A study of the water-gas shift reaction (see Problem 17.37) was made in which equilibrium was reached with $\mathbf{\left[}\mathbf{\text{CO}}\mathbf{\right]}\mathbf{=}$$\mathbf{\left[}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}}\mathbf{\right]}\mathbf{=}\mathbf{\left[}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\right]}\mathbf{=}\mathbf{0}\mathbf{.10}\mathbf{\text{M}}$ and$\mathbf{\left[}{\mathbf{\text{CO}}}_{\mathbf{2}}\mathbf{\right]}\mathbf{=}\mathbf{0}\mathbf{.40}\mathbf{\text{M}}$ . Afterdata-custom-editor="chemistry" $\mathbf{0}\mathbf{.60}\mathbf{}\mathbf{\text{mol}}$ ofdata-custom-editor="chemistry" ${\mathbf{H}}_{\mathbf{2}}$ is added to the$\mathbf{2}\mathbf{.0}$ -$\mathbf{L}$ container and equilibrium is re-established, what are the new concentrations of all the components?

Chemical equilibrium is the state in which no net change in the amounts of reactants and products occurs during a reversible chemical reaction.

$\begin{array}{c}\text{CO}\left(\mathrm{g}\right)+{\text{H}}_2\text{O}\left(\mathrm{g}\right)\rightleftharpoons {\text{CO}}_2\left(\mathrm{g}\right)+{\text{H}}_2\left(\mathrm{g}\right)\\ \left[\text{CO}\right]=0.10\mathrm{M}\\ \left[{\text{H}}_2\right]=0.10\mathrm{M}\\ \left[{\text{H}}_2\text{O}\right]=0.10\mathrm{M}\\ \left[{\text{CO}}_2\right]=0.40\end{array}$

The equilibrium constant for the given reaction is expressed as:

${\mathrm{K}}_{\text{c}}=\frac{\left[{\text{CO}}_2\right]{\text{H}}_2]}{\left[\text{CO}\right]{\text{H}}_2\text{O}]}$

As we have all needed values, we can calculate it:

data-custom-editor="chemistry" $\begin{array}{l}{\mathrm{K}}_{\text{c}}=\frac{\left[{\text{CO}}_2\right]\left[{\text{H}}_2\right]}{\left[\text{CO}\right]\left[{\text{H}}_2\text{O}\right]}\\ {\mathrm{K}}_{\text{c}}=\frac{0.4\mathrm{M}\cdot 0.1\mathrm{M}}{0.1\mathrm{M}\cdot 0.1\mathrm{M}}\\ {\mathrm{K}}_{\text{c}}=4\end{array}$

The concentration of data-custom-editor="chemistry" ${\mathrm{H}}_2$ added:

data-custom-editor="chemistry" $\begin{array}{c}\left[{\text{H}}_2\right]=\frac{\text{n}}{\text{V}}\\ \left[{\text{H}}_2\right]=\frac{0.6\text{mol}}{2\text{L}}\end{array}$

Therefore,

$\left[{\text{H}}_2\right]=0.3M$.

$\mathrm{x}$-the change in the concentration of ${\text{CO}}_2$.

The equilibrium concentrations by using$\mathrm{x}$:

data-custom-editor="chemistry" $\begin{array}{c}\text{CO}=0.10+x\\ {\text{H}}_2\text{O}=0.10+x\\ {\text{CO}}_2=0.40-x\\ {\text{H}}_2=0.40-x\end{array}$

Write the equation for the equilibrium constant and use the equilibrium concentrations to solve $\mathrm{x}$:

data-custom-editor="chemistry" $\begin{array}{c}{K}_{\text{c}}=\frac{\left[{\text{CO}}_2\right]\left[{\text{H}}_2\right]}{\left[\text{CO}\right]\left[{\text{H}}_2\text{O}\right]}\\ 4=\frac{{(0.4-x)}^2}{{(0.1+x)}^2}\sqrt{4.0}=\sqrt{\frac{{(0.4-x)}^2}{{(0.1+x)}^2}}\\ 2=\frac{(0.4-x)}{(0.1+x)}\\ 0.2+2x=0.4-x\\ 3x=0.4-0.2\\ 3x=0.2\\ x=\frac{0.2}{3}\end{array}$

Therefore, $\mathrm{x}=0.067$.

So, now we can calculate the equilibrium concentrations:

$\begin{array}{c}\left[\text{CO}\right]=\left[{\text{H}}_2\text{O}\right]\\ \left[\text{CO}\right]=\left[{\text{H}}_2\text{O}\right]=0.10+x\\ \left[\text{CO}\right]=\left[{\text{H}}_2\text{O}\right]=0.10+0.067\\ \left[\text{CO}\right]=\left[{\text{H}}_2\text{O}\right]=0.167M\\ \left[{\text{CO}}_2\right]=\left[{\text{H}}_2\right]\\ \left[{\text{CO}}_2\right]=\left[{\text{H}}_2\right]=0.40-x\\ \left[{\text{CO}}_2\right]=\left[{\text{H}}_2\right]=0.40-0.067\\ \left[{\text{CO}}_2\right]=\left[{\text{H}}_2\right]=0.33M.\end{array}$

Therefore, the new equilibrium concentrations:

$\left[\text{CO}\right]=\left[{\text{H}}_2\text{O}\right]=0.17M\backslash \left[{\text{CO}}_2\right]=\left[{\text{H}}_2\right]=0.33\text{M}$