1 Irreversible and Reversible Reactions. 2 Irreversible Reactions Chemical reactions that take place...
-
Upload
georgia-richard -
Category
Documents
-
view
230 -
download
0
Transcript of 1 Irreversible and Reversible Reactions. 2 Irreversible Reactions Chemical reactions that take place...
1
Irreversible Irreversible and and
Reversible Reversible ReactionsReactions
2
Irreversible ReactionsIrreversible 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 ReactionsIrreversible Reactions
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)
Examples : -
4
Q.1
2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)2Mg(s) + O2(g) 2MgO(s)
Conditions for reversible reactions : -• Closed reaction vessels to prevent escape of ga
ses• High temperature to favour the reversed proce
sses
5
Q.1
Conditions for reversible reactions : -• Concentration of HCl(aq) > 6 M
HCl(aq) + H2O(l) H3O+(aq) + Cl(aq)
6
Q.1
Conditions for reversible reactions : -
• Closed reaction vessels to prevent escape of Cl2
• Dilute OH(aq) at T 20C to prevent side reaction
3Cl2(g) + 6OH(aq) ClO3(aq) + 5Cl(aq) +
3H2O(l)Hot and concentra
ted
Cl2(g) + 2OH(aq) ClO(aq) + Cl(aq) + H2O(l)
7
Solid
Liquid
Vapour
Changes of physical phases are reversible
Reversible ProcessesReversible Processes
8
• Chemical reactions that can go in two opposite directions
• Incomplete reactions
Reversible ReactionsReversible Reactions
9
Q.2
CH3COOH(aq) + H2O(l) CH3COO(aq) + H3O+
(aq)NH3(aq) + H2O(l) NH4
+(aq) + OH(aq)
3H2(g) + N2(g) 2NH3(g)
CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l)
Cl2(aq) + H2O(l) HCl(aq) + HOCl(aq)
10
1. When HCl(aq) is added to CrO42(aq)
Observation : -
The yellow solution turns orange.
Examples of reversible reactionsExamples of reversible reactions
CrO42-(aq) + 2H+(aq) Cr2O7
2-(aq) + H2O(l)yellow orange
11
Examples of reversible reactionsExamples of reversible reactions
Interpretation : -
CrO42(aq) reacts with H+ to give Cr2O7
2(aq)
There is no further colour change when
rate of forward rx = rate of backward rx
CrO42-(aq) + 2H+(aq) Cr2O7
2-(aq) + H2O(l)yellow orange
1. When HCl(aq) is added to CrO42(aq)
12
2. When NaOH(aq) is added to Cr2O72(aq)
Observation : -
The orange solution turns yellow.
Examples of reversible reactionsExamples of reversible reactions
CrO42-(aq) + 2H+(aq) Cr2O7
2-(aq) + H2O(l)yellow orange
13
Examples of reversible reactionsExamples of reversible reactions
Interpretation : -
H+ ions are being removed by NaOH rate of forward rx rate of backward rx > rate of forward rx
a net change of Cr2O72(aq) to CrO4
2(aq)
CrO42-(aq) + 2H+(aq) Cr2O7
2-(aq) + H2O(l)yellow orange
2. When NaOH(aq) is added to Cr2O72(aq)
14
Examples of reversible reactionsExamples of reversible reactions
Interpretation : -
There is no further colour change when
rate of backward rx = rate of forward rx
CrO42-(aq) + 2H+(aq) Cr2O7
2-(aq) + H2O(l)yellow orange
2. When NaOH(aq) is added to Cr2O72(aq)
15
1.When H2O(l) is added to BiCl3(aq)
Observation : -
The colourless solution turns milky.
Examples of reversible reactionsExamples of reversible reactions
BiCl3(aq) + H2O(l) BiOCl(s) + 2HCl(aq)colourless White ppt
16
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
Examples of reversible reactionsExamples of reversible reactions
BiCl3(aq) + H2O(l) BiOCl(s) + 2HCl(aq)colourless White ppt
17
2. When HCl(aq) is added to BiOCl(s)
Observation : -
The milky solution becomes clear.
Examples of reversible reactionsExamples of reversible reactions
BiCl3(aq) + H2O(l) BiOCl(s) + 2HCl(aq)colourless White ppt
18
Interpretation : -
Examples of reversible reactionsExamples of reversible reactions
2. When HCl(aq) is added to BiOCl(s)
[HCl(aq)]
rate of backward rx > rate of forward rx
a net consumption of BiOCl(s)
BiCl3(aq) + H2O(l) BiOCl(s) + 2HCl(aq)colourless White ppt
19
Interpretation : -
Examples of reversible reactionsExamples of reversible reactions
2. When HCl(aq) is added to BiOCl(s)
There is no further change when
rate of forward rx = rate of backward rx
BiCl3(aq) + H2O(l) BiOCl(s) + 2HCl(aq)colourless White ppt
20
1. When NaOH(aq) is added to Br2(aq)
Prediction : -
The red-orange solution turns colourless.
Q.3Q.3
Red-orange
colourlessBr2(aq) + H2O(l) H+(aq) + Br-(aq) + HOBr(aq)
21
1. When NaOH(aq) is added to Br2(aq)
Interpretation : -
Before the addition,
rate of forward rx = rate of backward rx
Q.3Q.3
Red-orange
colourlessBr2(aq) + H2O(l) H+(aq) + Br-(aq) + HOBr(aq)
22
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)
Q.3Q.3
Red-orange
colourlessBr2(aq) + H2O(l) H+(aq) + Br-(aq) + HOBr(aq)
23
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
Q.3Q.3
Red-orange
colourlessBr2(aq) + H2O(l) H+(aq) + Br-(aq) + HOBr(aq)
24
2. When HCl(aq) is added
Prediction : -
The colourless solution turns red-orange.
Q.3Q.3
Red-orange
colourlessBr2(aq) + H2O(l) H+(aq) + Br-(aq) + HOBr(aq)
25
Interpretation : -
Addition of HCl increases [H+(aq)]
rate of backward rx > rate of forward rx
a net production of Br2(aq)
Q.3Q.3
Red-orange
colourlessBr2(aq) + H2O(l) H+(aq) + Br-(aq) + HOBr(aq)
2. When HCl(aq) is added
26
Interpretation : -
There is no further change of colour when
rate of forward rx = rate of backward rx
Q.3Q.3
Red-orange
colourlessBr2(aq) + H2O(l) H+(aq) + Br-(aq) + HOBr(aq)
2. When HCl(aq) is added
27
3.When AgNO3(aq) is added
Prediction : -
A pale yellow ppt is formed.
The red-orange solution turns colourless.
Q.3Q.3
Red-orange
colourlessBr2(aq) + H2O(l) H+(aq) + Br-(aq) + HOBr(aq)
28
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)
Q.3Q.3
Red-orange
colourlessBr2(aq) + H2O(l) H+(aq) + Br-(aq) + HOBr(aq)
3.When AgNO3(aq) is added
29
Interpretation : -
There is no further change of colour when
rate of forward rx = rate of backward rx
Q.3Q.3
Red-orange
colourlessBr2(aq) + H2O(l) H+(aq) + Br-(aq) + HOBr(aq)
3.When AgNO3(aq) is added
30
Phenolphthlein is a weak acid that ionizes slightly in water to give H3O+(aq)
Colourless Red
+ H3O+
(aq)
31
Colourless Red
+ H3O+
(aq)
What is the colour of phenolphthalein when pH < 8.3 ?
32
Colourless Red
+ H3O+
(aq)
When pH < 8.3 (e.g. deionized water),
The colourless form predominates
33
Colourless Red
+ H3O+
(aq)
When NaOH(aq) is added,
[H3O+(aq)]
rate of forward rx > rate of backward rx
a net production of the red form
34
Colourless Red
+ H3O+
(aq)
There is no further colour change when
rate of forward rx = rate of backward rx
35
Colourless Red
+ H3O+
(aq)
When pH > 10,
The red form predominates
36
Colourless Red
+ H3O+
(aq)
When 8.3 < pH < 10,
Both forms have similar concentrations
pink
37
For a reversible reaction,
Reversible reactions and dynamic Reversible reactions and dynamic equilibriumequilibrium
a state of dynamic equilibrium is said to be established when
rate of forward rx = rate of backward rx
reactants products
Apparently, there is no change in the concentrations of reactants and products.
Reactions continues at molecular level.
38
Dynamic Dynamic EquilibriumEquilibrium
An example of dynamic equilibrium
No change in the position of the girl
39
ALL chemical reactions are considered as reversible processes with different extents of completion.
Reversible reactions and chemical Reversible reactions and chemical equilibriumequilibrium
40
H < 0
Reversible reactions and chemical Reversible reactions and chemical equilibriumequilibrium
Ea Ea’
Forward rx is more complete than backward rx
At equil., k[reactant]eq = k’[product]eq
k >> k’
[reactant]eq << [product]eq Ea’ > Ea
41
H > 0
Ea > Ea’
Reversible reactions and chemical Reversible reactions and chemical equilibriumequilibrium
Ea
Ea’
Forward rx is less complete than backward rx
42
Chemical equilibrium is about how far a reaction can proceed.
Chemical kinetics is about how fast a reaction can proceed.
Chemical Equilibrium vs Chemical KineticsChemical Equilibrium vs Chemical Kinetics
43
The rate of rx depends on Ea or Ea’
The extent of completion of rx depends on H
Chemical Equilibrium vs Chemical KineticsChemical Equilibrium vs Chemical Kinetics
Ea
Ea’
44
Evidence for Dynamic Equilibrium
NaNO3(s) NaNO3(aq)
saturated
At fixed T, [NaNO3(aq)] is a constantAddition of 24NaNO3(s)
Detection of radioactivity in sat’d solution
Interchange of NaNO3 between the sat’d 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
H2(g) + I2(g) 2HI(g)
Q.4 (i)Q.4 (i)
Time taken to reach the equilibrium state
47
Q.4 (ii)Q.4 (ii)
H2(g) + I2(g) 2HI(g)
Time taken to reach the equilibrium state
The equilibrium concentrations need not be equal
48
Q.5Q.5Constant flame colour and temperature
49
Air
Fuel
CO2 and H2O
Steady state
Q.5Q.5
Open system
Not at equilibrium state
50
3. Equilibria can only be achieved in closed systems with no exchange of matter with their surroundings.
51
Q.6 A
Br2(g)
Br2(l)
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
Fe3O4(s) + 4H2(g)
3Fe(s) + 4H2O(g)
500C
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
KCl(s) KCl(aq)
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
CH3COOH(l) + C2H5OH(l)
CH3COOC2H5(l) + H2O(l)
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
H2(g) + I2(g) 2HI(g)
0.5 0.5 0
ninitial
nequil
n’initial
nequil
Q.7
0.78
0.5-0.78/2
= 0.11
0.5-0.78/2
= 0.11
nH = nI = 1.0 nHI = 1.0
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
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.
58
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.
59
Equilibrium Equilibrium LawLaw
60
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.
Equilibrium Equilibrium LawLaw
61
For the chemical system in For the chemical system in equilibrium,equilibrium,
Kc depends on temperature and the nature of reaction
Equilibrium constant expressed in concentration
aA + bB cC + dD
62
Equilibrium constant and reaction Equilibrium constant and reaction quotientquotient
aA + bB cC + dD
b[B]a[A]
d[D]c[C] Q
cReaction
quotient
Equilibrium constant
63
aA + bB cC + dD
b[B]a[A]
d[D]c[C] Q
cReaction
quotient
Equilibrium constant
Qc = Kc
the system is at equilibrium
64
aA + bB cC + dD
b[B]a[A]
d[D]c[C] Q
cReaction
quotient
Equilibrium constant
Qc > Kc
the system is NOT at equilibrium
The reaction proceeds from right to left until Qc = Kc.
65
aA + bB cC + dD
b[B]a[A]
d[D]c[C] Q
cReaction
quotient
Equilibrium constant
Qc < Kc
the system is NOT at equilibrium
The reaction proceeds from left to right until Qc = Kc.
66
aA + bB cC + dD
b[B]a[A]
d[D]c[C] Q
cReaction
quotient
Equilibrium constant
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
aA + bB cC + dD
b[B]a[A]
d[D]c[C] Q
cReaction
quotient
Equilibrium constant
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
A + B C [A][B]
[C]K
1C
units
mol1 dm3
C A + B [C]
[A][B]K
1-C mol dm3
1cK1
70
Relationship of Kc to the Stoichiometry of Equations
A + B C [A][B]
[C]K
1C
units
mol1 dm3
molx dm3x
xA + xB xC
xx
x
C [B][A][C]
K2 x
c )(K1
71
Relationship of Kc to the Stoichiometry of Equations
A + B C [A][B]
[C]K
1C
units
mol1 dm3
A + B C y
1y1
y1
y1
y1
y1
c
[B][A]
[C]K
3 y1c1
K y3
y1
dm mol
72
Q.9
A B
B C
C D
[A][B]
K1
[B][C]
K2
[C][D]
K3
[A][C]
K4 [B][C]
[A][B]
= K1K2
(1)
(2)
(3)
A C (4)
(4) = (1) + (2)
K4 = K1 K2
73
Q.9
A B
B C
C D
[A][B]
K1
[B][C]
K2
[C][D]
K3
[A][D]
K5
(1)
(2)
(3)
A D (5)
(5) = (1) + (2) + (3)
K5 = K1 K2 K3
[C][D]
[B][C]
[A][B]
= K1K2K3
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 2
1
21
34 KK
KK
21
62533
18670
)dmmol 10)(6.010(2.5dmmol 103.9
= 5.11015 mol2 dm6
75
DeterminatioDetermination of n of
Equilibrium Equilibrium ConstantsConstants
76
Equilibrium System: (TAS Expt. 11)Equilibrium System: (TAS Expt. 11)
CHCH33COOH(l) + CHCOOH(l) + CH33CHCH22OH(l) OH(l) CHCH33COOCHCOOCH22CHCH33(l) + (l) + HH22O(l)O(l)
eqm23eqm3
eqm2eqm323
OH(l)]CH[CHCOOH(l)][CH
O(l)][H(l)]CHCOOCH[CHcK
77
Equilibrium System: Equilibrium System:
CHCH33COOH(l) + CHCOOH(l) + CH33CHCH22OH(l) OH(l) CHCH33COOCHCOOCH22CHCH33(l) + (l) + HH22O(l)O(l)
0.1670.000H2O(l)
0.250 – 0.083
= 0.1670.000CH3COOCH2
CH3(aq)
0.0200.020H2SO4(l)
0.0830.250CH3CH2OH(aq)
0.1980.000
0.1980.000
0.0200.020
0.0980.296
0.0980.2960.0830.250CH3COOH(aq)
Amount at equilibrium
(mol)
Amount at beginning
(mol)
Amount at equilibrium
(mol)
Amount at beginning
(mol)
Experiment 2Experiment 1Reactant/Product
78
Equilibrium System: Equilibrium System:
CHCH33COOH(l) + CHCOOH(l) + CH33CHCH22OH(l) OH(l) CHCH33COOCHCOOCH22CHCH33(l) + (l) + HH22O(l)O(l)For experiment 1:
05.4)
V0.083)(
V0.083(
)V
0.167)(V
0.167(cK
1
79
Equilibrium System: Equilibrium System:
CHCH33COOH(l) + CHCOOH(l) + CH33CHCH22OH(l) OH(l) CHCH33COOCHCOOCH22CHCH33(l) + (l) + HH22O(l)O(l)For experiment 2:
08.4)
V0.098)(
V0.098(
)V
0.198)(V
0.198(cK
1
80
Equilibrium System: Equilibrium System:
CHCH33COOH(l) + CHCOOH(l) + CH33CHCH22OH(l) OH(l) CHCH33COOCHCOOCH22CHCH33(l) + (l) + HH22O(l)O(l)
2
K K
cK of value Average 2c
1c
24.08 4.05
= 4.065 (no unit)
81
Equilibrium System: Equilibrium System:
CHCH33COOH(l) + CHCOOH(l) + CH33CHCH22OH(l) OH(l) CHCH33COOCHCOOCH22CHCH33(l) + (l) + HH22O(l)O(l)• 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
With catalyst
Without catalyst
Same extent of completion
Same equilibrium composition
83
Equilibrium Equilibrium Constant in Constant in
Terms of Partial Terms of Partial PressuresPressures
84
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)
bB
aA
dD
cC
pPP
PPK
85
Relationship between Kc and Kp
PV = nRT
RTVn
P
= [Gas]RT
At fixed T,P [Gas]
86
bB
aA
dD
cC
pPP
PPK
Consider the equilibrium system :
aA(g) + bB(g) cC(g) + dD(g)
bbaa
ddcc
(RT)[B](RT)[A](RT)[D](RT)[C]
b)(ad)(cc(RT)K
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.
CO2(g) + H2(g) CO(g) + H2O(g)
Kp = 0.10 at 690 K
R = 0.082 atm dm3 K1 mol1
Find partial pressures of all gaseous components
88
CO2(g) + H2(g) CO(g) + H2O(g)
Initial no. of molesNo. of moles at equil.
0.50 0.50 0 0
0.50 - x 0.50 - x x x
))((
))((K
5.0x0.50
5.0x0.50
5.0x
5.0x
c 0.10Kp
x = 0.12
89
CO2(g) + H2(g) CO(g) + H2O(g)
No. of moles at equil.
0.50 - x 0.50 - x x x
0.38 0.38 0.12 0.12
nT = 0.38 + 0.38 + 0.12 + 0.12 = 1.00
VRTn
P TT
3
-1-13
dm 5.0K) )(690mol K dm atm mol)(0.082 (1.00
= 11.316 atm
90
CO2(g) + H2(g) CO(g) + H2O(g)
No. of moles at equil.
0.50 - x 0.50 - x x x
0.38 0.38 0.12 0.12
nT = 0.38 + 0.38 + 0.12 + 0.12 = 1.00
TCOCO PXP22
atm 11.3161.000.38
= 4.30 atm
THH PXP22
atm 11.3161.000.38
= 4.30 atm
OHCO 2PP atm 11.316
1.000.12
= 1.36 atm
91
Q.11
2SO2(g) + O2(g) 2SO3(g)
At fixed V & T, P n
2
2
2
2
O
SO
O
SO
n
n
P
P
Initial partial pressure
Partial pressure at equilibrium
3x x 0
1.5x
x – 1.5x/2 1.5x=
0.25x
= 3
92
Q.112SO2(g) + O2(g) 2SO3(g)
Partial pressure at equilibrium
1.5x
1.5x
0.25x
PT = 373 kPa
)(P)(P
)(PK
22
3
O2
SO
2SO
p (0.25x)(1.5x)
(1.5x)2
2
= 0.035 kPa1
= 1.5x + 0.25x + 1.5xx = 115 kPa
93
Q.12
PCl5(g) PCl3(g) + Cl2(g)
At fixed V & T, P n
Initial partial pressure
Partial pressure at equilibrium
x 0 0
0.14x
0.86x0.86x
PT = 101 kPa = 0.14x + 0.86x + 0.86x
x = 54.3 kPa
94
Q.12PCl5(g) PCl3(g) + Cl2(g)
Partial pressure at equilibrium
0.14x
0.86x0.86x
x = 54.3 kPa
5
23
PCl
ClPClp P
P PK
0.14x(0.86x)2
= 287 kPa
95
Q.13
N2(g) + O2(g) 2NO(g)
No. of moles at equilibrium
Initial no. of moles
2.0 1.0 0
2.0 - x 1.0 – x
2x
Concentration at equilibrium
2.0x2.0
2.0x1.0
2.02x
96
Q.13
N2(g) + O2(g) 2NO(g)
Concentration at equilibrium
2.0x2.0
2.0x1.0
2.02x
2.0x1.0
2.0x2.02.02x
K
2
c = Kp = 1.2 102
x = 0.073
[N2] = (2.0 – 0.073)/2.0 = 0.96 mol dm3
97
Q.13
N2(g) + O2(g) 2NO(g)
Concentration at equilibrium
2.0x2.0
2.0x1.0
2.02x
2.0x1.0
2.0x2.02.02x
K
2
c = Kp = 1.2 102(2.0)(1.0)
(2x)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
2.0x2.0
2.0x1.0
2.02x
2.0x1.0
2.0x2.02.02x
K
2
c = Kp = 1.2 102
x = 0.077
[N2] = (2.0 – 0.077)/2.0 = 0.96 mol dm3
(2.0)(1.0)(2x)2
99
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)
100
Homogeneous Equilibrium
CH3COOH(l) + C2H5OH(l)
CH3COOC2H5(l) + H2O(l)
CH3COOH(aq) + C2H5OH(aq)
CH3COOC2H5(l) + H2O(l)
Immiscible Two phases
Glacial ethanoic acid
Absolute alcohol
Dissolve both
products
101
Q.14
Cu2+(aq) + 4NH3(aq) Cu(NH3)4
2+(aq)
4eqm3eqm
2
eqm243
c (aq)][NH(aq)][Cu
(aq)])[Cu(NHK
102
Q.14
N2(g) + 3H2(g) 2NH3(g)
3HN
2NH
p
22
3
PP
PK
103
Q.14 CH3COOH(l) + C2H5OH(l)
CH3COOC2H5(l) + H2O(l)
eqm52eqm3
eqm2eqm523c OH(l)]H[CCOOH(l)][CH
O(l)][H(l)]HCOOC[CHK
104
Q.14
H+(aq) + OH(aq) H2O(l)
eqmeqm
eqm2c (aq)][OH(aq)][H
O(l)][HK
At pH 7, density of water 1000 g dm3
3
1
2 dm 1mol g 18
g 1000
O(l)][H
= 55.5 mol dm3
105
Q.14
H+(aq) + OH(aq) H2O(l)
[H2O(l)] 55.5 M
In large excess
a constant
eqmeqm
'c (aq)][OH(aq)][H
1K
eqmeqm
eqm2c (aq)][OH(aq)][H
O(l)][HK
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 cm3 at 298 K
Mass of 1 dm3 of 12.39 M HCl(aq)
= 1.19 g cm3 1000 cm3 = 1190 g
Mass of HCl present
= 12.39 mol (1 + 35.5) g mol1 = 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 cm3 at 298 K
Mass of water present
= (1190 – 452.2) g = 737.8 g
3
1
2 dm 1mol g 18.0
g 737.8
O(l)][H
= 41.0 M
108
Q.15(b)
3
1
2 dm 1mol g 18.0
g 737.8
O(l)][H
= 41.0 M< 55.5 M
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.
109
Q.15(b)
< 12.38 M
H+(aq) + OH(aq) H2O(l)
41.0 M
HCl(aq) H+(aq) + Cl(aq)12.38 M
110
Heterogeneous Equilibrium
Equilibrium systems involving two or more phases
H2O(l) H2O(g)
∵ Kp depends on temperature only
∴ at fixed T, vapour pressure of water (Kp) is a constant,
OHp 2PK
irrespective of the amountof water
present.
111
equil2
equil2c O(l)][H
O(g)][HK
∵ [H2O(l)]equil OH
OH
2
2
V
n
g 18ρ
g 18V
m
Vg 18
m
OH
OH
OH
OH
2
2
2
2
∴ [H2O(l)]equil = constant (at fixed T)
112
equil2
equil2c O(l)][H
O(g)][HK
equil2'cequil2c O(g)][HK O(l)][HK
RTO(g)][HP equil2OH2∵
RTKP 'cOH2
= 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
Conclusion : -
[X(s)] and [X(l)] do NOT appear in the equilibrium constant expressions of heterogeneous equilibria.
Changing the amount of a pure solid or a pure liquid in a heterogeneous equilibrium mixture does NOT disturb the equilibrium.
116
Q.16
Mg(s) + Cu2+(aq) Mg2+(aq) + Cu(s)
eqm2
eqm2
c (aq)][Cu
(aq)][MgK
117
Q.16
CaCO3(s) CaO(s) + CO2(g)
2COp PK
118
Q.16
Ag+(aq) + Cl(aq) AgCl(s)
eqmeqmc (aq)][Cl(aq)][Ag
1K
119
Q.16
Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(g)
4H
4OH
p
2
2
P
PK
120
Q.16
Br2(l) Br2(g)
2Brp PK
121
Partition Partition Equilibrium of a Equilibrium of a Solute Between Solute Between Two Immiscible Two Immiscible
SolventsSolvents
122
Partition (Distribution) EquilibriumPartition (Distribution) Equilibrium
• The equilibrium established when a non-volatile solute distributes itself between two immiscible liquids
A(solvent 2) A(solvent 1)
123
Water and hexane are immiscible with each other.
Partition (Distribution) EquilibriumPartition (Distribution) Equilibrium
Hexane
Water
124
I2 dissolves in both solvents to different extent.
Partition (Distribution) EquilibriumPartition (Distribution) Equilibrium
Hexane
Water
125
When dynamic equilibrium is established,
rate of movement = rate of movement
Partition (Distribution) EquilibriumPartition (Distribution) Equilibrium
Hexane
Water
126
Suppose the equilibrium concentrations of iodine in H2O and hexane are x and y respectively,
Hexane
Water
yx
KD
127
When dynamic equilibrium is established
the ratio of concentrations of iodine in water and in hexane is always a constant.
yx
KD
Suppose the equilibrium concentrations of iodine in H2O and hexane are x and y respectively,
KD : partition coefficient or
distribution coefficient
128
Changing the concentrations by the same extent does not affect the quotient.
0.5y0.5x
4y4x
3y3x
2y2x
yx
KD
∵ 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
The partition law can be represented by the following equation:
(no unit)
Units of concentration : -
Partition Partition CoefficientCoefficient
mol dm-3, mol cm-3, g dm-3, g cm-3
2 solvent in solute of ionconcentrat1 solvent in solute of ionconcentrat
KD
2 solvent
1 solvent
[solute][solute]
130
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
Partition Partition CoefficientCoefficient
1 solvent
2 solventD [solute]
[solute]K
2 solvent
1 solventD [solute]
[solute]K
131
• Depends on temperature ONLY.
• Not affected by the amount of solute added and the volumes of solvents used.
• TAS Experiment No. 12
Partition Partition CoefficientCoefficient
132
[CH3COOH]2-methylpropan-1-ol
[CH3COOH]water
ol1anmethylprop23water3 COOH][CH / COOH][CHslope
= KD
[CH3COOH]water [CH3COOH]2-methylpropan-1-
ol TAS Expt 12
133
Partition law holds truePartition 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)
134
Partition law holds truePartition law holds true
3. when the solute exists in the same form in both solvents.
C6H5COOH(benzene) C6H5COOH(aq)C2 C1
C1 and C2 are determined by titrating the acid in each solvent with standard sodium hydroxide solution.
i.e. no association or dissociation of solute
135
[C6H5COOH]water(C1
) / mol dm-3
[C6H5COOH]benzene(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
Not a constant
136
Interpretation : -Interpretation : -
• Benzoic acid tends to dimerize (associate) in non-polar solvent to give (C6H5COOH)2
Benzoic acid dimer
C
O
O H
C
O
OH
Violation of Partition law
• The solute does not have the same molecular form in both solvents
137
Interpretation : -Interpretation : -
2C6H5COOH(benzene) (C6H5COOH)2(benzene)
[C6H5COOH]total = [C6H5COOH]free + [C6H5COOH]associated
)1(2 C 221 C
C2 C2(1-) C2
= degree of association of benzoic acid
Determined by titration with NaOH
138
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.
Q.17(a)
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
)(1C2 α α221 C
2C6H5COOH(benzene) (C6H5COOH)2(benzene)
22
221
)](1[CC
Kα
α
)(1C2 α C6H5COOH(benzene) C6H5COOH(aq)
C1
)(1CC
K2
1D α
Q.17(c)
141
22
221
)](1[CC
Kα
α
2'2
2 CK2KC
)(1C α
α
2'
1
2
1D
CK
C)(1C
CK
α
'''D
2
1 KKKC
C
is a constant at fixed T
142
5.290.20
2.630.14
1.920.12
0.087
0.086
0.087
0.0860.4830.06
[C6H5COOH]benzene(C2) / mol dm-3
[C6H5COOH]water(C1) / mol dm-3
~constant
2
1
C
C
143
Applications of partition law
• Solvent extraction• Chromatography
Two classes of separation techniques based on partition law.
144
I2 in KI(aq)
I2 in hexane
I2 in KI(aq)
I2 in hexane
I2 in KI(aq)
I2 in hexane
I2 in KI(aq)
Colourless Hexane
Solvent extraction
I2 in KI(aq)
+ hexane
I2 in KI(aq)
I2 in hexane
At equilibrium,
rate of movement of I2 = rate of movement of I2
)(2
2
][
][
aqKI
hexane
I
IK
To remove I2 from an aqueous solution of KI, a suitable solvent is added.What feature should the solvent have?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.
By partition law,
145
Before shakin
g
After shakin
g
Solvent Solvent ExtractionExtractionHexane
layer
Aqueous layer
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
I2 + 2S2O3 2I + S4O6
2
Titrated with standard sodium thiosulphate solution
147
For the hexane layer, starch is not needed because the colour of I2 in hexane is intense enough to give a sharp end point.
Determination of I2 left in the KI solution
For the aqueous layer, starch is used as the indicator.
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.
Worked example
149
10g X in 25 cm3 aqueous solution
50g X in 40 cm3 ether solution
Worked example : -
M04.0
50
04.0M50
]X[ ether
water
etherD ]X[
]X[K By partition law,
M025.0
10
025.0M10
]X[ water
125.3
M025.010
M04.050
KD
(a) Calculate the partition coefficient of X between ether and water at 298 K.
M is the molecular mass of X
150
10g X in 25 cm3 aqueous solution
50g X in 40 cm3 ether solution
Worked example : -
125.32510
4050
K
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
5g of X in 30 cm3 aqueous solution
(5-x)g of X in 30 cm3 aqueous solution
xg of X in 30 cm3 ether solution30 cm3 ether
152
(b)(i)
5g of X in 30 cm3 aqueous solution
(5-x)g of X in 30 cm3 aqueous solution
xg of X in 30 cm3 ether solution30 cm3 ether
125.3K
79.35
xx
x
305
30
x
x
3.79 g of X could be extracted.
153
(b)(ii) First extraction
5g of X in 30 cm3 aqueous solution
15 cm3 etherx1g of X in 15 cm3 ether solution
(5-x1)g of X in 30 cm3 aqueous solution
125.3K
05.35
21
1
1
xx
x
30515
1
1
x
x
154
(b)(ii) Second extraction
(5-x1)g of X in 30 cm3 aqueous solution
15 cm3 etherx2g of X in 15 cm3 ether solution
(5-x1-x2)g of X in 30 cm3 aqueous solution
125.3K
19.105.35
22
2
2
xx
x
305
15
21
2
xx
x
155
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
156
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.
Important extraction processes : -
157
Q.18(a)
100 cm3 of 0.500 M ethanoic acid
200 cm3 alcohol
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;
Alcohol layer
Aqueous layer
158
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
0.2000.0500x
[acid]alcohol 0.100x)0.0500(1
[acid]water
0.2000.0500x
0.100x)0.0500(1
3.05K
Q.18(a)
x = 0.396 = 39.6%
159
Let x1, x2 be the fractions of ethanoic acid extracted to the alcohol layer in the 1st and 2nd extractions respectively.
Q.18(b)
1st extraction
0.247x3.05
0.1000.0500x
0.100)x0.0500(1
11
1
0.186x3.05
0.1000.0500x
0.100)xx0.0500(1
22
21
2nd extraction
% of acid extracted=0.247+0.186=0.433=43.3%
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.
1000.1yx
0.9y
][I][I
120Kwater2
solventX2
∴ 7.5 cm3 of solvent X is required
x = 7.5
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
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.
Chromatography
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.
Sample
Eluant
168
Column chromatographyThe 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
Suitable for large scale treatment of sample
For treatment of small quantities of samples, paper or thin layer chromatography is preferred.
Column chromatography
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 X(adsorbed water) X(solvent)
stationary phase mobile phase
171
The solvent moves up the filter paper by capillary action
Components are carried upward by the mobile solvent
Ascending chromatography
Paper chromatography
172
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
solvent by travelled distancespot by travelled distance R
f
175
filter paper
spot of coloured dye
solvent
separated colours
Using chromatography to separate the colours in a sweet.
176
b
dc
a
ab
(blue)Rf
ac
(red)Rf
ad
(green)Rf
Solvent front
177
separated colours
a chromatogram
178
Paper Paper ChromatographyChromatography
• 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 Paper ChromatographyChromatography
Amino acid Solvent
Mixture of phenol and ammonia
Mixture of butanol and ethanoi
c acidCystine 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 chromatographyStationary phase : -
Water adsorbed on a thin layer of solid adsorbent (silica gel or alumina).
Mobile phase : -
A suitable solvent X(adsorbed water) X(solvent)
stationary phase mobile phase
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 Factors Affecting Affecting
EquilibriumEquilibrium
185
THREE factors affecting chemical equilibria.
1. Changes in pressure
2. Changes in concentration
3. Changes in temperature
No effect on Kc or Kp
Alter the equilibrium position by changing the equilibrium constant
Alter the equilibrium position by changing the equilibrium composition
186
THREE ways of interpretation
1. Kc or Kp approach
2. Kinetic approach
3. Le Chatelier’s Principle
187
A(g) + B(g) C(g)
Effects of changes in pressure
P by reducing V
Equilibrium position shifts to the right
increase in pressure
decrease in pressure
P by increasing V
Equilibrium position shifts to the left
188
A(g) + B(g) C(g)
Interpretation : Kp approach
Pequil 1 PA PB PC
V21P 2PA 2PB 2PC
BA
Cp PP
PK
pp
BA
C
BA
Cp K
2
K
PPP
21
2P2P2P
Q
189
A(g) + B(g) C(g)
Interpretation : Kp approach
Pequil 1 PA PB PC
V21P 2PA 2PB 2PC
pp
BA
C
BA
Cp K
2
K
PPP
21
2P2P2P
Q
Equilibrium position shifts to the right until Qp = Kp
190
A(g) + B(g) C(g)
Interpretation : Kp approach
Pequil 1 PA PB PC
2 equilP 2PA - x 2PB – x 2PC + x
pBA
C
BA
Cp K
PPP
x2Px2Px2P
Q
Equilibrium position shifts to the right until Qp = Kp
191
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
Equilibrium position shifts to the right
192
Q.21
R1 = k1[A][B]
R-1 = k-1[C]
V½
A(g) + B(g) C(g)k1
k-1
R1’ = k1(2[A])(2[B]) = 4R1
R-1’ = k-1(2[C]) = 2R-1
More affected
193
Q.21
R1 = k1[A][B]
R-1 = k-1[C]
V2
A(g) + B(g) C(g)k1
k-1
R1’ = k1(½[A])(½[B]) = ¼R1
R-1’ = k-1(½[C]) = ½R-1
More affected
194
Le Chatelier’s PrincipleLe 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.
195
A(g) + B(g) C(g)
Q.22(a)
Change : PT
Response : PT
196
A(g) + B(g) C(g)
Q.22(b)/(c)
One mole
Two moles
197
A(g) + B(g) C(g)
Q.22(d)
One mole of gas exert less pressure.
One mole
Two moles
198
A(g) + B(g) C(g)
Q.22(e)
One mole of gas exert less pressure.
One mole
Two moles
The forward reaction lowers the pressure of the system.
199
A(g) + B(g) C(g)
Q.22(f)
One mole of gas exert less pressure.
One mole
Two moles
The forward reaction lowers the pressure of the system.
Equilibrium position shifts to the right.
200
A(g) + B(g) C(g)
Q.22(g)
Change : PT
Response : PT
201
N2O4(g) 2NO2(g)
pale yellow
brown
Sealed nozzle
N2O4(g), NO2(g)
Syringe
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
Sealed nozzle
N2O4(g), NO2(g)
Syringe
After a few seconds
The gas mixture turns palerbecause 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
Sealed nozzle
N2O4(g), NO2(g)
Syringe
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
Sealed nozzle
N2O4(g), NO2(g)
Syringe
After a few seconds
The gas mixture turns darker brown becausethe system the pressure by shifting the equilibrium position to the right (the side with more gas molecules).
205
H2(g) + CO2(g) H2O(g) + CO(g)
Q.23(a)/(b)
Cause : in PT by reducing VT
Cause : in PT by increasing VT
Effect : No effect on the equilibrium position
Effect : No effect on the equilibrium position
206
H2(g) + CO2(g) H2O(g) + CO(g)
Q.23(c)/(d)
Cause : in PT by increasing PCO
Effect : Equilibrium position shifts to the right
Cause : in PT by increasing 2HP
Effect : Equilibrium position shifts to the left
207
H2(g) + CO2(g) H2O(g) + CO(g)
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
PCl5(g) PCl3(g) + Cl2(g)
Q.23(a)/(b)
Cause : in PT by reducing VT
Cause : in PT by increasing VT
Effect : Equilibrium position shifts to the left
Effect : Equilibrium position shifts to the right
209
Q.23(c)/(d)
Effect : Equilibrium position shifts to the right
Cause : in PT by increasing 5PClP
PCl5(g) PCl3(g) + Cl2(g)
Effect : Equilibrium position shifts to the left
Cause : in PT by increasing2ClP
210
Q.23(e)
Cause : in PT by introducing He(g)
Effect : No effect on equilibrium position
PCl5(g) PCl3(g) + Cl2(g)
211
Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(g)
Q.23(a)/(b)
Cause : in PT by reducing VT
Cause : in PT by increasing VT
Effect : No effect on equilibrium position
Effect : No effect on equilibrium position
212
Q.23(c)/(d)
Effect : Equilibrium position shifts to the right
Cause : in PT by increasing 2HP
Effect : Equilibrium position shifts to the left
Cause : in PT by increasing OH2P
Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(g)
213
Q.23(e)
Cause : in PT by introducing He(g)
Effect : No effect on equilibrium position
Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(g)
214
H2(g) + CO2(g) H2O(g) + CO(g)
For the systems
Changing PT by altering VT has no effect on the equilibrium position
Interpretation : Kp approach
Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(g)
215
H2(g) + CO2(g) H2O(g) + CO(g)
For the systems
Pequil 12HP 2COP OH2
P COP
V21P
2H2P OH22P CO2P
2CO2P
22
2
COH
COOHp PP
PPK
22
2
COH
COOHp 2P2P
2P2PQ
= Kp
216
For the systems
Pequil 12HP OH2
P
V21P OH2
2P2H2P
4H
4OH
p
2
2
P
PK
4H
4OH
p
2
2
2P
2PQ
= Kp
Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(g)
217
H2(g) + CO2(g) H2O(g) + CO(g)
For the systems
Kinetic approach
The rates of forward and backward reactions are affected to the same extent.
Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(g)
218
H2(g) + CO2(g) H2O(g) + CO(g)
For the systems
By Le Chatelier’s principleSince the system has the same no. of gas molecules on either side,
Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(g)
No adjustment made by the system can reduce the change.
No shifting of equil. position
219
H2(g) + CO2(g) H2O(g) + CO(g)
For the systems
in PT by changing VT has no effect on the equilibrium position
PCl5(g) PCl3(g) + Cl2(g)
However, the partial pressures and thus the equilibrium composition change by altering VT
220
Q.24
CO2(aq) CO2(g)
Once the bottle is opened, CO2 escapes from the system and its partial pressure drops.
2COP in Decrease
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)
H2O(s) H2O(l)
222
H2O(s) H2O(l)
Q.25
More open More closely packed
Extremely high P
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
Bi3+(aq) + Cl(aq) + H2O(l) BiOCl(s) + 2H+(aq)
colourless
white ppt
Test 1 : Add HCl
Result : The white ppt disappears
The equil. position shifts to the left
224
Bi3+(aq) + Cl(aq) + H2O(l) BiOCl(s) + 2H+(aq)
colourless
white ppt
Interpretation : Kc approach
O(l)][H(aq)][Cl(aq)][Bi
(aq)][HK
equil2equilequil3
equil2
c*
Since H2O is in large excess
(aq)][Cl(aq)][Bi
(aq)][HK
equilequil3
equil2
c
225
Bi3+(aq) + Cl(aq) + H2O(l) BiOCl(s) + 2H+(aq)
colourless
white ppt
(aq)][Cl(aq)][Bi
(aq)][HK
equilequil3
equil2
c
Addition of HCl(aq)
Both [H+(aq)] and [Cl(aq)] to the same extent
c3
2
c K(aq)](aq)][Cl[Bi
(aq)][HQ
The equilibrium position shifts to the left to restore the Kc
226
Bi3+(aq) + Cl(aq) + H2O(l) BiOCl(s) + 2H+(aq)
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
Bi3+(aq) + Cl(aq) + H2O(l) BiOCl(s) + 2H+(aq)
Addition of HCl(aq)
Bi3+(aq) + 3Cl(aq) + H2O(l) BiOCl(s) + 2HCl(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
Bi3+(aq) + Cl(aq) + H2O(l) BiOCl(s) + 2H+(aq)
colourless
white ppt
Test 2 : Add large excess of H2O
Result : The white ppt reappears
The equil. position shifts to the right
229
Bi3+(aq) + Cl(aq) + H2O(l) BiOCl(s) + 2H+(aq)
colourless
white ppt
Interpretation : Kc approach
O(l)][H(aq)][Cl(aq)][Bi
(aq)][HK
equil2equilequil3
equil2
c*
Addition of large excess of H2O
*KO(l)](aq)][H(aq)][Cl[Bi
(aq)][HQ c
23
2
c*
Equilibrium position shifts to the right such that *Qc = *Kc
230
Bi3+(aq) + Cl(aq) + H2O(l) BiOCl(s) + 2H+(aq)
colourless
white ppt
Interpretation : Kinetic approach
[H2O(l)]
rate of forward rx > rate of backward rx
equilibrium position shifts to the right
231
Bi3+(aq) + Cl(aq) + H2O(l) BiOCl(s) + 2H+(aq)
colourless
white ppt
By Le Chatelier’s principle : -
[H2O(l)]
The system shifts to the right to reduce the water added.
232
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.
The Effect of Changes in Temperature on Equilibrium
233
Exothermic reaction:
Temperature (K)
500 600 700 800
Kp (atm-2) 90.0 3.0 0.3 0.04
Examples : -
N2(g) + 3H2(g) 2NH3(g) 10 mol kJ 92ΔH
Kp decreases as T increases
234
Exothermic reaction:
Examples : -
2C(graphite) + O2(g) 2CO(g)
Temperature (K)
298 500 700 900 1100
Kp (atm) 1.5 1048
3.1 1032
1.2 1026
3.1 1022
1.5 1020
10 mol kJ 211ΔH
Kp decreases as T increases
235
Endothermic reaction:
Examples : -
N2O4(g) 2NO2(g)
Temperature (K)
200 300 400 500
Kp (atm) 1.8 10-6 0.174 51 1510
10 mol kJ 58ΔH
Kp increases as T increases
236
Endothermic reaction:
Examples : -
N2(g) + O2(g) 2NO(g)
Temperature (K) 700 1100 1500
Kp (no unit) 5 10-13 4 10-8 1 10-
5
10 mol kJ 100ΔH
Kp increases as T increases
237
Van’t Hoff Equation
CRTΔH
lnKo
C'2.303RT
ΔHKlog
o
10
C and C’ are constants related to
oΔS
RΔS
Co
2.303R
ΔSC'
o
238
CRTΔH
lnKo
If the forward process is exothermic,
0ΔH o 0R
ΔH- and
o
T K
An increase in T shifts the equilibrium position to the left
(in the endothermic direction)
239
CRTΔH
lnKo
If the forward process is exothermic,
0ΔH o 0R
ΔH- and
o
T K
An decrease in T shifts the equilibrium position to the right
(in the exothermic direction)
240
CRTΔH
lnKo
If the forward process is endothermic,
0ΔH o 0R
ΔH- and
o
T K
An increase in T shifts the equilibrium position to the right
(in the endothermic direction)
241
CRTΔH
lnKo
If the forward process is endothermic,
0ΔH o 0R
ΔH- and
o
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 Chatelier’s principle
243
Q.26(a)
)(KT1 1
lnK
Forward reaction is exothermic
0R
ΔH-slope
o
Cintercept-y (If C > 0)
244
Q.26(a)
)(KT1 1
lnK
Forward reaction is exothermic
0R
ΔH-slope
o
Cintercept-y (If C < 0)
245
Q.26(a)
)(KT1 1
lnK
Forward reaction is endothermic
0R
ΔH-slope
o
Cintercept-y (If C > 0)
246
Q.26(a)
)(KT1 1
lnK
Forward reaction is endothermic
0R
ΔH-slope
o
Cintercept-y (If C < 0)
247
N2O4(g)(pale yellow) 2NO2(g)(brown)
50C
Q.27
Increase in T
Decrease in T
248
N2O4(g)(pale yellow) 2NO2(g)(brown)
Q.27(a) Increase in T
Decrease in T
0ΔH o
in T shift the equilibrium position to the right.
Thus, the forward reaction is endothermic
By Le Chatelier’s 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
TV K 273K 373
VV
K 273
K 373
33373K dm1.37 )dm (1
K 273K 373
V
250
Q.27(b)(ii)
∵ equilibrium position shifts to the right
total no. of moles of gas molecules
total volume of the system further
N2O4(g) 2NO2(g)
251
Q.27
T(K)
V(dm3
)
Ideal gas expansion
Actual in V
273
1.00
373
1.37
N2O4(g) 2NO2(g)
Increase in T
252
Interpretation of the Effects of Temperature Changes on Equilibrium in Terms of Chemical Kinetics
Ea
Ea’
AB+X
A+BX
Pote
ntia
l en
erg
y
Reaction co-ordinate
A–B + X A + B–X
0ΔH o
253
Ea
Ea’
AB+X
A+BX
Pote
ntia
l en
erg
y
Reaction co-ordinate
1a
1
2a
2
1
2
/RTET
/RTET
T
T
eA
eA
k
k
For the forward reaction (exothermic)
1a
2a
/RTE
/RTE
ee
)
T1
T1
(RE
12
a
e
21
12a
TRT)T(TE
e
> 1∵ Ea > 0 &
T2 – T1 > 0Rate as T
254
Ea
Ea’
AB+X
A+BX
Pote
ntia
l en
erg
y
Reaction co-ordinate
1a
1
2a
2
1
2
/RT'ET
/RT'ET
T
T
e'A
e'A
'k
'k
For the backward reaction (endothermic)
1a
2a
/RT'E
/RT'E
e
e
)
T1
T1
(R
'E
12
a
e
21
12a
TRT)T(T'E
e
21
12a
TRT)T(TE
e
∵ Ea’ > Ea &
T2 – T1 > 0
255
Conclusion :
The rate of endothermic reaction is affected more by temperature changes.
An in temperature the rates of endothermic and exothermic reactions to different extents.
256
Q.28
X(l) X(g) 0ΔH ovap
Prediction : -
Interpretation : -
An in T increases Kp
Thus, more X(l) evaporate until Qp = Kp
An in T shifts the equilibrium position to the right (in endothermic direction)
257
Q.28
X(s) X(g) 0ΔH osub
Prediction : -
Interpretation : -
An in T increases Kp
Thus, more X(s) sublime until Qp = Kp
An in T shifts the equilibrium position to the right (in endothermic direction)
258
C'2.303RT
ΔHKlog
o
10
if < 0 (forward reaction is exothermic)oH
log10K > 0 K > 1 (the exothermic reaction is more complete
and C’ is negligibly small
and the Extent of Completion of Reaction
oΔH
259
C'2.303RT
ΔHKlog
o
10
if > 0 (forward reaction is endothermic)
oH
log10K < 0 K < 1 (the endothermic reaction is less complete)
and C’ is negligibly small
and the Extent of Completion of Reaction
oΔH
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 mol1 and (ii) 100 kJ mol1 respectively.
(Given : R = 8.314 J K1 mol1)
C'2.303RT
ΔHKlog
o
10
2.303RTΔH o
K 298mol K J 8.3142.303
)mol J 1000 (-100-1-1-
-1
K 3 1017
(Units not known)
(i)
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 mol1 and (ii) 100 kJ mol1 respectively.
(Given : R = 8.314 J K1 mol1)
C'2.303RT
ΔHKlog
o
10
2.303RTΔH o
K 298mol K J 8.3142.303
)mol J 1000 (100-1-1-
-1
K 3 1018
(Units not known)
(ii)
262
Conclusion : -
Exothermic processes are Far More Complete than endothermic processes.
263
The total pressures of the following equilibrium system are 2.333104 Nm2 and 6.679104 Nm2 at 282.5 K and 298.1 K respectively.
Q.29
NH4HS(s) NH3(g) + H2S(g)
Since all gases arises from NH4HS(s)
TSHNH P21
PP23
264
NH4HS(s) NH3(g) + H2S(g)
At 282.5 K
eqmSHeqmNHp )(P)ln(PlnK23
2T )P
21
ln( C282.58.314
ΔH- o
eqmSHeqmNHp )'(P)'ln(PlnK23
' 2T )'P
21
ln( C298.18.314
ΔH- o
At 298.1 K
(1)
(2)
(2) – (1)2
T
T
P'P
ln
298.11
282.51
8.314ΔH o
2
4
4
102.333106.679
ln
298.11
282.51
8.314ΔH o
oΔH = +94.41 kJ mol1
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
Con
cen
tratio
n
Time
[A]
[B]
A BQ.30
Less time to attain equilibriumTime taken to attain equilibrium
t1t2
267
A(g) + B(g) C(g)Q.31
t1 Time
Rate
of re
actio
n
Forward reaction
Backward reaction
H > 0
1. in T
2. in PT by reducing VT
VT T (adiabatic compression)
e.g. expanding universe
268
A(g) + B(g) C(g)Q.31
t2 Time
Rate
of re
actio
n
Forward reaction
Backward reaction
H > 0
Adding a +ve catalyst
269
H2(g) + I2(g) 2HI(g)
Q.32 H < 0
t1 t2 t3 t4
Con
cen
tratio
n
[HI(g)]
[H2(g)]
[I2(g)]
Time
t1 : 1. adding a catalyst
2. in PT by adding an an inert gas at fixed VT
270 t1 t2 t3 t4
H2(g) + I2(g) 2HI(g)
Q.32 H < 0
Con
cen
tratio
n
[HI(g)]
[H2(g)]
[I2(g)]
Time
in PT by reducing VT has no effect on the equilibrium position but changes the equilibrium composition
271 t1 t2 t3 t4
H2(g) + I2(g) 2HI(g)
Q.32 H < 0
Con
cen
tratio
n
[HI(g)]
[H2(g)]
[I2(g)]
Time
t4 : in PT by reducing VT
272 t1 t2 t3 t4
H2(g) + I2(g) 2HI(g)
Q.32 H < 0
Con
cen
tratio
n
[HI(g)]
[H2(g)]
[I2(g)]
Time
t2 : in T at fixed VT
273 t1 t2 t3 t4
H2(g) + I2(g) 2HI(g)
Q.32 H < 0
Con
cen
tratio
n
[HI(g)]
[H2(g)]
[I2(g)]
Time
t3 : Input of H2(g) at fixed VT
274
CO(g) + 2H2(g) CH3OH(g)H < 0
Q.33
Kp Shifts to the left
in T
No effectShifts to the
right in PT by reduc
ing VT
Effect on Kp
Effect on equilibrium
positionChanges
275
CO(g) + 2H2(g) CH3OH(g)H < 0
Q.33
No effectShifts to the
right
No effectNo effect
Effect on Kp
Effect on equilibrium
positionChanges
Doubling PCO and
OHCH3P
OHCH3P
Doubling and 2HP
276
CO(g) + 2H2(g) CH3OH(g)H < 0
Q.33
No effectShifts to the
rightA little H2O(l) is
added
No effectNo effectA positive catalyst is
added
Effect on Kp
Effect on equilibrium
positionChanges
Soluble in water
277
A(g) + B(g) C(g)H = 0Q.34
No effectNo effect in T at fixed PT
No effectNo effect in T at fixed PT
Effect on Kp
Effect on equilibrium
positionChanges
278
A(g) + B(g) C(g)H = 0Q.34
No effectShifts to the
left in T at fixed VT
No effectShifts to the
right in T at fixed VT
Effect on Kp
Effect on equilibrium
positionChanges
279
Summary of the Effects of Changes Summary of the Effects of Changes of Various Factors on Equilibriumof 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
aA(g) + bB(g) cC(g) + dD(g)
280
Factor Equilibrium position
Equilibrium constant
Increase in pressure by reducing the volume of the
container
Shifts to right if (c + d) < (a + b)Shifts to left to
(a + b) < (c + d)No change if a + b = c + d
No change
Summary of the Effects of Changes Summary of the Effects of Changes of Various Factors on Equilibriumof Various Factors on Equilibrium
aA(g) + bB(g) cC(g) + dD(g)
Isothermal compression
281
Factor Equilibrium position
Equilibrium constant
Increase in temperature
Shifts to right if the forward reaction is
endothermicShifts to left if the forward reaction is exothermic
Kp if the forward reaction is endoth
ermic
Kp if the forward reaction is exother
mic
Summary of the Effects of Changes Summary of the Effects of Changes of Various Factors on Equilibriumof Various Factors on Equilibrium
aA(g) + bB(g) cC(g) + dD(g)
282
Factor Equilibrium position
Equilibrium constant
Addition of a catalyst
No change No change
Summary of the Effects of Changes Summary of the Effects of Changes of Various Factors on Equilibriumof Various Factors on Equilibrium
aA(g) + bB(g) cC(g) + dD(g)
283
16.1 Irreversible and Reversible Reactions (SB p.89)
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 ?
Back
Answer(a) A
(b) B
(c) A B
284
16.2 Dynamic Nature of Chemical Equilibrium (SB p.91)
List some characteristics of chemical equilibrium.
Back
AnswerSome 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)
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.
Back
AnswerChemical equilibrium is dynamic in nature. When a trace
amount of carbon monoxide labelled with radioactive carb
on-14 is added to the equilibrium system, the equilibrium
position shifts to the right. Therefore, radioactive carbon d
ioxide molecules are formed.
286
16.4 Equilibrium Law (SB p.94)
What is a closed system? Why can chemical equilibrium
only be established in a closed system?
Back
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)
Write the equilibrium expression (for Kc) and unit of equilibrium constants for the following equilibrium system.
(a) 2O3(g) 3O2(g)Answer
(a)
Unit of Kc: mol dm-3
2
eqm3
3
eqm2
c )]g(O[
)]g(O[K
288
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) 2NH3(g)Answer
(b)
Unit of Kc: mol-2 dm6
3
eqm2eqm2
2
eqm3
c )]g(H[)]g(N[
)]g(NH[K
289
16.4 Equilibrium Law (SB p.94)
Write the equilibrium expression (for Kc) and unit of equilibrium constants for the following equilibrium system.
(c) C(graphite) + H2O(g) CO(g) + H2(g)Answer
Back
(c)
Unit of Kc: mol dm-3
eqm2
eqm2eqm
c )]g(OH[
)]g(H[)]g(CO[K
290
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.
291
16.5 Determination of Equilibrium Constants (SB p.98)
(a) Calculate the equilibrium concentrations of Ag+(aq), Fe2+(aq) and Fe3+(aq).
Answer(a) Ag+(aq) + CNS–(aq) AgCNS(aq)
Number of moles of KCNS(aq)
=
= 3.05 10-4 mol
Number of moles of Ag+(aq) in 25 cm3 of the filtrat
e
at equilibrium = 3.05 10–4 mol
[Ag+(aq)]eqm =
= 0.012 2 mol dm-3
M 0.050dm10006.10 3
mol 10 25mol 10 3.05
3-
-4
292
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
=
= 0.05 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
33-
3-3-3
dm 10100)(100dm )10 (100 dm mol 0.100
293
16.5 Determination of Equilibrium Constants (SB p.98)
(b) Calculate the equilibrium constant (Kc).Answer
(b)
=
= 253.96 mol-1 dm3
eqmeqm
2
eqm
3
c )]aq(Ag[)]aq(Fe[
)]aq(Fe[K
3-3
3
dm mol 0.0122 dm mol 0.0122dm mol 0.0378
294
16.5 Determination of Equilibrium Constants (SB p.98)
(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.
Back
295
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
296
Let the volume of the reaction mixture be V dm3.
eqm2eqm2
2
eqm
c )]g(I[)]g(H[
)]g(HI[K
)V
xb)(
Vxa
(
)Vx2
( 2
)xb)(xa(x4 2
16.5 Determination of Equilibrium Constants (SB p.98)
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
Back
297
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
298
Let the volume of the reaction mixture be V dm3.
3
eqm2eqm2
2
eqm3
c )]g(H[)]g(N[
)]g(NH[K
3
2
)V
x3b)(
Vxa
(
)Vx2
(
3
3
2
2
)x3b(V
xaV
Vx4
2
3
2
V)x3b)(xa(
x4
16.5 Determination of Equilibrium Constants (SB p.99)
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
Back
299
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
300
16.5 Determination of Equilibrium Constants (SB p.100)
Initial concentration of Fe3+(aq) =
= 1.0 10-3 mol dm-3
Initial concentration of NCS-(aq) =
= 1.0 10-3 mol dm-3
Fe3+(aq) + NCS-(aq) [Fe(NC
S)]2+(aq)
At start: 1.0 10-3 M 1.0 10-3 M 0 M
Amount changed: -x M –x M x M
At equilibrium: (1.0 10-3 – x) M (1.0 10-3 – x) M x M
33-
3-333
dm 10 10) 01(dm 10 10 dm mol 102.0
33-
3-333
dm 10 10) 01(dm 10 10 dm mol 102.0
301
16.5 Determination of Equilibrium Constants (SB p.100)
From the given data, the equilibrium concentration of [Fe(NCS)]2+(a
q) 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
Kc =
= 189.3 dm3 mol-1
)1086.0)(1086.0(104.1
33
4
Back
302
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.
16.5 Determination of Equilibrium Constants (SB p.100)
What is the implication for an equilibrium reaction having an equilibrium constant much smaller than
1.0?
Back
Answer
303
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) PCl3(g) + Cl2(g)Answer
304
∵ At equilibrium, 0.002 mole of Cl2(g) was found in the flask.
X = 0.002 mol
Kc =
= = 0.072 mol dm-3
)]g(PCl[
)]g(Cl)][g(PCl[
5
23
3-
-33
dm mol 0.002) - (0.009dm mol 0.002dm mol 0.002)(0.250
16.5 Determination of Equilibrium Constants (SB p.100)
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
Back
305
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
306
16.6 Equilibrium Constant in Terms of Partial Pressures (SB p.101)
2NOBr(g) 2NO(g) + Br2(g)
The expression of Kp is:
Substituting the partial pressures into the expression, we have:
= 1 003.9 Nm–2
2
NOBr
Br
2
NO
p )P(
)P()P(K 22
2
2
p )246()300()450(
K
Back
307
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
308
16.6 Equilibrium Constant in Terms of Partial Pressures (SB p.102)
N2O4(g) 2NO2(g)
The expression of Kp is:
Substituting the value of Kp and the partial pressure of N2O4(g) into th
e expression,
= Kp ×
= 0.133 × 2.71
= 0.360 atm2
∴ = 0.600 atm
133.0P
PK
42
2
ON
2
NO
p
2
NO 2
P42ONP
2NOP Back
309
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
310
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
Mole fraction of H2(g) =
= 0.1
Mole fraction of I2(g) =
= 0.1
mol 1.6) 0.2 (0.2mol 0.2
mol 1.6) 0.2 (0.2mol 1.6
311
16.6 Equilibrium Constant in Terms of Partial Pressures (SB p.102)
Mole fraction of HI(g) =
= 0.8
Let P be the total pressure of the system.
= 64
mol 1.6) 0.2 (0.2mol 1.6
22 IH
2
HIp PP
PK
)P1.0)(P1.0(P)(0.8 2
Back
312
16.7 Equilibrium Position (SB p.103)
Determine the equilibrium constant (Kc) from the following data on the equilibrium system2SO2(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
313
16.7 Equilibrium Position (SB p.103)
The expression of Kc is:
From experiment 1:
= 3.94 dm3 mol-1
From experiment 2:
= 3.97 dm3 mol-1
Since Kc is a constant at a specific temperature, the values of Kc fro
m experiments 1 and 2 are very close, and the average value of Kc a
t 873 K is 3.955 dm3 mol–1.
eqm2
2
eqm2
2
eqm3
c )]g(O[)]g(SO[
(g)][SOK
30.160.13.62
K2
2
c
50.072.000.1
K2
2
c
Back
314
16.7 Equilibrium Position (SB p.104)
For the following reversible reaction:
CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l)
Calculate the equilibrium constant (Kc) using the following data.
(Assume that the equilibrium is established in a container of 1 dm3.)
Answer
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
315
16.7 Equilibrium Position (SB p.104)
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
eqm2
2
eqm2
2
eqm3
c )]g(O[)]g(SO[
(g)][SOK
4.12dm mol 0.33dm mol 0.33dm mol 0.67dm mol 0.67
K33
33
c
316
16.7 Equilibrium Position (SB p.104)
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
Since Kc is a constant at a specific temperature, the average value of
Kc from experiments 1 and 2 is 4.07.
4.02dm mol 3.07dm mol 0.07dm mol 0.93dm mol 0.93
K33
33
c
Back
317
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.
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?
30water[X]
neethoxyetha[X]
DK
Answer
318
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, the
n the mass of X left in water is (3.1 – a) g.
a = 3.0
3
neethoxyetha cm g 50a
[X]
3
water cm g 50
a-3.1[X]
50a - 3.1
50a
K D
50a - 3.1
50a
30 Back
319
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.
Answer
320
16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.106)
= 1.08 g cm-3
= 0.12 g cm-3
= 9
3ether cm 100g 108
]Y[
3water cm 50g 6
]Y[
water
etherD [Y]
[Y]K
3-
-3
cm g 0.12cm g 1.08
321
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 = 9
9 g of Y can be extracted using 100 cm3 of fresh ether.
100m - 10
100m
KD
100m - 10
100m
9
322
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 = 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
100
m - 1050
m
K1
1
D
100
m - 1050
m
91
1
323
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 = 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
100
m - 1.81850
m
K2
2
D
100
m - 1.81850
m
92
2
324
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
325
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:
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.
15)cm (g waterin Aof ionConcentrat
)cm (g thanetrichloroe-1,1,1 in Aof ionConcentratK
3-
-3
D
Answer
326
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 1,1,1-trichlor
oethane, then the mass of A left in 60 cm3 of aqueous layer is (6
– m).
m = 5.77 g
5.77 g of A is extracted using 100 cm3 of 1,1,1- trichlor
oethane.
60m6
100m
KD
60m6
100m
15
Back
327
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
328
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 th
e fact that they have different partition between the stationa
ry phase and the mobile phase. Therefore, they move upwa
rds to different extent. The Rf value is not related to the mas
s of the solute.
329
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
330
16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.110)
(b)
Back
331
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
332
16.9 Significances of Equilibrium Constants (SB p.111)
(a)
= 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 reache
d.
3
22
2
3c )]g(H)][g(N[
)]g(NH[Q
3
2
002.0001.0001.0
333
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
334
16.9 Significances of Equilibrium Constants (SB p.111)
(b)
= 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 reach
ed.
3
22
2
3c )]g(H)][g(N[
)]g(NH[Q
3
2
08.01001.0
Back
335
16.10 Factors Affecting Equilibrium (SB p.113)
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.
Back
336
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.
Answer(a) H2(g) + I2(g) 2HI(g)
3.50)22.0)(22.0(
)56.1()]g(I)][g(H[
)]g(HI[K
2
22
2
c
337
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
338
16.10 Factors Affecting Equilibrium (SB p.114)
(b) H2(g) + I2(g) 2HI(g)
At eqm: 0.22 mol dm-3 0.22 mol dm-3 1.56 mol dm-3
Now: (0.22 + 0.2) mol dm-3 0.22 mol dm-3 1.56 mol dm-3
Since the value of the reaction quotient (Qc) is less than that of the e
quilibrium constant (Kc), the system is not at equilibrium. In order to r
e-establish the equilibrium, the value of the reaction quotient should
be increased until it equals Kc. It can be predicted that more H2(g) a
nd I2(g) will react to form more HI(g).
34.26)22.0)(42.0(
)56.1()]g(I)][g(H[
)]g(HI[Q
2
22
2
c
339
16.10 Factors Affecting Equilibrium (SB p.114)
(b) H2(g) + I2(g) 2HI(g)
Now: (0.22 + 0.2) mol dm-3 0.22 mol dm-3 1.56 mol dm-3
At eqm: (0.42 - x) mol dm-3 (0.22 – x) mol dm-3 (1.56 + 2x) mol dm-3
Since Kc remains constant, we obtain:
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
)x22.0)(x42.0()x256.1(
)]g(I)][g(H[)]g(HI[
K2
22
2
c
)x22.0)(x42.0()x256.1(
2.502
Back
340
16.10 Factors Affecting Equilibrium (SB p.115)
Consider the following reaction at equilibrium:
2CrO42-(aq) + 2H+(aq) Cr2O7
2-(aq) + H2O(l)
Explain the changes of the graph at time t0, t1, t2 and t3 respectively.
Answer
341
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 equilibriu
m between them is established. At t2, when more H+(aq) is adde
d 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 f
orm more Cr2O72–(aq) and H2O(l).
Back
342
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.
343
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. Answer
(a)
Let the partial pressure of NO2(g) at equilibrium be p atm, then t
he partial pressure of N2O4(g) at equilibrium is (1 – p) atm.
p = 0.632 or –1.719 (rejected)
PNO2 = 0.632 atm
PN2O4 = (1 – 0.632) atm = 0.368 atm
eqm2
NO
eqmON
p )P(
)P(K
2
42
2pp1
92.0
344
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. Answer(b) Using the same method as in (a),
p = 1.028 or –2.115 (rejected)
PNO2 = 1.028 atm
PN2O4 = (2 – 1.028) atm = 0.972 atm
2pp2
92.0
345
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 tha
n doubled while PNO2 is less than doubled when the tot
al pressure increases from 1 atm to 2 atm. Thus, the e
quilibrium position shifts to the side with a smaller num
ber of molecules when the pressure increases.
346
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
(d) The impact of the increased pressure is reduced by shifti
ng the equilibrium position to the right-hand side of 2NO2
(g) N2O4(g). More NO2(g), which is brown in colo
ur, is used up. More N2O4(g), which is colourless, is forme
d. A colour change from brown to pale brown (or colourle
ss) can be observed.
347
16.10 Factors Affecting Equilibrium (SB p.117)
(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.
Answer(e) When the equilibrium mixture is put into hot water (the temperature
increases), the equilibrium will shift to the left and more NO2(g) will
be formed. Thus, the colour of the mixture will change to a darker
brown. When the equilibrium mixture is put into ice water (the temp
erature decreases), the equilibrium will shift to the right and more
N2O4(g) will be formed. As a result, the colour of the mixture will ch
ange to pale brown (or colourless).
Back
348
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. Answer(a)
)g(OH
3
)g(NO
)g(NO
2
)g(HNO
p
22
3
PP
PPK
349
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)?
Answer
(b) (i) Decrease
(ii) Increase
(iii) Increase
350
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?
Answer(c) The value of Kp will remain the same.
351
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. Answer
(a) atm 0.900P
PK
2
NO
ON
p
2
42
352
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
353
16.10 Factors Affecting Equilibrium (SB p.118)
(b) After compression to one-half the original volume, all t
he gas pressures will be doubled. Therefore, the partia
l pressures of N2O4(g) and NO2(g) will be 0.728 atm an
d 1.272 atm respectively.
2NO2(g) N2O4(g)
atm 0.450P
PQ
2
NO
ON
2
42
354
16.10 Factors Affecting Equilibrium (SB p.118)
(b) Since the value of the reaction quotient is less than tha
t of the equilibrium constant, the system is not at equili
brium. 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
2p )x2272.1()x728.0(
90.0K
355
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
356
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 Answer
357
16.10 Factors Affecting Equilibrium (SB p.120)
(a) The equilibrium position shifts to the left.
(b) The equilibrium position shifts to the left.
(c) The equilibrium position remains unchanged.
(d) The equilibrium position shifts to the right.
(e) The equilibrium poistion shifts to the left.
Back
358
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) Answer
359
16.10 Factors Affecting Equilibrium (SB p.121)
At 673 K,
At 773 K,
Combining (1) and (2),
H = 105 329 J mol-1
= -105.3 kJ mol-1
)1.........(67331.8
Httancons)106.1ln( 4
6424H
)104.1ln(5593
H)106.1ln( 54
)2.........(77331.8
Httancons)104.1ln( 4
Back
360
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) Answer
361
16.10 Factors Affecting Equilibrium (SB p.122)
+9.791.67 10-3
+7.442.00 10-3
+3.872.50 10-3
+1.362.86 10-3
-2.163.36 10-3
ln Kp1/T (K-1)
362
16.10 Factors Affecting Equilibrium (SB p.122)
A graph of ln Kp against produces a straight line with
slope .
T1
RH
363
16.10 Factors Affecting Equilibrium (SB p.122)
Slope =
= -7121.2
= -7121.2
H = 7121.2 8.314
= 59 206 J mol-1
= 59.2 kJ mol-1
310)67.100.2(79.944.7
RH
Back
364
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.
Answer
365
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.
366
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
367
16.10 Factors Affecting Equilibrium (SB p.122)
(b) The use of high pressure is as predicted in (a). This not only shifts
the equilibrium position to the right but also increases the rate of the
reaction. The use of catalysts shortens the time for the reaction to
reach the equilibrium while it has no effect on the equilibrium
constant.
The use of a high temperature is contradictory to the
prediction made in (a). It can be explained based on the rate of the
reaction which in turn determines the rate of manufacture of
ammonia. Although the equilibrium position shifts to the right at a
lower temperature, the rate of the reaction is very low (i.e. a longer
time is required to reach the equilibrium state). The use of a
moderate temperature is a compromise between the rate and the
yield of the reaction. At 450 °C, the reaction is reasonably fast and
the yield of ammonia is optimum.
Back