Consider an ionic compound, MX2, composed of generic metal M
Consider an ionic compound, denoted as MX2, which consists of a generic metal M and a generic gaseous halogen X. Given the following thermodynamic the enthalpy of formation of MX2 is ΔHf° = –957 kJ/mol, the enthalpy of sublimation of M is ΔHsub = 123 kJ/mol, the first and second ionization energies of M are IE1 = 757 kJ/mol and IE2 = 1451 kJ/mol, respectively, and the electron affinity of X is ΔHEA = –313 kJ/mol. Additionally, the bond energy of the diatomic halogen X2 is BE = 153 kJ/mol.
1 Answers
Feb 05, 2025
Lattice energy U(MX2) to calculate
Use Born-Haber(-Fajans) Cycle
http://en.wikipedia.org/wiki/Born%E2%80%93Haber_cy…
M(s) + X2 → MX2(s) ΔHf°MX2 = -957 kJ mol^-1
M(s) →ΔHsub(M)→ M(g) ΔHsub = 123kJ mol^-1
M(g) →ΔHIE1(M)→ M^+ + e⁻ 757 kJ mol^-1
M^+→ΔHIE2(M)→ M^2+ + e⁻ 1451 kJ mol^-1
X2 →ΔHBE(X2)→ 2X +153 kJ mol^-1
X + e⁻→ΔHEA(X)→ X^- -313 kJ mol^-1
ΔHfo = U + ΔHsub(M) + IE1 + IE2 + ΔHdissX2 + 2EA1(X)
-957= U + 123 + 757 + 1451 + 153 – 626 = U + 1858
U = -2815 kJ mol^-1 The –ve sign is often dropped
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