TOPIC 17
EQUILIBRIUM
17.1
THE EQUILIBRIUM LAW
By: Merinda Sautel
Alameda Int’l Jr/Sr High
School
Lakewood, CO
msautel@jeffco.k12.co.us
ESSENTIAL IDEA
The position of equilibrium can be quantified by
the equilibrium law. The equilibrium constant for
a particular reaction only depends on the
temperature.
NATURE OF SCIENCE (1.8 and 1.9)
Employing quantitative reasoning – experimentally
determined rate expressions for forward and backward
reactions can be deduced directly from the stoichiometric
equations and allow Le Chatelier’s principle to be applied.
THEORY OF KNOWLEDGE
The equilibrium law can be deduced by assuming that
the order of the forward reaction and backward reaction
matches the coefficients in the chemical equation. What
is the role of deductive reasoning in science?
We can use mathematics successfully to model
equilibrium systems. Is this because we create
mathematics to mirror reality or because the reality is
intrinsically mathematical?
Many problems in science can only be solved when
assumptions are made which simplify the mathematics.
What is the role of intuition in problem solving?
APPLICATION/SKILLS
Solution of homogeneous
equilibrium problems using the
expression for Kc.
A homogeneous system is one in which
all substances are in the same phase for
example all gases or all solutions.
Heterogeneous Equilibria
The position of a heterogeneous equilibrium
does not depend on the amounts of pure solids
or liquids present. Liquids and solids are
never a part of the K expression.
Write the equilibrium expression for
the reaction:
PCl5(s)  PCl3(l) + Cl2(g)
Pure
solid
Pure
liquid
 K  [C l2 ]
Remember the Equilibrium Constant Kc
jA + kB  lC + mD
l
m
[C ] [ D ]
K
j
k
[ A] [ B ]
Where Kc is the equilibrium constant,
and is unitless. The only thing that
can change Kc for a reaction is a
change in temperature.
Solving for Equilibrium
Concentration
Consider this reaction at some temperature:
H2O(g) + CO(g)  H2(g) + CO2(g)
K = 2.0
Assume you start with 8 moles of H2O and
6 moles of CO in a 1 dm3 container. How
many moles of H2O, CO, H2, and CO2 are
present at equilibrium?
Here, we learn about “ICE” – the most
important problem solving technique in
solving all equilibrium problems.
ICE TABLES
Only molar concentrations are used in ICE
tables.
ICE stands for
“Initial” concentrations,
“Change” in concentrations, and
“Equilibrium” concentrations.
Solving for Equilibrium
Concentration
H2O(g) + CO(g)  H2(g) + CO2(g)
K = 2.0
Step #1: We write the K expression for
the reaction. Always use concentrations in
ICE tables.
[ H 2 ][ CO 2 ]
2.0 
[ H 2 O ][ CO ]
Solving for Equilibrium Concentration
Step #2: We make an ICE table using the
initial concentrations. The “change” comes
from the coefficients.
H2O(g) + CO(g)  H2(g) + CO2(g)
Initial:
8
6
0
0
Change:
-x
-x
+x
+x
Equilibrium:
8-x
6-x
x
x
Solving for Equilibrium Concentration
Step #3: We plug equilibrium concentrations
into our equilibrium expression, and solve for x
H2O(g) + CO(g)  H2(g) + CO2(g)
Equilibrium:
8-x
6-x
x
( x )( x )
2.0 
(8  x )(6  x )
x4
x
ADDITIONAL COMMENTS
• Many equilibrium ICE tables have to be
solved using the quadratic equation.
• You will not be expected to use
calculations involving the quadratic
equation.
• When Kc is very small (10-3 or smaller),
you will be making an assumption that the
change in concentration of the reactants is
basically zero.
UNDERSTANDING/KEY IDEA
17.1.A
Le Chatelier’s principle for changes
in concentration can be explained
by the equilibrium law.
LeChatelier’s Principle
When a system at equilibrium is placed under stress,
the system will undergo a change in such a way as
to relieve that stress.
Translated: The system undergoes a temporary
shift in order to restore equilibrium.
1. When you add a substance or heat, the
system shifts to the opposite side.
2. When you take out a substance or heat, the
system shifts to the side of the take out.
3. When you increase pressure, the system
shifts to the side with the least number of
gaseous molecules.
KINETICS AND EQUILIBRIUM
• The equilibrium constant is the ratio of the rate
constants of the forward and backward
reactions.
• If “k” for the forward reaction is greater than “k”
for the backward reaction, then the reaction is
product favored and proceeds toward
completion. K is large.
• If “k” for the backward reaction is greater than
“k” for the forward reaction, then the reaction
has barely taken place. K is small.
• Using the relationship between k, the rate
constant, and K, the equilibrium constant,
we can add to our interpretation of how
equilibrium responds to changing
conditions.
CHANGING CONCENTRATION
• If you increase the concentration of the
reactant(s), you will increase the rate of
the forward reaction thus shifting
equilibrium to the right.
• If you increase the concentration of the
product(s), you will increase the rate of the
backward reaction thus shifting equilibrium
to the left.
• The value of the equilibrium constant, K,
stays constant since only temperature
affects K.
ADDING A CATALYST
• Adding a catalyst increases the rate
constants, k, of both the forward and
reverse reactions by the same amount so
the equilibrium constant is not affected.
CHANGING TEMPERATURE
• We know from kinetics and our studies of the
Arrhenius equation, k = Ae-Ea/RT, that as
temperature increases, the rate constant “k”
increases.
• From the potential energy diagrams, we know
that the activation energies of the forward and
backward reactions are different so their rate
constants would be different and are affected
differently by temperature.
• The ratio of the rate constants of the
forward and backward reactions is
temperature dependent.
• For an endothermic reaction, where Ea of
the forward reaction is greater than that of
the backward reaction, the increase in
temperature has a greater effect on
increasing “k” for the forward reaction so
Kc increases as temperature increases.
HOW DO YOU PROVE IT?
• Take the time with your partner and prove
the previous statement using the potential
energy diagrams and the
Arrhenius equation k = Ae-Ea/RT.
UNDERSTANDING/KEY IDEA
17.1.B
The position of equilibrium
corresponds to a maximum value
of entropy and a minimum value of
the Gibbs free energy.
• Different reactions have very different Kc
values reflecting a wide range of the
directions and extents of these reactions.
• Some reactions go almost all the way to
completion and others barely react at all.
• Gibbs free energy can help determine
these differences in reactions.
• ΔG◦ = negative
– Spontaneous
– Reaction proceeds in the forward direction
• ΔG◦ = positive
– Non-spontaneous
– Reaction proceeds in the backward direction
• ΔG◦ = zero
– Reaction is at equilibrium
• A reaction with a value of ΔG◦ that is large
and negative appears to occur
spontaneously and has an equilibrium
mixture with a high proportion of products.
• A reaction with a value of ΔG◦ that is large
and positive appears to be nonspontaneous and has an equilibrium
mixture with a high proportion of reactants.
ΔG neg
ΔG pos
Nonspontaneous
pixshark.com
ΔG neg
ΔG pos
Nonspontaneous
The position of equilibrium corresponds to a
maximum value of entropy for the system.
In other words, any system has the highest
possible value of entropy when free energy
is at a minimum (equilibrium).
UNDERSTANDING/KEY IDEA
17.1.C
The Gibb’s free energy change of a
reaction and the equilibrium constant
can both be used to measure the
position of an equilibrium reaction and
are related by the equation,
ΔG◦ = -RTlnK.
APPLICATION/SKILLS
Understand the relationship
between ΔG and the equilibrium
constant.
APPLICATION/SKILLS
Be able to do calculations with the
equation ΔG◦ = -RTlnK.
Kc AND THERMO DATA
• We now have 2 terms which relate to the
position of equilibrium.
– Kc, the equilibrium constant
– ΔG◦, the change in free energy
• From the equation, ΔG◦ = -RTlnK, we can
deduce relationships between the two
values, Kc and ΔG◦.
ΔG◦ negative
ln K positive
K>1
mostly products
ΔG◦ positive
ln K negative
K<1
mostly reactants
ΔG◦ = 0
ln K = 0
K=1
mixture of both
The equation ΔG◦ = -RTlnK can be used to solve for either
Gibb’s free energy or the equilibrium constant.
It is useful in situations where the K is difficult to measure
directly such as if the reaction is too slow to reach equilibrium
or when the component amounts are too small to measure.
The value of Gibb’s free energy is the reason that different
reactions have such different values for K.
Remember that Gibb’s free energy is also dependent upon the
change in enthalpy and change in entropy ΔG◦ = ΔH – TΔS.
Citations
International Baccalaureate Organization. Chemistry Guide, First
assessment 2016. Updated 2015.
Brown, Catrin, and Mike Ford. Higher Level Chemistry. 2nd ed.
N.p.: Pearson Baccalaureate, 2014. Print.
ISBN 978 1 447 95975 5
eBook 978 1 447 95976 2
Most of the information found in this power point comes directly
from this textbook.
The power point has been made to directly complement the Higher
Level Chemistry textbook by Brown and Ford and is used for direct
instructional purposes only.