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']
- (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]
- (b)(i)Explain why ΔH4 is greater than ΔH3.[1]
- (b)(ii)What names are given to the enthalpy changes ∆H6 and ΔΗ7? ΔH6 ΔH7[1]
- (c)Chlorine is in Group 17. Suggest the trend in the first electron affinity of the elements in Group 17. Explain your answer.[2]
- (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]
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Mark scheme
Marking points from the official mark scheme (Feb/Mar 2017). The mark or mark type is shown in brackets.
- [2] electron affinity: always negative; enthalpy change of atomisation: always positive; ionisation energy: always positive; lattice enthalpy: always negative
- [1] the second electron is removed from a (more) positively charged ion
- [1] ΔΗ6 is lattice (energy/enthalpy) AND ∆H7 is (energy/enthalpy of) formation
- [1] the electron affinity becomes less exothermic/negative down the Group 17
- [1] electron affinity depends (mainly) on the electron-nucleus distance which increases down Group 17
- [1] M1 correct use of ΔG = ΔH – TΔS
- [1] M2 ΔS = 26.9 – (32.7 + 102.5) = −108.3 JK⁻¹ mol⁻¹ OR –0.1083 kJ K⁻¹ mol⁻¹
- [1] M3 ΔG = −602 – (298 × (−0.1083)) = −570
- [1] M4 units: kJ mol⁻¹
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