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Hess' Law

What is the b...

Question

What is the bond enthalpy of $X e - F$ bond?
$X e F_{4} ⟶ {X e}^{+} (g) + F^{-} (g) + F_{2} (g) + F (g); Δ_{r} H = 292 k c a l / m o l$
Given : Ionization energy of $X e = 279 k c a l / m o l$
$B . E (F - F) = 38 k c a l / m o l$ , Electron affinity of $F = 85 k c a l / m o l$

24 k c a l / m o l

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34 k c a l / m o l

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8.5 k c a l / m o l

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none of these

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Solution

The correct option is B $34 k c a l / m o l$
The given reaction can be broken down into following steps.
$X e F_{4} \to X e + 4 F$
$X \to X e^{+}$ ionization of xenon.
$2 F \to F_{2}$ the bond enthalpy of fluorine
$F \to F^{-}$ Electron affinity of fluorine
Hence, the entahlpy change of the reaction is
$Δ H_{r} = 4 Δ H (X e - F) - (I . E X e + Δ H (F - F) + E . A (F))$
$292 = 4 Δ H (X e - F) - (- 279 + 38 + 35)$
$Δ H (X e - F) = 34 k c a l / m o l$

Suggest Corrections

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F_{2}

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X e F_{4} ⟶ X e^{+} + F^{-} + F_{2} + F; Δ H = ?

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34 k c a l m o l^{- 1}

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