AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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
${XeF}_{4} ⟶ {Xe}^{+} + F^{-} + F_{2} + F$ will be:

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$367 kcal / mole$

$425 kcal / mole$

$292 kcal / mole$

$392 kcal / mole$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

The given Value

${XeF}_{4} ⟶ Xe + 4 F Δ H = 34 \times 4 = 136$

$Xe ⟶ {Xe}^{-} + e^{-} Δ H = 279$

$F + e^{-} ⟶ F^{-} Δ H = - 85$

$F_{2} ⟶ 2 F Δ H = 38$

Totally the equation

${XeF}_{4} ⟶ {Xe}^{+} + F^{-} + F_{2} + F$

$Δ H = 136 + 279 - 85 - 38 = 292$

the enthalpy change for the reaction

${XeF}_{4} ⟶ {Xe}^{+} + F^{-} + F_{2} + F$ will be: $292 kcal / mole$

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!