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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 â–³H0f of NaCl(s) is -411 kJ/mol.What is its approximate lattice formation enthalpy?

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

## The correct option is B -788 kJ/molWe can use Born Haber cycle to find lattice energy △H0f = △Hsub+ IE +△Hdiss + EA + U = 108+496+122-349+ U=-411kJ/mol U = -788 kJ/mol

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