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Question

# The average Xe - F bond energy is 34 kcal/mol, first I.E. of Xe is 279 kcal/mol, electron affinity of F is 85 kcal/mol and bond dissociation energy of F2 is 38 kcal/mol. Then, the enthalpy change for the reaction XeF4→Xe++F−+F2+F will be

A
367 kcal/mol
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B
425 kcal/mol
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C
292 kcal/mol
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D
392 kcal/mol
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Solution

## The correct option is C 292 kcal/molThe given reaction is XeF4→Xe++F−+F2+F here, BEXe−F=34kcal/mol △Hioni[Xe→Xe+]=279 kcal/mol △EF−F=38 kcal/mol △Heg[F→F−]=85 kcal/mol The enthalpy change for the reaction is given by -△H=4BEXe−F+△Hioni[Xe→Xe+]−△EF−F−△Heg[F→F−] =4×34+279−85−38=136+279−123 =415−123=292 kcal/mol.

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