EQUILIBRIUM CONSTANTS (Keq)
One way to predict which direction is favored in a reversible chemical reaction
is to look at the concentrations of the reactants compared to the concentrations
of the products. Two Norwegian chemists, Cato Maximilian Guldberg and
Peter Waage proposed the Law of Mass Action as a general description of the
equilibrium condition. Guldberg and Waage mathematically compared the
ratio of products to reactants and took into consideration the coefficients of
each in the balanced chemical equation. The following generic equation will
be used to show how they set up their mass action expression (also called the
equilibrium expression).
mA + nB <---------------> p C + qD
In this reaction A and B represent the reactants and C and D represent the
products. The small letters in italics represent the coefficients in the balanced
equation. The mass-action expression is shown by multiplying the
concentrations of the products times each other and placing them over the
reactants times each other. The coefficients become exponents in the massaction expression. The mass-action expression for this reaction is shown
below.
[C]p x [D]q
-------------[A]m x [B]n
If the concentrations of the reactants and products are known, they can be
inserted into the mass-action expression and the equilibrium constant (Keq) can
be calculated. Concentration for calculating equilibrium constants is usually in
moles per liter (M). A large Keq tells us that the forward direction is favored in
a reaction and a small Keq tells us that the reverse direction is favored. A large
Keq is usually considered to be a number larger than 1 and a small Keq is
usually considered to be a number smaller than 1. The following example will
show how to set up the mass-action expression for a reaction and how to
calculate the Keq when the concentrations are known at equilibrium.
Example 1
Find the Keq for the following reaction and predict if the forward or reverse
direction is favored.
At equilibrium, the following concentrations were measured.
[HI] = 1.544M
[H2] = 0.228M
[I2] = 0.228M
H2(g) + I2(g) <---------------------> 2HI(g)
The mass-action expression for this reaction would be:
[HI]2
-----------[H2] x [I2]
To calculate the equilibrium constant we plug in the equilibrium concentrations
that were given in the problem and do the math.
[1.544]2
Keq = ------------------[0.228] [0.228]
= 45.9
The relatively large Keq tells us that the forward direction is favored at
equilibrium.
If you know the Keq for a reaction you can calculate what the equilibrium
concentrations would be for the reactants and products. The following example
shows how this can be done.
Example 2
Find the equilibrium concentration of N2 given the following information.
Keq = 1.20 x 10-2
[H2] = 1.25M
[NH3] = .450M
3H2(g) + N2(g) <---------------> 2NH3(g)
[NH3]2
Keq = 1.20 x 10-2 = ------------------[H2]3 [N2]
[.450]2
[N2] = ----------------------1.20 x 10-2 [1.25]3
=
=
[.450]2
------------------[1.25]3 [N2]
8.64 M
Example 3
In a 1.00 Liter container, 1 mole of H2O is decomposed according to the
equation
2H2O(g) <-----------------------> 2H2(g) + 02(g)
At the equilibrium point, 0.125 moles of O2 are present. Calculate the
concentrations of H2 and H20 in moles per liter and calculate the equilibrium
constant.
Since the H2 and O2 form in a 2:1 ratio (from the balanced equation), if there
are .125 moles of O2, there should be twice as much of the H2, or 2.50 moles.
We can figure out the number of moles of H2O left by applying the same logic.
For every 2 moles of H2 that form, 2 moles of H2O must react, so if we have
.250 moles of H2 then .250 moles of H2O must have reacted. Since we started
with 1 mole of H2O and .250 moles have reacted, then .750 moles of H2O must
remain in the container. The number of moles of each of the substances will be
the same as their concentrations in moles per liter since they are in a 1.00 L
container.
[O2] = .125M
[H2] = .250M
[H2O] = .750M
Now plug the concentrations into the mass-action expression for the reaction
and the equilibrium constant can be calculated.
[H2]2 [O2]
Keq = ------------------[H2O]2
=
[.250]2 [.125]
-------------------[.750]2
= 0.0139