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Transcript of 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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1
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
Hexamethylene diamine Adipic acid
------NH—(--CH2-)6 —NH – CO –( -CH2 -) 4 —CO----
Nylon 6.6 (polymide) n
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
Styrene Syndiotactic polystyrene
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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
Styrene Syndiotactic polystyrene
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
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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
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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
Hexamethylene diamine Adipic acid
------NH—(--CH2-)6 —NH – CO –( -CH2 -) 4 —CO----
Nylon 6.6 (polymide) n
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.
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• 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.
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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 bio- degradable.
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.
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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
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.
Monomer + Initiator + Chain transfer agent Polymer
(Liquid) (Mixed with monomer) (Mixed with monomer)
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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.
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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)
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• 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)
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VALLIAMMAI ENGINEERING COLLEGE
DEPARTMENT OF CHEMISTRY
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
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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
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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
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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)
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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.
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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
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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
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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.
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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
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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,
dG = TdS - PdV + PdV + VdP - TdS - SdTso, dG = VdP - SdT . . . . . . . .(6)
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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
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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),
dG = TdS - PdV + PdV + VdP - TdS - SdTso, dG = VdP - SdT . . . . . . . .(3)
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]
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= [ 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)
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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
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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)
Therefore it can be proved that
- [ ] ]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
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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)
Therefore it can be proved that
[ ] ]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.
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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.
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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
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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).
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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:
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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.
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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.
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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.
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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.
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‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
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.
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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 ’.”
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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
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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
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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
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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).
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The substances which exhibit phosphorescence are called phosphorescent substances. Example : ZnS, CaS, CaSO 4, BaSO4.
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(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
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-------------------------------------------------------------------------------------------------------------------------------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
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VALLIAMMAI ENGINEERING COLLEGE
DEPARTMENT OF CHEMISTRY
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
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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.
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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
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.
(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
1. Used for making resistance coils, heating elements in stoves & electric irons
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:
1. Used for making resistance coils, heating elements in stoves & electric irons
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
Types of Non Heat Treatable Stainless Steel
(a)
Magnetic Type
Composition: Chromium-12-22%; Carbon-0.35%
Properties:
1. Can be forged,rolled & machined2. 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 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
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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.
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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.
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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
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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
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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).
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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.
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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.
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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.
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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)
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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
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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)
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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.
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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.
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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
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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.
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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
iii. Electro-deposition
2. Bottom-up process.
Bottom-up process involves building-up of materials from the bottom by atom by atom or
molecule.
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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
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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.
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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.
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Methods:
i.Laser ablation
ii.CVD
iii. Electro-deposition
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
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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.
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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.