At 700 K, equilibrium constant for the reaction
$\mathrm{H}_{2(g)}+\mathrm{I}_{2(g)} \longleftrightarrow 2 \mathrm{HI}_{(g)}$
is $54.8$. If $0.5 \mathrm{molL}^{-1}$ of $\mathrm{HI}_{(g)}$ is present at equilibrium at $700 \mathrm{~K}$, what are the concentration of $\mathrm{H}_{2(g)}$ and $\mathrm{I}_{2(g)}$ assuming that we initially started with $\mathrm{HI}_{(g)}$ and allowed it to reach equilibrium at $700 \mathrm{~K}$ ?
It is given that equilibrium constant $K_{\mathrm{c}}$ for the reaction
$\mathrm{H}_{2(g)}+\mathrm{I}_{2(g)} \longleftrightarrow 2 \mathrm{HI}_{(g)}$ is 54.8.
Therefore, at equilibrium, the equilibrium constant $K_{\mathrm{C}}^{\prime}$ for the reaction
$2 \mathrm{HI}_{(\mathrm{g})} \longleftrightarrow \mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})}$ will be $\frac{1}{54.8} .$
$[\mathrm{HI}]=0.5 \mathrm{molL}^{-1}$
Let the concentrations of hydrogen and iodine at equilibrium be $x \mathrm{molL}^{-1}$
$\left[\mathrm{H}_{2}\right]=\left[\mathrm{I}_{2}\right]=x \mathrm{~mol} \mathrm{~L}^{-1} .$
Therefore, $\frac{\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]}{[\mathrm{HI}]^{2}}=K_{\mathrm{C}}^{\prime}$
$\Rightarrow \frac{x \times x}{(0.5)^{2}}=\frac{1}{54.8}$
$\Rightarrow x^{2}=\frac{0.25}{54.8}$
$\Rightarrow x=0.06754$
$x=0.068 \mathrm{molL}^{-1}$ (approximately)
Hence, at equilibrium, $\left[\mathrm{H}_{2}\right]=\left[\mathrm{I}_{2}\right]=0.068 \mathrm{~mol} \mathrm{~L}^{-1}$.