# Unattainable equilibrium

I read a question on a textbook that asked:

For the reaction

$$\ce{CaCO3(s) \rightleftharpoons CaO(s) + CO2(g)} \tag{1}$$

determine whether or not a mixture of $$\ce{CaCO3(s)}$$ and $$\ce{CO2(g)}$$ at a pressure greater than the value of $$K_p$$ can attain the above equilibrium.

The answer it gives is no, because the partial pressure of $$\ce{CO2}$$ won't be able to decrease to $$K_p$$.

However, I'm confused about what happens to the mixture if equilibrium isn't attained. Will it go to completion? What if it is a mixture of $$\ce{CaO}$$ and $$\ce{CO2}$$ at a pressure less than $$K_p$$?

• From my memories, I think the reaction stays out of equilibrium and just can't progress further so it stops. I think it's the also the case for dissolution before saturation. But maybe someone else can verify this because I'm not 100% sure. Commented Jan 14 at 13:33
• This is an equilibrium reaction. If CO2 (or CaO) were all removed more carbonate would form products until equilibrium is re-established and vice versa. Commented Jan 15 at 9:12
• I assume that we could formally take disappearing of CaO(s) as its activity going from 1 toward 0, being out of scope of the standard definition of equilibrium constants involving solids. Commented Jan 16 at 10:25

If the partial pressure of $$\ce{CO2}$$ is greater than $$K_p$$, some $$\ce{CaO}$$ will react with the excess of $$\ce{CO2}$$ and produce more $$\ce{CaCO3}$$. At the end, the system contains less $$\ce{CaO}$$ and more $$\ce{CaCO3}$$, so that the final pressure of $$\ce{CO2}$$ is equal to $$K_p$$

If the partial pressure of $$\ce{CO2}$$ is smaller than $$K_p$$, some $$\ce{CaCO3}$$ will get decomposed to produce more $$\ce{CaO}$$ and more $$\ce{CO2}$$, so that the final pressure of $$\ce{CO2}$$ is equal to $$K_p$$

Edit : I have admitted that the amounts of solids is at least an order of magnitude greater than the amount of gas. With this assumption, I will never run out of one of the solid species.

• Unless you run out of one of the solid species, in which case you don’t reach equilibrium.
– Karsten
Commented Jan 15 at 12:41
• Yes Karsten. But I have admitted that the amounts of solids is an order of magnitude greater than the amount of gas. It will be added in my text. Commented Jan 15 at 14:53
• The question says they are starting with a mixture of carbon dioxide and calcium carbonate. This implies that there is no calcium oxide at all, and making some is in the direction away from equilibrium. But I agree with your answer and it is important to know what happens when all three species are initially present.
– Karsten
Commented Jan 15 at 19:04

This is an interesting question! Lets maintain a CO2 pressure of one atmosphere at a temperature below 800C sufficient that CaCO3 will decompose to a pressure much less than 1 atmosphere say 0.1 atmosphere. If a sufficient amount of pure CaCO3 is introduced and maintained at that temperature would not a dynamic equilibrium be established between CaCO3 and CaO +CO2. If our definition of Kp is reasonable it seems that our definition of the activities of the solids is questionable. I suggest the possible formation of a nonstoichiometric compound so that the ratios of activities are such that the ratio will correspond to that needed to satisfy Keq.

A quick web search failed to find any studies of CaCO3 decomposition at lower temperatures and controlled CO2. This requires a library search. Possibly Xray diffraction of CaCO3 decomposed under high CO2 pressures has been studied.

• Why the downvote I known I went out on a limb. please explain the error in my thinking or if you have access to actual data. Commented Jan 15 at 21:29

In chemical reaction problems attaining equilibrium or completing a reaction means reaching a state at which the reaction ceases. Here it never started, how can it cease? The system begins in a state of dynamic equilibrium (the posted problem is poorly worded) and the net amounts of the species will not change. The reaction can be regarded as having already gone to completion - in the reverse direction.

The high pressure of $$\ce{CO2(g)} (> K_p)$$ prevents net decomposition of $$\ce{CaCO3(s)}$$. The term "net" is important because there is a dynamic equilibrium: $$\ce{CO2(g)}$$ might not interfere with the forward reaction and some $$\ce{CaCO3(s)}$$ may decompose, but $$\ce{CaO(s)}$$ thus produced will react back to carbonate, the high pressure of the gas exhausting all $$\ce{CaO(s)}$$.

On nomenclature: thermodynamic changes ($$\Delta G^\circ$$, $$\Delta H^\circ$$ etc) refer to processes between equilibrium states, an initial equilibrium state and a final equilibrium state. This is why the field is called equilibrium thermodynamics. For instance, a starting state of unmixed reagents (in separate containers, say) might be regarded as an equilibrium system (under the imposed constraints it is stationary, will not change). That is analogous to the meaning of equilibrium here. No change in the amounts of the species (or in T, p, etc) will occur under the given conditions.

Another post with a similar question explains how to analyze such a problem mathematically in terms of sums of chemical potentials of the components. It can be used to demonstrate that dissociation of carbonate will increase the total free energy:

\begin{align} \qquad &\ce{CaCO3(s) &<=>& CaO(s) + &CO2(g)}\\ \qquad &\ce{A &<=>& B + &C} \\ \mathrm{(initial)} \qquad & n_A &\qquad& 0 \quad &n_C\\ \mathrm{(final)} \qquad & n_A -\alpha &\qquad& \alpha \quad &n_C+\alpha \\ \mathrm{(change)} \qquad & -\alpha &\qquad& \alpha \quad &+\alpha\end{align}

The free energy change is

$$\Delta G = - αμ^∘ _A + αμ^∘ _B + αμ^∘ _C + RT (n_C + α) \ln p_{fin} - RT n_C \ln p_{ini} \\ = α\Delta G^∘ + α RT \ln p_{fin} + RT n_C \ln \left(\frac{p_{fin}}{p_{ini}}\right) \\ = - αRT \ln K_p + α RT \ln p_{fin} + RT n_C \ln \left(\frac{p_{fin}}{p_{ini}}\right) \\ = α RT \ln \left(\frac{p_{fin}}{K_p}\right) + RT n_C \ln \left(\frac{p_{fin}}{p_{ini}}\right)$$

If carbonate decomposes $$α>0$$ and $$p_{fin}> p_{ini}$$ so that $$\Delta G>0$$.

• Thank you. I also read that if there is some CaO and CO2 at the pressure of Kp they would react to form some CaCO3, thereby lowering the partial pressure to below Kp when equilibrium is reached. Is that incorrect then? Commented Jan 15 at 1:56
• That doesn't seem right. Where did you read this? Commented Jan 15 at 8:59
• "Chemistry: The Central Science" 12th ed. by Brown and others. It wasn't stated in the main text but was from an exercise. Commented Jan 15 at 9:23
• Again, that doesn't seem right. I can include a mathematical explanation later. Commented Jan 15 at 10:35