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

The reaction involves a decrease in the number of moles of gaseous products compared to the gaseous reactants, indicating that an increase in pressure favors the side of the reaction with fewer moles (more moles on the right-hand side).

To find the amount of carbon monoxide, we first calculate the total moles at equilibrium:

Total moles = 0.430 mol (CO) + 0.860 mol (H2) - 0.110 mol (CH3OH) = 1.180 mol

Using the reaction stoichiometry, we note the mole ratio: At equilibrium, if x is the amount of CO reacted, we have:

CO remaining = 0.430 - x H2 remaining = 0.860 - 2x

The equilibrium expression related to the total pressure can be used to find x after determining the total pressure contribution of each gas.

The equilibrium expression relates to the total pressure (P) calculated from the total moles:

o P_{CO} = (n_{CO}/n_{total}) imes P_{total}

By identifying the amounts, we find: P_{CO} = (0.320 mol / 1.07 mol) × 250 kPa = 74.5 kPa.

The equilibrium constant expression (Kc) is given by:

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

Concentrations should be expressed in mol/L.

Using the previously given partial pressures: Kc = (5.45 kPa) / (125 kPa * P_H2^2),

Reorganizing provides: P_H2^2 = (5.45 kPa) / (1.15 × 10^4 kPa^-2)

Calculating gives the value for P_H2.

The units for Kc' when relating concentrations will be kPa^-2.