# Why does calcium carbonate decompose into calcium oxide?

Why does calcium carbonate, when heated strongly, decompose into calcium oxide and carbon dioxide?

What is the mechanism? Can someone illustrate the means by which this happens? What bonds/associations are broken?

I'm guessing that one of the calcium ion has a strong intermolecular association with one of the partially negative oxygens, and this changes the bond character of the carbonate ion. Upon applying heat, this bond remains while carbon dioxide is off-gassed.

In other words, the positive charge on the calcium ion isolates the electron density to the nearest oxygen atom on the carbonate ion. This isolation of electron density away from the carbon and toward the calcium weakens the oxygen-carbon bond while strengthening the calcium ion-oxygen intermolecular association force.

• It is a matter of entropy $\ce{CaCO3 -> CaO + CO2 ^ }$. – Martin - マーチン Jun 13 '14 at 4:11
• Yes, I agree Martin, but why not carbon monoxide? That would also yield an increase in entropy. – Dissenter Jun 20 '14 at 13:15
• I think you approach this question in a very "organic-chemistrish" way. In this reaction the oxidation state of Ca and C do not change! – Greg Jun 20 '14 at 15:58

Some quick googling did not turn up a definitive reaction mechanism, and most of the papers I did find seemed to instead be focused on the overall reaction rate and controlling factors - probably because this is such an industrially important reaction and (at least according to this article) gas transport is the rate-limiting factor - this is common for heterogeneous reactions.

However, it is not too hard to propose a plausible mechanism. I would suggest the same mechanism you did - what is likely happening is that the oxygen near one calcium ion can pull some electron density away from the carbon, leaving $\ce{CO2}$ which diffuses away as a gas. The calcium ion stabilizes the new oxygen anion and forms a new, different lattice structure.

As @Martin points out, the reason that this process is thermodynamically favorable is that the products have a higher entropy. The standard Gibbs free energy of reaction (using data from Tro's Chemistry) is:

\begin{align} \Delta G &= \Delta H - T\Delta S \\ &= 179.2\space kJ/mol - T(0.1602\space J/mol*K) \end{align} Note that $\Delta H$ is positive, but so is $\Delta S$. At 298 K (room temp) $\Delta G = 131.3 \space kJ/mol$ - the reaction is not favorable.

The equilibrium temperature (ignoring temperature corrections) should be around 1120 K.

• What do you mean by temperature corrections? – Dissenter Jun 20 '14 at 3:47
• The enthalpy change is a function of temperature, so the real equilibrium temperature would be different. – thomij Jun 20 '14 at 4:16
• The real equilibrium temperature would be he calculated temperature plus the starting temperature - is that what you're saying? – Dissenter Jun 20 '14 at 5:29
• No, when you solve the equation for change in Gibbs free energy to find the equilibrium point, you are making an implicit assumption that change in enthalpy is independent of temperature. Since that is not true, the temperature you get is not exactly right. It is close in many cases, and can give a rough idea of the true equilibrium temperature. – thomij Jun 20 '14 at 12:44
• In otherwords you are assuming that the enthalpy function is isothermic? Dunno if I'm using the term correctly but I would love to learn! – Dissenter Jun 20 '14 at 17:24