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

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  1. 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

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