October 20, 2021, 09:07:31 AM
Forum Rules: Read This Before Posting


Topic: Why do we omit the concentrations of solids in calculating equilibrium constant?  (Read 1691 times)

0 Members and 1 Guest are viewing this topic.

Offline alihihg

  • New Member
  • **
  • Posts: 3
  • Mole Snacks: +0/-0
"All solids have a fixed density and constant concentration, so these are omitted" J. Green/S. Damji, Chemistry 3rd edition, pg. 184

so my question is why? Won't this significantly affect the value calculated from the equilibrium constant?

Offline magician4

  • Chemist
  • Full Member
  • *
  • Posts: 567
  • Mole Snacks: +70/-11
you're wondering why, and you are right to do so: this explanation (though leading to the "desired" result nevertheless) is complete rubbish.

of course the solid plays a role in equilibria involving named solid: else there won't be no equilibrium, would it?
and yes, the "surface" of the solid plays an important role for everything happening, and of course this is part for the respective K-value.

Fact of the matter however is, that the "concentration" of "surface" effectively  next to doesn't change in a relevant manner *) , and hence is already "incorporated" in the K-value as a fixed factor.


regards

Ingo



*)
this is an abbreviated way to explain at it. If you look at it more detailled, you'll see that it's in fact a number of interlocking trends, however mostly cancelling each other out in result. hence, the dependency of equilibria from the surface of named solid is quite small, so we neglect it.
... just like we do it with the real (!) concentration-change of water in reactions involving production/consumption of water  (think of pH calculations, for example)
There is a theory which states that if ever anybody discovers exactly what the Universe is for and why it is here, it will instantly disappear and be replaced by something even more bizarre and inexplicable. There is another theory which states that this has already happened.
(Douglas Adams)

Sponsored Links