The bond dissociation energy of X2,Y2 and XY are in the ratio of 1:0.5:1.∆H for the formation of XY is −200kJ∕mol. The bond dissociation energy of X2 will be
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∕mol BE(XY)=akJ∕mol We know that the formation reaction of XY is:
1
2
X2+
1
2
Y2⟶XY,∆H=−200kJ∕mol Using the enthalpy change equation: ∆rH=BE( Reactants )−BE( Products ) Substituting the bond dissociation energies into the equation gives: ∆rH=
1
2
BE(X2)+
1
2
BE(Y2)−BE(XY) Plugging in the known values: −200=
a
2
+
0.5a
2
−a Simplifying: −200=
a
2
+
0.25a
2
−a=
a
2
+
a
4
−a Combine the terms: −200=
a
2
+
0.25a
2
−
2a
2
−200=
0.5a+0.25a−2a
2
=
−0.75a
2
Solving for a : −200=
−0.75a
2
−200×2=−0.75a −400=−0.75a Dividing both sides by -0.75 : a=
400
0.75
=800kJ∕mol Thus, the bond dissociation energy of X2 is 800kJ∕mol.