When 3.0 moles of carbon monoxide and 1.0 moles of steam were mixed together in a container of fixed volume V, and in the presence of a suitable catalyst, the following chemical equilibrium was established at a temperature of 800 K - Leaving Cert Chemistry - Question 9 - 2020

The equilibrium constant (Kc) for the reaction is given by:

$K_c = \frac{[\text{CO}_2][\text{H}_2O]}{[\text{CO}][\text{H}_2]}$

Let x be the number of moles of CO produced at equilibrium:

Initially:

• CO: 3.0 mol
• H2: 1.0 mol

Change:

• CO: -x
• H2: -x
• CO2: +x
• H2O: +x

At equilibrium:

• CO: (3.0 - x) mol
• H2: (1.0 - x) mol
• CO2: x mol
• H2O: x mol

Setting up the Kc expression:

Given that Kc = 4.0 at 800 K:

$4.0 = \frac{x \cdot x}{(3.0 - x)(1.0 - x)}$

Solving this quadratic equation yields the number of moles of carbon monoxide in the equilibrium mixture.

i) The value of Kc will not change because Kc is only affected by temperature.

ii) The equilibrium yield of hydrogen will increase. According to Le Chatelier's principle, adding steam (a reactant) will shift the equilibrium to the right, favoring the formation of products, thus increasing the amount of hydrogen produced.

Increasing the temperature will generally decrease the value of Kc for exothermic reactions, such as the reaction in question. As temperature increases, the system shifts to favor endothermic reactions to absorb the added heat, hence reducing the concentration of products relative to reactants, which decreases Kc.