For the following equilibrium system, which of the following changes will form more CaCO3?

CO2(g) + Ca(OH)2(s) <--> CaCO3(s) + H2O(l)
deltaH(rxn) = -113 kJ

My choices are:
a) Decrease temperature at a constant pressure (no phase change)

B) Increase volume at a constant temperature

C) Increase partial pressure of CO2

D) Remove one-half of the initial CaCO3

I was thinking that it would be A, C, and D. Is that correct?

A and C are right. Removing 1/2 CaCO3 won't affect the equilibrium AS LONG AS there is a smidgen of CaCO3 solid present. A speck must be present to attain equilibrium; more or less than that has no effect.

Is that because it's a solid and not a gas?

To determine which changes will form more CaCO3 in the given equilibrium system, we need to consider Le Chatelier's principle. According to Le Chatelier's principle, a system at equilibrium will shift in a way that opposes the change applied to it.

a) Decreasing the temperature at a constant pressure: According to the given reaction, the reaction is exothermic (deltaH(rxn) = -113 kJ). Decreasing the temperature will favor the exothermic direction and promote the formation of CaCO3. Thus, this change will form more CaCO3.

b) Increasing volume at a constant temperature: Since the number of moles of gas (CO2) is the same on both sides of the reaction, changes in volume will not have a significant effect on the equilibrium position. Therefore, this change will not form more CaCO3.

c) Increasing the partial pressure of CO2: According to Le Chatelier's principle, increasing the concentration or partial pressure of a reactant will cause the equilibrium to shift in the direction that consumes more of that reactant. In this case, increasing the partial pressure of CO2 will favor the formation of CaCO3, as it consumes more CO2 to reach equilibrium. Thus, this change will form more CaCO3.

d) Removing one-half of the initial CaCO3: By removing CaCO3 from the system, the reaction will shift to compensate for the loss. According to Le Chatelier's principle, this will favor the forward reaction and promote the formation of more CaCO3 to establish equilibrium. Thus, this change will form more CaCO3.

Based on the analysis, the correct changes that will form more CaCO3 in the equilibrium system are options a), c), and d).

To determine which of the given changes will form more CaCO3, we need to consider Le Chatelier's principle. According to Le Chatelier's principle, if a stress is applied to a system at equilibrium, the system will shift in a way that reduces the stress.

Let's analyze each choice to see how it affects the equilibrium system:

a) Decrease temperature at a constant pressure (no phase change):
When the temperature decreases, the system will shift in the exothermic direction to compensate for the decrease in energy. The forward reaction is exothermic since deltaH(rxn) is negative. Therefore, decreasing the temperature will favor the formation of CaCO3. So, this choice is correct.

b) Increase volume at a constant temperature:
Changing the volume of a system at equilibrium without any phase change will not affect the position of the equilibrium. In this case, since there are no gas moles involved in the equilibrium equation, changing the volume does not have an effect. So, this choice can be eliminated.

c) Increase partial pressure of CO2:
According to Le Chatelier's principle, increasing the concentration of a reactant will shift the equilibrium towards the products. In this case, increasing the partial pressure of CO2 will favor the forward reaction, leading to more CaCO3 formation. So, this choice is correct.

d) Remove one-half of the initial CaCO3:
Removing CaCO3 from the system will disrupt the equilibrium. By removing one-half of the initial CaCO3, the system will try to restore the balance by shifting towards the side that forms CaCO3. So, this choice is correct.

Therefore, the correct choices are a) Decrease temperature at a constant pressure, c) Increase partial pressure of CO2, and d) Remove one-half of the initial CaCO3.