Chemistry - Thermochemistry Question with Solution | TestHub

Let the bond dissociation energy of $X_2=a\mathrm{kJ}/\mathrm{mol}$ $\text{ i.e. }\mathrm{BE}\left(X_2\right)=a\mathrm{kJ}/\mathrm{mol}$ Then, $\mathrm{BE}\left({\mathrm{Y}}_2\right)=0.5a$ and $\mathrm{BE}(XY)=a\mathrm{kJ}/\mathrm{mol}$ Given, $\frac{1}{2}{X}_2+\frac{1}{2}{Y}_2⟶XY,\Delta H=-200\mathrm{kJ}/\mathrm{mol}$ $\begin{array}{rl}{\Delta }_{r}H& =\mathrm{BE}(\text{ Rreactants })-\mathrm{BE}(\text{ Product })\\ {\Delta }_{r}H& =\left[\frac{1}{2}\mathrm{BE}\left(X_2\right)+\frac{1}{2}\mathrm{BE}\left(Y_2\right)\right]-\mathrm{BE}(XY)\\ \therefore -200& =\frac{a}{2}+\frac{0.5a}{2}-a\\ a& =\frac{200}{0.25}=800\mathrm{kJ}/\mathrm{mol}\end{array}$