Is it true that solids and liquids are not included in the expression of the equilibrium constant?

1 Answer
Feb 9, 2016

Answer:

Yes, that is correct.

Explanation:

The thing to remember about a heterogeneous equilibrium, which as you know involves substances in more than one phase, is that changing the amount of a solid will not affect the position of the equilibrium.

The same can be said for a pure liquid, with the mention here that the solutions in question must be rather dilute to begin with.

For your given equilibrium reaction

#"NH"_4"Cl"_text((s]) rightleftharpoons "NH"_text(3(g]) + "HCl"_text((g])#

adding more ammonium chloride, #"NH"_4"Cl"#, will not affect the position of the equilibrium. This is why the concentration of the solid is not included in the expressions for the two equilibrium constants #K_c# and #K_p#, which for this equilibrium will take the form

#K_c = overbrace(["NH"_3] * ["HCl"])^(color(purple)("concentrations"))" "# and #" "K_p = overbrace((NH_3) * (HCl))^(color(green)("partial pressures"))#

Now, without going into detail about this, the reason for why solids do not affect chemical equilibria can be traced back to their chemical activity.

In simple terms, chemical activity is a measure of how the concentration of a substance under some specific conditions compares with the concentration of that substance under standard conditions.

So, if a substance has the same concentration (you'll sometimes see this referred to as molar density) at some given conditions and at standard conditions, then its chemical activity is equal to #1#.

Excluding extreme changes in pressure and temperature, solids will always have the same concentration.

In your example, for a given temperature, the partial pressures and concentrations of the products will not depend on the amount of ammonium chloride present.

Once the equilibrium is established, adding or removing ammonium chloride will not change its chemical activity, i.e. its concentration, so the position of the equilibrium remains unchanged.