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 data:

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

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