Question

Given sublimation and ionization energy of Na are 108 kJ/mol and 496 kJ/mol respectively and bond dissociation energy required to make one mole chlorine atoms and its electron affinity energy are 122 kJ/mol and -349 kJ/mol.If $△H_f^0$ of NaCl(s) is -411 kJ/mol.What is its approximate lattice formation enthalpy?

• A. 286 kJ/mol
• B. -788 kJ/mol
• C. 388 kJ/mol
• D. -508 kJ /mol

Answer 1

The correct option is B -788 kJ/mol

We can use Born Haber cycle to find lattice energy
$△H_f^0$ = $△H_{sub}$+ IE +$△H_{diss}$ + EA + U = 108+496+122-349+ U=-411kJ/mol
U = -788 kJ/mol