Question

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

• A. $400\, kJ\, mol ^{-1}$
• B. $300\, kJ\, mol ^{-1}$
• C. $200\, kJ\, mol ^{-1}$
• D. $800\, kJ\, mol ^{-1}$

Answer 1

Answer: (D)

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

Answer 2

Formation of $X Y$ is shown as $X_{2}+Y_{2} \rightarrow 2 X Y$ $\Delta H=(B E)_{X-X}+(B E)_{Y-Y}-2(B E)_{X-Y}$ If $(B E)$ of $X-Y=a$ then $(B E)$ or $(X-X)=a$ and $(B E)$ of $(Y-Y)=\frac{a}{2}$ $\Delta H_{f}(X-Y)=-200\, k J$ $\therefore-400$ (for $2$ moles $XY$) $=a+\frac{a}{2}-2 a$ $-400=-\frac{a}{2}$ $a=+800\, kJ$ The bond dissociation energy of $X_{2}=800\, k J\, mol ^{-1}$