Chemistry Problem
posted by Anonymous .
Consider an ionic compound, MX, composed of generic metal M and generic halogen X.
The enthalpy of formation of MX is ÄHf° = –457 kJ/mol. The enthalpy of sublimation of M is ÄHsub = 121 kJ/mol. The ionization energy of M is IE = 433 kJ/mol. The electron affinity of X is EA = –313 kJ/mol. The bond energy of X2 is BE = 229 kJ/mol.
Determine the lattice energy of MX.
This is a problem on my homework assignment that I really don't understand. Please explain, it would help so much! Thanks!

You need to go through the "cycle" to see this.
M(s) ==> M(g) dH = 121 kJ
M(g) ==> M^+(g) + e dH = 433 kJ
1/2 X2(g) ==> X(g) dH = 229/2 kJ
X(g) + e ==> X^(g) dH = 313 kJ
M^+(g) + X^(g) ==> MX(s) dH = Ecrystal

Add these. M(g) cancels. M^+(g) cancels. e cancels, X^(g) cancels, and we have left
M(s) + 1/2 X2(g) ==> MX(s) dHf = 457 kJ
457=121 + 433+(229/2)+(313)+Ecrystal
And solve for Ecrystal.
Now for the confusing part.
Ecrystal is the lattice energy IF YOU DEFINE lattice energy as the heat liberated by combining the ions to form the solid crystal. But if you define lattice energy as the energy needed to break up the ionic compound into the ions then Ecrystal = lattice energy. About 813 kJ or +813 kJ for Ecrystal or Ecrystal. Look in your notes to see which way your prof defines lattice energy.
M^+(g) + X^(g) ==> MX(s) 813 kJ
MX(s) ==> M^+(g) + X^(g) +813 kJ