6
$\begingroup$

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?

$\endgroup$
6
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.