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S.K.P. Engineering College, Tiruvannamalai I SEM

Chemistry Department 1 Engineering Chemistry - I

SKP Engineering College

Tiruvannamalai – 606 611

A Course Material

on

Engineering Chemistry - I

By

Dr.R.Rajmohan,

Professor.

Chemistry Department



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Subject Code: CY6151

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CY6151 ENGINEERING CHEMISTRY - I L T P C 3 0 0 3

UNIT I POLYMER CHEMISTRY 9

Introduction: Classification of polymers – Natural and synthetic; Thermoplastic and

Thermosetting. Functionality – Degree of polymerization. Types and mechanism of

polymerization: Addition (Free Radical, cationic and anionic); condensation and

copolymerization. Properties of polymers: Tg, Tacticity, Molecular weight – weight

average, number average and polydispersity index. Techniques of polymerization: Bulk,

emulsion, solution and suspension. Preparation, properties and uses of Nylon 6,6, and

Epoxy resin.

UNIT II CHEMICAL THERMODYNAMICS 9

Terminology of thermodynamics - Second law: Entropy - entropy change for an ideal

gas, reversible and irreversible processes; entropy of phase transitions; Clausius

inequality. Free energy and work function: Helmholtz and Gibbs free energy functions

(problems); Criteria of spontaneity; Gibbs-Helmholtz equation (problems); Clausius-

Clapeyron equation; Maxwell relations – Van’t Hoff isotherm and isochore(problems).

UNIT III PHOTOCHEMISTRY AND SPECTROSCOPY 9

Photochemistry: Laws of photochemistry - Grotthuss–Draper law, Stark–Einstein law

and Lambert- Beer Law. Quantum efficiency – determination- Photo processes -

Internal Conversion, Inter-system crossing, Fluorescence, Phosphorescence,

Chemiluminescence and Photo-sensitization. Spectroscopy: Electromagnetic spectrum

- Absorption of radiation – Electronic, Vibrational and rotational transitions. UV-visible

and IR spectroscopy – principles, instrumentation (Block diagram only).

UNIT IV PHASE RULE AND ALLOYS 9

Phase rule: Introduction, definition of terms with examples, One Component System-

water system - Reduced phase rule - Two Component Systems- classification – lead-

silver system, zinc-magnesium system. Alloys: Introduction- Definition- Properties of

alloys- Significance of alloying, Functions and effect of alloying elements- Ferrous

alloys- Nichrome and Stainless steel – heat treatment of steel; Non-ferrous alloys –

brass and bronze.

UNIT V NANOCHEMISTRY 9

Basics - distinction between molecules, nanoparticles and bulk materials; size-

dependent properties. nanoparticles: nano cluster, nano rod, nanotube(CNT) and

nanowire. Synthesis: precipitation, thermolysis, hydrothermal, solvothermal,



electrodeposition, chemical vapour deposition, laser ablation; Properties and

applications

TOTAL :45 PERIODS

TEXT BOOKS

1. Jain P.C. and Monica Jain, “Engineering Chemistry”, Dhanpat Rai Publishing

Company (P) Ltd.,

New Delhi, 2010

2. Kannan P., Ravikrishnan A., “Engineering Chemistry”, Sri Krishna Hi-tech Publishing

Company

Pvt. Ltd. Chennai, 2009

REFERENCES

1. Dara S.S, Umare S.S, “Engineering Chemistry”, S. Chand & Company Ltd., New

Delhi 2010

2. Sivasankar B., “Engineering Chemistry”, Tata McGraw-Hill Publishing Company, Ltd.,

New Delhi, 2008.

3. Gowariker V.R. , Viswanathan N.V. and JayadevSreedhar, “Polymer Science”, New

Age

International P (Ltd.,), Chennai, 2006.

4. Ozin G. A. and Arsenault A. C., “Nanochemistry: A Chemical Approach to

Nanomaterials”, RSC Publishing, 2005.



CONTENTS

S.No Particulars Page

1 Unit – I 6

2 Unit – II 22

3 Unit – III 49

4 Unit – IV 74

5 Unit – V 98



Unit – I

Polymer Chemistry

Part - A

1. How are polymers classified on the basis of their tacticity? [CO1-L2-A.U Jan

2012]

a. Isotactic polymers

Here functional groups are arranged on the same side of the main chain. b. Syndiotactic polymers

Here functional groups are arranged in an alternating fashion. c. Atactic polymers

Here functional groups are arranged randomly. 2. What are thermosets? [CO1-L1-A.U June 2013]

Thermosets are plastics obtained by condensation polymerisation. Various

polymer chains are held together by strong covalent bonds (called crosslinks).

They get harden on heating and once harden, they cannot be softened again.

They are almost insoluble in organic solvents. 3. What are the factors affecting Tg [CO1-L2-A.U Jan 2015]

1. The value of Tg depends on (a) chain-length (b) extent of cross-linking (c) the barrier which hinders the internal rotation of the chain links.

2. The value of Tg of a given polymer varies with the rate of heating or cooling. 3. Below Tg the polymer is hard and brittle (like glass). 4. Tg of a liner polymer is sharp, because of the free movement of polymer

chains (chains are held by weak vanderWaals forces). 5. A cross-linked polymer does not possess any Tg because polymer chains

will not move. 4. What is number - average molecular mass? [CO1-L1]

It is the mass obtained by dividing the total mass of the polymer material with the

total number of molecules present. Mathematically it (Mn) is defined as n1M1 + n2M2+ n3M3 ....



Mn

5. What is poly dispersity index? [CO1-L1-May 2009, July 2010]

The ratio of the weight - average molecular weight Mw to that of number-average molecular weight Mn is known as polydispersity index (PDI).

PDI = Mw Mn

6. What are the advantages of solution polymerisation? [CO1-L1]

1. Heat control is easy.

2. Viscosity built up is negligible.

3. The mixture can be agitated easily. 7. What are the disadvantages of suspension polymerisation. [CO1-L1]

1. This method is applicable only for water insoluble monomers.

2. Control of particle size is difficult.. 8. What are the properties of epoxy resins? [CO1-L1]

1. They possess high chemical-resistance to water, acids, alkalis, various

solvents and other chemicals.

2. They are flexible, tough and possess very good heat resisting property. They possess excellent adhesion quality.

9.Define. Degree of polymerization. [CO1-L1-Dec 2014]

Numbers of monomers repeatly present in the polymers is called as Degree of polymerization. Example: Polystyrene containg n Numbers of styrene monomers present.n=1.2.3……etc

10. What are the Uses of Nylon 6,6. [CO1-L1]



(i) Nylon-6:6 is used for fibers, which are used in making socks, ladies shoes,

dresses, carpets, etc,.

(ii) Nylon-6 is mainly used for moulding purposes for gears, bearings, electrical

mountings, etc,. Nylon bearings work without any lubrication.

(iii) Nylons are used for making filaments for ropes, bristles for tooth-brushes, films

and tyre-cords.

Part - B

1. Classify the polymers with suitable examples.

[CO1-L2-Jan 2012]

Based on the source, polymers are classified into two types.

1. Natural polymers.

2. Synthetic polymers.

1. Natural polymers

The polymers obtained from nature (plants and animals) are called natural polymers.

Examples of natural polymers

(a) Starch

It is a polymer of glucose. It is a chief food reserve of plants.

(b) Cellulose

It is also a polymer of glucose. It is a chief structural material of the plants.

Both starch and cellulose are produced by plants during photosynthesis.

(c) Proteins



These are polymers of amino acids. They have generally 20 to 100 amino acids joined

together in a highly organized arrangement. These are building block of animals

d) Nucleic acids

These are polymers of various nucleotides. RNA and DNA are common nucleotides.

(e) Natural rubber

It is a polymer of unsaturated hydrocarbon, 2-methyl-1, 3 butadiene, called isoprene.

2. Synthetic polymers

The polymers which are prepared in the laboratories are called synthetic polymers.

These are also called man-made polymers.

Examples: PVC, polyethylene, nylon, teflon, etc.,

Types of synthetic polymers

(a) Organic Polymers

These are polymers containing hydrogen, oxygen, nitrogen, sulfur and halogen atoms

apart from carbon atoms.

Examples: Polyethylene, Polyvinyl alcohol, etc.,

(b) Elemento-organic (or) Hetero-organic Polymers

These are polymers composed of carbon atoms and hetero-atoms (like N, S & O). The

main chain consists of carbon atoms and whose side groups contain hetero atoms

linked directly to the C atoms in the chain.

Examples: Polysiloxanes, polytitoxanes.

(c) Inorganic Polymers

These are polymers containing no carbon atoms. The chains of these polymers are

composed of different atoms joined by chemical bonds.

Examples



Poly Silicon dioxide, Polyphosphoric acid:

2. Distinguish between thermoplastics and thermosetting plastics. [CO1-L2-June

2013]

Sl.No. Thermoplastic resins Thermosetting resins 1. They are formed by They are formed by

addition polymerisation. condensatin

polymerisation.

2. They consits of linear They consist of three

Longchainpolymers. dimentional network

structure.

3. All the polymer chains are

All the polymer chains are

Held together by weak linked by strong covalent

vanderWaals forces. bonds.

4. They are weak, soft and They are strong, hard and

less brittle. more brittle.

5. They soften on heating They do not soften on and harden on cooling. heating.

6. They can be remoulded. They cannotbe remoulded.

7. They have low molecular They have high molecular

weights. weights.

8. They Are solube in They are insoluble In organic solvents. organic solvents.

3. Describe the free radical mechanism of addition polymerization with a suitable

example. [CO1-L2-June 2013]



1. Initiation

It is considered to involve two reactions.

(a) First reaction involves production of free radicals by homolytic dissociation of an

initiator (or catalyst) to yield a '

I 2R

Initiator Freeradicals

Example of a commonly used thermal initiator Thermal initiator is a substance used to produce free radicals by homolytic

dissociation at high temperature.

70 90C

CH3COO OOC CH3 2CH3COO or 2R

Acetyl peroxide Free radicals

(b) Second reaction involves addition of this free radical to the first monomer to

produce chain initiating species.

H H

R

C

CH C R CH

Free radical 2 2

Y Y

(First monomer) chain initiating species

2. Propagation

It involves the growth of chain initiating species by successive addition of large

number of monomers.

H H H H

R CH C n CH C R — CH C — CH C

2 2 2 n 2

Y Y Y Y



The growing chain of the polymer is known as living polymer. 3. Termination Termination of the growing chain of polymer may occur either by coupling reaction

or disproportionation. (a) Coupling (or) Combination It involves the coupling of free radical of one chain end to another free radical

forming a macromolecule. H H H H

C

C CH R R CH

R CH C C CH R

2

2 2

2

Y Y Y Y

(b) Disproportionation

It involves transfer of a hydrogen atom of one radical centre to another radical centre, forming two macromolecules, one saturated and another unsaturated.

H H H H H H H H

R C

C

C C R

R C C H C C

R

H Y Y H Y Y H Unsaturated Saturated

macromolecule macromolecule

(Dead polymers)

The product of addition polymerisation is known as Dead polymer.

4. Distinguish between addition and condensation polymerization with an

example each. [CO1-L2-June 2012, June 2010,Jan 2011]

S.No.

Addition/chain Condensation/step

polymerisation polymerisation

1. The monomer must have The monomer must have

At least one multiple bond

At least two identical (or)



different functional groups.

Example: Example:

(i) Ethylene: CH2 = CH2 (i) Glycol: CH2 =OH

1

CH2 - OH

(ii) Acetylene: CH - CH (ii) 6-aminohexanoic acid

H2N (CH2)5 - COOH

2. Monomers add on to give Monomes condense to

a polymer and no other by

give polymer and

product isformed. byproducs such as

H2O, CH3OH are formed.

3. Number of monomeric Monomers disappear at the

uns decreases steadily early stage of reaction. throughut the reaction.

4. Moleculr weight of the Moleculr weight of the polymer is an integral polymer need not be an multiple of molecular integral multiple of

weight of monomer. monomer.

5. Hih molecular weight Moleculr weight of the

polymer is formed at once. polymer rises steadily

throughout the reaction.

6. Longer reaction times give

Longer reaction times are

higher yield, but have a essential to obtain high litte effect on molecular moleculr weight.

weight.

7. Thermoplastis are Thermosettig plastis are

produced.

produced.

Example polyethylene, Example Bakelite,



: :

PVC etc. urea-formaldehyde.

8. Homo-chain polymer is Hetero-chain polymer is obtained. obtained.

5. Brief about the following propertis of the polymers.

Tg, weight average molecular weight, tacticity and

polydispersi index. [CO1-L2- Dec 2012]

(i) Tg

Glass ransition temperature Tg is the temperature at

which the amorphous solid state is transformed to the melt state.

Below the glass transition temperature Tg the polymer is hard and above which it

is soft. The hard brittle state is called as glassy state and the soft flexible state is

called as rubbery or viscoelastic state. Thus the glass transition temperature is

an important property of a polymer and it decides whether the polymer behave

like glass or rubber.

Factors influencing Tg

(

a

)

1. chain-length (b) extent of

cross-linking (c)

the barrier which hinders the internal

rotation of

t

h

e chain links.

2. The value of Tg of a given polymer varies with the rate of heating or cooling.

3. Below Tg the polymer is hard and brittle (like glass).



4. A cross-linked polymer does not possess any Tg because polymer chains will not

move. Thus the polymers with a symmetrical structure are crystalline, while the

polymers with irregular chain back bone are amorphous.

(ii) Tacticity

The orientation of monomeric units or functional groups in a polymer molecule

can take place in an orderly (or) disorderly manner with respect to the main chain

is known as Tacticity. Tacticity do affect the physical properties of the polymer.

This orientation results in three types of stereo-regular polymers.

1. Isotactic polymer

If the functional groups are arranged on the same side of the main chain, the

polymer is called Isotactic polymer.

Example: Polystyrene

2. Syndiotactic polymer

If the functional groups are arranged in an alternating fashion, the polymer is

called Syndiotactic polymer.

3. Atactic polymer

If the functional groups are arranged randomly, the polymer is called Atactic



polymer.

(iii) Weight average molecular mass

In the averaging process, molecular weight of each individual species in multiplied

by the mass and not by the number. i.e.,

m M m M 2 m M

3

....

M 1 1 2 3

w

m1

m2

m3

Where,

m1 Mass of the monomer.

M1 Molecular weight of the monomer. Determination

Weight - average molecular mass is determined by light - scattering techniques and ultra-centrifugation techniques. (iv) Polydispersity index

The ratio of the weight - average molecular weight Mw to that of number-average molecular weight Mn is known as polydispersity index (PDI).

PDI Mw

Mn 6. Write notes on bulk, emulsion, solution and suspension polymerization

techniques. [CO1-L2-Dec 2012] 1. Bulk polymerisation

Bulk polymerisation is the simplest method of polymerisation. The monomer is taken in a flask as a liquid form and the initiator, chain transfer agents are dissolved in



it. The flask is placed in a thermostat under constant agitation and heated. Monomer +Initiator + Chain transfer agent ------------- Polymer

(Liquid) (mixed

with Monomer)

After a known period of time, the whole content is poured into a methanol (non-solvent) and the polymer is precipitated out. Advantages

1. It is quite simple and requires simple equipments.

2. Polymers are of high-purity obtained. Disadvantages

1. Mixing and control of heat is difficult.

2. Polymerisation is highly exothermic. 2. Suspension polymerisation

The water insoluble monomer is suspended in water as tiny droplet and a initiator

is dissolved in it by continuous agitation. The suspension (droplets) is prevented from

coagulation by using suspending agents like PVA, gelatin, methyl cellulose. Each

droplet of the monomer contains dissolved initiator. The whole content is taken in a flask and heated at constant temperature with vigorous agitation in a thermostat with nitrogen

atmosphere. After the end of 8 hrs, pearl-like polymers are obtained, which is filtered

and washed by water.

Monomer + Initiator + suspending agent --------------- Polymer

(Suspens

ion (Dissolved

(Suspended

in

in water) in monomer)

water as

beads)

Advantages

1. Products obtained is highly pure.

2. Isolation of product is very easy.



Disadvantages

1. This method is applicable only for water insoluble monomers.

2. Control of particle size is difficult.

Applications

1. Polystyrene beads are used as ion exchangers.

2. This technique is used in heterogeneous system.

Solution polymerisation

In solution polymerisation, the monomer, initiator and the chain transfer agents are

taken in a flask and dissolved in an inert solvent. The whole mixture is kept under

constant agitation. After required time, the polymer produced is precipitated by pouring

it in a suitable non-solvent.

Monomer Initiator

Chain transfer agent

Polymer

Dissolved in Dissolved in In solution inert solvent inert solvent

The solvent helps to control heat and reduces viscosity built up.

Examples: Poly acrylic acid, poly isobutylene and poly acrylo nitrile are prepared by

this method.

Advantages

1. Heat control is easy.

2. Viscosity built up is negligible.

3. The mixture can be agitated easily.

Disadvantages

1. The removal of last traces of solvent is difficult.



2. This polymerization requires solvent recovery and recycling.

3. It is difficult to get very high molecular weight polymer.

4. The polymer formed must be isolated from the solution either by evaporation of

the solvent or by precipitation in a non-solvent.

Applications

As the polymer is in solution form, it can be directly used as adhesives and coatings.

Emulsion polymerisation

Emulsion polymerisation is used for water insoluble monomer and water soluble initiator

like potassium persulphate. The monomer is dispersed in a large amount of water and

then emulsified by the addition of a soap. Then initiator is added. The whole content is

taken in a flask and heated at a constant temperature with vigorous agitation in a

thermostat with nitrogen atmosphere. After 4 to 6 hours, the pure polymer can be

isolated from the emulsion by addition of de-emulsifier like 3% solution of Al2 (SO4)3.

Monomer Initiator

Surfactant

Polymer

Dissolved in Water Emulsion inert solvent soluble in water

Advantages

1. The rate of polymerisation is high.

2. Heat can be easily controlled and hence viscosity built up is low.

3. High molecular weight polymer can be obtained.

Disadvantages

1. Polymer needs purification.

2. It is very difficult to remove entrapped emulsifier and de-emulsifiers.

3. It requires rapid agitation.



Applications

1. Emulsion polymerisation is used in large-scale production like water-based

paints, adhesives, plastics, etc.,

2. This method is also suitable for manufacturing tacky polymers like butadiene and

chloroprene.

3. Write the synthesis, properties and uses of nylon-6,6. [CO1-L2-Nov 2011, Jan 2010,Jan 2010] Preparation It is obtained by the polymerisation of adipic acid with hexamethylenediamine.

>

n [ H2N CH26 NH2 HOOC CH2 4

COOH ]

Hexamethylene diamine Adipic acid

O O

H H

N CH2 6 N C CH2 4 C n 2n H2O

Nylon 6:6 Properties of Nylon-6:6

(i) Nylon-6:6 is a less soft and stiff material when compared to nylon 6.

(ii)

(iii) It behaves as plastic as well as fiber.

(iv) It is translucent, white, horny and high melting polymers.

(v) It is insoluble in common organic solvents but soluble in phenol and formic acid.

(vi) It possesses good mechanical properties and fairly resistant to moisture.

(vii) It is characterised by combination of high strength, elasticity, toughness and

abrasion resistance. Uses of Nylons

(iii) Nylon-6:6 is used for fibers, which are used in making socks, ladies shoes, dresses, carpets, etc,.

(iv) Nylon-6 is mainly used for moulding purposes for gears, bearings, electrical

mountings, etc,. Nylon bearings work without any lubrication.



(iv) Nylons are used for making filaments for ropes, bristles for tooth-brushes, films and tyre-cords.

8. Write notes on epoxy resin.

[CO1-L1-Dec 2002, May 2001, Nov 2001]

These are cross-linked thermosetting resins. They are polyethers because the monomericunits

in the polymer have (v)

an ether type of structure i.e., R - O -= R.

Preparation

Epoxy polymer (or) Epoxy resins are prepared by condensing epichlorohydrin with

bisphenol.

The reactive epoxide and hydroxyl groups give a three dimensional cross-linked

structure. The value of n ranges from 1 to 20.

Properties



1. Due to the presence of stable ether linkage, epoxy resin possesses high

chemical-resistance to water, acids, alkalis, various solvents and other

chemicals.

2. They are flexible, tough and possess very good heat resisting property.

3. Because of the polar nature of the molecules, they possess excellent adhesion

quality.

Uses

1. Epoxy resins are used as surface coatings, adhesives like araldite, glass-fibre-

reinforced plastics.

2. These are applied over cotton, rayon and bleached fabrics to impart crease-

resistance and shrinkage control.

3. These are also used as laminating materials in electrical equipments. Moulds

made from epoxy resins are employed for the production of components for

aircrafts and automobiles.

Unit-II

Chemical Thermodynamics

Part A

1. State. Second law of thermodynamics. Represent by a mathematical equation.

[CO2-L2-Dec 2013]

It is impossible to construct a machine, which will transfer heat from a lower

temperature to a higher temperature.

It is mathematically stated as

dS dq

Trev

orS q

Trev

2. Define the term ‘standard free energy’. Illustrate with an example. [CO2-L2- May

2004]



are converted into the products in their standard states.” Thus

G

Gproducts

Greactants

The value of Delta Go can be calculated for a reaction from the standard free energies

Delta Gof

3. Calculate the change in entropy accompanying the isothermal expansion of 4

moles of an ideal gas at 300 K until its volume has increased three times. [CO2-

L3-Nov 2002]

Solution

Entropy change in an S 2.303 nR

log V2

isothermal expansion of an ideal gas

V

1

Given

V1 1, V2 3 ; n 4 ; R 1.987

S 2.303 4 1.987 log 3

1

8.733 cals . 4. Define Spontaneity. [CO2-L1-Dec 2001, Dec 2004]

It is defined as the tendency of a process to occur naturally is called the

spontaneity. 5. How does the entropy of system changes

(a) When a gas is liquefied (b) Ice is melted [CO2-L2- Dec 2003]

(a) When a gas is liquefied, entropy decreases.

(b) When ice is melted, entropy increases.

6. Predict whether the following reaction is spontaneous at



25C C

s

H

2O

l CO

g

H

2 g

H 31.4 kcal/mole and S 32 cal/deg at

25C [CO2-L2-June 2002] We know that,

G H TS

Given

H 31.4 k.cal; S 32 cals (or) 0.032 k.cals

T 25 273 298 K

on substituting these values in the above equation

G H TS

31.4 2980.032 31.4 9.536

G 21.9 k.cals .

G is positive the reaction is non-spontaneous. 7. What is the significance of free energy? [CO2-L2-May 2014, Dec 2015]

The decrease of free energy G ss at constant temperature and

pressure is equal to the useful work obtainable from the system. 8. Write Gibbs-Helmholtz equation? What is its application [CO2-L2- May 2003.

Dec2001] Gibbs-Helmholtz equation is

G H (or)

G

G T G H T

T

TP P

(i) Enthalpy change H for the cell reaction can be calculated.

(ii) Entropy change S can be calculated.



(iii) It is used to calculate H from values of free energy change at two different

temperature. 9. Give the relation between (i) H & G, (ii) Emf &

G. [CO2-L2-Dec.2001]

G H G T (ii)Gn FE

T

P

10. Give the reaction between the variations of equilibrium constant with temperature. [CO2-L2]

2 1 H T2 T1

log KP log KP

2.303R T T

1 2

PART - B

1. Derive an expression for entropy change of an ideal gas at constant

temperature. [CO2-L2- Nov 2002]

According to first law of thermodynamics

dE q PdV

(or) dE q w ... (1) [ PdV w ]

In a reversible isothermal expansion, there is no change of internal energy dE 0.

Equation (1) becomes,

qrev w 0

qrev w ... (2)

The work done in the expansion of n moles of a gas from volume V1 to V2 at constant temperature T is given by

w n RT

ln

V2 ... (3)

V

1



when equation (3) is introduced in equation (2), we get

q

n RT

ln

V2

rev V1

(or) TS n RT ln V2

V

1

S n R ln V2

V1

(or) S 2.303 nR

log V2 ... (4)

V

1

(or), . . . P V

1 RT ; V RT

1 1 P1

P V 2 RT ; V

2 RT

2 P2

S

2.303, nR

log P1

P

2

2. Calculate the change in entropy accompanying the isothermal expansion of 5

moles of an ideal gas at 330 K until its volume has increased six times. [CO2-H1-Dec 2013]

Entropy change in an isothermal V2

expansion of an ideal gas

S 2.303 nR log

V1

is given by

If weight of the substance is given, then

mole, n

Wt. of substance

mol. wt. of substance

Given: V1 1; V2 6; n 5; R 1.987 cals (known value)

S 2.303 5 1.987 log 6

1



2.303 5 1.987 0.778

17.8 cals/K .

3. Calculate the change in entropy accompanying the isothermal expansion of 4 moles of an ideal gas at 300

Solution

V2

Entropy change in an S 2.303 nR

log

isothermal expansion of an ideal gas

V

1

Given

V1 1; V2 3 ; n 4 ; R 1.987

S 2.303 4 1.987 log 3

1

8.733 cals /K .

4. Define Gibbs and Helmholtz free energies. How are these two thermodynamic

functions related to the maximum and useful work obtainable from a system

K until its volume has increased three times.

[CO2-H2- Nov 2002]

during the given change. [CO2-L2- Nov, 2002, June 2013] 1. Helmholtz Work Function A (or) Helmholtz Free Energy

It is also known that a part of internal energy of a system can be used at

constant temperature to do some useful work. This part of E

A

mathematically defined as

A E TS

Significance of the Work Function A

The work function is given by

A E TS ... (1)

A E TS

B S

S

qrev

T



(or) TS qrev

substituting equation (3) in (2), we get

A E qrev

But, according to first law of thermodynamics

E q w

(or) E q w

A w

(or) A wmax

(or) A wmax

where,

wmax maximum work.

Thus, the decrease of work function A of a

process at constant temperature gives a maximum work obtained from the system.

2. Gibbs Free Energy G (or) Thermodynamic Potential We know that a part of the total energy of a system is converted into work and the rest is unavailable. The part of the energy which is converted into useful work is called available energy. The isothermally available energy present in a

G hematically defined as

Available Total

Unavailable

energy

energy

energy

G

H

TS

Significance of Gibbs Free Energy G

Free energy (or) Gibbs free energy G otal available energy

present in a reversible system at constant temperature and pressure as useful

work. It is given by

G H TS ... (1) For a small change in a reversible system at constant temperature

G H TS ... (2)



G wmax PV

(or) G wmax PV

where,

PV work done by expansion (mechanical work)

wmax maximum work that is obtained from the system

(or) wmax PV network or useful work.

Hence, G wuseful

Thus, the decrease of free energy G of a process at constant temperature and

pressure is equal to the useful work obtainable from the system.

5. Derive Gibb’s Helmoltz equation.(or) Derive Gibb’s

Helmoltz equation and discuss its applications. [CO2-L2- May-2002, May 2005, Dec 2004, Jan. 2005 Dec 2012)

Consider the following relations,

G H TS Gibbs free energy

H E PV enthalpy

G E PV TS

But we know that the enthalpy change H is given as

H E PV ... (3)

substituting equation (3) in (2) we get

G E PV

TS ... (4)

But, A E TS or E A TS

Hence equation (4) may be written as

G A PV ... (5)

Since A wmax



For infinitesimal change,

dG dE P dV V dP T dS S dT ... (1)

But according to first and second law of thermodynamics,

dE dq P dV first law

dq T dS second law

dE TdS P dV ... (2)

Substituting equation (2) in (1) we get,

dG T dS P dV

P dV V dP T dS S

dT

dG V dP S

dT ... (3) Substituting equation (2) in (1) we get,

dG T dS P dV

P dV V dP T dS S

dT

dG V dP S

dT ... (3)

At constant pressure dP 0, equation (3) becomes

dG S dT

G S

Substituting equation (5) in G H TS, we get

G G H T

T P

G H

This is one form of the Gibb’s - Helmholtz equation.

For any two states of the system the equation (4) may be written as

dG1 S1dT initial state

dG2 S2dT final state To get the change

dG2 dG1 S2dT S1dT dG2 G1S2 S1 dT



dGS dT ... (7)

At constant pressure the equation (7) becomes

G ... (8)

S

T

P

But according to definition of free energy

G H TS

(or) S G H ... (9)

T

Substituting equation (9) in (8), we

get

G H G

T

T

P

(or) G

G H T

T

P

G ... (10)

G H T

T

P

A

may be derived.

Similarly A E T

T

V

This is another form of Gibbs - Helmholtz equation. Applications 1. Calculation of Emf of the cell The equation (2) is divided by nF and it becomes

H E

T E

nF

T

P H

E

E T



nF T

2. Calculation of Entropy Change S

H and S are related by the equation

G H TS

We know that G nFE

H can be calculated from the equation (3). Hence S can be calculated easily

from the above equation (5). 3. Gibbs-Helmholtz equation is applicable for a process occurring at constant pressure.

It is used to calculate H from the values of free energy change at two different

temperature. 4. For a reaction at constant volume the equation can be modified as

A

A E T T

v

6. The equilibrium constant for the reaction

N H 2 2NH at 400C is 1.64 10

4

and at

2 3

500C is

1.44 10

4. Calculate the heat of formation

of

1 mole of ammonia from its elements within given range

of temperature. [CO2-L2-May 2003]

We know that

2 1 H T2 T1

log KP log K

P 2.303

R

T T

1 2

K1P 1.64 10

4 atm ; K

2P 1.44 10

4 atm

T1 400 273 673 K; T2 500 273 773 K



R 1.987 cals; H ? Substituting the above values in the above equation.

log 1.44 10

4

log 1.64 10

4

H 773

673

773

2.303

1.987 673

3.84 3.785 4.576H

[ 1.92 10

4 ]

0.055 H 4.196 10

5

0.055

5 H 4.196 10 H 1310.8 cals .

H 1.3108

k.cals .

As per the stoichiometric equation the

1310.8 cals

heat of formation of 2 mole of

NH

3

Heat of formation of

1310.8

1 mole of NH3

2

655.4 cals or 0.6554 k.cals

7. Discuss the criteria for a spontaneous chemical reaction.

[CO2-L2- Nov2002] According to the second law of thermodynamics a process is said to be

Stotal is positive i.e., entropy of the universe (system + surroundings) increases. This criteria of spontaneity involves the entropy of surroundings, which is difficult to measure. So, we need a criteria, which does not involve the entropy of surroundings. The change in Gibbs free energy provides such a criteria.



We know that, the total entropy change Stotal is given by

qp

surroundings q

p

system

Hsystem

Since the surrounding is a large area, the temperature of the surroundings remains

constant, so we have

Ssurroundings

q

p

system

Hsystem ... (2)

T T

On substituting equation 2 in 1 we get

S S

system

Hsystem

Total T

Multiply the equation by T

T S

Total

T

S

system H

system TS H G

. .

. G H T S

Stotal Ssystem Ssurroundings ... (1)

If a reaction is carried out

at constant

T and P

The amount of heat transferred

S q

by the system to surroundings is

given by system

p system

The amount of heat taken by the S q

surroundings is equal to

surroundings p

system



Equation (4) is introduced in equation (3)

TSTotal

Hsystem

G

H

system

T Stotal G or T Stotal G

The equation (5) is the criterion for spontaneity interms of free energy of the system. Thus

When G ve G 0, the process is spontaneous.

When G 0, the process is in equilibrium.

When G ve G 0, the process is non-spontaneous.

8. The equilibrium constant for The reaction

N 3H 2NH is 1.64 10

4 atm and

2 2 3

0.144 10

4 atm at 400C and 500C,

respectivel

y

Calculate the heat of

reaction in terms of calories.

R 1.987

cals[CO2-L2-

Nov 2001, Dec 2011]

Solution

We know that

2 1 H

T2

T1

log KP log KP 2.303R T T

2

1

Given

K1

1.64 10

4

atm

;

K2

0.144 10 4 atm

P P

T1 400 273 673 K; T2 500 273 773

K

R 1.987 cals; H

?



Substituting the

values

in the above

equation.

log 0.144 10

4 log 1.64

10

4

H 773

673

773

2.303

1.987

67

3

4.84

3.785

H [ 1.92 10

4

]

4.576

1.055 H 4.196 10

5

1.055

H

4.196 10

5

H 25143 cals

H 25.143 k.cals .

9. Derive an expression to relate the equilibrium constant of a reaction with its

standard free energy change.

(or)

What is meant by Van’t Hoff isotherm? Derive an expression for the isotherm of a general reaction.

aA bB cC dD [CO2-L2-Dec 2011, Dec 2012]

Van’t Hoff isotherm gives a quantitative relationship between the free energy change and equilibrium constant. It can be derived as follows.

Consider a general reaction

aA bB cC dD We know that

G H TS HEPV



(or) G E PV TS

(or) dG dE P dV V dP T dS S dT ... (1)

But according to first and second law of thermodynamics

dE dq PdV I law

dq TdS II law

dE TdS PdV ... (2)

Substituting equation (2) in (1) it becomes

dG TdS PdV PdV VdP TdS SdT

dG V dP S dT ... (3)

dGT V dP

Free energy change for 1 mole of any gas at a constant temperature is given by

dG V dP

dG RT

dP ... (4)

P

[ ... PV RT; V

RT

]

P

On integrating the equation (4) becomes

dG RT dP

P

G G RT ln P

Where, G integration constant (standard free energy)

Let the free energies of A, B, C and D at their respective pressure PA, PB, PC and

PD are GA, GB, GC and GD respectively. Then the free energy change for the above

reaction is given by



G

G

product

G

reactant

[ c GC d GD ] [ a GA b GB ] ... (6)

Substituting the value of GA, GB, GC and GD from equation (5) in (6), we get,

G [ cGC c RT ln PC d GD d RT ln PD ]

[ a GA a RT ln PA b GB b RT ln PB ]

[ c GC d GD a GA b GB ] RT

ln

[ PC ]c[ PD

]d

[ PA ]a[ PB ]

b

(or) G G RT

ln

PCcPD

d .... (7)

PAaPB

b

We know that, at equilibrium, G 0.

0 G RT ln P

CcP

Dd

PAaPB

b eq

G RT ln Keq 0 ... (8)

The equation (8) may be written as

G RT ln Keq From equations (9) and (8), the equation (7) becomes

G RT ln Keq RT ln P

CcP

Dd

PAaPB

b

(or) G RT ln Keq RT

ln PC

cPD

d ... (10)

PAaPB

b

This expression is called Van’t Hoff isotherm.

10. Discuss the significance of Van’t Hoff equation with reference to equilibrium constant of a reaction.

(or)



What are the applications of Van’t Hoff equation? [CO2-L2-Dec 2001]

1. Equilibrium constant K and H of a reaction can be calculated using Van’t Hoff

equation.

2. A plot of log Kp versus 1/T will give a straight

line with slope H

From the slope, value 2.303

R

of H can be calculated.

i.e., H 2.303R slope 3. It is seen that if T2 T1, the quantity within the square brackets is positive. Kp2

Kp1, if H is a positive quantity, i.e., for an endothermic reaction the equilibrium constant

increases with temperature.

4. The magnitude of the equilibrium constant also depends on whether partial

pressures, molar concentrations or mole fractions are used.

5. The magnitude of the equilibrium constant depends on how the reaction is written

6. Two or more equilibria can be combined in order to get a new equilibrium.

11. By combining Van’t Hoff isotherm and Gibbs-Helmholtz equation illustrate the effect of temperature on equilibrium constant. [CO2-L2- May 2005]

The effect of temperature on equilibrium constant is quantitatively given by Van’t Hoff equation. It can be derived by combining the Van’t Hoff isotherm with Gibbs- Helmholtz equation as given below.

According to the Van’t Hoff isotherm, the standard free energy change G is related to the equilibrium constant (K) by the following equation.

G RT ln KP ... (1)

Differentiate equation (1) w.r.to temperature at constant pressure, we get

G R ln KP

RT

dln KP ... (2)

T

dT

P

When the equation (2) is multiplied by T, we get

T G RT ln K RT 2 dln KP

T P

dT

P

G is substituted for RT ln KP , we have



G G

RT

2 dln

KP

T

T

dT

P

Gibbs-Helmholtz’s equation for substance in their standard states may be written as

H

RT

2d

Tln

K

P

d

RT2dln K

(or) H P

dT

(or)

dln

KP H ... (7)

dT RT

2

The equation (7) is known as Van’t Hoff’s equation.

Since ent H

pressure of the reactants or products, the Van’t Hoff’s equation may be written as

dln

KP

H ... (8)

dT RT 2

The equation (8) is integrated between T1 and at

G G

2 dln KP...

(6)

T

T

H T

T

RT

dT

P

P

G G

RT 2dln

KP

T

T

T

T H

dT

P

P



T2

which the equilibrium constants are K1P and K

2p respectively and H is constant.

K2 H

T ... (9)

P 2 dT

dln KP R 2

KP1

T1 T

2 1 H 1 1

(or) ln KP ln KP

R T T

2 1

H 1

1 .... (10)

R T T

1 2

H T2 T1

R T T

1 2

2 1 H

T2

T1 ... (11)

(or) log KP log KP

2.303R T T

1 2

The equilibrium constant K2

P at temperature T2 can be calculated, if the equilibrium

constant K1

P at temperature T1 is known, provided the heat of the reaction H is

known. 12. Drive the clausius - clapeyron equation. Discuss its

applications. [CO2-L2]

Consider a system consisting of only 1 mole of a substance existing in two phases

A and B. The free energies of the substance in two phases A and B be GA and GB. Let the temperature and pressure of the system be T and P respectively. The system is in equilibrium, so there is no change in free energy i.e.,

GA GB ...(1)

If the temperature of the system be raised to T dT, the

pressure becomes P dP and the free energies become

GA dGA and GB d GB respectively. Then the equation (1) becomes



GA dGA GB dGB ...(2) We know that

G H TS (Gibbs free energy)

H E PV (Enthalpy)

G E PV TS ...(3)

dG dE PdV VdP TdS SdT ...(4)

But, dE dq dW [I law of Thermodynamics]

dq dE dW

dq dE PdV ...(5)

[ dW PdV]

We know that for reversible equation,

dq dS

T

dq TdS ...(6)

Substituting equation 6 in equation 5

TdS dE PdV

dE TdS PdV ...(7)

Substituting equation (7) in (4)

dG TdS PdV PdV VdP TdS SdT

dG VdP SdT ...(8) Here, the work done is due to volume change only, so equation 8 may be applied

to phase A as well as B

dGA VA dP SA dT ...(9)

dGB VB dP SB dT ...(10) Where,

VA and VB are the molar volumes of phases A & B respectively

SA and SB are their molar entropies.

Since GA GB, hence from equation (2)



dGA dGB ...(11)

Substituting equation (9) and (10) in equation (11)

VA dP SA dT VB dP SB dT ...(12)

SB dT SA dT VB dP VA dP ...(13)

dT SB SA dP VB VA ...(14)

SB SA dP ...(15)

V V dT

B A

SB SA

represents the change in entropy when 1 mole

of the substance passes from the initial phase A to the final phase B. It may be denoted

as S

Hence equation (15) becomes

S dP ...(16)

V V dT

B A

We know that entropy change S at constant T is

S q ...(17)

T

Substituting equation (17) in (16)

dP q ...(18)

dT T VB VA

This is called clausius-clapeyron equation

Applications of Clausius - Clapeyron equation 1. Calculation of latent heat of vapourization

If the vapour pressure of a liquid at two temperatures T1 and T2 be P1 and P2 respectively. The molar heat of

vapourisation HV can be calculated.



2. Calculation of boiling point (or) freezing point If the freezing point or the boiling point of a liquid at one pressure is known, it is

possible to calculate it at another pressure by using Clausius-Clapeyron equation. 3. Calculation of vapour pressure at another temperature

If latent heat of vapourization is known, it is possible to calculate the vapour pressure of a liquid at given temperature if the vapour pressure at another temperature is known. 4. Calculation of molar elevation constant

Molar elevation constant of a solvent can be calculated. 13. Derive all the four Maxwell relations. [CO2-L2-Dec 2011, Dec 2013, May 2014]

The various expressions connecting internal

energy E ,

H A G , with relevant

parameters such as pressure, temperature, volume and entropy may be given as (i) dE T dS P dV (ii) dH T dS V dP (iii) dA S dT P dV (iv) dG SdT V dP

From these expressions the Maxwell relations are obtained as follows. The combined form of first and second law is

dE T dS P dV ... (1) 2.26 Engineering Chemistry - I

If V is

constant

, so that

dV 0,

then

equatio

n (1) yields

E ... (2)

S

T

V

If S is

constant

, so that dS 0, then equation (1) yields

E P

... (3)



V S

Differentiating equation

(2) w.r.to V at constant S yields

2E

T ... (4)

S

V

V

S

Differentiating equation (3) w.r.to S at constant V yields

2E P ... (5)

V

S

S

V

It follows from equation (4) and (5) that

T P ... (6)

VS S V

Equation (6) is a Maxwell relation. 2.

Enthalpy is defined by

H E PV

dH dE P dV V dP

But, dE P dV T dS

dH T dS V dP ... (7)

If P is constant, so that

dP 0, then equation (7)

yields

H T

... (8)

S

P

If S is constant, so that dS 0, then equation (7)

yields

H V ... (9)



PS

Differentiating equation (8) w.r.to P, at constant S yields

2H T ...

(10)

SP

P S

Differentiating equation (9) w.r.to S at constant P yields

2H V ...

(11)

PS

S P


T V ... (12

)

P S S P

Equation (12) is also a Maxwell relation.

3. Helmholtz free energy (work function A) is given by

A E TS

dA dE T dS S dT But the combined first and second law of thermodynamics is

T dS dE P dV dE T dS P dV

dA S dT P dV ... (13)

If V is constant, so that

dV 0, then equation (13) yields

A S

...

(14

)



T

V

If T is constant, so that dT 0, then equation (13) yields

A P

...

(15

)

VT

Differentiating equation (14) w.r.to V, at constant T, yields

2A S ... (16)

TV

V

T

Differentiating equation (15) w.r.to T, at constant V, yields

2A P ... (17)

VT

T

V

If follows from equation (16) and (17) that

S P ... (18)

V

T TV

Equation (18) is also a Maxwell relation. 4.

Gibbs free energy G is defined as

G H TS

dG dH T dS S dT

But H E PV

dH dE P dV V dP

T dS V dP T dSdEP dV Thus, dG S dT V dP ... (19)

If P is constant, so that dP 0, then equation (19) yields Chemical Thermodynamics 2.29

G

S ...

(20

)



T P

If T is constant, so that dT 0, then

equation

(19) yields

G V

...

(21

)

P T

Differentiatin

g

equatio

n (20) w.r.to P at constant T, yields

2G

S ...

(22)

TP P T

Differentiating

equation (21) w.r.to T, at constant P yields

2G

V ...

(23)

PT

T

P


S V

P T TP

S V ... (24

)

(or)

P T T P

Equation (24) is also a Maxwell relation.

Thus equations (6), (12), (18), (24) are known as maxwell relations.



Unit iii

Photochemistry And Spectroscopy

Part A

1. Define Photochemistry. [CO3-L1-May 2006]

Photochemistry is the study of chemical reactions that are caused by absorption of light radiations.

2. What are the laws of Photochemistry? [CO3-L2]

a) The Grotthus-Draper law. b) The Stark- Einstein law of photochemical equivalence.

c) Lamberts law. d) Beer-Lamberts law.

3. State Grotthus – Draper law. [CO3-L1-Dec 2007]

This law states that “when light falls on any substance, only the fraction of incident light which is absorbed by the substance can bring about a chemical change”. It is also called the principle of photochemical activation.

4. State Stark – Einstein law. [CO3-L1-May 2013]

This law states that “In the primary photochemical process (first step) each reacting

molecule is activated by one quantum of effective light”. This law is also called the

principle of quantum activation.

5. How is energy of Einstein “E” related to wavelength? [CO3-L2-Dec 2006]

The energy E absorbed per mole of the reacting substance is given by

`E = NAhν =NAhc/λ

NA =Avagadro number = 6.023 x 1023 /mole h = Planks constant =



6.625 x 10-34joules.sec C = velocity of light = 3 x 108 m/sec

E = 6.023x1023/mole X 6.626 x10-34J.sec X 3x108m/s

λ

= 0.1197 J/mol

6. Define Beer - Lamberts law. [CO3-L1-Dec 2009]

This law states that “When a beam of monochromatic radiation is passed through a homogenous absorbing solution, the decrease in the intensity of the radiation dI with the thickness of the absorbing solution dx is directly proportional to the intensity of the incident radiation I as well as the concentration of the solution C.

7.What are the limitations of Beer-Lamberts law? [CO3-L2-Dec 2011] a) Beer- Lamberts law is obeyed only if the radiation used is monochromatic. b) It is applicable only for dilute solutions. c) The temperature of the system should not be allowed to vary to a larger

extent.

d) It is not applied to suspensions.

e) Deviations may contain if the solution contains impurities.

f) Deviations may also occur if the solution undergoes association or dissociation.

8. Define Quantum yield or Quantum Efficiency. [CO3-L1]

Quantum yield (Ф) is defined as “the number of molecules of the substance

undergoing photochemical reaction per quantum of radiation absorbed‟‟. The

quantum yield for photochemical reaction is defined as

Ф = No. of molecules reacting in a given time



No. of quanta (photons) of light absorbed in the same time

9.What are the causes of high quantum yield? [CO3-L2-Dec 2014]

i) The absorption of radiation in the primary step produces atoms or free radicals, which initiates a series of chain reactions.

ii) Formation of intermediate products which acts as catalyst.

iii) The reaction is exothermic so that the heat evolved may activate other molecules which react without absorption of additional radiation.

iv) The active molecules produced by primary absorption may collide with other molecules, thereby activating them, which in turn further activate other reacting molecules.

10.What are the causes of low quantum yield? [CO3-L2] Deactivation of the excited molecules before they form products. Excited molecules may lose energy by collision with non excited

molecules.

Molecules may receive insufficient amount of energy.

Primary photochemical reaction may get reversed. The dissociated fragments may recombine to give reactant which

results in low quantum yield.

12.What are primary and secondary photochemical reactions? [CO3-L1]

Primary process – the reacting molecules undergo activation by absorption of light.

A + h ν A*

13. Define Photo-Sensitization. [CO3-L1-May 2014]

In some photochemical reactions, the reactant molecules do not absorb radiation and no chemical reaction occurs. Certain reactions are made sensitive by the presence of small quantity of foreign substance which can absorb light and stimulate the reaction without taking part in it. The substance which absorbs light and induces a photochemical reaction without undergoing chemical reaction is known as photosensitiser and the phenomenon is known as Photo-sensitization.



Examples:-

Atomic Photosensitisers – Mercury, cadmium, Zinc

Molecular Photosensitisers – Benzophenone, Sulphur dioxide, Uranyl sulphate.

14. What is quenching? [CO3-L1]

During Photosensitization the foreign substance (sensitizer), absorbs light and gets excited. When the excited foreign substance collides with another substance it gets converted in to some other product due to the transfer of its energy to the colliding substance. This process is known as quenching.

15. What is fluorescence? [CO3-L1-Dec 2015]

Certain substances (atoms or molecules) when exposed to radiation of short wavelength (high frequency), emit light of different frequencies compared to those of incident radiations. This process is called fluorescence and stops as soon as the radiation is cut off. Actually decay period is very short i.e. 10-9 to 10-4 sec. The substance which exhibits fluorescence is called fluorescent substance.

Examples: - Fluorite (naturally occurring CaF2), Petroleum, Organic dyes like Eosin, fluorescein, Chlorophyll, Quinine sulphate solution, Vapours of Sodium, Iodine, Mercury.

16. What is Phosphorescence? [CO3-L1-May 2008]

Many substance (or molecules) when exposed to radiations of short wavelength (high frequency) continue to emit light for sometime (10-4 to few seconds) even after incident light is cut off. This phenomenon is called phosphorescence and is chiefly caused by Ultraviolet or Visible light. The substance which exhibits phosphorescence is called phosphorescent substance. Examples: - Zinc Sulphide, Alkaline earth sulphides (CaS, BaS, SrS).

17 Define internal conversion and intersystem crossing. [CO3-L1]

Internal conversion: - It is a type of transition which involves the return of the activated

molecule from higher

excited state to the lower excited state.



S3 S2 T3 T2

S2 S1 T2 T1

The energy of the activated molecule is dissipated in the form of heat through

molecular collisions. This process occurs in less than about 10-11 seconds. This is also

known as non-radiative transitions

Intersystem crossing: - The process in which the energy of the activated molecule is lost through transition between states of different spin multiplicity.Even though these transitions are forbidden they occur at relatively slow rates.

PART – B

1. Illustrate the Stark –Einstein law of Photochemical Equivalence. [CO3-L2]

This law is also called as principle of quantum activation. It states that “In the primary photochemical process each reacting molecule is activated by one quantum of effective light”. This law implies a 1:1 relationship between the number of reacting molecules and the number of quanta of light absorbed and is applicable only for the primary process.

The molecule A absorbs a photon of light and gets activated (primary step).The activated molecule (A*) then decomposes to yield B (secondary step).

Primary step: A+ hν A*

Secondary Step: A* B

Overall reaction: A + hν B



Suppose ν is the frequency of radiation absorbed then the corresponding

quantum of energy absorbed per molecule will be hν.The Energy absorbed per mole of

the reacting substance is called one

Einstein is given by `E = NAhν =NAhc/λ

NA = Avagadro number = 6.023X 1023 /mole

h = Planks constant = 6.625X10-34 J sec

C = velocity of light = 3.08X 108 m/sec

E = 6.023x1023

X 6.626x10-34

X 3x108

= 0.1196 J/ mol

Λ λ (m)

also we know 1 calorie = 4.184 J

E = 0.1196 J/mol= 2.859x10-2

cal/mol

4.184 J/cal x λ Λ

2. Derive the expression for Beer – Lamberts law. State its disadvantages. [CO3-L2- May 2010, Dec 2012]

Beer-Lamberts law states that when a beam of monochromatic radiation is passed through a solution of an absorbing substance, the rate of decrease of intensity of radiation (dI) with thickness of the absorbing solution (dx) is proportional to the intensity of incident radiation, I, as well as concentration of the solution, C.

It is mathematically represented as - dI/dx = kIC, where k = molar absorption co–efficient. Upon integration with in the limits, we get

dI/I = -kCdx

ln I/I0 = - kCx or 2.303log I/I0 = -kCx log I0/I = k /2.303 * Cx

A = ε Cx (Beer- Lamberts law)



Where, ε = k/2.303 = molar absorptivity co–efficient. Hence, the absorbance is directly proportional to molar concentration, C, and thickness (or) path length, x.

Application of Beer – Lambert Law

1.Determination of unknown concentration:

Let the absorbance and the concentration of standard solution be As and Cs, respectively, As = εCsx …… (1), and the absorbance and the concentration of unknown solution be Au and Cu, respectively. Au = εCux ……. (2)

Dividing equation (1) by equation (2), we get

As = Au

Cs Cu

Cu = Au * Cs/As

Thus the concentration of unknown solution is calculated.

Limitations:

a) Applicable only for monochromatic light and dilute solutions. b) Deviation occurs if the solution undergoes association or dissociation.

c) Valid only at constant temperature. d) It is not applied to suspensions.

e) Deviations may occur if the solution contains impurities.

3.Outline the experimental determination of quantum yield. [CO3-L3-June 2013]

There are two types of determination to find out the quantum yields of photochemical reactions.

(iv) Determination of number of moles of the light absorbing substance that react in a given time.



(v) Determination of number of photons of light of required wavelength absorbed by the same substance in the same time.

Determination of number of moles reacting:

The number of molecules reacted in a given time can be determined by the usual analytical techniques used in chemical kinetics.

The rate of the reaction is measured by pipetting small quantities of sample from the reaction mixture from time to time and the concentration of reactants are continuously measured by the usual volumetric methods (or) change in some physical property such as refractive index / absorption / optical rotation. From the data the amount and number of molecules reacted can be calculated.

Experimental determination of amount of photons absorbed:

A photochemical reaction takes place by the absorption of photons of light by the reacting molecules. In order to study the rate of reaction it is essential to determine the intensity of light absorbed by the reacting molecule.

An experimental arrangement for the study of rate of a photochemical reaction is shown in the figure.



Light radiation from a suitable source (tungsten filament or mercury vapour lamp) is rendered parallel by a lens. The parallel beam is then passed through a monochromator, which yields a beam of the desired wavelength only. The monochromatic light then enters the reaction cell containing the reacting mixture (made of quartz) which is immersed in a thermostat. Finally, the light transmitted from the cell is made to fall on a detector which measures the intensity of the transmitted radiation.

The intensity of radiation is generally measured with the help of actinometer. It contains 0.05 M oxalic acid and 0.01 M uranyl sulphate in water. On exposure to radiation the following reaction takes place.

UO2 2+

+ hν (UO22+

)*

UO22+

+( COOH ) 2 CO2 + CO + H2O + UO22+

The standard solution of uranyl oxalate is exposed to light radiation of definite frequency for a certain period of time. The extent of reaction can be determined by titrating the remaining oxalic acid with standard KMnO4 solution. The amount of oxalic acid consumed in the photochemical reaction is a measure of intensity of radiation (i.e. number of photons absorbed).

4. Explain the mechanism of photo physical process with the help of Jablonski diagram. [CO3-L2-Dec 2012, May 2013]

Spin multiplicity:-Most molecules posses an even number of paired electrons in ground state. The spin multiplicity of the state is given as 2S + 1, where S is the total spins of the electrons When the spins are paired Then S = s1+ s2 = +1/2 + (-1/2) = 0

Hence 2S + 1 = 2x0 + 1 =1, the spin multiplicity is 1.

The molecule is in singlet ground state.



Spin Orientation due to absorption of light

On absorption of a photon of energy hν, one of the paired electrons goes to a higher energy level (i.e. excited state).

The spin orientation of the two electrons may be either parallel or antiparallel. If the spins are parallel, then S = s1+ s2 = +1/2 + (+1/2) = 1

Hence 2S + 1= 2x1 +1 =3, the spin multiplicity is 3. The molecule is in triplet excited state.

If the spins are antiparallel, then S = s1+ s2 = +1/2 + (-1/2) = 0

Hence 2S+ 1= 2x 0+ 1 = 1, the spin multiplicity is 1.

The molecule is in singlet excited state. Depending upon the energy hυ of a photon, electron can jump to any higher electronic state and hence we get a series of

Singlet excited states S1, S2, S3, Triplet excited state T1, T2, T3…

S1, S2, S3… etc is called first singlet excited state, second singlet excited state, and third singlet excited state respectively.

T1, T2, T3… etc is called first triplet excited state, second triplet excited state, and third triplet excited state respectively.

According to Quantum mechanics the singlet excited state has higher energy than corresponding triplet excited state. Thus E (S1) > E (T1); E (S2) > E (T2) ; E(S3) > E(T3)



Consequences of light absorption: When light photon (hν) is absorbed by a molecule, the electron of the absorbing molecule may jump from S0 (Singlet ground state) to singlet excited state S1 or S2or S3.

Depending upon the

energy of the light photon

absorbed

The activated molecule returns to the ground state by dissipating its energy (hν) through any of the following type of process.

1.Non-radiative transitions:-Such a transition is from the higher excited states (S1, S3 or T2, T3) to the first excited state (S1 or T1).Since such a transition does not involve the emission of radiation, it is referred to as radiation less or non radioactive transitions.



2. Internal Conversion: - These transitions involve the return of the activated molecule from the higher excited state to the first excited states. The energy of the activated molecule is dissipated in the form of heat through molecular collisions.

This process occurs in less than about 10-11s

S3 S1 T3 T1

S2 S1 T2 T1

Inter system crossing: - (ISC)

The process in which the energy of the activated molecule is lost through transition between state of different spin multiplicity. Even though these transitions are forbidden, they occur at relatively slow rates

S2 T2

S1 T1

Radiative transition:-

These transitions involve the return of the activated molecule from the singlet

excited state S1 and the triplet excited state T1 to the singlet ground state S0. So,

such a transition is accompanied by emission of radiation.

1. When a molecule in the S1 state returns to the ground state S0, emission of radiation occurs in about 10-8 s and this process is known as fluorescence.

2. When a molecule in the T1 state returns to the ground state S0, emission of radiation occurs at a rather slow rate. This process is called

phosphorescence. Since the T1 to S1, transition is spectroscopically forbidden the life time of phosphorescence is much longer (about 10-3 and higher).



5. Explain the mechanism of energy transfer in photochemical reactions. Photosensitization and Quenching. [CO3-L2-Dec 2012, May 2014]

In some photochemical reactions, the reactant molecules do not absorb radiation and no chemical reaction occurs. Certain reactions are made sensitive by the presence of small quantities of foreign substance which can absorb light and stimulate the reaction without taking part in it. The substance reaction is which absorbs light and induces a photochemical reaction without undergoing chemical known as photosensitiser and the phenomenon is known as Photo-sensitization.

Examples:-

Atomic Photosensitisers – Mercury, cadmium, Zinc

Molecular Photosensitisers – Benzophenone, Sulphur dioxide, Uranyl sulphate.

During Photosensitization the foreign substance (sensitizer), absorbs light and gets excited. When the excited foreign substance collides with another substance it gets converted in to some other product due to the transfer of its energy to the colliding substance. This process is known as quenching.

Mechanism of photosensitization and Quenching:-

Consider a general donor – acceptor system in which only the donor D i.e. the sensitizer, absorbs the incident photon and the triplet state of the donor is higher than the triplet state of the acceptor A .i.e. the reactant. Absorption of the photon produces the singlet excited state of the donor,1D which via intersystem crossing (ISC) gives the triplet state of the donor 3D. This triplet excited state then collides with the acceptor producing the triplet excited state of the acceptor 3A and the ground state of the donor. If 3A gives the desired product, the mechanism is called photosensitization.

However if the desired products result from the excited state of the donor (3D), then A is called the quencher and the process is known as quenching.

The reactions for photosensitization and quenching may be represented as below,

D + hν 1D



1D 3D

3D + A D + 3A

3A products (photosensitization)

3D products (quenching)

The triplet excited state of the sensitizer or donor 1D must be higher in energy level than the triplet excited state of the reactant or acceptor (3A) so that the energy available

is

enough to raise the reactant molecule to its triplet state.

Examples of photosensitized reactions:-

Dissociation of H2 molecule:

Irradiation of a mixture of hydrogen gas and mercury vapour with light of wavelength 253.7 nm brings about dissociation of hydrogen molecule in to hydrogen atom.

Hg + hν Hg*

Hg* + hν H2* +



Hg

H2* 2H

Here mercury acts as photosensitiser.

Photosynthesis in plants

The most outstanding example for photosensitization is photosynthesis of carbohydrates in plants from H2O and CO2in which the green colouring matter of plants acts as photosensitiser. CO2 and H2O do not absorb any radiation in the visible region, but chlorophyll absorbs the radiation. It absorbs visible light in the wavelength range of 400-700 nm. The energy of light absorbed by chlorophyll is transferred to CO2 and H2O which then react to form (CH2O) n (cellulose /sugar).

6. Explain the mechanism of chemiluminescence with examples. [CO3-L2-Dec 2014]

Chemiluminescence is defined as the production of light by a chemical reaction and is thus the reverse of photochemical reaction. In order for a reaction to be chemiluminescent it must furnish sufficient energy to raise at least one of the reactants or intermediates to an electronically excited state so that as it returns to the ground state, it emits energy in the form of radiation.

Since the emission occurs at ordinary temperature, the emitted radiation is also known as “cold light”.

Examples of chemiluminescence reactions:

* The glow of phosphorus and its oxide, in which the oxide in its electronic excited state emits radiation.

* Oxidation of 5-mainophthalic hydrazines or cyclic hydrazines by H2O2 emits green light. 3.‟Cold light‟ emission by glow worms. This is an example of bioluminescence due to aerial oxidation of luciferon (a protein) in the presence of enzyme (luciferase).



SPECTROSCOPY

PART A

1.How does molecular spectrum arise? [CO3-L2-Dec 2005]

Molecular spectrum arises due to the interaction of electromagnetic radiation with molecule.

2.What are the differences between atomic spectra and molecular spectra? [CO3-L2-May 2003]

Atomic Spectra Molecular Spectra

Interaction of electromagnetic radiation

Interaction of electromagnetic radiation with

with atoms. molecules.

Line spectrum is obtained Complicated spectrum is obtained

It is due to electronic transition in an It is due to vibrational, rotational and

element. electronic transition in a molecule.

3. Calculate the number of modes of (IR) vibration for the following molecules. [CO3-H1-Dec 2009]

For a molecule contain N atoms, the number of vibration modes are given by:

(a) CH4 (Non linear molecule) = 3N- 6 = (3 x 5) – 6 = 9

(b) CO2 (linear molecule) = 3N- 5 = (3 x 3) – 5 = 4

(c) C2H6 (Non linear molecule) = 3N- 6

= (3 x 12) – 6 = 30

(d) HCl (linear molecule) = 3N- 5 = (3 x 2) – 5 = 1



4. What is Finger Print region? Mention its uses. [CO3-L2-May 2010]

The vibration spectral region at 1400 – 700 cm -1 gives very rich and intense absorption bands which is unique for each molecule. This region is termed as Finger Print Region.

Uses: It can be used to detect the presence of functional group and also to identify and characterize the molecule just as a finger print can be used to identify a person.

5. What are the differences between Chromophore and Auxochrome? [CO3-L2-May 2009]

S.No Chromophore Auxochrome

1. This group is responsible for the It does not impart colour, but when

colour of the compound. conjugated with a chromophore colour

is produced.

2. It does not form salt. It forms salt.

3. It contains at least one multiple It contains lone pair of electrons.

bond.

eg: -NO2, -NO , -N=N- eg: -OH , -NH2 , -NR2

6. Mention three applications of UV-Visible Spectroscopy. [CO3-L2]

Quantitative analysis: By using Beer-Lamberts law, the concentration of unknown solution can be determined.

Qualitative analysis: It is used for identification of aromatic compounds and conjugated olefins.

Detection of impurities: It is used for detecting impurities in organic compound

7. Name the transition responsible for molecular spectra. [CO3-L2] The molecular spectrum arises due to the following three transitions.



2. Rotational transitions.

3. Vibrational, transitions,

4. Electronic transitions.

8. What is the relation between wavelength, frequency and wave number? [CO3-L2] 1/λ = υ = υ/c

9.What are the factors which affect the absorbance of the photons by a molecule? [CO3-L2-Dec 2005]

a) The nature of the absorbing molecules

b) The concentration of the molecules

c) Path length of radiation

10. Define the term – Bathochromic shift, Hypsochromic shift. [CO3-L2-May 2007]

Bathochromic shift: shifting of absorption band towards a longer wavelength. Hypsochromic shift: shifting of absorption band towards a shorter wavelength.

PART B

1. What is principle of UV Spectroscopy? Explain its components and working. [CO3-L2-June 2013]

Principle:

UV – Visible spectra arises from the transition of valence electrons from the ground state to excited state by absorbing light from UV (100 – 400 nm) or visible region (400 – 750nm). It is a type of absorption spectra.

Components:

a) Radiation source: Hydrogen or Deuterium lamps are radiation source for UV region and tungsten filament lamp for visible region.



b) Monochromators: The monochromators is used to select a particular wavelength.

a) Cells: The cells are made up of quartz glass are used to hold the sample which should fulfill the following conditions.

1.They must be uniform in

construction. 2.The material

should be inert to solvents.

3.They must be transparent to UV and Visible light.

b) Detectors: The converts the radiation in to current which is directly proportional to the concentration of the solution.

Eg.. They are Photo multiplier tube or Photocells.

a) Recorder: The recorder records the signal from the detector and shows a display.

Working:

The Radiation from the source is allowed to pass through the monochromator. It

allow particular wavelength of light to pass through the exit slit. The beam of radiation

coming out of the slit is split into two parallel beams.

Block diagram



One half of the beam passes through the sample cell and another half passes through reference cell (only solvent). The detectors compare the intensities of the transmitted two beams of light. If the compound absorb light at a particular wavelength, the intensity of the transmitted beam through sample (I) will be less than that of reference beam (Io). The recorder produces the graph, which is a plot of wavelength of light Vs absorbance of light. This graph is known as adsorption spectrum.

2. Explain the principle, components and working of IR spectroscopy. [CO3-L2-Dec 2012]

Principle:

IR Spectra is produced by the absorption of energy by a molecule in the infrared region and the transitions occur between vibrational levels.

Range of IR radiation:

Near Infrared

= 12500 – 4000 cm-1

Infrared = 4000 – 667

cm-1

Far Infrared = 667 – 50

cm-

1

Components:

a. Radiation source: The main source of radiation is Nichrome wire or Nernst glower which is a filament containing oxides of Zr, Th, Ce held together with a binder.

b. Monochromators: It allows the light of the required wavelength to pass through light of other Sample cell: The cells must be transparent to IR radiation are used to hold the sample. Detector: Detectors are used to convert thermal radiant energy into electrical energy. Recorder: The recorder record the signal coming out from the detector

Block Diagram:



Working:

The radiation emitted by the source is split into two parallel beams. One of the beams passes through the sample and the other passes through the reference sample. When the two beams of light recombine they produce a signal which is measured by detector. The signal from the detector is recorded by the recorder.

3. Explain the molecular vibration in IR spectrum. Calculate the fundamental modes of vibration for non-linear and linear molecule. [CO3-L3-Dec 2008]

There are two kinds of fundamental vibrations in molecule.

1.Stretching vibrations: During stretching, the distance between two atoms decreases or increases, but bond angle remains unaltered.

2.Bending vibrations: During bending, bond angle increases or decreases but bond distance remains unaltered.

Types of Stretching and bending vibrations

The number of fundamental modes of vibration of a molecule can be calculated as follows. Non-linear molecule has 3N-6fundamental modes of „vibrations and a linear molecule has 3N-5 fundamental modes of vibrations where N is the number of atoms in a molecule.

Stretching vibrations: It classified into two types

(i) Symmetric stretching: The atoms of the molecule are moving in the same direction.

(ii) Asymmetric stretching: The atoms of the molecule are moving in opposite direction.

Asymmetric stretching requires more energy than the Symmetric stretching. Bending vibrations: It classified into two types.

a) In-plane bending



b) Out-of plane bending

Fundamental modes of vibration for non-linear and linear molecule

1.Methane (CH4) (N=5)

It is a non-linear molecule. Hence, 3N-6

= 3x5-6 = 9 Fundamental vibrational modes.

2.Benzene (C6H6) (N=12)

It is a non-linear molecule. Hence, 3N-6

= 3x12-6 = 30 Fundamental vibrational modes.

3.Water (H2O) (N=3) It is a non-linear molecule. Hence, 3N-6 = 3x9-6 = 3 Fundamental vibrational modes. 4. Carbon dioxide (CO2) (N=3) Hence, 3N-5 = 3x3-5 = 4 Fundamental vibrational modes.

4. Discuss the applications of UV-Spectroscopy. [CO3-L2-June 2010]

Qualitative analysis: Many types of compounds can be identified by comparing the UV spectrum of the sample with that of similar compounds available in reference books.

Testing the purity of the sample: UV Visible spectroscopy is the best method for detecting impurities in organic compounds because

a) The bands due to impurities are very intense.

b) Saturated compounds have weak absorption band and unsaturated compounds have strong absorption band.

The kinetics of chemical reaction: The progress of a chemical reaction can be easily followed by examining the UV-spectra of test solution at different time intervals. The concentration of either the reactant or product can be followed.

Determination of calcium in blood serum. Calcium in blood serum in indirectly determined by converting the Ca present in 1 ml of serum as calcium oxalate, and dissolving the calcium oxalate in dilute sulphuric acid and treated with ceric sulphate solution. The absorption of the excess Ce+4ion is measured at 315nm. The amount of Ca in blood serum is indirectly calculated from the amount of Ce+4ion consumed by the calcium oxalate.



Quantitative analysis: UV absorption spectroscopy is used for the quantitative determination of compounds based on Beer-Lambert‟s law.

5. Discuss the applications of IR spectroscopy. [CO3-L2-June 2008]

a. Identification of organic compounds. : IR spectroscopy is very useful in identification of organic compounds.

b. Identification of functional groups: The region from 4000 – 1400 cm-1 in IR spectrum is useful for functional group analysis.

Group Frequency range

C=O 1750 – 1600 cm-1

Free OH 3650 – 3300 cm-1

H bonded OH 3300 – 2800 cm-1

-C = C- 2250 – 2100 cm-1

-C = N 2260 cm-1

-NH2 3350 cm-1

-N = O 1600 – 1500 cm-1

-COOH 3300 – 2700 cm-1

a) Testing the purity of a sample: Presence of an impurity can be detected by their characteristic peaks.

b) Determination of symmetry of a molecule: The symmetry of a molecule whether linear or non-linear can be determined from the IR spectrum.

Example: IR spectra of H2O gives 3 peaks at 667 cm-1, 1330cm-1, 2349 cm-1. Since non-linear molecule should exhibit (3N-6) = 3 peaks, the compound is non linear.



e. Study of hydrogen bonding molecule: Intermolecular and Intramolecular hydrogen bonding can be differentiated by taking a series of spectra of a compound at different concentrations. Intermolecular hydrogen bonding decreases with dilution and the intensity of such peaks will also decrease. Intramolecular hydrogen bonding on the other hand will show no such change. This can be explained by taking the ortho and para nitrophenols.

f. Study of progress of a chemical reaction

The rate the reaction can be determined by taking IR spectra at regular intervals of time Example: Oxidation of secondary alcohol to ketone

Secondary alcohol gives the absorption band at 3570cm-1 due to - OH stretching slowly disappears and a new band appears at 1725cm-1 due to C=O stretching.

6. What are the different types of electronic transactions? Draw the energy level diagram for various transitions. [CO3-H1-June 2009]

n→π* transitions:

n→π* transitions are shown by unsaturated molecule containing hetero atoms like N, O & S.

It occurs due to the transition of non-bonding lone pair of electrons to the antibonding

orbitals.This transition shows a weak band, and occurs in longer wavelength with low intensity. Example aldehyde & ketone with no double bond shows a band in the range of 270-300nm and if it is existing with the double bond show a band in the range of 300-350nm.



n→π* transitions:

n→π* transitions are shown by unsaturated molecule containing hetero atoms like N, O

& S. It occurs due to the transition of non-bonding lone pair of electrons to the antibonding orbitals.This transition shows a weak band, and occurs in longer wavelength with low intensity. Example aldehyde and ketone with no double bond shows a band in the range of 270-300nm and if it is existing with the double bond show a band in the range of 300-350nm.

σ→σ* transitions:

This type of transitions occurs in the compounds having only single

bonds. Energy required for this transition is more and absorption band occurs in

the far UV region (120-136 nm) Example: Methane, Ethane.

n→σ* transitions:

These transitions occur in saturated compounds having lone pair of electrons. This transition occurs along with n→σ* transition .The energy required for this transition is less than σ→σ* so absorption band occurs at longer wavelength in the near UV region (180-200nm). Example: (CH3)3N for n→σ* transition- at 227nm, for σ→σ* transition at 99nm.

π→π* transitions:

These transitions occur in unsaturated compounds i.e. the compounds having double bond and triple bonds. Example: Ethylene molecule shows intense band at 174nm and weak band at 200nm. Both are due to π→π* transitions. According to selection rule, the intense band at 174nm is due to allowed transistion.If there is any alkyl substitution in olefins shifts the absorption band to longer wavelength. This effect is known is batho-chromic effect or red shift.



Unit IV

Phase Rule And Alloys Part A

1. State phase rule and explain the terms involved. [CO4-L2-May 2009]

If the equilibrium between any number of phases is not influenced by gravity, or

electrical, or magnetic forces but is influenced only by pressure, temperature and

concentration, then the number of degree of freedom (F) of the system is related to

number of components (C) and number of phases (P) by the following phase rule

equation.

2. Define phase (P) with suitable example. [CO4-L1-Dec 2009]

Phase is defined as, “any homogeneous physically distinct and mechanically

separable portion of a system which is separated from other parts of the system by

definite boundaries”.

3. What is meant by degree of freedom? [CO4-L1-Dec 2011]

Degree of freedom is defined as, “the minimum number of independent variable

factors such as temperature, pressure and concentration, which must be fixed in

order to define the system completely”

4. Define component with suitable example. [CO4-L1-May 2007]

Component is defined as, “the smallest number of independently variable

constituents, by means of which the composition of each phase can be expressed in

the form of a chemical equation”.

5. Solve the no. of phases of the following. [CO4-H1-Dec 2013]



i) Sulphur (monoclinic) ⇋Sulphur (rhombic)⇋Sulphur(liquid)

Ans: Three phase

ii) Water+Alcohol ⇋ Vapour

Ans: Two phase

6. Write the degree of freedom for

i) CuSO4 .5H2O(s) ⇋ CuSO4.H2O(s) + 4H2O(v) [CO4-H1-Dec 2014]

Ans: Number of components = 1

Number of phase = 2

F = C-P+2; 1-2+2; F=1

ii) PCl5(s) ⇋ PCl3(l) + Cl2(v)

Ans: Number of components = 2

Number of phase = 3

F = C-P+1; 2-3+1; F=0

7. What is condensed phase rule? State its significance. [CO4-L2-May 2011]

The system in which only the solid and liquid are considered and the gas phase

is ignored is called a condensed system. Since pressure kept constant, the phase

rule becomes

F’ = C – P + 1 This equation is called reduced phase rule.

8. What is the difference in the phase diagram of a system forming simple

eutectic and compound formation? [CO4-L2-June 2007]

A binary system consisting of two substances, which are completely miscible in

the liquid state, but completely immiscible in the solid state, is known as eutectic

(easy melt) system. They do not react chemically. Of the different mixtures of the

two substances, the mixture having the lowest melting point is known as the eutectic

mixture.

Example: Pb-Ag system



9. How many phases and components and degree of freedom present in the

following system? [CO4-L2-May 2008]

CaCO3(s) ⇋CaO(s) +CO2 (g)

In the above system there are one solid CaCO3, one solid CaO and CO2.

Therefore Number of phase=3, Number of component=2. (Because the composition

of each of the above phases can be expressed in terms of any two of the three

components present.)

The degree of freedom is given by the equation F=C-P+2, F=2-3+2=1.

10. State the merits and demerits of phase rule. [CO4-L1]

Uses (or) merits of phase rule

1. It is a convenient method of classifying the equilibrium states in terms of phases,

components and degree of freedom.

2. It helps in deciding whether the given number of substances remains in

equilibrium or not.

Limitations of phase rule

1. Phase rule can be applied for the systems in equilibrium.

2. Only three variables like P, T & C are considered, but not electrical, magnetic and

gravitational forces.

11.Define eutectic point, triple point and melting point. [CO4-L1-Dec 2010]

It is the temperature at which two solids and a liquid phase are in equilibrium.

Solid A + solid B Liquid

It is the temperature at which three phases are in equilibrium.

Solid liquid vapour

It is the temperature at which the solid and liquid phases, having the same

composition, are in equilibrium.

Solid A solid B



PART B

1. Draw a neat one component water system and explain in detail. [CO4-L2-

May 2007, Dec 2011]

i) Curve OA

The curve OA is called vaporisation curve, it represents the equilibrium between

water and vapour. At any point on the curve the following equilibrium will exist.

Water Water vapour

The degree of freedom of the system is one, i.e, univariant.

This is predicted by the phase rule.

F=C-P+2; F=1-2+2; F=1



This equilibrium (i.e. Line OA) will extend up to the critical temperature (347o C).

Beyond the critical temperature the equilibrium will disappear only water vapour will

exist.

(ii) Curve OB

The curve OB is called sublimation curve of ice, it represents the equilibrium

between ice and vapour. At any point on the curve the following equilibrium will exist.

ice vapour

The degree of freedom of the system is one, i.e. univariant. This is predicted by

the phase rule.

F = C – P + 2; F = 1-2=2 ; F=1

This equilibrium (line OB) will extend up to the absolute zero (-273o C), where no vapour

can be present and only ice will exist.

(iii) Curve OC

The curve OC is called melting point curve of ice, it represents the equilibrium

between the ice and water. At any point on the curve the following equilibrium will exist.

Ice water

The curve OC is slightly inclined towards pressure axis. This shows that melting

point of ice decreases with increase of pressure.

The degree of freedom of the system is one i.e., univariant.

(iv) Point O (triple point)

The three curves OA ,OB ,OC meet at a point “O” ,where three phases namely

solid, liquid and vapour are simultaneously at equilibrium .

This point is called triple point, at this point the following equilibrium will exist.

Ice water vapour

The degree of freedom of the system is zero i.e., non-variant. This is predicted by the

phase rule.

F=C-P+2; F=1-3+2=0



Temperature and pressure at the point “O” are 0.00750C and 4.58 mm respectively.

(v) Curve OB’: Metastable equilibrium

The curve OB’ is called vapour pressure curve of the super-cool water or

metastable equilibrium where the following equilibrium will exist.

Super-cool water vapour

Sometimes water can be cooled below O0C without the formation of ice, this

water is called super –cooled water. Super cooled water is unstable and it can be

converted in to solid by seeding or by slight disturbance.

(vi) Areas

Area AOC, BOC, AOB represents water, ice and vapour respectively .The

degree of the freedom of the system is two.i.e. Bivariant.

This is predicted by the phase rule

F=1-1+2; F=2

2. Explain the lead silver system with a suitable phase diagram. Give its

applications. [CO4-L2-April 2002, May 2003]



Since the system is studied at constant pressure, the vapour phase is ignored

and the condensed phase rule is rule is used.

F ‘= C-P+1

The phase diagram of lead –sliver system is shown in the figure

It contains lines,areas and the eutectic point.

curve AO

The curve AO is known as freezing point curve of sliver. Along the curve AO,

solid Ag and the melt are in equilibrium.

Solid Ag melt

According to reduced phase rule, F’=C-P+1

C=2, P=2, F’=1 - The system is univariant

curve BO

The curve BO is known as freezing point curve of lead. Along the curve BO, solid

Pb and the melt are in equilibrium.

Solid Pb melt


C=2, P=2, F’=1 - The system is univariant.

Point “ O ” (eutectic point)

The curves AO and BO meet at point ‘ O ‘ at a temperature of 303 o C ,where the

three phases are in equilibrium.

Solid Pb + Solid Ag melt


C=2, P=3, F’=1 - The system is non-variant.

The point “ O “ is called eutectic point or eutectic temperature and is

corresponding composition,97.4 % Pb and 2.6 % Ag ,is called eutectic composition.

Below this point the eutectic compound and the metal solidify.

Areas



The area above the line AOB has a single phase (molten Pb + Ag). According to

reduced phase rule, F’=C-P+1

C=2, P=1, F’=2 - The system is bi-variant.

The area below the line AO, OB and point “O” have two phases and hence the system

isunivariant.


C=2, P=2, F’=1 - The system is uni-variant.

The process of raising the relative proportion of Ag in the alloy is known as

pattinson’s process.

The Pattinson process was patented in 1833. It depended on well-known

material properties; essentially that lead and silver melt at different temperatures. The

equipment consisted of a row of about 8-9 iron pots, which could be heated from below.

Argentiferrous lead was charged to the central pot and melted. This was then allowed to

cool, as the lead solidified, it was skimmed off and moved to the next pot in one

direction, and the remaining metal was then transferred to the next pot in the opposite

direction. The process was repeated in the pots successively, and resulted in lead

accumulating in the pot at one end and silver in that at the other. The process was

economic for lead containing at least 250 grams of silver per ton.

3. With neat phase diagram, explain the salient features of Zn-Mg alloy

system. [CO4-L2-Dec 2011, May 2014]

http://en.wikipedia.org/wiki/Patent



The zinc - magnesium alloy system is a very good example for the formation of

compound with congruent melting point. A compound is said to possess congruent

melting-point, if it melts exactly at a constant temperature into a liquid having the same

composition as that of solid.

The phase diagram of Zn-Mg binary alloy system may be considered as a

combination of two phase diagrams of Pb-Ag system placed side-by-side.

1. Left side consists of Zn and MgZn2 system.

2. Right side consists of MgZn2 and Mg system.

Left side of the phase diagram

1. Curve AE1

The curve AE1 is freezing point curve of Zn. Point A is the melting point of pure

Zn (420ºC). The curve AE1 shows the melting point depression of Zn by the

successive addition of Mg. Along this curve AE1, solid Zn and the melt are in

equilibrium.

Solid Zn Melt

2. Point E1 (Eutectic point)

The point E1 is the eutectic point, where, three phases (solid Zn, solid MgZn2 and

theirmelt) are in equilibrium. The temperature at this point is 380ºC.

Solid Zn + Solid MgZn2 Melt

Right side of the phase diagram



3. Curve CE2

The curve CE2 is freezing point curve of Mg. Point C is the melting point of pure

Mg (650°C). The curve CE2 shows the melting point depression of Mg by the

successive addition of Zn. Along this curve CE2, solid Mg and the melt are in quilibrium.

Solid Mg Melt

4. Point E2 (Eutectic point)

The point E2 is the eutectic point, where, three phases (solid Mg, solid MgZn2 and

their melt) are in equilibrium. The temperature at this point is 347°C.

Solid Mg + Solid MgZn2 Melt

5. Curve E1 BE2

The curve E1 BE2 is freezing point curve of MgZn2. Along the curve, solid MgZn2 and

the melt are in equilibrium.

Solid MgZn2 Melt

6. Point B

The point ‘B’ is the melting-point of the compound MgZn2. The temperature at the

point is 590°C. Here the solid has the same composition as the liquid. So MgZn2 is said

to possess congruent melting point. The composition of MgZn2 is 33.3% Mg and Zn is

67.7 % (that is the ratio of MgZn2 and Zn is 1:2).

7. Areas

(a) Below the line AE1

The area below the line AE1 consists of solid Zn and the solution.

(b) Below the line CE2

the area below the line CE2 consists of solid Mg and the solution.

(c) Below the line E1BE2

The area below the line E1BE2 consists of solid MgZn2 and the solution.

(d) Below the line E1 and E2

They are below the point E1 and E2 consists of solid Zn + solid MgZn2 and

solid MgZn2+ solid Mg respectively.

(e) Below the line AE1BE2C

The area above the line AE1BE2C consists of only liquid phase.

ALLOYS



1. Mention any two advantages of alloy making. [CO4-L1-Dec 2006]

To increase the hardness o the metal.

To resist the corrosion of the metal

2. What is meant by quenching in heat treatment of metals? [CO4-L1]

It is the process of heating steel beyond the critical temperature and then

suddenly cooling it either in oil or brine-water or some other fluid.

3. What is normalizing? [CO4-L1-June 2007]

It is the process of heating steel to a definite temperature (above its higher critical

temperature) and allowing it to cool gradually in air.

4. Formulate the composition and uses of brass and bronze. [CO4-L2]

Brass

Cu = 60-90%

Zn = 10-40%

Uses

Forgings, rivets, hardwares, jewellery, flexible hoses.

Bronze

Cu = 80-95%

Sn = 5-20%

Uses

Manufacture of pumps, valves, coins and statues.

5. What is alnico? [CO4-L1-June 2012]

Al = 12%

Ni = 20%

Co = 6%

Fe = 62%

6. What is wood’s metal? [CO4-L1]



Wood’s metal: Wood’s metal is an alloy of lead, bismuth, tin and cadmium. Its

melting point at 60.50 C, which is for below the melting point of any of these

constituent metals .Alloying makes the metal easily fusible. Wood’s metal have low

melting point. Generally pure metal possess some useful properties such as high

melting point, high densities, malleability, and ductility. Good thermal and electrical

conductivity. Wood's metal is also used in the making of extracellular electrodes for

the electro-physiological recording of neural activity.

7. Alloys are more resistant to corrosion than pure metals. Why? [CO4-L2-Dec

2013]

Alloys are more resistant to corrosion: Metal, in pure form, are quite reactive and

easily corroded by atmospheric gases like hydrochloric acid, hydrogen sulphide,

sulfuric acid etc., thereby their life is reduced. But it a metal is alloyed, it is resist

corrosion.

Example :( i). Pure iron gets rusted, but when it is alloyed with carbon or

chromium (stainless steel), resists corrosion.

(ii). Stainless steel differs from carbon steel by the amount of chromium present.

Unprotected carbon steel rusts readily when exposed to air and moisture. This

iron oxide film (the rust) is active and accelerates corrosion by forming more iron

oxide.

8. How will you modify chemical activity and colour of the metal? [CO4-L2-

June 2009]

Modifying chemical activity: Chemical activity of metals (Copper, Nickel, iron

etc,.) can be increased or decreased by alloying.

Example: Sodium amalgam is less active than sodium, but aluminum is more

active than aluminum.

The dull coloured metals are improved by alloying with metals.



Example; Brass, an alloy of copper (red) and zinc (silver-white), is white in

colour.

9. How do you get good casting of metal? [CO4-L2]

Good casting of metal: casting involves pouring liquid metal into a mold, which

contains a hollow cavity of the desired shape, and then allowing it to cool and

solidify. The solidified part is also known as a casting, which is ejected or broken out

of the mold to complete the process. Some metals expand on solidification but are

soft and brittle. The addition of other metals produce alloys which are hard, fusible

and expand on solidification and thus give good casting.

Example; An alloy of lead with 5% tin and 2% antimony is used for casting

printing type, due to its good casting property.

Part B

1. What are the main purposes of alloying steel? [CO4-L2-May 2009]

To increase the hardness of the metal

Example: Gold and silver are soft metal they are alloyed with copper to make them hard

To lower the melting points of the metal

Example: Wood metal (an alloy of lead, bismuth, tin and cadmium) melts at 60.5⁰c

which is far below the melting points of any of these constituent metals

To resist the corrosion of the metal

Example: Pure iron rested but when it is alloyed with carbon chromium (stainless steel)

which resists corrosion

To modify chemical activity of the metal

Example: Sodium amalgam is less active than sodium but aluminium amalgam is more

active than aluminium

To modify the colour of the metal

Example: Brass an alloy of copper (red) and size (silver-white) is white colour.



To get good casting of metal

Example: An alloy of lead with 5% tin and 2% antimony is used forecasting printing type

due toits good casting property.

2. Discuss different heat treatment methods and their effects on alloys. [CO4-

L2-Dec 2009, May 2009, May 2011]

1 ANNEALING

Annealing means softening. This is done by heating the metal to high

temperature followed by slow cooling in a furnace.

Or

It is a heat treatment wherein a material is altered, causing changes in its

properties such as strength and hardness. It the process of heating solid metal to high

temperatures and cooling it slowly so that its particles arrange into a defined lattice.

Purpose of annealing

It increases the machinability

It also removes the imprisoned gases

Types of annealing

Annealing can be done in two types

Low temperature annealing (or) process annealing

High temperature annealing 9or) full annealing

Low temperature annealing (or) process annealing

It involves in heating steel to a temperature below the lower critical point followed

by slow cooling

Purpose

It improves machinability by reliving the internal stress or internal strain

It increases ductility and shock resistance

It reduce hardness



High temperature annealing (or) fault annealing

It involves in heating to a temperature about 30 to 50⁰C above the higher critical

temperature and holding it at that temperature for sufficient time to allow the internal

changes to take place and then cooled room temperature

The approximate annealing temperature of various grades of carbon steel are

1. Mild steel=840-870⁰c

2. Medium carbon steel=780-840⁰c

3. High carbon steel=760-780⁰c

Purpose

It increases the ductility and machinability

It makes the steel softer, together with an appreciable increases in its toughness

2 HARDENING (OR) QUENCHING

It is the process of heating steel beyond the critical temperature and then

suddenly cooling it either in oil or brine water or some other fluid.

The faster the rate of cooling harder will be the steel produced.

Medium and high carbon steel can be hardened but low carbon steel cannot

hardened.

Purpose

It increases its resistance to wear ability, to cut other metal and strength.

It increases abrasion resistance.

Used for making cutting tools.

3 TEMPERING

It is the process of heating the already hardened steel to a temperature lower

than its own hardening temperature & then cooling it slowly.

The reheating controls the development of the final properties

Thus,



(a)For retaining strength & hardness, reheating temperature should not exceed 400⁰C.

(b) For developing better ductility & toughness, reheating temperature should be within

400-600⁰C.

Purpose

It removes stress &strains that might have developed during quenching.

Increases toughness & ductility.

Used for cutting tools like blade, cutters etc.

4 NORMALISING:

It is the purpose of heating steel to a definite temperature (above its higher

critical temperature) & allowing it to cool gradually in air.

Purpose

Recovers homogeneity

Refines grain size.

Removes internal stresses

Increases toughness

Used in engineering works

NOTE: The difference between normalzsed & annealed steel are

Nnormalized steel will not be as soft as annealed steel.

Also normalizing takes much lesser time than annealing.

5 CARBURIZING

The mild steel article is taken in a cast iron box within containing small pieces of

charcoal (carbon material).

It is heated to about 900 to 950⁰C & allow it for sufficient time,so that the carbon

is absorbed to required depth .

The article is then allowed to cool slowly within the box itself.

The outer skin of the article is converted into high carbon steel containing about

0.8 to 1.2% carbon.



Purpose

To produce hard surface on steel article

6 NITRIDING

Nitriding is the process of heating the metal alloy in presence of ammonia to

about 550⁰C.

The nitrogen (obtained by the dissociation of ammonia) combines with the

surface of the alloy to form hard nitride.

Purpose

To get super-hard surface.

3. Give the composition and uses of the ferrous alloy. [CO4-L1-Dec 2010]

(i)NICHROME:

Nichrome is an alloy of nickel & chromium

Composition:

Nickel – 60%

Chromium – 12%

Iron – 26%

Manganese – 2%

Properties:

Good resistance to oxidation & heat

High melting point & electrical resistance

Withstand heat up to 1000-1100⁰C

USES:

Used for making resistance coils,heting elements in stoves & electric irons

Used in making parts of boilers,steam lines stills,gas turbines,aero engine

valves,retorts,annealing boxes.

(i) ALNICO:



Alnico is an alloy of aluminium-nickel-cobalt

Composition:

Aluminium – 8-12%

Nickel – 14-28%

Cobalt – 5-35%

Properties:

Excellent magnetic properties & high melting point

Magnetized to produce strong magnetic fields as high as 1500 gauss

Types of alnico alloys:

Alnico alloys are of two types

Isotropic alnico: It is effectively magnetized in any direction

Anisotropic alnico: It possesses proffered direction of magnetization. Anisotropic alnico

possesses greater magnetic capacity in their preferred orientation than isotropic alnico.

Uses

Used as permanent magnets in motors, generators, radio speaker’s

microphones, telephone receivers & galvanometers.

Stainless steels (or) corrosion resistant steels

These are alloy steels containing chromium together with other elements such as

nickel, molybdenum,etc.

Chromium-16% or more

Carbon-0.3-1.5%

PROPERTIES

Resist corrosion by atmospheric gases & also by other chemicals.

Protection against corrosion is due to the formation of dense, non-

porous,tough film of chromium oxide at the metal surface. If the film cracks, it

gets automatically healed up by atmospheric oxygen

TYPES OF STAINLESS STEEL:



1. HEAT TREATABLE STAINLESS STEEL

COMPOSITION

Carbon-1.2%

Chromium-less than 12-16%

PROPERTIES

Magnetic,tough & can be worked in cold condition

USES

Can be used up to 800⁰C

Good resistant towards weather & water

In making surgical instruments, scissors, blades and etc.

2. HEAT TREATABLE STAINLESS STEEL

PROPERTIES

Possess less strength at high temperature

Resistant to corrosion

TYPES OF NON HEAT TREATABLE STAINLESS STEEL

(a) MAGNETIC TYPE

COMPOSITION

Chromium-12-22%

Carbon-0.35%

PROPERTIES

Can be forged, rolled & machined

Resist corrosion

USES

Used in making chemical equipments& automobile parts.

(b) NON MAGNETIC TYPE

COMPOSITION



Chromium-18-26%

Nickel-8-21%

Carbon-0.15%

Total % of Cr & Ni is more than 23%.

EXAMPLE: 18/8 STAINLESS STEEL

COMPOSITION

Chromium-18%

Nickel-8%

PROPERTIES

It possess high Resistance to corrosion.

Corrosion resistance is increased by adding molybdenum

USES

In making household utensils,sinks,dental & surgical instruments.

NON FERROUS ALLOYS:

Do not contain iron as one of the main constituent.

Main constituents are copper,aluminum,lead,tin,etc.

PROPERTIES

Softness & good formability

Attractive (or) very good colours

Good electrical & magnetic properties

Low density & coefficient of friction

Corrosion resistance

4. Discuss the composition, characteristics and uses of non-ferrous alloys.

[CO4-L2-Nov 2009]

Brass is an important copper alloy, containing mainly copper and zinc. They

possess,

(i) Greater strength, durability and machinability than copper,



(ii) Lower melting points than Cu and Zn,

(iii) Good corrosion resistance and water resistance property.



BRONZE (COPPER ALLOY)

Bronze is also a copper alloy containing copper and tin.

They possess,

(i) Lower melting point than steel and are more readily produced from their constituent

metals.

(ii) Better heat and electrical conducting property than most of the steels,

(iii) non-oxidizing, corrosion resistance and wate resistance property.

IMPORTANT BRONZES THEIR PROPERTIES AND USES

Some important bronzes and their compositions, properties and uses are given in the following

table.



Unit 5

Nano Chemistry

Part - A

1. What are nano particles? [CO5-L1]

Nanoparticles are particles having size of which ranges from 1-50nm.

2. What are nano materials? [CO5-L1- NOV/DEC 2015]

Nanomaterials are the materials having components with size less than 100 nm at least

in one dimension.

3. Define nano-wires. [CO4-L1- MAY/JUNE 2014]

Nano-wire is a material having an aspect ratio ie., length to width ratio greater than 20.

They are also referred to as quantum wires.

4. What is nano –rod? [CO5-L1-NOV/DEC 2014]

Nano-rod is a material having an aspect ratio in the range 1 to 20 with short dimension

of the material being 10-100nm.

5. What are nano clusters? [CO5-L1]

Nano clusters constitute an intermediate state of matter between molecules and bulk

materials.

6. What are CNTs? [CO5-L1- JAN 2014]

Carbon nanotubes(CNT) are allotropes of carbon with a nanostructure having a length -

to-diameter ratio greater than 1,000,000.

7. Define nanochemistry. [CO5-L1-May 2010]

Nanochemistry is defined as the study of manipulation of materials at atomic molecular

and macromolecular scales.

8. Name the various methods of synthesis of nano-material. [CO5-L1]



Laser ablation

Chemical vapourdepostion

Precipitation

Electro-depostion

Thermolysis

9. What is CVD? [CO5-L1-May 2015]

CVD is Chemical Vapour deposition. It is a process of chemically reacting volatile

compound of a material with other gases, to produce a non-volatile solid that deposits

automatically on a suitably placed substrate.

10. Write any two important applications of gold nano particles in medicine. [CO5-

L2-NOV/DEC 2015]

Nano gold when mixed with different type microorganisms show different colours

which is used to identify the harmless and dangerous micro organisms.

Nano materials are used as drugs for the treatment of cancer and TB.

Gold nano particles injected in to the body reacts with the protein and the protein

concentration can be detected by passing laser beam externally.

11. Mention the application of Nano-wires. [CO5-L2-Dec 2009]

Nano-wires are used for enhancing mechanical properties of composites.

It is used to prepare active electronic components such as p-n junction

and logic gates.

12. What are the characteristics of Nano-rods? [CO5-L2]

Nano-rods are two-dimensional materials.

It also exhibits optical and electrical properties.

13. What is magic number? [CO5-L1-May 2009]

It is the number of atoms in the clusters of critical sizes with higher stability.

14. Mention some uses of CNTs. [CO5-L2-June 2011]



It is used in battery technology and in industries as catalyst.

It is used in composites, ICs.

CNTs are used effectively inside the human body for drug delivery.

15. Distinguish between bulk particles and nano-particles. [CO5-L2-Nov 2010]

Nano-particles Bulk particles

Size is less than 100 nm. Size is larger in micron size.

Collection of few molecules.

Collection of thousand of molecules.

Surface area is more.

Surface area is less.

.Strength, hardness are more.

Strength, hardness are less.

Part B

1.Explain the properties of Nano-materials. [CO5-L2-Jan 2009]

(i) Melting points:

They have lower melting points and reduced lattice constants.

(ii) Optical properties:

The change in optical properties is caused by two factors.

(i) The quantum confinement of electrons within the nano-particles increases

the energy level spacing.

(ii) Surface plasma resonance, which is smaller in size for nano-particles than

the wavelength of incident radiation.

(iii) Magnetic properties:

Bulk materials have ferro-magnetic properties. When the particle size is

reduced, they get super – paramagnetic properties.



(vi) Mechanical properties:

Mechanical properties of polymeric materials can be increased by the

addition of nano- fillers. Nano-materials are stronger, harder and more wear

resistant and corrosion resistant.

(v) Electrical properties:

(a) Electrical conductivity decreased in reduced dimension.

(b) Electrical conductivity increased in micro-structure.

(vi) Chemical properties:

(a) In heat treatment of nano material, diffusion of impurities, structural defects,

and dislocation increased.

(b) Increased perfection will have increased chemical properties.

2. Discuss the laser ablation, CVD and electrodeposition techniques for the

synthesis of nano-particles. [CO5-L2-June 2013]

Top Down / Physical / Hard Methods:

1. Laser Ablation:

Laser Ablation chamber :

High-power laser pulse is used to evaporate the matter from the target . The total

mass ablated from the target per laser pulse is referred to as the ablation rate.

Reaction setup:

When a beam of laser is allowed to irradiate the target, a supersonic jet of particle is

evaporated from the target surface. Simultaneously, an inert gas like Argon, Helium is

allowed into the reactor to sweep the evaporated particles from the furnace zone to the

colder collector.



The ablated species condense on the substrate placed opposite to the target. The

ablation process takes place in a vacuum chamber, or in the presence of some

background gas.

2. Chemical Vapour Deposition (CVD):

In this process volatile compound of a material chemically react with other gases, to

produce a non-volatile solid that deposit on a substrate.

CVD reaction requires activation energy to proceed .This energy can be provided by

several methods.

a) Thermal CVD: The reaction is activated by high temperature above 900o C .

b) Plasma CVD: The reaction is activated by plasma at temperature between 300-

700oC.

(c) Laser CVD: Pyrolysis occurs when laser thermal energy falls on an absorbing

substrate.



(d) Photo-laser CVD: The chemical reaction is induced by UV radiation, which has

sufficient photon energy, to break the chemical bond in the reactant molecules.

Various steps involved in synthesis of CVD are:

3. Transport of gaseous reactant to the surface.

4. Adsorption of gaseous reactant on the surface.

5. Catalysed reaction occurs on the surface.

6. Product diffuses to the growth site.

7. Nucleation and growth occurs on the growth site.

8. Desorption of reaction products away from the surface.

CVD Reactor: The CVD reactors are of two types

* Hot-wall CVD

* Cold-wall CVD

Hot wall CVD reactors are usually tubular in form and heating is

accomplished by surrounding the reactor with resistance elements.

In cold- wall CVD reactors, substrates are directly heated inductively by

graphite subsectors, while chamber walls are air or water cooled.

3. Electrodeposition:

Template assisted electro deposition is an important technique for synthesizing

metallic nanomaterials with controlled shape and size. Arrays of nano-structured

materials with, specific arrangements can be prepared by this method, using an active



template cathode in an electrochemical cell. The electrodeposition method consists of

an electrochemical cell. The cell usually contains a reference electrode, a specially

designed cathodes and an anode. The cathode substrate on which elelctrodeposition of

the nanostructures, can be made of either non metallic or metallic materials. By using

the surface of the cathode as a template, various desired nanostructures can be

synthesized for specific applications.

3. Describe the synthesis of nanomaterials by precipitation and thermolysis

method. [CO5-L2-Dec 2012]

Bottom - up methods /Chemical /soft methods:

It involves building up of nano materials from the bottom by atom by atom (=0.1

nm) or molecule by molecule or cluster by cluster. This method is carried out by the

following process:

Precipitation

Thermolysis

a) Solvothermal synthesis

b) Hydrothermal synthesis

A) Precipitation method:

Generally nano particles are synthesized by the precipitation reaction between

the reactants in presence of water soluble inorganic stabilizing agent.

Precipitation of BaSO4 Nano particles:

10 g of Sodium hexa meta phosphate (stabilizing agent) was dissolved in 80

ml of distilled water in 250ml beaker with constant stirring. Then 10ml of 1M sodium

sulphate solution was added followed by 10ml of 1M Barium Nitrate ( Ba (NO3)2)



solution .The resulting solution was stirred for 1 hour. The resulting precipitate was

then centrifuged, washed with distilled water and vacuum dried.

Ba(NO3)2 + Na2SO4 BaSO4 ↓+ 2NaNO3

In the absence of stabilizing agent, Bulk BaSO4 is obtained.

Precipitation by reduction :

Reduction of metal salt to the corresponding metal atoms .These atoms act

as nucleation centres leading to formation of atomic clusters. These clusters are

surrounded by stabilizing molecule that prevents the atoms agglomerating.

B. Thermolysis method:

Thermolysis is characterized by subjecting the metal precursors (usually

organometallic compounds in oxidation state zero) at high temperatures together

with a stabilising agent. Nano particles show an increase in size relating to the

temperature size.This is due to the elimination of stabilizing molecule, generating a

greater aggregation of the particles.

Hydrothermal synthesis



a.Hydrothermal synthesis:

It involves crystallization of substances from high temperature aqueous

solutions at high vapour pressure. Hydrothermal synthesis is usually performed below

the super critical temperature of water.(3740).

Method:

Hydrothermal synthesis is performed in an apparatus consisting of a steel pressure

vessel called autoclave in which a nutrient is supplied along with water. A gradient of

temperature is maintained at the opposite ends of the growth chamber, so that the

hotter end dissolves the nutrient and the cooler end cause seeds to take additional

growth.

b. Solvothermal synthesis :

Solvothermal synthesis involves the use of solvent under high temperature

(between 100o – 1000 oc) and moderate to high pressure (1 atm to 10,000 atm )

that facilitate the interaction of precursors during synthesis.

Method:

A solvent is mixed with certain metal precursors and the solution mixture is

placed in an autoclave kept at relatively high temperature and pressure in an oven to

carry out the crystal growth. The pressure generated in the vessel, due to the solvent

vapour , elevates the boiling point of the solvent.



Eg: Methanol, Ethanol, Toluene, Cyclohexane ,etc.

Solvothermal synthesis of Zinc oxide:

Zinc acetate dehydrate is dissolved in 2-propanol at 50 oC. Subsequently the

solution is cooled to 0oC and NaOH is added to form ZnO. The solution is then heated

to 65oC to allow ZnO growth for some period of time before a capping agent (1-

dodecanethiol) is injected to the suspension to arrest growth. The rod shaped ZnO

nano-crystal is obtained.

Uses:

Much geometry including thin film, bulk powder, and single crystals can be

prepared.

Thermodynamically stable novel materials can also be prepared easily.

4. Write a short note on Nano wires, Nano rods and Nano clusters. [CO5-L2-June 2013] A. Nano –wires:

Nano-wire is a material having an aspect ratio i.e. length to width ratio

greater than 20. Nano – wires are also referred to as “quantum wires”.

c) Nano –Wires of metals: Au ,Ni, Pt

d) Nano –Wires of semiconductors: Si, GaN

e) Nano –Wires of Insulators: SiO2, TiO2

f) Molecular Nano –Wires: DNA

Characteristics of Nano- Wires :

c) Nano- Wires are one – dimensional material.

d) Conductivity of a nano-wire is less than that of the corresponding bulk materials.



e) It exhibits distinct optical, chemical, thermal and electrical properties due to this

large surface area.

f) Silicon nano-wires show strong photoluminescence characteristics.

Synthesis of nanowires:

1. Template assisted synthesis:

The templates contain very small cylindrical pores or voids within the host

material and the empty spaces are filled with the chosen material to form nanowires.

Eg. Mesoporous Alumina.

2. VLS method:

This method is used for the production of single crystal of semiconductors of elemental

Silicon and Germanium. The mechanism involves a gas phase reaction followed by

anisotropic crystal growth (different properties in different directions). For example the

laser ablation and thermal evaporation of a solid target made of pure Si powder mixed

with metals (Fe, Co, Ni) catalyst at 1200o -1400o C followed by condensation on a

substrate maintained at 900o-1100o C facilitates the growth of long Si nanowires with

diameters in the range of 20-80 nm. Each wire consists of crystalline Si core encased

by an outer layer of silicon dioxide.

Applications of nanowires:

b) Nanowires are used to enhance mechanical properties of composites.

c) Semiconductor nanowires are used as components in making transistors, diodes,

logic gates and digital computing.

d) Nanowires find applications in high density data storage either as magnetic read

heads.

B. Nano Rods:



Nano rods are a material having an aspect ratio in the range 1 to 20 nm.

Characteristics of Nano rods:

It exhibits special optical and electrical properties.

Nano rods are two dimensional materials.

Synthesis of Nano rods:

Direct chemical

Synthesis:

Metal atoms are combined with ligands which act as shape control agents

which make bonds with different metal atoms of different strength to get nano rods.

Applications of nanorods:

Nano rods are used in display technology and micromechanical switches.

C. Nano clusters:

Nano clusters are multi atom particles of size intermediate between molecule

and bulk materials. The size of nano clusters range from 1-10nm .The atoms or

molecules in a clusters are bound by any of the forces like covalent, ionic, Vander

waal’s forces. When a gas condenses in to cluster of atoms the no. of atoms in these

clusters varies between a few to hundreds. Clusters of certain size called critical size

are more stable than others (200-103 atoms).

The no. of atoms in a cluster of critical size with higher stability is called

Magic number. Cluster of transition metal atoms have chemical, electronic,

magnetic properties which vary with number of constituent atoms.

Source of clusters:



Super sonic nozzle source:

The metal is vaporized in an oven and mixed with inert carrier gas at a pressure

of several atmospheres at a temperature of 75-1500K .Metal/ Carrier gas mixture is

allowed to pass through the nozzle in to high vacuum which produces a supersonic

beam of the mixture. The carrier gas produces large clusters. In the absence of carrier

gas small clusters are produced.

Gas aggregation source:

Vapours of metal atoms are introduced in to inert gas which is maintained at

high pressure and high temperature. The gas phase is supersaturated with metallic

species and aggregates. It acts as a seed and produces a continuous beam of nano

clusters.

5. Discuss the application of Nano materials in various fields. [CO5-L2-Jan 2009,

Dec 2011]

Applications of Nano materials:

A. Industries

1. As catalyst:



The nano materials have more no. of surface atoms which make them catalytically

active. For example, bulk gold chemically inert white nano gold possess excellent

catalytic property.

2. Water purification:

Dissolved salts and colour producing organic compounds can be filtered very easily

from water by using nano porous membrane having pores smaller than

10nm.Magnetic nano particles are used to remove heavy metal contamination from

waste water.

3. In Fabric industry:

Embedding of Nano particles on fabrics make them stain repellent. Socks embedded

with silver nano particles remove the bacteria and makes them odour free.

4. In automobiles:

Fuel consumption in automobiles can be reduced by using specially designed nano

particles as fuel additives. Incorporation of small amount of nano particles in car

bumpers make them stronger than steel.

5. In food industry:

Nano particles are used to make packing material and containers to store food.

6. In Solar cells:

Absorption of solar radiation in solar cells containing nano particles are higher than the

bulk materials.

B. Medicine :



1. Nano drugs:

Nano materials are used as drugs for the treatment of cancer and TB.

2. Nano medibots :

Nano particles act as nano medibots which identify and penetrate the cancer cells and

destroy them by supplying anti cancer drugs.

3. Gold coated Nano shells:

Gold coated nano shells containing silicon core are administered near the tumours.

When IR light is irradiated on the skin externally, the gold nano shells absorb the light

and convert in to heat which destroys the cancer cells responsible for tumour.

4. Gold nano particles as sensors:

Nano gold when mixed with different type microorganisms show different colours

which is used to identify the harmless and dangerous micro organisms.

5. Protein analysis:

Gold nano particles injected in to the body reacts with the protein and the protein

concentration can be detected by passing laser beam externally.

6. Gold nano shells for blood Immune assay:

Gold nano shells are used to detect WBC count in blood.

7. Gold nano shells in Imaging:

Gold nano shells is used as a scanning probe which gives the magnified image of cells

in a body.

8. Targetted drug delivery:



It involves slow and selective release of drug to the targeted organs.

C. Electronics:

Quantum wires have electrical conductivity

Nano radios are produced by using CNT

MOSFET (Metal Oxide Semiconductor Field Effect Transistor) is used

for Amplifying and switching electronic signals.

Nano wires are used to build transistors without p-n junction diode.

Transistor called NOMFET (Nano particle Organic Memory Field effect

transistor) is created by combining Gold nano particles with organic

molecules

D. Biomaterials:

Nano materials are used as bone cement and bone plates in hospitals

It is used as material for joint replacement

It is used in the manufacture of some components like heart valves,

contact lenses, dental implants etc.

6. Write a short note on CNT. [CO5-L1-June 2012]

Carbon NanoTubes ( CNTs) :

CNTs are allotropes of carbon in the form of molecular scale tubes made of

graphitic carbon with unique mechanical and electrical / electronic properties with

potential applications in electronic, information technology and medical fields.

Carbon nanotubes were discovered by Sumio Iijima 1991, while preparing

C60 molecules by the carbon arc process. They are also known as tubular

fullerenes and are cylindrical graphene sheets of sp2 bonded carbon atoms.

These nanotubes are single molecules measuring few nanometers in diameter

and several microns in length. Depending on the growth process, the length of the



tubes can be 100 nanometers to several microns (1 micron = 1000 nm), the diameter

varying from 1 to 20 nanometers.

Structure of CNTs:

Carbon nanotube is a tabular form of carbon with 1-3 nm diameters and a length

of few nm to microns. Each carbon atom in the carbon nanotubes is linked by the

covalent bond. They are two types of nanotubes.

1. Single walled carbon nanotubes (SWCNT) having a diameter of 1 nm. Three

types of SWCNTs are as follows:



2. Multi-walled nanotubes (MWNTs) or nested nano tubes consist of multiple

layers of graphite rolled to form tube shape.

Mutliwalled Nanotube

Synthesis of Carbon Nano Tubes:

Carbon nanotubes are prepared by the following methods.

a) Pyrolysis of hydrocarbons.

b) Laser evaporation

c) Carbon arc method.

d) Chemical vapour decomposition.

Pyrolysis of Hydrocarbon:

Carbon nano tubes are synthesized by the pyrolysis of hydrocarbons such as

acetylene at about 700oC in the presence of Fe-Silica or Fe-graphite catalyst under

inert conditions.

Laser Evaporation:



It involves vapourization of graphite target, containing small amount of cobalt and nickel,

by exposing it to an intense pulsed laser beam at higher temperature (1200oC) in a

quartz tube reactor. An inert gas such as argon is simultaneously allowed to pass into

the reactor to sweep the evaporated carbon atoms from the furnace to the colder

collector, on which they condense as carbon nanotubes.

Carbon arc Method:

It is carried out by applying direct current (60-100 A and 20 – 25 V) arc between

graphite electrodes of 10-20 µm diameters. The discharge vapourizes one of the

carbon rods and forms a small rod- shaped deposit on the other rod. The yield

depends on the uniformity of the plasma arc and the temperature. The deposits

contain 10 – 50 SWNTs, the average size about 1.4 nm in diameter and about 10 µm

in length.

Chemical Vapour Deposition:

It involves decomposition of vapour of hydrocarbons such as methane, acetylene,

ethylene, etc., at high temperatures (1100oC) in presence of metal nanoparticle

catalyst like nickel, cobalt, iron supported on MgO or Al2O3. Carbon atoms

produced by the decomposition condense on a cooler surface of the catalyst.

Properties of CNTs:

(1) Mechanical Properties:

Young’s modulus of carbon nano tubes is 10 times greater than that of steel.

CNTs have very structural defects in their walls, and hence, do not fracture on

bending. The tensile strength of CNTs is about about 20 times that of steel.

(2) Electrical Properties:



The electrical properties of CNTs vary between metallic to

semiconducting materials. The very high electrical conductivity of CNT is

due to the minimum defects in the structure.

(3) Thermal Conductivity:

The thermal conductivity of CNT is very high due to the vibration of covalent

bonds due to minimum defects in the structure.

Applications of Carbon NanoTubes:

Due to the unusual and unique properties of CNTs they find potential applications in the

following field:

Carbon nanotubes play an important role in the battery technology, because

some charge carriers can be successfully stored inside the nanotubes.

Multi-walled CNTs can be used as storage devices to store hydrogen gas in fuel

cells.

CNTs are used as catalyst in chemical reactions.CNTs can be used for drug

delivery within the body by placing the drugs within the tubes or by attaching the

drug to the sides of the tubes.

CNTs are used as light weight shielding materials to protect electronic

equipments from electromagnetic radiation.

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