7. For the equilibrium system below, which of the following would increase the concentration of 𝐻2𝑂(𝑔)?

𝐢(𝑠) + 𝐻2𝑂(𝑔) + 131π‘˜π½ ↔ 𝐢𝑂(𝑔) + 𝐻2(𝑔)

a. Increased temperature

b. Decreased pressure

c. Decreased [CO]

d. Decreased container volume

Remember, LeChatlier's Principle says that if a stress is applied to a reaction the reaction will naturally shift away from the applied stress to remove it. This will continue until a new equilibrium is established.

For your problem, in order increase the concentration of HOH, a stress would need to be applied to the product side of the reaction. Such would cause the reaction to shift left and, as a result, increase the HOH concentration. This would continue until the stress is removed and a new equilibrium would be established.

'D' choice is your best answer. Decreasing Volume of container increases Pressure (Boyles Law). The natural result is for the reaction to shift to the side of the equation having the LOWER moles of gas. In this case a left shift which increases the HOH concentration.

A => Increasing temp puts the stress on the reactant side and shifts reaction right. In future problems, look for the energy number in the equation. If temp is increased assume more weight is being added to that side causing the rxn to shift away from the energy number. If temperature is decreased, the opposite is true.

B => Decreasing Pressure causes a gas phase reaction to shift to the side have the higher moles of gas. Increasing pressure shifts reaction to lower moles of gas side.

C => Removing CO from the product side lightens the load on the product side and more reactants will react to replace what's been removed. In this case that would be a shift right to replace lost CO on the product side. This will continue until the concentration of CO has increased sufficiently to establish a new equilibrium. Hope this helps.

To determine which factors would increase the concentration of H2O(g) in the equilibrium system, we need to consider Le Chatelier's Principle.

Le Chatelier's Principle states that if a system at equilibrium is subjected to a change, it will adjust itself to counteract the change and restore equilibrium.

Let's analyze each option:

a. Increased temperature: According to Le Chatelier's Principle, increasing the temperature of an endothermic reaction (where heat is a reactant) will shift the equilibrium to the right, favoring the products. In this case, since the reaction consumes heat, increasing the temperature would cause the equilibrium to shift to the right, resulting in an increase in the concentration of H2O(g). So, option a would increase the concentration of H2O(g).

b. Decreased pressure: Decreasing the pressure of a system will have no effect on the equilibrium concentration of gases, as pressure only affects the number of moles of gas particles, not the concentration. Therefore, option b would not increase the concentration of H2O(g).

c. Decreased [CO]: If the concentration of CO(g) is decreased, according to Le Chatelier's Principle, the system will shift to the left to counteract the change. This will result in a decrease in the concentration of H2O(g) rather than an increase. Therefore, option c would not increase the concentration of H2O(g).

d. Decreased container volume: Decreasing the container volume will increase the pressure of the system. Again, as discussed in option b, pressure does not affect the concentration. So, option d would not increase the concentration of H2O(g).

Therefore, the only correct option to increase the concentration of H2O(g) in the equilibrium system is option a, increased temperature.

To determine which factor would increase the concentration of H2O in the equilibrium system, we need to consider Le Chatelier's principle. According to this principle, if a system at equilibrium is subjected to a change, it will shift in a direction that partially counteracts the effect of that change.

In the given equilibrium reaction:

C(s) + H2O(g) + 131kJ ↔ CO(g) + H2(g)

a. Increased temperature:
Higher temperatures favor endothermic reactions (reactions that absorb heat). In this equilibrium system, the forward reaction is endothermic, as denoted by the 131kJ energy being on the left side. Therefore, if the temperature is increased, the system will shift in the direction that absorbs heat, which is the forward reaction. Consequently, the concentration of H2O(g) will increase.

b. Decreased pressure:
The pressure does not directly affect the concentration of the gases involved in the equilibrium, except for the partial pressure of the reactants and products. However, because decreasing pressure shifts the equilibrium toward the side with more moles of gas, it would cause the system to shift to the left, reducing the concentration of H2O(g).

c. Decreased [CO]:
This option suggests decreasing the concentration of CO. According to Le Chatelier's principle, if the concentration of a reactant is decreased, the equilibrium will shift towards the side with more of that reactant. Since CO is on the right side of the equation, decreasing its concentration will shift the equilibrium to the right, resulting in an increase in the concentration of H2O(g).

d. Decreased container volume:
If the container volume is decreased, it will increase the pressure of the system. Similar to the earlier explanation for decreased pressure, increasing the pressure will shift the equilibrium towards the side with fewer moles of gas, which in this case is towards the left. Therefore, the concentration of H2O(g) would decrease as a result.

Based on the explanations given, c. Decreased [CO] is the correct choice as it would increase the concentration of H2O(g).