Given: ${\mathrm{\Delta }}_f{H}^{\circ }(\mathrm{UF},\mathrm{g})=92.0 \mathrm{kJ} {\mathrm{mol}}^{-1}$ and ${\Delta }_f{H}^{\circ }(\mathrm{U},\mathrm{g})=536 \mathrm{kJ} {\mathrm{mol}}^{-1}$. If bond energy of $F-F$ is $155 \mathrm{kJ} {\mathrm{mol}}^{-1}$, then the bond dissociation energy of  $\mathrm{U}-\mathrm{F}$ will be 

Solution:

Given information are as follows 

$\text{ (i) }\mathrm{U}\left(\mathrm{s}\right)+\frac{1}{2}{\mathrm{F}}_2\left(\mathrm{g}\right)\to \mathrm{UF}\left(\mathrm{g}\right){\mathrm{\Delta }}_{\mathrm{f}}{H}^{\circ }=92.0 \mathrm{kJ} {\mathrm{mol}}^{-1}$

$\left(\mathrm{ii}\right) \mathrm{U}\left(\mathrm{s}\right)\to \mathrm{U}\left(\mathrm{g}\right)$  ${\mathrm{\Delta }}_{\mathrm{f}}{H}^{\circ }=536 \mathrm{kJ} {\mathrm{mol}}^{-1}$

$\left(iii\right) \frac{1}{2}{\mathrm{F}}_2\left(\mathrm{g}\right)\to \mathrm{F}\left(\mathrm{g}\right)$   $\mathrm{\Delta }{H}^{\circ }=(1/2)\left(155 \mathrm{kJ} {\mathrm{mol}}^{-1}\right)$

For the bond dissociation energy, we have to calculate $\mathrm{\Delta }H$ for the reaction

$\mathrm{UF}\left(\mathrm{g}\right)\to \mathrm{U}\left(\mathrm{g}\right)+\mathrm{F}\left(\mathrm{g}\right)$

This reaction can be obtained by the following manipulations

$-\mathrm{Eq}\cdot \text{ (i) }+\mathrm{Eq}\cdot \left(\mathrm{ii}\right)+\mathrm{Eq}\cdot \text{ (iii) }$

Hence,

$\epsilon (\mathrm{U}-\mathrm{F})=[-92+536+(1/2\left)\right(155\left)\right] \mathrm{kJ} {\mathrm{mol}}^{-1}=521.5 \mathrm{kJ} {\mathrm{mol}}^{-1}$