Which part(s) of the reaction H2O(g) + C(s) ⇀↽ H2(g) + CO(g)

Question

Which part(s) of the reaction H2O(g) + C(s) ⇀↽ H2(g) + CO(g)

will be favored by an increase in the total pressure (resulting in compression)?
1. Neither is favored.
2. reactants
3. products
4. Unable to determine
Not sure how to do this

Answer 1

To determine which part(s) of the reaction will be favored by an increase in the total pressure, you need to consider the relationship between pressure and the number of gas molecules on each side of the equation.

In this reaction:
H2O(g) + C(s) ⇀↽ H2(g) + CO(g)

The reactants side consists of one molecule of water vapor (H2O(g)) and one molecule of solid carbon (C(s)). The products side consists of one molecule of hydrogen gas (H2(g)) and one molecule of carbon monoxide gas (CO(g)).

An increase in total pressure means an increase in the concentration of gas molecules. Based on Le Chatelier's principle, which states that a system at equilibrium will react to counteract any changes imposed on it, we can make some predictions:

1. If there is an increase in pressure, both the reactants and products sides will be affected. However, since the reactants side only has one gas molecule, and the products side has two gas molecules, it is more likely that the products will be favored.

2. Therefore, an increase in total pressure will generally favor the products side of the reaction, (H2(g) + CO(g)).

Based on this, the correct answer would be: 3. products are favored.

Answer 2

In very simple terms, Le Chatelier's Principle states that the equilibrium in a reaction will shift (meaning will move to the right or to the left) to undo whatever we do to it. So for this reaction you look on the left and see 1 mol gas. On the right there are two mols of gas. Therefore, if we add pressure the reaction will move so as to decrease the pressure. It can do that by moving to the left. H2O will increase, CO and H2 will decrease. Which answer meets these criteria?