Consider an ionic compound, MXMX, composed of generic metal MM and generic, gaseous halogen XX.
The enthalpy of formation of MXMX is Δ?∘f=−417ΔHf∘=−417 kJ/mol.
The enthalpy of sublimation of MM is Δ?sub=131ΔHsub=131 kJ/mol.
The ionization energy of MM is IE=417IE=417 kJ/mol.
The electron affinity of XX is Δ?EA=−303ΔHEA=−303 kJ/mol. (Refer to the hint).
The bond energy of X2X2 is BE=239BE=239 kJ/mol.
Determine the lattice energy of MXMX.
Answer:
Step 1: Explanation
The analysis of the formation of an ionic compound from its elements is commonly discussed in terms of a Born-Haber cycle, which breaks the overall process into a series of steps of known energy
Step 2: Calculation
Step 1: M(s) --> M(g) ------------> ( ΔHsub = 131 kJ/mol )
Step 2: 1/2X2(g) --> X(g) --------> ( BE = 239/2 kJ/mol ) = (119.5 kJ/mol )
[since given BE for X2 coversion but we need X hence EA will divide by 2]
Step 3: M(g) --> M+(g) + e- ----------> ( IE1 = 417 kJ/mol )
Step 4: X(g) + e- --> X-(g) -----------> ( ΔHEA = -303 kJ/mol )
Step 5: M+(g) +1/2 X-(g) --> MX ------------> ( ΔH0f = -417 kJ/mol )
Overall reaction: M(s)+ 1/2 X2(g) --> MX(s)
Overall energy
ΔH0f = U + ΔHsub + IE + BE + ΔHEA
-417 kJ/mol = U + 131 kJ/mo + 417 kJ/mol + 119.5 kJ/mol + (-303 kJ/mol)
U = -417 kJ/mol + 303kJ/mol - ( 131 kJ/mol + 417 kJ/mol + 119.5 kJ/mol )
lattice Energy (U)= -781.5 kJ /mol
Note: -ve sign is thermodynamically correct but most report lattice energies without -ve sign it is understood that it is -ve.
So change the sign according to your need
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