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A-LevelChemistryChemical energetics9701/42 · May/June 2025 · Q38 Marks

Define entropy.

  1. (a)Define entropy.[1]
  2. (b)(i)Fig. 3.1 shows how the entropy, S, of a pure substance changes with temperature, T. [Figure 3.1] Identify the process occurring at each of the temperatures T₁ and T2. T₁ ............................................ T₂ ............................................[1]
  3. (b)(ii)Explain why the entropy change, ΔS, at T2 is bigger than the entropy change at T₁.[1]
  4. (c)The equation for the reduction of iron(III) oxide by carbon monoxide at 450 °C is shown. Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) ΔGº = −36.2 kJ mol⁻¹ Table 3.1 shows the enthalpy of formation, ΔHᶠ, and the entropy, Sº, for some substances. [Table 3.1] Use the data in Table 3.1 to calculate the entropy, Sº, of carbon monoxide at 450 °C. Show your working. Sº of CO(g) = ................................. JK⁻¹ mol⁻¹ [3][3]
  5. (d)Iron(II) oxide can also be reduced to iron by carbon monoxide, as shown. FeO(s) + CO(g) → Fe(s) + CO2(g) ΔH° = −11.1 kJ mol⁻¹ ΔS° = −15.2 J K⁻¹ mol⁻¹ State the effect of increasing temperature on the feasibility of this reaction. Explain your answer.[2]

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Mark scheme

Marking points from the official mark scheme (May/June 2025). The mark or mark type is shown in brackets.

  1. [1] number of possible arrangements of particles/molecules AND energy in a system
  2. [1] T₁ melting OR solid to liquid
  3. [1] T2 boiling OR liquid to gas
  4. [1] change in disorder from a liquid to a gas is much bigger than solid to liquid OR change in intermolecular distance between liquid & gas is much bigger than solid to liquid
  5. [1] M1: ΔHᶠ = (3 × −393.5) − ((−824.2) + (3 × −110.5)) = −24.8 (kJ mol-1)
  6. [1] M2: ΔG = ΔH – TΔS AND T= 723 used
  7. [1] M3: (24800 – 36200) ÷ 723 = − [3 × 213.8 + 2 × 27.3) − (87.4) – (3x)] -15.8 = -608.6 + 3x, so 3x = 592.8, so x = 197.6/ 198 (J K-1 mol-1) min 3sf
  8. [1] • (as temperature increases) the reaction is less feasible
  9. [1] • as ΔG becomes less negative / (more) positive
  10. [1] • due to TΔS becoming more negative / -TΔS (becoming more) positive
  11. M1 / M2: any two of the above bullets [1] all three [2]

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About this question

Topic
Chemical energetics, A-Level Chemistry (syllabus 9701). Structured question, 8 marks.
Source
A-Level Chemistry, May/June 2025 Paper 4 Variant 2, question 3.

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