The average bond energy of the Xe¬F bonds in each fluoride is 132kJ/mol.
Given:
ΔH° of xenon difluoride (XeF2) = -105 kJ/mol
ΔH° of xenon tetrafluoride (XeF4)= -284 kJ/mol
ΔH° of xenon hexafluoride (XeF6) = -402 kJ/mol
The bond energy of Xe-F in XeF2 can be calculated as follows,
As we know that
ΔH° = ΔH°(bond formed) + ΔH°(bond broken)
The chemical reaction for the formation of XeF2 can be written in such a way,
Xe (g) + F2 (g) → XeF2 (g)
= [1 mol F2 (159 kJ/mol)] + [2(-Xe-F)] - 105 kJ/mol
= 159 kJ/mol + 2(-Xe-F) - 264 kJ/mol
= 2(-Xe-F)
Xe-F = 132 kJ/mol
Thus, we concluded that the average bond energy of the Xe¬F bonds in each fluoride is 132kJ/mol.
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