Equilibrium between barium carbonate, and barium oxide and carbon dioxide

The USNCO 2017 Question 33 is as follows:



> 
> Barium carbonate, $\ce{BaCO3}$, is stable at ambient temperatures, but decomposes to barium oxide and carbon dioxide at higher temperatures.
> 
> 
> $$\ce{BaCO3(s) <--> BaO(s) + CO2(g)}$$
> 
> 
> At a certain temperature, this system in in equilibrium in a closed system and contains appreciable amounts of all three compound. Which changes will lead to an increase in the pressure of $\ce{CO2}$ present at equilibrium?
> 
> 
> I. Adding more $\ce{BaCO3(s)}$
> 
> 
> II. Increasing the volume of the container
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> (A) I only
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> (B) II only
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> (C) Both I and II
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> (D) Neither I nor II
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> 
> 


What I understand so far:


* Concentrations of reactants and products at equilibrium are affected by:


	+ Increasing/decreasing concentrations
	+ Change in pressure/volume
	+ Change in temperature
* Temperature remains unchanged, so that factor can be ruled out
* (I) does not change concentration, as $\ce{BaCO3(s)}$ is considered to have a constant concentration, and it could be assumed that change in volume due to the addition of $\ce{BaCO3(s)}$ is negligible. Hence, (A) nor (C) are correct
* (II) If $\ce{CO2}$ was in a closed system itself, increasing the volume of its container will lead to a decrease of pressure and decrease of concentration.
* However, according to Le Châtelier’s Principle, the system should react by favouring the forward reaction, leading me to believe (D) is the correct answer


**How do I judge whether or not the increase in volume of the container can be balanced by the increased rate of the forward reaction, such that the pressure of $\ce{CO2}$ remains unchanged at the new equilibrium?**




Your conclusion is correct.


In the expression for equilibrium, we consider the activities for the solids to be 1. This ultimately leads to:$$K\_p=p\_{\ce{CO2}}$$


[Note that $K\_P=f\_{\ce{CO2}}$, where $f\_{\ce{CO2}}$ is the fugacity of $\ce{CO2}$. Under "ideal" conditions, $f\_{\ce{CO2}}=p\_{\ce{CO2}}$, where $p\_{\ce{CO\_2}}$ is the partial pressure due to $\ce{CO2}$.]


In other words, the equilibrium partial pressure of $\ce{CO2}$ will be constant. One may compare this to the aqueous tension in case of water VLE.


Increasing the volume would cause more $\ce{BaCO3}$ to dissociate, producing more $\ce{CO2}$. The ideal gas law tells us that: $$n\_{\ce{CO2}}\propto p\_{\ce{CO2}}\cdot V$$


Your equilibrium pressure due to $\ce{CO2}$ does not change.


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