Question

Consider an ionic compound, MX, composed of generic metal M and generic, gaseous halogen X.

The enthalpy of formation of MX is deltaHf = -475 kj/mol

The ethalpy of sublimation of M is deltaHsub = 119 kj/mol

The ionization energy of M is IE = 441 kj/mol

The electron affinity of X is deltaHEA = -341 kj/mol

The bond energy of X2 is BE = 181 kj/mol

Determine the lattice energy of MX.

Answer 1

M(s) -----------> M(g) delta H = 119 kj/mol

M(g) -----------> M+1(g) + 1e- IE = 441 kj/mol

1/2X2(g) --------> X(g) BE = 181/2 = 90.5 kj/mol

X(g) +1e- -----> X-(g) EA = -341 kj/mol

M(s) + 1/2X2(g) ----------> MX(s) delta Hf = -475 kj/mol

MX(s) -----------> M+(g) + X-(g) U(lattice energy) = ?

to get lattice energy we subtract all the energy values from delta Hf. So,

Lattice energy = -475 kj/mol - [119 + 441 + 90.5 -341 ]kj/mol

Lattice energy = -784.5 kj/mol