Consider the following equilibrium,

$$CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(g)$$

0.1 mol of CO along with a catalyst is present in a 2 dm$$^3$$ flask maintained at 500 K. Hydrogen is introduced into the flask until the pressure is 5 bar and 0.04 mol of $$CH_3OH$$ is formed. The $$K_p^0$$ is _______ $$\times 10^{-3}$$. (nearest integer)

Given $$R = 0.08$$ dm$$^3$$ bar K$$^{-1}$$ mol$$^{-1}$$

Assume only methanol is formed as the product and the system follows ideal gas behaviour.

The equilibrium to be studied is
$$CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(g) \qquad -(1)$$

For this reaction the equilibrium constant in terms of pressure is
$$K_p = \frac{P_{CH_3OH}}{P_{CO}\;P_{H_2}^{\,2}} \qquad -(2)$$

Data supplied at 500 K in a rigid 2 dm$$^3$$ flask:
  • initial moles of $$CO = 0.10$$ mol
  • hydrogen is added so that the final total pressure is 5 bar
  • at equilibrium $$0.04$$ mol of $$CH_3OH$$ has been formed.

1. Establish the equilibrium mole numbers

Let $$x$$ be the moles of $$CH_3OH$$ produced. It is given that $$x = 0.04$$ mol.
By the stoichiometry of (1):
  • $$CO$$ consumed $$= x = 0.04$$ mol
  • $$H_2$$ consumed $$= 2x = 0.08$$ mol

Hence, at equilibrium
$$n_{CO}^{eq} = 0.10 - 0.04 = 0.06 \text{ mol}$$
$$n_{CH_3OH}^{eq} = 0.04 \text{ mol}$$
$$n_{H_2}^{eq} =\;?$$ (to be found next)

2. Find the total number of moles at equilibrium

The flask is finally at 5 bar. Using the ideal-gas equation $$PV = nRT$$:
$$n_{tot}^{eq} = \frac{P\,V}{R\,T} = \frac{5 \times 2}{0.08 \times 500} = \frac{10}{40} = 0.25 \text{ mol} \qquad -(3)$$

Therefore
$$n_{H_2}^{eq} = n_{tot}^{eq} - n_{CO}^{eq} - n_{CH_3OH}^{eq} = 0.25 - 0.06 - 0.04 = 0.15 \text{ mol}$$

3. Calculate the partial pressures

Using $$P_i = \frac{n_i RT}{V}$$ with $$RT/V = \frac{0.08 \times 500}{2}=20$$ bar mol$$^{-1}$$:

$$P_{CO}=0.06 \times 20 = 1.2 \text{ bar}$$
$$P_{H_2}=0.15 \times 20 = 3.0 \text{ bar}$$
$$P_{CH_3OH}=0.04 \times 20 = 0.8 \text{ bar}$$

4. Evaluate $$K_p$$

Insert the partial pressures in (2):
$$K_p = \frac{0.8}{1.2 \,(3.0)^2} = \frac{0.8}{1.2 \times 9} = \frac{0.8}{10.8} = 7.407 \times 10^{-2}$$

5. Express in the asked form

$$K_p = 7.407 \times 10^{-2} = 74.07 \times 10^{-3}$$

To the nearest integer, $$K_p^0 = 74 \times 10^{-3}$$.