1. Write the reaction and initial moles

The reaction is: $\mathrm{CO(g)} + \mathrm{H_2O(g)} \rightleftharpoons \mathrm{CO_2(g)} + \mathrm{H_2(g)}$

Initially, in a $10\,\text{L}$ container:

• $\mathrm{CO} = 1$ mol
• $\mathrm{H_2O} = 1$ mol
• $\mathrm{CO_2} = 0$ mol
• $\mathrm{H_2} = 0$ mol

2. Use the given extent of reaction

It is given that $40\%$ of water by mass reacts.

Initial mass of water $= 1$ mole $\times 18 = 18\,\text{g}$.

Mass of water reacted: $0.40 \times 18 = 7.2\,\text{g}$

Moles of water reacted: $\frac{7.2}{18} = 0.4\,\text{mol}$

So, extent of reaction $x = 0.4$ mol.

3. Find equilibrium moles

Using stoichiometry:

$\mathrm{CO} + \mathrm{H_2O} \rightleftharpoons \mathrm{CO_2} + \mathrm{H_2}$

Change in moles:

• $\mathrm{CO}$ decreases by $x = 0.4$
• $\mathrm{H_2O}$ decreases by $0.4$
• $\mathrm{CO_2}$ increases by $0.4$
• $\mathrm{H_2}$ increases by $0.4$

Therefore, equilibrium moles are:

• $\mathrm{CO} = 1 - 0.4 = 0.6$
• $\mathrm{H_2O} = 1 - 0.4 = 0.6$
• $\mathrm{CO_2} = 0.4$
• $\mathrm{H_2} = 0.4$

4. Find equilibrium concentrations

Since volume $= 10\,\text{L}$,

$[\mathrm{CO}] = \frac{0.6}{10} = 0.06$ $[\mathrm{H_2O}] = \frac{0.6}{10} = 0.06$ $[\mathrm{CO_2}] = \frac{0.4}{10} = 0.04$ $[\mathrm{H_2}] = \frac{0.4}{10} = 0.04$

5. Calculate $K_c$

For the reaction, $K_c = \frac{[\mathrm{CO_2}][\mathrm{H_2}]}{[\mathrm{CO}][\mathrm{H_2O}]}$

Substitute values: $K_c = \frac{(0.04)(0.04)}{(0.06)(0.06)}$

$K_c = \frac{16 \times 10^{-4}}{36 \times 10^{-4}} = \frac{16}{36} = \frac{4}{9} \approx 0.444$

6. Compute $K_c \times 10^2$

$K_c \times 10^2 = 0.444 \times 100 = 44.4$

Nearest integer: $\boxed{44}$

7. Comparison with stored answer

Derived answer = $44$

Stored correct answer = $44$

So the derived answer agrees with the stored answer.