\begin{equation}\ce{CaCO3(s) -> CaO(s) + CO2(g)}\quad\Delta H^\circ = +556\ \mathrm{kJ/mol} \tag{endothermic}\end{equation}

How will the equilibrium partial pressure of $\ce{CO2}$ be affected if the volume is decreased?

So, I thought of course the moles will increase when you decrease the volume hence $pV = nRT$, thus $p = \frac{nRT}{V}$. Therefore, $p$ is inversely proportional to $V$. But, the right answer is NO CHANGE, can anyone help me to understand why?


2 Answers 2


The equilibrium constant for the reaction is

$$K=\frac{P_{\ce{CO2}}}{P^\circ}=e^{-\Delta G^\circ/RT}$$

Changing the volume changes neither $\Delta G^\circ$ or $T$ (at least this is implicitly implied), so $K$ and, hence, $P_{\ce{CO2}}$ does not change.

For an ideal gas


So if $V$ decreases then $n_{\ce{CO2}}$ must also decrease in order for $K$ to remain constant. So, when the equilibrium is perturbed by the volume-decrease the equilibrium shifts to the left to re-establish equilibrium, in accord with Le Chatelier's principle.


Okay let's recall what the Le Châtelier's principle is all about: (in short) whenever a system in equilibrium is disturbed the system will adjust itself in such a way that the effect of the change will be nullified.

Now let's look at your problem:

$\ce{CaCO3(s)->CaO(s) +CO2(g)}$ $\Delta H* = +556 \ce{KJ/mol}$ (endothermic)

Your ideas are correct, if we reduce the volume we are in fact increasing the pressure. (and in a way, this displeases our system)

Our system, naturally, reacts by shifting to a side which exerts a lower pressure or has fewer moles of gas (the left in this case).

Our system nullifies the effect of the stress applied to it.

Now because this shift occurs, we observe no change in the equilibrium partial pressure of carbon dioxide.

I hope this makes sense? Maybe, someone smarter than me can offer a clearer explanation.

  • $\begingroup$ Ok, we if we do know that the rxn goes to the left wouldnt the pressure of CO2 then decrease? i still don't get how it can remain the same and shift to the left at the same time as it kinda contradicts all the le chatelier problems i did before $\endgroup$
    – Jx1
    Jun 7, 2015 at 5:44
  • $\begingroup$ Okay. Think of it this way..you reduce the volume (i.e increase the pressure). Now the reaction shifts, pressure dec. (i.e original pressure is restored)? $\endgroup$
    – getafix
    Jun 7, 2015 at 6:00

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