# At 700 K, equilibrium constant for the reaction

Question:

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 $700 \mathrm{~K}$ ?

Solution:

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, $\left[\mathrm{H}_{2}\right]=\left[\mathrm{I}_{2}\right]=0.068 \mathrm{~mol} \mathrm{~L}^{-1}$.