Question

Consider an ionic compound, MX2, composed of generic metal M and generic, gaseous halogen X.

• The enthalpy of formation of MX2 is ΔH∘f=−831 kJ/mol.

• The enthalpy of sublimation of M is ΔHsub=129 kJ/mol.

• The first and second ionization energies of M are IE1=691 kJ/mol and IE2=1401 kJ/mol

• The electron affinity of X is ΔHEA=−307 kJ/mol.

• The bond energy of X2 is BE=153 kJ/mol.

Determine the lattice energy of MX2.

ΔHlattice=ΔHlattice= kJ/mol

Answer 1

By using Hess Law: Heat of a reaction remains same, reaction may occur in single step or multiple steps.

Lattice energy is -2591KJ

Please go through attached image for detailed explanation.

Mcs) + X(g) MX₂6) DHE - 831 KT bove reaction can be carried out in multiple steps as following The above sublimation a AH= 129 KJ First 1omisahan - and Tomsabon Mcg) - Mtcar AH=691 kJ mcg) DHE IU OH = 190 kJ > M Lattice formation > Mxls) Elechon gain Lattice ? Bond dissociation energy 209) > AH-153 KT 2 X 199 SH-307x2KJ =-614 KJ = -831 - By Using Hess Law, A 129 kJ + 691 KJ + 1401 KT + 153 KJ + (-614 KJ) PEBRO + DHL K 1760 kJ & DH Lattice = - 831 kJ St Lattice = - 83k5 - 17 60 kJ. AH Lattice - 2591 kJ