Dynamic Equilibrium

1
Dynamic Equilibrium
16.1 Irreversible and Reversible
Reactions 16.2 Dynamic Nature of Chemical
Equilibrium 16.3 Examples of Chemical
Equilibrium 16.4 Equilibrium Law 16.5 Determinati
on of Equilibrium Constants 16.6 Equilibrium
Constant in Terms of Partial Pressures 16.7 Equili
brium Position 16.8 Partition Equilibrium of a
Solute Between Two Immiscible Solvents 16.9 Signif
icances of Equilibrium Constants 16.10 Factors
Affecting Equilibrium
2
Irreversible and Reversible Reactions
3
Irreversible Reactions
16.1 Irreversible and Reversible Reactions (SB
p.88)

• Chemical reactions that take place in one
  direction only
• It goes on until at least one of the reactants is
  used up
• e.g. Reaction between Na and H2O to form NaOH
  and H2
• 2Na(s) 2H2O(l) ?? 2NaOH(aq) H2(g)

4
Reversible Reactions
16.1 Irreversible and Reversible Reactions (SB
p.88)

• Chemical reactions that can go in either way
• Do not proceed to completion
• e.g. Reaction of N2 and H2 to form NH3

5
Reversible Reactions
16.1 Irreversible and Reversible Reactions (SB
p.88)

• Reaction going from left to right is called
  forward reaction
• Reaction going from right to left is called
  backward reaction

6
16.1 Irreversible and Reversible Reactions (SB
p.89)
Thermal Decomposition of Calcium Carbonate

• CaCO3 is heated to 800 oC, it decomposes to form
  CaO and CO2
• CaO and CO2 react to form CaCO3

7
16.1 Irreversible and Reversible Reactions (SB
p.89)
Esterification

• Does not go to completion no matter how long the
  reaction mixture is heated
• Ethyl ethanoate reacts with water to give
  ethanoic acid and ethanol

8
16.1 Irreversible and Reversible Reactions (SB
p.89)
Esterification
The laboratory set-up for esterification under
reflux
9
Dynamic Nature of Chemical Equilibrium
10
16.2 Dynamic Nature of Chemical Equilibrium (SB
p.90)
Equilibrium State

• Both the forward and backward reactions proceed
  simultaneously
• After a certain period of time, a state is
  reached when there is no observable changes of
  the reaction mixture
• This state is known as equilibrium state

11
16.2 Dynamic Nature of Chemical Equilibrium (SB
p.90)
Chemical Equilibrium

• At the equilibrium state
• Rate of forward reaction
• Rate of backward reaction
• Concentrations of reactants and products remain
  constant

12
16.2 Dynamic Nature of Chemical Equilibrium (SB
p.90)
Dynamic Equilibrium
An example of dynamic equilibrium
13
Dynamic Equilibrium
16.2 Dynamic Nature of Chemical Equilibrium (SB
p.90)
Changes in concentrations of hydrogen, iodine and
hydrogen iodide with respect to time
14
16.2 Dynamic Nature of Chemical Equilibrium (SB
p.90)
Dynamic Equilibrium
Changes in the rates of forward and backward
reactions with respect to time
15
Examples of Chemical Equilibrium
16
16.3 Examples of Chemical Equilibrium (SB p.92)
Reaction of Bromine with Water

• Colour of solution is related to the amount of
  Br2(aq) which is yellowish brown

17
16.3 Examples of Chemical Equilibrium (SB p.92)
Reaction of Bromine with Water

• When an alkali is added, a colourless solution is
  observed
• Amount of Br2(aq) ?
• Equilibrium position shifts to the right (?
  H(aq) is removed)

18
16.3 Examples of Chemical Equilibrium (SB p.92)
Reaction of Bromine with Water

• When an acid is added, a yellowish brown solution
  is observed
• Amount of Br2(aq) ?
• Equilibrium position shifts to the left

19
16.3 Examples of Chemical Equilibrium (SB p.92)
Interconversion of Dichromate(VI) and
Chromate(VI) Ions

• When an alkali is added, the solution changes
  from orange to yellow
• Amount of Cr2O72-(aq) ?
• Amount of CrO42-(aq) ?
• Equilibrium position shifts to the right

20
16.3 Examples of Chemical Equilibrium (SB p.92)
Interconversion of Dichromate(VI) and
Chromate(VI) Ions

• When an acid is added, the solution changes back
  to orange
• Amount of Cr2O72-(aq) ?
• Amount of CrO42-(aq) ?
• Equilibrium position shifts to the left

21
Equilibrium Law
22
Equilibrium Law
16.4 Equilibrium Law (SB p.93)
(Constant at a fixed temperature)
23
Equilibrium Law
16.4 Equilibrium Law (SB p.93)
24
Equilibrium Law
16.4 Equilibrium Law (SB p.93)

• If the coefficients of the equation are
  multiplied by 2,

Kc2 (Kc1)2
25
Equilibrium Law
16.4 Equilibrium Law (SB p.93)
If the coefficients in a balanced chemical
equation are multiplied by a common factor n, the
new equilibrium constant will be the original one
to the power n.
26
Equilibrium Law
16.4 Equilibrium Law (SB p.94)

• If the chemical equation is reversed,

27
Equilibrium Law
16.4 Equilibrium Law (SB p.94)

• For chemical reactions involving reactants or
  products that are in solid or liquid states,
  their equilibrium concentrations are not included
  in the expression of Kc

28
16.4 Equilibrium Law (SB p.94)
Equilibrium Law

• Unit of Kc depends on the powers of the
  concentration of substances in the equilibrium
  law
• Related to the stoichiometric coefficients in the
  balanced equation

29
16.4 Equilibrium Law (SB p.94)
Equilibrium Law
no unit
30
Determination of Equilibrium Constants
31
16.5 Determination of Equilibrium Constants (SB
p.95)
32
16.5 Determination of Equilibrium Constants (SB
p.95)
Reactant/Product Experiment 1 Experiment 1 Experiment 2 Experiment 2
Reactant/Product Amount at beginning (mol) Amount at equilibrium (mol) Amount at beginning (mol) Amount at equilibrium (mol)
CH3COOH(aq) 0.250 0.083 0.296 0.098
CH3CH2OH(aq) 0.250 0.083 0.296 0.098
H2SO4(l) 0.020 0.020 0.020 0.020
CH3COOCH2CH3(aq) 0.000 0.167 0.000 0.198
H2O(l) 0.000 0.167 0.000 0.198
33
16.5 Determination of Equilibrium Constants (SB
p.96)
For experiment 1
34
16.5 Determination of Equilibrium Constants (SB
p.96)
For experiment 2
35
16.5 Determination of Equilibrium Constants (SB
p.96)
36
16.5 Determination of Equilibrium Constants (SB
p.96)

• Conc. H2SO4 acts as a positive catalyst
• The time required to reach the equilibrium
  becomes shorter

37
16.5 Determination of Equilibrium Constants (SB
p.96)
Effect of a positive catalyst on the time
required to reach the equilibrium
38
16.5 Determination of Equilibrium Constants (SB
p.97)

• Fe(NCS)2(aq) is blood-red in colour
• Its concentration can be determined by measuring
  the absorbance of a particular wavelength using a
  colorimeter

39
16.5 Determination of Equilibrium Constants (SB
p.97)

• Known amounts of Fe3(aq) and NCS-(aq) are mixed
  together
• Concentrations of Fe3(aq) (or NCS-(aq)) at
  equilibrium
• Initial conc. of Fe3(aq) (or NCS-(aq))
  Conc. of Fe(NCS)2(aq) formed

40
16.5 Determination of Equilibrium Constants (SB
p.97)
41
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42
Equilibrium Constant in Terms of Partial Pressures
43
16.6 Equilibrium Constant in Terms of Partial
Pressures (SB p.101)
Equilibrium Constant in Terms of Partial Pressures
For the following reaction
The equilibrium constant in terms of partial
pressures (Kp) is expressed as
44
(No Transcript)
45
Equilibrium Position
46
16.7 Equilibrium Position (SB p.102)
Equilibrium Position

• e.g. The Haber process has a Kc value of 0.062
  dm6 mol-2 at 500 oC

• Value of Kc is a function of temperature
• Not affected by the amount of gases that are
  mixed together at the start of reaction

47
16.7 Equilibrium Position (SB p.103)
Equilibrium Position
Expt Initial concentration (mol dm-3) Initial concentration (mol dm-3) Initial concentration (mol dm-3) Equilibrium concentration (mol dm-3) Equilibrium concentration (mol dm-3) Equilibrium concentration (mol dm-3) Kc at 500 oC (dm6 mol-2)
Expt N2(g) H2(g) NH3(g) N2(g) H2(g) NH3(g) Kc at 500 oC (dm6 mol-2)
1 1.0 1.0 0.0 0.921 0.762 0.159 0.062
2 2.0 1.0 3.0 2.40 3.20 2.20 0.062
48
16.7 Equilibrium Position (SB p.103)
Equilibrium Position

• Each specific set of equilibrium concentrations
  of the reaction mixture is known as equilibrium
  position

49
Partition Equilibrium of a Solute Between Two
Immiscible Solvents
50
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.104)
Partition Coefficient

• I2 is added to a beaker containing water and
  1,1,1-trichloroethane which are immiscible
• I2 will dissolve in both layers to different
  extent

51
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.104)
Partition Coefficient
When dynamic equilibrium is established ? the
concentrations of iodine in water and
1,1,1-trichloroethane reach a particular ratio
52
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.104)
Partition Coefficient
The partition law states that

• At a given temperature, the ratio of the
  concentrations of a solute in two immiscible
  solvents (solvent 1 and solvent 2) is constant
  when equilibrium has been reached
• This constant is known as the partition
  coefficient

53
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.104)
Partition Coefficient
The partition law can be represented by the
following equation
(no unit)
54
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.104)
Partition Coefficient

• Not affected by the amount of solute added and
  the volume of solvents
• Depend on temperature only
• Only apply for dilute solutions

55
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.104)
Partition Coefficient

• The solute does not have the same molecular form
  in both solvents
• ? Partition law does not apply
• e.g. distribution of ethanoic acid between water
  and benzene

56
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.104)
Solvent Extraction
Solvent extraction is a method of separating a
substance from a mixture, using the principle of
partition equilibrium of a solute between two
immiscible solvents
57
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.104)
Solvent Extraction
Iodine can be extracted from water by adding
hexane, shaking and separating the two layers in
a separating funnel
58
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.104)
59
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.108)
Paper Chromatography

• Paper chromatography is a method used to separate
  and analyze mixtures
• e.g. To separate a mixture of dyes

60
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.108)
Paper Chromatography
61
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.108)
Paper Chromatography

• Chromatography paper stationary phase
• Solvent mobile phase
• The solvent moves up the filter paper by
  capillary action

62
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.108)
Paper Chromatography

• The solvent moves across the sample spot of a
  mixture of dyes
• Partition of the dyes between the stationary
  phase and the mobile phase occurs

63
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.109)
Paper Chromatography

• The states of partition of the dyes depend on
• the tendency of the dyes to attach to the
  adsorbed water
• (b) the tendency of the dyes to dissolve in the
  solvent

64
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.109)
Paper Chromatography

• Different dyes have different states of partition
  between the mobile and stationary phases
• They will move upwards to different extent

65
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.109)
Paper Chromatography

• The solvent front (i.e. the highest position that
  the solvent moves up the paper) almost reaches
  the top of the paper
• The paper is removed and dried
• The chromatography paper obtained is called the
  chromatogram

66
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.109)
Paper Chromatography
The ratio of the distance travelled by a spot to
that travelled by the solvent is known as the Rf
value.
67
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.109)
Paper Chromatography
68
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.109)
Paper Chromatography

• The Rf value of any particular substance is about
  the same when using a particular solvent at a
  given temperature
• The Rf value of a substance differs in different
  solvents and at different temperatures

69
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.109)
Paper Chromatography
Amino acid Solvent Solvent
Amino acid Mixture of phenol and ammonia Mixture of butanol and ethanoic acid
Cystine 0.14 0.05
Glycine 0.42 0.18
Leucine 0.87 0.62
Rf values of some amino acids in two different
solvents at a given temperature
70
Significances of Equilibrium Constants
71
16.9 Significances of Equilibrium Constants (SB
p.110)
Extent of Reaction

• Large value of Kc or Kp (gtgt 1)
• the products dominate the reaction mixture at the
  equilibrium
• the equilibrium position lies mainly on the side
  of the products

72
16.9 Significances of Equilibrium Constants (SB
p.110)
Extent of Reaction

• Small value of Kc or Kp (ltlt 1)
• the reactants dominate the reaction mixture at
  the equilibrium
• the equilibrium position lies mainly on the side
  of the reactants

73
16.9 Significances of Equilibrium Constants (SB
p.110)
Extent of Reaction
Magnitude of Kc or Kp ? gives no clue to the time
required to reach the equilibrium
74
16.9 Significances of Equilibrium Constants (SB
p.110)
Reaction Quotient

• Reaction quotient (Q) is defined in the same way
  as K
• Except that the concentrations of reactants and
  products can be taken at any moment of the
  reaction
• Q is not a constant

75
16.9 Significances of Equilibrium Constants (SB
p.110)
Reaction Quotient

• To predict the direction of shifts in a chemical
  equilibrium
• ? compare Q with Kc or Kp

76
16.9 Significances of Equilibrium Constants (SB
p.111)
Reaction Quotient
77
16.9 Significances of Equilibrium Constants (SB
p.111)
Reaction Quotient
78
16.9 Significances of Equilibrium Constants (SB
p.111)
Reaction Quotient

• Q K
• ? The system is at equilibrium

79
16.9 Significances of Equilibrium Constants (SB
p.111)
Reaction Quotient

• Q gt K
• ? The system is not at equilibrium
• ? Concentrations or partial pressures of the
  products gt equilibrium values
• ? The reaction will proceed from the right
  (product side) to the left (reactant side)
  until equilibrium is reached

80
16.9 Significances of Equilibrium Constants (SB
p.111)
Reaction Quotient

• Q lt K
• ? The system is not at equilibrium
• ? Concentrations or partial pressures of the
  products lt equilibrium values
• ? The reaction will proceed from the left
  (reactant side) to the right (product side)
  until equilibrium is reached

81
Factors Affecting Equilibrium
82
16.10 Factors Affecting Equilibrium (SB p.112)
Le Chateliers Principle
Le Chateliers principle states that
If a system at equilibrium is subjected to a
change (e.g. concentration, pressure or
temperature), the equilibrium position of the
system will shift in a way to oppose (or reduce)
the effect of the change
83
16.10 Factors Affecting Equilibrium (SB p.112)
Effect of Changes in Concentration on Equilibrium
SO2(g)eqm 0.71 mol dm-3 O2(g)eqm 0.50 mol
dm-3 SO3(g)eqm 1.00 mol dm-3
84
16.10 Factors Affecting Equilibrium (SB p.112)
Effect of Changes in Concentration on Equilibrium
85
16.10 Factors Affecting Equilibrium (SB p.112)
Effect of Changes in Concentration on Equilibrium
When 0.20 mol dm-3 of O2(g) is added,
86
16.10 Factors Affecting Equilibrium (SB p.112)
Effect of Changes in Concentration on Equilibrium

• Qc lt Kc
• ? The equilibrium position will shift to the
  right (i.e. more SO3(g) will be formed) until
  equilibrium is re-established

87
16.10 Factors Affecting Equilibrium (SB p.113)
Effect of Changes in Concentration on Equilibrium
88
16.10 Factors Affecting Equilibrium (SB p.113)
Effect of Changes in Concentration on Equilibrium

• Changes in concentration only result in an
  adjustment of the equilibrium position of a
  system
• No change in the value of Kc or Kp

89
16.10 Factors Affecting Equilibrium (SB p.113)
90
16.10 Factors Affecting Equilibrium (SB p.115)
Effect of Changes in Pressure on Equilibrium

• 1. Addition of a gaseous reactant or product when
  the volume of the container is kept constant
• A gaseous reactant or product is added to an
  equilibrium system
• The partial pressure of the reactant or product
  will be increased
• The equilibrium position will shift in a
  direction to remove the additional species

91
16.10 Factors Affecting Equilibrium (SB p.115)
Effect of Changes in Pressure on Equilibrium

• 2. Removal of a gaseous reactant or product when
  the volume of the container is kept constant
• A gaseous reactant or product is removed from an
  equilibrium system
• The partial pressure of the reactant or product
  will be decreased
• The equilibrium position will shift in a
  direction to replace the loss

92
16.10 Factors Affecting Equilibrium (SB p.115)
Effect of Changes in Pressure on Equilibrium

• 3. Addition of an inert gas when the volume of
  the container is kept constant
• Addition of an inert gas increases the total
  pressure of a system
• No effect on the concentrations and the partial
  pressures of the reactants or the products in the
  system
• The equilibrium position of the system will not
  be changed

93
16.10 Factors Affecting Equilibrium (SB p.116)
Effect of Changes in Pressure on Equilibrium
94
16.10 Factors Affecting Equilibrium (SB p.116)
Effect of Changes in Pressure on Equilibrium
95
16.10 Factors Affecting Equilibrium (SB p.116)
Effect of Changes in Pressure on Equilibrium
96
16.10 Factors Affecting Equilibrium (SB p.116)
Effect of Changes in Pressure on Equilibrium

• When the volume of the container is reduced (i.e.
  pressure is increased)
• ? Qc lt Kc
• ? Equilibrium position will shift to the right
• ? More NH3(g) will be produced

97
16.10 Factors Affecting Equilibrium (SB p.116)
Effect of Changes in Pressure on Equilibrium

• When the volume of the container is increased
  (i.e. pressure is decreased)
• ? Qc gt Kc
• ? Equilibrium position will shift to the left
• ? More N2(g) and H2(g) will be produced

98
16.10 Factors Affecting Equilibrium (SB p.116)
Effect of Changes in Pressure on Equilibrium
99
16.10 Factors Affecting Equilibrium (SB p.117)
Effect of Changes in Pressure on Equilibrium

• When the volume of the container is reduced (i.e.
  pressure is increased)
• ? Qc gt Kc
• ? Equilibrium position will shift to the left
• ? More N2O4(g) will be produced

100
16.10 Factors Affecting Equilibrium (SB p.117)
Effect of Changes in Pressure on Equilibrium

• When the volume of the container is increased
  (i.e. pressure is decreased)
• ? Qc lt Kc
• ? Equilibrium position will shift to the right
• ? More NO2(g) will be produced

101
16.10 Factors Affecting Equilibrium (SB p.117)
Effect of Changes in Pressure on Equilibrium
It should be noted that 1. Le Chateliers
principle can be applied here to explain the
effect of changes in pressure on the equilibrium
position 2. The changes in pressure lead to the
adjustment of the equilibrium position but not
the value of Kc or Kp 3. The changes in pressure
only affect those equilibria that involve gaseous
reactants or products as solids and liquids are
incompressible
102
16.10 Factors Affecting Equilibrium (SB p.117)
103
16.10 Factors Affecting Equilibrium (SB p.119)
Effect of Changes in Temperature on Equilibrium
A change in temperature of an equilibrium system
results in an adjustment of the value of the
equilibrium constant.
104
16.10 Factors Affecting Equilibrium (SB p.119)
Effect of Changes in Temperature on Equilibrium
Temperature (K) 500 600 700 800
Kp (atm-2) 90.0 3.0 0.3 0.04
105
16.10 Factors Affecting Equilibrium (SB p.119)
Effect of Changes in Temperature on Equilibrium
Temperature (K) 298 500 700 900 1100
Kp (atm) 1.5 ? 1048 3.1 ? 1032 1.2 ? 1026 3.1 ? 1022 1.5 ? 1020
106
16.10 Factors Affecting Equilibrium (SB p.119)
Effect of Changes in Temperature on Equilibrium
Temperature (K) 200 300 400 500
Kp (atm) 1.8 ? 10-6 0.174 51 1510
107
16.10 Factors Affecting Equilibrium (SB p.119)
Effect of Changes in Temperature on Equilibrium
Temperature (K) 700 1100 1500
Kp (no unit) 5 ? 10-13 4 ? 10-8 1 ? 10-5
108
16.10 Factors Affecting Equilibrium (SB p.120)
Effect of Changes in Temperature on Equilibrium

• For exothermic reactions
• an increase in temperature leads to a decrease
  in the value of K
• To reach a new equilibrium position, more
  reactants will be formed

109
16.10 Factors Affecting Equilibrium (SB p.120)
Effect of Changes in Temperature on Equilibrium

• For endothermic reactions
• an increase in temperature leads to a increase
  in the value of K
• To reach a new equilibrium position, more
  products will be formed

110
16.10 Factors Affecting Equilibrium (SB p.120)
Effect of Changes in Temperature on Equilibrium
The temperature and the value of K can be
represented by the following equation
111
16.10 Factors Affecting Equilibrium (SB p.120)
Effect of Changes in Temperature on Equilibrium
Graph of ln K against 1/T for exothermic reactions
112
16.10 Factors Affecting Equilibrium (SB p.120)
Effect of Changes in Temperature on Equilibrium
Graph of ln K against 1/T for endothermic
reactions
113
16.10 Factors Affecting Equilibrium (SB p.121)
114
16.10 Factors Affecting Equilibrium (SB p.123)
Summary of the Effects of Changes of Various
Factors on Equilibrium
Factor Equilibrium position Equilibrium constant
Increase in concentration of reactants A or B Shifts to right No change
Increase in concentration of products C or D Shifts to left No change
115
16.10 Factors Affecting Equilibrium (SB p.123)
Summary of the Effects of Changes of Various
Factors on Equilibrium
Factor Equilibrium position Equilibrium constant
Increase in pressure by reducing the volume of the container Shifts to right if (c d) lt (a b) Shifts to left to (a b) lt (c d) No change if a b c d No change
116
16.10 Factors Affecting Equilibrium (SB p.123)
Summary of the Effects of Changes of Various
Factors on Equilibrium
Factor Equilibrium position Equilibrium constant
Increase in temperature Shifts to right if the forward reaction is endothermic Shifts to left if the forward reaction is endothermic No change
117
16.10 Factors Affecting Equilibrium (SB p.123)
Summary of the Effects of Changes of Various
Factors on Equilibrium
Factor Equilibrium position Equilibrium constant
Addition of a catalyst No change No change
118
The END
119
16.1 Irreversible and Reversible Reactions (SB
p.89)
Check Point 16-1
Back
Answer

1. A
2. B
3. A ?? B

120
16.2 Dynamic Nature of Chemical Equilibrium (SB
p.91)
Back
Let's Think 1
List some characteristics of chemical equilibrium.
Answer
Some characteristics of chemical equilibrium
include It can only be achieved in a closed
system. It can be achieved from either forward
or backward reactions. It is dynamic in
nature. The concentrations of all chemical
species present in a system at equilibrium state
remain constant as long as the reaction
conditions are unchanged.
121
16.3 Examples of Chemical Equilibrium (SB p.92)
Check Point 16-3
Back
Answer
Chemical equilibrium is dynamic in nature. When a
trace amount of carbon monoxide labelled with
radioactive carbon-14 is added to the equilibrium
system, the equilibrium position shifts to the
right. Therefore, radioactive carbon dioxide
molecules are formed.
122
16.4 Equilibrium Law (SB p.94)
Back
Let's Think 2
What is a closed system? Why can chemical
equilibrium only be established in a closed
system?
Answer
A closed system means that there is no transfer
of matter between the system and the
surroundings. If the system is open, some of the
reactants or products can enter or leave the
system. As a result, the equilibrium state can
never be reached.
123
16.4 Equilibrium Law (SB p.94)
Check Point 16-4
Answer
124
16.4 Equilibrium Law (SB p.94)
Check Point 16-4
Answer
125
16.4 Equilibrium Law (SB p.94)
Back
Check Point 16-4
Answer
126
16.5 Determination of Equilibrium Constants (SB
p.98)
Check Point 16-5A
127
16.5 Determination of Equilibrium Constants (SB
p.98)
Check Point 16-5A
(a) Calculate the equilibrium concentrations of
Ag(aq), Fe2(aq) and Fe3(aq).
Answer
128
16.5 Determination of Equilibrium Constants (SB
p.98)
Check Point 16-5A
129
16.5 Determination of Equilibrium Constants (SB
p.98)
Check Point 16-5A
(b) Calculate the equilibrium constant (Kc).
Answer
130
16.5 Determination of Equilibrium Constants (SB
p.98)
Check Point 16-5A
Back
(c) What is the significance of (i) using a dry
conical flask? (ii) allowing the mixture to
stand overnight?
Answer
(c) (i) The significance of using a dry conical
flask is to make sure the reaction mixture in the
conical flask is not diluted by the presence of
water. (ii) The reaction mixture is allowed to
stand overnight in order to give sufficient time
for the reaction mixture to reach the equilibrium
state.
131
16.5 Determination of Equilibrium Constants (SB
p.98)
Example 16-5A
Answer
132
16.5 Determination of Equilibrium Constants (SB
p.98)
Back
Example 16-5A
Reactant / Product Initial number of moles (mol) Change in number of moles (mol) Number of moles at equilibrium (mol)
H2(g) a -x a x
I2(g) b -x b x
HI(g) 0 2x 2x
133
16.5 Determination of Equilibrium Constants (SB
p.99)
Example 16-5B
Answer
134
16.5 Determination of Equilibrium Constants (SB
p.99)
Back
Example 16-5B
Reactant / Product Initial number of moles (mol) Change in number of moles (mol) Number of moles at equilibrium (mol)
N2(g) a -x a x
H2(g) b -3x b 3x
NH3(g) 0 2x 2x
135
16.5 Determination of Equilibrium Constants (SB
p.99)
Example 16-5C
Answer
136
16.5 Determination of Equilibrium Constants (SB
p.100)
Example 16-5C
137
16.5 Determination of Equilibrium Constants (SB
p.100)
Back
Example 16-5C
138
16.5 Determination of Equilibrium Constants (SB
p.100)
Back
Let's Think 3
What is the implication for an equilibrium
reaction having an equilibrium constant much
smaller than 1.0?
Answer
The equilibrium constant of a reaction is related
to the ratio of the concentration of products to
the concentration of reactants at equilibrium.
When the equilibrium constant of a reaction is
much greater than 1, the reaction goes nearly to
completion. Conversely, when the equilibrium
constant of a reaction is much smaller than 1,
the reaction hardly goes to completion.
139
16.5 Determination of Equilibrium Constants (SB
p.100)
Check Point 16-5B
Answer
140
16.5 Determination of Equilibrium Constants (SB
p.100)
Back
Check Point 16-5B
Reactant / Product Initial no. of moles (mol) Change in no. of moles (mol) No. of moles at equilibrium (mol)
PCl5(g) 0.009 -x 0.009 x
PCl3(g) 0.250 x 0.250 x
Cl2(g) 0 x x
141
16.6 Equilibrium Constant in Terms of Partial
Pressures (SB p.101)
Example 16-6A
Answer
142
16.6 Equilibrium Constant in Terms of Partial
Pressures (SB p.101)
Example 16-6A
Back
143
16.6 Equilibrium Constant in Terms of Partial
Pressures (SB p.102)
Example 16-6B
Answer
144
16.6 Equilibrium Constant in Terms of Partial
Pressures (SB p.102)
Example 16-6B
Back
145
16.6 Equilibrium Constant in Terms of Partial
Pressures (SB p.102)
Check Point 16-6
Answer
146
16.6 Equilibrium Constant in Terms of Partial
Pressures (SB p.102)
Check Point 16-6
147
16.6 Equilibrium Constant in Terms of Partial
Pressures (SB p.102)
Check Point 16-6
Back
148
16.7 Equilibrium Position (SB p.103)
Example 16-7
Experiment Equilibrium concentration (mol dm-3) Equilibrium concentration (mol dm-3) Equilibrium concentration (mol dm-3)
Experiment SO2(g) O2(g) SO3(g)
1 1.60 1.30 3.62
2 0.71 0.50 1.00
Answer
149
16.7 Equilibrium Position (SB p.103)
Example 16-7
Back
150
16.7 Equilibrium Position (SB p.104)
Check Point 16-7
Answer
Expt Initial no. of moles (mol) Initial no. of moles (mol) No. of moles at eqm (mol)
Expt CH3CHOOH(l) CH3CH2OH(l) CH3COOH(l)
1 1.00 1.00 0.33
2 1.00 4.00 0.07
151
16.7 Equilibrium Position (SB p.104)
Check Point 16-7
The equilibrium constant for the equilibrium is
expressed as For experiment 1
CH3COOH(l) CH3CH2OH(l) At start
1.00 mol 1.00 mol At eqm 0.33
mol 0.33 mol
CH3COOCH2CH3(l) H2O(l) At
start
0 mol 0 mol At eqm
(1.00 0.33) mol (1.00 0.33)
mol
0.67 mol 0.67 mol
152
16.7 Equilibrium Position (SB p.104)
Check Point 16-7
Back
153
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.104)
Example 16-8A
Answer
154
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.106)
Example 16-8A
Back
155
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.106)
Example 16-8B
At 298 K, 50 cm3 of an aqueous solution
containing 6 g of solute Y is in equilibrium with
100 cm3 of an ether solution containing 108 g of
Y. Calculate the mass of Y that could be
extracted from 100 cm3 of an aqueous solution
containing 10 g of Y by shaking it with (a) 100
cm3 of fresh ether at 298 K (b) 50 cm3 of fresh
ether twice at 298 K.
Answer
156
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.106)
Example 16-8B
157
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.106)
Example 16-8B
158
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.106)
Example 16-8B
159
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.106)
Example 16-8B
160
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.106)
Example 16-8B
? Total mass of Y extracted m1 m2
(8.182
1.487) g
9.669 g
Back
161
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.108)
Check Point 16-8A
Answer
162
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.108)
Check Point 16-8A
Back
163
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.110)
Check Point 16-8B

• A student wrote the following explanation for the
  different Rf values found in the separation of
  two amino acids, leucine (Rf value 0.5) and
  glycine (Rf value 0.3), by paper chromatography
  using a solvent containing 20 of water.
• Leucine is a much lighter molecule than
  glycine.
• Do you agree with this explanation? Explain your
  answer.

Answer
164
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.110)
Check Point 16-8B
(a) The difference in Rf value of leucine and
glycine is due to the fact that they have
different partition between the stationary phase
and the mobile phase. Therefore, they move
upwards to different extent. The Rf value is not
related to the mass of the solute.
165
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.110)
Check Point 16-8B
(b) Draw a diagram to show the expected
chromatogram of a mixture of A, B, C and D using
a solvent X, given that the Rf values of A, B, C
and D are 0.15, 0.40, 0.70 and 0.75 respectively.
Answer
166
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.110)
Check Point 16-8B
Back
167
16.9 Significances of Equilibrium Constants (SB
p.111)
Check Point 16-9
Answer
168
16.9 Significances of Equilibrium Constants (SB
p.111)
Check Point 16-9
169
16.9 Significances of Equilibrium Constants (SB
p.111)
Check Point 16-9
Answer
170
16.9 Significances of Equilibrium Constants (SB
p.111)
Check Point 16-9
Back

• (b)
• 0.001 95 mol-2 dm6
• Since Qc lt Kc, the reaction proceeds from the
  left (reactant side) to the right (product side)
  until the equilibrium is reached.

171
16.10 Factors Affecting Equilibrium (SB p.113)
Back
Example 16-10A
Answer
(a) According to Le Chateliers principle, the
equilibrium position will shift to the
right. (b) According to Le Chateliers principle,
the equilibrium position will shift to the left.
172
16.10 Factors Affecting Equilibrium (SB p.114)
Example 16-10B
Answer
173
16.10 Factors Affecting Equilibrium (SB p.114)
Example 16-10B
(b) If an additional 0.200 mol dm3 of H2(g) is
added to the above equilibrium mixture while
keeping volume and temperature constant, what
will happen? Calculate the equilibrium
concentrations of all species when equilibrium is
reached.
Answer
174
16.10 Factors Affecting Equilibrium (SB p.114)
Example 16-10B
175
16.10 Factors Affecting Equilibrium (SB p.114)
Back
Example 16-10B
176
16.10 Factors Affecting Equilibrium (SB p.115)
Check Point 16-10A
Answer
177
16.10 Factors Affecting Equilibrium (SB p.115)
Check Point 16-10A
When CrO42(aq) and H(aq) are mixed at t0, they
react continuously to form Cr2O72(aq) and H2O(l
). At t1, an equilibrium between them is
established. At t2, when more H(aq) is added to
the system, the equilibrium can no longer be
maintained. In order to attain the equilibrium
again (i.e. at t3), the additional H(aq) must be
removed by shifting the equilibrium to the right
to form more Cr2O72(aq) and H2O(l).
Back
178
16.10 Factors Affecting Equilibrium (SB p.117)
Example 16-10C
179
16.10 Factors Affecting Equilibrium (SB p.117)
Example 16-10C
(a) Calculate the partial pressures of NO2(g) and
N2O4(g) at equilibrium if the total pressure is 1
atm.
Answer
180
16.10 Factors Affecting Equilibrium (SB p.117)
Example 16-10C
(b) Calculate the partial pressures of NO2(g) and
N2O4(g) if the total pressure at equilibrium is 2
atm.
Answer
181
16.10 Factors Affecting Equilibrium (SB p.117)
Example 16-10C
(c) Compare the results of (a) and (b), and state
the effect of an increase in pressure on the
equilibrium.
Answer
(c) Comparing the results in (a) and (b), PN2O4
is more than doubled while PNO2 is less than
doubled when the total pressure increases from 1
atm to 2 atm. Thus, the equilibrium position
shifts to the side with a smaller number of
molecules when the pressure increases.
182
16.10 Factors Affecting Equilibrium (SB p.117)
Example 16-10C
(d) Explain why the brown colour of the
equilibrium mixture fades out when the pressure
of the equilibrium system is increased. Assume
there is no temperature change. (Hint The
colour of NO2(g) is dark brown and that of
N2O4(g) is pale brown or colourless.)
Answer
183
16.10 Factors Affecting Equilibrium (SB p.117)
Back
Example 16-10C
Answer
(e) When the equilibrium mixture is put into hot
water (the temperature increases), the
equilibrium will shift to the left and more
NO2(g) will be formed. Thus, the colour of the
mixture will change to a darker brown. When the
equilibrium mixture is put into ice water (the
temperature decreases), the equilibrium will
shift to the right and more N2O4(g) will be
formed. As a result, the colour of the mixture
will change to pale brown (or colourless).
184
16.10 Factors Affecting Equilibrium (SB p.118)
Check Point 16-10B
Answer
185
16.10 Factors Affecting Equilibrium (SB p.118)
Check Point 16-10B

• (i) Decrease
• (ii) Increase
• (iii) Increase

Answer
186
16.10 Factors Affecting Equilibrium (SB p.118)
Check Point 16-10B
Answer
(c) The value of Kp will remain the same.
187
16.10 Factors Affecting Equilibrium (SB p.118)
Check Point 16-10B
Answer
188
16.10 Factors Affecting Equilibrium (SB p.118)
Check Point 16-10B
2. The equilibrium partial pressures of N2O4(g)
and NO2(g) were found to be 0.364 atm and 0.636
atm respectively for the following reversible
reaction at 100 oC. 2NO2(g)
N2O4(g) (b) The vessel containing the
equilibrium mixture is compressed to one-half
original volume suddenly. Predict what would
happen. Calculate the equilibrium partial
pressures of N2O4(g) and NO2(g).
Answer
189
16.10 Factors Affecting Equilibrium (SB p.118)
Check Point 16-10B
190
16.10 Factors Affecting Equilibrium (SB p.118)
Check Point 16-10B
191
16.10 Factors Affecting Equilibrium (SB p.118)
Check Point 16-10B
(b) By solving the quadratic equation, x 0.143
8 or 1.405 9. ? x 0.143 8 ? PNO2 1.272 2
? 0.143 8 0.984 4 atm PN2O4 0.728 0.143 8
0.871 8 atm
Back
192
16.10 Factors Affecting Equilibrium (SB p.120)
Check Point 16-10C
Answer
193
16.10 Factors Affecting Equilibrium (SB p.120)
Check Point 16-10C
(a) The equilibrium position shifts to the
left. (b) The equilibrium position shifts to the
left. (c) The equilibrium position remains
unchanged. (d) The equilibrium position shifts to
the right. (e) The equilibrium poistion shifts to
the left.
Back
194
16.10 Factors Affecting Equilibrium (SB p.121)
Example 16-10D
Answer
195
16.10 Factors Affecting Equilibrium (SB p.121)
Example 16-10D
Back
196
16.10 Factors Affecting Equilibrium (SB p.121)
Example 16-10E
Determine graphically the enthalpy change of
formation of NO2(g) from N2O4(g) using the
following data
Temperature (K) Kp (atm)
298 0.115
350 3.89
400 47.9
500 1700
600 17 800
Answer
(Given R 8.314 J K1 mol1)
197
16.10 Factors Affecting Equilibrium (SB p.122)
Example 16-10E
198
16.10 Factors Affecting Equilibrium (SB p.122)
Example 16-10E
199
16.10 Factors Affecting Equilibrium (SB p.122)
Example 16-10E
Back
200
16.10 Factors Affecting Equilibrium (SB p.122)
Example 16-10F
Answer
201
16.10 Factors Affecting Equilibrium (SB p.122)
Example 16-10F

• Since the reaction is exothermic, a lower
  temperature will shift the equilibrium to the
  right-hand side and hence increase the yield of
  ammonia.
• As shown in the chemical equation, there are
  totally four nitrogen and hydrogen molecules on
  the left-hand side of the equation and only two
  ammonia molecules on the right-hand side. A
  higher pressure will shift the equilibrium
  position to the right and more ammonia will be
  produced. Also, increasing the concentration of
  the reactants (i.e. nitrogen and hydrogen) or
  removing the product (i.e. ammonia) from the
  reaction mixture will shift the equilibrium
  position to the right and thus the yield of
  ammonia will be increased.

202
16.10 Factors Affecting Equilibrium (SB p.122)
Example 16-10F
(b) The actual operating conditions of the Haber
process are a temperature of about 450 oC, a
pressure of about 400 atm and the presence of a
catalyst (e.g. iron). Justify the conditions used.
Answer
203
16.10 Factors Affecting Equilibrium (SB p.122)
Back
Example 16-10F
(b) The use of high pressure is as predicted in
(a). This not only shifts the equilibrium
position to the right but also increases the rate
of the reaction. The use of catalysts shortens
the time for the reaction to reach the
equilibrium while it has no effect on the
equilibrium constant. The use of a high
temperature is contradictory to the prediction
made in (a). It can be explained based on the
rate of the reaction which in turn determines the
rate of manufacture of ammonia. Although the
equilibrium position shifts to the right at a
lower temperature, the rate of the reaction is
very low (i.e. a longer time is required to reach
the equilibrium state). The use of a moderate
temperature is a compromise between the rate and
the yield of the reaction. At 450 C, the
reaction is reasonably fast and the yield of
ammonia is optimum.