Skip to main content
A-LevelChemistryChemical energetics9701/42 · Feb/Mar 2017 · Q210 Marks

Complete the table using ticks (√) to indicate whether the sign of each type of energy change, under standard conditions, is always positive, always negative or could be either positive or negative. [Table with columns: 'energy change', 'always positive', 'always negative', 'either positive or negative' and rows: 'electron affinity', 'enthalpy change of atomisation', 'ionisation energy', 'lattice energy']

  1. (a)Complete the table using ticks (√) to indicate whether the sign of each type of energy change, under standard conditions, is always positive, always negative or could be either positive or negative.[2]
  2. (b)(i)Explain why ΔH4 is greater than ΔH3.[1]
  3. (b)(ii)What names are given to the enthalpy changes ∆H6 and ΔΗ7? ΔH6 ΔH7[1]
  4. (c)Chlorine is in Group 17. Suggest the trend in the first electron affinity of the elements in Group 17. Explain your answer.[2]
  5. (d)The equation for the formation of magnesium oxide from its elements is shown. Mg(s) + ½O₂(g) → MgO(s) ΔH° = −602 kJ mol⁻¹ substance S°/J K⁻¹ mol⁻¹ Mg(s) 32.7 O₂(g) 205 MgO(s) 26.9 Use the equation and the data given in the table to calculate ΔG° for the reaction at 25°C. ΔG = units[4]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2017). The mark or mark type is shown in brackets.

  1. [2] electron affinity: always negative; enthalpy change of atomisation: always positive; ionisation energy: always positive; lattice enthalpy: always negative
  2. [1] the second electron is removed from a (more) positively charged ion
  3. [1] ΔΗ6 is lattice (energy/enthalpy) AND ∆H7 is (energy/enthalpy of) formation
  4. [1] the electron affinity becomes less exothermic/negative down the Group 17
  5. [1] electron affinity depends (mainly) on the electron-nucleus distance which increases down Group 17
  6. [1] M1 correct use of ΔG = ΔH – TΔS
  7. [1] M2 ΔS = 26.9 – (32.7 + 102.5) = −108.3 JK⁻¹ mol⁻¹ OR –0.1083 kJ K⁻¹ mol⁻¹
  8. [1] M3 ΔG = −602 – (298 × (−0.1083)) = −570
  9. [1] M4 units: kJ mol⁻¹

Write your answer and get it marked

On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.

Sign up free →

About this question

Topic
Chemical energetics, A-Level Chemistry (syllabus 9701). Structured question, 10 marks.
Source
A-Level Chemistry, Feb/Mar 2017 Paper 4 Variant 2, question 2.

Questions and mark schemes are from past examination papers and remain the copyright of the exam board. How Oracle Prep builds its content

Related Chemistry questions

© 2026 Oracle Prep