Question

Consider an ionic compound, MX2 , composed of generic metal M and generic, gaseous halogen X . The enthalpy of formation of MX2 is Δ?∘f=−677 kJ/mol. The enthalpy of sublimation of M is Δ?sub=161 kJ/mol. The first and second ionization energies of M are IE1=647 kJ/mol and IE2=1377 kJ/mol. The electron affinity of X is Δ?EA=−339 kJ/mol. (Refer to the hint). The bond energy of X2 is BE=221 kJ/mol. Determine the lattice energy of MX2 . Δ?lattice= kJ/mol

Answer 1

The enthalpy of formation of MX2 is Δ?∘f=−677 kJ/mol

M(s) + X2(g) ----------------> MX2(s)      Δ?∘f=−677 kJ/mol

The enthalpy of sublimation of M is Δ?sub=161 kJ/mol.

M(s) -------------------->M(g)     Δ?sub=161 kJ/mol.

The first ionisation energy

M(g) -----------------> M^+ (g) + e^-      IE1=647 kJ/mol

Second ionisation of energies

M^+ (g) ----------------> M^2+ (g) + e^-    IE2=1377 kJ/mol

The bond energy of X2 is BE=221 kJ/mol

X2(g) ---------------> 2X(g)                  Δ?BE=221 kJ/mol

The electron affinity of X is Δ?EA=−339 kJ/mol

2X(g) + 2e^- --------------> 2X^- (g)      Δ?EA=−339*2   = -678kJ/mol

Determine the lattice energy of MX2 . Δ?lattice= kJ/mol

M^2+ (g) + 2X^-(g) --------------> MX2(s)     Δ?lattice= kJ/mol

Δ?∘f    = Δ?sub + IE1   + IE2 + Δ?BE + Δ?EA + Δ?lattice

-677     = 161 + 647+1377+221--678 + Δ?lattice

Δ?lattice   = -3761KJ/mole   >>>>answer