The Born-Haber cycle for KCl is evaluated with the following data :
Δ _fH^Θ for KCl = -436.7 kJ mol⁻¹ ;
Δ _(sub)H^Θ for K = 89.2 kJ mol⁻¹ ;
Δ _(ionization)H^Θ for K = 419.0 kJ mol⁻¹ ;
Δ _(electron gain)H^Θ for Cl (g) = -348.6 kJ mol⁻¹ ;
Δ _(bond)H^Θ for Cl₂ = 243.0 kJ mol⁻¹
The magnitude of lattice enthalpy of KCl in kJ mol⁻¹ is _____________ (Nearest integer)

Solution

Chemistry ma’am · 4 steps

Build KCl from K and Cl₂ the long way round; the formation enthalpy must equal the sum of every step.

1. Set up the cycle. Making KCl(s) directly releases Δ_f H. The long route has five steps: sublime K, ionise K, break half a mole of Cl₂, add an electron to Cl, then let the gaseous ions form the lattice. By Hess's law both routes give the same total.

   Δ_f H = Δ_(sub)H + Δᵢ H + 1/2Δ_(bond)H + Δ_(eg)H + Δ_(lattice)H

2. Add the known steps. Only one Cl atom is needed per KCl, so take 1/2 × 243.0 = 121.5 kJ mol⁻¹. Then 89.2 + 419.0 + 121.5 - 348.6 = 281.1 kJ mol⁻¹.
3. Solve for lattice. Δ_(lattice)H = -436.7 - 281.1 = -717.8 kJ mol⁻¹. The negative sign shows that energy is released when the gaseous ions come together.
4. So. The magnitude of the lattice enthalpy of KCl is 717.8 kJ mol⁻¹, which rounds to 718.

The common slip: 839. Using the full bond enthalpy of Cl₂ (243.0) instead of half of it gives 839, not 718.

The idea to keep. In a Born-Haber cycle, match each step to one formula unit. For a diatomic gas like Cl₂ you need only half the bond enthalpy, and every sign must follow the direction of its step.

Answer from the NTA final answer key. Solution by PrepWiser.