What is the lattice energy of KCl if its enthalpy of formation is -437 kJ/mol, given: ΔH (K) = 89 kJ/mol, IE(K) = 419 kJ/mol, D(Cl₂) = 243 kJ/mol, EA(Cl) = -349 kJ/mol?
Correct answer: A. -717 kJ/mol
- A. -717 kJ/mol
- B. -437 kJ/mol
- C. -349 kJ/mol
- D. -243 kJ/mol
Explanation
Born-Haber cycle: ΔH = ΔH + IE + (1/2)D - EA - U. -437 = 89 + 419 + (243/2) - 349 - U. U ≈ -717 kJ/mol.
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