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& bond dissociation energy of F2 is 38 kcal/mol. Then, the enthalpy change for the reaction

XeF4$\longrightarrow$Xe+ + F- + F2 + F will be

• A. 367 kcal/mole
• B. 425 kcal/mole
• C. 292 kcal/mole
• D. 392 kcal/mole

Answer 1

Answer: (C)

292 kcal/mole

Answer 2

$\Delta$H = 4EXe–F + $\Delta$Hloni [Xe ® Xe+] – $\Delta$EF–F – $\Delta$Heg [F $\rightarrow$ F–]

= 4 × 34 + 279 – 85 – 38 = 136 + 279 – 123 = 415 – 213

= 292 kcal/ mole.