Question

Why don't pure solids or liquids shift equilibrium?

Answer 1

Because their concentrations are essentially constant. Concentrations only change easily for things that `(i)` have solute in them and `(ii)` are not rigid.

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An equilibrium constant in standard conditions is written for, e.g.

`aA + bB rightleftharpoons cC + dD`

as

`K^@ = (([C]//c^@)^c([D]//c^@)^d)/(([A]//c^@)^a([B]//c^@)^b)`

where these concentrations `[" "]` are equilibrium concentrations. `c^@ = "1 M"` is the standard state concentration for aqueous solutions, i.e. at `25^@ "C"` and `"1 atm"`. This does not apply for pure liquids and solids.

PURE LIQUIDS

For pure liquids... if there is no solute in solvent... it must be a pure liquid and thus it has no true concentration. One might say its solute concentration is `"0 M"`, but clearly that would make `K -> oo` or `K -> 0`, so we do something else instead.

We specify that its standard state is its molar density, `rho = "mols liquid"/"L liquid"`, and this is precisely its actual "concentration" if you calculated its "concentration" as `"mols"` of itself "dissolved" in a volume of itself.

Thus, `(["pure liquid"])/(["pure liquid"]^@) = (["pure liquid"])/(bar rho) = 1` and it essentially does not vary.

PURE SOLIDS

It is even more true for solids. Again, we use the standard state of its molar density, and if you calculated a "concentration", that would match the molar density, i.e.

`(["solid"])/(["solid"]^@) = (["solid"])/(bar rho) = 1`

And it again does not vary.