CaCO3(s) ⇄ CaO(s) + CO2(g) When heated, calcium carbonate decomposes according to the equation above — Equilibrium Chemistry Question
Problem Context
CaCO3(s) ⇄ CaO(s) + CO2(g)
When heated, calcium carbonate decomposes according to the equation above. In a study of the decomposition of calcium carbonate, a student added a 50.0 g sample of powdered CaCO3(s) to a 1.00 L rigid container. The student sealed the container, pumped out all the gases, then heated the container in an oven at 1100 K. As the container was heated, the total pressure of the CO2(g) in the container was measured over time. The data are plotted in the graph below.
[VISUAL]
The student repeated the experiment, but this time the student used a 100.0 g sample of powdered CaCO3(s). In this experiment, the final pressure in the container was 1.04 atm, which was the same final pressure as in the first experiment.
Model Answer
PV = nRT
n = PV / RT = (1.04 atm)(1.00 L) / ((0.0821 L atm / (mol K))(1100 K)) = 0.0115 mol CO2
1 point is earned for the proper setup using the ideal gas law and an answer consistent with the setup.
Model Answer
Do not agree with claim
Explanation I: In experiment 1, the moles of CaCO3 = 50.0 g / 100.09 g/mol = 0.500 mol CaCO3. If the reaction had gone to completion, 0.500 mol of CO2 would have been produced. From part (a) only 0.0115 mol was produced. Hence, the student’s claim was false.
Explanation II: The two different experiments (one with 50.0 g of CaCO3 and one with 100.0 g of CaCO3) reached the same constant, final pressure of 1.04 atm. Since increasing the amount of reactant did not produce more product, there is no way that all of the CaCO3 reacted. Instead, an equilibrium condition has been achieved and there must be some solid CaCO3 in the container.
1 point is earned for disagreement with the claim and for a correct justification using stoichiometry or a discussion of the creation of an equilibrium condition.
Model Answer
The final pressure would be equal to 1.04 atm. Equilibrium was reached in both experiments; the equilibrium pressure at this temperature is 1.04 atm. As the reaction shifts toward the reactant, the amount of CO2(g) in the container will decrease until the pressure returns to 1.04 atm.
1 point is earned for the correct answer with justification.
Model Answer
Yes. For the equilibrium reaction represented by the chemical equation in this problem, at a given temperature the equilibrium pressure of CO2 determines the equilibrium constant. Since the measured pressure of CO2 is also the equilibrium pressure of CO2, Kp = P_CO2 = 1.04.
Note: If the response in part (b) indicates “yes”, that all of the CaCO3(s) had decomposed, then the point can be earned by stating that the system did not reach equilibrium in either experiment and hence the value of Kp cannot be calculated from the data.
1 point is earned for correct explanation that is consistent with the student’s answer to part (b).