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 $F_2$ is 38 kcal/mol. Then, the enthalpy change for the reaction $Xe{F}_4\to X{e}^{+}+F^{-}+F_2+F$ will be

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

Answer 1

The correct option is C 292 kcal/mol

The given reaction is $Xe{F}_4\to X{e}^{+}+F^{-}+F_2+F$
here,
$B{E}_{Xe-F}=34kcal/mol$
$△H_{ioni}[Xe\to X{e}^{+}]=279kcal/mol$
$△E_{F-F}=38kcal/mol$
$△H_{eg}[F\to F^{-}]=85kcal/mol$

The enthalpy change for the reaction is given by -$△H=4B{E}_{Xe-F}+△H_{ioni}[Xe\to X{e}^{+}]-△E_{F-F}-△H_{eg}[F\to F^{-}]$
$=4\times 34+279-85-38=136+279-123$
$=415-123=292kcal/mol.$