Question

Calculate the lattice energy for KCl(s) using a Born-Haber cycle and the following information:

KCl(s) → K⁺(g) + Cl⁻ (g) ΔH = ?
K(s) + 1/2 Cl₂(g) → KCl(s) ΔH = -437.0 kJ/mol
K(s) → K(g) ΔH = +89 kJ/mol
K(g) → K⁺(g) + e⁻ ΔH = +419 kJ/mol
1/2 Cl₂(g) → Cl⁻(g) ΔH = +121 kJ/mol
Cl(g) + e→ Cl⁻(g) ΔH = -349 kJ/mol
O +796 kJ/mol
O +840 kJ/mol
O +152 kJ/mol
O +717 kJ/mol
O +650 kJ/mol

Answer 1

The lattice energy of KCl(s) is calculated using the Born-Haber cycle and given reaction enthalpies by adding sublimation, ionization, and dissociation energy and subtracting the electron affinity and formation enthalpy of KCl(s), resulting in a lattice energy of 1415 kJ/mol.

Step-by-step explanation:

To calculate the lattice energy for KCl using the Born-Haber cycle, we need to account for several steps, including the sublimation of potassium, the ionization of potassium, the dissociation of chlorine gas, and the combination of the potassium and chloride ions to form the ionic lattice. Given the reaction enthalpies for each step, we can apply Hess's Law to find the lattice energy:

• Sublimation of potassium (K(s) → K(g)): +89 kJ/mol
• Ionization of potassium (K(g) → K+ (g) + e−): +419 kJ/mol
• Dissociation of chlorine gas (1/2 Cl2(g) → Cl(g)): +121 kJ/mol
• Formation of chloride ion (Cl(g) + e− → Cl−(g)): -349 kJ/mol
• Formation of KCl(s) from elements (K(s) + 1/2 Cl2(g) → KCl(s)): -437 kJ/mol

The lattice energy of KCl(s) can be determined by summing these values while considering that the formation enthalpy is the inverse process of dissociating the lattice into gaseous ions:

1. Lattice energy of KCl(s) = Sublimation + Ionization + Dissociation + Electron Affinity - Formation enthalpy
2. Lattice energy of KCl(s) = 89 + 419 + 121 - (-349) - (-437)
3. Lattice energy of KCl(s) = 89 + 419 + 121 + 349 + 437
4. Lattice energy of KCl(s) = 1415 kJ/mol