Adding pure solid/liquid in heterogeneous equilibrium [duplicate]

If I have the heterogeneous equilibrium

$$\ce{aA(s) + bB(g) <=> cC(g) + dD(g)}$$

I understand that the equilibrium expression is

$$K_c = \frac{[\ce{C}]^c[\ce{D}]^d}{[\ce{B}]^b}$$

and that $$\ce{A}$$ is left out of the equilibrium expression since it is a pure solid and pure solids/liquids are left out of the equilibrium expression in heterogeneous equilibria.

However, does this mean that adding or removing substance $$\ce{A}$$ has no effect on the reaction at all? For instance, is it true that adding $$\ce{A}$$ does not produce more products and removing $$\ce{A}$$ does not produce more reactants (as Le Chatelier's principle would predict for a reactant included in the equilibrium expression, such as $$\ce{B}$$)?

On a side note, adding/removing the other substances could potentially affect $$[\ce{A}],$$ right? For example, adding $$\ce{C}$$ or $$\ce{D}$$ would cause $$[\ce{A}]$$ to increase (because more $$\ce{B}$$ would be formed and with that more $$\ce{A}$$ would be formed as well) and adding more of substance $$\ce{B}$$ would cause $$[\ce{A}]$$ to decrease (because $$\ce{B}$$ would be consumed and with it $$\ce{A}$$ would be consumed as well), correct?

• In order for an equilibrium to exist A must be in excess. – MaxW May 26 '19 at 22:53
• @MaxW Why is this? – Michael May 27 '19 at 2:13
• "Excess" is not quite the correct word. The idea is more that there needs to be sufficient A so that when the equilibrium is established then there is still some A remaining. // If there is no A remaining then there can't be an equilibrium. In that case A is the limiting reagent. – MaxW May 27 '19 at 4:35
• [A] does not change when the amount of solid increases or decreases (the distance of particles in the pure solid is a constant). It will become zero when the solid is used up, otherwise it is a constant. – Karsten Theis May 28 '19 at 15:01