A mixture of one mole of H$$_2$$O and 1 mole of CO is taken in a 10 litre container and heated to 725 K. At equilibrium 40% of water by mass reacts with carbon monoxide according to the equation:
$$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)$$
The equilibrium constant $$K_C \times 10^2$$ for the reaction is _______ (Nearest integer)

The reaction is: $$\text{CO}(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g)$$

Initial moles: CO = 1 mol, H$$_2$$O = 1 mol, CO$$_2$$ = 0, H$$_2$$ = 0

At equilibrium: 40% of water reacts, so 0.4 mol of H$$_2$$O reacts.

$$\text{CO} = 1 - 0.4 = 0.6 \text{ mol}$$

$$\text{H}_2\text{O} = 1 - 0.4 = 0.6 \text{ mol}$$

$$\text{CO}_2 = 0.4 \text{ mol}$$

$$\text{H}_2 = 0.4 \text{ mol}$$

Concentrations (Volume = 10 L):

$$[\text{CO}] = [\text{H}_2\text{O}] = \frac{0.6}{10} = 0.06 \text{ M}$$

$$[\text{CO}_2] = [\text{H}_2] = \frac{0.4}{10} = 0.04 \text{ M}$$

Equilibrium constant:

$$K_C = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]} = \frac{0.04 \times 0.04}{0.06 \times 0.06} = \frac{0.0016}{0.0036} = \frac{4}{9}$$

$$K_C \times 10^2 = \frac{4}{9} \times 100 = 44.44 \approx 44$$