Equilibrium between barium carbonate, and barium oxide and carbon dioxide

Question

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

(A) I only

(B) II only

(C) Both I and II

(D) Neither I nor II

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?

Answer 1

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.