You visited us 1 times! Enjoying our articles? Unlock Full Access!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

425 kcal/mol

No worries! We‘ve got your back. Try BYJU‘S free classes today!

292 kcal/mol

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

392 kcal/mol

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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

Suggest Corrections

Similar questions

F_{2}

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

F_{2}

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

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

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

C l - F

(61 k c a l m o l^{- 1})

F - F (38 k c a l m o l^{- 1})

C l - C l

(58 k c a l m o l^{- 1})

4.0 e V

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program