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Heat of Reaction

The average X...

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 $X e F_{4} \to X e^{+} + F^{-} + F_{2} + F$ will be

367 kcal/mol

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425 kcal/mol

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292 kcal/mol

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392 kcal/mol

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Solution

The correct option is C 292 kcal/mol
The given reaction is $X e F_{4} \to X e^{+} + F^{-} + F_{2} + F$
here,
$B E_{X e - F} = 34 k c a l / m o l$
$△ H_{i o n i} [X e \to X e^{+}] = 279 k c a l / m o l$
$△ E_{F - F} = 38 k c a l / m o l$
$△ H_{e g} [F \to F^{-}] = 85 k c a l / m o l$

The enthalpy change for the reaction is given by - $△ H = 4 B E_{X e - F} + △ H_{i o n i} [X e \to X e^{+}] - △ E_{F - F} - △ H_{e g} [F \to F^{-}]$
$= 4 \times 34 + 279 - 85 - 38 = 136 + 279 - 123$
$= 415 - 123 = 292 k c a l / m o l .$

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Similar questions

F_{2}

X e F_{4} \to X e^{+} + F^{-} + F_{2} + F

X e - F

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

X e = 279 k c a l / m o l

B . E (F - F) = 38 k c a l / m o l

F = 85 k c a l / m o l

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

X e - - F

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

{I E}_{1}

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

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

e_{F - - F} = 38 k c a l m o l^{- 1}

X e F_{2 (g)} + H_{2} (g) \to 2 H F_{(g)} + X e_{(g)}, Δ H^{o} = - 430 k J

\to H - H = 435 k J / m o l \to H - F = 565 k J / m o l

X e - F

Energy required to break all four C-H bonds in methane is 1662 kJ/mol. Calculate the average bond enthalpy for methane molecule.

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