Consider the following: Li(s) + 1/2 I2(g) LiI(s) H = 292 kJ. LiI(s) has a lattice

Question:

Consider the following:
Li(s) + 1/2 I2(g) → LiI(s) ΔH = –292 kJ.
LiI(s) has a lattice energy of –753 kJ/ mol.
The ionization energy of Li(g) is 520. kJ/ mol, the bond energy of I2(g) is 151 kJ/ mol, and the electron affinity of I(g) is -295 kJ/ mol. Use these data to determine the heat of sub-limation of Li(s).