If the bond dissociation energies of XY,X2 and Y2 (all diatomic molecules) are in the ratio of 1:1:0.5 and ΔfH for the formation of XY is −200kJmol−1. The bond dissociation energy of X2, will be

# If the bond dissociation energies of $\mathrm{XY},{\mathrm{X}}_{2}$ and ${\mathrm{Y}}_{2}$ (all diatomic molecules) are in the ratio of $1:1:0.5$ and ${\mathrm{\Delta }}_{\mathrm{f}}H$ for the formation of XY is $-200\mathrm{kJ}{\mathrm{mol}}^{-1}$. The bond dissociation energy of ${X}_{2}$, will be

1. A

$300\mathrm{kJ}{\mathrm{mol}}^{-1}$

2. B

$400\mathrm{kJ}{\mathrm{mol}}^{-1}$

3. C

$100\mathrm{kJ}{\mathrm{mol}}^{-1}$

4. D

$800\mathrm{kJ}{\mathrm{mol}}^{-1}$

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### Solution:

For the formation of XY, we have $\frac{1}{2}{X}_{2}\left(g\right)+\frac{1}{2}{Y}_{2}\left(g\right)⟶XY\left(g\right)$

In terms of bond enthalpies, we have

${\mathrm{\Delta }}_{\mathrm{f}}{H}^{\circ }=\frac{1}{2}{\epsilon }_{{\mathrm{X}}_{2}}+\frac{1}{2}{\epsilon }_{{\mathrm{Y}}_{2}}-{\epsilon }_{\mathrm{XY}}=\frac{1}{2}{\epsilon }_{{\mathrm{X}}_{2}}+\frac{1}{2}\left(\frac{1}{2}{\epsilon }_{{\mathrm{X}}_{2}}\right)-{\epsilon }_{{\mathrm{X}}_{2}}=-\frac{1}{4}{\epsilon }_{{\mathrm{X}}_{2}}$

Hence  ${\epsilon }_{{\mathrm{X}}_{2}}=-4\left({\mathrm{\Delta }}_{\mathrm{f}}{H}^{\circ }\right)=-4\left(-200\mathrm{kJ}{\mathrm{mol}}^{-1}\right)=800\mathrm{kJ}{\mathrm{mol}}^{-1}$  