ΔHfꝋ = ΔHatꝋ + ΔHatꝋ + IE + EA + ΔHlattꝋ
ΔHfꝋ = ΔH1ꝋ + ΔHlattꝋ
So, if we rearrange to calculate the lattice energy, the equation becomesΔHlattꝋ = ΔHfꝋ - ΔH1ꝋ

Answer

ΔHlattꝋ = ΔHfꝋ - ΔH1ꝋ
ΔHlattꝋ = ΔHfꝋ - [(ΔHatꝋ K) + (ΔHatꝋ Cl) + (IE1 K) + (EA1 Cl)]
ΔHlattꝋ = (-437) - [(+90) + (+122) + (+418) + (-349)] = -718 kJ mol-1

Answer

ΔHlattꝋ = ΔHfꝋ - ΔH1ꝋ
ΔHlattꝋ = ΔHfꝋ - [(ΔHatꝋ Mg) + (ΔHatꝋ O) + (IE1 Mg) + (IE2 Mg) + (EA1 O) + (EA2 O)]
ΔHlattꝋ = (-602) - [(+148) + (+248) + (+736) + (+1450) + (-142) + (+770)]= -3812 kJ mol-1
Make sure you use brackets when carrying out calculations using Born-Haber cycles as you may forget a +/- sign which will affect your final answer!
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