Problem: What is the enthalpy of sublimation for K, in kJ/mol?Given:Lattice energy of KCl = 699 kJ/molFirst ionization energy of K = 418.7 kJ/molElectron affinity of Cl = 349 kJ/molBond energy of Cl-Cl = 242.7 kJ/molEnthalpy of formation of KCl = -435.87 kJ/mol

FREE Expert Solution
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FREE Expert Solution

We’re being asked to calculate the heat of sublimation of potassium (K), given the lattice energy of KCl. 

Recall that lattice energy is the energy required to combine two gaseous ions into a solid ionic compound:

Mx+(g) + Ny–(g)  MyNx(s)

To calculate for lattice energy, we need to do the Born-Haber cycle for KCl

We start with the corresponding formation equation for KCl:

K(s) + ½ Cl2(g)  KCl(s), ΔH˚f

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Problem Details

What is the enthalpy of sublimation for K, in kJ/mol?


Lattice energy of KCl = 699 kJ/mol

First ionization energy of K = 418.7 kJ/mol

Electron affinity of Cl = 349 kJ/mol

Bond energy of Cl-Cl = 242.7 kJ/mol

Enthalpy of formation of KCl = -435.87 kJ/mol

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