Consider this reaction:

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

for this, $K_c=\left[\ce{CO2(g)}\right]$

For a given temperature, $K_c$ will remain fixed. That means, the amount of $\ce{CO2}$ released will be same, irrespective of amount of $\ce{CaCO3}$ taken.

But, that means, at the same temperature, irrespective of whether we start with 1 g of $\ce{CaCO3}$, or 10 kg of $\ce{CaCO3}$, the amount of $\ce{CO2}$ at equilibrium will be same(!) because, according to the $K_c$ expression, it should be constant.

This is against my intuition, because, it seems that if we start with 10 kg of $\ce{CaCO3}$, we should end up with a much larger amount of $\ce{CO2}$ at equilibrium than if we started with just 1 g of the carbonate.

Am I being mistaken somewhere, or is this is what will actually happen? I am confused.

  • $\begingroup$ Try putting it in terms of pressure ($K_p$) instead of concentration since $CO_2$ is a gas. So at equilibrium, the partial pressure of $CO_2$ just above the calcium carbonate would be constant for a given temp like you've mentioned. See en.wikipedia.org/wiki/Calcium_carbonate#Calcination_equilibrium for more info on this specific reaction. $\endgroup$
    – kaliaden
    Commented May 5, 2013 at 9:59
  • $\begingroup$ @kaliaden thanks for the link. But can you please try answering my question directly? the wiki page doesn't mention anything about my doubt- How can we end up with the same partial pressure of $CO_2$ for 10 kg as well as 1 gm of $CaCO_3$ ? that seems to defy common sense. $\endgroup$
    – nilanjana
    Commented May 5, 2013 at 16:08
  • $\begingroup$ Sorry about the late response. I haven't been on the site for a while now. I wasn't so sure about the answer which was why I had put it up as a comment. @perplexity has answered your question quite well below. $\endgroup$
    – kaliaden
    Commented May 19, 2013 at 8:51
  • $\begingroup$ nilanjana :Why you didn't get CaO(s) ,CaCO3(s) for the Kc expression? $\endgroup$ Commented Jun 17, 2015 at 15:29

1 Answer 1


First, as kaliaden commented, the true equilibrium constant

$$K(T,p) = a(\textrm{CaO,s}) a(\textrm{CO$_2$,g}) / a(\textrm{CaCO$_3$,s}) \approx p_\textrm{CO$_2$} / p^\circ \equiv K_p(T)$$

where $a$ are the activities, $p$ the partial pressure and $p^\circ$ is one bar. Thus $K$ basically only depends on temperature (unless you are interested in brutally large pressures, and we safely disregard the sublimation of CaCO$_3$ and the effect of the air). Obviously you can obtain

$$K_c(T) \approx K_p(T)/ RT = [\textrm{CO$_2$}]$$

Now your question is: how is it that this constant does not depend on the amount of matter that we have?

The fact is that as long as there is enough matter to reach the equilibrium (to saturate the air with carbon dioxide vapor) it doesn't matter how much CaCO$_3$ you start with. The reaction will proceed until reaching an equilibrium characterized by $K(T) \approx p_\textrm{CO$_2$} / p^\circ$. Obviously if you don't have enough CaCO$_3$ (for instance, a few micrograms) then everything will react and you will not reach the equilibrium.

  • 2
    $\begingroup$ If I may, the moral of the story is that thermodynamics is not useless. One can learn counterintuitive things from it! $\endgroup$ Commented May 10, 2013 at 2:40

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