To find the bond dissociation energy of
X2, let's assume the bond dissociation energy of
X2 is
a‌kJ∕mol Therefore,
BE(X2)=a‌kJ∕mol.
Given that the bond dissociation energy ratios are
1:0.5:1, then:
‌BE(Y2)=0.5a‌kJ∕mol‌BE(XY)=a‌kJ∕molWe know that the formation reaction of
XY is:
‌X2+‌Y2⟶XY,‌‌∆H=−200‌kJ∕molUsing the enthalpy change equation:
∆rH=BE(‌ Reactants ‌)−BE(‌ Products ‌)Substituting the bond dissociation energies into the equation gives:
∆rH=‌‌BE(X2)+‌‌BE(Y2)−BE(XY)Plugging in the known values:
−200=‌+‌−aSimplifying:
−200=‌+‌−a=‌+‌−aCombine the terms:
‌−200=‌+‌−‌‌−200=‌=‌Solving for
a :
‌−200=‌‌−200×2=−0.75a‌−400=−0.75aDividing both sides by -0.75 :
a=‌=800‌kJ∕molThus, the bond dissociation energy of
X2 is
800‌kJ∕mol.