equilibrium Kc (DSE level)

imagine what happens when the carbonate ion is placed next to a positive ion. The positive ion attracts the delocalised electrons in the carbonate ion towards itself. The carbonate ion becomes polarised.



If this is heated, the carbon dioxide breaks free to leave the metal oxide.

How much you need to heat the carbonate before that happens depends on how polarised the ion was. If it is highly polarised, you need less heat than if it is only slightly polarised.

The smaller the positive ion is, the higher the charge density, and the greater effect it will have on the carbonate ion. As the positive ions get bigger as you go down the Group, they have less effect on the carbonate ions near them. To compensate for that, you have to heat the compound more in order to persuade the carbon dioxide to break free and leave the metal oxide.

In other words, as you go down the Group, the carbonates become more thermally stable.



ionic radius (nm)

Mg2+ 0.065
Ca2+ 0.099
O2- 0.140
CO32- ?


The calculated enthalpy changes (in kJ mol-1) are given in the table. Figures to calculate the beryllium carbonate value weren't available. Remember that the reaction we are talking about is:



MgCO3 +117
CaCO3 +178
SrCO3 +235
BaCO3 +267

You can see that the reactions become more endothermic as you go down the Group. That's entirely what you would expect as the carbonates become more thermally stable. You have to supply increasing amounts of heat energy to make them decompose.


As a result it indicates that the Kc for thermal decomposition of CaCO3 is smaller than that of MgCO3.