Use the Born-Haber cycle to calculate the standard enthalpy of formation ( $\Delta{H}^{\circ}_f$ ) for LiCl(s) given the following data: $\Delta$ H(sublimation) Li = 155.2 kJ/mol
I₁ (Li) = 520 kJ/mol
Bond energy (Cl-Cl) = 242.7 kJ/mol
EA (Cl) = 349 kJ/mol
Lattice energy (LiCl(s) ) = 828 kJ/mol

Question

Use the Born-Haber cycle to calculate the standard enthalpy of formation (

\Delta{H}^{\circ}_f

) for LiCl(s) given the following data:

\Delta

H(sublimation) Li = 155.2 kJ/mol
I₁ (Li) = 520 kJ/mol
Bond energy (Cl-Cl) = 242.7 kJ/mol
EA (Cl) = 349 kJ/mol
Lattice energy (LiCl(s) ) = 828 kJ/mol

Quizplus · Accepted Answer

The Born-Haber cycle for LiCl(s) is as follows:
Li(s) → Li(g) ΔH(sublimation) = 155.2 kJ/mol
Li(g) → Li+(g) + e- ΔH1 = IE(Li) = 520 kJ/mol
Cl2(g) → 2Cl(g) ΔH2 = 2 × 1/2(Bond energy(Cl-Cl)) = 242.7 kJ/mol
Cl(g) + e- → Cl-(g) ΔH3 = EA(Cl) = 349 kJ/mol
Li+(g) + Cl-(g) → LiCl(s) ΔH4 = -Lattice energy = -828 kJ/mol
Summing up all the enthalpy changes gives:
ΔH(f) (LiCl) = ΔH(sublimation) + ΔH1 + ΔH2 + ΔH3 + ΔH4
ΔH(f) (LiCl) = 155.2 + 520 + 242.7 + 349 - 828
ΔH(f) (LiCl) = -380 kJ/mol, which is the standard enthalpy of formation (ΔH(f)) of LiCl(s). Therefore, the answer is D.

Use the Born-Haber Cycle to Calculate the Standard Enthalpy of Formation