When we first learned about the equilibrium constant
expression, we did not consider the physical states of the reactants and
products. We took the generic
equation

aA +
bB <---------->
cC +
dD
(1)

and derived the equilibrium constant expression

[C]_{eq}^{c}[D]_{eq}^{d
}K_{C} =
---------------
(2)

[A]_{eq}^{a}[B]_{eq}^{b}

In our earlier consideration of equilibrium, the only
significance we put on the physical state of the substances in the equation was
that if they were gases, we could consider an equilibrium constant that is
defined in terms of gas pressures instead of molar concentrations.
For example, taking substances A, B, C and D to all be gases,

aA(g) + bB(g) <----------> cC(g) + dD(g) (3)

we can write the pressure-based equilibrium constant

P_{C eq}^{c}
P_{D eq}^{d
}K_{P} =
----------------
(4)

P_{A eq}^{a} P_{B eq}^{b}

In deriving K_{C} (and similar arguments would hold
in a derivation of K_{P}) the idea was that constants were gathered on
the left side of the equation and quantities that could vary were gathered on
the right side of the equation. Recall
that the concentration-based equilibrium constant K_{C} was just the
ratio of the forward rate constant to the reverse rate constant:

k_{f
}K_{C} =
---
(5)

k_{r}

The forward and reverse rate constants are independent of
concentration, so K_{C} is likewise independent of concentration.
In equation (2), the individual concentrations can vary from one
equilibrium mixture to the next, but the mathematical result of substituting *all
*of those concentrations into the expression is always the same number.

But some concentrations never vary, so it makes sense to
just absorb those constant concentrations into the equilibrium constant, and
include on the right hand side of the equation, only those concentrations that
can truly vary. We will see that
when a substance is a gas or is dissolved in water (or any other solvent for
that matter – but water is the most common) its concentration can vary, but
when it is a liquid or a solid, its concentration can not vary.

Figure 1.
If 2 moles of gas occupies a volume of 2 liters,

then the concentration is 2 mol / 2 L = 1 mol/L.

If the 2 moles of gas is then compressed to

occupy a volume of 1 liter, 2 mol / 1 L = 2 mol/L.

Now consider the aqueous
(dissolved in water) state. You
know from experience that substances in solution can have a variable
concentration. If you put sugar in
your tea or coffee, you can adjust the sweetness to your personal taste by
choosing the amount of sugar you dissolve.
Two teaspoons of sugar in a glass of tea makes a sweeter drink than only
one teaspoon.

Unlike gases and substances in solution, liquids and solids have an essentially constant concentration. This is because liquids and solids are practically incompressible. The molecules in liquids and solids are very close together, with very little room to squeeze any closer. As a result, it takes an enormous pressure to cause even a tiny reduction in volume. In most circumstances, we can ignore any changes in volume for a solid or liquid substance.

Consider water as an example.
Perhaps you have learned in one of your science classes that the density
of water is 1.00 g/mL. Will this
value ever change? Well, to a tiny
extent, yes. But normally not
enough to worry about. Temperature
will have a small effect, because water will expand when heated and contract
when cooled. Since density is mass
divided by volume, changing the volume occupied by any given mass of water will
change its density. But the volume
changes are very small. If you put
some water in a pot on your stove and heat it all the way from room temperature
to boiling, you don’t notice any obvious increase in the water’s volume.
And because water (like other liquids and solids) is nearly
incompressible, ordinary atmospheric pressure changes are not going to affect
the volume of a water sample. Working
to 3 significant figures, we can safely say that water is always 1.00 g/mL.
And if water always has this density, then it always has the same molar
concentration. In fact, we can
calculate the molar concentration of any substance from its density. Here is the calculation for water:

g
1000 mL
1 mol
mol

1.00 ------- x
------------- x ------------
= 55.5
--------

mL
1 L
18.02 g
L

The only way water can have a
concentration other than 55.5 mol/L is if it has a density other than 1.00 g/mL.
So if we don’t have to consider any changes in density, then we don’t
have to consider any changes in molar concentration either.

Now lets revisit the generic
equilibrium reaction in equation 1, but this time lets assign some particular
physical states to the substances in that equation.

aA(s) + bB(aq) <----------> cC(g) + dD(aq) (6)

Up until now, these physical
states would not make any difference to us, and we would still write the
concentration-based equilibrium constant (K_{C}) as

[C]_{eq}^{c}[D]_{eq}^{d
}K_{C}^{’} =
--------------
(7)

[A]_{eq}^{a}[B]_{eq}^{b}

just as before.
The only difference between equation 7 and equation 2 is that I have
added a prime mark to K_{C} to
make it temporarily look different. You
will see momentarily why I chose to do this.

We now make the following
argument: Substance A is a solid, so its concentration is essentially constant.
Therefore, it belongs on the left side of the equation with the constant,
rather than the right side, with the things that can vary.
If we multiply both sides of equation 7 by [A]_{eq}^{a}
we get

[C]_{eq}^{c}[D]_{eq}^{d
}K_{C}^{’} ^{.} [A]_{eq}^{a}
= ----------------
(8)

[B]_{eq}^{b}

When the right side of equation
7 is multiplied by [A]_{eq}^{a}, it cancels out, therefore this
concentration does not appear on the right side of equation 8.
Notice that the left side of equation 8 contains two constants: K_{C}^{’}
and [A]_{eq}^{a}. Since
a constant multiplied by a constant is just another constant, we can declare K_{C}
(without the prime mark) to be equal to the product of K_{C}^{’}
and [A]_{eq}^{a}. That
is,

K_{C}
= K_{C}^{’}
^{.} [A]_{eq}^{a}
(9)

Substituting equation 9 in
equation 8 gives

[C]_{eq}^{c}[D]_{eq}^{d
}K_{C} =
-----------------
(10)

[B]_{eq}^{b}

Notice that the only difference between the right hand side of equation 10 and the right hand side of equation 2 is that the concentration of substance A is missing. This is precisely the one substance which was a solid. All the other substances were said to be either aqueous or gaseous. Of course, equation 10 was derived on the basis of substance A being the one and only substance to have a constant concentration. A different equation would result if a different substance (or substances) in the equation were considered to have a constant concentration. But we don’t need to re-derive an equation like equation 10 every time we change which substances have a constant concentration. In every case, the only difference between such an equation and equation 2 is that the substances with a constant concentration are omitted from the right hand side of the equation. Their constant concentration is considered to be “absorbed in the equilibrium constant”.

So the new rule we will now
always apply is:

*Write concentration-based equilibrium constant expressions in the
mathematical form you have learned about in the very beginning of your study of
equilibrium, except omit those substances that are solid or liquid.
Include only substances in the aqueous or gaseous state in your
concentration-based equilibrium constant (K _{C}).*

The above discussion has
focused mainly on concentration-based (K_{C}) equilibrium constants.
Equations 3 and 4 did consider pressure-based (K_{P}) equilibrium
constants, but in those equations, *all*
the substances were assumed to be gases. Perhaps
you were wondering how you would handle K_{P} if only *some*
of the substances were gases, but not all of them were.
The answer is that those substances that are in aqueous solution would
continue to be included in the equilibrium constant expression as molar
concentrations, with only the gases being entered as pressures in atmospheres.
Although we must enter the concentrations as moles per liter or
atmospheres as appropriate, we normally don’t enter the units.
That’s because thermodynamic equilibrium constants are dimensionless
numbers. We should actually be
using something called *activities*
rather than concentrations in moles per liters and pressures in atmospheres, but
the use of moles per liter and atmospheres gives – at least for our purposes
– a good approximation of the thermodynamic equilibrium constant.
And speaking of activities:

In equations 6 through 10, you saw how the
constant concentration of a solid (and the same argument would hold for a
liquid) can be used to exclude it from the equilibrium constant expression.
Here’s an even better explanation for why solids and liquids are
excluded: pure solids and liquids have an activity of 1.
Activities are dimensionless numbers, so a pure solid or liquid does not
change the value of an equilibrium constant.
Neither multiplying nor dividing by 1 changes anything.
That’s the *real* reason solids
and liquids don’t appear in equilibrium constant expressions.
However, the explanation put forth in the development of equations 6
through 10 makes more sense to students who are accustomed to thinking in terms
of molar concentrations rather than activities.

For the reaction in equation 6

aA(s) + bB(aq)
<----------> cC(g)
+ dD(aq)
(6)

the pressure-based (K_{P})
equilibrium constant expression is

P_{C eq}^{c}
[D]_{eq}^{d
}K_{P} =
------------------
(11)

[B]_{eq}^{b}

Only substance C is entered as a pressure, because only
substance C is a gas. Substances B
and D are in aqueous solution, and they continue to appear as molar
concentrations. Substance A does
not appear, because it is a solid.

To summarize, then, solids and liquid are omitted from both
the concentration-based (K_{C}) and pressure-based (K_{P})
equilibrium constant expressions. In
concentration-based (K_{C}) equilibrium constant expressions, both
aqueous and gaseous substances appear as molar concentrations.
In pressure-based (K_{P}) equilibrium constant expressions,
gaseous substances appear as pressures, and aqueous substances continue to
appear as molar concentrations.