Question

The bond dissociation energies of $X_2 , Y_2$ and $XY$ are in the ratio of $1 : 0.5 :1. \Delta H$ for the formation of $XY$ is - $200\, kJ\, mol^{-1}$. The bond dissociation energy of $X_2$ will be

• A. $400 \, kJ \, mol^{-1}$
• B. $200\, kJ \, mol^{-1}$
• C. $800\, kJ \, mol^{-1}$
• D. $100 \, kJ \, mol^{-1}$

Answer 1

Answer: (C)

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

Answer 2

The correct option is(C): $800\, kJ \, mol^{-1}$

The reaction for $\Delta_{ f } H ^{\circ}( XY )$
$\frac{1}{2} x_{2}(g)+\frac{1}{2} Y_{2}(g) \longrightarrow X Y(g)$

Bond energies of $X_{2}, Y_{2}$ and $X Y$ are $x, \frac{x}{2}, x$ respectively

$\Delta H=\left(\frac{x}{2}+\frac{x}{4}\right)-X=-200$

On solving, we get

$\Rightarrow-\frac{x}{2}+\frac{x}{4}=-200$
$\Rightarrow X=800 \,kJ / mole$