A-LevelChemistryGroup 29701/42 · May/June 2017 · Q110 Marks

Describe and explain the variation in the thermal stabilities of the carbonates of the Group 2 elements.

(a)(i)Describe and explain the variation in the thermal stabilities of the carbonates of the Group 2 elements.[3]
(a)(ii)Suggest and explain a reason why sodium carbonate is more stable to heat than magnesium carbonate.[1]
(b)(i)Write an equation for the decomposition of KHCO3.[1]
(b)(ii)Predict which of NaHCO3 or KHCO3 will decompose at the lower temperature. Explain your answer.[1]
(c)(i)Use the data in the table below, and relevant data from the Data Booklet, to calculate the lattice energy, ∆Hatt, of potassium oxide, K2O(s). energy change value/kJ mol-1 enthalpy change of atomisation of potassium, ΔΗ K(s) +89 electron affinity of O(g) -141 electron affinity of O-(g) +798 enthalpy change of formation of potassium oxide, ∆H K₂O(s) -361 ΔΗ = ... kJ mol-1[3]
(c)(ii)State whether the lattice energy of Na₂O would be more negative, less negative or the same as that of K₂O. Give reasons for your answer.[1]

Mark scheme

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

[1] increases down the group
[1] radius / size of (cat)ion/M²+ increases
[1] less polarisation / distortion of anion / carbonate ion / CO32-
[1] Na⁺ has smaller ionic charge and larger ionic radii OR the charge density of the Na⁺ is lower
[1] 2KHCO3→ K2CO3 + CO2 + H2O
[1] NaHCO3 because Na⁺ is smaller OR charge density Na⁺ is larger
[3] LE = ∆Hf − 2(∆Hat + IE) – ½(O=O) – (EA₁ + EA₂) = −361 – 2(89) – 2(418) – 496/2 – (−141+798) = -2280 (kJ mol⁻¹) correct answer scores [3]
[1] LE of Na2O will be more negative AND as Na(+) is smaller / larger charge density / smaller radii AND so greater attraction (between the ions) OR (ionic) bonds will be stronger

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