The water-gas shift reaction plays a central role in the chemical methods for obtaining cleaner fuels from coal:

${\mathbf{CO}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\rightleftharpoons }{\mathbf{CO}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}$

At a given temperature, Kp = 2.7. If 0.13 mol of CO, 0.56 mol of ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$, 0.62 mol of ${\mathbf{CO}}_{\mathbf{2}}$, and 0.43 mol of ${\mathbf{H}}_{\mathbf{2}}$ are put in a 2.0-L flask, in which direction, does the reaction proceed?

Rewrite the reaction as follows:

${\mathbf{CO}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\rightleftharpoons }{\mathbf{CO}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}$

The expression for thereaction quotientis ratio of concentrations of products to concentration of reactants, with each concentration term raised to power equal to its stoichiometric coefficient.

The expression for thereaction quotient $\left({\mathrm{Q}}_{\mathrm{c}}\right)$ is,

${\mathrm{Q}}_{\mathrm{c}}=\frac{\left[{\mathrm{CO}}_2\right]\left[{\mathrm{H}}_2\right]}{\left[\mathrm{CO}\right]\left[{\mathrm{H}}_2\mathrm{O}\right]}.........................\left(1\right)$

The number of CO is 0.13 mol, ${\mathrm{H}}_2\mathrm{O}$ is 0.56 mol, ${\mathrm{CO}}_2$is 0.62 mol and ${\mathrm{H}}_2$ is 0.43 mol. The volume of the flask is 2.0 L.

Divide the number of moles of each gas by the vessel volume, to obtain the concentration.

$$\begin{gathered} \left[{\mathrm{CO}}_2\right]=\frac{0.62\mathrm{mol}}{2.0\mathrm{L}} \\ =0.31\mathrm{M} \\ \left[{\mathrm{H}}_2\right]=\frac{0.43\mathrm{mol}}{2.0\mathrm{L}} \\ =0.215\mathrm{M} \\ \left[\mathrm{CO}\right]=\frac{0.13\mathrm{mol}}{2.0\mathrm{L}} \\ =0.065\mathrm{M} \\ \left[{\mathrm{H}}_2\mathrm{O}\right]=\frac{0.56\mathrm{mol}}{2.0\mathrm{L}} \\ =0.28\mathrm{M} \end{gathered}$$

Substitute the values in relation (1), and solve for $\left({\mathrm{Q}}_{\mathrm{c}}\right)$:

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{c}}=\frac{(0.31\mathrm{M})(0.215\mathrm{M})}{(0.065\mathrm{M})(0.28\mathrm{M})} \\ =3.7 \end{gathered}$$

The relation between equilibrium constants in terms of partial pressures $\left({\mathrm{K}}_{\mathrm{p}}\right)$and equilibrium constants in terms of concentration $\left({\mathrm{K}}_{\mathrm{c}}\right)$ is given below:

Here, R is gas constant $(0.0821\mathrm{L}\mathrm{atm}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1})$. T is absolute temperature and$∆{\mathrm{n}}_{\mathrm{gas}}$ is change in number of moles of gaseous reactants.

In the reaction the number of moles of gaseous products is 2 and number of moles of gaseous reactants is 2. So, the value of $∆{\mathrm{n}}_{\mathrm{gas}}$ is (2-2) or 0.

So, the value of the reaction is equal to $\left({\mathrm{K}}_{\mathrm{p}}\right)$or 2.7.

If the value of role="math" localid="1655095120341" $\left({\mathrm{Q}}_{\mathrm{c}}\right)$ is equal to role="math" localid="1655095187286" $\left({\mathrm{K}}_{\mathrm{c}}\right)$, the reaction will be at equilibrium. Suppose the value of $\left({\mathrm{Q}}_{\mathrm{c}}\right)$ is smaller than $\left({\mathrm{K}}_{\mathrm{c}}\right)$, the reaction proceed towards product side to attain equilibrium. Suppose the value of $\left({\mathrm{Q}}_{\mathrm{c}}\right)$ is greater than$\left({\mathrm{K}}_{\mathrm{c}}\right)$ , the reaction proceeds towards reactant side to attain the equilibrium.

Here, $\left({\mathrm{Q}}_{\mathrm{c}}\right)$ is greater than $\left({\mathrm{K}}_{\mathrm{c}}\right)$. Thus, the reaction shifts towards reactant.