Irreversible and
Reversible
Reactions
1
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
2
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)
3
Q.1
2Na(s) + 2H2O(l)
2Mg(s) + O2(g)
2NaOH(aq) + H2(g)
2MgO(s)
Conditions for reversible reactions : • Closed reaction vessels to prevent escape of gases
• High temperature to favour the reversed processes
4
Q.1
HCl(aq) + H2O(l)
H3O+(aq) + Cl(aq)
Conditions for reversible reactions : • Concentration of HCl(aq) > 6 M
5
Q.1
Cl2(g) + 2OH(aq)
ClO(aq) + Cl(aq) + H2O(l)
Conditions for reversible reactions : • Closed reaction vessels to prevent escape of Cl2
• Dilute OH(aq) at T  20C to prevent side reaction
3Cl2(g) + 6OH(aq)  ClO3(aq) + 5Cl(aq) + 3H2O(l)
Hot and
concentrated
6
Reversible Processes
Vapour
Solid
Liquid
Changes of physical phases are reversible
7
Reversible Reactions
• Chemical reactions that can go in two
opposite directions
• Incomplete reactions
8
Q.2
CH3COOH(aq) + H2O(l)
NH3(aq) + H2O(l)
Cl2(aq) + H2O(l)
CH3COO(aq) + H3O+(aq)
NH4+(aq) + OH(aq)
HCl(aq) + HOCl(aq)
CH3COOH(l) + C2H5OH(l)
3H2(g) + N2(g)
9
2NH3(g)
CH3COOC2H5(l) + H2O(l)
Examples of reversible reactions
CrO42-(aq) + 2H+(aq)
yellow
Cr2O72-(aq) + H2O(l)
orange
1. When HCl(aq) is added to CrO42(aq)
Observation : The yellow solution turns orange.
10
Examples of reversible reactions
CrO42-(aq) + 2H+(aq)
yellow
Cr2O72-(aq) + H2O(l)
orange
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
11
Examples of reversible reactions
CrO42-(aq) + 2H+(aq)
yellow
Cr2O72-(aq) + H2O(l)
orange
2. When NaOH(aq) is added to Cr2O72(aq)
Observation : The orange solution turns yellow.
12
Examples of reversible reactions
CrO42-(aq) + 2H+(aq)
yellow
Cr2O72-(aq) + H2O(l)
orange
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 > rate of forward rx
13
 a net change of Cr2O72(aq) to CrO42(aq)
Examples of reversible reactions
CrO42-(aq) + 2H+(aq)
yellow
Cr2O72-(aq) + H2O(l)
orange
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
14
Examples of reversible reactions
BiCl3(aq) + H2O(l)
colourless
BiOCl(s) + 2HCl(aq)
White ppt
1. When H2O(l) is added to BiCl3(aq)
Observation : The colourless solution turns milky.
15
Examples of reversible reactions
BiCl3(aq) + H2O(l)
colourless
BiOCl(s) + 2HCl(aq)
White ppt
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
16
Examples of reversible reactions
BiCl3(aq) + H2O(l)
colourless
BiOCl(s) + 2HCl(aq)
White ppt
2. When HCl(aq) is added to BiOCl(s)
Observation : The milky solution becomes clear.
17
Examples of reversible reactions
BiCl3(aq) + H2O(l)
colourless
BiOCl(s) + 2HCl(aq)
White ppt
2. When HCl(aq) is added to BiOCl(s)
Interpretation : [HCl(aq)] 
 rate of backward rx > rate of forward rx
 a net consumption of BiOCl(s)
18
Examples of reversible reactions
BiCl3(aq) + H2O(l)
colourless
BiOCl(s) + 2HCl(aq)
White ppt
2. When HCl(aq) is added to BiOCl(s)
Interpretation : There is no further change when
rate of forward rx = rate of backward rx
19
Q.3
Br2(aq) + H2O(l)
Red-orange
H+(aq) + Br-(aq) + HOBr(aq)
colourless
1. When NaOH(aq) is added to Br2(aq)
Prediction : The red-orange solution turns colourless.
20
Q.3
Br2(aq) + H2O(l)
Red-orange
H+(aq) + Br-(aq) + HOBr(aq)
colourless
1. When NaOH(aq) is added to Br2(aq)
Interpretation : Before the addition,
rate of forward rx = rate of backward rx
21
Q.3
Br2(aq) + H2O(l)
Red-orange
H+(aq) + Br-(aq) + HOBr(aq)
colourless
1. When NaOH(aq) is added to Br2(aq)
Interpretation : H+ ions are being removed by NaOH(aq)
 rate of forward rx > rate of backward rx
 a net consumption of Br2(aq)
22
Q.3
Br2(aq) + H2O(l)
Red-orange
H+(aq) + Br-(aq) + HOBr(aq)
colourless
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
23
Q.3
Br2(aq) + H2O(l)
H+(aq) + Br-(aq) + HOBr(aq)
Red-orange
colourless
2. When HCl(aq) is added
Prediction : The colourless solution turns red-orange.
24
Q.3
Br2(aq) + H2O(l)
H+(aq) + Br-(aq) + HOBr(aq)
Red-orange
colourless
2. When HCl(aq) is added
Interpretation : Addition of HCl increases [H+(aq)]
 rate of backward rx > rate of forward rx
 a net production of Br2(aq)
25
Q.3
Br2(aq) + H2O(l)
H+(aq) + Br-(aq) + HOBr(aq)
Red-orange
colourless
2. When HCl(aq) is added
Interpretation : There is no further change of colour when
rate of forward rx = rate of backward rx
26
Q.3
Br2(aq) + H2O(l)
H+(aq) + Br-(aq) + HOBr(aq)
Red-orange
colourless
3. When AgNO3(aq) is added
Prediction : A pale yellow ppt is formed.
The red-orange solution turns colourless.
27
Q.3
Br2(aq) + H2O(l)
Red-orange
H+(aq) + Br-(aq) + HOBr(aq)
colourless
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 < rate of forward rx
 a net consumption of Br2(aq)
28
Q.3
Br2(aq) + H2O(l)
Red-orange
H+(aq) + Br-(aq) + HOBr(aq)
colourless
3. When AgNO3(aq) is added
Interpretation : There is no further change of colour when
rate of forward rx = rate of backward rx
29
Phenolphthlein is a weak acid that ionizes
slightly in water to give H3O+(aq)
+ H3O+(aq)
Colourless
30
Red
What is the colour of phenolphthalein
when pH < 8.3 ?
+ H3O+(aq)
Colourless
31
Red
When pH < 8.3 (e.g. deionized water),
The colourless form predominates
+ H3O+(aq)
Colourless
32
Red
When NaOH(aq) is added,
[H3O+(aq)] 
 rate of forward rx > rate of backward rx
 a net production of the red form
+ H3O+(aq)
Colourless
33
Red
There is no further colour change when
rate of forward rx = rate of backward rx
+ H3O+(aq)
Colourless
34
Red
When pH > 10,
The red form predominates
+ H3O+(aq)
Colourless
35
Red
When 8.3 < pH < 10,
Both forms have similar concentrations  pink
+ H3O+(aq)
Colourless
36
Red
Reversible reactions and dynamic equilibrium
For a reversible reaction,
reactants
products
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.
37
Dynamic Equilibrium
No change in the
position of the girl
An example of dynamic equilibrium
38
Reversible reactions and chemical equilibrium
ALL chemical reactions are considered as
reversible processes with different extents
of completion.
39
Reversible reactions and chemical equilibrium
H < 0
At equil., k[reactant]eq = k’[product]eq
 E a’ > Ea
k >> k’  [reactant]eq << [product]eq
 Forward rx is more complete than backward rx
Ea
40
E a’
Reversible reactions and chemical equilibrium
H > 0
 Ea > Ea’
 Forward rx is less complete than backward rx
Ea
41
E a’
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.
42
Chemical Equilibrium vs Chemical Kinetics
The rate of rx depends on Ea or Ea’
The extent of completion of rx depends on H
Ea
43
E a’
Evidence for Dynamic Equilibrium
NaNO3(s)
NaNO3(aq)
saturated
At fixed T, [NaNO3(aq)] is a constant
Addition of
24NaNO
3(s)
 Detection of radioactivity in sat’d solution
 Interchange of NaNO3 between the sat’d
solution and the solid
44
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.
45
Q.4 (i)
H2(g) + I2(g)
2HI(g)
Time taken to
reach the
equilibrium state
46
Q.4 (ii)
H2(g) + I2(g)
2HI(g)
The equilibrium
concentrations
need not be
equal
Time taken to reach the
equilibrium state
47
Q.5
48
Constant flame colour and
temperature
Q.5
CO2 and H2O
Open system
 Not at equilibrium state
Air
Steady state
Fuel
49
3. Equilibria can only be achieved in closed
systems with no exchange of matter with
their surroundings.
50
Q.6 A
Observation : Br2(g)
Br2(l)
51
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.
Q.6 B
Observation : No observable change
Interpretation : Fe3O4(s) + 4H2(g)
500C
3Fe(s) + 4H2O(g)
52
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.
Q.6 C
Observation : The amount of solid KCl 
Interpretation : -
Water escapes by evaporation.
KCl(s)
53
[KCl(aq)] , making the rate of
precipitation greater than the
KCl(aq) rate of dissolution.
Q.6 D
Observation : -
A pleasant smell is detected.
The volume of the mixture 
CH3COOH(l) + C2H5OH(l)
CH3COOC2H5(l) + H2O(l)
54
Interpretation : The more volatile ester
escapes, causing a drop in
volume of both reactants.
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).
55
Q.7
nH = nI = 1.0  nHI = 1.0
H2(g)
+
I2(g)
2HI(g)
0.5
0
ninitial
0.5
nequil
0.5-0.78/2
0.5-0.78/2
= 0.11
= 0.11
n’initial 0
0
nequil (1.0-0.78)/2 (1.0-0.78)/2
= 0.11
56
= 0.11
0.78
1.0
0.78
Equilibrium position and equilibrium
composition
For a system with a more complete forward
reaction,
A + B
C
the equilibrium position is said to lie more to
the right hand side.
The equilibrium composition is richer in C,
i.e. [C]equil is much higher than [A]equil and [B]equil.
57
Equilibrium position and equilibrium
composition
For a system with a less complete forward
reaction,
A + B
C
the equilibrium position is said to lie more to the
left hand side.
The equilibrium composition is richer in A and B,
i.e. [A]equil and [B]equil are much higher than [C]equil.
58
Equilibrium Law
59
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.
60
For the chemical system in equilibrium,
aA + bB
cC + dD
Equilibrium constant
expressed in concentration
Kc depends on temperature and
the nature of reaction
61
Equilibrium constant and reaction quotient
aA + bB
cC + dD
Equilibrium
constant
Reaction
quotient
62
Qc 
c
d
[C] [D]
a
b
[A] [B]
aA + bB
cC + dD
Equilibrium
constant
Reaction
quotient
Qc
[C] c [D] d

[A] a [B]b
Q c = Kc
 the system is at equilibrium
63
aA + bB
cC + dD
Equilibrium
constant
Reaction
quotient
Qc
[C] c [D] d

[A] a [B]b
Qc > Kc
 the system is NOT at equilibrium
64
The reaction proceeds from right to
left until Qc = Kc.
aA + bB
cC + dD
Equilibrium
constant
Reaction
quotient
Qc
[C] c [D] d

[A] a [B]b
Qc < Kc
 the system is NOT at equilibrium
65
The reaction proceeds from left to
right until Qc = Kc.
aA + bB
cC + dD
Equilibrium
constant
Reaction
quotient
Qc
[C] c [D] d

[A] a [B]b
Large Kc
The forward reaction is more complete
The equilibrium position lies to the right.
66
The equilibrium mixture is richer in the
substances on the R.H.S. of the equation.
aA + bB
cC + dD
Equilibrium
constant
Reaction
quotient
Qc
[C] c [D] d

[A] a [B]b
Small Kc
The forward reaction is less complete
The equilibrium position lies to the left.
67
The equilibrium mixture is richer in the
substances on the L.H.S. of the equation.
Kc gives no indication about the rate
of reaction
Q.8
The rate of reaction depends on Ea
68
Relationship of Kc to the Stoichiometry of Equations
A+B
C
69
C
A+B
KC1
KC-1
[C]

[A][B]
[A][B]

[C]
1

Kc1
units
mol1 dm3
mol dm3
Relationship of Kc to the Stoichiometry of Equations
A+B
C
xA + xB
KC1
[C]

[A][B]
mol1 dm3
xC
[C]
x
KC2 
 (Kc1 )
x
x
[A] [B]
x
70
units
molx dm3x
Relationship of Kc to the Stoichiometry of Equations
A+B
1
A+
y
Kc3 
71
KC1
C
1
B
y
[C]

[A][B]
units
mol1 dm3
1
C
y
[C]
1
y
1
y
[A] [B]
1
y
 
 Kc1
1
y

mol
1
y
dm
3
y
Q.9
(1) A
(2) B
(3) C
(4) A
B
[B]
K1 
[A]
C
[C]
K2 
[B]
D
[D]
K3 
[C]
C
[C] [B] [C]
K4 


= K1K2
[A] [A] [B]
(4) = (1) + (2)  K4 = K1  K2
72
Q.9
(1) A
(2) B
(3) C
(5) A
D
B
[B]
K1 
[A]
C
[C]
K2 
[B]
D
[D]
K3 
[C]
[D] [B] [C] [D]
K5 



= K1K2K3
[A] [A] [B] [C]
(5) = (1) + (2) + (3)  K5 = K1  K2  K3
73
Q.10
(1) H2(g) + Cl2(g)
(2) N2(g) + 3H2(g)
2HCl(g)
2NH3(g)
(3) N2(g) + 4H2(g) + Cl2(g)
(4) NH3(g) + HCl(g)
2NH4Cl(s)
NH4Cl(s)
(4) = [(3) –1 (2) – (1)]  ½
1
70
6
18
 K3  2 
2
3.9 10 mol dm
  

K4  
33
5
2
6 
 K1K2   (2.5 10 )(6.0 10 mol dm ) 
= 5.11015 mol2 dm6
74
Determination
of Equilibrium
Constants
75
Equilibrium System: (TAS Expt. 11)
CH3COOH(l) + CH3CH2OH(l)
CH3COOCH2CH3(l) +
H2O(l)
Kc 
76
[CH3COOCH2CH3 (l)]eqm[H2O(l)]eqm
[CH3COOH(l)]eqm[CH3CH2OH(l)]eqm
Equilibrium System:
CH3COOH(l) + CH3CH2OH(l)
CH3COOCH2CH3(l) +
Reactant/
Experiment 1
Experiment 2
H2O(l)
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
CH3COOCH2
CH3(aq)
0.000
0.000
0.198
H2O(l)
0.000
0.250 – 0.083
= 0.167
0.167
0.000
0.198
77
Equilibrium System:
CH3COOH(l) + CH3CH2OH(l)
CH3COOCH2CH3(l) +
HFor
2O(l)
experiment 1:
(0.167 )(0.167 )
V  4.05
Kc  V
0.083 )(0.083 )
1 (
V
V
78
Equilibrium System:
CH3COOH(l) + CH3CH2OH(l)
CH3COOCH2CH3(l) +
HFor
2O(l)
experiment 2:
(0.198 )(0.198 )
V  4.08
Kc  V
0.098 )(0.098 )
1 (
V
V
79
Equilibrium System:
CH3COOH(l) + CH3CH2OH(l)
CH3COOCH2CH3(l) +
H2O(l)
K K
Average value of K c 
c
c
1
2
2
 4.05  4.08
2
= 4.065 (no unit)
80
Equilibrium System:
CH3COOH(l) + CH3CH2OH(l)
CH3COOCH2CH3(l) +
H2O(l)
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.
With
catalyst
Without
catalyst
Same extent of
completion
Same equilibrium
composition
82
Equilibrium
Constant in Terms
of Partial Pressures
83
For gaseous systems in dynamic equilibria,
it is more convenient to express the equilibrium
constants in terms of partial pressures.
aA(g) + bB(g)
cC(g) + dD(g)

PC  PD 
Kp 
a
b
PA  PB 
c
84
d
Relationship between Kc and Kp
PV = nRT
n
P   RT = [Gas]RT
V
At fixed T,
85
P  [Gas]
Consider the equilibrium system :
aA(g) + bB(g)
Kp 
c
d
a
b
PC PD
PA PB
cC(g) + dD(g)
[C]c (RT)c [D]d (RT)d

[A] a (RT)a [B]b (RT)b
 Kc (RT)(cd)(ab)
If a + b = c + d,
86
Kp = Kc
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.
CO2(g) + H2(g)
CO(g) + H2O(g)
Kp = 0.10 at 690 K
R = 0.082 atm dm3 K1 mol1
Find partial pressures of all gaseous components
87
CO2(g) + H2(g)
Initial no.
of moles
0.50
No. of moles 0.50 - x
at equil.
CO(g) + H2O(g)
0.50
0
0
0.50 - x
x
x
x
x
( 5.0
)( 5.0
)
Kc  0.50x 0.50x  Kp  0.10
( 5.0 )( 5.0 )
x = 0.12
88
CO2(g) + H2(g)
No. of moles 0.50 - x
at equil.
0.38
0.50 - x
0.38
CO(g) + H2O(g)
x
x
0.12
0.12
nT = 0.38 + 0.38 + 0.12 + 0.12 = 1.00
nT RT
PT 
V
(1.00 mol)(0.082atm dm3 K -1 mol-1 )(690K)

5.0 dm3
= 11.316 atm
89
CO2(g) + H2(g)
No. of moles 0.50 - x
at equil.
0.38
0.50 - x
0.38
CO(g) + H2O(g)
x
x
0.12
0.12
nT = 0.38 + 0.38 + 0.12 + 0.12 = 1.00
 0.38
PCO2  XCO2 PT  
  11.316 atm  = 4.30 atm
 1.00 
 0.38
PH2  XH2 PT  
  11.316 atm  = 4.30 atm
 1.00 
 0.12 
PCO  PH2O  
  11.316 atm  = 1.36 atm
 1.00 
90
Q.11
At fixed V & T,
PSO2
PO2

nSO2
nO2
Pn
=3
2SO2(g) + O2(g)
2SO3(g)
Initial partial
pressure
3x
x
0
Partial pressure
at equilibrium
1.5x
x – 1.5x/2
= 0.25x
1.5x
91
Q.11
2SO2(g) + O2(g)
Partial pressure
at equilibrium
1.5x
0.25x
2SO3(g)
1.5x
PT = 373 kPa = 1.5x + 0.25x + 1.5x
x = 115 kPa
(PSO3 ) 2
(1.5x)2
1
Kp 
=
0.035
kPa

(PSO ) 2 (PO ) (1.5x)2 (0.25x)
2
92
2
Q.12
At fixed V & T,
PCl5(g)
Initial partial
pressure
x
Partial pressure
at equilibrium
0.14x
Pn
PCl3(g) + Cl2(g)
0
0.86x
0
0.86x
PT = 101 kPa = 0.14x + 0.86x + 0.86x
x = 54.3 kPa
93
Q.12
Partial pressure
at equilibrium
PCl5(g)
PCl3(g) + Cl2(g)
0.86x
0.14x
0.86x
x = 54.3 kPa
Kp 
PPCl3  PCl2
PPCl5
(0.86x)

= 287 kPa
0.14x
2
94
Q.13
N2(g)
95
+
O2(g)
2NO(g)
Initial no. of
moles
2.0
1.0
0
No. of moles
at equilibrium
2.0 - x
1.0 – x
2x
Concentration
at equilibrium
2.0  x
2.0
1.0  x
2.0
2x
2.0
Q.13
N2(g)
Concentration
at equilibrium
2.0  x
2.0
+
O2(g)
1.0  x
2.0
2NO(g)
2x
2.0
2
 2x 


2.0 

Kc 
= Kp = 1.2  102
 2.0  x  1.0  x 



 2.0  2.0 
x = 0.073
[N2] = (2.0 – 0.073)/2.0 = 0.96 mol dm3
96
Q.13
N2(g)
Concentration
at equilibrium
2.0  x
2.0
+
O2(g)
2NO(g)
1.0  x
2.0
2
2x
2.0
 2x 


2
2.0
(2x)


Kc 

= Kp = 1.2  102
 2.0  x  1.0  x  (2.0)(1.0)



 2.0  2.0 
∵ x is small, 2.0 – x  2.0 and 1.0 – x  1.0
97
Q.13
N2(g)
Concentration
at equilibrium
2.0  x
2.0
+
O2(g)
1.0  x
2.0
2NO(g)
2x
2.0
2
 2x 


2
2.0
(2x)


Kc 

= Kp = 1.2  102
 2.0  x  1.0  x  (2.0)(1.0)



 2.0  2.0 
x = 0.077
[N2] = (2.0 – 0.077)/2.0 = 0.96 mol dm3
98
Homogeneous Equilibrium
Equilibrium system involving ONE phase only
N2(g) + 3H2(g)
2NH3(g)
Cl2(aq) + 2Br(aq)
2Cl(aq) + Br2(aq)
CH3COOH(l) + C2H5OH(l)
CH3COOC2H5(l) + H2O(l)
99
Homogeneous Equilibrium
Glacial
ethanoic acid
Absolute
alcohol
Dissolve both
products
CH3COOH(l) + C2H5OH(l)
CH3COOC2H5(l) + H2O(l)
CH3COOH(aq) + C2H5OH(aq)
CH3COOC2H5(l) + H2O(l)
Immiscible  Two phases
100
Q.14
Cu2+(aq) + 4NH3(aq)
Kc 
101
Cu(NH3)42+(aq)
[Cu(NH ) (aq)]eqm
2
3 4
[Cu (aq)]eqm [NH3 (aq)]
2
4
eqm
Q.14
N2(g) + 3H2(g)
Kp
2NH3(g)

P 

P P 
NH3
N2
102
2
H2
3
Q.14
CH3COOH(l) + C2H5OH(l)
CH3COOC2H5(l) + H2O(l)
Kc 
103
[CH3COOC2H5 (l)]eqm [H2O(l)]eqm
[CH3COOH(l)]eqm [C2H5OH(l)]eqm
Q.14
H+(aq) + OH(aq)
Kc 
H2O(l)
[H2O(l)]eqm
[H (aq)]eqm [OH (aq)]eqm
At pH 7, density of water  1000 g dm3
1000 g
18 g mol1
3
[H2O(l)] 
=
55.5
mol
dm
3
1 dm
104
Q.14
H+(aq) + OH(aq)
Kc 
H2O(l)
[H2O(l)]eqm
[H (aq)]eqm [OH (aq)]eqm
[H2O(l)]  55.5 M  In large excess
  a constant
1
K  

[H (aq)]eqm [OH (aq)]eqm
'
c
105
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
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 water present
= (1190 – 452.2) g = 737.8 g
737.8g
18.0 g mol1
[H2O(l)] 
= 41.0 M
3
1 dm
107
Q.15(b)
737.8g
18.0 g mol1
[H2O(l)] 
= 41.0 M < 55.5 M
3
1 dm
At very high acid concentrations, H2O is
NOT in large excess.
It is NOT justified to consider [H2O]equil as a
constant in ALL aqueous solutions.
108
Q.15(b)
HCl(aq)
H+(aq) + Cl(aq)
12.38 M
H+(aq) + OH(aq)
< 12.38 M
109
H2O(l)
41.0 M
Heterogeneous Equilibrium
Equilibrium systems involving two or more
phases
H2O(l)
H2O(g)
Kp  PH2O
∵ Kp depends on temperature only
∴ at fixed T, vapour pressure of water (Kp)
is a constant, irrespective of the amount
of water present.
110
Kc 
[H2O(g)]equil
[H2O(l)]equil
∵ [H2O(l)]equil

nH2O
VH2O
mH2O
18 g  mH2O 
ρ


 18 g 
VH2O  VH2O 
18 g
∴ [H2O(l)]equil   = constant (at fixed T)
111
Kc 
[H2O(g)]equil
[H2O(l)]equil
Kc [H2O(l)]equil  K  [H2O(g)]equil
'
c
PH2O  [H2O(g)]equil RT
 PH2O  K RT
'
c
= Kp (at fixed T)
112
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).
113
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.
114
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.
115
Q.16
Mg(s) + Cu2+(aq)
Kc 
116
Mg2+(aq) + Cu(s)
[Mg (aq)]eqm
2
[Cu (aq)]eqm
2
Q.16
CaCO3(s)
CaO(s) + CO2(g)
Kp  PCO2
117
Q.16
Ag+(aq) + Cl(aq)
AgCl(s)
1
Kc 


[Ag (aq)]eqm [Cl (aq)]eqm
118
Q.16
Fe3O4(s) + 4H2(g)
Kp
3Fe(s) + 4H2O(g)

P 

P 
H2O
H2
119
4
4
Q.16
Br2(l)
Br2(g)
Kp  PBr2
120
Partition Equilibrium
of a Solute Between
Two Immiscible
Solvents
121
Partition (Distribution) Equilibrium
•
The equilibrium established when a
non-volatile solute distributes itself
between two immiscible liquids
A(solvent 2)
122
A(solvent 1)
Partition (Distribution) Equilibrium
Water and hexane are immiscible with
each other.
Hexane
Water
123
Partition (Distribution) Equilibrium
I2 dissolves in both solvents to different
extent.
Hexane
Water
124
Partition (Distribution) Equilibrium
When dynamic equilibrium is established,
rate of  movement = rate of  movement
Hexane
Water
125
Suppose the equilibrium concentrations of iodine
in H2O and hexane are x and y respectively,
x
KD 
y
Hexane
Water
126
Suppose the equilibrium concentrations of iodine
in H2O and hexane are x and y respectively,
x
KD 
y
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
127
Changing the concentrations by the same extent
does not affect the quotient.
x 2x 3x 4x 0.5x
KD  



y 2y 3y 4y 0.5y
∵ 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.
128
Partition Coefficient
The partition law can be represented by
the following equation:
concentration of solute in solvent1
KD 
concentration of solute in solvent2
[solute]solvent 1

[solute]solvent 2
(no unit)
Units of concentration : -
mol dm-3, mol cm-3, g dm-3, g cm-3
129
Partition Coefficient
The partition coefficient of a solute
between solvent 2 and solvent 1 is given by
[solute]solvent 2
KD 
[solute]solvent 1
The partition coefficient of a solute
between solvent 1 and solvent 2 is given by
[solute]solvent 1
KD 
[solute]solvent 2
130
Partition Coefficient
• Depends on temperature ONLY.
• Not affected by the amount of solute
added and the volumes of solvents used.
• TAS Experiment No. 12
131
[CH3COOH]water
[CH3COOH]water
[CH3COOH]2-methylpropan-1-ol
TAS Expt 12
slope  [CH3COOH]water / [CH3COOH]2methylprop an1ol
= KD
[CH3COOH]2-methylpropan-1-ol
132
Partition law holds true
1. at fixed temperature
2. 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)
133
Partition law holds true
3. when the solute exists in the same
form in both solvents.
i.e. no association or dissociation of solute
C6H5COOH(benzene)
C2
C6H5COOH(aq)
C1
C1 and C2 are determined by titrating the
acid in each solvent with standard sodium
hydroxide solution.
134
[C6H5COOH]water(C1) [C6H5COOH]benzene(C2)
/ mol dm-3
/ 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
Not a constant
135
Interpretation : • Benzoic acid tends to dimerize (associate)
in non-polar solvent to give (C6H5COOH)2
H
O
O
C
C
O
H
O
Benzoic acid dimer
• The solute does not have the same
molecular form in both solvents
 Violation of Partition law
136
Interpretation : 2C6H5COOH(benzene)
C2 (1   )
(C6H5COOH)2(benzene)
1
2 C2
 = degree of association of benzoic acid
[C6H5COOH]total = [C6H5COOH]free + [C6H5COOH]associated
C2
C2(1-)
C2
Determined by titration with NaOH
137
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.
138
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).
139
Q.17(c)
2C6H5COOH(benzene)
(C6H5COOH)2(benzene)
C2 (1  α)
C2α
K
2
[C2 (1  α)]
1
2
C2α
1
2
C6H5COOH(benzene)
C2 (1  α)
C1
KD 
C2 (1  α)
140
C6H5COOH(aq)
C1
C2α
K
2
[C2 (1  α)]
1
2
 is a constant at fixed T
C2α
'
 C2 (1  α) 
 K C2
2K
C1
C1
KD 
 '
C2 (1  α) K C2

141
C1
C2
 KDK'  K''
[C6H5COOH]water(C1) [C6H5COOH]benzene(C2)
/ mol dm-3
/ mol dm-3
C1
C2
0.06
0.483
0.086
0.12
1.92
0.087
0.14
2.63
0.086
0.20
5.29
0.087
~constant
142
Applications of partition law
•
Solvent extraction
•
Chromatography
Two classes of separation
techniques based on partition law.
143
Solvent extraction
+ hexane
I2 in KI(aq)
IColourless
22 in hexane
Hexane
I22 in KI(aq)
To
It
Itdissolves
iscan
remove
immiscible
be recycled
II22from
but
with
not
an
easily
water.
KI.
aqueous
(e.g. by
solution
distillation)
of KI, a
What
feature
should
the
solvent
have?
At equilibrium,
suitable solvent is added.

Organic
Organic
solvents
(volatile)
solvents
preferred.
preferred.
rate of
movement
of I2 =are
rate
of  are
movement
of I2
By partition law,
144
[ I 2 ]hexane
K
[ I 2 ]KI ( aq )
Solvent Extraction
Hexane
layer
Aqueous
layer
Before
shaking
145
After
shaking
Iodine can be extracted from water by
adding hexane, shaking and separating the
two layers in a separating funnel
Determination of I2 left in both layer
Titrated with standard sodium thiosulphate
solution
I2 + 2S2O3 
146
2I + S4O62
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.
147
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.
Worked example
148
Worked example : 50g X in 40 cm3 ether solution
10g X in 25 cm3 aqueous solution
[ X]
ether of X between ether
(a) By partition
Calculatelaw,
the partition
K D coefficient
[X]water
and water at 298 K.
50
50
[X]ether  M 
0.04 0.04M
149
[X]water
10
10
 M 
0.025 0.025M
50
M is the molecular mass
ofMX
0
.
04
KD 
 3.125
10
0.025M
Worked example : 50g X in 40 cm3 ether solution
10g X in 25 cm3 aqueous solution
Or simply,
K
150
50
40
10
25
 3.125
(b)(i)
30 cm3 ether
5g of X in 30 cm3
aqueous solution
xg of X in 30 cm3
ether solution
(5-x)g of X in 30 cm3
aqueous solution
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
151
(b)(i)
30 cm3 ether
5g of X in 30 cm3
aqueous solution
xg of X in 30 cm3
ether solution
(5-x)g of X in 30 cm3
aqueous solution
K  3.125

x
30
5 x
30
x

 x  3.79
5 x
3.79 g of X could be extracted.
152
(b)(ii) First extraction
15 cm3 ether
5g of X in 30 cm3
aqueous solution
x1g of X in 15 cm3
ether solution
(5-x1)g of X in 30
cm3 aqueous solution
K  3.125

153
x1
15
5 x1
30
2 x1

 x1  3.05
5  x1
(b)(ii) Second extraction
15 cm3 ether
(5-x1)g of X in 30 cm3
aqueous solution
x2g of X in 15 cm3
ether solution
(5-x1-x2)g of X in 30
cm3 aqueous solution
K  3.125

154
x2
15
5 x1  x2
30
2 x2

 x2  1.19
5  3.05  x2
total mass of X extracted
= (3.05 + 1.19) g = 4.24 g > 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
155
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.
156
Q.18(a)
200 cm3 alcohol
Alcohol layer
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;
157
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
[acid] alcohol 
0.0500x
0.200
[acid]water
0.0500(1 x)
0.100
K  3.05 
0.0500x
0.200
158
0.0500(1 x)

0.100
x = 0.396 = 39.6%
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
0.0500(1 x1 )
0.100
 3.05  x1  0.247
0.0500x1
0.100
2nd extraction
0.0500(1 x1  x2 )
0.100
 3.05  x2  0.186
0.0500x2
0.100
% of acid extracted=0.247+0.186=0.433=43.3%
159
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.
[I2 ]solventX
K  120 
[I2 ]water
0.9y
 x
0.1y
100
x = 7.5
∴ 7.5 cm3 of solvent X is required
160
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.
161
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.
162
All chromatographic separations are
based upon differences in partition
coefficients of the components
between a stationary phase and a
mobile phase.
163
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.
164
The mobile phase is a second solvent
which seeps through the stationary
phase.
Three main types of chromatography
165
1.
Column chromatography
2.
Paper chromatography
3.
Thin layer chromatography
Column chromatography
Stationary phase : Water adsorbed on the
adsorbent (alumina or silica gel)
Mobile phase : A suitable solvent (eluant) that
seeps through the column
166
Column chromatography
Sample
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.
167
Eluant
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.
168
Column chromatography
Suitable for large scale
treatment of sample
For treatment of small
quantities of samples, paper
or thin layer chromatography
is preferred.
169
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
X(adsorbed water)
170
stationary phase
X(solvent)
mobile phase
Paper chromatography
The solvent moves up
the filter paper by
capillary action
Components are carried
upward by the mobile
solvent
 Ascending
chromatography
171
172
• Different dyes
have different KD
between the mobile
and stationary
phases
• They will move
upwards to
different extent
173
Paper chromatography
The components separated
can be identified by their
specific retardation factors,
Rf, which are calculated by
Rf  distance travelled by spot
distance travelled by solvent
174
filter
paper
separated
colours
spot of
coloured
dye
solvent
Using chromatography to separate the colours in a sweet.
175
Solvent front
b
176
c
d
a
b
Rf (blue) 
a
c
Rf (red) 
a
d
Rf (green) 
a
a chromatogram
separated
colours
177
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
178
Paper Chromatography
Amino acid
Solvent
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
179
Two-dimensional paper chromatography
180
Two-dimensional paper chromatography
All spots (except proline)
appears visible (purple)
when sprayed with
ninhydrin (a developing
agent)
181
Thin layer chromatography
Stationary phase : Water adsorbed on a thin
layer of solid adsorbent
(silica gel or alumina).
Mobile phase : A suitable solvent
X(adsorbed water)
stationary phase
182
X(solvent)
mobile phase
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
183
Factors Affecting
Equilibrium
184
THREE factors affecting chemical equilibria.
1. Changes in pressure
2. Changes in concentration
No effect on
Kc or Kp
Alter the equilibrium position by
changing the equilibrium composition
3. Changes in temperature
Alter the equilibrium position by
changing the equilibrium constant
185
THREE ways of interpretation
186
1.
Kc or Kp approach
2.
Kinetic approach
3.
Le Chatelier’s Principle
Effects of changes in pressure
increase in pressure
A(g)
+
B(g)
C(g)
decrease in pressure
P  by reducing V
Equilibrium position shifts to the right
P  by increasing V
187
Equilibrium position shifts to the left
Interpretation : Kp approach
A(g)
B(g)
Pequil 1
PA
PB
P1
2PA
2PB
2
188
+
V
C(g)
PC
Kp 
PAPB
PC
2PC
2PC
1  PC  Kp
   Kp
Qp 
 
2PA 2PB  2  PAPB  2
Interpretation : Kp approach
A(g)
Pequil 1
PA
+
B(g)
PB
C(g)
PC
Equilibrium position shifts to the right until
Qp = Kp
P1
2
189
V
2PA
2PB
2PC
2PC
1  PC  Kp
   Kp
Qp 
 
2PA 2PB  2  PAPB  2
Interpretation : Kp approach
A(g)
Pequil 1
PA
+
B(g)
PB
C(g)
PC
Equilibrium position shifts to the right until
Qp = Kp
Pequil 2
2PA - x
2PB – x
2PC + x
2PC  x
PC
Qp 

 Kp
2PA  x2PB  x PAPB
190
Interpretation : kinetic approach
A(g)
+
B(g)
C(g)
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
191
Equilibrium position shifts to the right
Q.21
A(g)
+
B(g)
k1
k-1
C(g)
R1 = k1[A][B]
R-1 = k-1[C]
V½
R1’ = k1(2[A])(2[B]) = 4R1
R-1’ = k-1(2[C]) = 2R-1
192
More
affected
Q.21
A(g)
+
B(g)
k1
k-1
C(g)
R1 = k1[A][B]
R-1 = k-1[C]
V2
R1’ = k1(½[A])(½[B]) = ¼R1
R-1’ = k-1(½[C]) = ½R-1
193
More
affected
Le Chatelier’s 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.
194
Q.22(a)
A(g)
+
B(g)
Change : PT 
Response : PT 
195
C(g)
Q.22(b)/(c)
A(g)
+
B(g)
Two moles
196
C(g)
One mole
Q.22(d)
A(g)
+
B(g)
Two moles
C(g)
One mole
One mole of gas exert less pressure.
197
Q.22(e)
A(g)
+
B(g)
Two moles
C(g)
One mole
One mole of gas exert less pressure.
The forward reaction lowers the pressure
of the system.
198
Q.22(f)
A(g)
+
B(g)
Two moles
C(g)
One mole
One mole of gas exert less pressure.
The forward reaction lowers the pressure
of the system.
Equilibrium position shifts to the right.
199
Q.22(g)
A(g)
+
B(g)
Change : PT 
Response : PT 
200
C(g)
N2O4(g)
pale yellow
2NO2(g)
brown
Sealed nozzle
Syringe
N2O4(g), NO2(g)
Immediately after pushing in the plunger
The gas mixture turns darker brown due to a
sudden increase in concentrations of both gases
201
N2O4(g)
pale yellow
Sealed nozzle
2NO2(g)
brown
Syringe
N2O4(g), NO2(g)
After a few seconds
202
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).
N2O4(g)
pale yellow
2NO2(g)
brown
Sealed nozzle
Syringe
N2O4(g), NO2(g)
Immediately after pulling out the plunger
The gas mixture turns paler due to a sudden
decrease in concentrations of both gases
203
N2O4(g)
pale yellow
Sealed nozzle
2NO2(g)
brown
Syringe
N2O4(g), NO2(g)
After a few seconds
204
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).
Q.23(a)/(b)
H2(g) + CO2(g)
H2O(g) + CO(g)
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
205
Q.23(c)/(d)
H2(g) + CO2(g)
H2O(g) + CO(g)
Cause :  in PT by increasing PH2
Effect : Equilibrium position shifts to the
right
Cause :  in PT by increasing PCO
Effect : Equilibrium position shifts to the left
206
Q.23(e)
H2(g) + CO2(g)
H2O(g) + CO(g)
Cause :  in PT by introducing He(g)
Effect : No effect on equilibrium position
Reason : The partial pressures of reactants
and products remain unchanged.
207
Q.23(a)/(b)
PCl5(g)
PCl3(g) + Cl2(g)
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
208
Q.23(c)/(d)
PCl5(g)
PCl3(g) + Cl2(g)
Cause :  in PT by increasing PPCl5
Effect : Equilibrium position shifts to the
right
Cause :  in PT by increasing PCl2
Effect : Equilibrium position shifts to the left
209
Q.23(e)
PCl5(g)
PCl3(g) + Cl2(g)
Cause :  in PT by introducing He(g)
Effect : No effect on equilibrium position
210
Q.23(a)/(b)
Fe3O4(s) + 4H2(g)
3Fe(s) + 4H2O(g)
Cause :  in PT by reducing VT
Effect : No effect on equilibrium position
Cause :  in PT by increasing VT
Effect : No effect on equilibrium position
211
Q.23(c)/(d)
Fe3O4(s) + 4H2(g)
3Fe(s) + 4H2O(g)
Cause :  in PT by increasing PH2
Effect : Equilibrium position shifts to the
right
Cause :  in PT by increasing PH2O
Effect : Equilibrium position shifts to the left
212
Q.23(e)
Fe3O4(s) + 4H2(g)
3Fe(s) + 4H2O(g)
Cause :  in PT by introducing He(g)
Effect : No effect on equilibrium position
213
For the systems
H2(g) + CO2(g)
Fe3O4(s) + 4H2(g)
H2O(g) + CO(g)
3Fe(s) + 4H2O(g)
Changing PT by altering VT has no effect on
the equilibrium position
Interpretation : Kp approach
214
For the systems
H2(g) + CO2(g)
H2O(g) + CO(g)
Pequil 1
PH2
PCO2
PH2O
PCO
P1
2PH2
2PCO2
2PH2O
2PCO
2
V
Kp 
PH2O PCO
PH2 PCO2
Qp

2P 2P 

2P 2P 
H2O
H2
= Kp
215
CO
CO2
For the systems
Fe3O4(s) + 4H2(g)
Pequil 1
P1
2
V
Kp
PH2
PH2O
2PH2
2PH2O

P 

P 
H2O
H2
3Fe(s) + 4H2O(g)
4
4
Qp

2P 

2P 
H2O
H2
= Kp
216
4
4
For the systems
H2(g) + CO2(g)
Fe3O4(s) + 4H2(g)
H2O(g) + CO(g)
3Fe(s) + 4H2O(g)
Kinetic approach
The rates of forward and backward reactions
are affected to the same extent.
217
For the systems
H2(g) + CO2(g)
H2O(g) + CO(g)
Fe3O4(s) + 4H2(g)
3Fe(s) + 4H2O(g)
By Le Chatelier’s 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
218
For the systems
H2(g) + CO2(g)
PCl5(g)
H2O(g) + CO(g)
PCl3(g) + Cl2(g)
 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
219
Q.24
Decreasein PCO2
CO2(aq)
CO2(g)
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.
220
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)
H2O(s)
221
H2O(l)
Q.25
Extremely high P
H2O(s)
More open
H2O(l)
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.
222
The Effect of Changes in Concentration on
Equilibrium
Bi3+(aq) + Cl(aq) + H2O(l)
colourless
BiOCl(s) + 2H+(aq)
white ppt
Test 1 : Add HCl
Result : The white ppt disappears
The equil. position shifts to the left
223
Bi3+(aq) + Cl(aq) + H2O(l)
BiOCl(s) + 2H+(aq)
colourless
white ppt
Interpretation : Kc approach
[H (aq)] equil
Kc  3

[Bi (aq)]equil [Cl (aq)]equil [H2O(l)]equil
*

2
Since H2O is in large excess
[H (aq)]2equil
Kc  3
[Bi (aq)]equil [Cl (aq)]equil
224
Bi3+(aq) + Cl(aq) + H2O(l)
BiOCl(s) + 2H+(aq)
colourless
white ppt
[H (aq)]2equil
Kc  3
[Bi (aq)]equil [Cl (aq)]equil
Addition of HCl(aq)
Both [H+(aq)] and [Cl(aq)]  to the same extent
[H (aq)]
Qc  3
 Kc

[Bi (aq)][Cl (aq)]

225
2
The equilibrium position shifts to the left to
restore the Kc
Bi3+(aq) + Cl(aq) + H2O(l)
colourless
BiOCl(s) + 2H+(aq)
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
226
Bi3+(aq) + Cl(aq) + H2O(l)
Bi3+(aq) + 3Cl(aq) + H2O(l)
BiOCl(s) + 2H+(aq)
BiOCl(s) + 2HCl(aq)
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
227
The Effect of Changes in Concentration on
Equilibrium
Bi3+(aq) + Cl(aq) + H2O(l)
colourless
BiOCl(s) + 2H+(aq)
white ppt
Test 2 : Add large excess of H2O
Result : The white ppt reappears
The equil. position shifts to the right
228
Bi3+(aq) + Cl(aq) + H2O(l)
colourless
BiOCl(s) + 2H+(aq)
white ppt
Interpretation : Kc approach
[H (aq)]2equil
Kc  3
[Bi (aq)]equil [Cl (aq)]equil [H2O(l)]equil
*
Addition of large excess of H2O
[H (aq)]2
Qc  3
 *Kc

[Bi (aq)][Cl (aq)][H2O(l)]
*
229
Equilibrium position shifts to the right such
that *Qc = *Kc
Bi3+(aq) + Cl(aq) + H2O(l)
colourless
BiOCl(s) + 2H+(aq)
white ppt
Interpretation : Kinetic approach
[H2O(l)] 
 rate of forward rx > rate of backward rx
 equilibrium position shifts to the right
230
Bi3+(aq) + Cl(aq) + H2O(l)
colourless
BiOCl(s) + 2H+(aq)
white ppt
By Le Chatelier’s principle : [H2O(l)] 
The system shifts to the right to reduce the
water added.
231
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.
232
Examples : -
Exothermic reaction:
N2(g) + 3H2(g)
2NH3(g)
ΔH0  92 kJ mol1
Temperature (K)
500
600
700
800
Kp (atm-2)
90.0
3.0
0.3
0.04
Kp decreases as T increases
233
Examples : -
Exothermic reaction:
2C(graphite) + O2(g)
2CO(g)
ΔH0  211 kJ mol1
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
234
Examples : -
Endothermic reaction:
N2O4(g)
2NO2(g)
ΔH0  58 kJ mol1
Temperature (K)
200
300
400
500
Kp (atm)
1.8  10-6
0.174
51
1510
Kp increases as T increases
235
Examples : -
Endothermic reaction:
N2(g) + O2(g)
2NO(g)
ΔH0  100 kJ mol1
Temperature (K)
700
1100
1500
Kp (no unit)
5  10-13
4  10-8
1  10-5
Kp increases as T increases
236
Van’t Hoff Equation
ΔHo
lnK  
C
RT
ΔH
log10K  
 C'
2.303RT
o
C and C’ are constants related to ΔS
o
o
ΔS
ΔS
C'
C
R
2.303R
237
o
ΔH
lnK  
C
RT
o
If the forward process is exothermic,
 ΔH
ΔH  0 and   R
TK
o
o

  0

An increase in T shifts the equilibrium
position to the left
(in the endothermic direction)
238
ΔH
lnK  
C
RT
o
If the forward process is exothermic,
 ΔH
ΔH  0 and   R
TK
o
o

  0

An decrease in T shifts the equilibrium
position to the right
(in the exothermic direction)
239
ΔH
lnK  
C
RT
o
If the forward process is endothermic,
 ΔH
ΔH  0 and   R
TK
o
o

  0

An increase in T shifts the equilibrium
position to the right
(in the endothermic direction)
240
ΔH
lnK  
C
RT
o
If the forward process is endothermic,
 ΔH
ΔH  0 and   R
TK
o
o

  0

An decrease in T shifts the equilibrium
position to the left
(in the exothermic direction)
241
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 Chatelier’s principle
242
Q.26(a)
Forward reaction is exothermic
lnK
ΔHo
slope  0
R
y - intercept  C
243
(If C > 0)
1 1
(K )
T
Q.26(a)
lnK
Forward reaction is exothermic
ΔHo
slope  0
R
y - intercept  C
244
(If C < 0)
1 1
(K )
T
Q.26(a)
Forward reaction is endothermic
lnK
y - intercept  C
(If C > 0)
ΔHo
slope  0
R
1 1
(K )
T
245
Q.26(a)
Forward reaction is endothermic
y - intercept  C
(If C < 0)
ΔHo
slope  0
R
lnK
246
1 1
(K )
T
Q.27
50C
Increase in T
N2O4(g)(pale yellow)
Decrease in T
247
2NO2(g)(brown)
Q.27(a)
Increase in T
N2O4(g)(pale yellow)
2NO2(g)(brown)
Decrease in T
 in T shift the equilibrium position to the right.
Thus, the forward reaction is endothermic
ΔH  0
o
By Le Chatelier’s principle, the system tends
to decrease the T by shifting in the
endothermic direction.
248
Q.27(b)(i)
Assume no change in equilibrium position
 n is fixed
PT inside the syringe = atomospheric pressure
 PT is fixed
VT
V373K 373K

V273K 273K
373 K
3
3
V373K 
(1 dm )  1.37 dm
273 K
249
Q.27(b)(ii)
N2O4(g)
2NO2(g)
∵ equilibrium position shifts to the right
 total no. of moles of gas molecules 
 total volume of the system  further
250
Q.27
V(dm3)
Increase in T
N2O4(g)
2NO2(g)
Actual  in V
1.37
Ideal gas expansion
1.00
251
273
373
T(K)
Interpretation of the Effects of Temperature
Changes on Equilibrium in Terms of Chemical Kinetics
A–B + X
A + B–X
Potential energy
Ea
ΔH  0
Ea’
AB+X
A+BX
252
Reaction co-ordinate
o
For the forward reaction (exothermic)
kT2
kT1

AT2 e  Ea/RT2
AT1 e
Ea /RT1
 Ea /RT2
e
 Ea/RT1  e
e
 Ea 1 1
(  )
R T2 T1
e
Ea (T2 T1 )
RT1T2
Potential energy
∵ Ea > 0 &
T2 – T1 > 0
Rate  as T 
Ea
Ea’
AB+X
A+BX
253
>1
Reaction co-ordinate
For the backward reaction (endothermic)
kT2 '
kT1 '

AT2 'e
AT1 'e
 Ea '/RT2
Ea '/RT1
 Ea '/RT2
e
 Ea '/RT1  e
e
 Ea ' 1 1
(  )
R T2 T1
e
Potential energy
e
Ea (T2 T1 )
RT1T2
∵ E a’ > E a &
Ea
Ea’
T2 – T1 > 0
AB+X
A+BX
254
Ea '(T2 T1 )
RT1T2
Reaction co-ordinate
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.
255
Q.28
X(l)
X(g)
ΔH
o
vap
0
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
256
Q.28
X(s)
X(g)
ΔH
o
sub
0
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
257
ΔHo and the Extent of Completion of Reaction
ΔH
log10K  
 C'
2.303RT
o
if Ho < 0 (forward reaction is exothermic)
and C’ is negligibly small
log10K > 0
K > 1 (the exothermic reaction is more complete
258
ΔHo and the Extent of Completion of Reaction
ΔH
log10K  
 C'
2.303RT
o
if Ho > 0 (forward reaction is endothermic)
and C’ is negligibly small
log10K < 0
K < 1 (the endothermic reaction is less complete)
259
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)
o
ΔH
log10K  
 C'
2.303RT
- (-100  1000 J mol-1 )
ΔHo


2.303RT 2.303 8.314 J K -1 mol-1  298K
260
K  3  1017
(Units not known)
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)
o
ΔH
log10K  
 C'
2.303RT
- (100  1000 J mol-1 )
ΔHo


2.303RT 2.303 8.314 J K -1 mol-1  298K
261
K  3  1018 (Units not known)
Conclusion : Exothermic processes are Far More Complete
than endothermic processes.
262
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.
NH4HS(s)
NH3(g) + H2S(g)
Since all gases arises from NH4HS(s)
PNH3  PH2S
263
1
 PT
2
NH4HS(s)
NH3(g) + H2S(g)
At 282.5 K
lnKp  ln(PNH3 )eqm (PH2S )eqm
o


ΔH
1
2
  C
 ln( PT )  
2
 8.314  282.5
At 298.1 K
(1)
o


ΔH
1
2
  C (2)
lnKp '  ln(PNH3 ')eqm (PH2S')eqm  ln( PT ')  
2
 8.314  298.1
(2) – (1)
2
 PT ' 
ΔHo  1
1 
ln  



 PT  8.314  282.5 298.1
 6.679 10
ln
4
2.333

10

4
264
2

ΔHo  1
1 
 



 8.314  282.5 298.1
ΔHo = +94.41 kJ mol1
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.
265
Q.30
A
B
Less time
Time
takentotoattain
attainequilibrium
equilibrium
Concentration
[A]
[B]
t2
266
t1
Time
Q.31
A(g) + B(g)
H > 0
Rate of reaction
Forward
reaction
1.  in T
2.  in PT by reducing VT
VT  T 
(adiabatic compression)
Backward
reaction
t1
267
C(g)
e.g. expanding universe
Time
Q.31
A(g) + B(g)
C(g)
H > 0
Rate of reaction
Forward
reaction
Adding a +ve catalyst
Backward
reaction
t2
268
Time
Q.32 H2(g) + I2(g)
2HI(g) H < 0
t1 : 1. adding a catalyst
Concentration
2.  in PT by adding an
an inert gas at fixed VT
[HI(g)]
[H2(g)]
[I2(g)]
269
t1
t2
t3
t4
Time
Q.32 H2(g) + I2(g)
2HI(g) H < 0
Concentration
 in PT by reducing VT has
no effect on the equilibrium
position but changes the
equilibrium composition
[HI(g)]
[H2(g)]
[I2(g)]
270
t1
t2
t3
t4
Time
Q.32 H2(g) + I2(g)
2HI(g) H < 0
t4 :  in PT by reducing VT
Concentration
[HI(g)]
[H2(g)]
[I2(g)]
271
t1
t2
t3
t4
Time
Q.32 H2(g) + I2(g)
2HI(g) H < 0
t2 :  in T at fixed VT
Concentration
[HI(g)]
[H2(g)]
[I2(g)]
272
t1
t2
t3
t4
Time
Q.32 H2(g) + I2(g)
2HI(g) H < 0
t3 : Input of H2(g) at fixed VT
Concentration
[HI(g)]
[H2(g)]
[I2(g)]
273
t1
t2
t3
t4
Time
Q.33
CO(g) + 2H2(g)
274
CH3OH(g)
H < 0
Changes
Effect on
equilibrium
position
Effect on Kp
 in PT by
reducing VT
Shifts to the
right
No effect
 in T
Shifts to the
left
Kp 
Q.33
CO(g) + 2H2(g)
Changes
Doubling PCO and
PCH3OH
Doubling PH2 and
PCH3OH
275
CH3OH(g)
H < 0
Effect on
equilibrium
position
Effect on Kp
No effect
No effect
Shifts to the
right
No effect
Soluble in water
Q.33
CO(g) + 2H2(g)
276
CH3OH(g)
H < 0
Changes
Effect on
equilibrium
position
Effect on Kp
A positive
catalyst is added
No effect
No effect
A little H2O(l) is
added
Shifts to the
right
No effect
Q.34 A(g) + B(g)
277
C(g)
H = 0
Changes
Effect on
equilibrium
position
Effect on Kp
 in T at fixed PT
No effect
No effect
 in T at fixed PT
No effect
No effect
Q.34 A(g) + B(g)
278
C(g)
H = 0
Changes
Effect on
equilibrium
position
Effect on Kp
 in T at fixed VT
Shifts to the
right
No effect
 in T at fixed VT
Shifts to the
left
No effect
Summary of the Effects of Changes
of Various Factors on Equilibrium
aA(g) + bB(g)
Factor
Increase in
concentration of
reactants A or B
Increase in
concentration of
products C or D
279
cC(g) + dD(g)
Equilibrium
position
Shifts to right
Equilibrium
constant
No change
Shifts to left
No change
Summary of the Effects of Changes
of Various Factors on Equilibrium
aA(g) + bB(g)
Factor
Increase in
pressure by
reducing the
volume of the
container
Isothermal
compression
280
cC(g) + dD(g)
Equilibrium
position
Shifts to right if
(c + d) < (a + b)
Shifts to left to
(a + b) < (c + d)
No change if
a+b=c+d
Equilibrium
constant
No change
Summary of the Effects of Changes
of Various Factors on Equilibrium
aA(g) + bB(g)
281
cC(g) + dD(g)
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
Summary of the Effects of Changes
of Various Factors on Equilibrium
aA(g) + bB(g)
Factor
Addition of a
catalyst
282
Equilibrium
position
No change
cC(g) + dD(g)
Equilibrium
constant
No change
16.1 Irreversible and Reversible Reactions (SB p.89)
Back
In the following reversible reaction:
A
B
(a) Give the letter that represents the reactant of the
forward reaction.
(b) Give the letter that represents the reactant of the
backward reaction.
(c) Which is the forward reaction, A  B or B  A ?
(a) A
(b) B
(c) A  B
283
Answer
16.2 Dynamic Nature of Chemical Equilibrium (SB p.91)
Back
List some characteristics of chemical equilibrium.
Answer
Some characteristics of chemical equilibrium include:
284
•
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.
16.3 Examples of Chemical Equilibrium (SB p.92)
Back
A trace amount of carbon monoxide labelled with
radioactive carbon-14 is added to the following
equilibrium system:
H2O(g) + CO(g)
H2(g) + CO2(g)
Explain why radioactive carbon dioxide molecules are
formed.
285
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.
Answer
16.4 Equilibrium Law (SB p.94)
Back
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.
286
16.4 Equilibrium Law (SB p.94)
Write the equilibrium expression (for Kc) and unit of
equilibrium constants for the following equilibrium
system.
(a) 2O3(g)
(a)
Kc 
3O2(g)
[O 2 (g)]3eqm
[O 3 (g)]2eqm
Unit of Kc: mol dm-3
287
Answer
16.4 Equilibrium Law (SB p.94)
Write the equilibrium expression (for Kc) and unit of
equilibrium constants for the following equilibrium
system.
(b) N2(g) + 3H2(g)
(b)
Kc 
[NH3 (g)]2eqm
[N2 (g)]eqm [H2 (g)]3eqm
Unit of Kc: mol-2 dm6
288
2NH3(g)
Answer
16.4 Equilibrium Law (SB p.94)
Back
Write the equilibrium expression (for Kc) and unit of
equilibrium constants for the following equilibrium
system.
(c) C(graphite) + H2O(g)
(c)
Kc 
[CO(g)]eqm [H2 (g)]eqm
[H2 O(g)]eqm
Unit of Kc: mol dm-3
289
CO(g) + H2(g)
Answer
16.5 Determination of Equilibrium Constants (SB p.98)
In the determination of the equilibrium constant (Kc) of:
Fe2+(aq) + Ag+(aq)
Fe3+(aq) + Ag( s)
100 cm3 of 0.100 M AgNO3(aq) and 100 cm3 of 0.100 M
FeSO4(aq) are mixed in a dry conical flask. The mixture is
then allowed to stand overnight and filtered. The
concentration of Ag+(aq) is found by titration. 25.00 cm3 of
the filtrate is titrated with 0.050 M KCNS(aq) and 6.10 cm3 of
the KCNS( aq) is required for complete reaction.
290
16.5 Determination of Equilibrium Constants (SB p.98)
(a) Calculate the equilibrium concentrations of Ag+(aq),
Fe2+(aq) and Fe3+(aq).
(a) Ag+(aq) + CNS–(aq)  AgCNS(aq)
Number of moles of KCNS(aq)
= 6.10 dm3  0.050 M
1000
Answer
= 3.05  10-4 mol
 Number of moles of Ag+(aq) in 25 cm3 of the filtrate
at equilibrium = 3.05  10–4 mol
-4
[Ag+(aq)]eqm = 3.05  10 mol
25  10 - 3 mol
291
= 0.012 2 mol dm-3
16.5 Determination of Equilibrium Constants (SB p.98)
(a)  Fe2+(aq) and Ag+(aq) are consumed at the same rate.
 [Fe2+(aq)]eqm = [Ag+(aq)]eqm = 0.012 2 mol dm–3
 [Fe2+(aq)]initial
0.100 mol dm -3  (100  10 -3 ) dm3
=
(100  100)  10 - 3 dm3
= 0.05 mol dm-3
[Fe3+(aq)]eqm = [Fe2+(aq)]initial – [Fe2+(aq)]eqm
= (0.05 – 0.012 2) mol dm-3
= 0.0378 mol dm-3
292
16.5 Determination of Equilibrium Constants (SB p.98)
(b) Calculate the equilibrium constant (Kc).
(b)
Kc 
[Fe3  (aq)]eqm
[Fe2  (aq)]eqm [ Ag (aq)]eqm
0.0378 mol dm 3
=
0.0122 mol dm 3  0.0122 mol dm-3
= 253.96 mol-1 dm3
293
Answer
16.5 Determination of Equilibrium Constants (SB p.98)
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.
294
16.5 Determination of Equilibrium Constants (SB p.98)
For the reversible reaction of hydrogen and iodine at
equilibrium:
H2(g) + I2(g)
2HI(g)
If the initial amount of H2(g) is a mol, I2(g) is b mol and
the amount of H2(g) or I2(g) reacted is x mol, express the
equilibrium constant (Kc) in terms of a, b and x.
Answer
295
16.5 Determination of Equilibrium Constants (SB p.98)
Back
Let the volume of the reaction mixture be V dm3.
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
Kc 
296
[HI(g)]
2
eqm
[H2 (g)]eqm [I2 (g)]eqm
2x 2
)
4x 2
V


ax bx
(a  x )(b  x )
(
)(
)
V
V
(
16.5 Determination of Equilibrium Constants (SB p.99)
For the Haber process,
N2(g) + 3H2(g)
2NH3(g)
If the initial amount of N2(g) is a mol, H2(g) is b mol
and the amount of N2(g) reacted is x mol, express the
equilibrium constant (Kc) in terms of a, b and x.
Answer
297
16.5 Determination of Equilibrium Constants (SB p.99)
Back
Let the volume of the reaction mixture be V dm3.
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
Kc 
298
[NH3 (g)]
2
eqm
[N2 (g)]eqm [H2 (g)]3eqm
2x 2
)
V

a  x b  3x 3
(
)(
)
V
V
(
4x 2
4x 2
V
V3
2


V
 2 

3
(a  x )(b  3 x )
V
a  x (b  3 x )3
16.5 Determination of Equilibrium Constants (SB p.99)
A student mixed 10 cm3 of 2.0 × 10–3 M Fe(NO3)3(aq) with
10 cm3 of 2.0 × 10–3 M KNCS(aq).
Fe3+(aq) + NCS–(aq)
[Fe(NCS)]2+(aq)
When the system reaches the equilibrium, the
concentration of [Fe(NCS)]2+(aq) is 1.4 × 10–4 M.
Determine the equilibrium constant (Kc) of the reaction.
Answer
299
16.5 Determination of Equilibrium Constants (SB p.100)
3
3
-3
3
Initial concentration of Fe3+(aq) = 2.0  10 mol dm  10  10 dm
(10  10)  10 - 3 dm3
= 1.0  10-3 mol dm-3
2.0  10 3 mol dm 3  10  10 -3 dm3
Initial concentration of NCS-(aq) =
(10  10)  10 - 3 dm3
= 1.0  10-3 mol dm-3
Fe3+(aq) +
NCS-(aq)
[Fe(NCS)]2+(aq)
At start:
1.0  10-3 M
1.0  10-3 M
0M
Amount changed:
-x M
–x M
xM
At equilibrium: (1.0  10-3 – x) M (1.0  10-3 – x) M
xM
300
16.5 Determination of Equilibrium Constants (SB p.100)
Back
From the given data, the equilibrium concentration of [Fe(NCS)]2+(aq)
is 1.4 × 10–4 M, thus x = 1.4 × 10–4 M.
∴ [Fe3+(aq)]eqm = (1.0 × 10–3 – 1.4 × 10–4) mol dm–3
= 0.86 × 10–3 mol dm–3
[NCS-(aq)]eqm = (1.0 × 10–3 – 1.4 × 10–4) mol dm–3
= 0.86 × 10–3 mol dm–3
1.4  10 4
Kc =
(0.86  10  3 )(0.86  10  3 )
= 189.3 dm3 mol-1
301
16.5 Determination of Equilibrium Constants (SB p.100)
Back
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.
302
16.5 Determination of Equilibrium Constants (SB p.100)
At 400 K, 0.250 mole of PCl3(g) and 0.009 mole of PCl5(g)
were mixed in a 1 dm3 flask. After the system was left
overnight, an equilibrium was established and 0.002
mole of chlorine gas was found in the flask. Determine
the equilibrium constant (Kc) of the reaction:
PCl5(g)
303
PCl3(g) + Cl2(g)
Answer
16.5 Determination of Equilibrium Constants (SB p.100)
Back
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
 At equilibrium, 0.002 mole of Cl2(g) was found in the flask.
 X = 0.002 mol
304
[PCl3 (g)][Cl2 (g)]
Kc =
[PCl5 (g)]
(0.250  0.002) mol dm 3  0.002 mol dm -3
=
= 0.072 mol dm-3
-3
(0.009 - 0.002) mol dm
16.6 Equilibrium Constant in Terms of Partial Pressures (SB p.101)
The following equilibrium reaction
2NOBr(g)
2NO(g) + Br2(g)
is studied at 298 K. The partial pressures of NOBr(g),
NO(g) and Br2(g) at equilibrium were found to be:
PNOBr = 246 Nm–2
PNO = 450 Nm–2
PBr2 = 300 Nm–2
Calculate the value of Kp for the reaction at 298 K.
Answer
305
16.6 Equilibrium Constant in Terms of Partial Pressures (SB p.101)
2NOBr(g)
2NO(g) + Br2(g)
The expression of Kp is:
(PNO )2 (PBr )
Kp 
(PNOBr )2
Substituting the partial pressures into the expression, we have:
2
2
( 450)2 (300)
Kp 
(246)2
= 1 003.9 Nm–2
306
Back
16.6 Equilibrium Constant in Terms of Partial Pressures (SB p.102)
The decomposition of dinitrogen tetroxide to form
nitrogen dioxide is a reversible reaction.
N2O4(g)
2NO2(g)
When the reaction reaches an equilibrium state, the
partial pressure of N2O4(g) was found to be 2.71 atm.
Calculate the partial pressure of NO2(g) at equilibrium
given that the value of Kp is 0.133 atm.
Answer
307
16.6 Equilibrium Constant in Terms of Partial Pressures (SB p.102)
N2O4(g)
2NO2(g)
The expression of Kp is:
Kp 
PNO
2
 0.133
2
PN O
Substituting the value of Kp and the partial pressure of N2O4(g) into
the expression,
2
PNO = Kp × PN O
2
2
2
4
= 0.133 × 2.71
= 0.360 atm2
∴ PNO = 0.600 atm
2
308
4
Back
16.6 Equilibrium Constant in Terms of Partial Pressures (SB p.102)
Equal amounts of hydrogen and iodine are allowed to
reach an equilibrium at 298 K:
H2(g) + I2(g)
2HI(g)
If 80% of the hydrogen is converted to hydrogen iodide
at the equilibrium, what is the value of Kp at this
temperature?
Answer
309
16.6 Equilibrium Constant in Terms of Partial Pressures (SB p.102)
Assume that the initial number of moles of H2(g) is 1 mol.
H2(g)
+
I2(g)
2HI(g)
At start:
1 mol
1 mol
0 mol
At equilibrium: (1 – 0.8) mol (1 – 0.8) mol
(0.8  2) mol
= 0.2 mol
= 0.2 mol
= 1.6 mol
0.2 mol
Mole fraction of H2(g) =
(0.2  0.2  1.6) mol
= 0.1
Mole fraction of I2(g) =
1.6 mol
(0.2  0.2  1.6) mol
= 0.1
310
16.6 Equilibrium Constant in Terms of Partial Pressures (SB p.102)
Mole fraction of HI(g) =
1.6 mol
(0.2  0.2  1.6) mol
= 0.8
Let P be the total pressure of the system.
PHI2
Kp 
PH PI
(0.8  P) 2

(0.1 P)(0.1 P)
= 64
2
311
2
Back
16.7 Equilibrium Position (SB p.103)
Determine the equilibrium constant (Kc) from the
following data on the equilibrium system
2SO2(g) + O2(g)
2SO3(g) at 873 K.
Experiment
Equilibrium concentration (mol dm-3)
[SO2(g)]
[O2(g)]
[SO3(g)]
1
1.60
1.30
3.62
2
0.71
0.50
1.00
Answer
312
16.7 Equilibrium Position (SB p.103)
Back
The expression of Kc is:
[SO3 (g)]2eqm
Kc 
[SO 2 (g)]2eqm [O 2 (g)]eqm
From experiment 1:
3.622
Kc 
1.60 2  1.30
= 3.94 dm3 mol-1
From experiment 2:
1.00 2
Kc 
0.72 2  0.50
313
= 3.97 dm3 mol-1
Since Kc is a constant at a specific temperature, the values of Kc
from experiments 1 and 2 are very close, and the average value of
Kc at 873 K is 3.955 dm3 mol–1.
16.7 Equilibrium Position (SB p.104)
For the following reversible reaction:
Answer
CH3COOH(l) + CH3CH2OH(l)
CH3COOCH2CH3(l) + H2O(l)
Calculate the equilibrium constant (Kc) using the
following data.
Expt
Initial no. of moles (mol)
No. of moles at
eqm (mol)
CH3CHOOH(l)
CH3CH2OH(l)
CH3COOH(l)
1
1.00
1.00
0.33
2
1.00
4.00
0.07
(Assume that the equilibrium is established in a
314 container of 1 dm3.)
16.7 Equilibrium Position (SB p.104)
The equilibrium constant for the equilibrium is expressed as:
Kc 
[SO3 (g)]2eqm
[SO 2 (g)]2eqm [O 2 (g)]eqm
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
315
0.67 mol dm 3  0.67 mol dm 3
Kc 
 4.12
3
3
0.33 mol dm  0.33 mol dm
16.7 Equilibrium Position (SB p.104)
Back
For experiment 1:
CH3COOH(l) + CH3CH2OH(l)
At start:
1.00 mol
4.00 mol
At eqm:
0.07 mol
(4.00 – 0.93) mol
= 3.07 mol
CH3COOCH2CH3(l) + H2O(l)
At start:
0 mol
0 mol
At eqm:
(1.00 – 0.07) mol (1.00 – 0.07) mol
= 0.93 mol
= 0.93 mol
316
0.93 mol dm 3  0.93 mol dm 3
Kc 
 4.02
3
3
0.07 mol dm  3.07 mol dm
Since Kc is a constant at a specific temperature, the average value of
Kc from experiments 1 and 2 is 4.07.
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.104)
An organic compound X has a partition coefficient of
30 in ethoxyethane and water.
[X]
ethoxyethane
K 
 30
D
[X]
water
There is 3.1 g of X in 50 cm3 of water. 50 cm3 of
ethoxyethane is then added to extract X from water.
How much X is extracted using ethoxyethane?
Answer
317
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.106)
Let a g be the mass of X extracted using 50 cm3 of ethoxyethane,
then the mass of X left in water is (3.1 – a) g.
a
[X]ethoxyetha ne 
g cm 3
50
3.1- a
[X] water 
g cm 3
50
a
 K D  50
3.1 - a
50
a
Back
 30  50
3.1 - a
50
318
a = 3.0
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.106)
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.
319
Answer
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.106)
[ Y] ether 
[ Y ] water
108 g
100 cm3
= 1.08 g cm-3
6g

50 cm3
= 0.12 g cm-3
[Y]ether
KD 
[Y]water
1.08 g cm-3

0.12 g cm- 3
=9
320
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.106)
(a) Let m g be the mass of Y extracted using 100 cm3 of ether, then
the mass of Y left in the aqueous layer is (10 – m) g.
m
 K D  100
10 - m
100
m
9  100
10 - m
100
m=9
 9 g of Y can be extracted using 100 cm3 of fresh ether.
321
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.106)
(b) Let m1 g be the mass of Y extracted using the first 50 cm3 of
ether, then the mass of Y left in the aqueous layer is (10 – m1) g.
m1
 K D  50
10 - m1
100
m1
9  50
10 - m1
100
m1 = 8.182
 Mass of Y extracted using the first 50 cm3 of ether = 8.182 g
Mass of Y left in the aqueous layer = (10 – 8.182) g = 1.818 g
322
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.106)
Let m2 g be the mass of Y extracted using the second 50 cm3 of ether,
then the mass of Y left in the aqueous layer is (1.818 – m2) g.
m2
50
KD 
1.818 - m 2
100
m2
50
9
1.818 - m 2
100
323
m2 = 1.487
 Mass of Y extracted using the second 50 cm3 of ether = 1.487 g
Mass of Y left in the aqueous layer = (1.818 – 1.487) g = 0.331 g
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.106)
∴ Total mass of Y extracted = m1 + m2
= (8.182 + 1.487) g
= 9.669 g
Back
324
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.108)
The partition coefficient (KD) of an unknown organic
compound A between 1,1,1-trichloroethane and water is
expressed as:
Concentration of A in 1,1,1 - trichloroethane (g cm -3 )
KD 
 15
-3
Concentration of A in water (g cm )
Calculate the mass of A that can be extracted from 60 cm3
of an aqueous solution initially containing 6 g of A using
100 cm3 of fresh 1,1,1-trichloroethane.
Answer
325
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.108)
Let m be the mass of A extracted using 100 cm3 of 1,1,1trichloroethane, then the mass of A left in 60 cm3 of aqueous
layer is (6 – m).
m
K D  100
6m
60
Back
m
15  100
6m
60

326
m = 5.77 g
5.77 g of A is extracted using 100 cm3 of 1,1,1trichloroethane.
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.110)
(a) 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
327
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.110)
(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.
328
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.110)
(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
329
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.110)
Back
(b)
330
16.9 Significances of Equilibrium Constants (SB p.111)
The reaction N2(g) + 3H2(g)
2NH3(g) has an
equilibrium constant of 0.062 dm6 mol–2 at 500 oC. Predict
the net chemical change, if there is any, for the following
concentrations of reactant(s) and product(s).
(a) [NH3(g)] = 0.001 mol dm–3, [N2(g)] = 0.001 mol dm–3
and [H2(g)] = 0.002 mol dm–3
Answer
331
16.9 Significances of Equilibrium Constants (SB p.111)
(a)
[NH3 (g)]2
Qc 
[N2 (g)][H2 (g)]3
0.0012

0.001 0.0023
= 125 000 mol-2 dm6
Since Qc > Kc, the reaction proceeds from the right (product
side) to the left (reactant side) until the equilibrium is
reached.
332
16.9 Significances of Equilibrium Constants (SB p.111)
The reaction N2(g) + 3H2(g)
2NH3(g) has an
equilibrium constant of 0.062 dm6 mol–2 at 500 oC. Predict
the net chemical change, if there is any, for the following
concentrations of reactant(s) and product(s).
(b) [NH3(g)] = 0.001 mol dm–3, [N2(g)] = 1 mol dm–3 and
[H2(g)] = 0.08 mol dm–3
Answer
333
16.9 Significances of Equilibrium Constants (SB p.111)
Back
(b)
[NH3 (g)]2
Qc 
[N2 (g)][H2 (g)]3
0.0012

1 0.08 3
= 0.001 95 mol-2 dm6
Since Qc < Kc, the reaction proceeds from the left (reactant
side) to the right (product side) until the equilibrium is
reached.
334
16.10 Factors Affecting Equilibrium (SB p.113)
Back
For the equilibrium system:
As4O6(s) + 6C(s)
As4(g) + 6CO(g)
predict how the equilibrium position will shift
in response to the following changes:
(a) removing CO(g)
(b) adding more As4(g)
Answer
(a) According to Le Chatelier’s principle, the equilibrium
position will shift to the right.
(b) According to Le Chatelier’s principle, the equilibrium
position will shift to the left.
335
16.10 Factors Affecting Equilibrium (SB p.114)
(a) For the reaction H2(g) + I2(g)
2HI(g), the
following data are determined at 490 oC,
[H2(g)] = 0.22 mol dm–3; [I2(g)] = 0.22 mol dm–3 and
[HI(g)] = 1.56 mol dm–3
Calculate the equilibrium constant (Kc) at 490 oC.
(a) H2(g) + I2(g)
2HI(g)
[HI(g)]2
(1.56)2
Kc 

 50.3
[H2 (g)][I2 (g)] (0.22)(0.22)
336
Answer
16.10 Factors Affecting Equilibrium (SB p.114)
(b) If an additional 0.200 mol dm–3 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
337
16.10 Factors Affecting Equilibrium (SB p.114)
(b)
H2(g)
+
At eqm: 0.22 mol dm-3
Now: (0.22 + 0.2) mol dm-3
I2(g)
0.22 mol dm-3
0.22 mol dm-3
2HI(g)
1.56 mol dm-3
1.56 mol dm-3
[HI(g)]2
(1.56)2
Qc 

 26.34
[H2 (g)][I2 (g)] (0.42)(0.22)
Since the value of the reaction quotient (Qc) is less than that of the
equilibrium constant (Kc), the system is not at equilibrium. In order to
re-establish the equilibrium, the value of the reaction quotient should
be increased until it equals Kc. It can be predicted that more H2(g)
and I2(g) will react to form more HI(g).
338
16.10 Factors Affecting Equilibrium (SB p.114)
(b)
339
Back
H2(g)
+
I2(g)
Now: (0.22 + 0.2) mol dm-3
0.22 mol dm-3
At eqm: (0.42 - x) mol dm-3 (0.22 – x) mol dm-3
2HI(g)
1.56 mol dm-3
(1.56 + 2x) mol dm-3
[HI(g)]2
(1.56  2x )2
Kc 

[H2 (g)][I2 (g)] (0.42  x )(0.22  x )
Since Kc remains constant, we obtain:
(1.56  2x )2
50.2 
(0.42  x )(0.22  x )
By solving the quadratic equation, x = 0.06 or 0.77.
If x equals 0.77, the concentration of H2(g) and I2(g) at equilibrium will
be negative. Therefore, the correct answer of x is 0.06.
∴ [H2(g)]eqm = 0.42 mol dm–3 - 0.06 mol dm–3 = 0.36 mol dm–3
[I2(g)]eqm = 0.22 mol dm–3 - 0.06 mol dm–3 = 0.16 mol dm–3
[HI(g)]eqm = 1.56 + 2  0.06 mol dm–3 = 1.68 mol dm–3
16.10 Factors Affecting Equilibrium (SB p.115)
Consider the following reaction at equilibrium:
2CrO42-(aq) + 2H+(aq)
Cr2O72-(aq) + H2O(l)
Explain the changes of the graph at time t0, t1, t2 and t3
respectively.
Answer
340
16.10 Factors Affecting Equilibrium (SB p.115)
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
341
16.10 Factors Affecting Equilibrium (SB p.117)
The diagram on the right
shows the effect of increasing
pressure on the equilibrium
2NO2(g)
N2O4(g).
The equilibrium constant Kp
for the reaction is 0.92 atm–1 at
a given temperature.
342
16.10 Factors Affecting Equilibrium (SB p.117)
(a) Calculate the partial pressures of NO2(g) and N2O4(g)
at equilibrium if the total pressure is 1 atm.
(a)
Kp 
(PN
2
O4
)eqm
Answer
(PNO )2 eqm
Let the partial pressure of NO2(g) at equilibrium be p atm, then
the partial pressure of N2O4(g) at equilibrium is (1 – p) atm.
1 p
0.92  2
p
p = 0.632 or –1.719 (rejected)
 PNO2 = 0.632 atm
PN2O4 = (1 – 0.632) atm = 0.368 atm
2
343
16.10 Factors Affecting Equilibrium (SB p.117)
(b) Calculate the partial pressures of NO2(g) and
N2O4(g) if the total pressure at equilibrium is 2 atm.
(b) Using the same method as in (a),
2p
p2
p = 1.028 or –2.115 (rejected)
 PNO2 = 1.028 atm
PN2O4 = (2 – 1.028) atm = 0.972 atm
0.92 
344
Answer
16.10 Factors Affecting Equilibrium (SB p.117)
(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.
345
16.10 Factors Affecting Equilibrium (SB p.117)
(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
346
(d) The impact of the increased pressure is reduced by
shifting the equilibrium position to the right-hand side of
2NO2(g)
N2O4(g). More NO2(g), which is brown in
colour, is used up. More N2O4(g), which is colourless, is
formed. A colour change from brown to pale brown (or
colourless) can be observed.
16.10 Factors Affecting Equilibrium (SB p.117)
Back
(e) Given that the enthalpy change for the reaction
2NO2(g)
N2O4(g) is –58 kJ, predict the colour
change when a glass syringe containing the
equilibrium mixture is put into a beaker of hot water
for about 30 seconds, and then a beaker of water with
a large amount of crushed ice for another 30 seconds.
347
(e) When the equilibrium mixture is put into hot water (the Answer
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).
16.10 Factors Affecting Equilibrium (SB p.118)
1. Consider the following reaction at equilibrium:
3NO2(g) + H2O(g)
2HNO3(g) + NO(g)
(a) Referring to the chemical equation above, write a
mathematical expression for the equilibrium
constant, Kp.
(a)
348
Kp 
2
HNO 3 ( g )
P
3
PNO
2
( g)
 PNO ( g )
 PH O( g )
2
Answer
16.10 Factors Affecting Equilibrium (SB p.118)
1. Consider the following reaction at equilibrium:
3NO2(g) + H2O(g)
2HNO3(g) + NO(g)
(b) If the partial pressure of H2O(g) is increased at
constant temperature, what changes, if any, occur in
the partial pressures of
(i) NO2(g)?
(ii) HNO3(g)?
(iii) NO(g)?
349
(b) (i) Decrease
(ii) Increase
(iii) Increase
Answer
16.10 Factors Affecting Equilibrium (SB p.118)
1. Consider the following reaction at equilibrium:
3NO2(g) + H2O(g)
2HNO3(g) + NO(g)
(c) If the partial pressure of H2O(g) is increased at
constant temperature, will the value of Kp increase,
decrease or remain the same?
(c) The value of Kp will remain the same.
350
Answer
16.10 Factors Affecting Equilibrium (SB p.118)
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)
(a) Calculate the equilibrium constant, Kp, for the
reaction.
(a)
351
Kp 
PN O
2
Answer
4
2
NO 2
P
 0.900 atm
16.10 Factors Affecting Equilibrium (SB p.118)
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
352
16.10 Factors Affecting Equilibrium (SB p.118)
(b) After compression to one-half the original volume, all
the gas pressures will be doubled. Therefore, the
partial pressures of N2O4(g) and NO2(g) will be 0.728
atm and 1.272 atm respectively.
2NO2(g)
N2O4(g)
Q
353
PN O
2
4
2
NO 2
P
 0.450 atm
16.10 Factors Affecting Equilibrium (SB p.118)
(b) Since the value of the reaction quotient is less than
that of the equilibrium constant, the system is not at
equilibrium. The reaction proceeds from the left to the
right until the equilibrium is reached. As a result, more
N2O4(g) will be formed.
2NO2(g)
N2O4(g)
At start:
1.272
0.728
At eqm:
1.272 – 2x
0.728 + x
K p  0.90 
354
(0.728  x )
(1.272  2x )2
16.10 Factors Affecting Equilibrium (SB p.118)
(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
355
16.10 Factors Affecting Equilibrium (SB p.120)
Predict how the equilibrium position is affected when the
equilibrium system
N2O4(g)
2NO2(g)
ΔH = +58 kJ
is subjected to the following changes:
(a) addition of NO2(g)
(b) removal of N2O4(g)
(c) addition of He( g)
(d) increase in volume of the container
(e) decrease in temperature
356
Answer
16.10 Factors Affecting Equilibrium (SB p.120)
(a)
(b)
(c)
(d)
(e)
The equilibrium position shifts to the left.
The equilibrium position shifts to the left.
The equilibrium position remains unchanged.
The equilibrium position shifts to the right.
The equilibrium poistion shifts to the left.
Back
357
16.10 Factors Affecting Equilibrium (SB p.121)
The equilibrium constant (Kp) of the following
reaction is 1.6 × 10–4 atm–2 at 673 K and 1.4 × 10–5 atm–2
at 773 K.
N2(g) + 3H2(g)
2NH3(g)
Determine the mean enthalpy change of formation of
1 mole of ammonia from its elements in the
temperature ranges from 673 K to 773 K.
(Given: R = 8.31 J K–1 mol–1)
358
Answer
16.10 Factors Affecting Equilibrium (SB p.121)
Back
At 673 K,
ln(1.6  10  4 )  cons tan t 
H
.........(1)
8.31 673
At 773 K,
ln(1.4  10  4 )  cons tan t 
H
.........(2)
8.31 773
Combining (1) and (2),
ln(1.6  10  4 ) 
359
H
H
 ln(1.4  10  5 ) 
5593
6424
H = 105 329 J mol-1
= -105.3 kJ mol-1
16.10 Factors Affecting Equilibrium (SB p.121)
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
(Given: R = 8.314 J K–1 mol–1)
360
Answer
16.10 Factors Affecting Equilibrium (SB p.122)
361
1/T (K-1)
ln Kp
3.36  10-3
-2.16
2.86  10-3
+1.36
2.50  10-3
+3.87
2.00  10-3
+7.44
1.67  10-3
+9.79
16.10 Factors Affecting Equilibrium (SB p.122)
A graph of ln Kp against
slope  H .
R
362
1
T
produces a straight line with
16.10 Factors Affecting Equilibrium (SB p.122)
7.44  9.79
Slope =
(2.00  1.67)  10  3
= -7121.2

H

R = -7121.2
H = 7121.2  8.314
= 59 206 J mol-1
= 59.2 kJ mol-1
Back
363
16.10 Factors Affecting Equilibrium (SB p.122)
Haber process is an important industrial process to
manufacture ammonia with the use of nitrogen and
hydrogen. Ammonia has numerous uses like making
fertilizers and explosives. The reaction between
nitrogen and hydrogen is a reversible reaction. It takes
place with release of thermal energy.
N2(g) + 3H2(g)
2NH3(g)
H = –92.6 kJ
(a) Based on your knowledge about “chemical
equilibrium”, predict the necessary conditions to
increase the yield of ammonia in the Haber process.
364
Answer
16.10 Factors Affecting Equilibrium (SB p.122)
(a) 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.
365
16.10 Factors Affecting Equilibrium (SB p.122)
(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
366
16.10 Factors Affecting Equilibrium (SB p.122)
Back
(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
367
ammonia is optimum.