Methanol is manufactured from carbon monoxide and hydrogen in the following equilibrium reaction - Leaving Cert Chemistry - Question 7 - 2022

At equilibrium, the concentrations of carbon monoxide (CO), hydrogen (H₂), and methanol (CH₃OH) remain constant over time. This means that the amount of each substance present does not change, even though both the forward and reverse reactions continue to occur.

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

$K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}$

where [ ] denotes the molar concentration of the substances at equilibrium.

Initially, 5.0 moles of CO and 10.0 moles of H₂ are present. Given that 25% of CO reacts:

• Moles of CO reacted = 0.25 × 5.0 = 1.25 mol
• Moles of CO at equilibrium = 5.0 - 1.25 = 3.75 mol
• Moles of H₂ reacted = 2 × 1.25 = 2.5 mol
• Moles of H₂ at equilibrium = 10.0 - 2.5 = 7.5 mol
• Moles of CH₃OH formed = 1.25 mol
• Total volume = 5.0 L

Now, we can find the concentrations:

• [[CO] = \frac{3.75 , \text{mol}}{5.0 , \text{L}} = 0.75 , \text{mol/L}]
• [[H_2] = \frac{7.5 , \text{mol}}{5.0 , \text{L}} = 1.5 , \text{mol/L}]
• [[CH_3OH] = \frac{1.25 , \text{mol}}{5.0 , \text{L}} = 0.25 , \text{mol/L}]

Substituting these values into the expression for Kc:

$K_c = \frac{0.25}{(0.75)(1.5^2)} = \frac{0.25}{(0.75)(2.25)} \approx 0.148$

An advantage of using a catalyst in an equilibrium reaction is that it accelerates the rate at which the equilibrium state is reached without affecting the position of the equilibrium. This means reactions can proceed more rapidly, allowing for faster production of desired products, such as methanol in this case.

Le Chatelier’s principle states that if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the system will adjust to counteract that change and establish a new equilibrium.

When more carbon monoxide is added, the equilibrium shifts to the right according to Le Chatelier’s principle, favoring the formation of methanol. This increases the yield of methanol as the system seeks to reduce the increased concentration of carbon monoxide.

The value of Kc remains unchanged when more carbon monoxide is added. Kc is a constant for a specific reaction at a given temperature, and while the concentrations of the reactants and products may change, the ratio defined by Kc does not.