = 395.4 − 219.5 = +175.9 J K⁻¹ mol⁻¹. Check: moles of gas go from 1 to 2, so a positive value is expected.
- (A) ignores the coefficient 2 on CO: 197.7 − 219.5.
- (B) subtracts the wrong way round (reactants minus products).
- (D) treats graphite as having S° = 0, as if it were an enthalpy of formation: 395.4 − 213.8. Elements have positive S° values.
Question 2 (multiple choice · core)
What is ΔS° for the reaction H₂(g) + Cl₂(g) → 2 HCl(g)?
- (A) +727.6 J K⁻¹ mol⁻¹
- (B) +20.0 J K⁻¹ mol⁻¹
- (C) −20.0 J K⁻¹ mol⁻¹
- (D) −166.9 J K⁻¹ mol⁻¹
Answer and explanation
Answer: (B). ΔS° = 2(186.9) − (130.7 + 223.1) = 373.8 − 353.8 = +20.0 J K⁻¹ mol⁻¹.
Moles of gas are equal (2 on each side), so Topic 9.1 predicts a small value whose sign cannot be found by counting. The calculation settles it: small and positive.
- (A) adds every value instead of subtracting reactants from products.
- (C) subtracts in the wrong order.
- (D) forgets the coefficient 2 on HCl: 186.9 − 353.8.
Question 3 (multiple choice · core)
For the combustion of methane, CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l), which value is ΔS° expressed in kJ K⁻¹ mol⁻¹?
- (A) −0.2429 kJ K⁻¹ mol⁻¹
- (B) −242.9 kJ K⁻¹ mol⁻¹
- (C) −0.0053 kJ K⁻¹ mol⁻¹
- (D) +0.2429 kJ K⁻¹ mol⁻¹
Answer and explanation
Answer: (A). Products: 213.8 + 2(70.0) = 353.8. Reactants: 186.3 + 2(205.2) = 596.7. ΔS° = 353.8 − 596.7 = −242.9 J K⁻¹ mol⁻¹. Dividing by 1000 gives −0.2429 kJ K⁻¹ mol⁻¹. Check: 3 mol of gas → 1 mol of gas, so negative.
- (B) has the right number in joules but the wrong unit. It is 1000 times too large in kJ.
- (C) uses H₂O(g) (188.8) instead of H₂O(l). That gives −5.3 J K⁻¹ mol⁻¹, but the equation shows liquid water.
- (D) has the sign reversed (reactants minus products).
Question 4 (calculation · core)
Ethanol burns completely in oxygen:
C₂H₅OH(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(l)
(a) Predict the sign of ΔS° without calculating. Give a reason. (b) Calculate ΔS° for the reaction. (c) Calculate the entropy change when 9.21 g of ethanol burns completely.
Worked solution and suggested Marlbridge rubric
(a) Negative. Moles of gas: 3 (all O₂) on the left; 2 (CO₂) on the right. Ethanol and water are liquids, so they are not counted.
(b) Products: 2(213.8) + 3(70.0) = 427.6 + 210.0 = 637.6 J K⁻¹ mol⁻¹. Reactants: 160.7 + 3(205.2) = 160.7 + 615.6 = 776.3 J K⁻¹ mol⁻¹. ΔS° = 637.6 − 776.3 = −138.7 J K⁻¹ mol⁻¹. This agrees with (a).
(c) M(C₂H₅OH) = 2(12.01) + 6(1.008) + 16.00 = 46.068 g mol⁻¹. n = 9.21 g ÷ 46.068 g mol⁻¹ = 0.19992 mol, which is 0.19992 mol of reaction (1 mol of ethanol per mole of reaction). ΔS = 0.19992 mol × (−138.7 J K⁻¹ mol⁻¹) = −27.7 J K⁻¹.
| Point | What earns it |
|---|---|
| 1 | (a) Negative, with a correct count of gas moles (3 → 2) |
| 1 | (b) Correct products and reactants sums, including 3 × S°(O₂) |
| 1 | (b) −138.7 J K⁻¹ mol⁻¹ with unit |
| 1 | (c) −27.7 J K⁻¹ (moles of ethanol × ΔS°); accept a value carried forward from (b) |
Common error: using H₂O(g) gives +217.7 J K⁻¹ mol⁻¹, which contradicts the prediction in (a). That disagreement is your signal to check the state symbols.
Question 5 (calculation · core)
For the reaction C₂H₄(g) + H₂(g) → C₂H₆(g), ΔS° = −120.8 J K⁻¹ mol⁻¹. Use this value and the data table to calculate the standard molar entropy of ethene, C₂H₄(g).
Worked solution and suggested Marlbridge rubric
- ΔS° = S°(C₂H₆) − [S°(C₂H₄) + S°(H₂)].
- −120.8 = 229.2 − S°(C₂H₄) − 130.7.
- S°(C₂H₄) = 229.2 − 130.7 + 120.8 = 219.3 J K⁻¹ mol⁻¹.
Check. A gas value between those of H₂ (130.7) and C₂H₆ (229.2) is sensible: ethene has two fewer atoms than ethane.
| Point | What earns it |
|---|---|
| 1 | Correct expression with C₂H₄ and H₂ both on the reactant side |
| 1 | 219.3 J K⁻¹ mol⁻¹ with unit |
Common error: substituting the entropy change as +120.8 gives S°(C₂H₄) = −22.3 J K⁻¹ mol⁻¹. A negative absolute entropy is impossible, so this answer should never be accepted.
Question 6 (constructed response · core)
Carbon monoxide burns in oxygen:
2 CO(g) + O₂(g) → 2 CO₂(g)
(a) Predict the sign of ΔS° and justify it. (b) Calculate ΔS° for the reaction. (c) Calculate the entropy change when 5.60 g of CO burns completely. (d) A student reports ΔS° = +32.2 J K⁻¹ mol⁻¹ for this reaction. Identify the error and explain how a sign check would have caught it.
Worked solution and suggested Marlbridge rubric
(a) Negative. Moles of gas: 2 + 1 = 3 on the left; 2 on the right. Fewer moles of gas means matter is less dispersed.
(b) Products: 2(213.8) = 427.6 J K⁻¹ mol⁻¹. Reactants: 2(197.7) + 205.2 = 600.6 J K⁻¹ mol⁻¹. ΔS° = 427.6 − 600.6 = −173.0 J K⁻¹ mol⁻¹.
(c) M(CO) = 12.01 + 16.00 = 28.01 g mol⁻¹; n(CO) = 5.60 ÷ 28.01 = 0.19993 mol. The equation uses 2 mol of CO per mole of reaction, so moles of reaction = 0.19993 ÷ 2 = 0.099964 mol. ΔS = 0.099964 mol × (−173.0 J K⁻¹ mol⁻¹) = −17.3 J K⁻¹.
(d) The student left out O₂, as if its S° were zero: 427.6 − 395.4 = +32.2. O₂ is an element, but elements have positive S° values. The prediction in (a) was negative (3 mol of gas → 2 mol), so a positive answer should have prompted a recheck.
| Point | What earns it |
|---|---|
| 1 | (a) Negative, with gas moles 3 → 2 |
| 1 | (b) −173.0 J K⁻¹ mol⁻¹, including S°(O₂) and both coefficients |
| 1 | (c) Moles of reaction found by halving moles of CO |
| 1 | (c) −17.3 J K⁻¹ (carry forward from (b) accepted) |
| 1 | (d) Identifies that O₂ was omitted or set to zero and states that the positive sign contradicts the moles-of-gas prediction |
Common error in (c): multiplying moles of CO directly by ΔS° gives −34.6 J K⁻¹, twice the correct value.
Question 7 (explanation · stretch)
Benzene evaporates easily at room temperature: C₆H₆(l) → C₆H₆(g).
(a) Calculate ΔS° for this process. (b) Explain the sign and size of your answer at the particle level. (c) A classmate says: “Hydrogen gas is an element in its standard state, so its S° must be zero, just like its enthalpy of formation.” Explain why this is wrong.
Model answer and suggested Marlbridge rubric
(a) ΔS° = 269.2 − 173.4 = +95.8 J K⁻¹ mol⁻¹.
(b) In the liquid, benzene molecules are held close together by intermolecular forces and can only move past one another. In the gas, they are far apart and move freely through a much larger volume. Matter becomes much more dispersed, so ΔS° is positive and large, as expected for liquid → gas.
(c) Enthalpies of formation are measured relative to the elements, so an element in its standard state is zero by definition. Entropy is measured from a true zero: a perfect crystal at 0 K. Hydrogen gas at 298 K has far more matter and energy dispersal than a perfect crystal at 0 K, so its absolute entropy is positive (130.7 J K⁻¹ mol⁻¹ in the table).
| Point | What earns it |
|---|---|
| 1 | (a) +95.8 J K⁻¹ mol⁻¹ |
| 1 | (b) Particle-level reason: molecules separate and spread through a larger volume (matter more dispersed) |
| 1 | (c) ΔH°f values are relative to elements (zero by definition), but S° is absolute, measured from zero for a perfect crystal at 0 K |
| 1 | (c) States that every substance at 298 K, element or compound, has positive S° |
Do not award (b) for “the gas is more disordered” without describing what the molecules are doing.
How did you do?
- Q1, Q2 or Q6(d) wrong: re-read the misconceptions about coefficients and elements in the study guide.
- Q3 wrong: practise the J-to-kJ conversion and check state symbols, as in Worked example 1.
- Q4(c) or Q6(c) wrong: remember that ΔS° is per mole of reaction; find moles of reaction first.
- Q5 wrong: follow Worked example 3, and reject any negative S°.
- Q7 incomplete: link the number to what the particles are doing.
Then tick off the topic checklist.