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Describe and explain the variation in the solubilities of the hydroxides of the Group 2 elements - CIE - A-Level Chemistry - Question 1 - 2017 - Paper 1

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Describe and explain the variation in the solubilities of the hydroxides of the Group 2 elements. The table lists the standard enthalpy changes of... show full transcript

Worked Solution & Example Answer:Describe and explain the variation in the solubilities of the hydroxides of the Group 2 elements - CIE - A-Level Chemistry - Question 1 - 2017 - Paper 1

Step 1

Describe and explain the variation in the solubilities of the hydroxides of the Group 2 elements.

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Answer

The solubility of hydroxides of Group 2 elements generally increases down the group. This is because the lattice enthalpy decreases as the ionic radius of the metal cation increases, making the dissolution process more favorable.

The enthalpy change of hydration also becomes more exothermic as the cation size increases, but this is outweighed by the decreasing lattice enthalpy.

The combination of decreasing lattice enthalpy and increasing hydration enthalpy favours solubility, resulting in more soluble hydroxides such as those of barium compared to magnesium.

Step 2

Use the data in the table to calculate the standard enthalpy change for reaction 1, ΔH_r°:

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Answer

To find ΔH_r°, we apply the formula:

ΔHr°=ΣΔHf°(products)ΣΔHf°(reactants)ΔH_r° = Σ ΔH_f°(products) - Σ ΔH_f°(reactants)

Given:

  • Formation of products:

    • BaCO₃ (s): -1216 kJ/mol
    • H₂O (l): -286 kJ/mol
  • Formation of reactants:

    • Ba(OH)₂ (aq) (unknown ΔH_f°)
    • CO₂ (g): -394 kJ/mol

Setting: Let ΔH_f(Ba(OH)₂) = y

Thus, we can substitute:

ΔHr°=(1216+286)(y+394)ΔH_r° = (-1216 + -286) - (y + -394)

Rearranging: ΔHr°=1502y+394ΔH_r° = -1502 - y + 394 ΔHr°=1108yΔH_r° = -1108 - y This will yield to Y being calculated as follows:

ΔHf(Ba(OH)2)=538ΔH_f(Ba(OH)₂) = 538

So, ΔHr°=1108538=1646ΔH_r° = -1108 - 538 = -1646 kJ/mol

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