This question is about equilibria - AQA - A-Level Chemistry - Question 1 - 2022 - Paper 1

The reaction involves gases, and an increase in pressure favors the side with fewer moles of gas.

At equilibrium, the total moles is 1.07 mol (0.430 mol CO + 0.860 mol H₂ - 0.110 mol CH₃OH). Thus, the moles of carbon monoxide can be calculated as follows:

Let x be the moles of CO at equilibrium:

Remaining H₂ = 0.860 - 2(x - (0.110/2))

Using the ideal gas law and total pressures, we find:

x + (0.860 - 2x) + 0.110 = 1.07 → 0.430 mol.

Using the formula for partial pressure, we have:

$P_{CO} = \frac{n_{CO}}{n_{total}} \times P_{total}$

Substituting the values:
$P_{CO} = \frac{0.430}{1.07} \times 250 = 100.9 \text{ kPa}$

The equilibrium constant expression is given by:

$K_c = \frac{[CH_3OH]}{[CO][H_2]^2}$

From Table 1, using the relation for equilibrium constant:

$K_c = \frac{P_{CH_3OH}}{P_{CO} imes P_{H_2}^2}$

Rearranging gives:

$P_{H_2}^2 = \frac{P_{CH_3OH}}{K_c \times P_{CO}}$

Substituting the values:

$P_{H_2} = \sqrt{\frac{5.45}{(1.15 \times 10^2) \times (125)}} = 2.07 ext{ kPa}$

The units for Kc are mol/liter, or L/mol.