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 $-100\,kJ\,mol^{- 1}.$ . The bond dissociation energy of $X_{2}$ will be_____ $kJ\,mol^{- 1}.$

Answer 1

The ratio of bond dissociation energies of $X_{2},Y_{2}$ and $XY$ is $1:0.5:1$
$\therefore $ Bond dissociation energies of $X_{2},Y_{2}$ and $XY$ are:
$\Delta H\left(\right.X-X\left.\right)=x$
$\Delta H\left(\right.Y-Y\left.\right)=0.5x$
$\Delta H\left(\right.X-Y\left.\right)=x$
The chemical equation for formation of $XY$ is :
$\frac{1}{2}X_{2}+\frac{1}{2}Y_{2} \rightarrow XY$
$\Delta H=\sum m\Delta H\left(\right.\text{ reactant bonds }\left.\right)$
$-\Sigma n\Delta H\left(\right.\text{ product bonds }\left.\right)$
$\therefore -100=\left[\frac{1}{2} \Delta H \left(\right. X - X \left.\right) + \frac{1}{2} \Delta H \left(\right. Y - Y \left.\right)\right]-\Delta H\left(\right.X-Y\left.\right)$ $-100=\left[\frac{1}{2} x + \frac{1}{2} \times 0 . 5 x\right]-x$
$-100=\left[\frac{x}{2} + \frac{x}{4}\right]-x$
$-100=\frac{6 x}{8}-x$
$-100=\frac{6 x - 8 x}{8}$
$-100=\frac{- 2 x}{8}$
$\therefore x=\frac{100 \times 8}{2}=400$
$\therefore $ Bond dissociation energy of $X_{2}=400\,kJ\,mol^{- 1}.$