Calculate the lattice energy of NaBr(s),given the following thermochemical equations,where $\Delta$IE and $\Delta$EA are enthalpy of ionization and electron attachment enthalpy,respectively.

Question

Quizplus · Accepted Answer

The lattice energy (ΔH lattice) of NaBr can be calculated using the following steps:
Na(s) → Na+(g) + e- (F1F1F1S1)
1/2Br2(g) + e- → Br-(g) (F1F1F10EA)
Na+(g) + Br-(g) → NaBr(s) (ΔH lattice)
The enthalpy change for this reaction can be calculated using the Born-Haber cycle:
ΔH lattice = -ΔH sub - ΔH IE - ΔH EA - ΔH f
where:
ΔH sub = sublimation enthalpy of Na (107.3 kJ/mol)
ΔH IE = ionization enthalpy of Na (496 kJ/mol)
ΔH EA = electron attachment enthalpy of Br (−324 kJ/mol)
ΔH f = standard enthalpy of formation of NaBr (-361 kJ/mol)
Substituting the values and using Hess's Law, we get:
ΔH lattice = -(-361 kJ/mol) - (496 kJ/mol) - (-324 kJ/mol) - (107.3 kJ/mol) = -751 kJ/mol

Therefore, the best choice is B, -751 kJ, which is the lattice energy of NaBr.

Calculate the Lattice Energy of NaBr(s),given the Following Thermochemical Equations,where Δ\DeltaΔ

Calculate the Lattice Energy of NaBr(s),given the Following Thermochemical Equations,where $\Delta$