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Let's take the following reaction that has reached equilibrium

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

What can be done to increase the yield of calcium oxide ?

My initial thought was to remove the carbon dioxide produced in the reaction mixture. I understand that this is correct since, according to Le Chatelier's principle, the reaction would shift to increase the yield of the products.

Why wouldn't adding more calcium carbonate lead to a shift to the products and hence lead to a greater yield of the products? Doesn't more reactant mean more product?

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marked as duplicate by Karsten Theis, Mathew Mahindaratne, Todd Minehardt, M.A.R., Tyberius Aug 26 at 14:32

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    $\begingroup$ Is that reaction a true equilibrium? The loss of carbon dioxide makes it somewhat irreversible... $\endgroup$ – NotEvans. Jun 18 '17 at 10:42
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    $\begingroup$ If you ran the reaction in a closed system, you could reach a state of true equilibrium under certain conditions of temperature and pressure. A complicating factor is that two of the components of the system are solids, which may be slow to react, especially if not in the form of very small particles. You can "shift the equilibrium" by removing or adding a component, but you no longer have a closed system, nor do you have a true equilibrium. The modified system can of course be closed again and brought to a new state of equilibrium, perhaps as a way to increase the yield of the desired produc $\endgroup$ – iad22agp Jun 18 '17 at 11:27
  • $\begingroup$ You could also leave the system closed, and get a higher yield of product by shifting the conditions (T, P) and letting the system re-equilibrate. $\endgroup$ – iad22agp Jun 18 '17 at 11:29
  • $\begingroup$ But why would adding more calcium carbonate not cause a shift , my textbook just states that's it's because it's a solid , so no shift means no greater yield of product by why is that so, shouldn't the yield be greater because of the greater amount of reactant $\endgroup$ – LM26 Jun 18 '17 at 11:45
  • $\begingroup$ Generally only the surface of a solid reacts so it matters whether there is, say, one large crystal or a fine powder. Also a product formed on the surface may inhibit any more reaction. $\endgroup$ – porphyrin Jun 18 '17 at 12:27
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I agree with your initial starting point, that the thermal decomposition of calcium carbonate into calcium oxide and carbon dioxide may be treated as an example of an heterogeneous equilibrium

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

that still is of relevance today

enter image description here

(source, Hughes et al. in Ind. Eng. Chem. Res. 2004, 43, 5529-5539).

and actually at around $\ce{900 ^\circ{}C}$, the equilibrium pressure of $\ce{CO2}$ over $\ce{CaCO3}$ and $\ce{CaO}$ would equal to $\pu{1 atm}$.

Your first reasoning, while keeping the other parameters of this equilibrium fixed ($T, p$), a removal of $\ce{CO2}$ may increase the yield of $\ce{CaO}$ since this eventually (macroscopically) leads to an exhaustive decomposition of $\ce{CaCO3}$ is correct.


For the second part, you have to balance the general idea about chemical equilibria (which in this case indeed means, keeping other parameters fixed, adding more $\ce{CaCO3}$ as starting material yields more $\ce{CO2}$) with the fact that this particular reaction is heterogeneous. Looking up the data for molar mass and density, (1, 2, and 3) allows you to estimate the volume one mole of each pure compound under standard conditions occupies:

| compound | molar mass |  density | molar volume |
|          |    [g/mol] | [g/cm^3] |   [cm^3/mol] |
|----------+------------+----------+--------------|
| CaCO3    |     100.09 |    2.711 |        36.92 |
| CaO      |      56.08 |     3.37 |        16.64 |
| CO2      |      44.01 | 0.001977 |     22261.00 |
|----------+------------+----------+--------------|

The value of the molar volume of $\ce{CO2}$ may remind you to the value of the volume occupied by the Ideal Gas at standard conditions ($\pu{22.7 L/mol}$) and demonstrates that the influence of $\ce{CO2}$ on the position of the equilibrium is much more important than the two other of the solids with their -- in comparison -- often negligible molar volumes. Hence, to underline that this equilibrium is this much pressure dependent, it is described by $K_{\text{p}}$, rather than by a normal $K$ value.

The general idea of influencing a chemical equilibrium along "an increase of partial volume (or partial pressure) of starting material(s) will increase the yield of the product(s)" better full-filled in the case of homogeneous reactions, like the synthesis of ammonia

$$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}. $$

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