dHf = HsubNa + IP + 1/2*BE + EA + 1/2*HsubI + Ecrystal
Solve for Ecryst, then U = lattice energy = -Ecrystal
dHf = heat formation
HsubNa = heat sublimation for Na
IP = ionization energy
BE = bond dissociation energy
EA = electron affinity
HsubI = heat sublimation I2
What is the lattice energy of NaI? Use the given information below.
Heat of formation for NaI = -288.0 kJ/mol
Heat of sublimation for Na = 107.3 kJ/mol
Ionization energy for Na = 496.0 kJ/mol
Bond dissociation energy for I2 = 149 kJ/mol
Electron affinity of I = -295.0 kJ/mol
Heat of sublimation for I2(s) = 48.30 kJ/mol
1 answer
1 answer