Use the born-haber cycle to calculate the lattice energy of kcl. (δhsub for potassium is 89.0 kj/mol, ie1 for potassium is 419 kj/mol, ea1 for chlorine is - 349 kj/mol, the bond energy of cl2 is 243 kj/mol, δh∘f for kcl is - 436.5 kj/mol.)

Given dа ta:

Sublimation of K

K (s) ↔ K (g) ΔH (sub) = 89.0 kj/mol

Ionization energy for K

K (s) → K⁺ + e⁻ IE (K) = 419 Kj/mol

Electron affinity for Cl

Cl (g) + e⁻ → Cl⁻ EA (Cl) = - 349 kj/mol

Bond energy for Cl₂

1/2Cl₂ (g) → Cl Bond energy = 243/2 = 121.5 kj/mol

Formation of KCl

K (s) + 1/2Cl₂ (g) → KCl (s) ΔHf = - 436.5 kJ/mol

To determine:

Lattice energy of KCl

K⁺ (g) + Cl⁻ (g) → KCl (s) U (KCl) = ?

Explanation:

The enthalpy of formation of KCl can be expressed in terms of the sum of all the above processes, i. e.

ΔHf (KCl) = U (KCl) + ΔH (sub) + IE (K) + 1/2 BE (Cl₂) + EA (Cl)

therefore:

U (KCl) = ΔHf (KCl) - [ΔH (sub) + IE (K) + 1/2 BE (Cl₂) + EA (Cl) ]

= - 436.5 - [89 + 419 + 243/2 - 349] = - 717 kJ/mol

Ans: the lattice energy of KCl = - 717 kj/mol