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A-LevelChemistryChemical energetics9701/43 · May/June 2022 · Q116 Marks

Answer all the questions in the spaces provided. (d) (i) The equation for the formation of a gaseous sulfate ion is shown. S(s) + 2O2(g) + 2e¯ → SO42-(g) ΔΗ = ΔΗ of SO42-(g) Calculate the standard enthalpy change of formation, ΔΗ, of SO42-(g). It may be helpful to draw a labelled energy cycle. Use relevant data from Table 1.1 in your calculations. Table 1.1 energy change value/kJmol-1 lattice energy of barium sulfate, BaSO4(s) -2469 standard enthalpy change of formation of barium sulfate -1473 standard enthalpy change of atomisation of barium +180 first ionisation energy of barium +503 second ionisation energy of barium +965 standard enthalpy change of atomisation of sulfur +279 standard enthalpy change for S(g) → S²-(g) +440 standard enthalpy change for O(g) → O2-(g) +657 O=O bond energy +496 (e) The reaction of solid hydrated barium hydroxide, Ba(OH)2•8H2O, with ammonium salts is endothermic. Ba(OH)2 8H2O(s) + 2NH4NO3(s) → 2NH3(g) + Ba(NO3)2(s) + 10H2O(1) ∆H = +132kJmol-1 AS = +616 JK-1 mol-1 (ii) Barium hydroxide reacts readily with ammonium chloride on mixing at room temperature. Ba(OH)2 8H2O(s) + 2NH4Cl(s) → 2NH3(g) + BaCl2•2H2O(s) + 8H2O(I) ∆H = +133kJ mol-1 Some relevant standard entropies are given in Table 1.2. Table 1.2 substance S/JK-1 mol-1 Ba(OH)2 8H2O(s) NH4Cl(s) 427 95 NH3(g) 192 BaCl2 2H2O(s) 203 H₂O(I) 70

  1. (a)The solubility of the Group 2 sulfates decreases down the group. Explain this trend.[3]
  2. (b)Describe what is observed when magnesium and barium are reacted separately with an excess of dilute sulfuric acid. magnesium barium[1]
  3. (c)The solubility product, Ksp, of BaSO4 is 1.08 × 10–10 mol² dm-6 at 298 K. Calculate the solubility of BaSO4 in g per 100 cm³ of solution. solubility of BaSO4 = ... g per 100 cm³ of solution[2]
  4. (d)(i)The equation for the formation of a gaseous sulfate ion is shown. S(s) + 2O2(g) + 2e¯ → SO42-(g) ΔΗ = ΔΗ of SO42-(g) Calculate the standard enthalpy change of formation, ΔΗ, of SO42-(g). It may be helpful to draw a labelled energy cycle. Use relevant data from Table 1.1 in your calculations. ∆H of SO42-(g) = ... kJ mol-1[3]
  5. (d)(ii)Suggest how the lattice energy of BaSO4(s) differs from the lattice energy of Cs₂SO₄(s). Explain your answer.[2]
  6. (e)(i)Calculate the minimum temperature at which the reaction of Ba(OH)2•8H2O with NH4NO3 becomes feasible. Show all your working. Ba(OH)2 8H2O(s) + 2NH4NO3(s) → 2NH3(g) + Ba(NO3)2(s) + 10H2O(1) ∆H = +132kJmol-1 AS = +616 JK-1 mol-1 temperature = ... °C[2]
  7. (e)(ii)Barium hydroxide reacts readily with ammonium chloride on mixing at room temperature. Ba(OH)2 8H2O(s) + 2NH4Cl(s) → 2NH3(g) + BaCl2•2H2O(s) + 8H2O(I) ∆H = +133kJ mol-1 Some relevant standard entropies are given in Table 1.2. Calculate the standard Gibbs free energy change, AG°, for this reaction at 25°C. AG = ... kJ mol-1[3]

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Mark scheme

Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.

  1. [3] M1 ∆Hlatt and ∆Hhyd decrease / both become less exothermic/less negative
  2. M2 Hlatt decreases / changes less/becomes less exothermic by a smaller extent OR AHhyd decreases / changes more/dominant factor
  3. M3 AHSol becomes less exothermic / less negative OR AHSol becomes (more) endothermic / (more) positive OR AHsol = ∆Ηhyd – ∆Hlatt expression AND reaction becomes less exothermic
  4. [1] Mg: fizzing
  5. Ba: (fizzing and) white solid/ppt forms
  6. [2] M1 solubility of BaSO4 = √1.08 × 10–10 = 1.04 × 10–5 (mol dm-³)
  7. M2 = 1.04 × 10-5 × 233.4 / 10 = 2.43 × 10–4 (g per 100 cm³ of solution) min 2sf
  8. [3] -1473 = 180 +503 +965 + ∆H°-2469
  9. ∆H of SO42-(g) = −652 kJ mol-1
  10. M1 correct five values used [1]
  11. M2 only correct five values used [1]
  12. M3 correct signs and evaluation [1]
  13. [2] • BaSO4 is more negative/bigger
  14. • as Ba2+ is smaller OR Ba2+ has a larger charge
  15. • stronger force of attraction between the ions
  16. One mark for two correct
  17. Two marks for all three correct
  18. [2] M1 AG = 0 so T = ∆H/S
  19. M2 T = 132/0.616 = 214.3 K
  20. T = -58.7 °C min 2sf
  21. [3] M1 ∆S® = (203 + (70 × 8) + (2 × 192)) – (427 – (2 × 95)) = +530 J K–1 mol–1
  22. M2 AG = ∆H – TAS
  23. M3 ∆G° = 133 – (298 × 0.530) = −24.9 kJ mol-¹ ecf 1dp min

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About this question

Topic
Chemical energetics, A-Level Chemistry (syllabus 9701). Structured question, 16 marks.
Source
A-Level Chemistry, May/June 2022 Paper 4 Variant 3, question 1.

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