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


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



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.




13


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)




14


Interpretation : -


rate of backward rx = rate of forward rx




15

1.When H2O(l) is added to BiCl3(aq)

Observation : -

The colourless solution turns milky.


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




17

2. When HCl(aq) is added to BiOCl(s)

Observation : -

The milky solution becomes clear.



18

Interpretation : -



[HCl(aq)]

rate of backward rx > rate of forward rx

a net consumption of BiOCl(s)


19

Interpretation : -



There is no further change when



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


Interpretation : -

Before the addition,


Q.3Q.3

Red-orange


22


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


23


Interpretation : -

There is no further change of colour when


Q.3Q.3

Red-orange


24

2. When HCl(aq) is added

Prediction : -

The colourless solution turns red-orange.

Q.3Q.3

Red-orange


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



26

Interpretation : -



Q.3Q.3

Red-orange



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


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



29

Interpretation : -



Q.3Q.3

Red-orange



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)]


a net production of the red form

34

Colourless Red

+ H3O+

(aq)



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


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


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’


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


Qc = Kc

the system is at equilibrium

64

aA + bB cC + dD

b[B]a[A]

d[D]c[C] Q

cReaction

quotient


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


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


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


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


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


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:


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


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


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



2

K K

cK of value Average 2c

1c

24.08 4.05

= 4.065 (no unit)

81


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 :


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)


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


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)


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)


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)


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)


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)


100

Homogeneous Equilibrium

CH3COOH(l) + C2H5OH(l)


CH3COOH(aq) + C2H5OH(aq)


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)


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


[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


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.


Hexane

Water

124

I2 dissolves in both solvents to different extent.


Hexane

Water

125

When dynamic equilibrium is established,

rate of movement = rate of movement


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


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


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)


(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


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


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.


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

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)


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)


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

Two moles

The forward reaction lowers the pressure of the system.

199

A(g) + B(g) C(g)

Q.22(f)


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


Q.23(a)/(b)





209

Q.23(c)/(d)


Cause : in PT by increasing 5PClP



Cause : in PT by increasing2ClP

210

Q.23(e)




211

Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(g)

Q.23(a)/(b)





212

Q.23(c)/(d)


Cause : in PT by increasing 2HP


Cause : in PT by increasing OH2P

Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(g)

213

Q.23(e)



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


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


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


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


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


colourless

white ppt

Kinetic approach : -

Both the rates of forward and backward reactions increase but the backward reaction increases more.

A net backward reaction is observed

The equilibrium position shifts to the left

227


Addition of HCl(aq)

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


colourless

white ppt

Test 2 : Add large excess of H2O

Result : The white ppt reappears

The equil. position shifts to the right

229


colourless

white ppt

Interpretation : Kc approach

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


colourless

white ppt

Interpretation : Kinetic approach

[H2O(l)]


equilibrium position shifts to the right

231


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


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


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


right

No effectNo effect

Effect on Kp


positionChanges

Doubling PCO and

OHCH3P

OHCH3P

Doubling and 2HP

276

CO(g) + 2H2(g) CH3OH(g)H < 0

Q.33


rightA little H2O(l) is

added

No effectNo effectA positive catalyst is

added

Effect on Kp


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


positionChanges

278

A(g) + B(g) C(g)H = 0Q.34


left in T at fixed VT


right in T at fixed VT

Effect on Kp


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


Increase in concentration of reactants A or B

Shifts to right No change

Increase in concentration of products C or D

Shifts to left No change


280



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



Isothermal compression

281



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



282



Addition of a catalyst

No change No change



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


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



(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



(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


(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


(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


(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


(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


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


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


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


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


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


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


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.


What is the implication for an equilibrium reaction having an equilibrium constant much smaller than

1.0?

Back

Answer

303


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


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


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


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


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


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


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


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


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


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


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


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


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


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


= 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


(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


(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


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


∴ Total mass of Y extracted = m1 + m2

= (8.182 + 1.487) g

= 9.669 g

Back

325


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


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


(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


(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


(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


(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


(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


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


(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


(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


(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


(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


(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


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


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


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


(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


(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


(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


(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


(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


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



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



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


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


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


(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


(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


(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


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


(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


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


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


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


+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


A graph of ln Kp against produces a straight line with

slope .

T1

RH

363


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


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


(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


(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


(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.

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1 Irreversible and Reversible Reactions. 2 Irreversible Reactions Chemical reactions that take place...

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