Question

What is the lattice energy of NaBr?

heat of formation of NaBr = -362 kJ/mol
heat of sublimation for Na = 107.30 kJ/mol
ionization energy for Na = 496 kJ/mol
bond dissociation energy for Br2 = 190 kJ/mol
electron affinity of Br = -325 kJ/mol
heat of vaporization for Br2 = 30.90 kJ/mol

Answer 1

Heat formation = Hsubl + IP + 1/2(bond diss) + electron affinity(a negative value) + 1/2(vaporization) + Ecrystal lattice.

Substitute and solve for E crystal.
If your text defines crystal lattice energy as -Ecrystal, then reverse the sign.
Something like +750 or so kJ/mol.

Answer 2

Answer