The estimated value for
is approximately -2957 kJ/mol.
The formation of magnesium fluoride
from its elements involves various steps:
1.
(First ionization energy)
2.
(Second ionization energy)
3.
(Bond dissociation of
)
4.
(Electron affinity of fluorine)
5.
(Enthalpy of sublimation for Mg)
6.

The enthalpy change for the formation of
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
can be calculated as follows:
![\[ \Delta H^\circ_f (\text{MgF}_2) = \text{Lattice energy} + \text{IE}_1(\text{Mg}) + \text{IE}_2(\text{Mg}) + \text{EA}(\text{F}) + \text{Bond energy} - \text{Sublimation energy} \]](https://img.qammunity.org/2024/formulas/chemistry/high-school/82f2pfa40jfr6w7h7jhohbppi29b86o3an.png)
Substitute the given values:
![\[ \Delta H^\circ_f (\text{MgF}_2) = -2913 + 735 + 1445 - 328 + 154 - 150 \]](https://img.qammunity.org/2024/formulas/chemistry/high-school/k3ah2rxlhrwpcxtqwgdcx3oh34jvr1u45f.png)
Sure, let's calculate the enthalpy change for the formation of magnesium fluoride
using the given values:
![\[\Delta H^\circ_f (\text{MgF}_2) = -2913 + 735 + 1445 - 328 + 154 - 150\]](https://img.qammunity.org/2024/formulas/chemistry/high-school/q9wq2jiqjswshxn55mjdw2o36xhseytq16.png)
![\[\Delta H^\circ_f (\text{MgF}_2) = -2957 \, \text{kJ/mol}\]](https://img.qammunity.org/2024/formulas/chemistry/high-school/i4oq968d1pl1nqztzcd0kq9h6uvmg3vkn3.png)