• # question_answer AB, ${{A}_{2}}$ and ${{B}_{2}}$are diatomic molecules. If the bond enthalpies of ${{A}_{2}},$AB and ${{B}_{2}}$are in the ratio $1:1\text{ }:0.5$ and enthalpy of formation of AB from ${{A}_{2}}$ and ${{B}_{2}}$is $-100kJ\,mo{{l}^{-1}}$. What is the bond energy of ${{A}_{2}}:$ : A)  $200\,kJ\,mo{{l}^{-1}}$                    B)  $100\,kJ\,mo{{l}^{-1}}$C)  $300\,kJ\,mo{{l}^{-1}}$       D)  $400\,kJ\,mo{{l}^{-1}}$

Let bond energy of ${{A}_{2}}$ be x then bond energy of AB is also x and bond energy of ${{B}_{2}}$ is $x/2$. Enthalpy of formation of AB is$~-100\text{ }KJ/mole$: ${{A}_{2}}+{{B}_{2}}\to 2AB;$      $\frac{1}{2}{{A}_{2}}+\frac{1}{2}{{B}_{2}}\to AB;\,\Delta =-100KJ$ or$-100=\left( \frac{x}{2}+\frac{x}{4} \right)-x$$\therefore$                               $-100=\frac{2x+x-4x}{4}\therefore x=400KJ$