chemistry enthalpy question

Question

Consider an ionic compound, MX2, composed of generic metal M and generic, gaseous halogen X.

 

The enthalpy of formation of MX2 is ΔHf° = –861 kJ/mol. The enthalpy of sublimation of M is ΔHsub = 151 kJ/mol. The first and second ionization energies of M are IE1 = 665 kJ/mol and IE2 = 1402 kJ/mol. The electron affinity of X is ΔHEA = –333 kJ/mol. The bond energy of X2 is BE = 191 kJ/mol.

 

Determine the lattice energy of MX2.

 

Answer 1

Following the path to go from M + X₂ to MX₂, we have the intermediate steps:

Make the metal gaseous => +151 kJ/mol

gain energy to ionize M to M⁺ => +665 kJ/mol

gain energy to ioinize M⁺ to M²⁺ => +1402 kJ/mol

decrease energy of X₂ bond => -191 kJ/mol

decrease energy for adding electrons to X => -333 kJ/mol

decrease energy to make lattice => -ΔH

 

By summing all intermediate steps, you will get the overall enthalpy of formation:

-861 = 151 + 665 + 1402 - 191 - 333 - ΔH

 

Rearranging:

ΔH = 2555 kJ/mol is the lattice energy