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AlladinOne [14]

1 year ago

13

"Inert" xenon actually forms many compounds, especially with highly electronegative fluorine. The ΔH° values for xenon difluorid

e, tetrafluoride, and hexafluoride are -2105, - 2284, and - 2402 kJ/mol, respectively. Find the average bond energy of the Xe¬F bonds in each fluoride.

1 answer:

svp [43]1 year ago

3 0

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.

learn more about bond energy:

brainly.com/question/11653058

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