If the bond enthalpies of X2,Y2 and XY are in the ratio 1:2:4 and delta Hformation of XY = -200kJ,find the bond enthalpy of Y2

Let us consider that,

Bond Enthalpies(BE) of X₂=a

 

Bond Enthalpies(BE) of Y₂=2a

 

Bond Enthalpies(BE) of XY=4a

 

Reaction can be written as:

 

1/2X₂+1/2Y₂ ...................> XY

 

ΔH=BE of Reactants - BE of products

 

=1/2 BE of X₂+1/2 BE of Y₂ - BE of XY

 

-200=(a/2+2a/2)-4a

 

a=80 kJ/mole