I have a very vague knowledge of thermodynamics, but let's take a reversible reaction:

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

Now if we increase the mass of the calcium carbonate, I understand that it would not increase the concentration by a very large approximation (since it's density does not increase by a very large approximation) so I understand that the equilibrium will not shift to the right, but the thing that confuses me is doesn't more reactant mean more product and a greater yield means a shift to the right? This also applies to a pure liquid reaction:

$$\ce{HCl(g) + H2O(l) <=> H3O+ + Cl-}$$

Let's assume that in the first reaction you have a mole ratio of the respective reactants in the form of 1:0.5 clearly water is a limiting reagent, however let's assume that we increase the mass of the pure liquid so that the mole ratio becomes 1:1 now we have more water available for the reaction to occur hence more product would occur, but according to le Chateliers principle this would not occur since a pure liquid does not cause a shift, rather if the reaction was at equilibrium it would remain as so. Why does this happen?

  • 1
    $\begingroup$ In your first example at constant temperature and in a vessel of fixed volume (to stop carbon dioxide escaping) there will be equilibrium as long as some solid carbonate is present. Adding more makes no difference, because, well, there is equilibrium. I don't follow you second example. You seem to have gaseous HCl above a solution of same? $\endgroup$
    – porphyrin
    Jun 13, 2017 at 8:34


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.