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PRINCIPLES OF CHEMISTRY II
CHEM 1212
CHAPTER 14
DR. AUGUSTINE OFORI AGYEMAN
Assistant professor of chemistry
Department of natural sciences
Clayton state university
CHAPTER 14
CHEMICAL EQUILIBRIUM
CHEMICAL EQUILIBRIUM
- Many reactions do not go to completion
- Amount of products formed or reactants consumed cannot be
predicted from stoichiometry alone
- These reactions achieve a condition of equilibrium
Equilibrium
- A state of balance between opposing processes
CHEMICAL EQUILIBRIUM
- Occurs when there is product build-up during a chemical reaction
- The product molecules interact with one another to
re-produce reactants
forward reaction
A + B
C + D
reverse reaction
Chemical Equilibrium
- When the rate of product formation (forward reaction)
is equal to the rate of reactant formation (reverse reaction)
CHEMICAL EQUILIBRIUM
- Reactant and product concentrations are usually not equal
- Such reactions are known as reversible reactions
- Forward reaction rate decreases with time as
reactants are used up
- Reverse reaction rate increases with time as
products are being formed
- Concentrations are reached when both forward
and reverse rates become equal
CHEMICAL EQUILIBRIUM
Homogeneous Equilibrium
- Substances are in the same phase
2NO2(g) + O2(g) ↔ 2NO3(g)
Heterogeneous Equilibrium
- Substances are in more than one phase
KOH(s) + SO3(g) ↔ KHSO4(s)
EQUILIBRIUM CONSTANT
- Describes the extent of reaction in a given system
- For a chemical reaction of the form
aA + bB ↔ cC + dD
- The equilibrium constant (Keq) is given by
K eq
[C]c [D] d

[A] a [B] b
EQUILIBRIUM CONSTANT
- [ ] denotes equilibrium concentration in mol/liter or M
- Product concentrations in the numerator
- Reactant concentrations in the denominator
- Concentrations are raised to the powers of the
respective coefficients (a, b, c, d)
EQUILIBRIUM CONSTANT
- Substances in solution (in mol/L) are written in
Keq expressions
- Pure solids, pure liquids (e.g. water), and solvents are not
included since their concentrations are constant
- Keq changes with change in temperature
- For exothermic forward reactions (heat released)
Keq decreases with increasing temperature
- For endothermic forward reactions (heat absorbed)
Keq increases with increasing temperature
EQUILIBRIUM CONSTANT
Large Keq
- Greater product concentrations than reactant concentrations
- Equilibrium position lies to the right
Small Keq
- Smaller product concentrations than reactant concentrations
- Equilibrium position lies to the left
Intermediate Keq (near unity)
- Both products and reactants are in significant amounts
- Equilibrium position lies neither to the right nor to the left
EQUILIBRIUM CONSTANT
- Longer arrows may be used to indicate the predominant
species
- Longer forward reaction arrow for large Keq
- Longer reverse reaction arrow for small Keq
CO2 + H2O
H2CO3
EQUILIBRIUM CONSTANT
- Any Keq expression refers to a particular form of a
chemical equation
- Consider the following
aA + bB ↔ cC + dD
[C]c [D] d
K
[A]a [B] b
- If the reaction is written in the reverse order
cC + dD ↔ aA + bB
[A]a [B] b
K' 
[C]c [D]d
EQUILIBRIUM CONSTANT
- Any Keq expression refers to a particular form of a
chemical equation
- This implies that
1
K' 
K
EQUILIBRIUM CONSTANT
- Any Keq expression refers to a particular form of a
chemical equation
- Consider the following
aA + bB ↔ cC + dD
[C]c [D] d
K
[A]a [B] b
- If the reaction is multiplied by a factor n
ncC + ndD ↔ naA + nbB
[A]na [B] nb
K" 
[C]nc[D] nd
EQUILIBRIUM CONSTANT
- Any Keq expression refers to a particular form of a
chemical equation
- This implies that
K'' = Kn
EQUILIBRIUM CONSTANT
- If two or more chemical equations are added to yield a new
equation
1. 2NO2 ↔ N2O4
2. N2O4 + O2 ↔ 2NO3
K1
K2
3. 2NO2 + O2 ↔ 2NO3
K3
- Reactions 1 and 2 are added to give reaction 3
- This implies that
K3 = K1K2
EQUILIBRIUM CONSTANT
At a given temperature, K = 1.3 x 10-2 for the reaction
N2(g) + 3H2(g) ↔ 2NH3(g)
a) Write an equilibrium expression for the reaction
b) Calculate the values of K for the following reactions
2NH3(g) ↔ N2(g) + 3H2(g)
1/2N2(g) + 3/2H2(g) ↔ NH3(g)
3N2(g) + 9H2(g) ↔ 6NH3(g)
[NH3]2
K
[N2][H2]3
K = 1/(1.3 x 10-2) = 77
K = (1.3 x 10-2)1/2 = 0.11
K = (1.3 x 10-2)3 = 2.2 x 10-6
PRESSURE AND CONCENTRATION
- The concentration of gases are generally expressed in terms
of their partial pressures
- Atmospheres (atm) or bars may be used in equilibrium
calculations involving gases
For the reaction below
2NO2(g) + O2(g) ↔ 2NO3(g)
[NO3]2
Kc 
[NO2]2 [O2]
(PNO3) 2
Kp 
(PNO2) 2 (PO2)
PRESSURE AND CONCENTRATION
- Kc is used for concentrations (mol/L)
- Kp is used for pressures (atm or bar)
- Keq is for general use when the differences are not important
PRESSURE AND CONCENTRATION
- From the ideal gas law
PV = nRT
- It is proved that
[n/V] = P/RT
- Pressure (in atm) is converted to concentration (in mol/L)
when divided by RT
R= 0.08206 L·atm/mol∙K
PRESSURE AND CONCENTRATION
- It is also be proved that
Kp = Kc(RT)∆n
∆n = change in number of moles of gas
= (number of moles of gaseous products) – (number of moles of gaseous reactants)
Example
2NO2(g) + O2(g) ↔ 2NO3(g)
∆n = 2 – (2 + 1) = -1
PRESSURE AND CONCENTRATION
For the reaction
2NO(g) + 2H2(g) ↔ N2(g) + 2H2O(g)
It is determined that, at equilibrium at a given temperature, the
concentrations are as follows:
[NO(g)] = 8.1 x 10-3 M, [H2(g)] = 4.1 x 10-5 M,
[N2(g)] = 5.3 x 10-2 M and [H2O(g)] = 2.9 x 10-3 M
Calculate the equilibrium constant for the reaction at this
temperature
4.0 x 106
PRESSURE AND CONCENTRATION
At a particular temperature, a 3.0-L flask contains 2.4 mol Cl2, 1.0
mol NOCl, and 4.5 x 10-3 mol NO. Calculate the equilibrium
constant at this temperature for the reaction
2NOCl(g) ↔ 2NO(g) + Cl2(g)
1.7 x 10-5
PRESSURE AND CONCENTRATION
The following equilibrium pressures at 27 oC were observed for
the reaction
2NO2(g) ↔ 2NO(g) + O2(g)
PNO2 = 0.55 atm, PNO = 6.5 x 10-5 atm, PO2 = 4.5 x 10-5 atm
Calculate the value for the equilibrium constant Kp at this
temperature
Also calculate Kc
6.3 x 10-13
2.6 x 10-14
PRESSURE AND CONCENTRATION
At 54 oC, the equilibrium concentrations are
[CH3OH(g)] = 0.15 M, [CO(g)] = 0.24 M, and [H2(g)] = 1.1 M
for the reaction
CH3OH(g) ↔ CO(g) + 2H2(g)
Calculate the value for the equilibrium constant Kp at this
temperature
1.4 x 103
REACTION QUOTIENT (Q)
- Used when the system is not in equilibrium
- Concentrations at the time rather than equilibrium concentrations
are used
Used to determine whether
- more products will form
- more reactants will form
or
- system is at equilibrium
REACTION QUOTIENT (Q)
- Consider the following which is not at equilibrium
aA + bB ↔ cC + dD
[C]c [D] d
Q
[A]a [B] b
REACTION QUOTIENT (Q)
- Q tells which direction the reaction proceeds to reach equilibrium
- Concentrations change to bring Q closer to Keq
If Q < Keq
- Reaction proceeds to the right
- Concentration of products will increase
- Concentration of reactants will decrease
- Q will increase to get closer to Keq
REACTION QUOTIENT (Q)
- Q tells which direction the reaction proceeds to reach equilibrium
- Concentrations change to bring Q closer to Keq
If Q > Keq
- Reaction proceeds to the left
- Concentration of products will decrease
- Concentration of reactants will increase
- Q will decrease to get closer to Keq
REACTION QUOTIENT (Q)
The Keq is 2.4 x 103 at a certain temperature for the reaction
2NO(g) ↔ N2(g) + O2(g)
For which of the following sets of conditions is the system at
equilibrium?
For those that are not at equilibrium, in which direction will the
system shift?
a) A 1.0-L flask contains 0.024 mol NO, 2.0 mol N2, 2.6 mol O2
b) A 2.0-L flask contains 0.032 mol NO, 0.62 mol N2, 4.0 mol O2
c) A 3.0-L flask contains 0.060 mol NO, 2.4 mol N2, 1.7 mol O2
a) Q > Keq , reaction shifts left
b) Q = Keq , reaction at equilibrium
c) Q < Keq , reaction shifts right
LE CHATELIER’S PRICIPLE
- If a stress (change of conditions) is applied to a system in
equilibrium the system will readjust (change the equilibrium
position) in the direction that best reduces the stress imposed
on the system
- Change of such conditions as
Concentration, Pressure, Temperature
- If more products form as a result of the applied stress
the equilibrium is said to have shifted to the right
- If more reactants form as a result of the applied stress
the equilibrium is said to have shifted to the left
LE CHATELIER’S PRICIPLE
Concentration Changes
For a reaction mixture at equilibrium
- Addition of reactant(s) shifts the equilibrium position to the right
- Removal of product(s) shifts the equilibrium position to the right
- Addition of product(s) shifts the equilibrium position to the left
- Removal of reactant(s) shifts the equilibrium position to the left
LE CHATELIER’S PRICIPLE
Volume and Pressure Changes
- Gases must be involved in the chemical reaction
- The total number of moles of the gaseous state must change
- Equilibrium is shifted in the direction of fewer moles
- Volume and pressure are inversely related
- Decrease in volume implies increase in pressure
- Increase in volume implies decrease in pressure
LE CHATELIER’S PRICIPLE
Volume and Pressure Changes
Higher moles of gaseous reactants than products (∆n is negative)
- Increase in pressure shifts the equilibrium position to the right
- Decrease in pressure shifts the equilibrium position to the left
Higher moles of gaseous products than reactants (∆n is positive)
- Increase in pressure shifts the equilibrium position to the left
- Decrease in pressure shifts the equilibrium position to the right
LE CHATELIER’S PRICIPLE
Volume and Pressure Changes
No change in equilibrium position occurs if
- There is no reactant nor product in the gaseous state
- Number of moles of gaseous reactants equals number of
moles of gaseous products (∆n is zero)
- Pressure is increased by adding a nonreactive (inert) gas
LE CHATELIER’S PRICIPLE
Temperature Changes
Exothermic Reactions
- Heat is a product
- Increase in temperature shifts the equilibrium position to the left
- Decrease in temperature shifts the equilibrium position to the right
N2(g) + 3H2(g) ↔ 2NH3(g)
∆H = -92 kJ
LE CHATELIER’S PRICIPLE
Temperature Changes
Endothermic Reactions
- Heat is a reactant
- Increase in temperature shifts the equilibrium position to the right
- Decrease in temperature shifts the equilibrium position to the left
H2(g) + I2(g) ↔ 2HI(g)
∆H = +52 kJ
LE CHATELIER’S PRICIPLE
Addition of Catalysts
- Catalysts do not change equilibrium positions
- Catalysts speed up both forward and reverse reactions so have
no net effect
- Catalysts allow equilibrium to be established more quickly by
lowering the activation energy
LE CHATELIER’S PRICIPLE
In which direction will the position of the equilibrium
2HI(g) ↔ H2(g) + I2(g)
be shifted for each of the following changes?
HI(g) is added
HI(g) is removed
H2(g) is added
H2(g) removed
I2(g) is added
I2(g) removed
Some Ar(g) is added
The volume of the container is tripled
The temperature is decreased (the reaction is exothermic)
EQUILIBRIUM CALCULATIONS
To determine the equilibrium constant from
the concentrations of species
- Write the balanced chemical equation
- Complete the iCe table
- Write the equilibrium constant expression
- Substitute values using the iCe table
- Solve for Keq
EQUILIBRIUM CALCULATIONS
The iCe table
aA + bB ↔
cC + dD
intital concentration, M
Change in concentration, M
equilibrium concentration, M
Fill in iCe concentrations (mol/L or M) for all speices
EQUILIBRIUM CALCULATIONS
A 1.00-L flask was filled with 2.00 mol gaseous SO2 and
2.00 mol gaseous NO2 and heated. After equilibrium is
reached, it was found that 1.30 mol gaseous NO
was present. Calculate Keq for the reaction if it
occurs under these conditions.
SO2(g) + NO2(g) ↔ SO3(g) + NO(g)
3.4
EQUILIBRIUM CALCULATIONS
To determine the concentrations of species from
the equilibrium constant
- Intitial concentrations of reactants are usually known
- Keq is also known
- Concentrations of products are written using a variable
- Fill in the iCe table
- Write the equilibrium expression
- Solve for the unknown variable
EQUILIBRIUM CALCULATIONS
At 35 oC, Keq = 1.6 x 10-5 for the reaction
2NOCl(g) ↔ 2NO(g) + Cl2(g)
Calculate the concentrations of all species at equilibrium
if the initial amounts are
a) 1.0 M NOCl
b) 2.0 mol NOCl and 1.0 mol Cl2 in a 1.0-L flask
a) [NOCl] = 1.0 M, [NO] = 0.032 M, [Cl2] = 0.016 M
b) [NOCl] = 2.0 M, [NO] = 0.0080 M, [Cl2] = 1.0 M
HETEROGENEOUS EQUILIBRIA
- Substances are in more than one phase
- Solids and pure liquids are excluded in equilibrium calculations
because their concentrations are constant
Example
CaCO3(s) ↔ CaO(s) + CO2(g)
Kc = [CO2]
Kp = PCO2
HETEROGENEOUS EQUILIBRIA
Consider the following reaction at a certain temperature
4Fe(s) + 3O2(g) ↔ 2Fe2O3(s)
An equilibrium mixture contains 1.0 mol Fe, 1.0 x 10-3 mol O2,
and 2.0 mol Fe2O3 all in a 2.0-L container.
Calculate Keq for this reaction.
8.0 x 109
SOLUBILITY EQUILIBRIA
- Reactions involving dissolution of solids into solution
and
formation of solids from solution
- Precipitation reactions
Example
Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq)
Net Ionic Equation
Pb2+(aq) + 2I-(aq) → PbI2(s)
SOLUBILITY PRODUCT CONSTANT
- Ksp is the equilibrium constant for the reaction in which
a solid salt (an ionic compound) dissolves to give its
constituent ions in solution
- Solid is in its standard state so is omitted from the Ksp expression
Example
PbI2(s) ↔ Pb2+ + 2I-
Ksp = [Pb2+][I-]2 = 7.9 x 10-9
SOLUBILITY PRODUCT CONSTANT
- The solution is said to be saturated if all the solid is
capable of being dissolved
- Electrostatic interaction causes the formation of some other
types of ions (ion pairing) but is not discussed here
SOLUBILITY PRODUCT CONSTANT
Use is made of the iCe table for Ksp calculations
PbI2(s) ↔ Pb2+ + 2Ii
solid
0
C
-s
+s
e
solid
s
Ksp = (s)(2s)2 = 4s3
0
+2s
2s
SOLUBILITY PRODUCT CONSTANT
Calculate the solubility product constant (Ksp) if the
solubility of silver sulfate (Ag2SO4) is 1.0 x 10-2 M
Ag2SO4(s) ↔ 2Ag+ + SO42i
solid
0
0
C
-s
+2s
+s
e
solid
2s
s
Ksp = (2s)2(s) = 4s3 = 4(1.0 x 10-2)3 = 4.0 x 10-6
SOLUBILITY PRODUCT CONSTANT
The solubility product constant (Ksp) for copper(II) iodate
Cu(IO3)2 is 7.4 x 10-8. Calculate the solubility of this
compound in water
Cu(IO3)2(s) ↔ Cu2+ + 2IO3i
solid
0
C
-s
+s
e
solid
s
Ksp = (s)(2s)2
7.4 x 10-8 = 4s3
s = 2.6 x 10-3 M
0
+2s
2s
SOLUBILITY PRODUCT CONSTANT
Common Ion Effect
- Supposing x1 M NaI is added to PbI2
PbI2(s) ↔ Pb2+ + 2Ii
solid
0
C
-s
+s
e
solid
s
x1
+2s
x1 + 2s
- Concentration of I- has contributions from both PbI2 and NaI
SOLUBILITY PRODUCT CONSTANT
Common Ion Effect
Ksp = [Pb2+][I-]2 = (s)(x1 + 2s)2
- A salt is less soluble if one of its constituent ions is already
present in the solution
- If 2s if far less than x1 (2s <<< x1)
That is if 2s ≤ 5% of x1
2s can be ignored
Implies Ksp ≈ (s)(x1)2
SOLUBILITY PRODUCT CONSTANT
Common Ion Effect
The solubility product constant (Ksp) for copper(II) iodate
Cu(IO3)2 is 7.4 x 10-8. Calculate the solubility of this
compound in a 0.10 M copper(II) nitrate solution
Cu(IO3)2(s) ↔ Cu2+ + 2IO3i
solid
0.10 M
C
-s
+s
e
solid
0.10 + s
7.4 x 10-8 = (0.10 + s)(2s)2
7.4 x 10-8 ≈ (0.10)(2s)2
s = 4.3 x 10-4 M
0
+2s
2s
SOLUBILITY PRODUCT CONSTANT
Formation of a Precipitate
Consider the following
PbI2(s) ↔ Pb2+ + 2IKsp = 7.9 x 10-9
The reaction quotient
Q = [Pb2+][I-]2
Compare Q and Ksp
If Q > Ksp, solid forms (precipitation occurs)
If Q < Ksp, no solid forms
SOLUBILITY PRODUCT CONSTANT
Formation of a Precipitate
Q < Ksp
No solid present
Q > Ksp
Solid present
Q = Ksp
Equilibrium
SOLUBILITY PRODUCT CONSTANT
Precipitation Titration of Mixtures
- The less soluble precipitates first (as Q becomes greater than Ksp)
- Smaller Ksp implies less soluble
Consider a mixture of PbI2 and PbCl2
PbI2(s) ↔ Pb2+ + 2IKsp = 7.9 x 10-9
(precipitates first)
PbCl2(s) ↔ Pb2+ + 2Cl-
Ksp = 1.7 x 10-5