CY6151 ENGINEERING CHEMISTRY-I NOTES

UNIT I POLYMER CHEMISTRY Part – A

1. Define polymer and polymerization.

Polymers are macromolecules formed by repeated linkage of large number of small molecules called monomers. e.g.

nCH2 = CH2 (-CH2 – CH2) - n

Ethylene (Monomer) Polyethylene (Polymer)

Polymerization is the chemical process in which large numbers of monomers combine

together, to form a polymer with or without the elimination of simple molecules like water, HCl,

etc.

2. Define condensation polymerization with suitable example.

It is a reaction between simple polar groups containing monomers that yield polymers with

the elimination of small molecules like H2O, HCl, etc.

n H2N-(CH2)6-- NH2 + n HOOC (CH2)4 COOH

Hexamethylene diamine Adipic acid

-[HN – (CH2)6 – NH – C – (CH2)4 – C] n--

+ 2nH2O || || O O

Nylon 6:6 (Polyamide)

3. Define hetero chain polymer.

If the back bone of the polymer chain is made up of different atoms it is called heterochain

polymer, e.g., polyesters, Nylon 6:6. Polyesters contain carbon and oxygen atoms in the main

chain.

Explain functionality of a monomer with a suitable example.

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

CH2 = CH2 (ethylene) – Functionality is 2 (Two bonding sites are due to the presence of

one double bond in the monomer. Therefore ethylene is a bifunctional monomer).

5. What is degree of polymerization?

The number of repeating units (n) in a polymer chain is known as the degree of polymerization. It is represented by the following relationship,

Degree of polymerization (n) = Molecular weight of the polymeric network

Molecular weight of the repeating unit 6. What is addition polymerization?

It is a reaction that yields polymer which is an exact multiple of the original monomer

molecule. The monomer contains one or more double bonds. This reaction is initiated by the

application of heat, light, pressure, or catalyst, for breaking down the double bonds of monomers.

In addition polymerization there is no elimination of any molecule. (e.g) Polyethylene is produced from ethylene:

nCH2 = CH2 Heat / Pressure n (– CH2 – CH2 -)

Ethylene Catalyst

- (CH2 – CH2) n--

Polyethylene

7. Explain co-polymerization with an example.

It is the joint polymerization in which two (or more) different moments combine to give a

co-polymer. Co-polymerization is mainly carried out to vary the properties of polymers such as

hardness, strength, rigidity, heat resistance, etc.

n CH2 = CH – CH = CH2 + nCH2 = CH

(Butadiene) C6H5 (Styrene)

(-CH2 - CH = CH - CH2 - CH2 – CH) n -

C6H5

(SBR)

8. Why thermosetting plastics cannot be remoulded?

Thermosetting Plastic consists of three dimensional network structure where the polymer

chains are connected by strong covalent bonds by cross linking. They get hardened on heating and

cannot be remoulded.

9. What is 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

10. Define Tg.

Glass transition temperature (Tg) is the temperature at which the amorphous solid state (glassy state) is transformed to the melt state.

11. What is poly dispersity index?

The ratio of the weight – average molecular weight (M w) to that of number – average molecular weight (M n) is known as polydispersity index (PDI).

PDI = M w/ M n 12. How are the polymers classified on the basis of their tacticity? 1. Isotactic polymers

Here functional groups are arranged on the same side of the main chain. 2. Syndiotactic

polymers

Here functional groups are arranged in an alternating fashion. 3. Atactic polymers Here functional groups are arranged randomly.

1. What are the advantages of solution polymerization? Exothermic heat evolved during polymerization is controlled easily. Viscosity built up is negligible. The mixture can be agitated easily.

2. What are the factors affecting Tg?

The value of Tg depends on chain length, extent of cross linking, the barrier which hinders the internal rotation of the chain links. The values of Tg of a given polymer varies with the rate of heating or cooling.

3. Below Tg the polymer is hard and brittle.

Part – B 1. Explain the mechanism of free radical polymerization. Free radical mechanism

Polymerization by free radical mechanism occurs by the following three steps,

namely, initiation, propagation, and termination.

1. Chain Initiation

1. Production of a pair of primary free radicals (R.) by homolytic cleavage of an initiator:

I 2R●

Initiator Free radicals

Heat is used to produce free radicals by homolytic cleavage of weaker O-O bond of the peroxide molecule

(1) CH3COO – OOCCH3 70- 90oC 2CH3COO

● or (R

●)

Acetyl Peroxide free radicals

(2) C6H5COO – OOCC6H5 80 – 95oC 2C6H5COO

● or (R

●)

Benzoyl Peroxide

2. Addition of this free radical to the first monomer to produce chain-initiating species:

H H

| |

R●

+ CH2 = C R – CH2 – C ●

| |

Y Y

(Free radical) (First monomer) (Chain-initiating species)

2. Chain Propagation:

Propagation step consist of the growth of the chain-initiating species by successive addition of large numbers of monomer molecules:

H H H H

| | | |

R-CH2 -C● + n CH2 = C R – (CH2 – C)n – CH2– C

●

| | | |

Y Y Y Y

3. Chain Termination:

Termination of the long living radical polymer may occur either by coupling reaction

or by disproportionation

a) By Coupling or Combination:

Coupling of free radical of one chain end to another free radical chain leads to a macro

molecule.

H H H H

R — CH2 — C ●

+ ●

C — CH2 – R R— CH2 — C — C — CH2 — R

Y Y Y Y

Macromolecule

(Dead Polymer)

b) By Disproportionation:

Transfer of a hydrogen atom of one radical centre to another radical centre forming two

macromolecules, one saturated and another unsaturated.

3. Distinguish thermoplastics and thermosetting plastics with suitable examples.

Thermoplastic resin Thermosetting resins

They are formed by addition polymerization They are formed by condensation

Polymerization

They consist of linear long chain polymers They consist of three-dimensional network

structure

Polymer chains are held together by weak Polymer chains are held by strong

Vander Waals forces covalent bonds

These polymer are usually soluble in organic They are usually insoluble in organic

solvent solvents

They can be remoulded and recycled They cannot be remoulded or recycled

They can be softened on heating and hardened on They get hardened on heating and they

cooling reversibly. cannot be softened on reheating

2. Distinguish between addition and condensation polymerization

Addition Polymerization Condensation Polymerization

The monomer should have at least one The monomer should have at least two

multiple bond, e.g., identical functional groups

CH2 = CH2 (ethylene) CH2 – OH

| (glycol)

H H H H

R –CH2 – C●

+ ●

C—CH2 –R R — C = C + H— CH— CH2— R

Y Y

Unsaturated Saturated

macromolecule macromolecule

CH2 – OH

H2 N(CH2)6 COOH

(6-aminohexanoic acid)

Monomer adds to give a polymer and no Monomers condense to give a polymer and

other by-product is formed. by-products such as H2O, CH3OH, are

formed.

Number of monomeric units decreases Monomers disappear at the early stage of

steadily throughout the reaction as there is reaction as oligomers are formed first and

steady growth of polymer chain. then polymers are produced

High molecular weight polymer is formed Molecular weight of the polymer rises

at once. steadily throughout the reaction.

Usually thermoplastics are produced, Usually thermosetting plastics are

e.g., polyethylene, PVC, etc. produced, e.g., Bakelite, urea-

formaldehyde.

This follows a free radical mechanism. This follows the mechanism of

condensation reaction

Homochain polymer is obtained. Hetero-chain polymer is obtained.

3. Give the preparation, properties, and uses of:

1. Nylon – 6:6 It is obtained by the polymerization of adipic acid with hexamethylene diamine:

H2N (CH2)6 NH2 + HOOC (CH2)4 COOH

Hexamethylenediamine Adipicacid

Condensation polymerization 21

N – (CH2)6 – N – C - (CH2)4 – C -–– -

H H O O

n Polyhexamethylene adipate)

Properties

1. Nylon is tough, strong, and easily mouldable. 2. It has good chemical resistance and low co-efficient of friction. 3. Insoluble in common organic solvents but soluble in phenol and formic acid. 4. It has high thermal stability.

Uses

1. Nylon is used to make textile fibers for use in dresses, socks, carpets, etc. 2. It is used for making filaments for ropes, tooth brush, etc. 3. It is used for making tyres, watch straps. 4. It is a good substitute for metal in gears and bearings.

Epoxy resin: Preparation:

Epoxy resin is prepared by condensing epichlorohydrin with bisphenol.

CH3

n HO C OH + Cl – CH2 – CH – CH2

CH3 O

Bisphenol Alkaline catalyst

- nHCl

CH3

O C O CH2 CH CH2

CH3 OH n

Epoxy resin

22

Properties

1. It has high chemical – resistance to water, acids, alkalis, various solvents and other chemicals. 2. They are flexible, tough and possess very good heat resisting property. 3. Due to the polar nature of the molecules, they possess excellent adhesion quality. Uses: iii) It is used as surface coatings, adhesives like araldite, glass- fiber reinforced Plastics. iv) It is applied over cotton, rayon and bleached fabrics to impart crease-resistance

and shrinkage control. 3. It is used as laminating material in electrical equipments.

4. Write notes on bulk and suspension polymerization techniques.

Bulk polymerization The monomer is taken in 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) (Hydrogen atom donor)

The reaction is slow but becomes fast as the temperature rises. After a known period of time, the whole content is poured into methanol and the polymer is precipitated out. Examples: polystyrene, PVC, PMA.

Advantages:

1. It is simple and requires simple equipments.

2. The polymer has high optical clarity.

3. High-purity polymers are obtained.

Disadvantages:

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

of heat is difficult. 23

2. Polymerization is highly exothermic

Uses: It is used in casting formulations, adhesives, plasticizers and lubricant and additives.

Suspension Polymerization:

It is used only for water insoluble monomers. This polymerization reaction is carried out for heterogeneous systems.

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

it by continuous agitation. The suspension is prevented from coagulation by using suspending agents

like PVA, gelatin, methyl cellulose. The whole content is taken in a flask and heated at constant

temperature with vigorous agitation in a thermostat in nitrogen atmosphere. After eight hours pearl-

like polymers are obtained, which is filtered and washed by water.

Monomer + Initiator + suspending agent polymer

(Suspension (dissolved (suspended in

in water) in monomer) water as beads)

Examples: polystyrene, polystyrene- divinyl benzene. Advantages:

1. Products are highly pure.

2. This method is more economical.

3. Efficient heat control.

4. Isolation of product is easy.

Disadvantages:

1. It is only applicable for water insoluble monomers.

2. Control of particle size is difficult. Uses: 1. it is used as ion exchangers.

2. This technique is used in heterogeneous system. 1. How are solutions and emulsion polymerization carried out? Mention some applications

Solution polymerization:

The monomer, initiator and the chain transfer agents (Hydrogen atom donor) are taken in

a flask and dissolved in an inert solvent. The whole mixture is kept under constant agitation. After

some time, the polymer is precipitated by pouring it in a suitable non-solvent.

Monomer + Initiator + chain transfer agent polymer

(Dissolved in (Dissolved in (In solution)

inert solvent) inert solvent)

Examples: Polyacrylic acid, polyisobutylene and polyacrylonitrile. Advantages:

1. Heat control is easy.

2. Viscosity built up is negligible.

3. The mixture can be easily agitated Disadvantages:

1. It requires solvent recovery and recycling.

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

3. The polymer formed is isolated from the solution either by evaporation of the solvent or by precipitation in a non- solvent.

Applications:

It can be directly used as adhesives and coatings. Emulsion polymerization:

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

soap. Then initiator is added. The whole content is taken in a flask and heated at a constant

temperature with vigorous agitation in a thermostat with nitrogen atmosphere. After 4 to 6 hours,

the pure polymer can be isolated from the emulsion by addition of de-emulsifier like 3% solution

of Al2(SO4)3.

Monomer + Initiator + surfactant polymer (Dissolved in (water soluble) (Emulsion in water) inert solvent)

Advantages: 1. The rate of polymerization is high. 2. Heat can be easily controlled and hence viscosity is low.

25

3. High molecular weight polymer can be obtained.

Disadvantages: 1. It is very difficult to remove entrapped emulsifier. 2. Polymer needs purification. 3. It requires rapid agitation

Applications: 1. It is used in production like water based paints, adhesives, plastics, etc.,

2. It is used for the manufacturing polymers like butadiene and chloroprene.

6. Explain the mechanism of anionic polymerization with an example.

This type of polymerization takes place when electron withdrawing groups like

Cl, CN is present in a monomer. These groups stabilize the carbanion.

Examples: Vinyl chloride, styrene, acrylonitrile.

The catalyst used to initiate the reaction are Lewis bases like NaNH2, LiNH2 etc.,

(i) Initiation

It involves the formation of chain initiating species.

KNH2 K+ + NH2 -

H H H H

NH2 - + C = C H2N – C – C

-

H CN H CN

ii) Propagation

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

large number of monomers and the negative charge simultaneously shifts to the newly added

monomer.

H H H H H H H H

NH2 - C –C

- + n C = C H2N (C – C) n – C – C

- ( Growing chain)

H CN H CN H CN H CN

(iii) Termination

Termination of the growing chain occurs by adding ammonia.

H H H H

NH2

C – C) n

C - C - + H

NH2

(

H CN H CN

H H H H

NH2 (C – C ) n – C – C – H + NH2 -

H CN H CN

7. Explain the mechanism of cationic polymerization with example Cationic Polymerisation:

This type of polymerization takes place when electron donating groups like CH3, C6H5

are present in monomers. The catalyst used to initiate the reaction is Lewis acid like AlCl3, BF3

with co-catalyst water.

Cationic Polymerization occurs in three major steps.

1. Initiation 2. Propagation 3. Termination

Initiation:

It involves the formation of chain initiating species.

AlCl3 + H2O H+. AlCl3OH

-

Catalyst Co-catalyst

H H H H

27

C = C + H+.AlCl3OH

- H– C – C

+. AlCl3OH

-

H C6H5 HC6H5

Chain initiating species

Propagation:

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

Monomers.

H H H H

H – C – C +.

AlCl3OH-+

n C = C

H C6H5 H C6H5

H H H H

H – (C – C)n – C – C +.

AlCl3OH-

H C6H5 H C6H5

Growing Chain

Termination

Termination of the growing chain involves removal of the catalyst by the addition of proton to the counter

ion AlCl3OH-

H H H H HH H H

H – ( C – C )n –C – C +.

AlCl3OH- H– ( C – C ) n – C = C + H

+.AlCl3OH

-

H

C6H5 H C6H5 HC6H5 C6H5

Macromolecule

8. Explain Glass transition temperature (T g) and mention its factors influencing it.

Glass transition temperature is the temperature at which the amorphous solid state is transformed

into glass state. Below the glass transition temperature (T g) the polymer is hard and above which is

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

viscoelastic state. Thus the glass transition temperature is an important property of a polymer.

28

Factors influencing T g

1. Molecular weight: The glass transition temperature of a polymer is influenced by its

molecular weight. As the molecular weight increases the Tg also increases. 2. Flexibility: A free rotational motion of the polymer chain imparts flexibility to the polymer.

As flexibility decreases the Tg of the polymer increases. 3. Crystallinity: Higher the crystallinity, larger is the Tg value of a polymer. In crystalline polymer,

the linear or stereo-regular chains are lined up parallel to each other and are held by strong

cohesive forces. This leads to a high Tg value of the polymer. 4. Plasticisers: This when added to polymer improves the flexibility and reduces brittleness and

Tg.

9. Discuss about Stereo specific polymer (or) Tacticity

The special orientation of groups in a polymer molecule can be in an order (or) disorder with respect to main chain and this is known as Tacticity. The three types of stereo - regular polymers are

Isotactic polymer: If the groups attached to the carbon are arranged on the same side of the main chain, the polymer is called Isotactic polymer.

Syndiotactic polymer: If the groups attached to the carbon chain are arranged in an alternating side of the main chain, the polymer is called Syndiotactic polymer.

Atactic polymer: If the groups attached top the carbon chain are arranged randomly, the polymer is called Atactic polymer. 12. Explain about number average and weight average concept of Molecular weight of a

Polymer.

Assume that there are total n number of molecules in a polymer sample and ni of them have

M1 molecular weight; n2 have M2 molecular weight and so on till we get ni having Mi

molecular weight.

Now we have total number of molecules (n) is given by

n = n1 + n2 ------ + ni = ∑ni

Number of Polymer molecules in fraction 1 = n1

Number fraction of fraction 1 = n1

∑ni

Polymer Molecular weight contribution by fraction 1 = n1M1

∑ni Similarly molecular weight contribution by other fractions will be

= n2M2 , n3M3, ….. niMi

∑ni ∑ni ∑ni Number average molecular weight of the whole polymer will be given by

Similarly, Total weight of the polymer = W = ∑ niMi

Weight of polymer fraction 1 = W1 = n1M1

Weight of polymer fraction of fraction 1 = n1M = niMi

W ∑ni Mi

Polymer Molecular weight contribution by the other fractions will be

n2M22 , n3M3

2 ….. niMi

2

∑niMi ∑niMi ∑niMi The weight average molecular weight for all polymer fractions will be then (Mw)

n1M12 + n2M2

2 +…..+ niMi

2 = ∑ niMi

2

∑niMi ∑niMi ∑niMi ∑niMi

12. Explain the Polydispersity index of a polymer

A simple chemical compound contains molecules of same molecular weight, such

system is called mono dispersed system. For example water contains H2O molecules of

molecular weight of 18. A polymer which contains molecules of different molecular

weight is called as polydispersed system. This is due to the variation in the degree of

polymerization. The polydispersity of a polymer can be well understood by a simple

molecular weight distribution curve (ni Vs Mi).

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

molecular weight (M n) is known as polydispersity index (PDI). The PDI is always more than 1

PDI = M w/ M n

Unit – II CHEMICAL THERMODYNAMICS

PART-A 1. Define System

A system is defined as any specified portion of matter under study which is separated from

the rest of the universe with a bounding surface. A system may consists of one (or) more

ssubstances.

2. Define Surroundings

The rest of the universe which might be in a position to exchange energy and matter with

the system is called surrounding. Ex. Surroundings imply air (or) a water bath in which a

system under examination is immersed.

3. Define Isolated System

A system which cannot exchange both energy and matter with its surroundings is called and isolated system.

Example: Consider a system consisting 100ml of water in contact with its vapour in a

closed vessel. Since the vessel is closed no matter (liquid or vapour) can enter or leave the

vessel. Further if the vessel is also insulated (as shown in the fig) the vessel cannot lose or

gain heat from the surroundings.

4. What is an Isothermal Processes.

Those processes in which the temperature remains fixed are termed as isothermal process.

This is achieved by placing the system in a thermostat i.e. dT=0

5. Define Cyclic Processes

When initial and final states of a system in a process are the same the process is called a

cyclic process. In this process, a system undergoes various processes and returns back to

its initial state. i.e. dE=0, dH=0.Define Intensive and extensive property Intensive

property The property which do not depend on the amount of substance but

depend only on nature of the susbstance present in the system are called intensive

property. Example: Temperature, pressure, concentration and density.

Extensive property

The proeperty which depend on the amount of substance present in the system is called

extensive property.

Example: Mass, volume , energy, Internal energy (E), enthalpy(H), entropy(S), Free Energy(G).

8. Define Internal Energy Change (∆E) Internal Energy Change (∆E) Change in internal energy depend on initial and final states of the system and does not depend on intermediate states. Thus, the internal energy change (∆E) is defined as the difference in the internal energies of initial and final states of the system. Thus, For Chemical reaction

9. Define Enthalpy Enthalpy is defined as “the sum of internal energy and pressure-volume energy of a system under a particular set of conditions”.

i.e. H = E + PV (1)

10. Define Enthalpy change Enthalpy Change (∆H) Like internal energy, enthalpy also cannot be measured, but change in enthalpy (∆H) can be measured from the differences in the enthalpy of initial and final state of the system. Definition The enthalpy change (∆H) is defined as the difference in the enthalpy of initial and final states of the system. Thus,

∆H = H2 – H1 (2) But for chemical reaction

Substituting the values of H2 and H1 from equation(1) to equation(2) We get

= (E2 – E1) ― (P2V2 – P1V1) ∆H = ∆E − P∆V (3)

Equation (3) is the equation for change in enthalpy at constant pressure

11. State the law of conservation of energy. The law of conservation of energy states that “ Energy can neither be created or destroyed but it can be transformed from one form to another form “.

12. Give the mathematical expressions for the First Law of thermodynamics The mathematical expression of First Law of thermodynamics is

∆E = q – w (or) dE = q − P∆V

13. State II law of thermodynamics. Represent by a mathematical equation. Clausius

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

temperature (cold body) to a higher temperature (hot body).

It is mathematically stated as

14. Define Entropy It is a measure of randomness or disorderness in a molecular system

15. Define the term ‘standard free energy’. Illustrate with an example. It is defined as “The free energy change for a process at 25⁰C in which the reactants are

converted into the products in their standard states.”. Thus, The value of ∆G⁰ can be calculated for a reaction from the standard free energies of formation

(∆G⁰f)

16. Calculate the change in entropy accompanying the isothermal expansion of 4 moles of an ideal gas at 300K until its volume has increased three times. Entropy change in an isothermal expansion of an ideal gas Given:

V1 = 1, V2 = 3; n=4; R=1.987cals

=8.733 cals

17. Define Spontaneity It is defined as the tendency of a process to occur naturally is called the spontaneity.

3. How does the entropy of system changes

When a gas is liquefied Ice is melted (a) When a gas is liquefied, entropy decreases (b) When ice is melted, entropy increases

19. What are the conditions for a process to be spontaneous based on the relation

∆G=∆H−T∆S. ∆H = negative & ∆S = positive ∆H = negative & ∆S = negative at low temperature ∆H = negative & ∆S = positive at high

temperature

20. Predict whether the following reaction is spontaneous at 25⁰C C(s) + H2O(l) CO(g) + H2(g)

∆H=31.4kcal/mole and ∆S=32cal/deg at 25⁰C We know that

∆G=∆H−T∆S Given

∆H=31.4kcal; ∆S=32cals (or) 0.032k.cals , T=25 + 273 = 298K

On substituting these values in the above equation

∆G=∆H−T∆S = 31.4 – 298(0.032) = 31.4 – 9.536

∆G=21.9k.cals. Since, ∆G is positive rhe reaction is non-spontaneous.

21. ∆G for a reaction at 300K is −16.0kcal, ∆H for the reaction is −10.0k.cal. What is the entropy change for this reaction. Solution We know that

∆G = ∆H − T∆S Given:

∆H = −10.0 k.cals; ∆G = −16.0 k.cals; T = 300 K

20cals.K-1 2. What is the influence of ∆H and ∆S values in the spontaneity of the reaction?

b) If ∆H is negative and ∆S is positive then the process will be spontaneous as ∆G is negative.

c) When both ∆H and ∆S are positive, at higher temperature, ∆G will be negative and the

process will be spontaneous.

3. What is Clausius inequality?

The cyclic integral of is always less than (or) equal to zero. (or) The amount of heat transferred to the system divided by its temperature is equal to or less than zero is known as clausius inequality.

24. According to Clausius, what is value for reversible heat engine and irreversible heat engine.

(i) For reversible heat engine (ii) For irreversible heat engine

25. What is Helmholtz work function (A)? The part of the internal energy of a system can be used at constant temperature to do useful

work. This part of internal energy which is isothermally available is called work function of the

system A=E – TS

26. What is the significance of free energy? The decrease of free energy (−∆G) of a process at constant temperature and pressure is equal to the useful work obtainable from the system.

27. Write Gibbs-Helmholtz equation? What is its application? Gibbs-Helmholtz equation is

(or) Applications

(i) Enthalpy change (∆H) for the cell reaction can be calculated (ii) Entropy change (∆S) can be calculated (iii) It is used to calculate ∆H from values of free energy change at two different

temperature. 28. Give the relation between (i) ∆H & ∆G, (ii) Emf & ∆G

(ii)−∆G=nFE

29. What is the significance of Gibbs Helmholtz equation? What are its applications. Significance: It relates the free energy change (∆G) to the enthalpy change (∆H) and the rate of change of free energy with temperature at constant pressure. Applications:

1. Calulation of enthalpy change (∆H) for the cell reaction

2. Calculation of Emf of the cell

3. alculation of free energy change (∆G) ∆G=∆H−T∆S

30. Prove that (∂S/∂P)T = −(∂V/∂T)P Gibbs free energy G is defined as

G = H−TS dG = dH−TdS−SdT

But H = E + PV

dH = 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

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

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

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

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

2 Write the four Maxwell relations? 5. What is the significance of Maxwell relations

i Using these relations tedious experimental work can be reduced into simple paper and pencil exercise.

Maxwell relations are also very useful to deduce many other thermodynamics

relations viz. Clapeyran equation, thermodynamic equation of state etc.,

33. What is Van’t Hoff isotherm . Give its relationship? A quantitative relationship between the free energy change and equilibrium constant is known as Van’t Hoff isotherm.

2. Give some important significance of Van’t Hoff equation

The magnitude of the equilibrium constant also depends on whether partial pressures, molar concentrations or mole fractions are used.

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

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

3. Give the reaction between the variation of equilibrium constant with temperature.

4. The value of equilibrium constant for a reaction is found to be 10,000 at 25⁰C.

Calculate ∆G⁰ for the reaction. We know that

∆G⁰=−RTlnK Given:

K=10,000; T=25 + 273 = 298K; R=8.314JK-1mol-1 On substituting these value in the above equation

∆G⁰=−RTlnK ∆G⁰=−8.314 X 298 X 2.303log10,000

=−5705.8 X 4 = −22823.2J/mole

(or) = −22823.2 X 1.987 / 8.314 −5456.6 cals.

2. Give the expressions for: a) integrated form of the Van’t Hoff equation b) Gibbs-Helmholtz equation c) Van’t Hoff isotherm

Integrated form of Van’t Hoff equation

Gibbs-Helmholtz equation

Van’t Hoff Isotherm

PART-B

1. Derive an expression for entropy change of an ideal gas at constant temperature. Entropy change in Isothermal expansion of an ideal gas

In isothermal expansion of an ideal gas carried out reversibly there will be no change in

internal energy i.e. ∆E = 0 and hence, from first law of thermodynamics (i.e. ∆E ═ q + w)

qrev = −w (1) In such a case the workdone in the expansion of n moles of a gas from volume V1 to V2 at constant temperature T is given by

We know

P1V1 = RT; V1 = RT / P1

P2V2 = RT; V2 = RT / P2

Thus, the equation (5) is known as the entropy change of an ideal gas

2. ∆G and ∆H for a reaction at 300K are −16.0k.cal and −10.k.cal respectively.

Calculate the free energy at 330K Solution: The free energy change is expressed as

∆G = ∆H − T∆S ∆G = −16.0k.cals; ∆H = −10.0k.cals ; T = 300 K

The free energy at 330 K ∆G = ∆H − T∆S ∆G = −10.0 – (330 X 0.02)

= −10.0−6.6 = −16.6k.cals

3. Define Gibb’s and Helmholtz free energies. How are these two thermodynamic

functions related to the maximum and useful work obtainable from a system during the given change. Definition Helmholtz Free Energy

It is known that a part of internal energy of a system can be used at constant temperature to

do some useful work. This part of internal energy (E) which is isothermally available is

called “Work Function” (A) of the system. It is mathematically defined as A = E – TS

Significance of the Work Function (A) (or) A related to maximum work obtained The work function is given by

A = E – TS (1) For a small change in a reversible system at constant temperature ∆A = ∆E − T∆S (2)

But, entropy change (∆S) is given as

(or) T∆S = qrev (3)

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

∆A = ∆E – qrev (4) But, according to first law of thermodynamics

∆E = q – w

∆E – q = −w (5)

Comparing equation (4) and (5)

∆A = −w

∆A = −wmax

−∆A = wmax where,

wmax = maximum work

Thus,the decrease of work function (−∆A) of a process at constant temperature gives maximum work obtainable from the system.

Gibb’s Free Energy (G) (OR) Thermodynamic Potential Definition

A part of the total energy of the system is converted into work and the rest is unavailable.

The part of the energy which is converted into useful work is called available energy. The

isothermally available energy present in a system is called free energy (G). It is mathematically defined as

G = H – TS

Significance of Gibbs Free Energy (G) (or) G related to useful work

Free energy (or) Gibb’s free energy (G) is the total available energy present in a reversible

system at constant temperature and pressure as useful work. It is given by G = H – TS (1)

For a small change in a reversible system at constant temperature ∆G = ∆H − T∆S (2)

But, we know that the enthalpy change ∆H is given as

∆H = ∆E + P∆V (3)

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

∆G = ∆E + P∆V − T∆S (4)

But,∆A = ∆E − T∆S (or) ∆E = ∆A + T∆S

Hence, equation (4) may be written as

∆G = ∆A + P∆V (5)

Since, ∆A = −wmax ∆G = −wmax + P∆V −∆G = wmax − P∆V

where

P∆V= work done by expansion (mechanical work) wmax = maximum work that is

obtained from the system (or) wmax−P∆V = network or useful work

Hence, −∆G= wuseful Thus, the decrease of free energy (−∆G) of a process at constant temperature and

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

4. Discuss the criteria for a spontaneous chemical reaction

According to the second law of thermodynamics a process is said to be spontaneous only when

∆Stotal is positive.i.e. entropy of the universe (system + surroundings) increases. This criteria of

spontaneity involves the entropy of surroundings which is difficult to measure. So, we need a

criteria which does not involve the entropy of surroundings. The change in Gibbs free energy

provides such a criteria.

We know that the total entropy change (∆Stotal) is given by ∆Stotal = ∆Ssystem + ∆Ssurrounding (1)

If a reaction is carried out at constant T and P The amount of heat transferred by the system to surroundings is given by

∆Ssystem = (−qp)system The amount of heat taken by the surroundings is equal to

∆Ssurrounding = (−qp)system (qp)surroundings = −(qp)system = −∆Hsystem

Since, the surrounding is a large area, the temperature of the surroundings remains constant, so we have

On substituting equation (2) in equation (1), we get

Multiplying the equation by T

We know ∆G = ∆H − T∆S

T∆S = ∆H − ∆G (4) Substituting equation (4) in equation (3), we get

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

when ∆G = −ve (∆G<0), the process is spontaneous when ∆G = 0 , the process is in equilibrium when ∆G = +ve (∆G>0), the process is non-spontaneous

5. Derive Gibbs Helmholtz equation and discuss its application Gibb’s- Helmholtz

equation (or) Relation between ∆G and ∆H

Consider the following relations

G = H – TS (Gibbs free energy)

H = E + PV (enthalpy)

Therefore, G = E + PV – TS

For infinitesimal change,

dG = dE + PdV + VdP – TdS – SdT (1)

But according to first and second law thermodynamics,

First Law , dE = dq – PdV

Second Law , dq = Tds

Therefore, dE = Tds – PdV (2)

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

dG = Tds – PdV + PdV + VdP – TdS – SdT

dG = VdP – SdT (3)

At constant pressure dP=0 and equation (3) becomes

dG = – SdT (4)

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

This is one form of the Gibb’s – Helmholtz equation For any two states of the system the equation (4) may be written as

dG1 = −S1dT (Initial State)

dG2 = −S2dT (Final state) To get the change

dG2 − dG1 = −S2dT –( −S1dT)

d(G2 – G1) = −(S2 – S1)dT d(∆G) = −∆SdT (7)

At constant pressure the equation (7) becomes But according to definition of free energy

∆G = ∆H − T∆S Comparing equation (8) and (9) , we get

Similarly,

This is another form of Gibbs-Helmholtz equation

Gibb’s-Helmholtz equation relates the free energy changoe (∆G) to the enthalpy

change (∆H) and the rate of change of free energy with temperature at constant

pressure.

Applications of Gibb’s-Helmholtz Equation 1. Caluclation of Enthalpy Change (∆H) for the Cell Reaction (Galvanic

Cell)

If a cell yields nF Coulombs of electricity in a reversible manner, it must be equal to the decrease in the free energy, then

−∆G⁰ = nFE⁰ (1) where

n=No: of electrons involved in the process F= Faraday = 96500 Coulombs E⁰ = Energy in volts

Hence, Gibb’s-Helmholtz equation is written as 2. Calculation of Emf of the cell The equation (2) is divided by nF and it becomes

3. Calculation of Entropy Change (∆S)

∆H and ∆S are related by the equation

∆G=∆H−T∆S (5)

We know that ∆G⁰=−nFE⁰ ∆H can be calculated from the equation (3). Hence, ∆S can be calculated easily from the above equation (5) 4. Gibb’s-Helmholtz equation is applicable for a process occurring at constant pressure. It is

used to calculate ∆H from the values of free energy change at two different temperature. 5. For a reaction at constant volume the equation can be modified as

6. The free energy change, ∆G for process is −138kJ at 30⁰C and −135kJ at 40⁰C.

Calculate the change in enthalpy accompanying the process at

35⁰C. Solution According to Gibbs-Helmholtz equation

Here, ∂(∆G) = ∆G2 − ∆G1 and ∂T = T2 – T1

Given: ∆G1 = −138kJ; ∆G2 = −135kJ

T1 = 30⁰C = 30 + 273 = 303 K

T2 = 40⁰C = 40 + 273 = 313 K

The change in free energy ∆G at 35⁰C may be taken as the average value of ∆G at 30⁰C and 40⁰C Substituting these values in the Gibbs-Helmholtz equation it becomes

= −136.5 – 308(0.3) ∆H = −228.9 kJ

7. Gibbs free energy of a reaction at 300K and 310K are −29.k.cal and −29.5k.cal respectively. Determine its ∆H and ∆S at 300K Solution According to Gibbs’-Helmholtz equation

Here, ∂(∆G) = ∆G2 − ∆G1 and ∂T = T2 – T1 Given:

∆G1 = −29 k.cal (at 300 K) ∆G2 = −29.5 k.cal (at 310 K)

T 1 = 300 K ; T2 = 310 K (i) Calculation of ∆H

∆H = −29− [ 300 X −0.05 ] = −29 + 15 = −14 k.cals.

3. Calculation of ∆S

We know that from Gibb’s-Helmholtz equation

∆S = 0.05k.cals ∆S = 0.05k.cals

8. The free energy change ∆G for a reaction is found to be −3.138 k.cal at 300 K and

for it is −14.39 cal/degree. Find ∆H for the reaction at 300 K

Solution: According to Gibb’s Helmholtz equation

Given:

∆G = −3.138 k.cals or 3138 cals; T = 300 K

On substituting these values in the above equation

= −3138 −300(−14.39) = −3138 + 4317

∆H = 1179 cals (or) 1.179 k.cals

7. Explain the significance of Gibbs Free Energy(G)

Free energy (or) Gibbs free energy (G) is the total available energy present in a reversible system at constant temperature and pressure as useful work. It is given by

G=H−TS (1) For a small change in a reversible system at constant temperature ∆G=∆H−T∆S (2)

But we know that the enthalpy change ∆H is given as

∆H=∆E + P∆V (3)

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

∆G=∆E + P∆V −T∆S (4)

But, ∆A=∆E−T∆S (or) ∆E=∆A + T∆S

Hence, equation (4) may be written as

∆G=∆A + P∆V (5)

Since, ∆A=−wmax ∆G=−wmax + P∆V (or) −∆G=wmax−P∆V

where

P∆V= work done by expansion (mechanical work) wmax = maximum work that is obtained from the system

(or) wmax−P∆V = network or useful work Hence, −∆G= wuseful Thus, the decrease of free energy (−∆G) of a process at constant temperature and

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

10. Derive all the four Maxwell relations. Thermodynamic relations(or) Maxwell Relation

The various expressions connecting internal energy (E), enthalpy(H), Helmholtz free

energy(A) and Gibbs free energy (G) with relevant parameters such as pressure,

Temperature, volume and entropy may be given as (i) dE = TdS – PdV (ii) dH = TdS + VdP (iii) dA = −SdT – PdV (iv) dG = −SdT + VdP

From these expressions the Maxwell relations are obtained as follows.

(i) The combined form of first and second law is

dE = TdS – PdV (1)

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

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

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

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

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

Equation (6) is a Maxwell relation.

(ii) Enthalpy is defined by

H = E + PV dH = dE + PdV + VdP but dE + PdV = TdS

dH = TdS + VdP (7) If P is constant so that dP═0, then equation (7) yields

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

Differentiating equation (8) w.r.to P , at constant S yields Differentiatng equation (9) w.r.to S, at constant P yields It follows from equation (10) and (11) that Equation (12) is also a Maxwell relation.

(iii) Helmholtz free energy (work function A) is given by

A = E−TS dA = dE−TdS−SdT

But the combined first and second law of thermodynamics is TdS = dE + PdV

dE = TdS−PdV

Therefore, dA = −SdT−PdV (13) If V is constant, so that dV=0, then equation (13) yields If T is constant so that dT=0, then equation (13) yields Differentiating equation (14) w.r.to V, at constant T, yields

Differentiating equation (15) w.r.to T, at constat V, yields It follows from equation (16) and (17) that Equation (18) is also Maxwell relation

(iv) Gibbs free energy G is defined as

G = H−TS

dG = dH−TdS−SdT But H = E + PV

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

Thus, dG = −SdT + VdP (19) If P is constant, so that dP=0, then equation (19) yields If T is constant, so that dT=0, then equation (19) yields Differentiating equation (20) w.r.to P at constant T, yields

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

It follows from equation (22) and (23) that

Equation (24) is also Maxwell relation Thus, equation (6),(12),(18),(24) are known as Maxwell Relations.

11. What is meant by Van’t Hoff isotherm? Derive an expression for the

isotherm of a general reaction.

aA + bB cC + dD VAN’T HOFF ISOTHERM

Van’t Hoff isotherm gives a quantitative relation for the free energy change and equilibrium constant accompanying a chemical reaction. It can be derived as follows: Consider a general reaction

aA + bB cC + dD We know that

G = H – TS G = E + PV – TS (Since, H = E + PV)

(or) dG = dE + PdV + VdP –TdS – SdT (1)

But dq = dE + PdV and dS = dq/T

So, the above equation (1) becomes

dG = VdP – SdT (2)

At constant temperature equation (2) becomes

dG)T = V.dP

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

dG = V.dP

On integrating equation (3) becomes

G = G⁰ + RTlnP (4) where G° = integration constant (standard free energy change) Let the free energies of A,B,C and D at their respective pressure PA, PB, PC and PD are GA,

GB, GC and GD respectively. Then the free energy change for the above reaction is given by

= [ cGC + dGD] − [aGA + bGB] (5)

substituting the values of GA, GB, GC and GD in equation (5) from (4), we get,

∆G = [cG⁰C + cRTlnPC + dG⁰D + dRTlnPD ] – [aG⁰A + aRTlnPA + bG⁰B + bRTlnPB ] (or)

We know, at equilibrium, ∆G=0

∆G⁰ + RTlnKeq (7)

The equation (7) may be written as

∆G⁰ = − RTlnKeq (8) From equation (7) and (8),we get (or)

This expression is called Van’t Hoff isotherm.

12. Derive Clapeyron-Clausius equation. Discuss its applications. Consider a system consisting of only 1 mole of a substance existing I two phases A and B. The

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

GA=GB (1) If the temperature of the system may be raised to T + dT, the pressure becomes P + dP and

the free energies become GA + dGA and GB + dGB respectively. Then, the equation one becomes

GA + dGA = GB + dGB (2)

We know that

G=H−TS (Gibss free energy)

H=E+PV (Enthalpy)

G=E+PV−TS (3)

For infinitsemal small change

dG=dE+PdV+VdP−TdS−SdT (4)

But, dE=dq−dW [ First Law of Thermodynamics]

dq=dE+PdV (5) [since, dW=PdV]

We know that for reversible equation

dq/T = dS

dq=TdS (6)

Here, the work done is due to volume change only, so equation (8) may be applied to phase A & B

dGA = VAdP−SAdT (9)

dGB = VBdP−SBdT (10) where,

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

SA and SB are their molar entropies

Since, GA = GB hence, from equation (2)

dGA = dGB (11) Substituting equation (9) and (10) in equation (11)

VAdP−SAdT = VBdP−SBdT (12)

SBdT−SAdT = VBdP− VAdP (13)

dT(SB−SA) = dP(VB−VA) (14)

SB−SA represents the change in entropy when 1 mole of the substance passes from the initial

phase A to the final phase B. It may be denoted as ∆S Hence, equation (15) becomes

Substituting equation (6) in equation (5)

TdS=dE + PdV

dE=TdS−PdV (7) Substituting equation (7) in equation (4)

dG= TdS−PdV +PdV+VdP−TdS−SdT

dG= VdP−SdT (8)

Substituting equation (17) in equation (16) we get, This is the Clapeyron-clausius equation Applications of Clapeyron-clausius equation 1. Calculation of Latent Heat of vapourization If the vapour pressure of a liquid at two temperatures T1 and T2 be P1 and P2 respectively.

The molar heat of vapourisation ∆HV can be calculated. 2. Calculation of boiling point (or) freezing point If the freezing point or boiling point of the liquid at one pressure is known, it is possible to calculate it at another pressure by using Clapeyron-clausius equation. 3. Calculation of vapour pressure at another temperature If latent heat of vapourisation is known, it is possible to calculate the vapour pressure of a liquid at given temperature if the vapour pressure at another temperature is

known. 4.

Molar elevation constant of a solvent can be calculated.

13. Derive an expression for the variation of equilibrium constant of a reaction with

temperature. (OR) (i) By combining Van’t Hoff isotherm and Gibb’s-Helmholtz equation illustrate the

effect of temperature on equilibrium constant. (ii) What are the applications of Van’t Hoff Equation?

Variation of equilibrium constant with temperature (OR) Van’t Hoff Equation (OR) Van’t Hoff Isochore 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 Gibb’s-Helmholtz

equation as given below.

According to the Van’t Hoff isotherm, the standard free energy change (∆G⁰) is related to the equilibrium constant (K) by the following equation

∆G⁰ = − RTlnKP (1) Differentiating equation (1) w.r. to temperature at constant pressure, we get

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

∆G⁰ is substituted for − RTlnKP , we have

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

Equation (5) is introduced in equation (4)

Calculation of molar elevation constant

This equation (7) is known as Van’t Hoff’s eqution. Since, enthalpy change (∆H) does not vary appreciably with change in partial pressure of the reactants or

products, the Van’t Hoff’s equation may be written as

The equation (8) is integrated between T1 and T2 at which the equilibrium constants are K1p and K2

p

respectively and ∆H is constant

This equilibrium constant K2p at temperature T2 can be calculated, if the equilibrium constant K1

p at

temperature T1 is known provided the heat of the reaction (∆H) is known.. Application or Significance of Van’t Hoff Equation 1. Equilibrium constant (K) and ∆H of a reaction can be calculated using Van’t

Hoff equation.

2. A plot of logKp versus 1/T will give a straight line with slope = −∆H⁰/2.303R. From the slope value of ∆H⁰ can be calculated. i.e. ∆H⁰ = −2.303R X slope

3. It is seen that if T2 > T1, the quantity within the square brackets is positive. (Kp)2>(Kp)1, if ∆H is a

positive quantity i.e. for an endothermic reaction the equilibrium constant increases with remperature.

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

concentration or mole fractions are used. 5. Two or more equilibria can be combined in order to get a new equilibrium.

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

14. The vapour pressure of water at 100⁰C is 760mm. What will be the vapour pressure at 95⁰C?.

The heat of vapourisation of water in this temperature range is 41.27kJ per mole. Solution: We know that

Given: P2 = 760mm; T2 = 100 + 273 = 373 K P1 = ? ; T1 = 95 + 273 = 368 K ∆H = 41.27kJ/mole; R=8.314JK-1mol-1 Since, ∆H value is given in kJ, R value should be in kJ

R =( 8.314 / 1000) kJK-1mol-1 = 0.008314kJmol-1

Substituting these values in the above equation = 2155.41 [0.0000364]

2.881 – logP1 = 0.07846

– logP1 = 0.07846 −2.881

logP1 = 2.881 – 0.07846

logP1 = 2.8025

P1 = A.log(2.8025)

P1 = 634.6 mm

15. The equilibrium constant for the reaction N2 2NH 3 at 400⁰C is 1.64 X 10-4 and at 500⁰C is 0.144 X 10-4 atm. Calculate the heat of formation of 1

mole of ammonia from its elements within given range of temperature. Solution: We know that Given:

K1p = 1.64 X 10-4 atm ; K2

p = 0.144 X 10-4 atm T1 = 400 + 273 = 673 K; T2 = 500 + 273 = 773 K R = 1.987 cals; ∆H = ?

Substituting these values in the above equation

−1.055 = ∆H X 4.196 X 10-5

∆H = −25143 cals ∆H = −25.143 k.cals As per the stoichiometric equation the heat of formation of 2 mole of NH3 = −25143 cals Therefore, heat of formation of 1 mole of NH3 = (−25143 / 2)

= −12571 cals (or) −12.571 k.cals

16. he equilibrium constant KP for the reaction, N2 2NH3 is 1.64 X 10-4 atm at 400⁰C. Calculate the equilibrium constant at 500⁰C, if ∆H = −25.140

k.cals or −105185.8 Joules ( R = 1.987 calories or 8.314 JK-1mol-1) [ 1 calorie = 4.184J] Solution We know that Given:

K1p = 1.64 X 10-4 atm ; T1 = 400 + 273 = 673 K

T2 = 500 + 273 = 773 K ; ∆H = −105185.8 J

R = 8.314 JK-1mol-1 ; K2p = ?

Substituting these values in the above equation

= −3.785 −5493.6 [ 0.0000192 ] = −3.785 – 0.1055

K2p = A.log (−3.8905)

Therefore, the equilibrium constant at 500⁰C K2P = 1.287 X 10-4 atm

17. The equilibrium constant for a reaction at 500 and 700⁰C are 1.64 X 10-4 and 0.64 X 10 -4

respectively. Calculate the enthalpy of a reaction assuming it to be a constant over this temperature range.

Solution We know that Given:

K1P = 1.64 X 10 -4 atm ; K2

P = 0.64 X 10-4 atm ; T1 = 500 + 273 = 773 K T2 = 700 + 273 =

973 K; R = 8.314 JK-1mol-1 Substituting these values in the above equation

−0.409 = ∆H X 1.3888 X 10-5 ∆H = −29450 J ∆H = −29.450 kJ

UNIT IV

PHOTOCHEMISTRY AND SPECTROSCOPY PART A 1. Define Photochemistry. Photochemistry is the study of chemical reactions that are caused by absorption of light radiations.

What are the laws of Photochemistry? 4. The Grotthus-Draper law.

5. The Stark- Einstein law of photochemical equivalence.

6. Lamberts law.

7. Beer-Lamberts law.

3. State Grotthus – Draper law. This law states that “when light falls on any substance, only the fraction of incident light which is

absorbed by the substance can bring about a chemical change”. It is also called the principle of photochemical activation.

4. State Stark – Einstein law. This law states that “In the primary photochemical process (first step) each reacting molecule is

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

5. How is energy of Einstein “E” related to wavelength? The energy E absorbed per mole of the reacting substance is given by

`E = NAhν =NAhc/λ

NA =Avagadro number = 6.023 x 1023

/mole

h = Planks constant = 6.625 x 10-34

joules.sec

C = velocity of light = 3 x 108 m/sec

E = 6.023x1023

/mole X 6.626 x10-34

J.sec X 3x108m/sec = 0.1197 J/mol

λ m λ

6. Define Beer - Lamberts law. This law states that “When a beam of monochromatic radiation is passed through a homogenous

absorbing solution, the decrease in the intensity of the radiation dI with the thickness of the absorbing solution dx is directly proportional to the intensity of the incident radiation I as well as the concentration of the solution C.

λ What are the limitations of Beer-Lamberts law?

Beer- Lamberts law is obeyed only if the radiation used is monochromatic.

It is applicable only for dilute solutions.

The temperature of the system should not be allowed to vary to a larger extent.

It is not applied to suspensions. Deviations may contain if the solution contains impurities.

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

λ Define Quantum yield or Quantum Efficiency. Quantum yield (Ф) is defined as “the number of molecules of the substance undergoing

photochemical reaction per quantum of radiation absorbed‟‟. The quantum yield for photochemical reaction is defined as

Ф = No. of molecules reacting in a given time

No. of quanta (photons) of light absorbed in the same time 9. What are the causes of high quantum yield?

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

Formation of intermediate products which acts as catalyst.

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

The active molecules produced by primary absorption may collide with other molecules, thereby

activating them, which in turn further activate other reacting molecules.

10. What are the causes of low quantum yield?

Deactivation of the excited molecules before they form products.

Excited molecules may lose energy by collision with non excited molecules.

Molecules may receive insufficient amount of energy. Primary photochemical reaction may get reversed.

The dissociated fragments may recombine to give reactant which results in low quantum yield.

11. What are primary and secondary photochemical reactions? Primary process – the reacting molecules undergo activation by absorption of light.

A + h ν A* Secondary process – the activated molecules undergo a photochemical change. A* B 12. Define Photo-Sensitization.

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

Atomic Photosensitisers – Mercury, cadmium, Zinc

Molecular Photosensitisers – Benzophenone, Sulphur dioxide, Uranyl sulphate.

13. What is quenching? During Photosensitization the foreign substance (sensitizer), absorbs light and gets excited. When

the excited foreign substance collides with another substance it gets converted in to some other product due to the transfer of its energy to the colliding substance. This process is known as quenching.

14. What is fluorescence? Certain substances (atoms or molecules) when exposed to radiation of short wavelength (high

frequency), emit light of different frequencies compared to those of incident radiations. This process is called fluorescence and stops as soon as the radiation is cut off. Actually decay period is very short i.e.

10-9

to 10-4

sec. The substance which exhibits fluorescence is called fluorescent substance. Examples: - Fluorite (naturally occurring CaF2), Petroleum, Organic dyes like Eosin, fluorescein, Chlorophyll, Quinine sulphate solution, Vapours of Sodium, Iodine, Mercury. 15. What is Phosphorescence?

Many substance (or molecules) when exposed to radiations of short wavelength (high frequency)

continue to emit light for sometime (10-4

to few seconds) even after incident light is cut off. This phenomenon is called phosphorescence and is chiefly caused by Ultraviolet or Visible light. The substance which exhibits phosphorescence is called phosphorescent substance. Examples: - Zinc Sulphide, Alkaline earth sulphides (CaS, BaS, SrS).

16. Define internal conversion and intersystem crossing. Internal conversion: - It is a type of transition which involves the return of the activated

molecule from higher excited state to the lower excited state.

S3 S2 T3 T2 S2 S1 T2 T1

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

collisions. This process occurs in less than about 10-11

seconds. This is also known as non-radiative transitions

Intersystem crossing: - The process in which the energy of the activated molecule is lost through transition between states of different spin multiplicity.

S2 T2

S1 T1 Even though these transitions are forbidden they occur at relatively slow rates.

PART – B 1. Illustrate the Stark –Einstein law of Photochemical Equivalence.

This law is also called as principle of quantum activation. This law states that “In the primary photochemical process each reacting molecule is activated by

one quantum of effective light”. This law implies a 1:1 relationship between the number of reacting molecules and the number of quanta of light absorbed and is applicable only for the primary process.

Primary step: A+ hν A*

Secondary Step: A* B

Overall reaction: A + hν B

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

Suppose ν is the frequency of radiation absorbed then the corresponding quantum of energy absorbed per molecule will be hν. The Energy absorbed per mole of the reacting substance is called one Einstein is given by `E = NAhν

=NAhc/λ

NA = Avagadro number = 6.023X 1023

/mole

h = Planks constant = 6.625X10-34

J sec

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

E = 6.023x1023

X 6.626x10-34

X 3x108 = 0.1196 J/ mol

λ λ (m)

also we know 1 calorie = 4.184 J

E = 0.1196 J/mol = 2.859x10-2

cal/mol

4.184 J/cal x λ λ

2. Derive the expression for Beer – Lamberts law. State its disadvantages. Beer-Lambert‟s law states that when a beam of monochromatic radiation is passed through a

solution of an absorbing substance, the rate of decrease of intensity of radiation (dI) with thickness of the absorbing solution (dx) is proportional to the intensity of incident radiation, I, as well as concentration of the solution, C. It is mathematically represented as - dI/dx = kIC, where k = molar absorption co–efficient. Upon integration with in the limits, we get

I x

dI/I = - kCdx

I0 0

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

A = Cx (Beer- Lamberts law)

Where, ε = k/2.303 = molar absorptivity co–efficient. Hence, the absorbance is directly proportional to molar concentration, C, and thickness (or) path length, x. Application of Beer – Lambert Law Determination of unknown concentration: Let the absorbance and the concentration of standard solution be As and Cs, respectively, As = εCsx ……

(1), and the absorbance and the concentration of unknown solution be Au and Cu, respectively. Au = εCux

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

As = Au

Cs Cu

Cu = Au * Cs

As Thus the concentration of unknown solution is calculated. Limitations:

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

Valid only at constant temperature.

It is not applied to suspensions.

Deviations may occur if the solution contains impurities.

3. Outline the experimental determination of quantum yield.

There are two types of determination to find out the quantum yields of photochemical reactions. Determination of number of moles of the light absorbing substance that react in a given time.

Determination of number of photons of light of required wavelength absorbed by the same

substance in the same time.

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

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

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

An experimental arrangement for the study of rate of a photochemical reaction is shown in the radiation from a suitable source (tungsten filament or mercury vapour lamp) is rendered parallel by a lens. The parallel beam is then passed through a monochromator, which yields a beam of the desired wavelength only. The monochromatic light then enters the reaction cell containing the reacting mixture (made of quartz) which is immersed in a thermostat. Finally, the light transmitted from the cell is made to fall on a detector which measures the intensity of the transmitted radiation. The intensity of radiation is generally measured with the help of actinometer. It contains 0.05 M oxalic acid and 0.01 M uranyl sulphate in water. On exposure to radiation the following reaction takes place.

UO22+

+ hν (UO22+

)*

UO22+

+ COOH CO2 + CO + H2O + UO22+

COOH

The standard solution of uranyl oxalate is exposed to light radiation of definite frequency for a certain

period of time. The extent of reaction can be determined by titrating the remaining oxalic acid with standard

KMnO4 solution. The amount of oxalic acid consumed in the photochemical reaction is a measure of intensity of radiation (i.e. number of photons absorbed).

3. Explain the mechanism of photochemical decomposition of HI. Comment on the reason for low quantum yield.

The photochemical decomposition of HI is found to take place in the wavelength range of 2070Å - 2820Å.The following mechanism has been suggested for the photochemical decomposition of HI. One molecule of HI absorbs one quantum of radiation (primary process)

HI + hν k1 H + I Rate = k1Iabs

H + HI k2 H2 + I Rate = k2 [H] [HI]

I + I k3 I2 Rate = k3 [I]2

The overall rate of decomposition of HI is given by adding the rate equation (i) & (ii) -d[HI] =

k1Iabs + k2 [H] [HI] (1)

dt

Applying steady state approximation to [H] ,we have

d [H] = k1Iabs - k2 [H] [HI] =0

dt

k1Iabs = k2 [H] [HI] (2)

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

-d[HI] = k1Iabs + k1Iabs = 2 k1Iabs

dt

The quantum yield of the reaction is given by

Ф = Rate of disappearance of HI = -d [HI]/ dt = 2 k1Iabs = 2.0

Rate of absorption of light k1Iabs k1Iabs Actually the quantum yield of the reaction falls below 2.0 after sometime. This is due to the fact

that sufficient I2 is produced in step (iii), the following reaction, regenerating, HI is also set up. H + I2 HI + I

4. Explain the mechanism of photo physical process with the help of Jablonski diagram. Spin multiplicity:-Most molecules posses an even number of paired electrons in ground state. The spin multiplicity of the state is given as 2S + 1, where S is the total spins of the electrons When the spins are paired

Then S = s1+ s2 = +1/2 + (-1/2) = 0 Hence 2S + 1 = 2x0 + 1 =1, the spin multiplicity is 1. The molecule is in singlet ground state.

Spin Orientation due to absorption of light

On absorption of a photon of energy hν, one of the paired electrons goes to a higher energy level (i.e. excited state). The spin orientation of the two electrons may be either parallel or antiparallel. If the spins are parallel, then

S = s1+ s2 = +1/2 + (+1/2) = 1 Hence 2S + 1= 2x1 +1 =3, the spin multiplicity is 3. The molecule is in triplet excited state. If the spins are antiparallel, then

S = s1+ s2 = +1/2 + (-1/2) = 0 Hence 2S+ 1= 2x 0+ 1 = 1, the spin multiplicity is 1. The molecule is in singlet excited state. Depending upon the energy hυ of a photon, electron can jump to any higher electronic state and hence we get a series of Singlet excited states S1, S2, S3, Triplet excited state T1, T2, T3… S1, S2, S3… etc is called first singlet excited state, second singlet excited state, and third singlet excited state respectively. T1, T2, T3… etc is called first triplet excited state, second triplet excited state, and third triplet excited state respectively. According to Quantum mechanics the singlet excited state has higher energy than corresponding

triplet excited state. Thus E (S1) > E (T1); E (S2) > E (T2) ; E(S3) > E(T3) Consequences of light absorption: When light photon (hν) is absorbed by a molecule, the electron of the absorbing

molecule may jump from S0 (Singlet ground state) to singlet excited state S1 or S2

or S3.

Depending upon the energy of the light photon absorbed, for each singlet excited state (S1, S2, S3 etc)

there is a corresponding triplet state respectively. (T1, T2, T3etc).The molecule is said to be activated, whether it is in singlet excited state or triplet excited state.

M0 + hν M* M0 = Molecule in the ground state M*=Molecule in the activated state. The activated molecule returns to the ground state by dissipating its energy (hν) through any of the following type of process.

1.Non-radiative transitions:-Such a transition is from the higher excited states (S1, S3 or T2, T3) to the

first excited state (S1 or T1).Since such a transition does not involve the emission of radiation, it is

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

This process occurs in less than about 10-11

s

S3 S1 T3 T1

S2 S1 T2 T1

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

S2 T2

S1 T1

Radiative transition:- These transitions involve the return of the activated molecule from the singlet excited state S1 and the

triplet excited state T1 to the singlet ground state S0. So, such a transition is accompanied by emission of radiation. When a molecule in the S1 state returns to the ground state S0, emission of radiation occurs in about 10

-

8 s and this process is known as fluorescence.

When a molecule in the T1 state returns to the ground state S0, emission of radiation occurs at a rather

slow rate. This process is called phosphorescence. Since the T1 to S1, transition is spectroscopically

forbidden the life time of phosphorescence is much longer (about 10-3

and higher).

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

Examples:- Atomic Photosensitisers – Mercury, cadmium, Zinc

Molecular Photosensitisers – Benzophenone, Sulphur dioxide, Uranyl sulphate. During Photosensitization the foreign substance (sensitizer), absorbs light and gets excited. When the excited foreign substance collides with another substance it gets converted in to some other product due to the transfer of its energy to the colliding substance. This process is known as quenching. Mechanism of photosensitization and Quenching:-

Consider a general donor – acceptor system in which only the donor D i.e. the sensitizer, absorbs the incident photon and the triplet state of the donor is higher than the triplet state of the acceptor A .i.e.

the reactant. Absorption of the photon produces the singlet excited state of the donor,1D which via

intersystem crossing (ISC) gives the triplet state of the donor 3D. This triplet excited state then collides

with the acceptor producing the triplet excited state of the acceptor 3A and the ground state of the donor.

If 3A gives the desired product, the mechanism is called photosensitization.

However if the desired products result from the excited state of the donor (3D), then A is called the

quencher and the process is known as quenching. The reactions for photosensitization and quenching may be represented as below,

D + hν 1D

1D 3D

3D + A D +

3A

3A products (photosensitization)

3D products (quenching) The triplet excited state of the sensitizer or donor

1D must be higher in energy level than the triplet excited

state of the reactant or acceptor (3A) so that the energy available is enough to raise the reactant molecule

to its triplet state.

Examples of photosensitized reactions:-Dissociation of H2 molecule: Irradiation of a mixture of hydrogen gas and mercury vapour with light of wavelength 253.7 nm

brings about dissociation of hydrogen molecule in to hydrogen atom. Hg + hν Hg*

Hg* + hν H2* + Hg

H2* 2H Here mercury acts as photosensitiser. Photosynthesis in plants:-

The most outstanding example for photosensitization is photosynthesis of carbohydrates in plants

from H2O and CO2in which the green colouring matter of plants acts as photosensitiser. CO2 and H2O do not absorb any radiation in the visible region, but chlorophyll absorbs the radiation. It absorbs visible light

in the wavelength range of 400-700 nm. The energy of light absorbed by chlorophyll is transferred to CO2

and H2O which then react to form (CH2O) n (cellulose /sugar).

6. Explain the mechanism of chemiluminescence with examples. Chemiluminescence is defined as the production of light by a chemical reaction and is thus the reverse

of photochemical reaction. In order for a reaction to be chemiluminescent it must furnish sufficient energy to raise at least one of the reactants or intermediates to an electronically excited state so that as it returns to the ground state, it emits energy in the form of radiation. Since the emission occurs at ordinary temperature, the emitted radiation is also known as “cold light”. Examples of chemiluminescence reactions: v) The glow of phosphorus and its oxide, in which the oxide in its electronic excited state emits radiation. vi) Oxidation of 5-mainophthalic hydrazines or cyclic hydrazines by H2O2 emits green light. 3.‟Cold light‟ emission by glow worms. This is an example of bioluminescence due to aerial oxidation of luciferon (a protein) in the presence of enzyme (luciferase).

SPECTROSCOPY

PART A 1) How does molecular spectrum arise?

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

2.What are the differences between atomic spectra and molecular spectra?

Atomic Spectra Molecular Spectra

Interaction of electromagnetic radiation Interaction of electromagnetic radiation with with atoms. molecules.

Line spectrum is obtained Complicated spectrum is obtained

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

element. electronic transition in a molecule.

3. Calculate the number of modes of (IR) vibration for the following molecules.

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

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

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

(c) C2H6 (Non linear molecule) = 3N- 6 = (3 x 12) – 6 = 30 (d) HCl (linear molecule) = 3N- 5 = (3 x 2) – 5 = 1

4. What is Finger Print region? Mention its uses.

The vibration spectral region at 1400 – 700 cm -1

gives very rich and intense absorption bands which is unique for each molecule. This region is termed as Finger Print Region. Uses: It can be used to detect the presence of functional group and also to identify and characterize the molecule just as a finger print can be used to identify a person.

5. What are the differences between Chromophore and Auxochrome?

S.No Chromophore Auxochrome

1. This group is responsible for the It does not impart colour, but when colour of the compound. conjugated with a chromophore colour

is produced.

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

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

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

6. Mention three applications of UV-Visible Spectroscopy. i. Quantitative analysis: By using Beer-Lamberts law, the concentration of unknown solution can be determined. 1. Qualitative analysis: It is used for identification of aromatic compounds and conjugated olefins. 2. Detection of impurities: It is used for detecting impurities in organic compound

7. Name the transition responsible for molecular spectra. The molecular spectrum arises due to the following three transitions. Rotational transitions

Vibrational transitions Electronic transitions.

8. What is the relation between wavelength, frequency and wave number?

1/λ = υ = υ/c 1. What are the factors which affect the absorbance of the photons by a molecule?

The nature of the absorbing molecules

The concentration of the molecules

Path length of radiation 75

10. Define the term – Bathochromic shift, Hypsochromic shift Bathochromic shift: shifting of absorption band towards a longer wavelength. Hypsochromic shift: shifting of absorption band towards a shorter wavelength.

PART B

1. What is principle of UV Spectroscopy? Explain its components and working. Principle: UV – Visible spectra arises from the transition of valence electrons from the ground state to

excited state by absorbing light from UV (100 – 400 nm) or visible region (400 – 750nm). It is a type of

absorption spectra. Components: Radiation source: Hydrogen or Deuterium lamps are radiation source for UV region and tungsten filament lamp for visible region. Monochromators: The monochromators is used to select a particular wavelength. Cells: The cells are made up of quartz glass are used to hold the sample which should fulfill the following conditions. They must be uniform in construction. The material should be inert to solvents. They must be transparent to UV and Visible light. Detectors: The converts the radiation in to current which is directly proportional to the concentration of the solution Eg.. They are Photo multiplier tube or Photocells.

Recorder: The recorder records the signal from the detector and shows a display. Working: The Radiation from the source is allowed to pass through the monochromator. It allow particular wavelength of light to pass through the exit slit. The beam of radiation coming out of the slit is split into two parallel beams. One half of the beam passes through the sample cell and another half passes through reference cell (only solvent). The detectors compare the intensities of the transmitted two beams of light. If the compound absorb light at a particular wavelength, the intensity of the transmitted beam through sample (I) will be

less than that of reference beam (Io). The recorder produces the graph,

which is a plot of wavelength of light Vs absorbance of light. This graph is known as adsorption spectrum. 8. Explain the principle, components and working of IR spectroscopy. Principle:

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

Near Infrared = 12500 – 4000 cm-1

Infrared = 4000 – 667 cm-1

Far Infrared = 667 – 50 cm-1

Components: Radiation source: The main source of radiation is Nichrome wire or Nernst glower which is a filament containing oxides of Zr, Th, Ce held together with a binder. Monochromators: It allows the light of the required wavelength to pass through light of other Sample cell: The cells must be transparent to IR radiation are used to hold the sample. Detector: Detectors are used to convert thermal radiant energy into electrical energy. Recorder: The recorder record the signal coming out from the detector

Working: The radiation emitted by the source is split into two parallel beams. One of the beams passes through the

sample and the other passes through the reference sample. When the two beams of light recombine they produce a signal which is measured by detector. The signal from the detector is recorded by the recorder.

9. Explain the molecular vibration in IR spectrum. Calculate the fundamental modes of vibration for non-linear and linear molecule. There are two kinds of fundamental vibrations in molecule. Stretching vibrations: During stretching, the distance between two atoms decreases or increases, but bond angle remains unaltered. Bending vibrations: During bending, bond angle increases or decreases but bond distance remains unaltered. Types of Stretching and bending vibrations The number of fundamental modes of vibration of a molecule can be calculated as follows. Non-linear

molecule has 3N-6fundamental modes of „vibrations and a linear molecule has 3N-5 fundamental modes of vibrations where N is the number of atoms in a molecule.. Stretching vibrations: It classified into two types (i) Symmetric stretching: The atoms of the molecule are moving in the same direction. (ii) Asymmetric stretching: The atoms of the molecule are moving in opposite direction.

Asymmetric stretching requires more energy than the Symmetric stretching. Bending vibrations: It classified into two types. In-plane bending Out-of plane bending Fundamental modes of vibration for non-linear and linear molecule

1. Methane (CH4) (N=5) It is a non-linear molecule. Hence, 3N-6 = 3x5-6 = 9 Fundamental vibrational modes.

2. Benzene (C6H6) (N=12) It is a non-linear molecule. Hence, 3N-6 = 3x12-6 = 30 Fundamental vibrational modes.

2. Water (H2O) (N=3) It is a non-linear molecule. Hence, 3N-6 = 3x9-6 = 3 Fundamental vibrational

modes. (i) Symmetric stretching (ii) Asymmetric stretching (ii) Bending vibration

υ1 = 3652 cm-1

υ2 = 3756 cm-1

υ1 = 1596 cm-1

4. Carbon dioxide (CO2) (N=3)

It is a linear molecule. Hence, 3N-5 = 3x3-5 = 4 Fundamental vibrational modes.

(i) (ii) (iii) (iv)

Symmetric stretching Asymmetric stretching In-plane Bending Out-plane bending

υ1 = 1340 cm-1

υ2 = 2350 cm-1

υ3 = 666 cm-1

υ4 = 666 cm-1

10. Discuss the applications of UV-Spectroscopy. Qualitative analysis: Many types of compounds can be identified by comparing the UV spectrum of the sample with that of similar compounds available in reference books. Testing the purity of the sample: UV Visible spectroscopy is the best method for detecting impurities in organic compounds because

The bands due to impurities are very intense.

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

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

Determination of calcium in blood serum. Calcium in blood serum in indirectly determined by converting the Ca present in 1 ml of serum as calcium oxalate, and dissolving the calcium oxalate in dilute

sulphuric acid and treated with ceric sulphate solution. The absorption of the excess Ce+4

ion is measured

at 315nm. The amount of Ca in blood serum is indirectly calculated from the amount of Ce+4

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

11. Discuss the applications of IR spectroscopy.

Identification of organic compounds. : IR spectroscopy is very useful in identification of organic compounds. Identification of functional groups: The region from 4000 – 1400 cm

-1 in IR spectrum is useful for

functional group analysis.

Group Frequency range

C=O 1750 – 1600 cm-1

Free OH 3650 – 3300 cm-1

H bonded OH 3300 – 2800 cm-1

-C = C- 2250 – 2100 cm-1

-C = N 2260 cm-1

-NH2 3350 cm-1

-N = O 1600 – 1500 cm-1

-COOH 3300 – 2700 cm-1

Testing the purity of a sample: Presence of an impurity can be detected by their characteristic peaks. Determination of symmetry of a molecule: The symmetry of a molecule whether linear or non-linear can be determined from the IR spectrum. Example: IR spectra of H2O gives 3 peaks at 667 cm

-1, 1330cm

-1, 2349 cm

-1. Since non-linear molecule

should exhibit (3N-6) = 3 peaks, the compound is non linear. Study of hydrogen bonding molecule: Intermolecular and Intramolecular hydrogen bonding can be differentiated by taking a series of spectra of a compound at different concentrations. Intermolecular hydrogen bonding decreases with dilution and the intensity of such peaks will also decrease. Intramolecular hydrogen bonding on the other hand will show no such change. This can be explained by taking the ortho and para nitrophenols. Study of progress of a chemical reaction The rate the reaction can be determined by taking IR spectra at regular intervals of time Example: Oxidation of secondary alcohol to ketone Secondary alcohol gives the absorption band at 3570cm

-1 due to - OH stretching slowly disappears and a

new band appears at 1725cm-1

due to C=O stretching.

12. What are the different types of electronic transactions? Draw the energy level diagram for

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

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

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

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

& S. It occurs due to the transition of non-bonding lone pair of electrons to the antibonding orbitals.This

transition shows a weak band, and occurs in longer wavelength with low intensity. Example aldehyde and

ketone with no double bond shows a band in the range of 270-300nm and if it is existing with the double bond show a band in the range of 300-350nm. σ→σ* transitions:

This type of transitions occurs in the compounds having only single bonds. Energy required for this transition is more and absorption band occurs in the far UV region (120-136 nm) Example: Methane, Ethane n→σ* transitions:

These transitions occur in saturated compounds having lone pair of electrons. This transition occurs along with n→σ* transition .The energy required for this transition is less than σ→σ* so absorption

band occurs at longer wavelength in the near UV region (180-200nm). Example: (CH3)3N for n→σ* transition- at 227nm, for σ→σ* transition at 99nm. π→π* transitions:

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

UNIT IV

PHASE RULE AND ALLOYS

Chemical reactions are of two types

1. Irreversible reaction homogeneous

2. Reversible reaction – it’s of two types

A. homogeneous reversible reaction heterogeneous B.heterogeneous reversible reaction --- its behaviour can be studied by PHASE RULE given by Willard Gibbs (1874). Phase rule

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

Explanation 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.

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

b.Liquid phase

It depends on the number of liquids present and their miscibilities. 8. If two liquids are immiscible, they will form three separate phases two liquid phase and one

vapour phase. For example: benzene-water. 9. If tow liquids are miscible, they will form one liquid phase and one vapour phase. For example:

alcohol – water. C .Solid phase

Every solid constitutes a separate phase

For example:

(i) Water system ------- three phases

(ii) Rhombic sulphur (s) monoclinic sulphur (s) ----- two phase

iii) Sugar solution in water ----- one phase

iv) CuSO4.5H2O(s) CuSO4.3H2O(s) + 2H2O(g) ---- three phases.

2. Component (C)

“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”.

For example:

i) Water system ---- one component ( H2O )

ii) An aqueous system of NaCl --- two component ( NaCl , H2O )

iii) PCl5(s) PCl3 (l) + Cl2 (g) --- two component ,three phases

λ CuSO4.5H2O(s) CuSO4.3H2O(s) + 2H2O(g) ---- three phases,two

component 3. Degree of freedom(F)

“The minimum number of independent variable factors such as temperature, pressure and concentration, which much be fixed in order to define the system completely”.

i) Water system

Ice (s) water (l) vapour (g) F = Non variant (or) zero variant

ii) Ice (s) water (l) F = univariant (one)

iii) For a gaseous mixture of N2 and H2, we must state both the pressure and temperature. Hence,the system is bivariant.

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 APPLICATIONS OF PHASE RULE TO ONE COMPONENT SYSTEM The water system:

Water exists in three possible phases namely solid, liquid and vapour. Hence there

can be three forms of equilibria.

Solid Liquid

Liquid Vapour

Solid Vapour Each of the above equilibrium involves two phases. The phase diagram for the water system is shown in the figure.

This phase diagram contains curves, areas, and triple.

(i)Curve OA

The curve OA is called vaporisation curve, it represents the equilibrium between water and vapour. At any point on the curve the following equilibrium will exist.

Water Water vapour

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

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

This equilibrium (i.e. Line OA) will extend up to the critical temperature (347o C). Beyond the critical temperature the equilibrium will disappear only water vapour will exist. (ii) Curve OB

The curve OB is called sublimation curve of ice, it represents the equilibrium between ice and vapour. At any point on the curve the following equilibrium will exist.

ICE VAPOUR

The degree of freedom of the system is one, i.e. univariant. This is predicted by the phase rule. F = C – P + 2; F = 1-2=2 ; F=1

This equilibrium (line OB) will extend up to the absolute zero (-273o C), where no vapour can be present and only ice will exist. iii) Curve OC

The curve OC is called melting point curve of ice, it represents the equilibrium between the ice and water.

At any point on the curve the following equilibrium will exist. Ice water The curve OC is slightly inclined towards pressure axis. This shows that melting point of ice

decreases with increase of pressure. The degree of freedom of the system is one i.e., univariant. iv) point O (triple point) The three curves OA ,OB ,OC meet at a point “O” ,where three phases namely solid ,liquid and vapour are simultaneously at equilibrium . This point is called triple point, at this point the following equilibrium will exist. Ice water vapour

The degree of freedom of the system is zero i.e., nonvariant.This is predicted by the phase rule. F=C-P+2; F=1-3+2=0

Temperature and pressure at the point “O” are 0.0075 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 vapour

Sometimes water can be cooled below O oC without the formation of ice, this water is called super –cooled water. Super cooled water is unstable and it can be converted in to solid by seeding or by slight disturbance. vi) Areas

Area AOC, BOC, AOB represents water, ice and vapour respectively .The degree of the freedom of the system is two.i.e. Bivariant. This is predicted by the phase rule F=C-P=2; F=1-1+2; F=2

Two component alloy system or multi component equilibria Reduced phase rule or condensed system

The system in which only the solid and liquid are considered and the gas phase is ignored is called a condensed system.since pressure kept constant, the phase rule becomes

F’ = C – P + 1

This equation is called reduced phase rule. Classification of two component system

Based on the solubility and reactive ablity, the two component systems are classified in to three types.

4. Simple eutectic formation - A binary system consisting of two substances, which are completely miscible in the liquid state, but completely immiscible in the solid state, is known as eutectic (easy melt) system. They do not react chemically. Of the different mixtures of the two substances, the mixture having the lowest melting point is known as the eutectic mixture. 5. a) formation of compound with congruent melting point 6. Formation of compound with incongruent melting point

Formation of solid solution

BINARY ALLOY SYSTEM OR THE SIMPLE EUTECTIC SYSTEM The Lead – Sliver system

Since the system is studied at constant pressure,the vapour phase is ignorned and the condensed phase rule is rule is used. F I= C-P+1

The phase diagram of lead –sliver system is shown in the figure It contains lines,areas and the eutectic point.

i) curve AO The curve AO is known as freezing point curve of sliver. Along the curve AO, solid Ag and the melt are in equilibrium. Solid Ag melt According to reduced phase rule

F’=C-P+1

C=2

P=2

F’=1

The system is univariant ii) curve BO

The curve BO is known as freezing point curve of lead . Along the curve BO, solid Pb and the melt are in equilibrium. Solid Pb melt According to reduced phase rule

F’=C-P+1

C=2

P=2

F’=1

The system is univariant. v) 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 rule

F’=C-P+1

C=2

P=3

F’=1

The system is non-variant.

The point “ O “ is called eutectic point or eutectic temperature and is corresponding composition,97.4 % Pb and 2.6 % Ag ,is called eutectic composition.below this point the eutectic compound and the metal solidfy.

vii) Areas The area above the line AOB has a single phase( molten Pb + Ag ). According to reduced phase rule F’=C-P+1 C=2 P=1 F’=2 The system is bi-variant. The area below the line AO ,OB and point “O” have two phases and hence the system is univariant. According to reduced phase rule F’=C-P+1 C=2 P=2 F’=1 The system is uni-variant.

The process of raising the relative proportion of Ag in the alloy is known as pattinson’s process. The Pattinson process was patented in 1833. It depended on well-known material properties;

essentially that lead and silver melt at different temperatures. The equipment consisted of a row of about 8-9 iron pots, which could be heated from below. 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 in one direction, and the remaining metal was then transferred to the next pot in the opposite direction. The process was repeated in the pots successively, and resulted in lead accumulating in the pot at one end and silver in that at the other. The process was economic for lead containing at least 250 grams of silver per ton. Uses of eutectic system

1.suitable alloy composition can be predicted with the help of eutectic systems. 2.eutectic systems are used in preparing solders ,used for joining two metal pieces together.

Melting point It is the temperature at which the solid and liquid phases, having the same composition ,are in

equilibrium. Solid A solid B

Eutectic point It is the temperature at which two solids and a liquid phase are in equilibrium . Solid A + solid B Liquid Triple point It is the temperature at which three phases are in equilibrium. Solid liquid vapour By definition , All the eutectic points are melting points, but all the melting points need not be eutectic points. ll ly , all the eutectic points are triple points ,but all the triple points need not be eutectic points. Uses (or) merits of phase rule

1. It is a convenient method of classifying the equilibrium states in terms of phases ,components and degree of freedom. 2. It helps in deciding whether the given number of substances remain in equilibrium or not. Limitations of phase rule

1.phase rule can be applied for the systems in equilibrium. 2.only three variables like P,T & C are considered ,but not electrical, magnetic and gravitational forces.

ALLOYS

Definition

An alloy is defined as “homogeneous solid solution of two or more different element one of which at least is essentially a metal”. Alloy containing Hg as a constituent element are called amalgams. Properties of alloys

2) Alloy are harder less malleable and possess lower melting point than their component metals

3) Alloys possess low electrical conductivity 4) Alloys resist corrosion and the action of acids

Importance or need of making alloys

3. To increase the hardness of the metal Example Gold and silver are soft metal they are alloyed with copper to make them hard

2. To lower the melting points of the metal Example

Wood metal (an alloy of lead, bismuth, tin and cadmium) melts at 60.5⁰c which is far below the melting points of any of these constituent metals

3. To resist the corrosion of the metal

Example

Pure iron rested but when it is alloyed with carbon chromium (stainless steel) which resists corrosion

4. To modify chemical activity of the metal Example

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

5. To modify the colour of the metal Example

Brass an alloy of copper (red) and size (silver-white) is white colour. 6. To get good casting of metal Example

An alloy of lead with 5% tin and 2% antimony is used fro casting printing type due toits good casting property

Functions (or) effects of alloying elements

Addition 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 steels

Some important alloying element and their functions are given in table

CLASSIFICATION (OR) TYPES OF ALLOYS

ALLOYS

FERROUS ALLOYS NON-FERROUS ALLOYS

(i)Nichrome

(i) Solder

(ii) Alnico

(ii) Brass

(iii)Stainless steel (iii) Bronze

FERROUS ALLOYS OR ALLOY STEELS

Ferrous alloys are the type of steels in which the elements like Al,B,Cr,Co,Cu,Mn are present in sufficient quantities, in addition to carbon & iron.

PROPERTIES

2. High yield point & strength 3. Sufficient formability,ductility & weldability 4. Corrosion & abrasion resistant 5. Less distortion & cracking 6. High temperature strength

IMPORTANT FERROUS ALLOYS (i)NICHROME

Nichrome is an alloy of nickel & chromium

COMPOSITION

Nickel – 60% Chromium – 12% Iron – 26%

Manganese – 2%

PROPERTIES

3. Good resistance to oxidation & heat 4. High melting point & electrical resistance

5. Withstand heat up to 1000-1100⁰C USES

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

2. Used in making parts of boilers,steam lines stills,gas turbines,aero engine valves,retorts,annealing boxes.

(ii)ALNICO

Alnico is an alloy of aluminium-nickel-cobalt . COMPOSITION

Aluminium – 8-12% Nickel – 14-28% Cobalt – 5-35%

PROPERTIES

1. Excellent magnetic properties & high melting point 2. Magnetized to produce strong magnetic fields as high as 1500 gauss

TYPES OF ALNICO ALLOYS Alnico alloys are of two types

1. ISOTROPIC ALNICO

It is effectively magnetized in any direction 2.ANISOTROPIC ALNICO

It possess preffered direction of magnetization. Anisotropic alnico possesses greater magnetic capacity in their preffered

orientation than isotropic alnico. USES

2. Used as permanent magnets in motors,generators,radio speakers microphones,telephone receivers & galvanometers.

(iii)STAINLESS STEELS (or)CORROSIOPN RESISTANT STEELS

These are alloy steels containing chromium together with other elements such as nickel,molybdenum,etc.

Chromium-16% or more Carbon-0.3-1.5%

PROPERTIES

1. Resist corrosion by atmospheric gases & also by other chemicals. 2. Protection against corrosion is due to the formation of dense, non-porous,tough

film of chromium oxide at the metal surface. If the film cracks, it gets automatically healed up by atmospheric oxygen

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 800⁰C 2. Good resistant towards weather & water

4. In making surgical instruments,scissors,blades,etc. 2.HEAT TREATABLE STAINLESS STEEL PROPERTIES

Possess less strength at high temperature Resistant to corrosion

TYPES OF NON HEAT TREATABLE STAINLESS STEEL

(a)MAGNETIC TYPE

COMPOSITION

Chromium-12-22%

Carbon-0.35%

PROPERTIES

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

In making household utensils,sinks,dental & surgical instruments. NON FERROUS ALLOYS

Do not contain iron as one of the main constituent. Main constituents are copper,aluminium,lead,tin,etc.

PROPERTIES

1. Softness & good formability 2. Attractive (or) very good colours 3. Good electrical & magnetic properties 4. Low density & coefficient of friction 5. Corrosion resistance

IMPORTANT NON FERROUS ALLOYS 1. COPPER ALLOYS (BRASS) Brass contains mainly copper & zinc

PROPERTIES

Greater strength, durability & machinability Lower melting points than Cu & Zn Good corrosion resistance & water resistance property

2.BRONZE(COPPER ALLOY) Bronze contains copper & tin

PROPERTIES

Lower melting point Better heat & electrical conducting property Non-oxidizing,corrosion resistance & water resistance property.

3.SOLDERS Solders are low- melting alloys of tin & lead

PROPERTIES

Solder is melted to join metallic surfaces ,especially in the fields of electronic &

plumbing

USES

1. Used in electrical industry 2. Alloy with 50% tin is general-purpose solder 3. For sealing automotive radiator cores. 4. As fuses for fire-extinguishing equipments,boiler plugs,etc.

Heat treatment of alloys (steel) Heat treatment is defined as” the process of heating and cooling of solid steel article under

carefully controlled condition”. During heat treatment certain physical properties are altered without altering its chemical composition

Objectives (or) purpose of heat treatment Heat treatment causes

i. Improvement in magnetic and electrical properties ii. Refinement of grain structure iii. Removal of the imprisoned trapped gases iv. Removal of internal stress v. Improves fatique and corrosion resistance

Types of heat treatment of alloys (steel) 1. Annealing

Annealing 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 machinability ii. 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

1. It improves mashinability by reliving the internal stress or internal strain 2. It increases ductility and shock resistance 3. 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

are

1. Mild steel=840-870⁰c

2. Medium carbon steel=780-840⁰c

3. High carbon steel=760-780⁰c Purpose

1. It increases the ductility and machinability 2. It makes the steel softer, together with an appreciable increases in its toughness

2.Hardening (or) quenching

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

cooling it either in oil or brine water or some other fluid. The faster the rate of cooling harder will be the steel produced. Medium and high carbon steel can be hardened but low carbon steel cannot

hardened Purpose

1. It increases its resistance to wear ability ,to cut other metal and strength . 2. It increases abrasion resistance. 3. Used for making cutting tools. 3. TEMPERING

It is the process of heating the already hardened steel to a temperature lower than its own hardening temperature & then cooling it slowly.

The reheating controls the development of the final properties Thus,

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

400⁰C. (b) For developing better ductility & toughness, reheating temperature should be

within 400-600⁰C. Purpose

1. It removes stress &strains that might have developed during quenching. 2. Increased toughness & ductility.

3. Used for cutting tools like blade,cutters etc. 4. NORMALISING

It is the purpose of heating steel to a definite temperature (above its higher critical temperature) & allowing it to cool gradually in air. Purpose

1. Recovers homogeneity 2. Refines grains. 3. Removes internal stresses 4. Increases toughness 5. Used in engineering works

NOTE: The difference between normalised & annealed steel are

1. A normaled steel will not be as soft as annealed steel. 2. Also normalizing takes much lesser time than annealing.

5.CARBURIZING

The mild steel article is taken in a cast iron box within containing small pieces of charcoal(carbon material).

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

The article is then allowed to cool slowly within the box itself. The outer skin of the article is converted into high carbon steel containing

about 0.8 to 1.2% carbon. Purpose

To produce hard surface on steel article 6.NITRIDING

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

to about 550⁰C. The nitrogen (obtained by the dissociation of ammonia) combines with the

surface of the alloy to form hard nitride. Purpose

To get super-hard surface.

UNIT V

NANO CHEMISTRY

PART-A

1. What are nano particles?

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

2.What are nano materials?

Nanomaterials are the materials having components with size less than 100 nm at least in one dimension.

3. Define nano-wires.

Nano-wire is a material having an aspect ratio ie., length to width ratio greater than 20. They are also referred to as

quantum wires.

4. What is nano –rod?

Nano- rod is a material having an aspect ratio in the range 1 to 20 with short dimension of the material being 10-

100nm.

5. What are nano clusters?

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

6. What are CNTs?

Carbon nanotubes (CNT) are allotropes of carbon with a nanostructure having a length -to-diameter ratio greater

than 1,000,000.

7. Define nanochemistry.

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

scales.

8.Name the various methods of synthesis of nano-material.

Laser ablation

Chemical vapour depostion

Precipitation

Electro- depostion

Thermolysis

9.What is CVD?

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

other gases, to produce a non-volatile solid that deposits automatically on a suitably placed substrate.

10.Mention the application of Nano-wires.

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

It is used to prepare active electronic components such as p-n junction and logic gates.

11.What are the characteristics of Nano-rods?

Nano-rods are two-dimensional materials.

It also exhibits optical and electrical properties.

12. What is magic number?

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

13.Mention some uses of CNTs.

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

It is used in composites, ICs.

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

PART-B

1.Explain the properties of Nano-materials.

(i) Melting points:

They have lower melting points and reduced lattice constants.

(ii) Optical properties:

The change in optical properties is caused by two factors.

(a)The quantum confinement of electrons within the nano-particles increases the energy level spacing.

(b)Surface plasma resonance, which is smaller in size for nano-particles than the wavelength of incident radiation.

(iii) Magnetic properties:

Bulk materials have ferro-magnetic properties. When the particle size is reduced ,they get super –

paramagnetic properties.

(iv)Mechanical properties:

Mechanical properties of polymeric materials can be increased by the addition of nano-

fillers.

Nano-materials are stronger, harder and more wear resistant and corrosion resistant.

(v)Electrical properties:

(a)Electrical conductivity decreased in reduced dimension.

(b)Electrical conductivity increased in micro-structure.

(vi) Chemical properties:

In heat treatment of nano material, diffusion of impurities, structural defects, and dislocation increased.

Increased perfection will have increased chemical properties.

2. Discuss the laser ablation ,CVD and electrodeposition techniques for the synthesis ofnano-particles.

Top Down / Physical / Hard Methods:1. Laser Ablation:

Laser Ablation chamber :

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

laser pulse is referred to as the ablation rate.

Reaction setup:

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

surface. Simultaneously, an inert gas like Argon, Helium is allowed into the reactor to sweep the evaporated particles

from the furnace zone to the colder collector.

The ablated species condense on the substrate placed opposite to the target. The ablation process takes place in a

vacuum chamber, or in the presence of some background ga

2.Chemical Vapour Deposition (CVD):

In this process volatile compound of a material chemically react with other gases, to produce a non-volatile solid

that deposit on a substrate.

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

a) Thermal CVD:

The reaction is activated by high temperature above 900o C . b) Plasma CVD:

The reaction is activated by plasma at temperature between 300-700oC.

(c) Laser CVD:

Pyrolysis occurs when laser thermal energy falls on an absorbing substrate.

(d) Photo-laser CVD:

The chemical reaction is induced by UV radiation, which has sufficient photon energy, to break the chemical bond

in the reactant molecules.

Various steps involved in synthesis of CVD are:

1.Transport of gaseous reactant to the surface.

2.Adsorption of gaseous reactant on the surface.

3.Catalysed reaction occurs on the surface.

4.Product diffuses to the growth site.

5.Nucleation and growth occurs on the growth site.

6.Desorption of reaction products away from the surface.

CVD Reactor:

The CVD reactors are of two types

*Hot-wall CVD

*Cold-wall CVD

Hot wall CVD reactors are usually tubular in form and heating is accomplished by surrounding the reactor with

resistance elements.

In cold- wall CVD reactors, substrates are directly heated inductively by graphite subsectors, while chamber walls

are air or water cooled.

3.ELECTRODEPOSITION:

Template assisted electro deposition is an important technique for synthesizing metallic nanomaterials with

controlled shape and size. Arrays of nano-structured materials with, specific arrangements can be prepared by this

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

electrochemical cell. The cell usually contains a reference electrode, a specially designed cathodes and an anode. The

cathode substrate on which elelctrodeposition of the nanostructures, can be made of either non metallic or metallic

materials. By using the surface of the cathode as a template, various desired nanostructures can be synthesized for specific

applications.

3.Describe the synthesis of nanomaterials by precipitation and thermolysis method.

Bottom - up methods /Chemical /soft methods:

It involves building up of nano materials from the bottom by atom by atom (=0.1 nm) or molecule by molecule or

cluster by cluster. This method is carried out by the following process:

1.Precipitation

2.Thermolysis

a)Solvothermal synthesis

b)Hydrothermal synthesis

1. Precipitation method :

Generally nano particles are synthesized by the precipitation reaction between the reactants in presence of water

soluble inorganic stabilizing agent.

Precipitation of BaSO4 Nano particles:

10 g of Sodium hexa meta phosphate (stabilizing agent) was dissolved in 80 ml of distilled water in 250ml beaker

with constant stirring. Then 10ml of 1M sodium sulphate solution was added followed by 10ml of 1M Barium Nitrate (

Ba (NO3)2 )solution .The resulting solution was stirred for 1 hour. The resulting precipitate was then centrifuged, washed

with distilled water and vacuum dried.

Ba(NO3)2 + Na2SO4 BaSO4 ↓+ 2NaNO3 In the absence of stabilizing agent, Bulk BaSO4 is obtained.

Precipitation by reduction :

Reduction of metal salt to the corresponding metal atoms .These atoms act as nucleation centres leading to

formation of atomic clusters. These clusters are surrounded by stabilizing molecule that prevents the atoms

agglomerating.

2. Thermolysis method:

Thermolysis is characterized by subjecting the metal precursors (usually organometallic compounds in oxidation

state zero) at high temperatures together with a stabilising agent. Nano particles show an increase in size relating to the

temperature size.

This is due to the elimination of stabilizing molecule, generating a greater aggregation of the particles.

Hydrothermal synthesis

a.Hydrothermal synthesis:

It involves crystallization of substances from high temperature aqueous solutions at high vapour pressure.

Hydrothermal synthesis is usually performed below the super critical temperature of water.(3740).

Method:

Hydrothermal synthesis is performed in an apparatus consisting of a steel pressure vessel called autoclave in which a

nutrient is supplied along with water. A gradient of temperature is maintained at the opposite ends of the growth chamber,

so that the hotter end dissolves the nutrient and the cooler end cause seeds to take additional growth.

b. Solvothermal synthesis :

Solvothermal synthesis involves the use of solvent under high temperature (between 100o– 1000oc) and moderate to

high pressure (1 atm to 10,000 atm ) that facilitate the interaction of precursors during synthesis.

Method:

A solvent is mixed with certain metal precursors and the solution mixture is placed in an autoclave kept at

relatively high temperature and pressure in an oven to carry out the crystal growth. The pressure generated in the vessel,

due to the solvent vapour , elevates the boiling point of the solvent.

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

Solvothermal synthesis of Zinc oxide:

Zinc acetate dehydrate is dissolved in 2-propanol at 50oC. Subsequently the solution is cooled to 0oC and NaOH is

added to form ZnO. The solution is then heated to 65oC to allow ZnO growth for some period of time before a capping

agent (1-dodecanethiol) is injected to the suspension to arrest growth. The rod shaped ZnO nano-crystal is obtained.

Uses:

→Much geometry including thin film, bulk powder, and single crystals can be prepared.

→Thermodynamically stable novel materials can also be prepared easily.

4.Write a short note on Nano wires, Nano rods and Nano clusters.

Nano –wires:

Nano-wire is a material having an aspect ratio i.e. length to width ratio greater than 20. Nano –wires are also

referred to as “quantum wires”.

Nano –Wires of metals: Au ,N i,Pt

Nano –Wires of semiconductors: Si, GaN

Nano –Wires of Insulators: SiO2, TiO2

Molecular Nano –Wires: DNA

Characteristics of Nano- Wires :

>Nano- Wires are one – dimensional material.

>Conductivity of a nano-wire is less than that of the corresponding bulk materials.

>It exhibits distinct optical, chemical, thermal and electrical properties due to this large surface area.

>Silicon nano-wires show strong photoluminescence characteristics.

Synthesis of nanowires:

1. Template assisted synthesis:

The templates contain very small cylindrical pores or voids within the host material and the empty spaces are filled

with the chosen material to form nanowires.

Eg. Mesoporous Alumina

2.VLS method:

This method is used for the production of single crystal of semiconductors of elemental Silicon and Germanium.

The mechanism involves a gas phase reaction followed by anisotropic crystal growth (different properties in different

directions). For example the laser ablation and thermal evaporation of a solid target made of pure Si powder mixed with

metals (Fe, Co, Ni) catalyst at 1200o -1400o C followed by condensation on a substrate maintained at 900o-1100o C

facilitates the growth of long Si nanowires with diameters in the range of 20-80 nm. Each wire consists of crystalline Si

core encased by an outer layer of silicon dioxide.

Applications of nanowires:

Nanowires are used to enhance mechanical properties of composites.

Semiconductor nanowires are used as components in making transistors, diodes, logic gates and digital computing.

Nanowires find applications in high density data storage either as magnetic read heads.

Nano Rods:

Nano rods are a material having an aspect ratio in the range 1 to 20 nm.

Characteristics of Nano rods:

It exhibits special optical and electrical properties. Nano rods are two dimensional materials.

Synthesis of Nano rods: Direct chemical synthesis:

Metal atoms are combined with ligands which act as shape control agents which make bonds with different metal

atoms of different strength to get nano rods.

Applications of nanorods:

Nano rods are used in display technology and micromechanical switches.

Nano clusters:

Nano clusters are multi atom particles of size intermediate between molecule and bulk materials. The size of nano

clusters range from 1-10nm .The atoms or molecules in a clusters are bound by any of the forces like covalent, ionic,

Vander waal’s forces. When a gas condenses in to cluster of atoms the no. of atoms in these clusters varies between a few

to hundreds. Clusters of certain size called critical size are more stable than others (200-103 atoms).

The no. of atoms in a cluster of critical size with higher stability is called Magic number. Cluster of transition

metal atoms have chemical, electronic, magnetic properties which vary with number of constituent atoms.

Source of clusters:

Super sonic nozzle source:

The metal is vaporized in an oven and mixed with inert carrier gas at a pressure of several atmospheres at a

temperature of 75-1500K .Metal/ Carrier gas mixture is allowed to pass through the nozzle in to high vacuum which

produces a supersonic beam of the mixture. The carrier gas produces large clusters. In the absence of carrier gas small

clusters are produced.

Gas aggregation source:

Vapours of metal atoms are introduced in to inert gas which is maintained at high pressure and high temperature.

The gas phase is supersaturated with metallic species and aggregates. It acts as a seed and produces a continuous beam of

nano clusters.

5. Discuss the application of Nano materials in various fields.Applications of Nano materials:

Industries

1. As catalyst:

The nano materials have more no. of surface atoms which make them catalytically active. For example, bulk gold

chemically inert white nano gold possess excellent catalytic property.

2. Water purification:

Dissolved salts and colour producing organic compounds can be filtered very easily from water by using nano porous

membrane having pores smaller than 10nm.Magnetic nano particles are used to remove heavy metal contamination from

waste water.

3. In Fabric industry:

Embedding of Nano particles on fabrics make them stain repellent. Socks embedded with silver nano particles

remove the bacteria and makes them odour free.

4. In automobiles:

Fuel consumption in automobiles can be reduced by using specially designed nano particles as fuel additives.

Incorporation of small amount of nano particles in car bumpers make them stronger than steel

5. In food industry:

Nano particles are used to make packing material and containers to store food.

6. In Solar cells:

Absorption of solar radiation in solar cells containing nano particles are higher than the bulk materials.

Medicine :

1. Nano drugs:

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

2. Nano medibots :

Nano particles act as nano medibots which identify and penetrate the cancer cells and destroy them by supplying anti

cancer drugs

3.Gold coated Nano shells:

Gold coated nano shells containing silicon core are administered near the tumours .

When IR light is irradiated on the skin externally, the gold nano shells absorb the light and convert in to heat which

destroys the cancer cells responsible for tumour.

4. Gold nano particles as sensors:

Nano gold when mixed with different type microorganisms show different colours which is used to identify the

harmless and dangerous micro organisms.

5.Protein analysis:

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

concentration can be detected by passing laser beam externally.

6. Gold nano shells for blood Immune assay:

Gold nano shells are used to detect WBC count in blood.

7. Gold nano shells in Imaging:

Gold nano shells is used as a scanning probe which gives the magnified image of cells in a body

8. Targetted drug delivery:

It involves slow and selective release of drug to the targeted organs.

Electronics:

1.Quantum wires have electrical conductivity

2.Nano radios are produced by using CNT

3.MOSFET (Metal Oxide Semiconductor Field Effect Transistor) is used for Amplifying and switching electronic signals.

4.Nano wires are used to build transistors without p-n junction diode.

5.Transistor called NOMFET (Nano particle Organic Memory Field effect transistor) is created by combining Gold nano

particles with organic molecules

Biomaterials:

1.Nano materials are used as bone cement and bone plates in hospitals

2.It is used as material for joint replacement

3.It is used in the manufacture of some components like heart valves, contact lenses, dental implants etc.

6.Write a short note on CNT.

Carbon NanoTubes ( CNTs) :

CNTs are allotropes of carbon in the form of molecular scale tubes made of graphitic carbon with unique

mechanical and electrical / electronic properties with potential applications in electronic, information technology and

medical fields.

Carbon nanotubes were discovered by Sumio Iijima 1991, while preparing C60molecules by the carbon arc

process. They are also known as tubular fullerenes and are cylindrical graphene sheets of sp2 bonded carbon atoms.

These nanotubes are single molecules measuring few nanometers in diameter and several microns in length. Depending

on the growth process, the length of the tubes can be 100 nanometers to several microns (1 micron = 1000 nm), the

diameter varying from 1 to 20 nanometers

Structure of CNTs:

Carbon nanotube is a tabular form of carbon with 1-3 nm diameters and a length of few nm to microns. Each carbon

atom in the carbon nanotubes is linked by the covalent bond. They are two types of nanotubes.

1. Single walled carbon nanotubes (SWCNT) having a diameter of 1 nm. Three types of SWCNTs are as follows:

2. Multi-walled nanotubes (MWNTs) or nested nanotubes consist of multiple layers of graphite rolled to form tube

shape.

MutliwalledNanotube

Synthesis of Carbon Nano Tubes:

Carbon nanotubes are prepared by the following methods.

Pyrolysis of hydrocarbons.

Laser evaporation

Carbon arc method.

Chemical vapour decomposition.

Pyrolysis of Hydrocarbon:

Carbon nano tubes are synthesized by the pyrolysis of hydrocarbons such as acetylene at about 700oC in the

presence of Fe-Silica or Fe-graphite catalyst under inert conditions.

Laser Evaporation:

It involves vapourization of graphite target, containing small amount of cobalt and nickel, by exposing it to

an intense pulsed laser beam at higher temperature (1200oC) in a quartz tube reactor. An inert gas such as argon is

simultaneously allowed to pass into the reactor to sweep the evaporated carbon atoms from the furnace to the colder

collector, on which they condense as carbon nanotubes.

Carbon arc Method:

It is carried out by applying direct current (60-100 A and 20 – 25 V) arc between graphite electrodes of 10-20 µm

diameters. The discharge vapourizes one of the carbon rods and forms a small rod- shaped deposit on the other rod. The

yield depends on the uniformity

of the plasma arc and the temperature. The deposits contain 10 – 50 SWNTs, the average size about 1.4 nm in diameter

and about 10 µm in length.

Chemical Vapour Deposition:

It involves decomposition of vapour of hydrocarbons such as methane, acetylene, ethylene, etc., at high

temperatures (1100oC) in presence of metal nanoparticle catalyst like nickel, cobalt, iron supported on MgO or Al2O3.

Carbon atoms produced by the decomposition condense on a cooler surface of the catalyst.

Properties of CNTs:

(1) Mechanical Properties:

Young’s modulus of carbon nano tubes is 10 times greater than that of steel. CNTs have very structural defects in

their walls, and hence, do not fracture on bending. The tensile strength of CNTs is about about 20 times that of steel.

(2)Electrical Properties:

The electrical properties of CNTs vary between metallic to semiconducting materials.

The very high electrical conductivity of CNT is due to the minimum defects in the structure.

(3)Thermal Conductivity:

The thermal conductivity of CNT is very high due to the vibration of covalent bonds

due to minimum defects in the structure.

Applications of Carbon NanoTubes:

Due to the unusual and unique properties of CNTs they find potential applications in the following field:

Carbon nanotubes play an important role in the battery technology, because some charge carriers can be successfully

stored inside the nanotubes.

Multi-walled CNTs can be used as storage devices to store hydrogen gas in fuel cells.

CNTs are used as catalyst in chemical reactions.

CNTs can be used for drug delivery within the body by placing the drugs within the tubes or by attaching the drug to the

sides of the tubes.

CNTs are used as light weight shielding materials to protect electronic equipments from electromagnetic radiation.