State the meaning of the term enthalpy change - AQA - A-Level Chemistry - Question 1 - 2021 - Paper 1

1. Ca(s) → Ca²⁺(g) + 2e⁻

2. Cl2(g) → 2Cl⁻(g)

3. CaCl2(s) → Ca²⁺(g) + 2Cl⁻(g)

Using the Born-Haber cycle, we can express the enthalpy change for the formation of CaCl2(s) as:

$\Delta H_{formation} = \Delta H_{atomization} (Ca) + \Delta H_{ionization} (Ca) + \Delta H_{atomization} (Cl) + \Delta H_{electron\ affinity} (Cl) + \Delta H_{lattice}$

Substituting values from Table 1:

$-795 = 193 + 590 + 121 + (-364) + \Delta H_{lattice}$

Rearranging gives:

$\Delta H_{lattice} = -795 - (193 + 590 + 121 - 364)$

Calculating:

$\Delta H_{lattice} = -795 - 540 = -255 \, kJ \, mol^{-1}$

The equation for the enthalpy of solution of magnesium chloride is:

We know:

$\Delta H_{solution} = \Delta H_{lattice} + \Delta H_{hydration}$

From Table 2:

$\Delta H_{lattice} = +2493 \, kJ \, mol^{-1}$

Hydration energies:

• For $Mg^{2+}(aq)$: -1920 kJ/mol
• For $Cl^{-}(aq)$: 2 * -364 kJ/mol = -728 kJ/mol

Calculating hydration energy:

$\Delta H_{hydration} = -1920 - 728 = -2648 \, kJ \, mol^{-1}$

Thus,

$\Delta H_{solution} = 2493 - 2648 = -155 \, kJ \, mol^{-1}$

The enthalpy of hydration of $Ca^{2+}(g)$ is greater due to the larger charge density of $Ca^{2+}$ compared to $Mg^{2+}$. This causes stronger ion-dipole interactions with water molecules, leading to a more exothermic release of energy during hydration.