NOTES: CHAPTER 13 - EQUILIBRIUM
I. The Equilibrium Condition

chemical equilibrium
o the state where the concentrations of all reactants and products remains constant with time

dynamic equilibrium not static
o for any chemical rxn the forward and reverse rxns occur, under equilibrium the rates of the
forward and reverse rxns are equal, leading to the concentrations remaining constant. However,
since both forward and reverse rxns are ongoing leading to a constant destruction and formation
of reactants and products
II. Equilibrium Expressions

equilibrium expressions represent the law of mass action (a general description of the equilibrium
condition)

If aA + bB
⇌ cC + dD then K 
[C ] c [ D] d
where K is the equilibrium constant
[ A] a [ B] b
NOTE: (1) Phase matters. Pure liquids and solids are excluded from the equilibrium expression (see
sample exercise 13.6)
(2) If a rxn is reversed, the equilibrium expression is “flipped”. Such that K reversed  K   K1
(3) Since the coef’s become exponents in the equilibrium expression, when the balanced
equilibrium rxn is multiplied by a constant (n), then the equilibrium constant is raised to that
“n” power. This means that “K” becomes “Kn”
(4) K values are typically written without units. (this is unlike k values in kinetics)
(5) If K > 1 then the amount of products is greater than the amount of reactants and the forward
rxn (products) is favored.
(6) If K < 1 then the amount of reactants is greater than the amount of products and the reverse
rxn (reactants) is favored.

See example exercises 13.1, 13.2, 13.3
III. Pressure-Based Equilibrium Expressions

gas phase equilibria conditions lead to equilibrium constants based upon pressure rather concentrations.

Since PV = nRT and molarity is M = Vn , then P =
n
V
RT. Thus the molar concentration of a gas is directly
related to the pressure of the gas. The individual pressure of the gas in an equilibrium system would be
the partial pressure of the gas.
[C ] c [ D] d
PCc  PDd

Combining K 

The two equilibrium constants K and Kp are related via Kp = K(RT)n where n is (c +d) - (a +b)

See example exercises 13.4, 13.5
[ A] a [ B] b
with P =
n
V
RT leads to the equilibrium expression K p 
PAa  PBb
.
IV. Applications of the equilibrium constant and Determination of equilibrium concentrations
The value of the equilibrium constant allows us to calculate not only the extent of rxn, but also the equilibrium
concentrations of the various chemical species given the initial concentrations of the chemical species.
Equilibrium rxns do not go to completion. By this, I mean that if the reactants A + B are mixed together they are
not 100% converted to C + D.
This entails the calculation of the reaction quotient, Q, where Q 
[C ] c [ D] d
[ A] a [ B] b
.
NOTES: CHAPTER 13 - EQUILIBRIUM
When Q = K, then the reaction is at equilibrium.
When Q  K, then either the forward or reverse reaction rate will be advantaged and the reaction will appear to
“shift” resulting in the formation/destruction of chemical species until the equilibrium condition (forward rate =
reverse rate) is established.
When Q < K then the reaction will shift to the right creating more products and consuming reactants.
When Q > K then the reaction will shift to the left creating more reactants and consuming products.
ICE Tables
Equilibrium concentrations can be determined by evaluating the Initial conditions, the Change (using Q to find
loss/gain of reactants and products) and the Equilibrium conditions
note: for simplicity we will assume all coef’s are 1, thus all mol/mol ratios are 1/1 and  that Q < K


So for the equilibrium rxn aA + bB
⇌ cC + dD where
K
[C ] c [ D] d
[ A] a [ B] b
, the following applies
Chemical
species
Initial
Concentrations
Change
Equilibrium
Concentrations
A
[A] 
-x
[A]eq = [A]  - x
B
[B] 
-x
[B]eq = [B]  - x
C
[C] 
+x
[C] eq = [C]  + x
D
[D] 
+x
[D]eq = [D]  + x
Substituting the information in the table into the equilibrium expression leads to the following:
K
[C ]  x c [ D]  x d
[ A]  x a [ B]  x b
One can use the value of K and the initial concentrations to solve for the value of x. Once x has been
determined then the equilibrium concentrations can be calculated through simple subtraction.

NOTE: If the coef’s in the equilibrium reaction are not “1”, then the ICE table can be most readily
determined by dividing to ensure that the given chemical species coef is “1”.

NOTE: If Q > K, then the shift is opposite resulting in -x becoming +x and +x becoming -x within the
ICE table and the resulting equilibrium expression.
c

So for the equilibrium rxn
a
a
A
⇌
c
a
C +
d
a
D where
K
d
[C ] a [ D] a
a
, the following applies
[ A] a
Chemical
species
Initial
Concentrations
Change
Equilibrium
Concentrations
A
[A] 
-x
[A]eq = [A]  - x
C
[C] 
+
c
a
x
[C] eq = [C]  +
c
a
x
D
[D] 
+
d
a
x
[D]eq = [D]  +
d
a
x
NOTES: CHAPTER 13 - EQUILIBRIUM
SOLVING for the VALUE of x

Often an equilibrium expression results in the formation of an exponential equation containing x. Solving
such an exponential equation would result in more than one value for x. e.g
calculate “reasonable” equilibrium concentrations using those values.
4   2 One must
For example, if [A]  = 8 and x =  2, then using [A]eq = [A]  - x would result in [A]eq 6 or 10. Since
“A” is being consumed there should be less of A at equilibrium so [A] eq must be 6 and not 10.
SHORTCUT
If the the value of x is very small - less than 5% of the initial concentrations. This results in the
addition/subtraction of x to an nonzero initial concentration being negligible. {This often occurs when
the equilibrium constant is very small, K  10-3 or very large K  10+3}
[A]eq = [A]  - x 
[A]eq = [A] 
For you “hardcore math/calculator types” do not forget that we are
working in the real world, which means we have limitations (errors) to our
measurements and thus significant figures. In other words, 2.0 - 0.01 = 2.0
EXAMPLE
Question:
Solution:
Given the system, A ⇌ C + 2D where K is 1.0 x 10-6, a student places 0.20 M of A in to the
rxn vessel, calculate the equilibrium concentrations of A, C and D.
Overall, we need to know the value of the change “x” so that we can add/subtract it to the initial
amounts to determine the equilibrium concentrations.
First since we only have A, Q < K. So we will use up A and make C and D.
Second, we write an ICE table.
Chemical
species
A
Initial
Concentrations
Change
Equilibrium
Concentrations
-x
0.20 - x
+x
0+x
2
1
0 + 2x
mol
L
0.20
C
0
D
0
+
x
Third, write the equilibrium expression and solve for x.
K
[C ]  x c [ D]  2 x d
[ A]  x a

1.0 10 6 
[0  x 1 0  2 x 2
0.20  x 1
since K is small, x is small, so 0.20 - x = 0.20 and we have
1.0 10 6 
[0  x 1 0  2 x 2
0.20 1
 1.0  10 6 
4x 3
0.20 

0.0079 = x {0.0037-.20 is 1.9 %}
Third, using the Equilibrium concentrations column of the ICE table, we calculate
[A]eq = 0.20 - 0.0037 = 0.20 M
[C]eq = 0+ 0.0037
= 0.0037 M
[D]eq = 0 + 2(0.0037) = 0.0074 M
NOTES: CHAPTER 13 - EQUILIBRIUM
V. le Châtelier’s Principle
le Chatelier’s Principle
“When a stress is applied to an equilibrium system, the system will shift so as to relieve
that stress.”
To fully understand le Chatelier’s principle we have to understand: stress and shift.
stress - A change in reaction conditions (concentration, temp, pres).
Remember that changes in concentration and temp are directly reflected in the reaction rate.
shift - At equilibrium the rates of the forward and reverse directions are equal, but when a stress
applied the stress gives an advantage to either the forward/reverse direction and a disadvantage
to the other direction. The results in an increase in the concentration of all chemical species that
are produced in the advantaged direction and a decrease in the concentration of all chemical
species that are produced in the disadvantaged direction. The “shift” is the direction in which
the advantaged direction is going.
EXAMPLE
Question:
Solution:
Given the system, A + B ⇌ C + D + heat, determine the shift and resulting changes in
concentration when the following stresses are applied: (a) an increase in [C]
(b) a decrease in [C]
(a) Increasing [C] results in an increase in the rate of the reverse direction and no impact in the
forward direction. Since the reverse direction’s rate has increase the [C] + [D] will decrease
and the [A] + [B] will increase.
(b) Decreasing [C] results in a decrease in the rate of the reverse direction and no impact in the
forward direction. Since the reverse direction’s rate has decreased the [C] + [D] will
increase and the [A] + [B] will decrease.
VI. Types of Problems
Calculating K given equilibrium concentrations
Calculating K from K p , and vice versa
Calculating equilibrium concentrations given initial concentrations
Determining Shifts Based on Le Châtelier’s Principle
Calculating Equilibrium Concentrations Based on Shifts
VII. Website for Tutorial
http://www.chem.uncc.edu/faculty/murphy/1252/Chapter15/sld001.htm
IX. Demo/Lab
Fe3++SCN-FeSCN2+
Fe3+ is yellow. SCN- is colorless. FeSCN2+ is deep red. Stresses on the system
change the color of the solution. If the concentration of FeSCN2+ increases, the
solution would become a deeper red. If the concentration of FeSCN2+ decreases,
the solution would become a lighter red.
http://www2.ucdsb.on.ca/tiss/stretton/CHEM2/equilab1.htm