Question

Use the following data to estimate ΔH∘ f for magnesium fluoride. Mg(s)+F₂(g)→MgF₂(s) Lattice energy =−2913 kJ/mol First ionization energy of Mg=735 kJ/mol Second ionization energy of Mg=1445 kJ/mol Electron affinity of F=−328 kJ/mol Bond energy of F2=154 kJ/mol Enthalpy of sublimation for Mg=150.kJ/mol Select one or more: a. −1085 kJ/mol b. −292 kJ c. −411 kJ/mol d. 167 kJ/mol

Answer 1

The estimated value for
`\( \Delta H^\circ_f (\text{MgF}_2) \)` is approximately -2957 kJ/mol.

The formation of magnesium fluoride
`(\( \text{MgF}_2 \))` from its elements involves various steps:

1.
`\( \text{Mg}(s) \) - > \( \text{Mg}^(2+)(g) + 2e^- \)` (First ionization energy)

2.
`\( \text{Mg}^(2+)(g) \) - > \( \text{Mg}^(2+)(g) \)` (Second ionization energy)

3.
`\( \text{F}_2(g) \) - > \( 2\text{F}(g) \)` (Bond dissociation of
`\( \text{F}_2 \)`)

4.
`\( \text{F}(g) + e^- \) - > \( \text{F}^- (g) \)` (Electron affinity of fluorine)

5.
`\( \text{Mg}(s) \) - > \( \text{Mg}(g) \)` (Enthalpy of sublimation for Mg)

6.
`\( \text{Mg}(g) + 2\text{F}(g) \) - > \( \text{MgF}_2(s) \) (Formation of \( \text{MgF}_2 \))`

The enthalpy change for the formation of
`\( \text{MgF}_2 \)` can be determined using Hess's Law by considering these steps.

Given:

Lattice energy = -2913 kJ/mol

First ionization energy of Mg = 735 kJ/mol

Second ionization energy of Mg = 1445 kJ/mol

Electron affinity of F = -328 kJ/mol

Bond energy of F2 = 154 kJ/mol

Enthalpy of sublimation for Mg = 150 kJ/mol

The enthalpy change for the formation of
`\( \text{MgF}_2 \)` can be calculated as follows:

`\[ \Delta H^\circ_f (\text{MgF}_2) = \text{Lattice energy} + \text{IE}_1(\text{Mg}) + \text{IE}_2(\text{Mg}) + \text{EA}(\text{F}) + \text{Bond energy} - \text{Sublimation energy} \]`

Substitute the given values:

`\[ \Delta H^\circ_f (\text{MgF}_2) = -2913 + 735 + 1445 - 328 + 154 - 150 \]`

Sure, let's calculate the enthalpy change for the formation of magnesium fluoride
`(\( \Delta H^\circ_f (\text{MgF}_2) \))` using the given values:

`\[\Delta H^\circ_f (\text{MgF}_2) = -2913 + 735 + 1445 - 328 + 154 - 150\]`

`\[\Delta H^\circ_f (\text{MgF}_2) = -2957 \, \text{kJ/mol}\]`