$$\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$$

How will the amount of each element change if the pressure is increased?

My guess is that the equilibrium will shift to the left, because the right side has 1 gas molecule. I know my solution contradicts with Le Chatelier's principles, but there is no gas on the left. So how will the the amount of the elements on the right decrease and the solid on the left increase since pressure only has an effect on gases.


2 Answers 2


Consider the equilibrium expression for this reaction. $$\ce{CaCO3(s) <=> CaO(s) + CO2(g)}\quad K_c^{'} = \frac{[\ce{CaO}] [\ce{CO2}]}{ [\ce{CaCO3}]}$$ The concentrations of solids and liquids are constant. They are the molar densities. Since $\ce{[CaO]}$ and $\ce{[CaCO3]}$ don't change, they are moved to the left hand side and "folded into" the equilibrium constant.

\begin{align} K_c^{'} &\cdot \frac{\ce{[CaCO3]}}{\ce{[CaO]}} = \ce{[CO2]} \\ K_c &= \ce{[CO2]} \end{align}

Therefore, as long as solid $\ce{CaO}$ and solid $\ce{CaCO3}$ are present along with $\ce{CO2}$ gas there will be an equilibrium. Only changes to the concentration of $\ce{CO2}$ will cause a shift in the equilibrium.

You asked how will the amounts change if the pressure is increased. The pressure of $\ce{CO2}$ is increased by either adding more $\ce{CO2}$ or by reducing the volume of the container. Adding more $\ce{CO2}$ will increase the concentration of $\ce{CO2}$ momentarily, which will shift the equilibrium to the left, using up some $\ce{CaO}$ and making $\ce{CaCO3}$. The pressure of $\ce{CO2}$ can also be increased by reducing the volume of the container. Again, the concentration of $\ce{CO2}$ is increased, which increases the reaction with $\ce{CaO}$ to make additional $\ce{CaCO3}$. The $\ce{CO2}$ pressure gets closer to what it was originally, as predicted by Le Chatelier's principle. Since the molar densities of CaO and $\ce{CaCO3}$ are constant, they don't appear in the equilibrium expression. That is why only changes to the pressure (concentration) of $\ce{CO2}$ affect the position of the equilibrium.

If the pressure in the container is increased by adding an inert or non-reacting gas, nothing happens to the amounts of $\ce{CO2}$, $\ce{CaO}$ or $\ce{CaCO3}$. The added gas won't affect the partial pressure of $\ce{CO2}$.


I don't think that your guess contradicts Le Chatelier's Principle, which states that introducing a change will prompt the reaction to oppose that change.

Increasing the pressure will lead to a decrease in $\ce{CO2}$ as it converts back into the starting reactant, which will lower the pressure as the gas is converted into $\ce{CaCO3}$.


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