Engineering Chemistry i Qbwithans annauniv

7/23/2019 Engineering Chemistry i Qbwithans annauniv

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VALLIAMMAI ENGINEERING COLLEGE S.R.M Nagar, Kattankulathur – 603203

DEPARTMENT OF CHEMISTRY

CY-6151-ENGINEERING CHEMISTRY-I

Question Bank with Answers(2015-2016)

Unit Title Page No.I Polymer Chemistry 1-30

II Chemical Thermodynamics 31-48

III Photochemistry & Spectroscopy 49-73

IV Phase Rule & Alloys 74-94

V Nanochemistry 95-120



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VALLIAMMAI ENGINEERING COLLEGEDEPARTMENT OF CHEMISTRY

CY 6151- ENGINEERING CHEMISTRY – IUNIT I – POLYMER CHEMISTRY

PART – A (2 MARKS)

1.

What is meant by polymerization?Polymerization is a process in which large number of small molecules called monomers

combine to give a bigger molecule called polymer with or without elimination of small

molecules like water, HCl etc.,

2. Define degree of polymerization?The number of repeating units in a polymer chain is known as degree of polymerization.

It is represented as

Molecular weight of the polymer

Degree of polymerization (n) = -----------------------------------------------

Molecular weight of the monomer

3.

Differentiate homochain and heterochain polymers? Give examples?

Homochain polymers are those polymers which contain only carbon atoms in

their long linear chain. Ex: Polyethylene.

Heterochain polymers are those polymers which contain hetero atoms like

Oxygen,Nitrogen Sulphur etc., other than carbon atoms in their long linear chain.Ex: Nylon6,6

4. Explain functionality with a suitable example.

The number of bonding sites or reactive sites or functional groups present in a monomer is

known as its functionality.

Ex: Hexa methylene diamine. The functionality for this monomer is two because it contains two

functional groups.It is called as a bifunctional monomer.

5. What is meant by 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.It is of three types namely Isotactic, syndiotactic and atactic polymer.

6. What are syndiotactic polymers?The functional groups are arranged in an alternating fashion respect to that of the main

chain , then the polymer is called Syndiotactic polymer.

Ex: Polystyrene formationH H

n CH = CH2 → - C – CH2 – C – CH2 – C – CH2 – C –

H H n

Styrene Syndiotactic polystyrene



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7. How copolymerization is carried out? Give an example.

The joint polymerization in which two or more different monomers combine to give highmolecular weight polymers is known as co polymerization.

It is mainly carried out to vary the properties of polymers such as hardness,

strength,rigidity, heat resistance etc.,

Ex: n(CH2=CH-CH=CH2) + n CH2=CH →----CH2-CH=CH-CH2-CH2-CH-----

n

Butadiene Styrene Strene butadiene rubber - SBR

8. Define glass transition temperature.

The temperature below which a polymer is a glassy brittle solid and above it a soft elastic

substance is known as glass transition temperature.

It can be determined by using dialatometer or differential scanning calorimeter.

9. How solution polymerization is carried out?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. The polymer

produced is precipitated by pouring it in a suitable non-solvent.

Monomer + Initiator + Chain Transfer agent → polymer

10. Distinguish between addition and condensation polymerization.

S.NO ADDITION POLYMERIZATION CONDENSATIONPOLYMERIZATION

1 The monomer must have atleast one multiple bond.

The monomer must have atleast twodifferent functional groups.

2 Monomers add on to give a polymer and noother by product is formed.

Monomers condense to give a polymerand by-products such as H2O, CH3OH.

3 Molecular weight of the polymer is anintegral multiple of molecular weight ofmonomer.

Molecular weight of the polymer neednot be an integral multiple of monomer.

4 High molecular weight polymer is formed atonce.

Molecular weight of the polymer risesthroughout the reaction.

11.Define poly dispersity index.

The ratio of the weight –average molecular weight ( w) to that of number –average

molecular weight ( n) is known as polydispersity index.(PDI).

PDI = w/ n

As w is always greater than n, the ratio is greater than 1 or equal to 1



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PDI = w/ n≥ 1

12.Mention the various techniques of polymerization.

• Bulk polymerization

• Solution polymerization

•

Suspension polymerization

• Emulsion polymerization

13.Explain condensation polymerization reaction with an example.

It is a reaction between simple polar groups containing monomers with the formation of

polymer and elimination of small molecules like H2o, HCl,etc.

Example: n H2 N-(-- CH2-)6 —NH2 + n HOOC-(--CH 2-)4 —COOH

Hexamethylene diamine Adipic acid

------NH—(--CH2-)6 —NH – CO –( -CH2 -) 4 —CO----

Nylon 6.6 (polymide) n

In some cases condensation polymerization takes place without the elimination of small

molecules like H2O, HCl,etc., but by just the opening of cyclic compounds.

14.What are the advantages and disadvantages of plastics?

ADVANTAGE:

• They are light in weight

•

They possess low melting point.

• They can be easily moulded and have excellent finishing.

• They possess very good strength and toughness.

DISADVANTAGE:

• They have high softness.

• They undergo embrittlement at low temperature.

•

They are Non –biodegradable.• They undergo deformation under load.



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15.What are thermo and thermosetting plastics? Give example.

THERMO PLASTIC:

Thermo plastics are prepared by addition polymerization. They are straight chain(or)

slightly branched polymers and various chains are held together by weak Vander Waal’s forces

of attraction.

Thermoplastics can be softened on heating and hardened on cooling. They are generally soluble

in organic solvents.

Example: Polyethylene. PVC.

THERMOSETTING PLASTICS:

Thermosetting plastics are prepared by condensation polymerization.

Various polymers are held together by strong covalent bonds(called cross linked).

Thermosetting plastics are harden on heating and once harden., they cannot be softened again.

They are almost insoluble in organic solvents,

Example: Bakelite, polyester.

16.What do you understand by disproportion of polymer chains?

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

R----CH2 – CH-----CH 2---CHO + 0CH – CH2 ----CH – CH 2 -------R

Y n Y Y Y n

R----CH2 – CH-----CH 2---CH2 + CH = CH - ----CH – CH2 -------R

Y n Y Y Y n

Saturated macromolecule Unsaturated macromolecule

The product of addition polymerization is known as Dead polymer



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17.What is AIBN? Mention its role in polymerization reaction?

AIBN is Azo bis isobutylonitirile , Initiators are compounds which produce free radicals by the

hemolytic dissociation.If the hemolytic dissociation is carried out at high temperature they are

called thermal initiators.

CN CN CN

H3C –C – N = N – C – CH3 50 – 70 0C 2CH3 – C O (or) 2R 0

CH3 CH3 CH3

2,2 – Azobis isobutylonitrile(AIBN) Free radicals

18.State the classification of polymers?

Based on the Preparation , polymers are broadly classified into three types,

Types of Polymerisation:

i)Addition polymerization.

ii)Condensation polymerization.

iii)Co-Polymerisation.

Polymer

Natural Synthetic

Inorganic Organic Inorganic Organic

Example example example example

Clay(silicates) proteins,RNA,DNA Silicones PE,PP,Polyester.

19.What are epoxy resins? States its preparation.

There are cross linked thermosetting resins.They are polymers because the monomericunits in the polymer have an ether type of structure.R-O-R.



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PREPARATION: Epoxy (or)Epoxy resins are prepared by condensing epichlorohydrin with

disphenol.

CH3

n HO - C6H5 - C –C6H5 –OH + Cl – CH 2 – CH – CH 2

CH3 O

Bisphenol Epichlorohydrin

CH3

---- O - C6H5 - C –C6H5 –O – CH 2 – CH – CH 2-----

CH3 OH n Epoxy resin

20.How is Nylon 6.6 prepared? State its properties and uses,

PREPARATION: Nylon 6.6 is manufactured by solution polymerization by condensing

hexamethylene diamine and adipic acid in toluene solvent at higher temperature in an inert

atmosphere .

n H2 N-(-- CH2-)6 —NH2 + n HOOC-(--CH 2-)4 —COOH


------NH—(--CH2-)6 —NH – CO –( -CH2 -) 4 —CO----


PROPERTIES OF NYLON 6.6 :

• It is a horny translucent material.

• Its melting point is high (264oC)

•

Nylon 6.6 is a less soft and stiff material when compared to nylon 6.• It is insoluble in common organic solvents but soluble in formic acid and cresol.

• It does not absorb water and hence can be dried easily.

APPLICATIONS OF NYLON 6.6 :

• It is used for making socks, dress materials and ropes.

• The majority of the woven fibre are used in the manufacture of tyre cards.



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• It is used as an engineering plastics.

• It is used in ball bearing , mountings, electrical connections, etc.

•

It is also used to make wheels, bearing. Which can run without the use of lubricants.

PART – B

1. I) Differentiate thermoplastics and thermosetting plastics.

S.NO THERMO PLASTICS THERMOSETTINGPLASTICS

1. They are formed by addition polymerization

They are formed by condensation polymerization

2. They consist of linear long chain polymers.

They consist of three dimensionalnetwork structure.

3. All the polymer chains are held together by weak Vanderwaals forces.

All the polymer chains are linked by strong covalent bonds.

4. They are weak,soft and less brittle. They are strong, hard and more brittle.

5. They soften on heating and harden oncooling.

They do not soften on heating.

6. They can be remoulded. They cannot be remoulded.7. They have low molecular weights. They have high molecular

weights.8. They are soluble in organic solvents. They are insoluble in organis

solvents.9. Ex : Polyethylene, PVC etc., Ex : Bakelite, Ureaformaldehyde

etc.,

ii) Explain the mechanism of free radical polymerization of polyvinyl chloride.Free radical mechanism:Free radical polymerization involves three major steps1.Initiation

2.Propagation3. Termination

1. Initiation:

Initiation involves two reactions

a) First reaction:

It involves production of free radicals by homolytic dissociation of an initiator

or catalyst to yield a pair of free radicals.



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Initiator Free radical

Initiators are compounds which produce free radicals by the hemolytic

dissociation. If the hemolytic dissociation is carried out at high temperature

they are called thermal initiators.

Examples:1.

Acetyl peroxide Free radical

2.

Benzyl peroxide Free radical

CN CN CN

3.

H3C- C- N = N – C - CH3 → 2 CH 3 – C ̊ (OR) 2R ̊

CH3 CH3 CH3

Azobisisobutylonitrile (AIBN) Free radical

b) Second reaction:

Second reaction involves addition of this free radical to the first

monomer to produce chain initiating species.

R ̊ + CH2 = CH → R – CH2 – CH ̊

Cl Cl

2.Propagation:

It involves the growth of chain initiating species by the successive

addition of large number of monomers.

R – CH2 – CH ̊+ CH2 = CH → R – CH2 – CH – CH2 - CH ̊

Cl Cl Cl Cl

R – CH2 - CH ̊+ n CH2 = CH → R – CH2 – CH – CH2 - CH ̊

Cl Cl Cl n Cl

Growing chain

3.Termination:

Termination of the growing chain of the polymer occurs either bycoupling reaction or disproportionation.

(a) Coupling orCombination:



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It involves coupling of free radical of one chain end to another

free radical to form macromolecule .(dead polymer)

R - CH2 – CH – CH2 - CH ̊ + ˚CH – CH2 – CH – CH2 – R

Cl n Cl Cl Cl n↓

R - CH2 – CH – CH2 – CH - CH – CH2 – CH – CH2 – R

Cl n Cl Cl Cl n

↓

Macromolecule (Dead polymer)

(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.

R - CH2 – CH – CH2 - CH ̊ + ˚CH – CH2 – CH – CH2 – R

Cl n Cl Cl Cl n

↓

R - CH2 – CH – CH2 – CH2 + CH = CH – CH – CH2 – R

Cl n Cl Cl Cl n

Saturated macromolecule Unsaturated macromolecule

The products of addition polymerization is known as dead polymers.

2.i) Explain the mechanism of cationic polymerization.

CATIONIC POLYMERIZATION:

Cationic polymerization occurs in three major steps

1. Initiation



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2. Propagation

3. TerminationThis Polymerization takes place when electron donating groups like CH3,

C6H5 are present in a monomer. These groups stabilize the carbonium ion

formation. It is carried out at low temperature and in a non polar solvent.

Ex: Styrene, Isobutylene, Isoprene.Catalysts: The catalysts used to initiate the reaction are

Lewis acids like AlCl3, BF3 with a co- catalyst like HCl, H2O.

1. Initiation:

The catalyst initiates polymerization by the addition of H+ ion to the

monomer to form chain initiating species.

AlCl3 + HCl → H + AlCl -4 ≡ H + R -

H+ R - + CH2 = CH → H – CH 2 – CH + R -

X X

Where X = electron donating groups2. Propagation:

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

large number of monomers and the positive charge simultaneously shifts to the

newly added monomer.

H – CH2 – CH + R - + n CH 2 = CH → H - CH2 – CH – CH2 – CH + R -

X X X n X

3.

Termination:

Terminatin of the growing chain involves removal of proton fromthe polymer chain.

H - CH2 – CH – CH 2 – CH + R - H – CH 2 – CH – CH = CH + H + R -

X n X X X

ii) Explain how molecular weight of a polymer is calculated by number average

method.

Number average molecular mass :



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Number average molecular mass is the mass obtained by dividing the total mass of

the polymer material with the total number of molecules present. Mathematically it isexpressed as

Calculation of

In computing the number average molecular weight it is assumed that each fraction

contributes to the average molecular weight in proportion to its number.

Let n1, n2, n3 …… ni be the number of polymer molecules in 1 st, 2nd,3rd ……..i th

fractions respectively.

Let M1,M2,M3 ….M i be the molecular weight of the polymers in the respectivefractions.

The number fraction of the 1st fraction =

Molecular weight contribution by the 1st fraction to the average molecular weight =

X M1

Molecular weight contribution by the 1st, 2nd,3rd …. Ith fraction to the average

molecular weight is given by

X M1 + X M 2 + X M 3 ……………. X M i

X M1 + X M 2 + X M 3 ……………. X M i

Determination of

1. Mn is determined by the end group analysis and measurement of colligative

properties like freezing point depression, elevation of boiling point, osmotic

pressure etc.,2. Gel permeation chromatography is also used.

Significance (or) properties of

1. It is a good index of physical properties such as impact and tensile strength.

2. It is also a good index of other properties like flow.

3. i) Explain the mechanism of anionic polymerization.

ANIONIC POLYMERIZATION:



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Anionicpolymerization occurs in three major steps

1.Initiation

2.Propagation

3.Termination

This Polymerization takes place when electron withdrawing groups like Cl-,

CN- are present in a monomer. These groups stabilize the carbanion formation.

Ex: Styrene, Vinyl chloride, Acrylo nitrile.

Catalysts: The catalysts used to initiate the reaction are

Lewis base like KNH2, NaNH2, LiNH2.

1.Initiation:

The catalyst initiates polymerization by the addition of NH2- ion to the

monomer to form chain initiating species.

KNH2 → K + + NH 2

K + NH 2 + CH 2 = CH → H 2 N – CH2 – CH K +

Y Y

Chain initiating species

Where Y = electron withdrawing groups2.Propagation:

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

large number of monomers and the negative charge simultaneously shifts to thenewly added monomer.

H2 N –CH2 – CH K + + nCH 2 = CH → H 2 N -CH2 – CH–CH2 CH K +

Y Y Y n Y

Growing chain

3.Termination:

Terminatin of the growing chain involves removal of proton from

the polymer chain.

2HN - CH2 CH –CH 2 –CH K + + H –NH 2→ H 2 N –CH2 –CH – CH 2 – CH2 + NH2

K +

Y n Y Y n Y

Saturated dead polymer



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Living polymer:

When anionic polymerization is carried out using a catalyst inan inert solvent, chain termination does not takes place and the polymer formed

contains terminal ion pair. When these polymer samples are again mixed with

fresh monomer the polymer sample grows. These polymers are referred to as

living polymers.

ii) Explain how molecular weight of a polymer is calculated by weight average

method.

Weight average Molecular mass- :

Weight average molecular mass depends not only on the number of particles, but also

on the molecular size. So,in averaging process molecular weight of each individual

species is multiplied by the mass and not by the number.

Mathematically it is defined as =

Calculation of

In comparing the weight average molecular weight it is assumed that each fraction

contributes to the average molecular weight in proportion to its number.

Let n1,n2,n3…….. ni be the number of polymer molecules present in the 1st , 2 nd,3rd………ith fractions respectively.

Let M1,M2,M3 ……..Mi be the molecular weight of the polymer in respective

fractions.

The weight fraction of the firstfraction =

=

Molecular weight contribution by the 1st fraction to the average molecular weight =

X M1

Molecular weight contribution by the 1st,2nd,3rd…….ith fraction to the average

molecular weight is given by

= X M1 + XM 2 + XM 3………… XMi



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= + + + …………

=

Determination of

1. is determined by light – scattering techniques and ultra- centrifugation

techniques.

2. Gel permeation chromatography is also used.

Significance : is always greater than

3. I) Give a detailed account on techniques of polymerization.

Techniques of polymerization:

The following methods are generally used for the polymerization reaction.

1.Bulk Polymerization

Bulk polymerization is the simplest method of polymerization. The monomer is takenin 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)The reaction is slow but becomes fast as the temperature rises. After a known period

of time, the whole content is moulded into desired object.

Ex: Polystyrene, PVC, PMA, are prepared by this method.

Advantages

1. It is quite simple and requires simple equipments.

2. Polymers of high purity obtained.

3. Excess monomer can be removed by evaporation , as the

monomer is a solvent.

4.

The polymer obtained has high optical clarity.

Disadvantages

1. During polymerization viscosity of the medium increases hence

mixing and control of heat is difficult.

2. Polymerization is highly exothermic.

3. Difficult to remove last traces of monomers and initiators.



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Applications

1. The polymers obtained by this method are used in castingformulations.

2. Low molecular weight polymers obtained by this method are used

as adhesives, plasticizers and lubricant additives.

2,Solution Polymerization

In solution polymerization 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 inert solvent) (In solution)

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

Ex: Polyacrylic acid, polyisobutylene and polyacrylonitrile

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.

3.Suspension or pearl polymerization:

Suspension polymerization is used only for water insoluble monomers.This polymerization reaction is carried out in heterogeneous systems. At

the end of polymerization polymer is separated out as spherical beads or

pearls. This method is also called pearl polymerization.

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

initiator is dissolved in it by continuous agitation. The suspension is

prevented from coagulation by using suspending agents like PVA, gelatin,methyl cellulose. Each droplet of the monomer contains dissolved



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initiator. The whole content is taken in a flask ans heated at constant

temperature with vigorous agitationin a thermostat with nitrogenatmosphere. After the end of 8 hrs, pearl like polymers are obtained which

is filtered and washed by water.

Monomer + Initiator + Suspending agent → Polymer

(Suspended (Dissolved (Suspended inIn water) in monomer) water as beads)

Ex: Polystyrene, Polystyrene – divinyl benzene etc.,

Advantages:

Since water is used as a solvent this method is more

economical.

Products obtained is highly pure.

Isolation of products is very easy.

Efficient heat control.

Viscosity built up of polymer is negligible.

Disadvantages:

o This method is applicable only for water insoluble

monomers.o Control of particle size is difficult,

Applications:o Polystyrene beads are used as ion exchangers.

o

this technique is used in heterogeneous systems.

4.Emulsion polymerization:

Emulsion polymerization is used for water insoluble monomer and water soluble

initiator like potassium persulphate.

PROCESS:

The monomer is dispersed in a large amount of water and then emulsified by the

addition of a soap. Then initiator os 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 hrs 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 soluble)(Emulsion in water)

Inertsolvent)

Ex: Polyvinyl acetate, PVC etc.,

Advantages:

The rate of polymerization is high.



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

High molecular weight polymer can be obtained.Disadvantages:

Polymer needs purification.

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

It requires rapid agitation.Applications:

Emulsion polymerization is used in large scale production like water-

based paints, adhesives, plastics etc.,

This method is also suitable for manuf acturing tacky polymers like

butadiene and chloroprene.

ii) Explain various functionality of a polymer with example and state its

significance.Funtionality:

The number of bonding sites or reactive sites or functional groups present in amonomer is known as its functionality.

Ex: CH2=CH2 , H2N---(--CH2--)6--- NH2 ----Funtionality is 2

(Ethylene) (Hexamethylene diamine)

CH2-OH

CH – OH

--- Funtionality is 3

CH2 – OH

(Glycerol)

Significance:

Bifuntional monomers:

Bifunctional monomers i.e, functionality of the monomer is 2 mainly

form linear or straight chain polymer. Each monomeric unit in thelinear chain is linked by strong covalent bonds but the different chains

are held together by weak Vanderwaal’s forces of attraction. Therfore

there is no restriction to movement of one chain over another. This

type of polymers are soft and flexible and possess less strength, lowheat resistance. These are soluble in organic solvents.

Mixed functional monomers:

When a trifuctional monomer i.e, functionality of the monomer is 3 is

mixed in small amounts with a bifunctional monomer they form

branched chain polymer.

The movement of polymer chain in branched polymer is more restricted

than of straight chain polymers.



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Poly fuctional monomers:Polyfuctional monomers form cross-linked polymer i.e., three

dimensional network polymer. All the monomers in the polymer are

connected to each other by strong covalent bonds. Therefore the

movement of polymer chain is totally restricted. This type of polymersare hard and brittle and possess very high strength and heat resistance.

They are insoluble in almost all organic solvents.

4. i) What are stereospecific polymers? Explain its various types.

STEREOSPECIFIC POLYMER (OR) TACTICITY:The orientation of monomeric units or f unctional 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. ISOTACTIC POLYMER:

If the functional groups are arranged on the same side of the

main chain thepolymer is called Isotactic polymer.

Ex: Polystyrene

H H H H

n CH = CH2 → - C – CH2 – C – CH2 – C – CH2 – C –

n

Styrene Isotactic polystyrene

SYNDIOTACTIC POLYMER:

If the functioan groups are arranged in an alternating fashion the polymer is called

Syndiotactic polymer.

H H

n CH = CH2 → - C – CH2 – C – CH2 – C – CH2 – C – H H n




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ATACTIC POLYMER:

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

H H

n CH = CH2 → - C – CH2 – C – CH2 – C – CH2 – C – H H n

Styrene Atactic polystyrene

ii) How are polymers classified? Explain.

Polymers are giant molecules of high molecular weight formed by the repeatedlinking of large number of small molecules called monomers.

Polymers are classified as

Based on Source:

1. Natural Polymers- The polymers obtained naturally are called natural

polymers.

Ex: Starch, Cellulose, Proteins etc.,

2. Synthetic Polymers- The polymers which are man made are called

synthetic polymers.

Ex: Polyethylene, PVC etc.,

Based on Structure:

1. Linear Polymers – They are straight chain polymers and do not have crosslinkages or branches.

Ex: Nylons,etc.,2. Branched Polymers – They are branched polymers with side linkages .

Ex: Low density polyethylene,etc.,3. Three dimensional network polymers – They have cross linkages due to

polyfunctionality of the monomer.

Ex: Bakelite, Urea formaldehyde etc.,

Based on synthesis:

1. Addition polymers(or) Chain growth polymers – Formed due to mere

addition of monomeric unit without elimination of small molecules .

Ex: PE, PVC,etc.,

2. Condensation polymers (or) Step wise polymers - Formed due to mere

addition of monomeric unit with the elimination of small molecules.

Ex: Polyester, Bakelite , etc.



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Based on Molecular forces:

1. Thermoplastics- They are straight chain or slightly branched polymers andvarious chains are held together by weak vanderwaal’s forces of attraction,

Ex: Polyethylene, PVC, etc.,

2.

Thermosetting plastics – They are cross linked polymers and variouschains are held together by strong covalent bonds.

Ex: Polyester, Bakelite, etc.,

6.Explain the following properties of polymers i) Glass transition temperature ii)Stereospecific polymer (or) tacticity.

i)GLASS TRANSITION TEMPERATURE:

Rubber is an elastic substance at room temperature. But when it is cooled to -79oc it becomes aglossy brittle solid, which when struck, crumbles to a powder

Glossy brittle solid - 79oc Rubber

Thus , for every polymer there exists a temperature below which it is a glossy brittle solid andabove which it is a soft elastic substance. This temperature is known as glass transition

temperature.

DETERMINATION OF TG:

Glass transition temperature can be determined using

• A dialatometer, which measure the change in specific volume with temperature.Using the

plot of specific volume with temperature, Tg can be determined.



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• Tg can also be determined using differential scanning calorimeter(DSC)

FACTORS AFFECTING Tg.:

MOLECULAR WEIGHT:

Generally, Tg of all polymers increases with increase in molecular weight. But themolecular weight beyond 20,000, has no effect on Tg.

EFFECT OF SIDE GROUP:

Presence of side group hinders the free rotation about the C-C bond of polymer

chain and hence increases the Tg.

BRANCHING AND CROSS- LINKING:

When branches and cross- linking increases in polymer chain, Tg. increases.

INTERMOLECULAR FORCES:

Presence of large number of polar groups in the polymer chain increasesintermolecular forces of attraction,which restricts the mobility. This leads to an increasein Tg.

PLASTICIZERS:Addition of plasticizer , to the polymer, decreases the value of Tg..

CRYSTALLINITY:When crystallinity increases, the Tg of polymer also increases. This is due to the

compact arrangement of polymer chain.

EFFECT OF HEATING AND COOLING:

When the rate of heating increases , the Tg decreases. But, when the rate ofcooling increases, the Tg increases.

SIGNIFICANCE (OR) IMPORTANCE OF Tg.:

• Tg value gives an idea of heat capacity, thermal expansion, refractive index,

mechanical and electrical properties of the polymer.

• Tg value of the polymer is used to decide whether the polymer behaves like glass

or rubber at the use temperature.

• Tg value is used to measure the flexibility of a polymer.



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• Tg value is useful in choosing the correct temperature during the moulding

process of a polymer.

ii)STEREOSPECIFIC OLYMER(OR)TACTICITY:

The orientation of monomeric units of 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.

ISOTACTIC POLYMER :

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

called Isotactic polymer.

Example: Polystyrene

CH=CH2 H H H H

n C6H5 ------------ C - CH2 - C – CH2 – C – CH 2 – C ------- ------

C6H5 C 6H5 C6H5 C6H5 n

Styrene Isotactic polystyrene

SYNDIOTACTIC POLYMER:

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

CH=CH2 H C 6H5 H C 6H5

n C6H5 ------------ C - CH2 - C – CH2 – C – CH 2- C ------- ------

C6H5 H C6H5 H n


ATACTIC POLYMER:

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

CH=CH2 H H C 6H5 C 6H5

n C6H5 ------------ C - CH2 - C – CH2 – C – CH 2 - C ------- ------

C6H5 C 6H5 H H n



23

Styrene Atactic polystyrene

iii)POLYDISPERSITY INDEX(PDI):

The ratio of the weight –average molecular weight ( w) to that of number –

average molecular weight ( n) is known as polydispersity index.(PDI).

PDI = w/ n

As w is always greater than n, the ratio is greater than 1 or equal to 1

PDI = w/ n≥ 1

SIGNIFICANCE OF PDI:

PDI is helpful to know whether a polymer system has narrow or broad distribution.

• Low PDI value suggests narrow distribution of molecular weight.

•

High PDI value suggests broad distribution of molecular weight.

• Medium values suggests moderate distribution of molecular weight.

When a graph is plotted between molecular mass of the polymer (M) to the weight fraction of the

polymer (Wi), the following graph is obtained. The Gel permeation chromatogram gives the

pattern of distribution of molecular weight of a polymer.

The graph shows that the weight average molecular weight ( w) is greater than the number

average molecular weight ( n). The ( w) is always greater than ( n) . v is in between the



24

two values of ( w) and ( n), but close to ( w). The ratio ( w) / ( n)is referred to as

polydispersity index (PDI).

7.Write the preparation ,properties and uses of i)Nylon 6.6 ii) Epoxy resin

i)Nylon6.6:

PREPARATION:

Nylon 6.6 is manufactured by solution polymerization by condensing hexamethylene

diamine and adipic acid in toluene solvent at higher temperature in an inert atmosphere .

n H2 N-(-- CH2-)6 —NH2 + n HOOC-(--CH 2-)4 —COOH


------NH—(--CH2-)6 —NH – CO –( -CH2 -) 4 —CO----


PROPERTIES:

• It is a horny translucent material.

•

Its melting point is high (264oC)• Nylon 6.6 is a less soft and stiff material when compared to nylon 6.

• It is insoluble in common organic solvents but soluble in formic acid and cresol.

• It does not absorb water and hence can be dried easily.

• Both fibre and plastics have high tensile strength and dimensional stability.

• It shows good impact strength due to the large number of flexible groups.

• As I/d ratio of the polymer filament is very high and there is high intermolecular

attraction

between the molecules due to H- bonds between the carbonyl and amide –NH-group it

can act asa very good fibers.

USES:

AS FIBRE:

• The majority of the woven fibre are used in the manufacture of tyre cards.



25

• It is used for making socks, dress materials and ropes.

• It is blended with wool to make wool more resistant to abrasion.

AS PLASTIC:

• It is used as an engineering plastic.

•

It is used to make various parts due to its high tensile strength, good impact strength,good dimensional stability and its resistance to abrasion.

• It is used in ball bearing, mountings, electrical connections,etc.

• It is also used to make wheels, bearing, etc, which can run without the use of lubricants.

ii)EPOXY RESINS (OR)EPOXIDE POLYMERS:

These are cross- linked thermosetting resins. They are polyethers because the monomeric units in

the polymer have an ether type of structure i.e R-O-R.

PREPARATION:

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

CH3

n HO - C6H5 - C – C6H5 –OH + Cl – CH 2 – CH – CH 2

CH3 O

Bisphenol Epichlorohydrin

CH3

---- O - C6H5 - C –C6H5 –O – CH 2 – CH – CH 2-----

CH3 OH n

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

value of n ranges from 1 to 20.

The product obtained is treated with alkali to get an epoxide.

PROPERTIES:

• Due to the presence of stable ether linkage, epoxy resin possess high chemical- resistance

to water, acids, alkalis, various solvents and other chemicals.

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

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

• The resin has excellent adhesion quality.



26

USES:

• Epoxy resins are used as surface coatings, adhesives like araldite, glass- fibre reinforced

plastics.

• These are applied over cotton, rayon and bleached fabrics to impart crease-resistance and

shrinkage control.

•

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.

8.i)What are plastics? Explain its advantages and disadvantages.

Plastics are high molecular weight organic materials, that can be moulded into anydesired shape by the application of heat and pressure in the presence of a catalyst.

ADVANTAGES OF PLASTICS OVER OTHER MATERIALS:

•

They are light in weight.

• They possess low melting point.

• They can be easily moulded and have excellent finishing.

• They possess very good strength and toughness.

• They possess good shock absorption capacity.

• They are corrosion resistance and chemically inert.

• They have low co –efficient of thermal expansion and possess good thermal and

electrical insulating property.

• They are very good water – resistant and possess good adhesiveness.

DISADVANTAGES OF PLASTICS:

• They have high softness.

• They undergo embrittlement at low temperature.

• They undergo deformation under load.

• They possess low heat- resistant and poor ductility.

• Combustibility is high.

• They undergo degradation upon exposure to heat and UV-radiation.

• They are Non biodegradable.

ii)Explain the bulk polymerization technique and mention the name of polymer that canbe prepared by this technique.

Bulk polymerization is the simplest method of polymerization. The monomer is taken in a flask

as a liquid from and the initiator, chain transfer agents are dissolved in it. The flask is placed in athermostat under constant agitation and heated.



27

The reaction is slow but becomes fast as the temperature rises. After a known period of

time , the whole content is moulded into desired object.

EXAMPLE: Polystyrene, PVC, PMA are prepared by this method.

ADVANTGES:

• It is quite simple and requires simple equipments.

• Polymers are of high- purity obtained.

• As the monomer is solvent , excess monomer can be removed by evaporation.

• The polymer has high optical clarity.

DISADVANTAGES:

• During polymerization, viscosity of the medium increases hence mixing and control of

heat is difficult.

• Polymerization is highly exothermic.

• Difficult to remove last traces of monomers and initiators.

APPLICATIONS:

• The polymers obtained by this method are used in casting formulations.

• Low molecular weight polymers, obtained by this method are used as adhesives ,

plasticizers and lubricant additives.

9.i)explain the mechanism of Condensation polymerization in detail.

It is a reaction between simple polar groups containing monomers with the formation of

polymer and elimination of small molecules like H2o, HCl,etc.

Example: n H2 N-(-- CH2-)6 —NH2 + n HOOC-(--CH 2-)4 —COOH


------NH—(--CH2-)6 —NH – CO –( -CH2 -) 4 —CO----


In some cases condensation polymerization takes place without the elimination of small

molecules like H2o, HCl,etc., but by just the opening of cyclic compounds.

Monomer + Initiator + Chain transfer agent Polymer

(Liquid) (Mixed with monomer) (Mixed with monomer)



28

ii)Explain the nomenclature of polymers.

NOMENCLATURE OF POLYMERS:

HOMO POLYMER :

A polymer containing same (or only one )type of monomers (C) is known as homopolymer.

Example: Polyethylene, Polyvinylchloride.

-C-C-C-C-C-

n CH2 = CH 2 ---(--CH 2 –CH 2--)n---

TYPES OF HOMOPOLYMER :

The homo polymers are subdivided into the following three types based on the manner in which

monomers are arranged,

i)Linear homo polymer

ii) Branched homopolymer

iii) Cross linked homo polymer

HETEROPOLYMER (OR) COPOLYMER

A polymer containing more than one type of monomers (C and H) is known as copolymer or

hetero polymers.

Example: Polyamide (nylon), Polyester (terylene),SBR Rubber

-A-B-A-B-A-B-A-

TYPES OF CO POLYMER:

The copolymers are subdivided into three types as

i)Random copolymer

ii)Block copolymer

iii)Graft copolymer

HOMOCHAIN POLYMER :

If the main chain of a polymer is made up of same species of atoms, it is called homochain

polymer.



29

Example: Polythene , Polyvinyl chloride.

-C-C-C-C-C-C-C-

HETERO CHAIN POLYMER:

If the main chain of a polymer is made up of different atoms, it is called heterochain polymer.

Example: Nylons, Terylene .

-C-C-O-C-C-O-C-C-

10.i)Describe the Emulsion polymerization technique. Give two examples.

Emulsion polymerization 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. The initiator is added. The whole content is taken in a flask and heated at a constanttemperature 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.

EXAMPLE:

Polyvinyl acetate, PVC,are prepared by this method.

ADVANTAGES:

• The rate of polymerization is high. • Heat can be easily controlled and hence viscosity built up is low.

• High molecular weight polymer can be obtained.

DISADVANTAGES:

•

Polymer needs purification.

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

• It requires rapid agitation.

APPLICATIONS:

• Emulsion polymerization is used in large- scale production like water based paints,

adhesives, plastics, etc.

Monomer + Initiator + Surfactant Polymer

(Dissolved in inert solvent) (Water soluble) (Emulsion in water)



30

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

chloroprene.

ii) Describe the suspension polymerization technique. Give two examples.

SUSPENSION POLYMERISATION (OR) PEARL POLYMERISATION:

Suspension polymerization is used only for water insoluble monomers. This

polymerization reaction is carried out in heterogeneous systems.

At the end of polymerization, polymer is separated out as spherical beads or pearls. This

method is also called pearl polymerization.

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 8hrs, pearl- like polymers are obtained, which is filtered and washed by water.

EXAMPLES:

Polystyrene, Polystyrene- divinyl benzene are prepared by this method.

ADVANTAGES:

• Since water is used as a solvent, this method is more economical.

• Products obtained is highly pure.

• Isolation of product is very easy.

• Efficient heat control.

• Viscosity build up of polymer is negligible.

DISADVANTAGES:

• This method is applicable only for water insoluble monomers.

• Control of particle size is difficult.

APPLICATIONS:

• Polystyrene beads are used as ion exchangers.

• This technique is used in heterogeneous system

Monomer + Initiator + Suspending agent Polymer

(Suspension in water) (Dissolved in monomer) (Suspended in water as beads)



31

VALLIAMMAI ENGINEERING COLLEGE


UNIT 2 CHEMICAL THERMODYNAMICS

PART A (2 MARKS)

1.

Distinguish open, closed and isolated system.Type Open System Closed System Isolated SystemDefinition This is a system that

can exchange bothmass and energywith thesurroundings

This is a system thatcan exchange energy but not mass with thesurroundings

This is a system thatcannot exchange both mass andenergy with thesurroundings

Example Boiling water kept ina beaker, waterescapes out in theform of vapour andit loses heat to the

surroundings.

a gas enclosed in acylinder made of amaterial which is agood conductor

a gas enclosed in acylinder made of amaterial which is a perfect insulator

2. Compare adiabatic process and isothermal process.Type Adiabatic process Isothermal process

Definition

A process that is carried outwithout changing the totalheat energy of the systemeven though other properties may change

A process that is carriedout at constant temperatureeven though other properties may change, ie.,ΔT=0

Temperature Vary (dT ≠ 0) Constant (dT=0)

Process

The system is thermallyinsulated from thesurroundings

The system is usually keptin contact with a constanttemperature bath(thermostat) and theconstant temperature ismaintained

Example

Air (mixture of gases)undergoes adiabatic heatingwhen compressed andadiabatic cooling whenexpanded

Melting of ice

3. Explain thermodynamically reversible process.

Entropy is a measure of the degree of randomness or disorder in a molecular system.i.e., dQrev/T=dS

4. What are the limitations of 1st law of thermodynamics?(OR)What is the need for 2nd law of thermodynamics?

1. The second law predicts the feasibility of a process2. It explains why it is not possible to convert heat into an equivalent amount of work3. The second law is able to predict the direction of energy transformed and also the



32

direction of spontaneous process.The above concepts are not satisified by the 1st law of thermodynamics.

5. State any two statements of 2nd law of thermodynamics.

1. Heat cannot pass from a colder to a hotter body without the aid of an external agency.

2. All natural process are spontaneous processes and tend to proceed in one direction andcannot be reversed.

6. Explain 2nd law of thermodynamics in terms of entropy.Even though the total energy of the universe remain constant, the total entropy of the

universe is increasing

7. Calculate the ∆S for fusion of ice. Given that T = 0ºC, ∆Hf =80Cal/g.Latent heat of fusion on one gram of ice = 80 Cals/gm (or) 80x18 cals/molei.e., ΔHf =1440 Cal

Entropy of fusion of ice = ΔHf /Tf = 1440/273 = 5.275 cals/mole/deg

8. What is a spontaneous process? What are its criteria?A spontaneous process is the time-evolution of a system in which it releases free energyand moves to a lower, more thermodynamically stable energy state.i.e., ΔStotal = ΔSsystem + ΔSsurroundings

Criteria: When ΔH is negative (i.e., exothermic process) and ΔS is positive, then the process will be spontaneous as ΔG is negative.

9. What is work function? Give its significance.The part of internal energy which is isothermally available is called work function (A) of

the system.Significance: Decrease in the function (i.e., -ΔA) gives the maximum work obtained from

the system during the given change. This is also referred as Helmholtz free energy.

10. Explain Gibb’s free energy with example.The isothermally available energy present in a system is called free energy (G). Thequantity G is due to Gibbs and is called Gibb’s free energy.Example: - ΔG is a measure of the maximum net work that can be obtained from a

system at constant temperature and pressure.

11. State and explain the significance of Gibb’s Helmholtz equation.1. This equation relates the free energy change (ΔG) to the enthalpy change (ΔH) and therate of change of free energy with temperature at constant pressure

2. It helps to understand the nature of the chemical reaction as,(i) ΔG = -ve, the reaction is spontaneous(ii) ΔG = 0, the reaction is in equilibrium

12. For a process having ∆S = +ve , What is the condition for spontaneityat all temperature.At low temperature it is non-spontaneous & ΔG is endothermicAt high temperature it is spontaneous & ΔG is endothermic



33

13. Write a VantHoff’s isotherm equation for the following process

n1 A + n 2 b n3 C + n 4 D

14. From Vanthoff isochore explain Lechatlier’s principle.

15. Bring out the deduction ofdE=Tds-Pdv

16. Calculate for a system with P1 = 1 atm, P 2 = 10 atm, T 1 = 273k,

T2 = 373 k.Solution: from clausius-clapeyron eq, logP2/P1 =ΔH/2.303R[(T2-T1)/T1T2]ΔHv = logP2/P1 x 2.303R[(T2-T1)/T1T2]ΔHv = log10/1 X [20303x8.314{(373-273)/(373x273)ΔHv = 0.0051 cal/mol/deg

17. What is the ∆S for a isochoric process for one mole of a gas with Cv = 1.7 cal/g with initial temperature273 k to final temperature 373 k.

ΔS = 2.303 CvlogT2/T1

ΔS = 2.303 x 1.7 log373/273ΔS = 0.00194

18. In what way ∆Go is useful in calculating equilibrium constant.No. Value of ΔG Nature of the process 1 ΔG = -ve or ΔG<0 spontaneous2 ΔG = 0 equilibrium3 ΔG = +ve or ΔG>0 Non-spontaneous

19. How is ∆Go is useful in electrochemistry calculations.

(i) ΔG = ΔH-TΔS(ii) ΔG = -nFE



34

20. Distinguish intensive and extensive property with example.

Type Intensive property Extensive property

Definition Properties that do notdepend upon the amount of

the substances present

Properties that depend uponthe amount of the

substances present Example Temperature, density,viscosity,, refractiveindex..etc

Volume, enthalpy, entropy,free energy.. etc

PART – B1. i)On the basis of thermodynamics derive the relation

Using Maxwell’s relations the above equations may be derived.

a)

The combined form of first and second law isdE= TdS-PdV . . . . . . .(1)If V is constant, so that dV=0, then equation (1) yields

[ ]V = T . . . . . . . .(2)

If S is constant, so that dS=0 , then equation (1) yields

[ ]S = -P . . . . . . . . .(3)

Differentiating equation (2) w.r.t. V at constant S yields

= [ ]S . . . . . . .(4)Differentiating equation (3) w.r.t. S at constant V yields

= -[ ]V . . . . . . .(5)

Therefore it can be proved that

[ ] ]V . . . . . . .(6)

b)

Enthalpy is defined byH=E+PVdH= dE+PdV=VdPBut, dE +PdV =TdSdH= TdS+VdP . . . . . . .. (1)If P is constant, so that dP=0, then equation (1) yields

[ ]P = T . . . . . .. . .(2)

If S is constant, so that dS=0 , then equation (1) yields

[ ]S = V . . . . .. . . .(3)



35

Differentiating equation (2) w.r.t. P at constant S yields

= [ ]S . . . . . . . . . (4 )

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

= [ ]P . . . .. . . . .(5)

Therefore it can be proved that[ ] ]P

ii) Discuss the various conditions for spontaneity and equilibrium stateof a system.

Spontaneity is defined as the tendency of a process to occur naturally .According to the second law of thermodynamics a process is said to be spontaneous only when S total

Is positive i.e., entropy of the universe(system + surrounding) increases. We need a criteria which does notinvolve entropy of the surroundings(difficult to measure). The change in Gibbs free energy provides sucha criteria.

The total entropy change S total is given byS total = S system + S surrounding . . . . . . . . . .(1)

If a reaction is carried out at constant T and PThe amount of heat transferred by the system to the surrounding is S system = (-q p) system

The amount of heat taken by the surrounding is S surroundings= (-q p) system

So, q psurroundings = (-q p) system = - H system

Since the surroundings is a large area, the temperature of the surroundings remains constant, so we haveS surrounding = (-q p) system = - H system

T TOn substituting eqn 2 in 1 we get

S total = S system - H system

TMultiply the equation by T

S total =T S system - H system . . . . . . . . .(3)

S = H - G ( since G = H - S ) . . . . . . .(4)

S total = H system - G - H system

- S total = G or S total = G . . . . . . . .(5)

This equation is the criterion for spontaneity interms of free energy of the system. ThusWhen G = -ve( G < 0) the process is spontaneous

When G = 0the process is equilibrium

When G = +ve( G >0) the process is non- spontaneous.



36

H S G= H - T CONCLUSION

- (exothermic) + _ Spontaneous change

- (exothermic) - - ( at low T Spontaneous change

+(at high T) Non-Spontaneous change

+(endothermic) + +(at lowT) Non-Spontaneous change

-(at highT) Spontaneous change

+(endothermic) - + Forbidden in the forward reacti

2 i)Derive the Clausius inequality.

Clausius inequality states that the cyclic integral of is always is less than or equal to zero

0

Where = Differential heat transfer at the system boundary during a cycle

T = Absolute temperature at the boundary= integration over the entire cycle

Consider two heat engines, one a reversible heat engine and the other one an irreversible engine. Forthe purpose of developing an clausius inequality we assume that both the engines absorb the sameamount of heat QH from the heat source having a temperature of TH. Both the engines reject the heatto a heat sink at a temperature TL. Applying the first law of thermodynamics to both the engines,

Wrev = QH - QL.revWirrev = QH - Q L.irrev

Since the reversible engine is more efficient than the irreversible engine, it must reject less heat(QL.rev) to the thermal sink TL than that of rejected by the irreversible engine(QL.irrev). So, thereversible heat engine produces more work than the irreversible heat engine for the same heat inputQH.

Wrev = QH - QL.rev > Wirrev = QH - Q L.irrev

FOR REVERSIBLE HEAT ENGINE (CARNOT)Consider first the reversible heat engine. The reversible heat transfer can only occur isothermally (atconstantT), so the cyclic integral of the heat transfer divided by the temperature can be written as

= -

Since =

= 0



37

FOR IRREVERSIBLE HEAT ENGINEThe two heat engines have the same value of heat transfer from the hermal source QH. But heat

rejection QL is more in irreversible engine than the reversible one

QL.irrev > QL.rev

= - < 0.

Thus for any heat engine we obtain the calusius inequality

0

ii) Derive that ∆S > 0 for an irreversible process.Consider a system maintained at a higher temperature T1 and its surrounding maintained at a lower

temperature T2. If q amount of heat passes irreversibly from the system to surroundings, then,Decrease in entropy of the system S system = -

Increase in entropy of the surroundings S surroundings = +

Hence, net change in the entropy is given byS total = S system + S surrounding

S total = - +

= q [ - ]

=q [ ]

Since T1 > T 2, T1 - T 2 is positive. Hence

S total = positive

∆Stotal > 0

3. i) Derive the entropy change for an isothermal reversible expansionof an ideal gas.

According to the first law of thermodynamicsdE = q - PdV

(or) dE = q - w . . . . . . . . .(1) ( PdV = w)In a reversible isothermal expansion, there is no change of internal energy i.e., dE = 0So, equation (1) becomes,

qrev - w = 0qrev = w . . . . . . . . . . . .(2)

The work done in the expansion of n moles of a gas from volume V1 to V 2 at constant temperatureT is given by



38

W= nRT ln . . . . . . . .(3)

When eqn (3) is introduced in eqn (2), we get

qrev = nRT ln ( S = )

or T S = nRT ln

S = nR ln

S =2.303 nR log . . . . . . . . .(4)

P1V1 = RT ; V 1 =

P2V2 = RT ; V 2 =

S =2.303 nR log

ii) Discuss the various statements for second law of thermodynamics with the mathematical expressionfor it and the significance of entropy.

Various statements of second law of thermodynamics are as follows:i) Clausius Statement: It is impossible to construct a machine which can transfer heat from a cold body

to a hot body, unless some external work is done on the machine.ii) Kelvin statement: It is impossible to take heat from a hot body and convert it completely into work by

a cyclic process without transferring a part of heat to a cold body.iii) In terms of Entropy: A spontaneous process is always accompanied by an increase in entropy of the

universe

Entropy: Entropy is ameasure of degree of disorder or randomness in a molecular system. It is alsoconsidered as a measure of unavailable form of energy.

Solid Liquid Vapour

Disorderliness increasesEntropy increases

Mathematical statement for Entropy.

S =

Significance of entropy:

i)

Entropy and unavailable energy: When heat is supplied to the system some portion is used andsome of it goes as waste (unavailable). Second law states that entropy is a measure of unavailableenergy. Hence entropy is the unavailable energy per unit temperature.

Entropy = Unavailable energy

Temperatureii)

Entropy and randomness: Entropy is a measure of randomness in a asystem. Increase in entropymeans change from ordered state to disordered state.



39

iii) Entropy and probability: An irreversible process tend to proceed from less probable state to more probable state. Since entropy increases in a spontaneous process entropy may be defined as afunction of probability of the thermodynamic state.

4i) Derive Gibb’s Helmholtz equation . Mention its applications

Consider the following relations,G = H - TS (Gibbs free energy ) . . . . . . . . (1)H = E + PV ( enthalpy) . . . . . . . . (2)

Substituting equation (2) in (1) it becomesG = E + PV - TS . . . . . . . . . (3)

For infinitesimal change,dG = dE + PdV + VdP - TdS - SdT . . . . . . . (4)

But according to first and second law of thermodynamics,dE = dq - PdV (first law)

dq = TdS (second law)so, dE = TdS - PdV . . . . . . . . . .(5)Substituting equation (5) in (4) we get,

dG = TdS - PdV + PdV + VdP - TdS - SdTso, dG = VdP - SdT . . . . . . . .(6)

At constant pressure dP=0,equation (6) becomesdG = -SdT . . . . . . . . . . . (7) ( )

( ) p = -s . . . . . . (8)

Substituting equation (8) in eqn (1) we get,

G = H + T ( ) p

G - T( ) p = H . . . . . . . . . . (9)

This is one form of Gibbs-Helmholts equation.For any two states of the system the equation(7) may be written as

dG 1 = -S 1dT (initial state)dG 2= -S2dT (final state)

To get the change

dG 2 - dG 1 = -S 2dT - (-S1dT)

d(G 2 -G 1 ) = - (S 2-S1)dTd G) = . . . . . . . (10)

At constant pressure the equation (10) becomes

( ) p = - S . . . . . . .(11)

But according to definition of free energyG = H - T S



40

Or - s = . . . . . . . . . (12)

Substituting equation (12) in (11) , we get

= ( ) p

Or G - H = T ( ) p

So G = H - T ( ) p

This is another form of Gibbs-Helmholtz equation.Significance of Gibbs-Heimholtz equation

1. Gibb’s – Helmholtz equation relates the free energy change to the enthalpy change and therate of change of free energy with temperature at constant pressure.

2. It helps in understanding the nature of the chemical reaction( spontaneous or not)

ii) For the reactionH2 + ½ O2→H2O , ∆H =-68.32, ∆S = - 56.69

Calculate the value of free energy change at 25ºC.

∆G = ∆H – T ∆ S = - 68.32 – 298 (-56.69)= - 68.32 + 16,893.62= 16,825.3kJ

The reaction is not feasible since ∆G value is positive.

5 i) Derive Clausius Clayperon equation. Mention its significance.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 G B. Let the temperatureand pressure of the system be T and P respectively. The system is in equilibrium, so there isno change in free energy i.e.,

GA = G B . . . . . . . . . . (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 G B + dG B respectively. Then the equation(1) becomesGA + dG A = G B + dGB . . . . . . . . (2)

We know thatG = H- TS ( Gibbs free energy)

So H = E + PV (Enthalpy)Substituting equation of (H) in (G) it becomes

G = E + PV - TS . . . . . . . . . (3)For infinitesimal change,

dG = dE + PdV + VdP - TdS - SdT . . . . . . . (4)But according to first and second law of thermodynamics,

dE = dq - PdV (first law)dq = TdS (second law)

so, dE = TdS - PdV . . . . . . . . . .(5)Substituting equation (5) in (4) we get,




41

Here the work done is due to volume change only, so equation 6 may be applied to phase A and phase B

dGA = V AdP - SAdT . . . . . . . . (7)

dGB = V BdP - SBdT . . . . . . . . .(8)Where, VA and VB are the molar volumes of phases A and B respectively. S A and SB are their molar

entropies.Since, GA = G B , hence from equation (2)dGA = dG B . . . . . . . . .(9)

Substituting equation (7) and (8) in (9)

VAdP - SAdT = VBdP - SBdT . . . . . . . . .(10)

SBdT - SAdT = VBdP - VAdP . . . . . . . .(11)

(SB - S A ) dT = (V B - VA)dP . . . . . . . . (12)

= . . . . . . . . .(!3)

SB-SA represents the change in entropy when 1 mole of the substance passes from theinitial phase A to the final phase B. It may be denoted as S

= . . . . . . . . . .(14)

We know that entropy change ( S) at constant T is

S = . . . . . . . . .(15)

Substituting equation (15) in (16)

=

This is the clausius-clapeyron equation.

APPLICATIONS:1. Calculation of latent heat of vaporization2. Calculation of boiling point or freezing point3. Calculation of capour pressure at another temperature4. Calculation of molar elevation constant.

ii) Calculate the change in Gibb’s free energy for a process at 100ºC , ∆H = 120 KCal, ∆S = 1.2 K .Comment on the feasibility.

G = H - T S

Given: ∆H = 120 K Cal∆S = 1.2 K Cal

T= 100ºC = 100+273 =373 K



42

G = 120 – 373(1.2)

= - 327.6 k.cals.Since G is –ve , the reaction is feasible.

6 i) Derive Vant Hoff’s isothermVan’t Hoff isotherm gives a quantitative relationship between the free energy change and

equilibrium constant. It can be derived as follows.Consider the general reaction

aA + bB cC + dD

Consider the following relations,G = H - TS (Gibbs free energy )H = E + PV ( enthalpy)

So, G = E + PV - TSFor infinitesimal change,

dG = dE + PdV + VdP - TdS - SdT . . . . . . . (1)But according to first and second law of thermodynamics,

dE = dq - PdV (first law)dq = TdS (second law)

so, dE = TdS - PdV . . . . . . . . . .. (2)Substituting these equations in (1),


At constant temperature, dT=0, equation(3) becomes(dG)T = V.dP

So, free energy change for 1 mole of any gas at a constant temperature is given by

(dG) = V.dP

dG = RT . . . . . . . . .(4)

(since,PV = RT, V = )

On integratin the equation(4) becomes

∫ dG = RT ∫

G = G + RT ln P . . . . . . . .(5)

Where G = integration constant (standard free energy)

Let the free energies of A,B,C and D at their respective pressure P A, PB, PC and P D are GA, GB,GC and G D respectively. Then the free energy change for the above reaction is given by

G = Σ G product - Σ G reactant

= [ cGc+ dGD ] – [aG A + bG B] . . . . . . . .(6)

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

G = [ cG c+ cRT ln Pc+ dG D + dRT ln P D] – [aG A + aRT ln P A + bG B + bRT ln P B]



43

= [ cG c+ dG D - aG A - bG B] + RT ln [PC]c [PD]d . . . . . . .. . . . .(7)

[PA]a [PB] b

We know that, at equilibrium, G = 0,

So, 0 = G + RT ln

G + RT ln K eq = 0 . . . . . (8) (since =Keq)

The equation (8) may be written asG = - RT ln K eq . . . . . . . .(9)

Substituting equation (9) in (7) we getG = - RT ln K eq + RT ln [PC]c [PD]d

[PA]a [PB] b

- G = RT ln K eq - RT ln [PC]c [PD]d

[PA]a [PB] b

This expression is called Van’t Hoff isotherm.

ii) Calculate the entropy change in the evaporation of one mole of water at 100ºC. Latent heat of

vapourisation at 100ºC is 540 Cal/g- S for vaporization, - SV = =

L = 540 ; T= 100+273 = 373K- HV = Molal latent heat of vaporization (L)

= 540cals/gm- SV = 540/373 = 1.22e.u.

7. i) Derive an expression for the variation of equilibrium constant of a reaction with temperature.

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.

Accordingly to the Van’t Hoff isotherm the standatd free energy change

G = - RT ln Kp . . . . . . . . . .(1)

Differentiating equation(1) w.r.t.temperature at constant pressure, we get

= - R ln Kp – RT . . . . . . . . . (2)



44

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

T = -RT in Kp – RT2 . . . . . . . . .(3)

Substitute (1) in (3),

T = G – RT2

. . . . . . . . . . .(4)On rearranging equation (4) we get

G = RT2 + T . . . . . . . . .(5)

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

G = H +T . . . . . . . (6)

Comparing equation (6) and(5) we get,

H = RT2

Or = . . . . . . . . .(7)

The equation is known as Van’t Hoff equationThe equation (8) is integrated between T1 and T 2 at which the equilibrium constants

areKp1 and Kp 2 respectively and H is constant.

= .. . . . . . . . (8)

= . . . . . . . . . (9)

ln Kp2 – ln Kp 1 = [ - ]

= [ - ] . . . . . . . . (10)

= [ ]

ln Kp2 – ln Kp 1 = [ ] . . . . . . . . . (11)

The equilibrium constant Kp2

at temperature T 2 can be calculated, if the equilibriumconstant Kp1 at temperature T1 is known, provided the heat of the reaction H is known.

ii) k p for N2 + 3H2 - NH3 is 1.64 x 10-4 atm and 0.144x10-4 atm at 400 o c and 500oc respectively.Calculate the heat of the reaction. Given R = 1.987 Cal/mole

log Kp2 – log Kp1 = [ ]

Given: Kp2 = 0.144x10-4 atm , Kp1 = 1.64 x 10-4 atm , T2 = 500oc, T1 = 400 o cR = 1.987 Cal/mole



45

Log 0.144x10-4 - log 1.64 x 10-4 = [ ]

-1.05 = (0.000192)

H = - 25,025cals.

8. i) Derive the following relation

a.

G = H – TSdG = dH – TdS – SdT

But, H = E + PVdH = dE + PdV + VdP

= TdS + VdP (since TdS = dE + PdV)

Thus, dG = -SdT + VdP . . . . . . . .(1)If P is constant, so that dP = 0 then equation (1) yields

[ ]P = -S . . . . . .. . .(2)

If T is constant, so that dT=0 , then equation (1) yields

[ ]T = V . . . . .. . . .(3)

Differentiating equation (2) w.r.t. P at constant T yields

=- [ ]T . . . . . . . . . (4 )

Differentiating equation (3) w.r.t. T at constant P yields

= [ ]P . . . .. . . . .(5)


- [ ] ]P

Or [ ] ]P

b.

Helmholtz free energy is given byA = E – TS

dA = dE – TdS – SdTBut the combined I and II law of thermodynamics

TdS = dE + PdV



46

dE = TdS – PdVSo, dA = - SdT – PdV .. . . . . . .(1)

If V is constant, so that dV = 0 then equation (1) yields

[ ]V = -S . . . . . .. . .(2)

If T is constant, so that dT=0 , then equation (1) yields

[ ]T = -P . . . . .. . . .(3)

Differentiating equation (2) w.r.t. V at constant T yields

=- [ ]T . . . . . . . . . (4 )

Differentiating equation (3) w.r.t. T at constant V yields

= -[ ]V . . . .. . . . .(5)


[ ] ]V

ii) Distinguish betweena) Thermodynamically reversible and irreversible process

S.No Reversible process Irreversible process

1. Driving force and opposing forcediffer by an small amount

Driving force and opposing force differ bya large amount

2. It is a slow process It is a rapid process

The work obtained is more The work obtained is less

4. It can be reversed by any of thethermodynamic variables

It cannot be reversed

5. It is a unreal process It is a real process

b) Isothermal and adiabatic process

S.No Isothermal process Adiabatic process

1 In this process the temperature

remains constant

In this process the temperature varies

2 Heat can flow into and out of thesystem

No heat can flow into or out of thesystem

3 This is achieved by placing thesystem in a thermostat dT = 0

This is achieved by carrying the processin an insulated container dq = 0

9. i) What is a system? Discuss the various types of system.A thermodynamic system is defined as the part of the universe, which is selected for theoretical

and experimental investigation. A system usually has a definite amount of a specific substance.



47

Various types of systems are:1. Isolated system: A system which cannot exchange both energy and matter with its

surroundings is called an isolated system. It has no mechanical or thermal contactwith surroundings

2. Closed system: A system which can exchange energy but not matter with itssurroundings is called closed system.

3.

Open system: A system which can exchange energy as well as matter with itssurroundings is called a open system.4. Homogeneous system: When a system is completely uniform throughout, it is called

homogeneous system.5. Heterogeneous system: When a system is not uniform throughout, which consists of

two or more phases, it is called a heterogeneous system.ii) The equilibrium constant for the reaction

N2 + 3 H 2 2NH3 at 500°C is 1.644 X 10-4 and at 700°C is 0.64 X 10 -4 atm. Calculate the

enthalpy of a reaction.

log Kp2 – log Kp1 = [ ]

Kp

1

= 1.644 X 10

-4

, Kp

2

= 0.64 X 10

-4

T1 = 500 + 273 = 773 K and T =m700+ 273 = 973 K

R = 8.314 J/K/mol

So, log [0.64 X 10 -4] – log[1.644 X 10-4] = [ ]

-4.194 – (- 3.785) = H X 26591 X 10 -4

19.147-4.194 = H X 1.3888 X 10 -5

H = -29450 J OR – 29.450 KJ

10. i) Describe a) Extensive property b) Intensive property c) Macroscopic property

1. Extensive property: The properties which depend on the amount of substance present in thesystem are called extensive properties. These properties change depending on how much of thesubstance is added or removed. The value of the additive property is proportional to the size ofthe system. For example if the size is increased, then the property will also increase. Extensive properties include: energy, entropy, mass, length, particle number, number of moles, volume,magnetic momet, weight and electrical charge.

These properties are directly proportional to the size and the quantity of the substance. Forexample: if the amount of water increases, the weight of the water will also increase; the morethe water, the heavier it will be. Another example: the energy it would take to melt an ice cube is

proportional to its size. The energy it would take to melt an ice cube differs from the energy thatwould be required to melt an iceberg.

2. Intensive property: The properties which donot depend on the amount of substance butdepend only on the nature of the substance present in the system are called intensive properties.Eg: temperature, pressure, density. These properties do not change when more of a substance isadded or some of the substance is removed. Intensive properties include: density, color,viscosity, electrical resistivity, spectral absorption, hardness, melting point/boiling point, pressure, ductility, elasticity, malleability, magnetization, concentration, temperature andmagnetic field.



48

These properties do not change if the size of the quantity of the substance changes. For example:the hardness of a diamond does not change, no matter how many times the diamond is cut. Thecolor of the salt does not change no matter how much of it is added to the original amount. Theseall describe the intensive properties of the diamond and salt.

3. Macroscopic properties: The properties associated with a macroscopic system ( large

number of particles) are called macroscopic properties.

Eg. Density, viscosity,volume, etc.

ii) Calculate the ∆G when one mole of the ideal gas expands reversibly isothermally at 37ºCfrom an initial volume of 55dm3 to 1000 dm3.

∆G = nRT ln V1/V2= 1* 8.314 *310 *2.303 log 55/1000= 5,935.6 log55/1000= 5,935.6 * 0.0017 = 10.09kJ



49

UNIT – 3 PHOTOCHEMISTRY AND SPECTROSCOPY

PART – A (2 MARKS)

1. What is Photochemistry?

It is the science of the chemical effects of radiations, whose wavelengths range from 2000 Å to8000 Å, which lie in the visible and ultraviolet region. Simple reactions involvingcombination,

decomposition, polymerization, oxidation and reduction can be brought about by exposure to

such radiations (lower energy).

2. What are dark reactions?The chemical reactions, which take place in the absence of light are called dark reactions.

3.

What are the differences between photochemical and thermal reactions?

4. Write the statement of Grotthus-Draper Law.Grotthus -Draper law states that only the light, which is absorbed by a substance, can bringabout

a photochemical change.

5. State Stark-Einstein law of photochemical equivalence.Stark-Einstein law of photochemical equivalence states that, in a primary photochemical process

(first step) each molecule is activated by the absorption of one quantum of radiation

(onephoton).



50

6.

State Lambertz law.Lambert’s law states that, “when a beam of monochromatic radiation is passed through a

homogeneous absorbing medium the rate of decrease of intensity of the radiation ‘dI’ with

thickness of absorbing medium ‘dx’ is proportional to the intensity of the incident radiation ‘ I ’.

Where, k = absorption coefficient.

7. Define Beer-Lambertz law. According to this law, “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 the

concentration of the solution ‘C ’.”

A = εCx

8. Define quantum yield.

Quantum yield (or) efficiency (Ф) is defined as, “the number of molecules of reactants reacted or

products formed per quantum of light absorbed”, ie.,

9. Define Photosensitization. The foreign substance, which absorbs the radiation and transfers the absorbed energy to the

reactants, is called a photosensitizer. This process is called photosensitized reaction (or) photosensitization.Example

(a) Atomic photosensitizers: Mercury, cadmium, zinc (b) Molecular photosensitizers: Benzophenone, sulphur dioxide

10. What is quenching?When the foreign substance in its excited state collides with another substance it gets convertedinto some other product due to the transfer of its energy to the colliding substance. This process isknown as quenching.

11.

What is fluorescence?The emission of radiation due to the transition from singlet excited state, S1 to ground state S 0 iscalled fluorescence (S1 S 0). This is an allowed transition and occurs in about 10-8 sec.

Example for fluorescent substances:Fluorite (naturally occurring CaF2) petroleum, organic dyes like eosin, fluorescein, ultramarineand vapours of Na, Hg and I2.

12. What is phosphorescence?The emission of radiation due to the transition from the triplet excited state T1 to the ground stateS0 is called phosphorescence (T 1 S 0).This transition is slow and forbidden transition.Example for fluorescent substances: ZnS, alkaline earth sulphides and sulphates of Ca, Ba &Sr.

13. What are IC and ISC?INTERNAL CONVERSION:



51

These transitions involve the return of the activated molecule from the higher excited states to thefirst excited states, ie.,

(OR)

This process is called internal conversion (IC) and occurs in less than about 10-11 second.

INTER SYSTEM CROSSING:These transitions involve the return of the activated molecules from the states of different spins,ie.,different multiplicity, ie.,

These transitions are forbidden, occurs relatively at slow rates.14.

What is Chemiluminescence?If light is emitted at ordinary temperature, as a result of chemical reactions, the phenomenon isknown as Chemiluminescence. Thus, it is the reverse of a photochemical reaction.As the emission occurs at ordinary temperature, the emitted radiation is also known as “coldlight“.EXAMPLE:

Bioluminescence: Emission of “cold light” by fire flies (glow-worm) due to the aerial oxidation ofluciferon(a protein) in the presence of enzyme (luciferase).

15. What is meant by the term absorption spectroscopy?When a beam of electromagnetic radiation is allowed to fall on a molecule in the ground state, themolecule absorbs photon of energy hv and undergoes a transition from the lower energy level tothe higher energy level.The measurement of this decrease in the intensity of radiation is the basisof absorption spectroscopy. The spectrum thus obtained is called the absorption spectrum.

E2

h ν

E1

16.

Differentiate chromophores and auxochromes?Give examples.



52

17. What are the various types of electronic transitions?

18. What is finger print region? Mention its important uses?The vibrational spectral (IR spectra) region at 1400-700cm-1 gives very rich and intense

absorption bands. This region is termed as fingerprint region. The region 4000-1430cm-1 isknown as group frequency region.Uses of Fingerprint RegionFingerprint region can be used to detect the presence of functional group and also to identify andcharacterize the molecule just as a fingerprint can be used to identify a person.

19. Define the term; Bathochromic shift.Shifting of absorption to a longer wavelength is called bathochromic shift or red shift.

20. Calculate the energy per mole of light having wavelength of 85nm.We know that

E = (NAhc)/λ Given values

λ = 85nm = 85x10-9 m

Known valuesh = 6.62 x 10-34Js; c = 3 x 108ms-1; NA = 6.023 x 1023 mol-1

Sub these values in the above equationE = (6.023 x 1023 mol-1)(6.62 x 10-34Js)(3 x 108ms-1)

85x10-9 mE = 1.407 x 106J mol-1 or 1.407 x 103KJ mol-1

UNIT – 3 PHOTOCHEMISTRY AND SPECTROSCOPY

PART – B (16 MARKS)

10.

(i) With the help of jablonski diagram, explain radiative and non-radiative pathways for anelectronic transition.



53

Types of Transitions The activated molecules return to the ground state by emitting its energy through the following

general types of transitions.

1.Non-Radiative Transitions

These transitions do not involve the emission of any radiations, so these are also known as non-

radiative or radiationless transitions. Non-radiative transitions involve the following two

transitions.

(a) Internal conversion (IC)These transitions involve the return of the activated molecule from the higher excited states to the

first excited states, ie.,

(OR)

The energy of the activated molecule is given out in the form of heat through molecular

collisions. This process is called internal conversion (IC) and occurs in less than about 10-11

second.

(b) Inter system crossing (ISC)The molecule may also lose energy by another process called inter system crossing (ISC).

These transitions involve the return of the activated molecules from the states of different spins,

ie., different multiplicity, ie.,

These transitions are forbidden, occurs relatively at slow rates.

2.Radiative TransitionsThese transitions involve the return of activated molecules from the singlet excited state

S1 and triplet excited state T 1 to the ground state S 0. These transitions are accompanied by the

emission of radiations. Thus, radiative transitions involve the following two radiations.(a) Fluorescence The emission of radiation due to the transition from singlet excited state, S1 to ground

state S 0 is called fluorescence (S 1 S 0). This transition is allowed transition and occurs in about

10-8 second.

(b) Phosphorescence The emission of radiation due to the transition from the triplet excited state T1 to the

ground state S0 is called phosphorescence (T 1 S 0).This transition is slow and forbidden

transition.



54

3.QUENCHING OF FLUORESCENCEThe fluorescence may be quenched (stopped), when the excited molecule collides with a

normal molecule before it fluoresces. During quenching, the energy of the excited molecule gets

transferred to the molecule with which it collides. Quenching occurs in two ways

(a) Internal quenching

Quenching may also occur, when the molecule changes from the singlet excited state tothe triplet excited state. This phenomenon is called internal quenching.

( b) External quenching Quenching may also occur from the addition of an external substance, which absorbs

energy from the excited molecule. This phenomenon is called external quenching.

---------------------------------------------------------------------------------------------------------------------(ii) How quantum efficiency is determined experimentally? Explain.

The number of molecules reacted in a given time can be determined by the usual

analytical techniques, used in chemical kinetics

Measurement of Rate of Reaction

The rate of reaction is measured by the usual methods.

Small quantities of the samples are pipetted out from the reaction mixture from time to time and

the concentration of the reactants are continuously measured by the usual volumetric methods

(or) the change in some physical property such as refractive index (or) absorption (or) optical

rotation.

From the data, the amount and hence number of molecules can be calculated.

Experimental Determination of amount of Photons Absorbed

A photochemical reaction occurs by the absorption of photons of the light by the reactantmolecules.

Therefore, it is essential to determine the intensity of the light absorbed.

An experimental set up for the study of photochemical reaction is illustrated in the following fig.

Radiation emitted from a source of light (L) (Sunlight, tungsten filament, mercury vapor lamp)is passed through the lens, which produces parallel beams.

The parallel beams are then passed through a filter (or) monochromatic ‘ B’, which yields a beam of the desired (one) wavelength only.

The lightfrom the monochromatic (monochromatic light) is allowed to enter into the reaction cell

‘C’ immersed in a thermostat, containing the reaction mixture. The part of the light that is not absorbed fall on a detector ‘ X ’, which measures the intensity ofradiation.

Among the so manydetector, the most frequently employed is the chemical actinometer.The Chemical ActinometerA chemical actinometer is a device used to measure the amount of radiation absorbed by thesystem in a photochemical reaction. Using chemical actinometer, the rate of a chemical reactioncan be measured easilyUranyl Oxalate actinometerUranyl oxalate actinometer is a commonly used chemical actinometer. It consists of 0.05 Moxalic acid and 0.01 M uranylsulphate in water.



55

‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐

2.(i) Distinguish between

(a) Fluorescence and Phosphorescence

(b) Thermal and Photochemical reactionsRefer Part A question no. 3

-------------------------------------------------------------------------------------------------------------------------------(ii) What are the causes of high and low quantum yield? Define the same.High (or) Low Quantum YieldThe quantum efficiency varies from zero to 106. If a reaction obeys the Einstein law, one molecule isdecomposed per photon, the quantum yieldФ = 1.1. High Quantum Yield

When two or more molecules are decomposed per photon, the quantum yieldФ>1 and the

reaction has a high quantum yield.2. Low Quantum Yield

When the number of molecules decomposed is less than one per photon, the quantum yield Ф>1and the reaction has a low quantum yield.Conditions for High and Low Quantum YieldThe reacting molecules should fulfill the following conditions.

1.All the reactant molecules should be initially in the same energy state and hence equallyreactive.

2.The reactivity of the molecules should be temperature independent.



56

3.The molecules in the activated state should be largely unstable and decompose to form the products.

Causes (or) Reasons for high Quantum yield1.Absorption of radiations in the primary step involves production of atoms or free radicals, which initiatea series of chain reactions.2.Formation of intermediate products will act as a catalyst.

3.If the reactions are exothermic, the heat evolved may activate other molecules without absorbing theadditional quanta of radiation.4. The active molecules, produced after absorption of radiation, may collide with other molecules andactivate them which inturn activate other reacting molecules.EXAMPLE(i). Decomposition of HIIn the primary reaction, one HI molecule absorbs a photon and dissociated to produce one H and one I.This is followed by the secondary reaction as shown below.

The overall reaction shows that the two HI are decomposed for one photon .Thus, the quantum yield (Ф)= 2.Causes (or) Reasons for Low Quantum yield1. Excited molecules may get deactivated before they form products.2. Excited molecules may lose their energy by collisions with non-excited molecules.3. Molecules may not receive sufficient energy to enable them to react.4. The primary photochemical reaction may be reversed.5. Recombination of dissociated fragments will give low quantum yield. EXAMPLE:

-------------------------------------------------------------------------------------------------------------------------------3.(i) What is the statement, expressions and the limitations of Beer-Lambertz law?Beer’s law (or) Beer-Lambert’s Law

Beer extended the above equation (2) to solutions of compound in transparent solvent.

According to this law, “when a beam of monochromatic radiation is passed through a solution ofan 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 the concentration of the solution ‘C ’.”



57

Thus, the absorbance (A) is directly proportional to molar concentration (C) and thickness (or) path length(x).APPLICATION OF BEER-LAMBERT'S LAW Determination of unknown concentration

LIMITATIONS OF BEER-LAMBERT'S LAW

1. Beer-Lambert’s law is not obeyed if the radiation used is not monochromatic.2. It is applicable only for dilute solutions.3. The temperature of the system should not be allowed to vary to a large extent.4. It is not applied to suspensions.5. Deviation may occur, if the solution contains impurities.6. Deviation also occurs if the solution undergoes polymerization (or) dissociation.-------------------------------------------------------------------------------------------------------------------------------(ii)A monochromatic light is passed through a cell of 1 cm length. The intensity is reduced by10%.Ifthe same radiation is passed through the same solution in a cell of length 8 cm, what is the



58

transmittance? Calculate the length of the cell in order to have 20% absorption.I0 = 100%, I=90%because reduction in intensity is 10%.

A1 = εcl 1 ---------- (1) ; A2 = εcl 2 -------------- (2)Equation (1) & (2) become as (A1/A2) = (l 2/l 1) ; A2 = (l 2/l 1)A1 ; A2 = (8/1) log (100/90) = 0.3661

A2 = - log T = log (1/T) = 0.3661 ; (1/T) = Antilog 0.3661 = 2.323 ; T = (1 / 2.323) = 0.4305 (Or)

T = 43.05 %

Next,

l 1 = 1 cm ; A1 = 10%; l 2 = x cm ; A2 = 20 %

(A1/A2) = (l 2/l 1) ; (A1/A2) = (l 2/l 1) ; (A1/A2) l 1 = l 2 ; (10/20)1 = l 2 = 2 cm

l 2 = 2 cm

4.(i) Explain about Chemiluminescence and Photosensitization with suitable examples.PHOTOSENSITIZATION:

The foreign substance, which absorbs the radiation and transfers the absorbed energy to thereactants, is called a photosensitizer. This process is called photosensitized reaction (or) photosensitization. Example:(a) Atomic photosensitizers: Mercury, Cadmium, Zinc.(b) Molecular photosensitizers: Benzophenone, Sulphur dioxide.

In a donor-acceptor system, only the donor D (ie., the sensitizer) absorbs the incident photon. When the donor absorbs the photon, it gets excited from ground state (S0) to singlet state (S1);

Then the donor, via inter system crossing (ISC), gives the triplet excited state (T1 or 3D). Thetriplet state of the donor is higher than the triplet state of the acceptor (A).

This triplet excited state of the donor then collides with the acceptor produces the triplet excitedstate of the acceptor (3A) and returns to the ground state (S0).

If the triplet excited state of the acceptor (3A) gives the desired products, the mechanism is called photosensitization



59

CHEMILUMINESCENCE:If light is emitted at ordinary temperature, as a result of chemical reactions, the phenomenon is

known as Chemiluminescence. Thus, it is the reverse of a photochemical reaction. As the emission occursat ordinary temperature, the emitted radiation is also known as “cold light“.Explanation:

In a Chemiluminescent reaction, the energy released during the chemical reaction makes the product molecule electronically excited. The excited molecule then emits radiation, as it returns to the

ground state.Examples:1. Glow of phosphorus and its oxide, in which the oxide in its excited electronic state emits light.2. Oxidation of 5-aminophthalic hydrazides or cyclic hydrazides by H2O2, emits bright green light.3. Bioluminescence: Emission of “cold light” by fire flies (glow-worm) due to the aerial oxidation ofluciferon (a protein) in the presence of enzyme (luciferase).

MECHANISM:Mechanism of Chemiluminescence can be explained by considering anion-cation reactions.

ILLUSTRATION:

The aromatic anion ((Ar -) contains two paired electrons in the bonding molecular orbital (BMO)and one unpaired electron in the antibonding molecular orbital (ABMO). The ABMO of thearomatic cation (Ar +) + is empty. When the electron is transferred from the ABMO of the anion((Ar -) to the ABMO of the cation ((Ar +) + ), the singlet excited state

1Ar* is formed. The excited state can be deactivated by the emission of photon hv.-------------------------------------------------------------------------------------------------------------------------------(ii) Calculate the energy associated with (a) one photon, (b) one Einstein of radiation of wavelength8000Å.Solution: One photon/quantum energy, E = (hc) / λ

h = 6.62 x 10-34

Js; c = 3 x 108

ms-1

λ = 8000 Å = 8000 x 10-10

m

(6.62 x 10-34Js)(3 x 108ms-1)E = ------------------------------------------------

8 x 10-7mE = 2.4825 x 10-19 J

One Einstein energy, E = (NA hc) / λ

NA = 6.023 x 1023 mol-1

E = (6.023 x 1023 mol-1) x (2.4825 x 10-19 J)



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E = 14.9521 x 104 J mol-1 -------------------------------------------------------------------------------------------------------------------------------5. (i) (a) How do atomic spectra differ from molecular spectra ? (b) How do emission spectra differ fromabsorption spectra?

Sl.No. Emission spectra Absorption spectra

1. The molecule comesdown from theexcited state to theground state with theemission of photonsof energy hv

The molecule absorbs photon of energy hvand undergoes atransition from thelower energy level tothe higher energy level

2.E1

hv

E0

E1 hv

E0 -------------------------------------------------------------------------------------------------------------------------------

(ii) What are electromagnetic spectrum and explain the characteristics of it.

The entire range over which electromagnetic radiation exists is known as electromagneticspectrum.

The electromagnetic spectrum covers larger range of wavelength. Following figure 4.8 shows a diagrammatic representation of the electromagnetic spectrum. A logarithmic scale is used in this representation.

The divisions between the different spectral regions indicate the origin of radiation. The limits indicated in figure are arbitrary.

Characteristics of Electromagnetic Spectrum The major characteristics of various spectral regions are described as follows:



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-------------------------------------------------------------------------------------------------------------------------------

6. (i) Explain in detail about the Rotational, Vibrational and Electronic transitions.The molecular spectra arises from the following three types of transitions, viz.,

(i) Rotational transition.(ii) Vibrational transition.(iii) Electronic transition.These transitions are accompanied by the absorbance of electromagnetic radiation. ENERGY LEVEL DIAGRAM

The energy level diagrams, showing different transition in molecules, are shown in the followingdiagram



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ROTATIONAL (MICROWAVE) SPECTRA (OR) ROTATIONAL TRANSITION

VIBRATIONAL (INFRARED) SPECTRA (OR) VIBRATIONAL TRANSITION

Vibrational energy arises due to the to and fro motion of the molecule ie., stretching, contracting

and bending of covalent bonds in a molecule.

Vibrational spectra and Vibrational-Rotational spectra results from transitions between thevibrational energy levels of a molecule, due to the absorption of radiation in the infrared region.

IR spectra occur in the spectral range of 500 - 4000 cm-1. Since vibrational transition isaccompanied by rotational transition, vibrational spectra is also termed as vibrational rotationalspectra.

EXAMPLE:

Electronic spectra (or) Electronic transition

-------------------------------------------------------------------------------------------------------------------------------(ii) Explain the various changes occurring during absorption of radiation and what are thefactors affecting it.

Absorption of RadiationWhen electromagnetic radiation is passed through a matter, the following changes occur.1. As the photons of electromagnetic radiations are absorbed by the matter, electronic transition,vibrational changes (or) rotational changes may occur. After absorption, molecules get excited from the



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ground state to excited state. Then they liberate energy quickly in the form of heat (or) re-emitelectromagnetic radiation.2. But in some cases, the portion of electromagnetic radiation, which passes into the matter, instead of being absorbed may be scattered (or) reflected (or) re-emitted.3. When the electromagnetic radiation is absorbed (or) scattered, it may undergo changes in polarisationor orientation.4. In some cases the molecules absorbs radiation and get excited.

(a) Fluorescence If the excited molecules re-emits the radiation almost instantaneously (within 10-8 seconds), it is

called fluorescence.(b) Phosphorescence

If the excited molecules re-emit the radiation after sometime (slowly), it is called phosphorescence. Factors Affecting Absorbance The fractions of photons being absorbed by the matter depends on,1. The nature of the absorbing molecules. 2. The concentration of the molecules.

If the concentrations of the molecules are more, the absorbed photons will be more.3. The length of the path of the radiation through the matter.

If the length of the path is long, the larger number of molecules are exposed and hencegreater the photons will be absorbed

-------------------------------------------------------------------------------------------------------------------------------7. (i) Explain the principle of IR spectroscopy and discuss the functions of various components in IRspectrophotometer.INFRARED SPECTROSCOPY

Principle

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

IR spectroscopy is also known as Vibrational spectroscopy.Range of Infrared Radiation

The range in the electromagnetic spectrum extending from 12500 to 50cm-1 (0.8 to 200µ) iscommonly referred to as the infrared. This region is further divided into three sub regions

Sources of IR: Electrically heated rod of rare-earth oxidesINSTRUMENTATION:I.Components:

1.

Radiation source 2. Monochromator

3.

Sample Cell 4. Detector5. Recorder

Radiation Source:The main sources of IR radiation

(a)

Nichrome wire.(b) Nernst glower, which is a filament containing oxides of Zr, Th, Ce, held together

with a binder.When they are heated electrically at 1200 to 2000°C, they glow and produce IR radiation.

Monochromator



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It allows the light of the required wave length to pass through, but absorbs the light of otherwavelength.Sample Cell

The cell, holding the test sample, must be transparent to IR radiation.Detector

IR detectors generally convert thermal radiant energy into electrical energy. There are so manydetectors, of which the followings are important.

•

Photoconductivity cell.• Thermocouple.•

Pyroelectric detectors.Recorder

The recorder records the signal coming out from the detector.

II.Working of IR Spectrophotometer

The radiation emitted by the source is split into two identical beams having equal intensity. One of the beams passes through the sample and the other through the reference sample.

When the sample cell contains the sample, the half-beam travelling through it becomes lessintense.

When the two half beams (one coming from the reference and the other from the sample)recombine, they produce an oscillating signal, which is measured by the detector.

The signal from the detector is passed to the recording unit and recorded.

-------------------------------------------------------------------------------------------------------------------------------

(ii) Discuss the applications of IR spectroscopy.1. Identity of the compound can be established The IR spectrum of the compound is compared with that of known compounds.

From the resemblance of the two spectra, the nature of the compound can be established. This is because a particular group of atoms gives a characteristic absorption band in the IR

spectrum.Example:

2. Detection of Functional GroupsIn a given environment, a certain functional group will absorb IR energy of very nearly the samewavelength in all molecules.

Example:



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3.

Testing purity of a SamplePure sample will give a sharp and well – resolved absorption bands. But impure sample will give a

broad and poorly resolved absorption bands. Thus by comparison with IR spectra of pure compound, presence of impurity can be detected.4. Study of progress of a chemical reaction

The progress of a chemical reaction can be easily followed by examining the IR spectrum of testsolution at different time intervals(i) Progress of oxidation of secondary alcohol to ketone is studied by geeting IR Spectra of test

solution at different time intervals.

The secondary alcohol absorbs at 2.8µ (~3570cm-1) due to O – H stetching. As the reaction proceeds this band slowly disappears and a new band near 5. 8µ (~ 1725cm-1) due to C=Ostretching appears.

(ii) Similarly, the progress of any chromatographic separations can be readily monitored byexamining the IR spectra of the selected fractions.

5.

Determination of shape or Symmetry of a Molecule

Whether the molecule is linear (or) non-linear (bend molecule) can be found out by IR spectra.IR spectra of NO2 gives three peaks at 750, 1323 and 1616cm -1.

According to the following calculations,(i) For a non-linear molecule = (3n-6) = 3 peaks(ii) For a linear molecule = (3n-5) = 4 peaksSince the spectra shows only 3 peaks, it is confirmed that NO2 molecule is a non linear (bend)molecule.

6. To study TautomerismTautomeric equilibria can be studied with the help of IR spectroscopy.The common systems such as keto-enol, lacto – lactum, and mercapto-thioamide, contain a grouplike C=O, - OH, - NH (or) C=S. these groups show a characteristic absorption band in the IRSpectrum, which enable us to find at which form predominates in the equilibrium.

7. Industrial Applications(a) Determination of structure of chemical products

During the polymerisation, the bulk polymer structure, can be determined using IR spectra.(b) Determination of molecular weight

Molecular weight, of a compound can be determined by measuring end group concentrations,using IR spectroscopy.

(c) Crystallinity

The physical structure like crystallinity can be studied through changes in IR spectra.The absorption band at 934cm-1 is for crystalline nylon 6:6



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The absorption band at 1238cm-1 is for amorphous nylon 6:68. Isomers can be identified in the fingerprint region

Similar molecules may show very similar spectra in the group frequency region (4000-1430cm-1). But, they show some differences in the fingerprint region (1400-700cm-1).

The IR spectra of the above three isomers in the group frequency region are almost similar. But

these are identified from their IR spectra in the fingerprint region, due to three different absorptions.

9. Determination of hydrogen bonding in a molecule

To detect the hydrogen bond and also distinguish between inter and intra molecular hydrogen

bonding present in a compound, a series of IR spectra of the compound at different dilutions are taken.

On dilution, it is observed that the peak at 3630cm-1 becomes more intense as the concentration of

the free – OH group increases. At the same time broader peak at 3500-3200cm-1 becomes less intense and

disappears at larger dilutions.

The above explanation is applicable to –OH groups involved in Hydrogen bonding. If the –OH group is

involved in intramolecular hydrogen bonding, dilution will not affect the intensity of the peak.

10. Determination of AromaticityThe difference in the wavelengths of C – H bonds in different environments can be used to

determine.(a) The relative proportions of saturated and unsaturated rings present in hydrocarbon.(b)

The % of aromatic compounds or olefins in the mixtures.



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8. (i) Discuss the principle, instrumentation and working mechanism of UV-Visible spectroscopy.

VISIBLE AND ULTRAVIOLET (UV) SPECTROSCOPY Principle

INSTRUMENTATION I Components The various components of a visible UV spectrometer are as follows.1. Radiation Source

In visible - UV spectrometers, the most commonly used radiation sources are hydrogen (or)deuterium lamps. Requirements of a radiation source

a. It must be stable and supply continuous radiation. b. It must be of sufficient intensity.

2.MonochromatorsThe monochromator is used to disperse the radiation according to the wavelength. The essential

elements of a monochromator are an entrance slit, a dispersing element and an exit slit. The dispersingelement may be a prism or grating (or) a filter.

3.Cells (Sample Cell and Reference Cell)The cells, containing samples or reference for analysis, should fulfil the following conditions.

i. They must be uniform in construction.ii. The material of construction should be inert to solvents.

iii.

They must transmit the light of the wavelength used.4.Detectors

i. There are three common types of detectors used in visible UVspectrophotometers.

ii.

They are Barrier layer cell, Photomultiplier tube and Photocell.iii.

The detector converts the radiation, falling on which, into current.iv. The current is directly proportional to the concentration of the solution.

5.Recording Systemi.

The signal from the detector is finally received by the recording system.ii. The recording is done by recorder pen.

II Working of visible and UV Spectrophotometer • The radiation from the source is allowed to pass through the

monochromator unit.



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•

The monochromator allows a narrow range of wavelength to pass throughan exit slit.

• The beam of radiation coming out of the monochromator is split into twoequal beams.

• One-half of the beams (the sample beam) is directed to pass through atransparent cell containing a solution of the compound to be analysed.

•

Another half (the reference beam) is directed to pass through anidentical cell that contains only the solvent.•

The instrument is designed in such a way that it can compare theintensities of the two beams.

-------------------------------------------------------------------------------------------------------------------------------(ii) Discuss the applications of UV-Visible spectroscopy.

1. Predicting relationship between different groups

2. Qualitative AnalysisUV absorption spectroscopy is used for characterizing and identification of aromatic

compounds and conjugated olefins by comparing the UV absorption spectrum of the sample withthe same of known compounds available in reference books.

3. Detection of ImpuritiesUV absorption spectroscopy is the best method for detecting impurities in organic

compounds, because(i) The bands due to impurities are very intense.(ii) Saturated compounds have little absorption band and unsaturated compounds have strongabsorption band.

4. Quantitative Analysis

Determination of substances: UV absorption spectroscopy is used for the quantitativedetermination of compounds, which absorbs UV. This determination is based on Beer’s law



69

First, absorbances (optical densities) of the different solutions of known concentrations are

measured.

Then the graph is plotted between absorbance vs concentration (calibration curve).

A straight line is obtained. Then absorbance of unknown solution is measured.

From the graph the concentration of unknown substance is found out.

5. Determination of Molecular Weight

Molecular weight of a compound can be determined if it can be converted into a suitablederivative, which gives an absorption band.

6. Dissociation constants of Acids and Bases

7.

Study of Tautomeric EquilibriumThe percentage of various keto and enol forms present in a tautomeric equilibrium can be

determined by measuring the strength of the respective absorption bands using UV spectroscopy.

Example : Ethylacetoacetate

8. Studying kinetics of Chemical ReactionsKinetics of chemical reactions can be studied using UV spectroscopy by following the

change in concentration of a product or a reactant with time during the reaction.

9.

Determination of calcium in blood serum

--------------------------------------------------------------------------------------------------------------------------9. (i) Write notes on a) Finger print region b) Phosphorescence.

(a) Fingerprint Region The vibrational spectral (IR spectra) region at 1400-700cm-1 gives very rich and intense

absorption bands. This region is termed as fingerprint region. The region 4000-1430cm-1 isknown as group frequency region.Uses of Fingerprint Region

Fingerprint region can be used to detect the presence of functional group and also toidentify and characterize the molecule just as a fingerprint can be used to identify a person.

b) PhosphorescenceWhen a molecules or atom absorbs radiation or higher frequency the emission of radiation is

continuous for some time even after the incident radiation is cut off. This process is calledPhosphorescence or delayed fluorescence (10-9 to 10 -4 sec).



70

The substances which exhibit phosphorescence are called phosphorescent substances. Example : ZnS, CaS, CaSO 4, BaSO4.

-------------------------------------------------------------------------------------------------------------------------------

(ii) Explain the types of stretching and bending vibrations with suitable examples.Types of stretching and bending Vibrations

The number of fundamental (or) normal vibrational modes of a molecule can be

calculated as follows.For Non-linear molecule:A non-linear molecule containing ‘n’ atoms has (3n-6) fundamental vibrational modes

EXAMPLES:

For linear moleculesA linear molecule containing ‘n’ atoms (3n-5) fundamental vibrational modes

EXAMPLES:

Illustrations

1.Water



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2.Carbondioxide



72

-------------------------------------------------------------------------------------------------------------------------------10. (i) State the following (a) Hypsochromic shift, (b) Hyper chromic shift,(c) Hypochromic shift, (d) Bathochromic shift

Some important definitions related to change in wavelength and intensity

-------------------------------------------------------------------------------------------------------------------------------



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(ii) Differentiate Chromophore from Auxochrome



74

VALLIAMMAI ENGINEERING COLLEGE


CY6151- ENGINEERING CHEMISTRY-1

UNIT-4 PHASE RULE AND ALLOYS

PART-A (2 MARKS)

1. Define phase. In what way does it differ from ‘state of matter’?Any homogeneous physically distinct and mechanically separable portion of a system

which is separated from other parts of the system by definite boundaries.States of matter are solids, liquids, gases, and plasma. The state is the form taken by

matter at a given temperature and pressure.

2. Define a component. In what way does it differ from a constituent?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.Oneoftheset oftheminimumnumberofchemicalconstituentsby which everyphaseofagivensy

stemcanbedescribed.

3. Define degree of freedom. What is the degree of freedom of a given quantity of a gas?The minimum number of independent variable factors such as temperature, pressure and

concentration, which much be fixed in order to define the system completely.

The degree of freedom of a given quantity of a gas is bivariant (F = 1-1+2 = 2).

4. Calculate the no. of phases of the followingi) Sulphur(monoclinic)Sulphur (rhombic)Sulphur (liquid)ii) Water+Alcohol Vapour

Ans:i) Sulphur(monoclinic) Sulphur (rhombic) Sulphur (liquid) – Three phasesii) Water + Alcohol Vapour – Two phases

5.

Calculate the degree of freedom fori)

CuSO4 .5H 2O(s) CuSO 4.H2O(s) + 4H2O(v)ii) PCl5(s) PCl 3(l) + Cl2(v)

Ans:i) CuSO4 .5H 2O(s) CuSO 4.H2O(s) + 4H2O(v)

F=C-P+2; F=2-2+2; F=2ii) PCl5(s) PCl 3(l) + Cl2(v)

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

6. Calculate the no.of components for



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i) CuSO4 .5H 2O(s) CuSO 4.H2O(s) + 4H2O(v)ii) PCl5(s) PCl 3(v) + Cl2(v)

Ans:i) CuSO4 .5H 2O(s) CuSO 4.H2O(s) + 4H2O(v) - Three phases, Two componentsii) PCl5(s) PCl 3(v) + Cl2(v) - Two components ,Three phases

7.

What is phase rule? Explain.Ans: 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.F = C-P+2

8. State the merits and demerits of phase rule.Ans: Uses (or) merits of phase rule1. 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 rule1. The phase rule can be applied for the systems in equilibrium.2. The three variables like P,T & C are only considered, but not electrical, magnetic and

gravitational forces.

9. What is condensed phase rule? State its significance.Ans: The system in which only the solid and liquid are considered and the gas phase is ignoredis called a condensed system. Since pressure kept constant, the phase rule becomes F’ = C – P+ 1.This equation is called reduced phase rule.

10.

What is meant by triple point? State its characteristics.Ans: It is the temperature at which three phases namely solid, liquid and vapour aresimultaneously at equilibrium. This point is called triple point, at this point the followingequilibrium will exist.

Solid Liquid Vapour

The degree of freedom of the system is zero i.e., nonvariant. This is predicted by the phase

rule.

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

11. What is eutectic point? Mention its characteristics.Ans: It is the temperature at which two solids and a liquid phase are in equilibrium.

Solid A + Solid Liquid

According to reduced phase rule,

F’=C-P+1

C=2, P=3, therefore F’=1



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The system is non-variant. Below this point the eutectic compound and the metal solidify.

12. What is the difference in the phase diagram of a system forming simple eutectic andcompound formation?Simple Eutectic is a unique mixture of two solids which has the lowest melting point. Since it

is completely immiscible in the solid state, it is a mixture and not a compound.But a compound formation is completely miscible in the solid state, it is a not a mixture.

13. What is thermal analysis? Mention its uses.Ans: Thermal analysis is a method involving a study of the cooling curves of variouscompositions of a system during solidification. The form of the cooling curve indicates thecomposition of the solid.Ex: 1. Cooling curve of a pure solid.

Ex: 2. Cooling curve of a mixture A + B.

Uses of cooling curves:

i) Percentage purity of the compounds can be noted from the cooling curve.

ii) The behavior of the compounds can be clearly understood from the cooling curve.

iii) The procedure of thermal analysis can be used to derive the phase diagram of any two

component system.

14. Mention the differences between triple point and eutectic point.Triple point:

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

Solid Liquid Vapour

Eutectic point:

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

Solid A + Solid B Liquid

By definition,

All the eutectic points are melting points, but all the melting points need not be eutectic

points.

Similarly, all the eutectic points are triple points, but all the triple points need not be

eutectic points.

15. Distinguish melting point, boiling point and triple point.Melting point:

It is the temperature at which the solid and liquid phases are in equilibrium.Solid Liquid

Boiling point:

It is the temperature at which the liquid and vapour phases are in equilibrium.

Liquid Vapour

Triple point:

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

Solid Liquid Vapour



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16. What is congruent melting point? Give an example.Ans: A compound is said to possess congruent melting point, if it melts exactly at a constanttemperature into a liquid having the same composition as that of solid.

Example: Mg(Zn2) (Zinc-Magnesiumalloy system).

17. What is an alloy? Give example for ferrous and nonferrous alloy.Ans: An alloy is defined as “homogeneous solid solution of two or more different element oneof which at least is essentially a metal”. An alloy containing Hg as a constituent element iscalled amalgam.Examples:

Ferrous alloy: Nichrome, Stainless steels Nonferrous alloy: Copper alloys (brass), bronze (copper alloy)

18. Give the composition and uses of brass and bronze.i) Brass:Composition: Cu - 60-90%; Zn - 40-10%

Uses:i) Low cost jewelleries.ii) Rivets, Screws, Musical instruments, Cartridge cases, etc.,

ii) Bronze:Composition: Cu - 80-95%; Sn - 20-9%Uses:i) Pumps, valves, statues, wires, etc.,ii) Bushes, utensils, jewellery, photo frames, etc.,

19. State the significance of increasing the carbon content in steel.

Increasing the Carbon content in steel, the melting point of the Fe-C alloy is progressivelydepressed.

20. Write the composition of Dutch metal and Gun metal?Dutch metal: Cu - 80 %; Zn - 20%Gun metal: Cu - 85%; Zn - 4%; Sn - 8%; Pb - 3%

PART-B (16 MARKS)1. (i) State Phase rule and explain the terms involved in it.

The number of degree of freedom (F) of the system is related to number of components (C) and numberof phases (P) by the following phase rule equation.

F = C-P+2Explanation or meaning of terms

1. Phase(P)Any homogeneous physically distinct and mechanically separable portion of a system which is

separated from other parts of the system by definite boundaries.a. Gaseous phase

All gases are completely miscible and there is no boundary between one gas and the other.For example: air – single phase



78

b. Liquid phaseIt depends on the number of liquids present and their miscibilities.

i. If two liquids are immiscible, they will form three separate phases two liquid phase and onevapour phase. For example: benzene-water.

ii.

If tow liquids are miscible, they will form one liquid phase and one vapour phase. For example:alcohol – water.c. Solid phaseEvery solid constitutes a separate phaseFor example:(i) Water system – three phases(ii) Rhombic Sulphur (s) Monoclinic Sulphur s –Two phases (iii) Sugar solution in water – one phases (iv) CuSO4.5H2O(s) CuSO4.3H2O(s) + 2H2O(g) ---- three phases2. Component (C)

The smallest number of independently variable constituents, by means of which the compositionof each phase can be expressed in the form of a chemical equation.(i) Water system ---one component(ii)

An aqueous system of NaCl --- two component ( NaCl , H2O )(iii)

PCl5(s) PCl3(l)Cl2(g)---two component ,three phases

(iv)

CuSO4.5H2O(s) CuSO4.3H2O(s) + 2H2O(g) ---- three phases3. Degree of freedom (F)“The minimum number of independent variable factors such as temperature, pressure andconcentration, which much be fixed in order to define the system completely”.i) Water systemIce (s) water (l) vapour (g)F = Non variant (or) zero variantii) Ice (s) water (l)F = univariant (one)iii)

For a gaseous mixture of N2 and H 2, we must state both the pressure and temperature.Hence, the system is bivariant.

4.

Phase diagram:Phase diagram is a graph obtained by plotting one degree of freedom against another.Types of phase diagrams

(i)P-T Diagram:used for one component system(ii) T-C Diagram:used for two component system

(ii) Explain thermal analysis. Mention its uses.

Thermal analysis is a method involving a study of the cooling curves of various compositions of a

system during solidification. The form of the cooling curve indicates the composition of the solid.

Ex: 1. Cooling curve of a pure solid.

Ex: 2. Cooling curve of a mixture A + B.

A cooling curve is a line graph that represents the change of phase of matter, typically from agas toa solid or a liquid to a solid. The independent variable (X-axis) is time and the dependent variable (Y-

axis) is temperature.

Below is an example of a cooling curve.



79

The initial point of the graph is the starting temperature of the matter, here noted as the "pouring

temperature". When the phase change occurs there is a "thermal arrest", that is the temperature stays

constant. This is because the matter has more internal energy as a liquid or gas than in the state that it is

cooling to. The amount of energy required for a phase change is known as latent heat. The "cooling

rate" is the slope of the cooling curve at any point.

A Pure substance in the fused or liquid state is allowed to cool slowly. The temperature is noted at

different times. When represented graphically the rate of cooling will be a continuous from ‘a’ to ‘b’.

When the freezing point is reached and solid making its appearance there will be a break in thecontinuity of the cooling curve. The temperature will thereafter remain constant until the liquid is

completely solidified. Thereafter the fall in temperature wil again becomes continuous.

a. Cooling curve of a pure substances b. Cooling curve of a mixture

If a mixture of two solids in the fused state is cooled slowly we get a cooling curve. Here also first a

continuous cooling curve will be obtained as long as the mixture is in the liquid state.

When a solid phase begins to form there will be a break in the cooling curve .But the temperature

will not remain constant unlike in the case of cooling of a purified substance. The temperature will

decrease continuously but at a different rate. The fall of temperature will continue till the mixture

forms a eutectic and the eutectic point is reached.

The temperature will thereafter remain constant until solidification is complete. Thereafter the fallof temperature will become uniform, but the rate of fall will be different from that for a pure substance.

Uses of cooling curves

i) Percentage purity of the compounds can be noted from the cooling curve.

ii) The behavior of the compounds can be clearly understood from the cooling curve.

iii) The procedure of thermal analysis can be used to derive the phase diagram of any two

component system.



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2. (i) Draw a neat one component water system and explain in detail.

The water system:Water exists in three possible phases namely solid, liquid and vapour. Hence therecan be three

forms ofequilibrium.

SolidLiquidLiquid VapourSolidVapour

Each of the above equilibrium involves two phases. The phase diagram for the water system isshown in the figure.

This phase diagram contains curves, areas, and triple.

(i) Curve OAThe curve OA is called vaporization 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=1This equilibrium (i.e. Line OA) will extend up to the critical temperature (347oC). Beyond the

critical temperature the equilibrium will disappear only water vapour will exist.

(ii) Curve OBThe 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



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The curve OC is called melting point curve of ice, it represents the equilibrium between the iceand water. At any point on the curve the following equilibrium will exist.

Ice waterThe 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 vapourThe degree of freedom of the system is zero i.e., nonvariant.This is predicted by the phase rule.

F=C-P+2; F=1-3+2=0Temperature and pressure at the point “O” are 0.0075 oC 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 vapourSometimes water can be cooled below 0oC 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 byseeding or by slight disturbance.

vi) AreasArea 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=C-P=2; F=1-1+2; F=2

(ii) Describe Pattinson’s process of desilverisation of lead.

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; essentiallythat 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. Agentiferous 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 inone 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 oneend and silver in that at the other. The process was economic for lead containing at least 250 grams ofsilver per ton.

3. (i) Discuss the phase diagram of a two component system with congruent melting point.The Zn-Mg 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 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.The phase diagram of Zn-Mg system is shown in the figure. It contains two parts,

i) Left side consists Zn and MgZn2 systemii) Right side consists of MgZn2 and Mg system



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Left side of the phase diagram

i) Curve AE1 The curve AE1 is known as freezing point curve of Zn.Point A is the melting point of pure Zn (420oC).The curve AE1 shows melting point depression of Zn by the successive addition of Mg.Along the curve AE1, solid Zn and the melt are in equilibrium.

Solid Zn Meltii) Point E1 (Eutectic point)

The curve E1 is the eutectic point, where three phases solid Zn, solid MgZn 2 and their melt are inequilibrium. The temperature at this point is 380oC

Solid Zn + Solid MgZn2 Melt

Right side of the phase diagram

iii) Curve CE2 The curve CE2 is known as freezing point curve of Mg.Point C is the melting point of pure Mg (420oC).The curve CE2 shows melting point depression of Mg by the successive addition of Zn.Along the curve CE2, solid Mg and the melt are in equilibrium.

Solid Mg Meltiv)

Point E2 (Eutectic point)The curve E2 is the eutectic point, where three phases solid Mg, solid MgZn 2 and their melt are inequilibrium. The temperature at this point is 347oC

Solid Mg + Solid MgZn2 Melt

v)

Curve E1 BE2 The curve E1 BE2 is known as freezing point curve of MgZn 2.Along the curve, solid MgZn2 and the melt are in equilibrium.

Solid MgZn2 Meltvi) Point B

The point B is the melting point of the compound MgZn2. The temperature at the point is 590oC.Here the solid has the same composition as the liquid. So MgZn2 is said to possess congruentmelting point. The composition of MgZn2 is 33.3% Mg and Zn is 67.7% (i.e., the ratio of Mg andZn is 1:2).

vii) Areas



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(a) Below the line AE 1

The area below the line AE1consists of solid Znand the solution.(b) Below the line CE 2

The area below the line CE 2 consists of solid Mg and the solution.(c) Below the line E 1 BE 2

The area below the line E 1 BE 2 consists of solid MgZn 2 and the solution.(d) Below the point E 1 and E 2

The area below the point E1 and E 2 consists of solid Zn + solid MgZn2 and solid Mgrespectively.

(e) Above the line AE 1 BE 2C

The area above the line AE 1 BE 2C consists of only liquid phase.

(ii) Explain the heat treatment processes, i) Annealing ii) Tempering iii) Hardening.

1. AnnealingAnnealing means softening. This is done by heating the metal to high temperature followed by

slow cooling in a furnace.

Purpose of annealing:

i. It increases the machinabilityii. It also removes the imprisoned gases

Types of annealing

Annealing can be done in two types

i. Low temperature annealing (or) process annealing

ii. 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:

i.

It improves machinability by reliving the internal stress or internal strain.

ii.

It increases ductility and shock resistance.

iii.

It reduce hardness

(i) 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 is,

1. Mild steel=840-870⁰c

2. Medium carbon steel=780-840⁰c

3. High carbon steel=760-780⁰c

Purpose:i. It increases the ductility and machinability.

ii.

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

2. 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,



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(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:

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

ii. Increased toughness & ductility.

Used for cutting tools like blade, cutters etc.

3.

Case HardeningCase hardening is a simple method of hardening steel. It is less complex than hardening and

tempering. This technique is used for steels with low carbon content. Carbon is added to the outer

surface of the steel, to a depth of approximately 0.03mm. One advantage of this method of hardening

steel is that the inner core is left untouched and so still processes properties such as flexibility and is

still relatively soft.

4. (i) Explain the lead silver system with a suitable phase diagram.

The Lead – Sliver system is studied at constant pressure, the vapour phase is ignored and thecondensed phase rule is rule is used.

F’= C-P+1The phase diagram of lead –sliver system is shown in the figure It contains lines, areas and the

eutectic point.

viii) Curve AOThe curve AO is known as freezing point curve of sliver.Along the curve AO, solid Ag and the melt are in equilibrium.

Solid Ag MeltAccording to reduced phase ruleF’=C-P+1C=2P=2F’=1, The system is univariant

ix) Curve BOThe 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



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According to reduced phase ruleF’=C-P+1C=2P=2F’=1, The system is univariant.

x)

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 + Soild Ag Melt

According to reduced phase ruleF’=C-P+1C=2P=3F’=1The system is non-variant.The point “ O “ is called eutectic point or eutectic temperature and is correspondingcomposition,97.4 % Pb and 2.6 % Ag ,is called eutectic composition. Below this point the eutecticcompound and the metal solidify.

xi) Areas

The area above the line AOB has a single phase (molten Pb + Ag ). According to reduced phaseruleF’=C-P+1 C=2P=1F’=2The system is bi-variant.The area below the line AO, OB and point “O” have two phases and hence the system is univariant.According to reduced phase rule F’=C-P+1C=2P=2F’=1The system is uni-variant.

(ii) Explain the heat treatment processes, i) Nitriding ii) Normalizing iii) Carburizing. Nitriding:i) Nitriding is the process of heating the metal alloy in presence of ammonia to about 550⁰C.ii)

The nitrogen (obtained by the dissociation of ammonia) combines with the surface of the alloyto form hard nitride.

PurposeTo get super-hard surface.

Normalizing:It is the purpose of heating steel to a definite temperature (above its higher critical temperature)&

allowing it to cool gradually in air. Purpose

1.

Recovers homogeneity.2.

Refines grains.3. Removes internal stresses.4. Increases toughness.5. Used in engineering works.

Note: The difference between normalized & annealed steel are 1. Normaled steel will not be as soft as annealed steel.

2. Also normalizing takes much lesser time than annealing. Carburizing

i) The mild steel article is taken in a cast iron box within containing small pieces of charcoal(carbon material).



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ii) It is heated to about 900 to 950⁰C & allows it for sufficient time, so that the carbon is absorbedto required depth.iii) The article is then allowed to cool slowly within the box itself.iv)

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.

5. (i) Draw a neat Zn-Mg system and explain in detail.The Zn-Mg 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 at a constant temperature intoa 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 phasediagrams of Pb-Ag system placed side-by-side.

The phase diagram of Zn-Mg system is shown in the figure. It contains two parts,iii) Left side consists Zn and MgZn2 systemiv) Right side consists of MgZn2 and Mg system

Left side of the phase diagram

xii) Curve AE1 The curve AE1 is known as freezing point curve of Zn.Point A is the melting point of pure Zn (420oC).The curve AE1 shows melting point depression of Zn by the successive addition of Mg.Along the curve AE1, solid Zn and the melt are in equilibrium.

Solid Zn Meltxiii) Point E1 (Eutectic point)

The curve E1 is the eutectic point, where three phases solid Zn, solid MgZn 2 and their melt are inequilibrium. The temperature at this point is 380oC

Solid Zn + Solid MgZn2 Melt

Right side of the phase diagram

xiv) Curve CE2 The curve CE2 is known as freezing point curve of Mg.



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Point C is the melting point of pure Mg (420oC).The curve CE2 shows melting point depression of Mg by the successive addition of Zn.Along the curve CE2, solid Mg and the melt are in equilibrium.

Solid Mg Meltxv)

Point E2 (Eutectic point)The curve E2 is the eutectic point, where three phases solid Mg, solid MgZn 2 and their melt are inequilibrium. The temperature at this point is 347oC

Solid Mg + Solid MgZn2 Meltxvi)

Curve E1 BE2 The curve E1 BE2 is known as freezing point curve of MgZn 2.Along the curve, solid MgZn2 and the melt are in equilibrium.

Solid MgZn2 Meltxvii) Point B

The point B is the melting point of the compound MgZn2. The temperature at the point is 590oC.Here the solid has the same composition as the liquid. So MgZn2 is said to possess congruentmelting point. The composition of MgZn2 is 33.3% Mg and Zn is 67.7% (i.e., the ratio of Mg andZn is 1:2).

xviii) Areas(f) Below the line AE 1

The area below the line AE1 consists of solid Zn and the solution.(g) Below the line CE 2

The area below the line CE 2 consists of solid Mg and the solution.(h) Below the line E 1 BE 2

The area below the line E 1 BE 2 consists of solid MgZn 2 and the solution.(i) Below the point E 1 and E 2

The area below the point E1 and E 2 consists of solid Zn + solid MgZn2 and solid Mgrespectively.

(j) Above the line AE 1 BE 2C

The area above the line AE 1 BE 2C consists of only liquid phase.

(iii) What are the types of alloys? Discuss the purpose of making alloys.

ALLOYS

FERROUS

ALLOYS

NON-FERROUS

ALLOYS

(i) Nichrome(ii) Alnico (i) Brass

(iii)Stainless steel (ii) Bronze

Importance or need of making alloys1.

To increase the hardness of the metal ExampleGold and silver are soft metal they are alloyed with copper to make them hard

2. To lower the melting points of the metal ExampleWood metal (an alloy of lead, bismuth, tin and cadmium) melts at 60.5⁰c which is far below themelting points of any of these constituent metals

3. To resist the corrosion of the metalExamplePure iron rested but when it is alloyed with carbon chromium (stainless steel) which resists



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corrosion4. To modify chemical activity of the metal

ExampleSodium amalgam is less active than sodium but aluminium amalgam is more active thanaluminium

5. To modify the colour of the metal ExampleBrass an alloy of copper (red) and size (silver-white) is white colour.

6. To get good casting of metal ExampleAn alloy of lead with 5% tin and 2% antimony is used from casting printing type due toits goodcasting property

Functions (or) effects of alloying elementsAddition of small amount of certain metals such as Ni, Cr, Mo, Mn, Si, v and Al impart special

properties like hardness, tensile strength, resistance to corrosion and coefficient of expansion on steel.

Such products are known as special steel or alloy steelsSome important alloying element and their functions are given in table

6.

(i) Discuss the effect of Ni, Cr and Mn in the alloying of steel.i)

Nonmagnetic type steel:

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%

Properties1. Resistance to corrosion.

2. Corrosion resistance is increased by adding molybdenum

Uses

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

ii) Nichrome

Nichrome is an alloy of nickel & chromium

Composition

Nickel – 60%

Chromium – 12%

Iron – 26%Manganese – 2%

Properties

1. Good resistance to oxidation & heat

2. High melting point & electrical resistance

3. Withstand heat up to 1000-1100⁰C

Uses

1. Used for making resistance coils, heating elements in stoves & electric irons

2. Used in making parts of boilers, steam lines stills, gasturbines, aero engine



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valves,retorts,annealing boxes.

(ii)

Discuss a) Heat resisting alloy steel, b) Magnetic steel

a)

Non - Heat Treatable Stainless Steel

Properties

Possess less strength at high temperatureResistant to corrosion

Types of Non Heat Treatable Stainless Steel

(a)

Magnetic Type

Composition: Chromium-12-22%; Carbon-0.35%

Properties:

1. Can be forged, rolled & machined

2. 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:

1. Resistance to corrosion.

2. Corrosion resistance is increased by adding molybdenum

Uses:

They are useful in making household utensils, sinks, dental & surgical instruments.

7.

(i) What is stainless steel? Describe the different types of stainless steel.Stainless Steels (or) Corrosion Resistant Steels

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

molybdenum, etc.

Composition

Chromium-16% or more; Carbon-0.3-1.5%

Properties

1.

Resist corrosion by atmospheric gases & also by other chemicals.

2.

Protection against corrosion is due to the formation of dense, nonporous, 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. Stainless steel

2. Heat treatable stainless steel

3. Non heat treatable stainless steel

4.

Magnetic type nonmagnetic type



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1. Heat Treatable Stainless Steel

Composition

Carbon-1.2%; Chromium-less than 12-16%

Properties

Magnetic, tough & can be worked in cold condition

Uses

1. Can be used up to 800⁰C2. Good resistant towards weather & water

3. In making surgical instruments, scissors, blades, etc.

2. Non - Heat Treatable Stainless Steel

Properties

Possess less strength at high temperature

Resistant to corrosion


a. Magnetic Type


Properties:1. Can be forged, rolled & machined

2. Resist corrosion

Uses:


b. Non Magnetic Type


than 23%.

Example: 18/8 STAINLESS STEEL


Properties:

1. Resistance to corrosion.2. Corrosion resistance is increased by adding molybdenum

Uses:

They are useful in making household utensils, sinks, dental & surgical instruments.

(ii) What are Non-ferrous alloys? What are its applications? Write about any two Non-ferrous

alloys.

Nonferrous alloys

Do not contain iron as one of the main constituent.

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

Properties1. Softness & good formability

2. Attractive (or) very good colours

3. Good electrical & magnetic properties

4. Low density & coefficient of friction

5. Corrosion resistance

Important Nonferrous Alloys

1. Copper Alloys (Brass)



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Brass contains mainly copper & zinc

Compositon: Cu - 90 %; Zn - 10%

Properties

• Greater strength, durability & machinability

• Lower melting points than Cu & Zn

•

Good corrosion resistance & water resistance property

Uses:Rivets, screws and forging hardwares and jewellery.

2.

Bronze(Copper Alloy)Bronze contains copper & tin

Compositon: Cu: 80-95%; Sn: 20-5 %

Properties

•

Lower melting point

• Better heat & electrical conducting property

•

Non-oxidizing,corrosion resistance & water resistance property.

Uses:

Pumps, valves, coins, statues and utensils.

(ii) Discuss the composition, characteristics and uses of German silver and Gun metal.German silver:

Composition:Cu: 25 - 50 %; Zn: 10 - 35 %; Sn: 5 - 35 %

Characteristics:

1) It looks like silver.

2) It is ductile and malleable

3) It possesses good strength and corrosion resistance to water.

Uses:

Utensils bolts, screws, ornaments and coinage decorative articles, table wares.

Gun metal:

Composition: Cu - 85 %; Zn - 4 %; Sn - 8 %; Pb - 3%

Characteristics:

1) Hard and tough.

2) Strong to resist the force explosion.

Uses:

Foundry works, heavy load bearings, steam plants and water fittings.

8.

(i) Discuss Nichrome with its composition and applications

Nichrome

Nichrome is an alloy of nickel and chromium.

Composition

Nickel – 60%; Chromium – 12%; Iron – 26%; Manganese – 2%

Properties

1. It shows good resistance to oxidation & heat.

2. It possesses high melting point & electrical resistance.

3. It can withstand heat up to 1000-1100⁰C.



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Uses


2. Used in making parts of boilers, steam lines stills, gasturbines, and aero-

enginevalves,retorts,annealing boxes.

(ii) What are ferrous alloys? Give their propertiesFerrous Alloys or Alloy Steels

Ferrous alloys are the type of steels in which the elements like Al, B, Cr, Co, Cu, Mn are presentin sufficient quantities, in addition to carbon & iron.

Properties:

1. It possesses high yield point & strength.

2. It possesses sufficient formability, ductility & weldability.

3. They are sufficiently corrosion & abrasion resistant.

4. Less distortion & cracking.

5. High temperature strength.

9.

(i) Discuss the composition, properties and uses of any two ferrous alloys.a)

Nichrome

Nichrome is an alloy of nickel and chromium.

Composition:

Nickel – 60%; Chromium – 12%; Iron – 26%; Manganese – 2%

Properties:

1. It shows good resistance to oxidation & heat.

2. It possesses high melting point & electrical resistance.

3. It can withstand heat up to 1000-1100⁰C.

Uses:


2. Used in making parts of boilers, steam lines stills, gas turbines, and aero-engine valves, retorts,

annealing boxes.

b) Stainless Steels (or) Corrosion Resistant Steels

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

molybdenum, etc.

Composition

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, nonporous, 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. Stainless steel

2. Heat treatablestainless steel

3.

Non heat treatablestainless steel

4.

Magnetic type nonmagnetictype

1. Heat Treatable Stainless Steel



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Composition

Carbon-1.2%; Chromium-less than 12-16%

Properties

Magnetic, tough& can be worked in cold condition

Uses

1. Can be used up to 800⁰C

2. Good resistant towards weather & water3. In making surgical instruments,scissors,blades,etc.

2.

Non - Heat Treatable Stainless Steel

Properties

Possess less strength at high temperature

Resistant to corrosion


(a)

Magnetic Type


Properties:

1. Can be forged,rolled & machined2. Resist corrosion

Uses:


(b) Non Magnetic Type


than 23%.

Example:18/8 STAINLESS STEEL


Properties:

1. Resistance to corrosion.

2. Corrosion resistance is increased by adding molybdenumUses:

They are used in making household utensils,sinks,dental & surgical instruments.

(ii) Mention the limitations of Phase rule.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.

10. (i) What is condensed phase rule? What is the number of degrees of freedom at theEutectic point

The system in which only the solid and liquid are considered and the gas phase is ignoredis called a condensed system. Since pressure kept constant, the phase rule becomesF’ = C – P + 1.This equation is called reduced phase rule.

For example,

Solid A + Solid Liquid

According to reduced phase rule,

F’=C-P+1



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C=2, P=3, therefore F’=1

The system is non-variant. Below this point the eutectic compound and the metal solidify.

(ii) Discuss the composition, properties and uses of various types of Brass and Bronze.Brass:Compositon: Cu - 90%; Zn - 10%

It is an alloy of copper. It contains mainly copper and zinc.Properties:

1) Lower the melting point the Cu and Zn.

2) Good corrosion and water resistance property.

3) Greater strength and durability.

Uses:

Rivets, screws and forging hardwares and jewellery.

Bronze:Compositon: Cu: 80-95%; Sn: 20-5 %

Properties:

1) Tough and strong

2) Corrosion resistance property.

Uses:

Pumps, valves, coins, statues and utensils.



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Unit V

Nanochemistry

PART-A (2 MARKS)

1. What is nanochemistry?

Nanochemistry is the phenomena and manipulation of materials at atomic molecular

macromolecular scales. It is the branch of nanoscience with deals with the study of synthesis

characterization and with its nano materials and chemical applications.

2. What are the differences between nanoparticles and bulk materials?

S.

No

Nanoparticle Bulk materials

1 The size of nanoparticlesare less than 100nm in

diameter

Bulk materials are larger inmicron size

2 Nanoparticle are collection

of few molecules that is less

than 100nm

Bulk materials contains

thousands of molecules

3 Surface area of Nanparticlesis more

Surface area of Bulkmaterials is less

4 Hardness of Nanomaterialsis 5 times more

Hardness of Bulk materialis 5 times less

5 Strength of Nanomaterials

is 3-5 times higher

Strength of Bulk material 3-

5 times less

6 Nanoparticles possesses

size dependent properties

Bulk material possess

constant physical properties

7 Nanoparticles due to its

size

Bulk material due to

unexpected size

3. What is meant by size dependent property of nanoparticles?

Nanomaterials (size range 1 x 10-9 m) possess relatively larger surface area compared to

their bulk materials (size range 1x10-6 m) and more active and also inert for some other cases.

Nearly all the properties of depend on their size. The reason for which is mere reduction in their

grain size. These grains contain only few atoms within each grain but large number of atoms



96

present in the grain boundaries. Decrease in grain size will increase in grain boundaries or

interfaces. So the physical and chemical properties of nonmaterial are significantly altered.

4. What are nanowires?

Nanowire is one dimensional cylindrical solid material having an aspect ratio length towidth ratio greater than 20 nm. Diameter of the nanowire ranges from 10-100 nm.

Examples:

1. Metallic nanowires Au, Ni, and Pt.

2. Nanowires of semiconductors InP, Si, and GaN.

3. Nanowires of insulators SiO2 and TiO 2.

4. Molecular Nanowires DNA.

5. What are nanoclusters?

Nanoclusters are fine aggregates of atoms (1000 more atoms) or molecules and the size

of which ranges from 0.1 to 10 nm. Of all the nanomaterials, nanoclusters are the smallest sized

nanomaterials because of their close packing arrangement of atoms. Examples: CdS, ZnO, SiO2

etc.,

6. What are nanorods? Mention their specific applications.

Nanorod is one dimensional cylindrical solid material having an aspect ratio length to

width ratio less than 20 nm. Examples: Zinc oxides and Au nanorods.

1. Nanorods find application in display technologies.

2. It is also used in the manufacturing of micro mechanical switches.

3. Nanorods are used in an applied electric field, micro electro mechanical systems, etc.,

7. Write any four nanomaterials.

1. Nanoparticles - (Nacl)n

2. Nanowires - DNA

3. Nanorod - Zinc oxides and Cadmium Sulphide

4. Nanoclusters - CdS and ZnO

8. Mention some characteristic properties of nonmaterials.

Melting Points: Nanomaterials have a significantly lower melting point and appreciable

reduced lattice constants. This is due to huge fraction of surface atoms in the total amount of

atoms.



97

Optical properties: Optical properties of nanomaterials are different from bulk forms.

Reduction of material dimensions has pronounced effects on the optical properties.

Magnetic properties: Magnetic properties of nanomaterial are different from that of

bulk materials. Ferro-magnetic behavior of bulk materials disappear, when the particlesize is

reduced and transfers to super-paramagnetics. This is due to the huge surface area. Mechanical properties: The nanomaterials have fewer defects compared to bulk

materials, which increases the mechanical strength.

9. Mention a few applications of nanomaterials.

1. Nanomaterials are used as nanodrugs for the cancer and TB therapy.

2. Nanotechnology is used in the production of laboratories on a chip.

3. Nanoparticles function as nano-medibots that release anti-cancer.

4. Nanofilteration makes use of nano-porous membranes having pores smaller than

10nm.

5. Dissolved solids and color producing organic compounds can be filtered very easily

from water.

10. What are carbon nanotubes?

Carbon nanotube is tubular form of carbon with 1-3 nm diameters and a length of few nm

to microns. When graphite sheets are rolled into a cylinder, their edges join ton each other form

carbon nanotubes, Each carbon atom in the carbon nanotubes is linked by covalent bonds, But

the number of nanotubes align into ropes and are held together by weak vander Wall’s forces.

Types of Carbon nanotubes:

Depending upon the way in which graphite sheets are rolled two types of CNTs are

formed.

Singled-walled nanotubes (SWNTs)

Multi-walled nanotubes (MWNTs)

11. Distinguish between Single-walled carbon nanotube (SWCNT) and Multi-walled carbon

nanotube (MWCNT).

S.No. SWCNT MWCNT

1 Single layer of graphite. Multi layer of graphite.



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2 Catalyst is required for

synthesis.

Catalyst is not required.

3 Bulk synthesis is difficult. Bulk synthesis is easy.

4 Purity is poor. Purity is high.

12. What is laser ablation?

This method involves vaporization of target material containing small amount of catalyst

(nickel or cobalt) by passing an intense pulsed laser beam at higher temperature to about 120oC.

13. Write note on Chemical Vapor Deposition (CVD).

It is a process of chemically reacting volatile compound of material with other gases, to

produce a non volatile solid that deposits automatically on a suitably placed substrate.

14. Write note on Electro-Deposition.

Electro-Deposition is an electrochemical method in which ions from the solution are

deposited at the surface of the cathode. Template assisted electro-deposition is an important

technique for synthesizing metallic nanomaterials with controlled shape and size.

15. Write the applications of nanowires and nanoclusters.

Nanowires

Used for enhancing mechanical properties of composites.

Used to prepare active electronic compounds such as p-n junction and logic gates.

Used in digital computing and high density data storage media.

Nanoclusters

Used as catalyst, nano based chemical sensors and light emitting diode in quantum

computers.

16. How nano-particles prepared by precipitation method?

Nano-particles prepared by precipitation reaction between the reactants in the presence of

water soluble inorganic stabilizing agent.

Ex. BaSO4 nano particle prepared sodium hexameta-phosphate as stabilizing agent

Ba(NO3)2 + Na 2SO4 → BaSO 4↓ + 2NaNO 3

17. What is the role of CNT in the H 2 – O 2 fuel cell?



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Carbon nanotube is tubular form of carbon with 1-3 nm diameters and a length of few nm

to microns. It is used for storage of H2 gas in the fuel cell.

18. What are the applications of nanomaterials in Electronics?

Used as quantum wire, in integrated memory circuit, in transistor.

Used in digital computing and high density data storage media.

19. Distinguish between top-down and bottom-up approach in the preparation of

nanomaterial.

S.No Top-down Bottom-up

1 Conversion of bulk materials

into smaller particles of nanoscale structure.

Involves building-up of

materials from the bottom byatom by atom or molecule.

2 Methods:

i. Laser ablation

ii. CVD

iii. Electro-deposition

Methods

i. Precipitation

ii.Thermolysis

20. Write short note on Hydrothermal and Solvothermal synthesis of nanoparticles.

Hydrothermal

It involves crystallization of substances from high temperature aqueous solutions at high

vapor pressure. It performed below super critical temperature of water 3740C.

Solvothermal

It involves the use of solvent under high temperature 100-1000oC and moderate to high

pressure 1-10,000 atm that facilitate the interaction of precursors during synthesis.

PART-B (8 MARKS)

1. (i) Discuss the size dependent properties of nanomaterilas.



100

Nearly all the properties of depend on their size .The reason for which is mere reduction

in their grain size (1x10-9 m). The properties like hardness, strength, ductility, melting point anddensities vary for nanomaterials as given in following chart.

Reason: Nanomaterials possess relatively larger surface area compared to their bulk materials

(size range 1x10-6 m) more active and inert for some other cases.

These grains contain only few atoms within each grain but large number of atoms present

in the grain boundaries. Decrease in grain size will increase in grain boundaries or interfaces. Sothe physical and chemical properties of nonmaterial are significantly altered.

The defect configurations affect the properties of nanomaterials. Generally presence of

increased fractions of defects increases the mechanical and chemical properties of nanomaterials.

Examples

1. Nano-crystalline ceramics are tougher and stronger than those with coarse grains.

2. Nano-sized metals exhibit significant decreases in toughness and increase in yield

strength.

3. A piece of gold is fairly gold colored materials, but gold nanoparticles are in deep redcolor.

4. Silver nanoparticles of silver react rapidly with dilute hydrochloric acid because of the

large surface area to volume ratio.

(ii) Distinguish molecules, nanoparticles and bulkmaterials

S.

NO

Properti

es

Molecu

les

Nanoparti

cles

Bulkmater

ials

1 Size of

the

Size is

much

Size is

larger

Size is

much



101

particle smaller

1x10-12

m

than

molecules

but

smaller

than bulk

materials1x10-9 m

larger than

the

molecules

and

nanopartil

ces1x10-6 m

2 Magnitu

de of

constitut

ing

particles

Few

Angstro

ms (Ao)

(10-10

m)

Angstrom

s to

nanometer

(10-10 m

to10-10)

Microns

to

millimeter

3 Numberofconstitut

ing particles

Twoatomsfor

molecules

2 toseveralthousands

Infinite

4 Electron

ic

structure

Confine

d

confined continuou

s

5 Geometr

icstructure

Well-

definedstructur

e and

predicta

ble

Well-

definedstructure

and

predictabl

e

Crystal

structuredecides

6 Example NaCl,

HCl

(NaCl)n Gold bar

& Silve

bar



102

2. (i) Explain laser ablation method of preparing nanoparticles.

Laser evaporation

It involves vaporization of graphite target, containing small amount of cobalt and nickel,

by exposing it to an intense pulsed laser beam at higher temperature (12000C) in a quartz tube

reactor. An inert gas such as argon is simultaneously allowed to pass into the reactor to seep the

evaporated carbon atoms from the furnace to the colder copper collector, on which they

condense as carbon nanotubes.

Uses

Nanotubes having a diameter of 10 to 20 nm and 100µm can be produced by this

method.

Ceramic particles and coating can be produced.

Other materials like silicon, carbon can also be converted into nanoparticles by

this method.

Advantages of laser ablation

or



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1. It is very easy to operate

2. The amount of heat required is less

3. It ids eco-friendly method because no solvent is used

4. The product, obtained by this method, is stable

5.

This process is economical(ii) Discuss precipitation process with example in preparing nanoparticles.

Nano-particles prepared by precipitation reaction between the reactants in the presence of

water soluble inorganic stabilizing agent.

Precipitation of BaSO4 Nano-Particles

10 gm of sodium hexameta-phosphate (stabilizing agent) was dissolved in 80 ml of

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

was added followed by 10 ml of 1M Ba(NO3)2 Solution. The resulting solution was stirred for1hr. Precipitation occurs slowly. The resulting Precipitate was then centrifuged, washed with

distilled water and vacuum dried.

Ba(NO3)2 + Na 2SO4 → BaSO 4↓ + 2NaNO 3

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

Precipitation by reduction

Reduction of metal salt to the corresponding metal atoms

These clusters are surrounded by stabilizing molecule that prevents the atoms

agglomerating.

Examples

3. (i) Describe the hydrothermal synthesis of nanoparticles.

Hydrothermal synthesis

It involves crystallizations of substances from high temperature aqueous solutions at high

vapor pressure. Hydrothermal synthesis is usually performed below the super critical temperature

of water (374 oC).



104

Method

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

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

maintained at the opposite ends of the growth chamber, so that the hotter end dissolves the

nutrient and the cooler end causes seeds to take additional growth.

(ii) Explain solvothermal process for the preparation of nanoparticles.

Solvothermal synthesis

Solvothermal synthesis involves the use of solvent under high temperature (between 1000C to 1000 0C) and moderate to high pressure (1atm to 10000 atm) that facilitate the interaction

of precursors during synthesis.

Method

A solvent like ethanol, methanol, and 2-propanol is mixed with certain metal precursors

and the solution mixture is placed in 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

vapor elevates the boiling point of the solvent.

Example: Sovothermal synthesis of Zinc oxide

Zinc acetate dehydrate is dissolved in 2-propanol at 500C .Subsequently, the solution is

cooled to 00C and NaOH is added to precipitate ZnO .The solution is then heated to 650C to

allow ZnO growth for some period of time .Then a capping agent (1-dodecanethiol) is injected

into the suspension to arrest the growth. The rod shaped ZnO nano-crystal is obtained.



105

4. (i) Discuss the electrical properties of nanomaterials/CNT.

Electrical properties

CNT act as semiconductors or metallic conductors depending upon the diameter and the

chirality of the tube. Chirality refers to the helical type of the tube. The metallic tube are said to

have’ arm-chair’ structure.

Plot of energy gaps of the CNT’S versus the reciprocal of the diameter gives a straight

line.

Single-walled nanotubes are mostly used for miniaturizing electronics beyond the microelectromechanical scale that is currently the basis of modern electronics. The most basic building block of these systems of these systems is the electric wire, and SWNTs can be excellent

conductors.The measured conductance G is equal to 1/V, where V stands for the voltage difference

between the tip and the other parts of CNT. This is a measure of the local electronic density of

the energy states .Higher the density of the states closer the energy levels.



106

Scanning tunneling microscopy (STM) is used to investigate the above study .When the

current ( in Nano amperes ) is measured for a single walled nanotube (SWNT) across two metalelectrodes voltage and plotted versus the voltage ,it (in milli volts )shows a step like feature .

Metallic conductivity of CNT is very high carrying a billion amperes square cm becauseof the very low resistance. Another advantage is, high current does not heat the CNTs. They also

have high thermal conductivity .The magnetic resistance change of resistance by the application

of external magnetic field is exhibited at low temperatures (similar to Zeeman Effect in atomicspectra).

Plot of magnetic field dependence on resistance shows decrease in resistance with

decrease in temperature. This is because under applied magnetic field, additional energy levels

are introduced in between already existing energy levels. Hence conduction increases while

resistance decreases.

(ii) Describe the synthesis, properties and applications of carbon nanorods.

Nanorod is one dimensional cylindrical solid material having an aspect ratio length to

width ratio less than 20 nm. Examples: Zin oxides, Au rod

Synthesis of Nano rods ( Electro-deposition of Gold on Silver)

Nano rods are produced by direct chemical synthesis .A combination of ligands act as

shape control agents and bond to different face to the Nano rods with different strength. This

allows different nano rods to grow at different rates producing elongated objects. Many of the

above nanorods are not manufactured due to lack of commercial demand.

Properties of nanorods

1. Nano rods are three-dimensional materials.



107

2. It exhibits optical and electrical properties.

Applications

1. Nano rods find application display technologies

2. It is also used in the, manufacturing of micro mechanical switches.

3.

Nano rods are used in an applied electric field, micro electro mechanical systems, etc.4. Nano rodsalong with noble metal nanoparticles function as theragnostic agents.

5. They are used in energy harvesting and light emitting devices.

6. Nanorods have used as cancer therapeutics.

5. (i) Write note on carbon nanotubes and its properties?

Carbon nanotube (CNTs)

Carbon nanotube is tubular form of carbon with 1-3nm diameter and a length of few nm

to microns. Generally carbon in the solid phase exists in different allotropic forms like graphite,

diamond, fullerenes and nano tubes. Carbon nanotubes tubular forms of carbon. When graphite

sheets are rolled into a cylinder, their edges join ton each other form carbon nanotubes, each

carbon atom in the carbon nanotubes is linked by covalent bonds, but the number of nanotubes

align into ropes and are held together by weak Vander Walls forces.

Structure or types of Carbon nanotubes

Depending upon the way in which graphite sheets are rolled two types of CNTs are

formed.

Singled-walled nanotubes (SWCNTs)

Multi-walled nanotubes (MWCNTs)



108

Singled-walled Carbon nanotubes (SWCNTs): SWCNTs consist of one tube of graphite .It is

one-atom thick having a diameter of 2 nm and a length of 100um. SWCNTs are very important,

because they exhibit important electrical properties .It is an excellent conductor. There are kinds

of nanotubes are resulted, based on the orientation of the hexagon lattice.

Arm-chair Structure

The lines of hexagons are parallel to the axis of the nanotube.

OR



109

Zig-Zag Structure

The lines of carbon bonds are down the Centre.

Chiral nanotubes

It exhibits twist or spiral around the nanotubes.

It has been confirmed that arm-chair carbon nanotubes are metallic while Zig-Zag and chiral

nanotubes are semiconducting.

Multi-walled Carbon nanotubes (MWCNTs)

MWCNTs (nested nanotubes) consist of multiple layers of graphite rolled in on

themselves to form a tube shape. It exhibits both metallic and semiconducting properties. It is

used storing fuels such as hydrogen and methane.

Properties of CNTs

1. CNTs are very strong; withstand extreme strain in tension and possess elastic

flexibility.

2. The atoms in a nano-tube are continuously vibrating back and forth.

3.

It is highly conducting and behaves like metallic or semiconducting materials.

4. It has very high thermal conductivity and kinetic properties.

(ii) Discuss Chemical Vopour Deposition (CVD) method for the synthesis of nanomaterials.

Chemical Vapour Deposition (CVD)



110

This process involves conversion of gaseous molecules into solid Nano materials in the

form of tubes, wires or thin films .First the solid materials are converted into gaseous molecules

and then deposited as Nano materials.

CNT preparation

The CVD reactor consists of a higher temperature vacuum furnace maintained at inertatmosphere .The solid substrate containing catalyst like nickel, cobalt, iron supported on a

substrate material like, silica, quarts is kept inside the furnace. The hydrocarbons such as

ethylene, acetylene and nitrogen are connected to the furnace .Carbon atoms, produced by the

decomposition at 1000 0C, condense on the cooler surface of the catalyst.

As this process is continuous, CNT is produced continuously.

Types of CVD Reactors

Generally the CVD reactors are of two types

Hot-wall CVD

Hot-wall CVD reactors are usually tubular in form. Heating is done by surrounding the reactor

with resistance elements.

Cold-wall CVD

In Cold-wall CVD reactors, substrates are directly heated inductively while chamber walls are air(or) water cooled.



111

Advantages of CVD

1. Nano materials, produced by this method are highly pure

It is economical

2.

Nano materaials, produced by this method are defect free

3. As it is simple experiment, mass production in industry can be done without major

difficulties

6. (i) Explain Electo-deposition method for the synthesis of nanomaterial.

Electro-Deposition is an electrochemical method in which ions from the solution are

deposited at the surface of the cathode. Template assisted electro-deposition is an important

technique for synthesizing metallic nanomaterials with controlled shape and size. Array of nano

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

template as cathode. Process of Electro-deposition

The cell consists of reference electrode, specially designed cathode and anode. All the

electrodes are connected with battery through a voltmeter and dipped in an electrolytic solution

of suitable metal as shown in figure. When the current is passed through the electrodes of

template the metal ions from solution enter in to the pores and gets reduced at the cathode,

resulting in the growth of nanowire inside the pores of the template.

EX: Electro-deposition of Gold on Silver

Nano structured gold can be prepared by the electro deposition technique using gold

sheets as an anode and silver plate as cathode. An array of alumina template is kept over the

cathode as shown in the figure and AuCl3 is used as on electrolyte.



112

When the current of required strength is applied through the electrodes, Au+ ions diffuse

into pores of alumina templates and get reduced at the cathode resulting in the growth of

nanowires or nanaorods inside the pores of alumina templates.

Advantages:

1.This method is cheap and fast

2.Complex shape objects can be coated3. The film or wire obtained is uniform

4. Nano wire of Ni,Co,Cu and Au can be made.

6. (ii) Discuss the vibration properties of CNTs with suitable diagram.

There are two types of vibrations

1. The first mode of vibration is perpendicular to the axis, increases and decreases in

diameter of CNT.

2. The second one is parallel to the axis, length of CNT is changes



113

7. (i) What are nanoclusters and nanowires? Explain their properties and applications.

Nanoclusters

Nanoclusters are fine aggregates of atoms or molecules and the size of which ranges from

0.1 to 10 nm. Of all the nano materials, nanoclusters are the smallest sized nano materials

because of their close packing arrangement of atoms.

EX: Cds, ZnO etc.,

Magic number is the number of atoms present in the clusters of critical sizes with higher stability

Production of nanocluster

Nanoclusters can be produced from atomic or molecular constituents or from the bulk

materials either by bottm up process or top down process as shown in figure.

Properties of nanoclusters

1.

Less reactivity, Low melting point and confined electronic structure. Applications of nanoclusters

Used as catalyst, nano based chemical sensors and light emitting diode in quantumcomputers.



114

7. (ii) Discuss various types of synthesis involved in the preparation of nanomaterials.

The following two approaches are used for the synthesis of nanomaterials

1. Top-down process or physical or hard methods.

2. Bottom-up process or chemical or soft methods.

1. Top-down process.

Top-down process involves conversion of bulk materials into smaller particles of nano scale

structure.

Methods:

i. Laser ablation

ii. CVD


2. Bottom-up process.

Bottom-up process involves building-up of materials from the bottom by atom by atom or

molecule.



115

Methods

i. Pricipitation

ii.Thermolysis

a. Solvothermolysis b. Hydrothermolysis.

8. (I )How are carbon nanotubes are synthesized explain in detail.

Carbon nanotubes can be synthesized by any one of the following methods

1. Pyrolysis

2. Laser evaporation

3. Carbon arc method

4.

Chemical Vapor deposition

1. Pyrolysis

Carbon nanotubes are synthesized by the pyrolysis of hydrocarbons such as acetylene at

about 700oC in the presence of Fe-graphite catalyst under inert conditions.

2. Laser evaporation

It involves vaporization of graphite target, containing small amount of cobalt and nickel,

by exposing it to an intense pulsed laser beam at high temperature 1200oC in quartz tube reactor.

An inert gas such as argon is simultaneously allowed to pass into the reactor to sweep the

evaporated carbon atom from the furnace to the colder copper collector, on which they condense

as carbon nanotubes.

3. Carbon arc method



116

It is carried out by applying direct current 60-100 A and 20-25V arc between graphite

electrodes of 10-20µm diameter.

4. Chemical Vapor Deposition

It involves decomposition of vapor of hydrocarbons such as methane, acetylene etc., at

high temperatures 1100o

C in the presence of metal nana particles catalysts like nickel, cobalt,iron supported on MgO or Al2O3.Carbon atoms produced by the decomposition condense on a

cooler surface of catalyst.

8. (ii) Explain various applications of carbon nanotubes.

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

2. It is also used as light weight shielding materials for protecting electronic equipments.

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

4. It is used in composites, ICs

5. It is also act as efficient catalysts for some chemical reactions.

6. It acts as a very good biosensor. Due to its chemical inertness carbon nano tubes are

used to detect many molecules presence in the blood.

7. It also used in water softening process as a filter.

8. used in H2-O2 fuel cell

9. (i) Explain the various properties of nanomaterials.

1. Melting point

Nano-materials have a significantly lower melting point and appreciable reduced lattice

constants. This is due to huge fraction of surface atoms in the total amount of atoms.

2. Optical properties

Reduction of materials has pronounced effects on the optical properties. Optical

properties of nano-materials are different from bulk forms.

The change in optical properties is caused by two factors1. The quantum confinement of electrons within the nano particles increases the energy level

spacing.

Ex: The optical absorption peak of a semiconductor nano-particles shifts to a short wavelength,

due to an increased band gap

2. Surface plasma resonance, which is due to smaller size of nano particles then the wave length

of incident radiation.



117

Ex: The color of metallic nano-particles may change with their sizes due to surface plasma

resonance.

3. Magnetic properties

Magnetic properties of nano materials are different from that of bulk materials.

Ferro-magnetic behavior of bulk materialsFerro-magnetic behavior of bulk materials disappear, when the particle size reduced and

transferred to super-paramagnetic. This is due to the huge surface area.

4. Mechanical properties

The nano-materials have fewer defects compared to bulk materials, which increases the

mechanical strength.

i. Mechanical properties of polymeric materials can be increased by addition of nano fillers.

ii. As nano materials are stronger, harder and more wear resistant and corrosion resistant, they

are used in spark plugs.

5. Electrical properties

i. Electrical conductivity decreases with a reduced dimension due to increased surface

scattering. However, it can be increased due to better ordering in micro-structure.

Ex: Polymeric fifibres

ii. Nanocrystalline materials are used as very good separator plates in batteries, because

they can hold more energy than the bulk materials.

Ex: Nano Crystalline Nickel-metal hydride batteries.

9. (ii) Explain top–down and bottom–up approach for nanomaterial preparation with

examples.

1.

Top-down process or Physical or Hard methods.

2. Bottom-up process or chemical or soft methods.

1. Top-down process.

Top-down process involves conversion of bulk materials into smaller particles of nano scalestructure.



118

Methods:

i.Laser ablation

ii.CVD


2. Bottom-up process.

Bottom-up process involves building-up of materials from the bottom by atom by atom or

molecule.

Methods

i.Pricipitation

ii.Thermolysis a. Solvothermolysis b. Hydrothermolysis.

10. (i) Outline medicinal and industrial application of nanomaterials.

1.Medicine

1. Nano drugs

Nano materials are used as nano drugs for the cancer and TB therapy.

2. Laboratories on a chip

Nano technology is used in the production of laboratories on a chip.

3. Nano-medibots

Nanoparticles function as nano-medibots that release anti-cancer drug and treat cancer.

4. Gold- coated nanoshells

It converts light in to heat, enabling the destruction of tumors.

5. Gold nanoparticles as sensors

Gold nanoparticles undergo color change during the transition of nanoparticles.6. Protien analysis

Protein analysis can also be done using nanomaterials.

7. Gold nano shells in imaging

Optical properties of the gold nano shells are utilized for both imaging and therapy.

8. Gold nano shells for blood



119

Gold nano shells are used for blood immune assay.

9. Targeted drug delivery using gold nanoparticles.

It involves slow and selective release of drug to the targeted organs.

10. Repairing work

Nanotechnology is used to partially repair neurological damage. Industries

1. as Catalyst

It depends on the surface area of the material. As nano-particles have an appreciable

fraction of their atom at the surface, its catalytic activity is good.

Ex: Bulk gold is chemically inert but nano has good catalytic property.

2. in water purification

Nano-filtration makes use of nano-porous membranes having pores smaller then 10nm.

Dissolved solids and color producing organic compounds can be filtered vary easily from water.

Magnetic nano-particles are effective in removing heavy metal contamination from waste

water.

3. in fabric industry

The production of smart- clothing is possible by putting a nano coating on the fabric.

i. Embedding of nano particles on fabric make stain repellent.

ii. Socks with embedded silver nano-particles fill all the bacteria and makes it odor free.

4. in automobiles

Disincorporation of small amount of nano-particles in car bumpers can make them

stronger than steel.

ii. Specially designed nano-particles are used as fuel additive to lower consumption in

vehicles.

5. in food industries

The inclusion of nano-particles in food contact materials can be used to generate novel

type of backing material and containers.6. in energy sector

In solar power, nano-technology reduces the cost of photovoltaic cells by 10 to 100 times.

10. (ii) Give an account of electronics and biomaterials applications of nanomaterials.

In electronics

1. Quqntum wire found to have high electrical conductivity.



2. The integrated memory circuits have been found to be effective devices.

3. A transistor, Called NOMFET [Nano particle Organic Memory Field Effect Transistor] is

created by combining gold nanoparticles with organic molecules.

4. Nano wires are used to build transistors without p-n junction

5. Nano radios are other important devices, using carbon nano-tubes.6. MOSFED [Metal Oxide Semi conductor Field Effect Transistor], performs both as switches

and as amplifiers.

Bio-materials [Biology]

1. Nano materials are used as bone cement and bone plates in hospitals.

2. It also used as a material for joint replacements

3. Nano technology is being used to develop miniature video camera attached to a blind person’s

glasses.

4. Nano materials are also used in the manufacture of some components like heart valves and

contact lenses.

5. Nano materials are also used in dental implant and breast implants.

6. CNTS are used as light weight shielding materials for protecting electronic equipments against

electromagnetic radiation.

Engineering Chemistry i Qbwithans annauniv

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