In a heterogeneous equilibrium, the concentrations of pure solids and pure liquids are ignored in the equilibrium expression for the reaction because

Question

Options:

 

they are not required for the equilibrium to be established.

their concentrations remain constant and do not affect the expression.

they do not participate in the reaction.

 

OR
they need not be present in the equilibrium mixture.

Answer 1

If a reaction occurs in more than one phase, that reaction will occur in a heterogeneous system.

 

In general, we write equilibrium constant expressions in terms of molar concentration or gas phase pressures.

The position of the equilibrium state of a system does not depend on the amount of liquid or solid in the reaction, provided that some exists.
Pure liquids and solids do not appear in the equilibrium expression. 

To understand why this is true, consider a simple reaction like the vaporization of water:

H2O(l) < - > H2O(g)
At a given temperature, the vapor pressure of water is a constant no matter how much liquid water there is. Thus, the position of the equilibrium does not change when you add or subtract water, and so the equilibrium expression will have no term for liquid water.

This example helps explain why the term for the solvent never shows up in equilibrium constant expressions for reactions that are done in that solvent. Only the solutes appear.

 

So, I think option 2 is the answer you want.