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There are three parts to my question.

Part one:

  • Let's assume we have a reversible reaction at equilibrium as follows : Calcium carbonate (solid) goes into calcium oxide(solid) and carbon dioxide(gas).
  • Let's assume that I add more calcium carbonate i.e I add more solid.
  • Now I understand that there would be no shift in the position of the equilibrium since we added a solid therefore we induced a very negligible change in concentration , but what I am confused about is doesn't more reactant mean more product and doesn't more product symbolise a shift to the right?

Part two:

  • I have read that the equilibrium constant $Kc$ and $Kp$ are really just close approximations of the true thermodynamic constant which uses activities hence the activity of a solid and pure liquid is always 1 .
  • However upon further research I discovered that for an esterification reaction the equilibrium constant is 4 -First of all, the above reaction involves all pure liquids and hence the thermodynamic equilibrium constant would be one ,however why stick to a value of 4 why not easily say it is one since all activities are 1.

Part three:

  • If we do take pure liquids in the $Kc$ value how do you measure the concentration of such a species, wouldn't the value be very high.My intution tells me that pure liquids are only taken when a reaction only involves pure liquids only , but they are not taken in heterogeneous reactions or when the liquid involved acts as a solvent as well.However in a case of a heterogeneous reaction only involving solids and liquids what will your equilibrium constant look like .

Lastly I have no knowledge of thermodynamics and I am still a high school student .

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  • $\begingroup$ If there are three parts of a question, than it is easier to tackle each of them as an individual question. Otherwise, the question may be considered as too broad to be answered. $\endgroup$
    – Buttonwood
    Jun 15 '17 at 22:11
  • $\begingroup$ Regarding your first part, although more reactant means more product, the percentage yield is the same and thus, there is no shift in the position of equilibrium. $\endgroup$ Jun 16 '17 at 0:48