Consider an ionic compound

Question

Consider an ionic compound, M*X_{2} composed of generic metal M and generic, gaseous halogen X.

• The enthalpy of formation of M*X_{2}*is*Delta * H_{1} ^ 2 = - 997kJ / m * ol

• The enthalpy of sublimation of M is Delta H mb =119 kJ/mol.

• The first and second ionization energies of Mare I*E_{1} = 781kJ / m * ol and I*E_{2} = 1386kJ / m * ol

• The electron affinity of X is Delta H EA =-305 kJ/mol. (Refer to the hint.)

• The bond energy of X_{2} is BE = 211kJ / m * ol

Determine the lattice energy of M*X_{2}

H lattice =

I got 2075 but that's wrong

Answer 1

So, this is a problem that deals with the Born Haber cycle.

We begin with the knowledge that

∆H_formation = ∆H_sublimation + Ionization energy + Bond energy + Electron affinity + Lattice energy.

Next, we set up a scheme as follows for MX₂

M(s) + X₂(g) ==> MX₂ ... ∆H_formation = -997

M(s) ==> M(g) ... ∆H_sublimation = 119

M(g) ==>M⁺(g) ... IE₁ = -781 (note: use negative value)

M⁺(g) ==> M²⁺(g) ... IE₂ = -1386 (note: use negative value)

X₂(g) ==> 2X(g) ... Bond energy = 211 (this is for both )

2X(g) ==> 2X^-(g) ... Bond energy

M²⁺(g) + 2X^-(g) ==> MX₂ ... LE = ? (we are going to solve for Lattice energy)

∆H_formation = ∆H_sublimation + Ionization energy + Bond energy + Electron affinity + Lattice energy

-997 = 119 + (-781) + (-1386) + 211 + LE

-997 = -1837 + LE

LE = 840 kJ/mole