BITSAT 2024 Shift 1 Paper

To find the bond dissociation energy of $X_2$, let's assume the bond dissociation energy of $X_2$ is $a\,\mathrm{kJ}/\mathrm{mol}$ Therefore, $\mathrm{BE}(X_2)=a\,\mathrm{kJ}/\mathrm{mol}$.Given that the bond dissociation energy ratios are $1: 0.5: 1$, then:$\;\mathrm{BE}(Y_2)=0.5a\,\mathrm{kJ}/\mathrm{mol}$$\;\mathrm{BE}(XY)=a\,\mathrm{kJ}/\mathrm{mol}$We know that the formation reaction of $X Y$ is:$\;\frac{1}{2}X_2+\;\frac{1}{2}Y_2\longrightarrow XY,\;\;\Delta H=-200\,\mathrm{kJ}/\mathrm{mol}$Using the enthalpy change equation:$\Delta_r H=\mathrm{BE}(\;\text{ Reactants }\;) - \mathrm{BE}(\;\text{ Products }\;)$Substituting the bond dissociation energies into the equation gives:$\Delta_r H=\;\frac{1}{2}\mathrm{BE}(X_2)+\;\frac{1}{2}\mathrm{BE}(Y_2)-\mathrm{BE}(XY)$Plugging in the known values:$-200=\;\frac{a}{2}+\;\frac{0.5a}{2}-a$Simplifying:$-200=\;\frac{a}{2}+\;\frac{0.25a}{2}-a=\;\frac{a}{2}+\;\frac{a}{4}-a$Combine the terms:$\;-200=\;\frac{a}{2}+\;\frac{0.25a}{2}-\;\frac{2a}{2}$$\;-200=\;\frac{0.5a+0.25a-2a}{2}=\;\frac{-0.75a}{2}$Solving for $a$ :$\;-200=\;\frac{-0.75a}{2}$$\;-200 \times 2 = -0.75a$$\;-400=-0.75a$Dividing both sides by -0.75 :$a=\;\frac{400}{0.75}=800\,\mathrm{kJ}/\mathrm{mol}$Thus, the bond dissociation energy of $X_2$ is $800\,\mathrm{kJ}/\mathrm{mol}$.