Use the Born-Haber cycle to calculate the standard enthalpy of formation ($\Delta$H )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 A)440 kJ/mol B)320 kJ/mol C)-260 kJ/mol D)-380 kJ/mol E)-1420 kJ/mol

Question

Quizplus · Accepted Answer

Step 1: 
Li(s) → Li(g) ΔH = 155.2 kJ/mol (sublimation energy)

Step 2: 
Li(g) → Li+(g) + e- ΔH = 520 kJ/mol (ionization energy)

Step 3: 
1/2 Cl2 → Cl(g) ΔH = bond energy = 242.7 kJ/mol

Step 4: 
Cl(g) + e- → Cl -(g) ΔH = -349 kJ/mol (electron affinity of Cl)

Step 5: 
Li+(g) + Cl -(g) → LiCl(s) ΔH = -828 kJ/mol (lattice energy)

Using the Born-Haber cycle equation:

ΔH°f (LiCl) = ΔHsub + ΔHIE + (1/2)ΔHBE + ΔHEA + ΔHLattice

ΔH°f (LiCl) = 155.2 kJ/mol + 520 kJ/mol + (1/2)(242.7 kJ/mol) - 349 kJ/mol - 828 kJ/mol 
ΔH°f (LiCl) = -379.95 kJ/mol ≈ -380 kJ/mol

Therefore, the standard enthalpy of formation for LiCl(s) is -380 kJ/mol. Thus, the answer is D.

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