For the system

H2(g) + CO2(g) <---> H2O(g) + CO(g)
at equilibrium, the addition of H2(g) would cause (according to LeChatelier’s principle)
1. more H2O(g) and CO(g) to form.
2. only more CO2(g) to form.
3. only more H2O(g) to form.
4. no change in amounts of products or reactants.
5. only more CO(g) to form.

Adding more H2 means the reaction will shift to use the added H2 which means it will shift to the right. If it shifts to the right more H2O and more CO will form.

Well, let me put on my clown nose and answer this question. According to LeChatelier's principle, when you add more H2(g), it's like inviting more H2 molecules to join the party. And you know what happens when there's a party? Things start to get crowded! So, option 1 seems like the logical choice. More H2O(g) and CO(g) will form to make some space for all those extra H2 molecules. It's like H2 is the life of the party, making everyone else squeeze in.

According to Le Chatelier's principle, when an additional reactant is added to a reaction at equilibrium, the system will respond by shifting the reaction in the direction that consumes or uses up the added reactant.

In this case, if H2(g) is added, the system will respond by shifting the equilibrium to consume the additional H2(g). Looking at the balanced equation, we can see that consuming H2(g) involves the forward reaction, which produces H2O(g) and CO(g). Therefore, the addition of H2(g) would cause more H2O(g) and CO(g) to form.

So, the correct answer is 1. more H2O(g) and CO(g) to form.

To determine the effect of adding H2(g) on the equilibrium system according to Le Chatelier’s principle, we need to consider the reaction stoichiometry and the changes that occur when a reactant or product is added.

In this system, H2(g) and CO2(g) are reactants, while H2O(g) and CO(g) are products. Let's break down the reaction:

H2(g) + CO2(g) ↔ H2O(g) + CO(g)

When H2(g) is added, it increases the concentration of one of the reactants. According to Le Chatelier’s principle, the system will shift in a way that reduces the concentration change caused by the addition. In other words, the equilibrium will try to counteract the increase in H2(g) concentration.

To analyze the options:

1. more H2O(g) and CO(g) to form: This option suggests that both products will increase. However, adding H2(g) does not directly affect the concentration of H2O(g) or CO(g), so this is an incorrect choice.

2. only more CO2(g) to form: This option suggests that the concentration of CO2(g) will increase. Since adding H2(g) does not directly affect the concentration of CO2(g), this is also an incorrect choice.

3. only more H2O(g) to form: This option suggests that the concentration of H2O(g) will increase. Adding H2(g) does not directly affect the concentration of H2O(g), so this is an incorrect choice.

4. no change in amounts of products or reactants: This option suggests that adding H2(g) will not cause any change in the concentrations of products or reactants. According to Le Chatelier’s principle, since H2(g) is a reactant, the system will shift in a way that opposes the increase. Therefore, this is an incorrect choice.

5. only more CO(g) to form: This option suggests that the concentration of CO(g) will increase. Since adding H2(g) increases the concentration of one reactant (H2), the system will shift in a way that consumes the added H2. This shift will result in the formation of more of the product that contains H2 (CO). Therefore, this is the correct choice.

Therefore, the correct answer is: Option 5. Only more CO(g) will form.