Dynamic equilibria

1
Irreversible and Reversible Reactions
2
Irreversible Reactions

• Chemical reactions that take place in one
  direction only
• It goes on until at least one of the reactants is
  used up
• ? complete reaction

3
Irreversible Reactions
Examples -
2Na(s) 2H2O(l) ? 2NaOH(aq) H2(g)
HCl(aq) H2O(l) ? H3O(aq) Cl?(aq)
2Mg(s) O2(g) ? 2MgO(s)
Cl2(g) 2OH?(aq) ? ClO?(aq) Cl?(aq) H2O(l)
4
Q.1

• Conditions for reversible reactions -
• Closed reaction vessels to prevent escape of
  gases
• High temperature to favour the reversed processes

5
Q.1

• Conditions for reversible reactions -
• Concentration of HCl(aq) gt 6 M

6
Q.1

• Conditions for reversible reactions -
• Closed reaction vessels to prevent escape of Cl2
• Dilute OH?(aq) at T ? 20?C to prevent side
  reaction

7
Reversible Processes
8
Reversible Reactions

• Chemical reactions that can go in two opposite
  directions
• Incomplete reactions

9
Q.2
10
Examples of reversible reactions
1. When HCl(aq) is added to CrO42?(aq)
Observation -
The yellow solution turns orange.
11
Examples of reversible reactions
1. When HCl(aq) is added to CrO42?(aq)
Interpretation -
CrO42?(aq) reacts with H to give
Cr2O72?(aq) There is no further colour change
when rate of forward rx rate of backward rx
12
Examples of reversible reactions
2. When NaOH(aq) is added to Cr2O72?(aq)
Observation -
The orange solution turns yellow.
13
Examples of reversible reactions
2. When NaOH(aq) is added to Cr2O72?(aq)
Interpretation -

• H ions are being removed by NaOH
• rate of forward rx ?
• rate of backward rx gt rate of forward rx
• ? a net change of Cr2O72?(aq) to CrO42?(aq)

14
Examples of reversible reactions
2. When NaOH(aq) is added to Cr2O72?(aq)
Interpretation -
There is no further colour change when rate of
backward rx rate of forward rx
15
Examples of reversible reactions
1. When H2O(l) is added to BiCl3(aq)
Observation -
The colourless solution turns milky.
16
Examples of reversible reactions
1. When H2O(l) is added to BiCl3(aq)
Interpretation -
BiCl3(aq) reacts with H2O(l) to give
BiOCl(s) There is no further change when rate of
forward rx rate of backward rx
17
Examples of reversible reactions
2. When HCl(aq) is added to BiOCl(s)
Observation -
The milky solution becomes clear.
18
Examples of reversible reactions
2. When HCl(aq) is added to BiOCl(s)
Interpretation -
HCl(aq) ? ? rate of backward rx gt rate of
forward rx ? a net consumption of BiOCl(s)
19
Examples of reversible reactions
2. When HCl(aq) is added to BiOCl(s)
Interpretation -
There is no further change when rate of forward
rx rate of backward rx
20
Q.3
1. When NaOH(aq) is added to Br2(aq)
Prediction -
The red-orange solution turns colourless.
21
Q.3
1. When NaOH(aq) is added to Br2(aq)
Interpretation -
Before the addition, rate of forward rx rate
of backward rx
22
Q.3
1. When NaOH(aq) is added to Br2(aq)
Interpretation -
H ions are being removed by NaOH(aq) ? rate of
forward rx gt rate of backward rx ? a net
consumption of Br2(aq)
23
Q.3
1. When NaOH(aq) is added to Br2(aq)
Interpretation -
There is no further change of colour when rate of
forward rx rate of backward rx
24
Q.3
2. When HCl(aq) is added
Prediction -
The colourless solution turns red-orange.
25
Q.3
2. When HCl(aq) is added
Interpretation -
Addition of HCl increases H(aq) ? rate of
backward rx gt rate of forward rx ? a net
production of Br2(aq)
26
Q.3
2. When HCl(aq) is added
Interpretation -
There is no further change of colour when rate of
forward rx rate of backward rx
27
Q.3
3. When AgNO3(aq) is added
Prediction -
A pale yellow ppt is formed. The red-orange
solution turns colourless.
28
Q.3
3. When AgNO3(aq) is added
Interpretation -
Ag(aq) react with Br-(aq) to give pale yellow
ppt of AgBr(s). ? rate of backward rx lt rate of
forward rx ? a net consumption of Br2(aq)
29
Q.3
3. When AgNO3(aq) is added
Interpretation -
There is no further change of colour when rate of
forward rx rate of backward rx
30
Phenolphthlein is a weak acid that ionizes
slightly in water to give H3O(aq)
Colourless
Red
31
What is the colour of phenolphthalein
when pH lt 8.3 ?
Colourless
Red
32
When pH lt 8.3 (e.g. deionized water), The
colourless form predominates
Colourless
Red
33
When NaOH(aq) is added, H3O(aq) ? ? rate of
forward rx gt rate of backward rx ? a net
production of the red form
Colourless
Red
34
There is no further colour change when rate of
forward rx rate of backward rx
Colourless
Red
35
When pH gt 10, The red form predominates
Colourless
Red
36
When 8.3 lt pH lt 10, Both forms have similar
concentrations
? pink
Colourless
Red
37
Reversible reactions and dynamic equilibrium
For a reversible reaction,
a state of dynamic equilibrium is said to be
established when rate of forward rx rate of
backward rx
Apparently, there is no change in the
concentrations of reactants and
products. Reactions continues at molecular level.
38
Dynamic Equilibrium
No change in the position of the girl
39
Reversible reactions and chemical equilibrium
ALL chemical reactions are considered as
reversible processes with different extents of
completion.
40
Reversible reactions and chemical equilibrium
?H lt 0
At equil., kreactanteq kproducteq
k gtgt k
? reactanteq ltlt producteq
? Ea gt Ea
? Forward rx is more complete than backward rx
41
Reversible reactions and chemical equilibrium
?H gt 0 ? Ea gt Ea
? Forward rx is less complete than backward rx
42
Chemical Equilibrium vs Chemical Kinetics
Chemical equilibrium is about how far a reaction
can proceed. Chemical kinetics is about how fast
a reaction can proceed.
43
Chemical Equilibrium vs Chemical Kinetics
The rate of rx depends on Ea or Ea The extent of
completion of rx depends on ?H
44
Evidence for Dynamic Equilibrium
At fixed T, NaNO3(aq) is a constant
Addition of 24NaNO3(s) ? Detection of
radioactivity in satd solution ? Interchange of
NaNO3 between the satd solution and the solid
45
Features of Chemical Equilibria

1. A system in chemical equilibrium consists of a
   forward reaction and a backward reaction both
   proceeding at the same rate.
2. All macroscopic properties (such as temperature,
   pressure, concentration, density, colour, etc.)
   of an equilibrium system remain unchanged.

46
Q.4 (i)
47
Q.4 (ii)
The equilibrium concentrations need not be equal
48
Q.5
Constant flame colour and temperature
49
Q.5
Open system ? Not at equilibrium state
Steady state
50
3. Equilibria can only be achieved in closed
systems with no exchange of matter with their
surroundings.
51
Q.6 A
Observation - The brown vapour escapes until
all brown liquid disappears
Interpretation - Br2 escapes from the system.
Thus, the rate of condensation is always less
than the rate of evaporation.
52
Q.6 B
Observation - No observable change
Interpretation - H2(g) and H2O(g) escape from
the system leaving only Fe3O4(s) and Fe(s).
Thus, both forward and backward reactions stop
due to absence of reactants.
53
Q.6 C
Observation - The amount of solid KCl ?
Interpretation - Water escapes by
evaporation. KCl(aq) ?, making the rate of
precipitation greater than the rate of
dissolution.
54
Q.6 D
Observation - A pleasant smell is detected.
The volume of the mixture ?
Interpretation - The more volatile ester
escapes, causing a drop in volume of both
reactants.
55
4. The state of equilibrium can be attained from
either the forward or the backward direction.
5. The equilibrium composition under a given set
of conditions is independent of the direction
from which the equilibrium is approached. In
other words, the same set of equilibrium
concentrations of reactants and products can be
obtained from either side of the reversible
reaction under the same set of conditions (See
Q.7).
56
nH nI 1.0
? nHI 1.0
Q.7
0.78
0.5-0.78/2 0.11
0.5-0.78/2 0.11
0 0 1.0
0.78
(1.0-0.78)/2 0.11
(1.0-0.78)/2 0.11
57
Equilibrium position and equilibrium composition
the equilibrium position is said to lie more to
the right hand side. The equilibrium composition
is richer in C, i.e. Cequil is much higher than
Aequil and Bequil.
58
Equilibrium position and equilibrium composition
the equilibrium position is said to lie more to
the left hand side. The equilibrium composition
is richer in A and B, i.e. Aequil and Bequil
are much higher than Cequil.
59
Equilibrium Law
60
Equilibrium Law
For any chemical system in dynamic equilibrium,
the concentrations or partial pressures of all
the substances present are related to one another
by a mathematical expression which is always a
constant at fixed temperature.
61
For the chemical system in equilibrium,
Kc depends on temperature and the nature of
reaction
62
Equilibrium constant and reaction quotient
Reaction quotient
63
Equilibrium constant
Reaction quotient
Qc Kc ? the system is at equilibrium
64
Equilibrium constant
Reaction quotient
Qc gt Kc
? the system is NOT at equilibrium
The reaction proceeds from right to left until Qc
Kc.
65
Equilibrium constant
Reaction quotient
Qc lt Kc
? the system is NOT at equilibrium
The reaction proceeds from left to right until Qc
Kc.
66
Equilibrium constant
Reaction quotient
Large Kc
The forward reaction is more complete
The equilibrium position lies to the right.
The equilibrium mixture is richer in the
substances on the R.H.S. of the equation.
67
Equilibrium constant
Reaction quotient
Small Kc
The forward reaction is less complete
The equilibrium position lies to the left.
The equilibrium mixture is richer in the
substances on the L.H.S. of the equation.
68
Kc gives no indication about the rate of reaction
Q.8
The rate of reaction depends on Ea
69
Relationship of Kc to the Stoichiometry of
Equations
units
mol?1 dm3
mol dm?3
70
Relationship of Kc to the Stoichiometry of
Equations
units
mol?1 dm3
A B C
mol?x dm3x
71
Relationship of Kc to the Stoichiometry of
Equations
units
mol?1 dm3
A B C
72
Q.9
(1)
A B
(2)
B C
(3)
C D
K1K2
(4) (1) (2)
? K4 K1 ? K2
73
Q.9
(1)
A B
(2)
B C
(3)
C D
K1K2K3
(5) (1) (2) (3)
? K5 K1 ? K2 ? K3
74
Q.10
(1) H2(g) Cl2(g) 2HCl(g)
(2) N2(g) 3H2(g) 2NH3(g)
(3) N2(g) 4H2(g) Cl2(g) 2NH4Cl(s)
(4) NH3(g) HCl(g) NH4Cl(s)
(4) (3) (2) (1) ? ½
5.1?1015 mol?2 dm6
75
Determination of Equilibrium Constants
76
(No Transcript)
77
Experiment 2
Experiment 1
Reactant/Product
Amount at equilibrium (mol)
Amount at beginning (mol)
Amount at equilibrium (mol)
Amount at beginning (mol)
0.083
0.250
CH3COOH(aq)
0.083
0.250
CH3CH2OH(aq)
0.020
0.020
H2SO4(l)
0.250 0.083 0.167
0.000
CH3COOCH2CH3(aq)
0.167
0.000
H2O(l)
78
For experiment 1
79
For experiment 2
80
(No Transcript)
81

• Conc. H2SO4 acts as a positive catalyst
• It can shorten the time taken to reach the state
  of equilibrium but has no effect on the extent of
  completion of the reaction.

82
Same extent of completion Same equilibrium
composition
83
Equilibrium Constant in Terms of Partial Pressures
84
For gaseous systems in dynamic equilibria, it is
more convenient to express the equilibrium
constants in terms of partial pressures.
85
Relationship between Kc and Kp
PV nRT
GasRT
At fixed T, P ? Gas
86
If a b c d, Kp Kc
87
Simple Calculations Involving Kc and Kp
0.50 mole of CO2(g) and 0.50 mole of H2(g) are
mixed in a 5.0 dm3 flask at 690 K and are allowed
to establish the following equilibrium.
Kp 0.10 at 690 K R 0.082 atm dm3 K?1 mol?1
Find partial pressures of all gaseous components
88
Initial no. of moles
No. of moles at equil.
x 0.12
89
No. of moles at equil.
nT 0.38 0.38 0.12 0.12 1.00
11.316 atm
90
No. of moles at equil.
nT 0.38 0.38 0.12 0.12 1.00
4.30 atm
4.30 atm
1.36 atm
91
Q.11
At fixed V T,
P ? n
3
Initial partial pressure
3x
x
0
Partial pressure at equilibrium
1.5x
x 1.5x/2
1.5x
0.25x
92
Q.11
Partial pressure at equilibrium
0.25x
1.5x
1.5x
PT 373 kPa
1.5x 0.25x 1.5x
x 115 kPa
0.035 kPa?1
93
Q.12
At fixed V T,
P ? n
Initial partial pressure
x
0
0
Partial pressure at equilibrium
0.14x
0.86x
0.86x
PT 101 kPa
0.14x 0.86x 0.86x
x 54.3 kPa
94
Q.12
Partial pressure at equilibrium
0.14x
0.86x
0.86x
x 54.3 kPa
287 kPa
95
Q.13
N2(g) O2(g) 2NO(g)
No. of moles at equilibrium
2.0 - x
1.0 x
2x
Concentration at equilibrium
96
Q.13
N2(g) O2(g) 2NO(g)
Concentration at equilibrium
Kp 1.2 ? 10?2
x 0.073
N2 (2.0 0.073)/2.0 0.96 mol dm?3
97
Q.13
N2(g) O2(g) 2NO(g)
Concentration at equilibrium
Kp 1.2 ? 10?2
? x is small, 2.0 x ? 2.0 and 1.0 x ? 1.0
98
Q.13
N2(g) O2(g) 2NO(g)
Concentration at equilibrium
Kp 1.2 ? 10?2
x 0.077
N2 (2.0 0.077)/2.0 0.96 mol dm?3
99
Homogeneous Equilibrium
Equilibrium system involving ONE phase only
100
Homogeneous Equilibrium
Glacial ethanoic acid
Absolute alcohol
101
Q.14
Cu2(aq) 4NH3(aq) Cu(NH3)42(aq)
102
Q.14
103
Q.14
104
Q.14
H(aq) OH?(aq) H2O(l)
At pH 7, density of water ? 1000 g dm?3
55.5 mol dm?3
105
Q.14
H(aq) OH?(aq) H2O(l)
? In large excess ? ? a constant
H2O(l) ? 55.5 M
106
Q.15(a)
Calculate the molarity of water in 12.39 M
hydrochloric acid. Given Density of 12.39 M
hydrochloric acid is 1.19 g cm?3 at 298 K
Mass of 1 dm3 of 12.39 M HCl(aq) 1.19 g cm?3 ?
1000 cm3 1190 g Mass of HCl present 12.39 mol
? (1 35.5) g mol?1 452.2 g
107
Q.15(a)
Calculate the molarity of water in 12.39 M
hydrochloric acid. Given Density of 12.39 M
hydrochloric acid is 1.19 g cm?3 at 298 K
Mass of water present (1190 452.2) g 737.8 g
41.0 M
108
Q.15(b)
41.0 M
lt 55.5 M
At very high acid concentrations, H2O is NOT in
large excess.
It is NOT justified to consider H2Oequil as a
constant in ALL aqueous solutions.
109
Q.15(b)
12.38 M
lt 12.38 M
41.0 M
110
Heterogeneous Equilibrium
Equilibrium systems involving two or more phases
? Kp depends on temperature only ? at fixed T,
vapour pressure of water (Kp) is a constant,
irrespective of the amount
of water present.
111
? H2O(l)equil ? ? constant (at fixed T)
112
Kp (at fixed T)
113
In a solution, and in a gas, the concentration
changes as the particles (molecules, atoms or
ions) become closer together or further apart.
In a solid or a liquid, the particles are at
fixed distance from one another
this means that the concentration is also
fixed.
In effect, the concentration of a solid or a
liquid is equivalent to its density (also known
as the effective reacting concentration).
114
In heterogeneous equilibria, the effective
reacting concentration of a pure liquid or a pure
solid is a constant and is independent of the
amount of liquid or solid present
Since collisions of reacting particles occur at
the boundary of phases.
A change in the surface area of a solid or a
liquid (by changing the amount) affects the rates
of forward and backward reactions to the same
extent.
115
Changing the amount of a pure solid or a pure
liquid in a heterogeneous equilibrium mixture
does NOT disturb the equilibrium.
Conclusion -
X(s) and X(l) do NOT appear in the
equilibrium constant expressions of heterogeneous
equilibria.
116
Q.16
Mg(s) Cu2(aq) Mg2(aq) Cu(s)
117
Q.16
CaCO3(s) CaO(s) CO2(g)
118
Q.16
Ag(aq) Cl?(aq) AgCl(s)
119
Q.16
Fe3O4(s) 4H2(g) 3Fe(s) 4H2O(g)
120
Q.16
Br2(l) Br2(g)
121
Partition Equilibrium of a Solute Between Two
Immiscible Solvents
122
Partition (Distribution) Equilibrium

• The equilibrium established when a non-volatile
  solute distributes itself between two immiscible
  liquids

123
Partition (Distribution) Equilibrium
Water and hexane are immiscible with each other.
124
Partition (Distribution) Equilibrium
I2 dissolves in both solvents to different extent.
125
Partition (Distribution) Equilibrium
When dynamic equilibrium is established, rate of
? movement rate of ? movement
126
Suppose the equilibrium concentrations of iodine
in H2O and hexane are x and y respectively,
127
Suppose the equilibrium concentrations of iodine
in H2O and hexane are x and y respectively,
When dynamic equilibrium is established the ratio
of concentrations of iodine in water and in
hexane is always a constant.
KD partition coefficient or distribution
coefficient
128
Changing the concentrations by the same extent
does not affect the quotient.
? the rates of the two opposite processes are
affected to the same extent.
However, changing the concentrations by the same
extent changes the colour intensity of the
solutions.
129
Partition Coefficient
The partition law can be represented by the
following equation
(no unit)
Units of concentration -
mol dm-3, mol cm-3, g dm-3, g cm-3
130
Partition Coefficient
The partition coefficient of a solute between
solvent 2 and solvent 1 is given by
The partition coefficient of a solute between
solvent 1 and solvent 2 is given by
131
Partition Coefficient

• Depends on temperature ONLY.
• Not affected by the amount of solute added and
  the volumes of solvents used.
• TAS Experiment No. 12

132
CH3COOHwater
CH3COOH2-methylpropan-1-ol
133
Partition law holds true

• at fixed temperature
• for dilute solutions ONLY
• For concentrated solutions, interactions between
  solvent and solute have to be considered and the
  concentration terms should be expressed by
  activity (effective concentration)

134
Partition law holds true
3. when the solute exists in the same form in
both solvents.
i.e. no association or dissociation of solute
C1 and C2 are determined by titrating the acid in
each solvent with standard sodium hydroxide
solution.
135
C6H5COOHwater(C1) / mol dm-3 C6H5COOHbenzene(C2) / mol dm-3 C1/C2
0.06 0.483 0.124
0.12 1.92 0.063
0.14 2.63 0.053
0.20 5.29 0.038
136
Interpretation -

• Benzoic acid tends to dimerize (associate) in
  non-polar solvent to give (C6H5COOH)2

• The solute does not have the same molecular form
  in both solvents

? Violation of Partition law
137
Interpretation -
? degree of association of benzoic acid
C6H5COOHtotal C6H5COOHfree
C6H5COOHassociated
C2
C2(1-?)
C2?
Determined by titration with NaOH
138
Q.17(a)
The interaction between benzoic acid and benzene
molecules are weaker than the hydrogen bonds
formed between benzoic acid molecules. Thus
benzoic acids tend to form dimers when dissolved
in benzene. In aqueous solution, benzoic acid
molecules form strong H-bond with H2O molecules
rather than forming dimer.
139
Q.17(b)
In aqueous solution, there is no association as
explained in (a). Also, dissociation of acid can
be ignored since benzoic acid is a weak acid
(Ka 6.3 ? 10-5 mol
dm-3).
140
Q.17(c)
C1
141
? is a constant at fixed T
142
C6H5COOHbenzene(C2) / mol dm-3
C6H5COOHwater(C1) / mol dm-3
0.483
0.06
1.92
0.12
2.63
0.14
5.29
0.20
143
Applications of partition law

• Solvent extraction
• Chromatography

Two classes of separation techniques based on
partition law.
144
Solvent extraction
To remove I2 from an aqueous solution of KI, a
suitable solvent is added.
It is immiscible with water. ? Organic solvents
are preferred.
It dissolves I2 but not KI. ? Organic solvents
are preferred.
It can be recycled easily (e.g. by distillation)
? Organic (volatile) solvents are preferred.
What feature should the solvent have?
At equilibrium, rate of ? movement of I2 rate
of ? movement of I2
By partition law,
145
Solvent Extraction
Iodine can be extracted from water by adding
hexane, shaking and separating the two layers in
a separating funnel
146
Determination of I2 left in both layer
Titrated with standard sodium thiosulphate
solution
I2 2S2O3? ? 2I? S4O62?
147
Determination of I2 left in the KI solution
For the aqueous layer, starch is used as the
indicator.
For the hexane layer, starch is not needed
because the colour of I2 in hexane is intense
enough to give a sharp end point.
148
In solvent extraction, it is more efficient (but
more time-consuming) to use the solvent in
portions for repeated extractions than to use it
all in one extraction.
149
Worked example -
(a) Calculate the partition coefficient of X
between ether and water at 298 K.
M is the molecular mass of X
150
Worked example -
Or simply,
151
(b)(i)
Determine the mass of X that could be extracted
by shaking a 30 cm3 aqueous solution containing 5
g of X with a single 30 cm3 portion of ether at
298 K
152
(b)(i)
3.79 g of X could be extracted.
153
(b)(ii) First extraction
154
(b)(ii) Second extraction
155
total mass of X extracted (3.05 1.19) g
4.24 g gt 3.79 g.
Repeated extractions using smaller portions of
solvent are more efficient than a single
extraction using larger portion of
solvent. However, the former is more
time-consuming
156
Important extraction processes -
1. Products from organic synthesis, if
contaminated with water, can be purified by
shaking with a suitable organic
solvent. 2. Caffeine in coffee beans can be
extracted by Supercritical carbon dioxide fluid
(decaffeinated coffee) 2. Impurities such as
sodium chloride and sodium chlorate present in
sodium hydroxide solution can be removed by
extracting the solution with liquid ammonia.
Purified sodium hydroxide is the raw material
for making soap, artificial fibre, etc.
157
Q.18(a)
Alcohol layer
200 cm3 alcohol
Aqueous layer
100 cm3 of 0.500 M ethanoic acid
Calculate the of ethanoic acid extracted at 298
K by shaking 100 cm3 of a 0.500 M aqueous
solution of ethanoic acid with 200 cm3 of
2-methylpropan-1-ol
158
Q.18(a)
Let x be the fraction of ethanoic acid extracted
to the alcohol layer
No. of moles of acid in the original solution
0.500 ? 0.100 0.0500 mol
x 0.396 39.6
159
Q.18(b)
Let x1, x2 be the fractions of ethanoic acid
extracted to the alcohol layer in the 1st and 2nd
extractions respectively.
1st extraction
2nd extraction
of acid extracted0.2470.1860.43343.3
160
Q.19
Let x cm3 be the volume of solvent X required to
extract 90 of iodine from the aqueous solution
and y be the no. of moles of iodine in the
original aqueous solution.
x 7.5
? 7.5 cm3 of solvent X is required
161
Chromatography
A family of analytical techniques for separating
the components of a mixture. Derived from the
Greek root chroma, meaning colour, because the
original chromatographic separations involved
coloured substances.
162
Chromatography
In chromatography, repeated extractions are
carried out successively in one operation
(compared with fractional distillation in which
repeated distillations are performed) which
results, (as shown in the worked example and
Q.18), in an effective separation of components.
163
All chromatographic separations are based upon
differences in partition coefficients of the
components between a stationary phase and a
mobile phase.
164
The stationary phase is a solvent (often H2O)
adsorbed (bonded to the surface) on a solid.
The solid may be paper or a solid such as alumina
or silica gel, which has been packed into a
column or spread on a glass plate.
The mobile phase is a second solvent which seeps
through the stationary phase.
165
Three main types of chromatography 1. Column
chromatography 2.     Paper chromatography 3.    
 Thin layer chromatography
166
Column chromatography
Stationary phase - Water adsorbed on the
adsorbent (alumina or silica gel)
Mobile phase - A suitable solvent (eluant) that
seeps through the column
167
Column chromatography
Partition of components takes place repeatedly
between the two phases as the components are
carried down the column by the eluant.
The components are separated into different bands
according to their partition coefficients.
168
Column chromatography
The component with the highest coefficient
between mobile phase and stationary phase is
carried down the column by the mobile phase most
quickly and comes out first.
169
Column chromatography
Suitable for large scale treatment of sample For
treatment of small quantities of samples, paper
or thin layer chromatography is preferred.
170
Paper chromatography

• Stationary phase -
• Water adsorbed on paper.
• Mobile phase -
• A suitable solvent
• The best solvent for a particular separation
  should be worked out by trials-and-errors

171
Paper chromatography
The solvent moves up the filter paper by
capillary action Components are carried upward by
the mobile solvent ? Ascending chromatography
172
(No Transcript)
173

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

174
Paper chromatography
The components separated can be identified by
their specific retardation factors, Rf, which are
calculated by  
175
separated colours
Using chromatography to separate the colours in a
sweet.
176
(No Transcript)
177
separated colours
178
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

179
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
180
Two-dimensional paper chromatography
181
Two-dimensional paper chromatography
All spots (except proline) appears visible
(purple) when sprayed with ninhydrin (a
developing agent)
182
Thin layer chromatography
Stationary phase - Water adsorbed on a thin
layer of solid adsorbent (silica gel or
alumina). Mobile phase - A suitable solvent
183
Q.20
Suggest any advantage of thin layer
chromatography over paper chromatography. A
variety of different adsorbents can be used. The
thin layer is more compact than paper,
more equilibrations can be achieved in a few
centimetres (no. of extraction ?). ? A microscope
slide is long enough to provide effective
separation
184
Factors Affecting Equilibrium
185
THREE factors affecting chemical equilibria.
1. Changes in pressure 2. Changes in concentration
Alter the equilibrium position by changing the
equilibrium composition
3. Changes in temperature
Alter the equilibrium position by changing the
equilibrium constant
186
THREE ways of interpretation
1. Kc or Kp approach 2. Kinetic approach 3. Le
Chateliers Principle
187
Effects of changes in pressure
P ? by reducing V Equilibrium position shifts to
the right
P ? by increasing V Equilibrium position shifts
to the left
188
Interpretation Kp approach
Pequil 1
189
Interpretation Kp approach
Pequil 1
Equilibrium position shifts to the right until Qp
Kp
190
Interpretation Kp approach
Pequil 1
Equilibrium position shifts to the right until Qp
Kp
2PC x
191
Interpretation kinetic approach
Both the rates of forward and backward reactions
are increased by doubling the partial pressures
of all gaseous components of the system.
However, the rate of forward reaction is
increased more. There is a net forward
reaction Equilibrium position shifts to the right
192
Q.21
k1
k-1
R1 k1AB R-1 k-1C
V?½
R1 k1(2A)(2B) 4R1 R-1 k-1(2C) 2R-1
193
Q.21
k1
k-1
R1 k1AB R-1 k-1C
V?2
R1 k1(½A)(½B) ¼R1 R-1 k-1(½C) ½R-1
194
Le Chateliers Principle
If a system at equilibrium is forced to change,
the equilibrium position of the system will shift
in a way to reduce (or oppose) the effect of the
change.
195
Q.22(a)
Change PT ? Response PT ?
196
Q.22(b)/(c)
One mole
Two moles
197
Q.22(d)
One mole
Two moles
One mole of gas exert less pressure.
198
Q.22(e)
One mole
Two moles
One mole of gas exert less pressure.
The forward reaction lowers the pressure of the
system.
199
Q.22(f)
A(g) B(g) C(g)
One mole
Two moles
One mole of gas exert less pressure.
The forward reaction lowers the pressure of the
system.
Equilibrium position shifts to the right.
200
Q.22(g)
Change PT ? Response PT ?
201
N2O4(g) 2NO2(g)
pale yellow
brown
Immediately after pushing in the plunger The gas
mixture turns darker brown due to a sudden
increase in concentrations of both gases
202
N2O4(g) 2NO2(g)
pale yellow
brown
After a few seconds The gas mixture turns paler
because the system reduces the pressure by
shifting the equilibrium position to the left
(the side with less gas molecules).
203
N2O4(g) 2NO2(g)
pale yellow
brown
Immediately after pulling out the plunger The gas
mixture turns paler due to a sudden decrease in
concentrations of both gases
204
N2O4(g) 2NO2(g)
pale yellow
brown
After a few seconds The gas mixture turns darker
brown because
the system ? the pressure by shifting the
equilibrium position to the right (the side with
more gas molecules).
205
Q.23(a)/(b)
Cause ? in PT by reducing VT
Effect No effect on the equilibrium position
Cause ? in PT by increasing VT
Effect No effect on the equilibrium position
206
Q.23(c)/(d)
Effect Equilibrium position shifts to the right
Cause ? in PT by increasing PCO
Effect Equilibrium position shifts to the left
207
Q.23(e)
Cause ? in PT by introducing He(g)
Effect No effect on equilibrium position
Reason The partial pressures of reactants and
products remain unchanged.
208
Q.23(a)/(b)
Cause ? in PT by reducing VT
Effect Equilibrium position shifts to the left
Cause ? in PT by increasing VT
Effect Equilibrium position shifts to the right
209
Q.23(c)/(d)
Effect Equilibrium position shifts to the right
Effect Equilibrium position shifts to the left
210
Q.23(e)
Cause ? in PT by introducing He(g)
Effect No effect on equilibrium position
211
Q.23(a)/(b)
Cause ? in PT by reducing VT
Effect No effect on equilibrium position
Cause ? in PT by increasing VT
Effect No effect on equilibrium position
212
Q.23(c)/(d)
Effect Equilibrium position shifts to the right
Effect Equilibrium position shifts to the left
213
Q.23(e)
Cause ? in PT by introducing He(g)
Effect No effect on equilibrium position
214
For the systems
Changing PT by altering VT has no effect on the
equilibrium position
Interpretation Kp approach
215
For the systems
H2(g) CO2(g) H2O(g) CO(g)
Pequil 1
Kp
216
For the systems
Kp
217
For the systems
Kinetic approach The rates of forward and
backward reactions are affected to the same
extent.
218
For the systems
By Le Chateliers principle
Since the system has the same no. of gas
molecules on either side,
No adjustment made by the system can reduce the
change.
? No shifting of equil. position
219
For the systems
? in PT by changing VT has no effect on the
equilibrium position
However, the partial pressures and thus the
equilibrium composition change by altering VT
220
Q.24
Once the bottle is opened, CO2 escapes from the
system and its partial pressure drops.
The system responds by releasing CO2 from the
aqueous solution.
221
Effects of pressure changes on equilibrium
systems involving ONLY solids and/or liquids are
negligible since solids and liquids are
incompressible (with fixed density at fixed T)
222
Q.25
Extremely high P
More open
More closely packed
The great increase in pressure causes the more
open structure of ice to collapse to give the
more closely packed structure of liquid water.
223
The Effect of Changes in Concentration on
Equilibrium
colourless
white ppt
Test 1 Add HCl
Result The white ppt disappears The equil.
position shifts to the left
224
colourless
white ppt
Interpretation Kc approach
Since H2O is in large excess
225
colourless
white ppt
Addition of HCl(aq) Both H(aq) and Cl?(aq) ?
to the same extent
The equilibrium position shifts to the left to
restore the Kc
226
colourless
white ppt
Kinetic approach -
Both the rates of forward and backward reactions
increase but the backward reaction increases
more. ? A net backward reaction is observed ? The
equilibrium position shifts to the left
227
Addition of HCl(aq)
The system responds in such a way as to reduce
the amount of HCl added ? the equilibrium
position shifts to the left
228
The Effect of Changes in Concentration on
Equilibrium
colourless
white ppt
Test 2 Add large excess of H2O
Result The white ppt reappears The equil.
position shifts to the right
229
colourless
white ppt
Interpretation Kc approach
Addition of large excess of H2O
Equilibrium position shifts to the right such
that Qc Kc
230
colourless
white ppt
Interpretation Kinetic approach

• H2O(l) ?
• ? rate of forward rx gt rate of backward rx
• ? equilibrium position shifts to the right

231
colourless
white ppt
By Le Chateliers principle -
H2O(l) ? The system shifts to the right to
reduce the water added.
232
The Effect of Changes in Temperature on
Equilibrium
A change in temperature of an equilibrium system
results in an adjustment of the equilibrium
system to a new equilibrium position with a new
equilibrium constant.
233
Examples -
Exothermic reaction
Temperature (K) 500 600 700 800
Kp (atm-2) 90.0 3.0 0.3 0.04
Kp decreases as T increases
234
Examples -
Exothermic reaction
Temperature (K) 298 500 700 900 1100
Kp (atm) 1.5 ? 1048 3.1 ? 1032 1.2 ? 1026 3.1 ? 1022 1.5 ? 1020
Kp decreases as T increases
235
Examples -
Endothermic reaction
Temperature (K) 200 300 400 500
Kp (atm) 1.8 ? 10-6 0.174 51 1510
Kp increases as T increases
236
Examples -
Endothermic reaction
Temperature (K) 700 1100 1500
Kp (no unit) 5 ? 10-13 4 ? 10-8 1 ? 10-5
Kp increases as T increases
237
Vant Hoff Equation
238
If the forward process is exothermic,
T ? ? K ? An increase in T shifts the equilibrium
position to the left
(in the endothermic direction)
239
If the forward process is exothermic,
T ? ? K ? An decrease in T shifts the equilibrium
position to the right
(in the exothermic direction)
240
If the forward process is endothermic,
T ? ? K ? An increase in T shifts the equilibrium
position to the right
(in the endothermic direction)
241
If the forward process is endothermic,
T ? ? K ? An decrease in T shifts the equilibrium
position to the left
(in the exothermic direction)
242
Conclusion
1. An increase in temperature shifts the
equilibrium position in the endothermic
direction. 2. A decrease in temperature shifts
the equilibrium position in the exothermic
direction.
Consistent with Le Chateliers principle
243
Q.26(a)
Forward reaction is exothermic
lnK
(If C gt 0)
244
Q.26(a)
Forward reaction is exothermic
lnK
(If C lt 0)
245
Q.26(a)
Forward reaction is endothermic
lnK
(If C gt 0)
246
Q.26(a)
Forward reaction is endothermic
(If C lt 0)
lnK
247
Q.27
248
Q.27(a)
? in T shift the equilibrium position to the
right. Thus, the forward reaction is endothermic
By Le Chateliers principle, the system tends to
decrease the T by shifting in the endothermic
direction.
249
Q.27(b)(i)
Assume no change in equilibrium position ? n is
fixed PT inside the syringe atomospheric
pressure ? PT is fixed
250
Q.27(b)(ii)
? equilibrium position shifts to the right ?
total no. of moles of gas molecules ? ? total
volume of the system ? further
251
Q.27
V(dm3)
Ideal gas expansion
252
Interpretation of the Effects of Temperature
Changes on Equilibrium in Terms of Chemical
Kinetics
253
For the forward reaction (exothermic)
gt 1
? Ea gt 0 T2 T1 gt 0
Rate ? as T ?
254
For the backward reaction (endothermic)
? Ea gt Ea T2 T1 gt 0
255
Conclusion
An ? in temperature ? the rates of endothermic
and exothermic reactions to different extents.
The rate of endothermic reaction is affected more
by temperature changes.
256
Q.28
Prediction -
An ? in T shifts the equilibrium position to the
right (in endothermic direction)
Interpretation -
An ? in T increases Kp Thus, more X(l) evaporate
until Qp Kp
257
Q.28
Prediction -
An ? in T shifts the equilibrium position to the
right (in endothermic direction)
Interpretation -
An ? in T increases Kp Thus, more X(s) sublime
until Qp Kp
258
and C is negligibly small
log10K gt 0
K gt 1 (the exothermic reaction is more complete
259
and C is negligibly small
log10K lt 0
K lt 1 (the endothermic reaction is less
complete)
260
Example -
Estimate the values of K at 298 K for the
equilibrium systems in which the ?H of the
forward reactions are (i) 100 kJ mol?1 and (ii)
100 kJ mol?1 respectively. (Given R 8.314 J
K?1 mol?1)
(i)
(Units not known)
K ? 3 ? 1017
261
Example -
Estimate the values of K at 298 K for the
equilibrium systems in which the ?H of the
forward reactions are (i) 100 kJ mol?1 and (ii)
100 kJ mol?1 respectively. (Given R 8.314 J
K?1 mol?1)
(ii)
(Units not known)
K ? 3 ? 10?18
262
Conclusion - Exothermic processes are Far More
Complete than endothermic processes.
263
Q.29
The total pressures of the following equilibrium
system are 2.333?104 Nm?2 and 6.679?104 Nm?2 at
282.5 K and 298.1 K respectively.
Since all gases arises from NH4HS(s)
264
At 282.5 K
(1)
At 298.1 K
(2)
(2) (1)
265
Effects of catalysts on Equilibrium
It can be shown that catalysts have no effect on
the equilibrium position since they affect the
rates of both forward and backward reactions to
the same extent. (Refer to Notes on Chemical
Kinetics, p.37 Q.29)
A catalyst has no effect on the equilibrium
position but can change the time taken to attain
the equilibrium state.
266
Q.30
Less time to attain equilibrium
Time taken to attain equilibrium
Concentration
Time
267
Q.31
?H gt 0

• ? in T
• ? in PT by reducing VT
• VT ? T ? (adiabatic compression)

e.g. expanding universe
268
Q.31
?H gt 0
Adding a ve catalyst
269
Q.32
?H lt 0
t1 1. adding a catalyst 2. ? in PT by adding
an an inert gas at fixed VT
270
Q.32
?H lt 0
? in PT by reducing VT has no effect on the
equilibrium position but changes the equilibrium
composition
t1 t2 t3 t4
271
Q.32
?H lt 0
t4 ? in PT by reducing VT
t1 t2 t3 t4
272
Q.32
?H lt 0
t2 ? in T at fixed VT
t1 t2 t3 t4
273
Q.32
?H lt 0
t3 Input of H2(g) at fixed VT
t1 t2 t3 t4
274
Q.33
Effect on Kp
Effect on equilibrium position
Changes
No effect
Shifts to the right
? in PT by reducing VT
Kp ?
Shifts to the left
? in T
275
Q.33
Effect on Kp
Effect on equilibrium position
Changes
No effect
No effect
No effect
Shifts to the right
276
Q.33
Effect on Kp
Effect on equilibrium position
Changes
No effect
No effect
A positive catalyst is added
No effect
Shifts to the right
A little H2O(l) is added
277
Q.34
Effect on Kp
Effect on equilibrium position
Changes
No effect
No effect
? in T at fixed PT
No effect
No effect
? in T at fixed PT
278
Q.34
Effect on Kp
Effect on equilibrium position
Changes
No effect
Shifts to the right
? in T at fixed VT
No effect
Shifts to the left
? in T at fixed VT
279
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
280
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
Isothermal compression
281
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 exothermic Kp ? if the forward reaction is endothermic Kp ? if the forward reaction is exothermic
282
Summary of the Effects of Changes of Various
Factors on Equilibrium
Factor Equilibrium position Equilibrium constant
Addition of a catalyst No change No change
283
16.1 Irreversible and Reversible Reactions (SB
p.89)
Check Point 16-1
Back
Answer

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

284
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.
285
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.
286
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.
287
16.4 Equilibrium Law (SB p.94)
Check Point 16-4
Answer
288
16.4 Equilibrium Law (SB p.94)
Check Point 16-4
Answer
289
16.4 Equilibrium Law (SB p.94)
Back
Check Point 16-4
Answer
290
16.5 Determination of Equilibrium Constants (SB
p.98)
Check Point 16-5A
291
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
292
16.5 Determination of Equilibrium Constants (SB
p.98)
Check Point 16-5A
293
16.5 Determination of Equilibrium Constants (SB
p.98)
Check Point 16-5A
(b) Calculate the equilibrium constant (Kc).
Answer
294
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.
295
16.5 Determination of Equilibrium Constants (SB
p.98)
Example 16-5A
Answer
296
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
297
16.5 Determination of Equilibrium Constants (SB
p.99)
Example 16-5B
Answer
298
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
299
16.5 Determination of Equilibrium Constants (SB
p.99)
Example 16-5C
Answer
300
16.5 Determination of Equilibrium Constants (SB
p.100)
Example 16-5C
301
16.5 Determination of Equilibrium Constants (SB
p.100)
Back
Example 16-5C
302
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.
303
16.5 Determination of Equilibrium Constants (SB
p.100)
Check Point 16-5B
Answer
304
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
305
16.6 Equilibrium Constant in Terms of Partial
Pressures (SB p.101)
Example 16-6A
Answer
306
16.6 Equilibrium Constant in Terms of Partial
Pressures (SB p.101)
Example 16-6A
Back
307
16.6 Equilibrium Constant in Terms of Partial
Pressures (SB p.102)
Example 16-6B
Answer
308
16.6 Equilibrium Constant in Terms of Partial
Pressures (SB p.102)
Example 16-6B
Back
309
16.6 Equilibrium Constant in Terms of Partial
Pressures (SB p.102)
Check Point 16-6
Answer
310
16.6 Equilibrium Constant in Terms of Partial
Pressures (SB p.102)
Check Point 16-6
311
16.6 Equilibrium Constant in Terms of Partial
Pressures (SB p.102)
Check Point 16-6
Back
312
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
313
16.7 Equilibrium Position (SB p.103)
Example 16-7
Back
314
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
315
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
316
16.7 Equilibrium Position (SB p.104)
Check Point 16-7
Back
317
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.104)
Example 16-8A
Answer
318
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.106)
Example 16-8A
Back
319
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
320
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.106)
Example 16-8B
321
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.106)
Example 16-8B
322
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.106)
Example 16-8B
323
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.106)
Example 16-8B
324
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
325
16.8 Partition Equilibrium of a Solute Between
Two Immiscible Solvents (SB p.108)
Check Point 16-8A
Answer