A-LevelChemistryGroup 29701/41 · May/June 2017 · Q113 Marks

The table lists the standard enthalpy changes of formation, ΔΗ, for some compounds and aqueous ions. ``` species ΔH_f/kJ mol⁻¹ Ba²⁺(aq) -538 OH⁻(aq) -230 CO₂(g) -394 BaCO₃(s) -1216 H₂O(l) -286 ```

(a)Describe and explain the variation in the solubilities of the hydroxides of the Group 2 elements.[4]
(b)(i)Reaction 1 occurs when CO₂(g) is bubbled through an aqueous solution of Ba(OH)₂. Use the data in the table to calculate the standard enthalpy change for reaction 1, ΔΗᵣ₁. Ba(OH)₂(aq) + CO₂(g) → BaCO₃(s) + H₂O(l) reaction 1 ΔΗᵣ₁ = ______________ kJ mol⁻¹[2]
(b)(ii)If CO₂(g) is bubbled through an aqueous solution of Ba(OH)₂ for a long time, the precipitated BaCO₃(s) dissolves, as shown in reaction 2. BaCO₃(s) + CO₂(g) + H₂O(l) → Ba(HCO₃)₂(aq) reaction 2 The standard enthalpy change for reaction 2, ΔΗᵣ₂, = −26 kJ mol⁻¹. Use this information and the data in the table to calculate the standard enthalpy change of formation of the HCO₃⁻(aq) ion. ΔH_fHCO₃⁻(aq) = ______________ kJ mol⁻¹[2]
(b)(iii)The overall process is shown by reaction 3. Use your answer to (ii), and the data given in the table, to calculate the standard enthalpy change for reaction 3, ΔΗᵣ₃. Ba(OH)₂(aq) + 2CO₂(g) → Ba(HCO₃)₂(aq) reaction 3 ΔΗᵣ₃ = ______________ kJ mol⁻¹[1]
(b)(iv)How would the value of ∆Hᵣ₃ compare with the value of ∆Hᵣ₄ for the similar reaction with Ca(OH)₂(aq) as shown in reaction 4? Explain your answer. Ca(OH)₂(aq) + 2CO₂(g) → Ca(HCO₃)₂(aq) reaction 4[2]
(c)The standard entropy change for reaction 1 is ∆Sᵣ₁. Suggest, with a reason, how the standard entropy change for reaction 3 might compare with ∆Sᵣ₁.[2]

Mark scheme

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

[1] solubility increases down the group
[1] ∆Hlatt and ∆Hhyd both decrease
[1] or ∆Hlatt and ∆Hhyd both become less exothermic / more endothermic
[1] ∆Hlatt decreases / changes more (than ∆Hhyd as OH¯ being smaller than M²⁺)
[1] ΔHsol becomes more exothermic / more negative / less endothermic / less positive
[1] ∆Hr1 = -(538 + 2x230 + 394) = −(1216 + 286)
[1] ∆Hr1 = -110
[1] let ∆H(HCO3(aq)) = y
[1] 2y - 538 = -1216 – 394 – 286 – 26
[1] y = -692
[1] ∆Hr3 = -538 – 2(230 + 394) = −538 – 2(692)
[1] ∆Hr3 = -136
[1] ∆Hr3 will be identical to ∆Hr4, / unchanged
[1] as the reaction is the same, or: 2OH⁻(aq) + 2CO₂(g) → 2HCO₃⁻(aq) or metal ions stay in solution/metal ions are unchanged / are spectators
[1] more gaseous moles are being consumed (in reaction 3) or more CO₂ moles are being consumed (in reaction 3)
[1] ∆S is therefore expected to be more negative/less positive for reaction 3.