Question

Calculate the enthalpy change (in kcal) for the reaction- $XeF _{4} \longrightarrow Xe ^{+} F ^{-}+ F _{2}+ F$
The average $Xe - F$ bond enthalpy is $34 \,k\,Cal / mol$, first I.E. of Xe is $279\, k\,Cal / mol$, electron gain enthalpy of $F$ is $-85\, k\,Cal / mol$ and the bond dissociation enthalpy of $F _{2}$ is $38\, k\,Cal / mol :$

• A. 292 kCal mol ^{-1}
• B. 383 kCal mol$^{-1}$
• C. 521 k Cal mol $^{-1}$
• D. 528 k Cal mol$ ^{-1}$

Answer 1

Answer: (A)

We need to calculate the $\Delta H$ for the following reaction-

$XeF _{4} \longrightarrow Xe ^{+}+ F ^{-}+ F _{2}+ F$

Energy Absorbed:

$4 \times 34=136 kcal / mol$



Energy released:

85 kcal/mol (E.G.E)



$\Delta H=415-123=292$