This question is about enthalpy changes for calcium chloride and magnesium chloride - AQA - A-Level Chemistry - Question 1 - 2021 - Paper 1

1. Ca(g)
2. Cl2(g)
3. CaCl2(s)

To calculate the enthalpy of lattice dissociation (LE), we can rearrange the Born-Haber cycle:

$LE = \text{Enthalpy of formation} + \text{Enthalpy of atomisation of Ca} + \text{1st Ionisation energy of Ca} + \text{2nd Ionisation energy of Ca} + \text{Enthalpy of atomisation of Cl} + \text{Electron affinity of Cl}$

Substituting the values from Table 1:

$LE = -795 + 193 + 590 + 1150 + 121 - 364 = +224$

Thus, the enthalpy of lattice dissociation of calcium chloride is +224 kJ mol^{-1}.

The enthalpy of solution can be represented by the following equation:

$\text{MgCl}_2(s) \rightarrow \text{Mg}^{2+}(aq) + 2 \text{Cl}^-(aq)$

Using the answer from Question 01.4, we have:

$\text{Enthalpy of solution} = \text{Enthalpy of lattice dissociation} + \text{Enthalpy of hydration of Mg}^{2+} + 2 \times \text{Enthalpy of hydration of Cl}$

Substituting the values from Table 2:

$\text{Enthalpy of solution} = +2493 + (-1920) + 2(-364)$

$= 2493 - 1920 - 728 = -155 kJ mol^{-1}$

The enthalpy of hydration of Ca^{2+}(g) is less exothermic than that of Mg^{2+}(g) because Ca^{2+} has a larger ionic radius compared to Mg^{2+}. As a result, the force of attraction between Ca^{2+} and water is weaker, leading to less energy released during hydration.