If the bond dissociation energies of XY,X_{2}\, and \, Y_2 (all diatomic molecules) are in the ratio of 1:1:0.5 and \Delta H_{f}  for the formation of XY is -200 kJ mol-1 . The bond  dissociation energy of X_2 will be 

  • Option 1)

    100 kJ mol-1

  • Option 2)

    200 kJ mol-1

  • Option 3)

    800 kJ mol-1

  • Option 4)

    400 kJ mol-1

 

Answers (1)

As we learnt in 

Bond dissociation enthalpy -

It is the average of enthalpy required to dissociate the said bond present in different gaseous compound in to free atoms in gaseous state.

- wherein

N_{2}+Bond\, Energy\rightarrow 2N

 

 X2 + Y2\rightarrow2XY, \DeltaH = 2(-200)= - 400

Let x be the bond dissociation energy of X2 then

\DeltaH = -400 = Ex-x + Ey-y -2Ex-y

= x + 0.5x - 2x = - 0.5x

x=\frac{400}{0.5}=800 KJ mol^{-1}


Option 1)

100 kJ mol-1

this is incorrect option

Option 2)

200 kJ mol-1

this is incorrect option

Option 3)

800 kJ mol-1

this is correct option

Option 4)

400 kJ mol-1

this is incorrect option

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