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Q.

Calculate the bond enthalpy of Xe –F bond as given in the equation
${XeF}_{4} (g) \to {Xe}^{+} (g) + F^{-} (g) + F_{2} (g) + F (g)$
$Δ_{r} H = 292 K cal {mol}^{- 1}$
Ionisation energy of Xe = 279 K cal / mol
Bond energy (F–F) = 38 Kcal/mol
Electron affinity of F = –85 Kcal/mol

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a

34 kcal/mol

b

3.4 kcal/mol

c

8.5 kcal/mol

d

24 kcal/mol

answer is B.

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Detailed Solution

${XeF}_{4} (g) \to {Xe}^{+} (g) + F^{-} (g) + F_{2} (g) + F (g); ∆ H_{r} = 292 Kcal / mole$

${BE}_{Xe - F} = \frac{∆ H_{r} - IE - EA - {BE}_{F_{2}}}{4} = \frac{292 - 279 + 38 + 85}{4} = \frac{136}{4} = 34 Kcal / mole$

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Calculate the bond enthalpy of Xe –F bond as given in the equationXeF4(g)→Xe+(g)+F−(g)+F2(g)+F(g)ΔrH=292K cal mol−1Ionisation energy of Xe = 279 K cal / mol Bond energy (F–F) = 38 Kcal/molElectron affinity of F = –85 Kcal/mol