A-LevelChemistryChemical periodicity9701/41 · May/June 2018 · Q115 Marks

Sodium oxide, Na2O, is a white crystalline solid with a high melting point.

(a)Write an equation for the reaction of sodium with oxygen, forming sodium oxide. Include state symbols.[2]
(b)Explain why sodium oxide has a high melting point.[2]
(c)(i)When sodium oxide reacts with water an alkaline solution is obtained. Explain why the solution obtained is alkaline. You should use the Brønsted-Lowry theory of acids and bases in your answer.[2]
(c)(ii)Calculate the pH of the solution obtained when 3.10g of sodium oxide are added to 400 cm³ of water.[3]
(d)Use the data below, and other suitable data from the Data Booklet, to calculate the lattice energy of sodium oxide, ΔHlatt Na2O(s). energy change | value/kJ mol-1 --- | --- standard enthalpy change of formation of sodium oxide, ΔH°f Na2O(s) | -416 standard enthalpy change of atomisation of sodium, ΔH°at Na(s) | +109 electron affinity of O(g) | -142 electron affinity of O-(g) | +844 ΔHlatt Na2O(s) =[4]
(e)State how ΔHlatt Na2S(s) differs from ΔHlatt Na2O(s). Indicate this by placing a tick (√) in the appropriate box in the table. ΔHlatt Na2S(s) is more exothermic than ΔHlatt Na2O(s) ΔHlatt Na2S(s) is the same as ΔHlatt Na2O(s) ΔHlatt Na2S(s) is less exothermic than ΔHlatt Na2O(s) Explain your answer.[2]

Mark scheme

Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.

[1] 4Na(s) + O2(g) → 2Na2O(s)
[1] state symbols
[1] giant ionic
[1] strong bond / attraction between AND positive and negative ions / anions and cations / Na+ and O2– / oppositely charged ions
[1] the reaction produces sodium hydroxide / hydroxide ions / OH− ions
[1] the hydroxide ions can receive / accept H+ ions / protons
[1] Calculation of Na2O moles 3.10 g / 62 OR 0.05
[1] Calculation of [OH−] 0.05 × (2 / 0.400) = 0.25 mol dm−3
[1] Calculation of pH –log 0.25 = 0.60 14 – 0.60 = 13.40
[1] use of (2 x 109) or 218 and (2 × 494) or 988
[1] use of (0.5 x 496) or 248
[1] use of 416, 142, 844
[1] evaluation of expression correctly ΔHlatt = –416 – (2 × 109) – (0.5 × 496) – (2 × 494) – (–142 + 844) = –2572
[1] the lattice energy of Na2S is less exothermic
[1] the sulfide ion is larger than the oxide ion / S2– larger than O2– / ionic radii quoted 0.184 nm and 0.140 nm AND less attraction (between the ions) / bonds are weaker

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