Using ΔHf(NaCl) = −411, ΔHsub(Na) = +108, IE(Na) = +496, ½BDE(Cl₂) = +121 and EA(Cl) = −349 kJ mol⁻¹, what is the lattice energy of NaCl?
Correct answer B. −787 kJ mol⁻¹
Explanation
Born–Haber: lattice energy = ΔHf − (ΔHsub + IE + ½BDE + EA) = −411 − (108 + 496 + 121 − 349) = −411 − 376 = −787 kJ mol⁻¹.
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