Equilibrium: Le Chatelier's Principle and CaCO3 Decomposition

In summary, the conversation discusses a Le Chatelier problem about the equilibrium shift in the decomposition of CaCO3. The question asks about the direction of equilibrium shift if the volume of the container was halved. The person initially thought the equilibrium would not shift due to different states of the reactants, but the correct answer is that the equilibrium would shift left to reduce pressure. This is because the equilibrium is determined by the CO2 gas partial pressure and halving the container volume would require half of the CO2 to react with CaO to re-establish the equilibrium partial pressure.
  • #1
Zayn
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Homework Statement



I had a question on a test and I got the wrong answer, but I still don't understand why. It was a Le Chatelier problem about the equilibrium shift, and it described the decomposition of CaCO3

CaCO3(s) <->CaO(s) + CO2(g)

The question asked something along the lines of "If the volume of the container was halved, in what direction would equilibrium shift?"

Homework Equations



N/A

The Attempt at a Solution



I thought it wouldn't shift at all because the CO2 is in a different state than the CaCO3 and CaO, and because one of the ways you can change an equilibrium system without disturbing the equilibrium was to have reactants in different states, but apparently I was wrong. Can you please explain to me why the equilibrium would shift left even if there are no gas entities? Is it just to reduce pressure because there is 1 reactant entity for every 2 product entities? Thanks.
 
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  • #2
The equilibrium is determined solely by the equilibrium value of the CO2 gas partial pressure. If the container volume is halved, half of the CO2 will have to react with CaO to re-establish the equilibrium partial pressure.
 
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