For homogeneous equilibrium, why are liquids and solids included in the equilibrium constant (when they aren't in heterogeneous equilibria)? In a heterogeneous reaction (where the states are varied) we do not include liquids and solids in the equilibrium equation because their concentrations do not change.

E.g. Chemguide.co.uk



However, when it is a homogenous equation we DO include solids and liquids. 

E.g. Chemguide.co.uk


1) Why do we include liquids and solids in the Kc equation in homogeneous equilibrium equations? I get that if we didn't there would be nothing on the right hand side of Kc=.... , but why is it physically different to the case where there are varied phases?

2) What if there was an equation that involved only liquids and solids  (if this is possible)? As this is a heterogeneous equation would we still not include solids and liquids? If so how would you write the Kc equation where there is nothing but solids and liquids?
Chemguide is simplified for A-levels and therefore in this case is strictly speaking incorrect.

The equilibrium constant $K$ is defined as a product of activities. I described this in a previous answer here.

The crux of the matter is that the activity of a pure solid or pure liquid is equal to 1, which means that it can be omitted from the expression for $K$ without affecting the value.

In your first reaction

$$\ce{H2O(g) + C(s) &lt;=&gt; H2(g) + CO(g)}$$

the chunks of carbon in the reaction are necessarily pure because they don't mix with the gases.

In the second reaction (yes, I am lazy, please feel free to edit for me)

$$\ce{EtOAc(l) + H2O(l) &lt;=&gt; AcOH(l) + EtOH(l)}$$

none of the liquids are pure, hence their activities deviate from unity.

Just as a final example, in the dissociation of a weak acid

$$\ce{HA(aq) + H2O(l) &lt;=&gt; H3O+(aq) + A-(aq)}$$

water is omitted from the expression for $K_\mathrm{a}$ because water, as the solvent, is in large excess over $\ce{HA}$ and is therefore effectively "pure".