Question

Use the following data to estimate ΔHf° 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 of Mg 150. kJ/mol

Answer 1

Answer:

The correct answer is -1085 KJ/mol

Explanation:

To calculate the formation enthalphy of a compound by knowing its lattice energy, you have to draw the Born-Haber cycle step by step until you obtain each element in its gaseous ions. Find attached the correspondent Born-Haber cycle.

In the cycle, Mg(s) is sublimated (ΔHsub= 150 KJ/mol) to Mg(g) and then atoms are ionizated twice (first ionization: ΔH1PI= 735 KJ/mol, second ionization= 1445 KJ/mol) to give the magnesium ions in gaseous state.

By other hand, the covalent bonds in F₂(g) are broken into 2 F(g) (Edis= 154 KJ/mol) and then they are ionizated to give the fluor ions in gaseous state 2 F⁻(g) (2 x ΔHafinity=-328 KJ/mol). The ions together form the solid by lattice energy (ΔElat=-2913 KJ/mol).

The formation enthalphy of MgF₂ is:

ΔHºf= ΔHsub + Edis + ΔH1PI + ΔH2PI + (2 x ΔHaffinity) + ΔElat

ΔHºf= 150 KJ/mol + 154 KJ/mol + 735 KJ/mol + 1445 KJ/mol + (2 x (-328 KJ/mol) + (-2913 KJ/mol).

ΔHºf= -1085 KJ/mol