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QUESTION 23 Using the provided bond dissociation energies, determine the change in enthalpy (AH) in kilojoules for the follow
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Answer #1

The given reaction is     Y2        +      2 X2 --------->           2YX2

                                      Y=Y               2 X-X                      2 X-Y-X

The change in enthalpy, \Delta H of the reaction is given

\DeltaHrxn = Bond dissociation energy of reactants - bond dissociation energy of products

        = Bond dissociation energy of Y=Y + 2* Bond dissociation energy of (X-X) - 4*Bond dissociation energy of (x-y)

          = 449 + 2*272 - 4*174 = 297 KJ/ mole

The change in enthalpy, \Delta H of the reaction is 297 KJ/mole

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