Question

31. XeF (g) + H2(g) 2HF(g) + Xe(g); AH = – 430 kJ
using the following bond energies : H-H = 435 kJ/mol, H-F = 565 kJ/mol
Calculate the average bond energy of Xe-F in XeF2​

Answer 1

The average bond energy of Xe-F in $XeF_2​$ is 132.5kJ

Explanation:

The balanced chemical reaction is,

$XeF_2(g)+H_2(g)\rightarrow 2HF(g)+Xe(g)$   $\Delta H=-430kJ$

The expression for enthalpy change is,

$\Delta H=\sum [n\times B.E(reactant)]-\sum [n\times B.E(product)]$

$\Delta H=[(n_{XeF_2}\times B.E_{XeF})+(n_{H_2}\times B.E_{H_2}) ]-[(n_{HF}\times B.E_{HF})+(n_{Xe}\times B.E_{Xe})]$

$\Delta H=[(2\times B.E_{Xe-F}+(1\times B.E_{H-H}) ]-[(2\times B.E_{H-F})+(1\times 0)]$

where,

n = number of moles

Now put all the given values in this expression, we get

$-430kJ=[(2\times B.E_{Xe-F})+(1\times 435)]-[(2\times 565)]$

$B.E_{Xe-F}=132.5kJ$

Therefore, the average bond energy of Xe-F in $XeF_2$ is 132.5kJ

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