# Le Chatelier's principle with pure solids and liquids

In the following reaction:

$$\ce{A(s) <=> B(g) + C(g)}$$

since adding a pure solid is not supposed to shift the equilibrium in any way, does that mean that I can essentially remove all of the solid except for 1 molecule, and the equilibrium will not shift at all since the concentration of the solid stays the same? Similarly, does this mean that I can essentially fill up the container with the solid, and no extra B and C will be produced?

## 1 Answer

Your first statement is right. You just need a certain amount of solid or pure liquid for the reaction to occur, just like how putting in more solid in a saturated solution doesn't increase dissolution, while you need solid in there to call it an equilibrium between aqueous and solid phases. However, the amount of solid or liquid does change the kinetics of the reaction: the rate at which it proceeds to equilibrium.

But if you put in some solid A, the Q is 0, so some B and C will form. It's the same reason why putting in a sugar cube in water increases aqueous sugar concentration: there is an equilibrium between solid and aqueous components, it's just that the constant means the maximum amount of aqueous concentration.