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 :

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

Tardigrade Admin · Accepted Answer

We need to calculate the Δ H for the following reaction- XeF 4 longrightarrow Xe ++ F -+ F 2+ F Energy Absorbed: 4 × 34=136 kcal / mol

Energy released: 85 kcal/mol (E.G.E)

Δ H=415-123=292

292 kCal mol ^{-1}

383 kCal mol $^{- 1}$

521 k Cal mol $^{- 1}$

528 k Cal mol $^{- 1}$

We need to calculate the $Δ H$ for the following reaction-

$X e F_{4} ⟶ X e^{+} + F^{-} + F_{2} + F$

Energy Absorbed:

$4 \times 34 = 136 k c a l / m o l$

Energy released:

85 kcal/mol (E.G.E)

$Δ H = 415 - 123 = 292$

Q. Calculate the enthalpy change (in kcal) for the reaction- XeF4​⟶Xe+F−+F2​+F The average Xe−F bond enthalpy is 34kCal/mol, first I.E. of Xe is 279kCal/mol, electron gain enthalpy of F is −85kCal/mol and the bond dissociation enthalpy of F2​ is 38kCal/mol:

292 kCal mol ^{-1}

383 kCal mol−1

521 k Cal mol −1

528 k Cal mol−1

We need to calculate the ΔH for the following reaction- XeF4​⟶Xe++F−+F2​+F Energy Absorbed: 4×34=136kcal/mol Energy released: 85 kcal/mol (E.G.E) ΔH=415−123=292

383 kCal mol $^{- 1}$

521 k Cal mol $^{- 1}$

528 k Cal mol $^{- 1}$

We need to calculate the $Δ H$ for the following reaction-

$X e F_{4} ⟶ X e^{+} + F^{-} + F_{2} + F$

Energy Absorbed:

$4 \times 34 = 136 k c a l / m o l$

Energy released:

85 kcal/mol (E.G.E)

$Δ H = 415 - 123 = 292$