CHAPTER 14
Chemical Equilibrium
14.1: Equilibrium Constant, Keq
Objective:
(1) To write the equilibrium constant expression for a
chemical reaction.

Reversible Reactions and Equilibrium


Reversible Reaction: A chemical reactions in which
products re-form the original reactants.
Arrows that point in opposite directions are used to
indicate a reaction is reversible.


Example: H2(g) + I2(g)
2HI(g)
Chemical Equilibrium: A state of balance in which the
rate of a forward reaction equals the rate of the reverse
reactions and the concentrations of products and reactants
remain unchanged.
Equilibrium Constant, Keq

Equilibrium Constant, Keq: a number that relates
that concentrations of starting materials and
products of a reversible chemical reaction to one
another at a given temperature.
aA + bB  cC + dD
c
d
[C ] [ D]
K eq 
a
b
[ A] [ B]
coefficient
concentration
Writing an Equilibrium Constant
Expression
•
•
•
•
Step 1: Balance the chemical equation.
Step 2: Set up your Keq expression with the products
on the top of a fraction and the reactants on the
bottom of a fraction.
Step 3: Raise each substance's concentration to the
power equal to the substance’s coefficient in the
balanced equation.
Note: Solids (s) and pure liquids (l) are not used in
the expression because their concentrations do not
change.
Example
Write the equilibrium constant expression for the following
reaction:
CaCO3(s) + CO2(aq) + H2O(l)
Ca2+(aq) + 2HCO3-(aq)

2
 2
3
[Ca ][HCO ]
K eq 
[CO2 ]
Practice
Write the equilibrium constant expression for the following
chemical reactions at equilibrium (don’t forget to balance
the equation):
1.) H2CO3(aq) + H2O(l)
HCO3-(aq) + H3O+ (aq)

2.) COCl2 (g)
3.) CO(g)
CO(g) + Cl2 (g)
C(s) + CO2 (g)
Answers
HCO3-(aq) + H3O+ (aq)
1.) H2CO3(aq) + H2O(l)
[CO][Cl2 ]
K eq 
[COCl2 ]
2.) COCl2 (g)
CO(g) + Cl2 (g)


3
[ H 3O ][HCO ]
K eq 
[ H 2CO3 ]
3.) 2CO(g)
C(s) + CO2 (g)
[CO2 ]
K eq 
2
[CO ]
14.1: Equilibrium Constant, Keq
Objective:
(1) To calculate the equilibrium constant.

What does the Keq tell us?



Keq < 1 Favors Reactants
Keq = 1 Same amount of Reactants and Products
Keq > 1 Favors Products
Practice: Determine if the following Keq values favor the
reactants, products, or neither.
1.) Keq = 0.02
2.) Keq = 1
3.) Keq = 50

Calculating Keq




Step 1: Write the balanced chemical equation.
Step 2: Set up your Keq expression.
Step 3: Substitute concentrations.
Step 4: Calculate!
Example
An aqueous solution of carbonic acid reacts to reach
equilibrium as described below:
H2CO3(aq) + H2O(l)
HCO3-(aq) + H3O+ (aq)
The solution contains the following solute concentrations:
H2CO3 = 3.3 x 10-2 M; HCO3- = 1.19 x 10-4 M; H3O+
= 1.19 x 10-4 M. Determine the Keq.

[ H 3O  ][HCO3 ] (1.19x104 )(1.19x104 )
7
K eq 


4
.
29
x
10
2
[ H 2CO3 ]
(3.3x10 )
Note: Keq does not have units!
Practice
1.a. Calculate the equilibrium constant for the
following reaction:
COCl2(g)
CO(g) + Cl2(g)
[CO] = 0.0178 M
[Cl2] = 0.0178 M
[COCl2] = 0.00740 M
b. Are the reactants for products favored?
Practice
2.a. For the system involving dinitrogen tetraoxide
and nitrogen dioxide at equilibrium at a
temperature of 100⁰C, the product concentration of
N2O4 is 4.0 x 10-2 M and the reactant concentration
of NO2 is 1.4x 10-1 M. What is the Keq value for
this reaction?
NO2(g)
N2O4(g)
b. Are the reactants or products favored?
Practice
3.a. An equilibrium mixture at 852 K is found to
contain 3.61 x 10-3 M of SO2, 6.11 x 10-4 M of O2,
and 1.01 x 10-2 M of SO3. Calculate the
equilibrium constant for the reaction.
SO2 (g) + O2 (g)
SO3 (g)
b. Are the reactants or products favored?
Calculating Concentrations from Keq
4. Keq for the equilibrium below is 1.8 x 10-5 at a
temperature of 25⁰C. Calculate [NH3] when
[NH4+] and [OH-] are 3.5 x 10-4 M.
NH3(aq) + H2O(l)
NH4+ (aq) + OH-(aq)

K eq 
[ NH 4 ][OH  ]
[ NH 3 ]
4
4
(
3
.
5
x
10
)(
3
.
5
x
10
)
5
1.8x10 
[ NH 3 ]
[NH3] = 6.8 x10-3 M
Practice
5. a. If the equilibrium constant is 1.65 x 10-3 at
2027⁰C for the reaction below, what is the
equilibrium concentration of NO when [N2] = 1.8 x
10-3 M and [O2] = 4.2 x 10-3 M.
N2(g) + O2(g)
NO(g)
b. Are the reactants for products favored?
Practice
6.a. At 600⁰C, the Keq for the reaction below is 4.32
when [SO3] = 0.260 M and [O2] = 0.045 M.
Calculate the equilibrium concentration for sulfur
dioxide.
SO2(g) + O2(g)
SO3(g)
b. Are the reactants or products favored?
14.2 Solubility Product Constant, Ksp
Objective:
(1) To calculate the solubility product constant, Ksp.

Solubility

The maximum concentration of a salt in an aqueous
solution is called the solubility of the salt in water.
Solubility Product Constant, Ksp


Solubility Product Constant, Ksp: the equilibrium
constant for a solid that is in equilibrium with the
solid’s dissolved ions.
How much of a partially soluble salt will dissolve?
AaBb (s)
aA (aq) + bB (aq)
Ksp  [ A] [B]
a
b
Calculating Ksp
The lower the value of Ksp, the less soluble the
substance.
 Practice:
Rank the following substances from least soluble to
most soluble:

Salt
Ag2CO3
BaSO4
Ca3(PO4)2
CuS
Ksp
8.4 x 10-12
1.1 x 10-10
2.1 x 10-33
1.3 x 10-36
Calculating Ksp
The lower the value of Ksp, the less soluble the
substance.
 Practice:
Rank the following substances from least soluble to
most soluble:

Salt
Ag2CO3
BaSO4
Ca3(PO4)2
CuS
Ksp
8.4 x 10-12
1.1 x 10-10
2.1 x 10-33
1.3 x 10-36
CuS
Ca3(PO4)2
Ag2CO3
BaSO4
Least soluble
Most soluble
Calculating Ksp




Step 1: Write and Balance the equation.
Step 2: Determine the concentration of the ions.
Step 3: Write the solubility product expression.
Step 4: Substitute values and calculate.
Example

Most parts of oceans are nearly saturated with
calcium fluoride. A saturated solution of CaF2 at
25⁰C has a solubility of 3.4 x 10-4 M. Calculate the
solubility product constant for CaF2.
CaF2(s)
Ca2+(aq) + F-(aq)
Solution
Balance equation: CaF2(s)
Ca2+(aq) + 2F-(aq)
2.
Determine Concentrations:
CaF2(s)
Ca2+(aq) + 2F-(aq)
3.4 x 10-4
3.4 x 10-4 6.8 x 10-4
3. Write solubility product expression:
Ksp  [Ca2 ][F  ]2
4. Substitute values and calculate:
1.
Ksp  [Ca 2 ][F  ]2  (3.4x104 )(6.8x104 )2  1.6x1010
Note: Ksp does not have units!
Practice
1. Copper(I) bromide is dissolved in water to
saturation at 25⁰C. The concentration of Cu+ and
Br- ions in solution is 7.9 x 10-5 M. Calculate the Ksp
for copper(I) bromide at this temperature.
Practice
2. What is the Ksp value for calcium phosphate at 298
K if the concentrations in a solution at equilibrium
with excess solid are 3.42 x 10-7 M for Ca2+ and
2.28 x 10-7 M for PO43- ions?
Practice
3. If a saturated solution of silver chloride contains an
AgCl concentration of 1.34 x 10-5 M, what is the
solubility product constant?
Practice
4. A saturated solution of magnesium fluoride contains
a MgCl2 concentration of 1.19x10-3 M. What is the
Ksp for magnesium fluoride?
Calculating Concentration from Ksp
5. What is the concentration of Ca2+ in a saturated
solution of CaF2 if the concentration of F- is 2.20 x
10-3M and Ksp = 5.30 x 10-9.
Practice
6. What is the concentration of Al3+ in a saturated
solution of Al(OH)3 if the OH- concentration is 7.90
x 10-9 M. Ksp = 1.30 x 10-33.
Practice: Chem 331
7. The Ksp for lead(II) iodide is 7.08 x 10-9 at 25⁰C.
What is the molar concentration of PbI2 in a
saturated solution?
The Ksp for lead(II) iodide is 7.08 x 10-9 at 25⁰C.
What is the molar concentration of PbI2 in a
saturated solution?
Step 1: Write and Balance Equation
PbI2 (s)
Pb2+ (aq) + 2I- (aq)
The Ksp for lead(II) iodide is 7.08 x 10-9 at 25⁰C.
What is the molar concentration of PbI2 in a
saturated solution?
Step 1: Write and Balance Equation
PbI2 (s)
Pb2+ (aq) + 2I- (aq)
Step 2: Write the Ksp expression
K  [ Pb2 ][I  ]2
sp
The Ksp for lead(II) iodide is 7.08 x 10-9 at 25⁰C.
What is the molar concentration of PbI2 in a
saturated solution?
Step 1: Write and Balance Equation
PbI2 (s)
Pb2+ (aq) + 2I- (aq)
Step 2: Write the Ksp expression
K  [ Pb2 ][I  ]2
sp
Step 3: Assign x values to concentrations
PbI2 (s)
Pb2+ (aq) + 2I- (aq)
x
x
2x
The Ksp for lead(II) iodide is 7.08 x 10-9 at 25⁰C.
What is the molar concentration of PbI2 in a
saturated solution?
Step 1: Write and Balance Equation
PbI2 (s)
Pb2+ (aq) + 2I- (aq)
Step 2: Write the Ksp expression
K  [ Pb2 ][I  ]2
sp
Step 3: Assign x values to concentrations
PbI2 (s)
Pb2+ (aq) + 2I- (aq)
x
x
2x
Step 4: Solve
Ksp  [Pb2 ][I  ]2  [ x][2x]2  ( x)(4x2 )  4x3  7.08x109
x = [PbI2] = 1.21 x 10-3 M
Practice
8. The Ksp of calcium sulfate is 9.1 x 10-6. What is the
molar concentration of calcium sulfate in a
saturated solution?
Practice
9. The Ksp of CdF2 is 6.4 x 10-3. What is the molar
concentration of cadmium fluoride in a saturated
solution?
14.3 LeChatelier’s Principle
Objective:
(1) To use LeChatelier’s Principle to determine how a
system at equilibrium will respond to an external
stress.

LeChatelier’s Principle

LeChatelier’s Principle: When a system at
equilibrium is disturbed, the system adjusts in a way
to reduce the change.
There are 3 possible disturbances:
Change in (1) concentration, (2) temperature, or (3)
pressure

1. Change in Concentration




Increase concentration of reactant  Equilibrium
shifts toward products
Decrease concentration of reactant  Equilibrium
shifts toward reactants
Increase concentration of product  Equilibrium
shifts toward reactants
Decrease concentration of product  Equilibrium
shifts toward products
Use the following reaction to answer the questions
below:
H2 (g) + I2 (g)
2HI (g)
In which direction (left or right) does the equilibrium
shift in each of the following situations:
1.) Increase H2
2.) Decrease I2
3.) Increase HI
4.) Decrease HI

Use the following reaction to answer the questions
below:
H2 (g) + I2 (g)
2HI (g)
In which direction (left or right) does the equilibrium
shift in each of the following situations:
1.) Increase H2
RIGHT
2.) Decrease I2
LEFT
3.) Increase HI
LEFT
4.) Decrease HI RIGHT

2. Change in Temperature

Think of heat as a reactant or product
Exothermic: heat is a product
Endothermic: heat is a reactant

For an exothermic reaction:


temperature  equilibrium favors reactants
 Decreasing temperature  equilibrium favors products
 Increasing

For an endothermic reaction
temperature  equilibrium favors products
 Decreasing temperature  equilibrium favors reactants
 Increasing
Use the following reaction to answer the questions
below:
2SO3(g) + CO2 (g) + heat
CS2 (g) + 4O2(g)

In which direction (left or right) does the equilibrium shift
in each of the following situations:
1.) Increase the temperature
2.) Decrease the temperature
Use the following reaction to answer the questions
below:
2SO3(g) + CO2 (g) + heat
CS2 (g) + 4O2(g)

In which direction (left or right) does the equilibrium shift
in each of the following situations:
1.) Increase the temperature RIGHT
2.) Decrease the temperature LEFT
3. Change in Pressure



Only affects gases!
Increasing pressure  Equilibrium shifts toward the
side with fewer moles of gas
Decreasing pressure  Equilibrium shifts toward
the side with more moles of gas
Use the following reaction to answer the questions
below:
2SO3(g) + CO2 (g) + heat
CS2 (g) + 4O2(g)

In which direction (left or right) does the equilibrium shift
in each of the following situations:
1.) Increase the pressure
2.) Decrease the pressure
Use the following reaction to answer the questions
below:
2SO3(g) + CO2 (g) + heat
CS2 (g) + 4O2(g)

In which direction (left or right) does the equilibrium shift
in each of the following situations:
1.) Increase the pressure LEFT (3 moles gas)
2.) Decrease the pressure
RIGHT (5 moles gas)
Use the following reaction to answer the questions
below:
H2 (g) + I2 (g)
2HI (g)
In which direction (left or right) does the equilibrium
shift in each of the following situations:

1.) Increase Pressure
2.) Decrease Pressure
Use the following reaction to answer the questions
below:
H2 (g) + I2 (g)
2HI (g)
In which direction (left or right) does the equilibrium
shift in each of the following situations:

1.) Increase Pressure
2.) Decrease Pressure
NO CHANGE
NO CHANGE
Practice
What direction will the equilibrium shift (left or
right) in the reaction:
___POCl3(g)
___PCl3(g) + ___O2 (g) + heat

1.) Add PCl3
2.) Increase Pressure
3.) Increase Temperature
Practice
What direction will the equilibrium shift (left or
right) in the reaction:
_2_POCl3(g)
_2_PCl3(g) + _1_O2 (g) + heat

1.) Add PCl3
2.) Increase Pressure
3.) Increase Temperature
LEFT
LEFT
LEFT