19.1. The Nature of Chemical Equilibrium http://www.ck12.org
Heterogeneous Equilibria
So far, we have been assuming that a decrease in the amount of a reactant or product also decreases its concentration.
This is true for gases, where the concentration is equal to the number of moles divided by the volume of the container.
Because the container generally does not change size over the course of the reaction, fewer moles of gas means a
lower concentration. This is also true for substances that are dissolved in water or another solvent. Assuming the
volume of the solvent does not change appreciably during the reaction, fewer moles of solute in the same amount of
solvent results in a lower concentration.
However, for solid or liquid reaction components, this is not necessarily true. Consider the following reaction:
NH 4 CO 2 NH 2 (s) 2NH 3 (g)+CO 2 (g)
The "concentration" of a solid is essentially another way of measuring its density. A certain number of moles are
present for a given volume of the solid, and this value does not depend on how much of the solid is present. As the
reaction proceeds, the amount of solid reactant will decrease, but so will the volume it occupies. A similar argument
can be made for liquids. As a result, the concentrations of solids and liquids are essentially constant over the course
of a reaction.
Now, let’s look at the equilibrium constant expression that could be generated for this reaction. Following the rules
that we have given so far, it would take the following form:
Keq= [NH^3 ]
(^2) [CO 2 ]
[NH 4 CO 2 NH 2 ]
However, because the concentration of solid ammonium carbamate is constant during the reaction, we could poten-
tially simplify this expression as follows:
(Keq)[NH 4 CO 2 NH 2 ] = [NH 3 ]^2 [CO 2 ]
Keq′ = [NH 3 ]^2 [CO 2 ]
where K’eqis the product of the original equilibrium constant and the constant concentration of the solid reactant.
By convention, the simpler form is used by chemists. Whenever solid or liquid reactants or products are part of a
chemical reaction, their concentrations arenotincluded in the equilibrium constant expression, and it is understood
that the constant concentration of the missing reaction components is incorporated into the value of any measured
equilibrium constants.
Example 19.2
Write the equilibrium constant expression for the following reaction:
2Mg(s)+O 2 (g) 2MgO(s)
Answer:
Because both Mg and MgO are present as solids, they are not included in the equilibrium constant expression. The
only remaining component is oxygen gas. Because oxygen is a reactant, it will appear in the denominator.
Keq=[O^1
2 ]
Concentration vs. Partial Pressure
Let’s take a look at one way in which the ideal gas law can be rearranged. We are most familiar with the following
form:
PV=nRT
Dividing both sides by volume and rearranging the variables, we could also write this law as follows: