Question

Methanol can be prepared from carbon monoxide and hydrogen at high temperature and pressure in the presence of a suitable catalyst.

2H₂(g)+CO(g) ⇌ CH₃ OH(g) ΔH=-90.2kJ

What will happen to the concentrations of H₂ CO and CH₃OH at equilibrium if CO is removed?

A. [H₂] decreases, [CO] decreases, [CH₃OH] increases
B. [H₂] increases, [CO] decreases, [CH₃OH] increases
C. [H₂} increases, [CO] decreases, [CH₃OH] decreases

Answer 1

Removal of CO from the equilibrium mixture will result in a leftward shift, causing an increase in H₂ concentration and a decrease in CO and CH₃OH concentrations.

Step-by-step explanation:

When carbon monoxide (CO) is removed from the equilibrium mixture of the reaction 2H₂(g) + CO(g) ⇌ CH₃OH(g), with an enthalpy change (ΔH) of -90.2 kJ, the system will adjust according to Le Chatelier's Principle. Since CO is part of the reactants, removing it disturbs the equilibrium, causing the reaction to shift to the left to increase the concentration of CO.

This shift results in the consumption of CH₃OH and production of more CO and H₂, thus the concentration of H₂ will increase, CO will decrease, and CH₃OH will decrease. Therefore, the correct answer is: C. [H₂} increases, [CO] decreases, [CH₃OH] decreases.