PANIMALAR INSTITUTE OF TECHNOLOGY (Jaisakthi … I.pdf(b) Effect of side group The presence of side...

PANIMALAR INSTITUTE OF TECHNOGY – DEPARTMENT OF CHEMISTRY

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PANIMALAR INSTITUTE OF TECHNOLOGY

(Jaisakthi Educational Trust)

Chennai – 600 123

Engineering Chemistry I (CY6151) - Questions Bank with Answer

(2013 – Regulation)

Common to all branches


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UNIT – I

POLYMER CHEMISTRY

PART A

1. Differentiate between thermoplastic and thermosetting polymers. [AU, Man/Jun, 2016]

S.No. Thermoplastic resins Thermosetting resins

1 They are formed by addition

polymerization

They are formed by condensation

polymerization

2 They consists of linear long chain

polymers

They consists of three dimensional

network structure

3

All the polymer chains are held

together by weak van der waals

forces

All the polymer chains are linked by

strong covalent bonds

4 They are weak, soft and less

brittle They are strong, hard and more brittle

2. How are polymers classified? (any one method) [AU, Nov/Dec, 2015]

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

1. Natural Polymers

2. Synthetic polymers


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3. Write the preparation of nylon 6,6 with the relevant reaction. [AU, Nov/Dec, 2015]

Nylon 6,6 is prepared by solution polymerization by condensing hexamethylene diamine

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

4. Define poly-dispersity index. [AU, Apr/May, 2015]

The ratio of the weight-average molecular weight (Mw) to that of number-average molecular

weight (Mn) is known as poly-dispersity index (PDI).

PDI = Mw / Mn

5. What prediction can you make about the PDI ratio, if the polymer is monodisperse and

polydisperse?

If the polymers is monodisperse then PDI = 1, and if the polymers is polydisperse then PDI is

always greater than 1.

6. What is the functionality of (a) Propylene (b) phenol towards polymerization? [AU, Apr/May,

2015]

The functionality of Propylene is 2

The functionality of Phenol is 1

7. What are thermoplastics? [AU, Nov/Dec, 2014]

Thermoplastics are plastics obtained by addition polymerization. They are straight chain or

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

of attraction. It can be softened on heating and hardened on cooling. They are generally soluble

in organic solvents.

8. What are copolymers? [AU, Nov/Dec, 2014]

It is the joint polymerization in which two or more different monomers combine to give a

polymer.

9. Mention any two uses of nylon 6.6. [AU, May/June, 2014]

a. The majority of the woven fibers are used in the manufacture of tyre cards.

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


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10. What is functionality of polymers? [AU, May/June, 2014]

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

known as the functionality of polymers.

11. Mention any two uses of epoxy resins. [AU, Nov/Dec, 2014]

a. Epoxy resins are used as surface coatings, adhesives like araldite, glass-fiber reinforced

plastics.

b. Epoxy resins are also used as laminating materials in electrical equipments.

12. What is meant by degree of polymerization? [AU, Jan, 2014]

The number of repeating units (n) in a polymer chain is known as degree of

polymerization. It is represented by the following relationship.

Degree of Polymerization =

13. How polymers are classified on the basis of their tacticity? [AUCEG, July, 2013]

Based on the orientation of monomeric units or functional groups in a polymeric molecule,

polymers are classified into 3 types,

1. Isotactic polymer

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

2. Syndiotactic polymer

Here functional groups are arranged in an alternating fashion

3. Atactic polymer

Here functional groups are arranged randomly

14. What are the different types of polymerization?

Polymerization reactions are classified into 3 types,

1. Addition polymerization

2. Condensation polymerization

3. Co- polymerization

15. What is the nature of molecules that undergo cationic and anionic polymerization?

For cationic polymerization the molecule must contain electron withdrawing groups like (-

CH3, -C6H5) for anionic polymerization the molecule must contain electron donating groups like

(-CN, -Cl-)

Molecular weight of the polymer network

Molecular weight of the repeating unit


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PART B

1. Bring out the difference between thermoplastic and thermosetting resins. [AU, Nov/Dec, 2015,

AU, Nov/Dec, 2014, AU, Apr/May, 2015]

S.No. Thermoplastic resins Thermosetting resins

1 They are formed by addition

polymerization

They are formed by condensation

polymerization

2 They consists of linear long chain

polymers

They consists of three dimensional

network structure

3

All the polymer chains are held

together by weak van der waals

forces

All the polymer chains are linked by

strong covalent bonds

4 They are weak, soft and less

brittle They are strong, hard and more brittle

5 They soften on heating and

harden on cooling They do not soften on heating

6 They can be remoulded They cannot by remoulded

7 They have low molecular weights They have high molecular weights

8 They are soluble in organic

solvents They are insoluble in organic solvents

2. Describe the mechanism of free radical polymerization. [May/June, 2014, AU, Nov/Dec, 2015, AU,

Apr/May, 2015]

Free radical polymerization occurs in three major steps.

(i) Initiation

(ii) Propagation

(iii) Termination

i)Initiation

(a) First reaction

First reaction involves production of free radical by hemolytic dissociation of an

initiator (or catalyst) to yield a pair of free radicals (R.)

R - R → 2R.

(b) Second reaction

Second reaction involves addition of this free radical to the first monomer to produce

chain initiating species.


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(ii)Propagation

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

of

monomers

iii)Termination

Termination of the growing chain of the polymer occurs either by coupling reaction

or disproportionation.

(a) Coupling or Combination

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

macromolecule (dead polymer)

(b) Disproportionation

It involves transfer of a hydrogen atom of one radical center to another radical center

forming two macromolecules, one saturated and another unsaturated.

The products of addition polymerization are known as Dead polymers


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3. What is co-polymerization? Describe the different types of copolymerization. [AU, Nov/Dec,

2015] It is the joint polymerization in which two or more different monomers combine to give a

polymer. High molecular weight polymers, obtained by co-polymerization are called

copolymers. Co-polymerization is mainly carried out to vary the properties of polymers such

as hardness, strength, rigidity, heat resistance etc.

Examples

Butadiene and styrene copolymerize to give SBR rubber.

Vinyl chloride and vinyl acetate copolymerize to give polyvinyl chloride-co-vinyl acetate

Examples:

4. Explain the term glass transition temperature. What are the factors influencing Tg? [AU,

Nov/Dec, 2015]

The glass transition temperature is the temperature range in which the thermosetting polymer

changes from hard, rigid or “glassy” state to the rubbery state.

Factors affecting Tg

(a) Molecular weight

Generally, Tg of all polymers increases with increase in molecular weight.


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(b) Effect of side group

The presence of side group hinders the free rotation about the C-C bond of polymer chain

and hence increases the Tg.

(c) Branching and cross-linking

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

(d) Plasticizers

Addition of plasticizers, to the polymer, decreases the value of Tg.

5. Explain the following. [AU, Nov/Dec, 2014, AU, Apr/May, 2015]

a) Emulsion Polymerization

b) Solution Polymerization

a. Emulsion polymerization

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

like potassium persulphate.

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

of soap. The 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 hour, 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)

b. Solution polymerization

In solution polymerization, the monomer, initiator and the chain transfer agents

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

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

in a suitable non-solvent.

Monomer + Initiator + Chain transfer agent → Polymer

(Dissolved in (Dissolved in (In solution)

inert solvent) inert solvent)

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

6. Discuss bulk polymerization technique. Mention any two polymers synthesized by this

technique. [AU, May/June, 2014]

In this technique the monomer is taken in flask as a liquid form and the initiator, chain transfer agents

are dissolved in it. The flask is placed in a thermostat under constant agitation and heated.

Monomer + Initiator + Chain transfer agent → Polymer (Liquid) (Mixed with (Mixed with

monomer) monomer)


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After a known period of time, the whole content is moulded into desired object.

The polymers synthesized under these techniques are: Polystyrene, PVC.

Applications:

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

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

plasticizers and lubricant additives.

7. Write the preparation and uses of epoxide resin. [AU, May/June, 2014]

Preparation:

Epoxide polymers are prepared by condensing epichlorohydrin with bisphenol - A

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

The value of n ranges from 1 to 20. The product obtained is treated with alkali to get an

epoxide.

Uses:

c. Epoxy resins are used as surface coatings, adhesives like araldite, glass-fiber reinforced

plastics.


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

shrinkage control.

e. Epoxy resins are also used as laminating materials in electrical equipments.

f. Moulds made from epoxy resins are employed for the production of components for

aircrafts and automobiles.

8. Discuss the preparation, properties and uses of Nylon 6:6. [AU, Nov/Dec, 2014]

Preparation:

Nylon 6,6 is prepared by solution polymerization by condensing hexamethylene diamine and adipic

acid in toluene at higher temperature in an inert atmosphere.

.

Properties:

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

ii) It is insoluble in common organic solvents, but soluble in formic acid and cresol.

iii) Both fibre and plastic have high tensile strength and dimensional stability

iv) It shows good impact strength due to the large numbers or flexible groups.

Uses:

As fibre:

i) The majority of woven fibre are used in the manufacture of tyre cards

ii) It is used for making socks, dress materials and rope

iii) It is blended with wool to make wool more resistance to abrasion

As plastic:

i) It is used as an engineering plastic

ii) It is used to make various parts due to its high tensile strength, good impact strength,

good dimensional stability and its resistance to abrasion.

iii) It is used in ball bearings, mountings, electrical connections etc.

iv) It is also used to make wheels, bearings, etc., which can run without the use of

lubricants.


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UNIT-II

CHEMICAL THERMODYNAMICS

PART A

1. State Clausius and Kelvin statements of second law of thermodynamics. (A.U.Jan16)

1.Clasius statement: It is impossible to construct a machine which can transfer heat from a cold body to a hot

body, unless some external work is done on the machine.

2. Kelvin statement It is impossible to take heat from a hot body and convert it completely into work by a cyclic

process without transferring a part of heat to a cold body.

2. What happens to the entropy change when (A.U.Dec 2015)

(i) Ice is converted into water at room temperature?

Entropy increases

(ii) I2 vapour is sublimated to I2 solid?

Entropy decreases

3. What happens to entropy of the following? (A.U.May 2015)

(a) A gaseous nitrogen is converted to liquid nitrogen.

Entropy decreases

(b) Solid iodine is sublimed to its vapour.

Entropy increases

4. Calculate the entropy change involved in converting one mole of water at 373 K to its vapour

at the same temperature. (Molar heat of vapourisation of water = 40.66 KJ K-1

mole-1

(A.U.May 2015)

Solution

We know that ∆s for vapourisation, ∆ sv= ∆ Hv / T

Given

∆ Hv =40.66 KJ K-1

mole-1; T=373K

∆ SV = 40.66

373

= 0.10900 KJ K-1

mole-1

(Or)

= 109.008 J K-1

mole-1

5. What is the significance of decrease in free energy? (A.U.Dec 2014)

The decrease of free energy (-∆G) of a process at constant temperature and pressure is equal to the

useful work obtainable from the system.

-∆G = Wuseful work


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6. State second law of thermodynamics. Represent by a mathematical equation.

(A.U.Jun 2014, Dec 2014,2015)

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

higher temperature.

It is mathematically stated as

dS = dqrev (or) ∆S = qrev

T T

7. Define entropy for an ideal gas. (A.U.Jun 2014,Jan 2014)

Entropy is a measure of degree of disorder or randomness in a molecular system. It is also

considered as a measure of unavailable form of energy.

8. Calculate the entropy change for the reversible isothermal expansion of 10 moles of an ideal

gas to 50 times its original volume at 298K. (A.U.Jan 2014)

Solution: The entropy change is given by

V2

∆S = 2.303 nR log

V1

Given

n= 10; V1 = 1lit; V2 = 50lit; R=1.987 cal

∆S = 2.303×10×1.987 log 50

1

∆S = 45.761 × 1.699 = 77.75 cals/K

(or)

∆S = 77.75 × 4.184 = 325.3 J/K

9. Define the term ‘standard free energy’. Illustrate with an example. (A.U.May 2004)

It is defined as, “The free energy change for a process at 25oC in which the reactants are converted

into the products in their standard states.” Thus

∆GO =

∑GO

(products) - ∑GO

(reactants)

The value of ∆GO

can be calculated for a reaction from the standard free energies of formation (∆GO

f )

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

∆G = ∆H – T∆S. (A.U.May 2005)

∆H = negative & ∆S = positive.

∆H = negative & ∆S = negative at low temperature.

∆H = positive & ∆S = positive at high temperature.

11. What is the influence of ∆H and ∆S values in the spontaneity of the reaction?

(A.U.Dec 2001)

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


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

will be spontaneous.

12. What is Clausius inequality?

The cyclic integral of TQ / 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.

0/TQ

13. 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

14. Write Gibbs-Helmholtz equation? What is its application? (A.U.May 2003)

Gibbs-Helmholtz equation is

PP TGTHGorHTGTG /)()(/

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 temperatures.

15. The value of equilibrium constant for a reaction is found to be 10,000 at 25oC. Calculate ∆G

o for

the reaction. (A.U.May 2003)

We know that ∆Go = - RT ln K

Given:

K = 10,000; T=25+273 = 298 K; R = 8.314 J/K/mole

On substituting these value in the above equation

∆Go = -8.314 × 298 × 2.303 log 10,000

= -5705.8 × 4

= - 22823.2 J/mole

(or)

= -22823.2 ×1.9877 = - 5456.6 cals.


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PART B

1. Derive Gibb’s Helmoltz equation and discuss its application .(May-05,Jan -05 ,Dec-12,Jan -14

,May-14,May -15 & June-16)


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2.Van’t Hoff Equation or Van’t Hoff Ishchore or Variation of Equilibrium Constant with

Temperature

Theeffect of temperature on equilibrium constant is quantitatively given by van‟t Hoff equation. It c

an be derived by combinig the Van‟t Hoff isotherm with Gibb‟s Helmholtz equation as given below.

According to the Van‟t Hoff isotherm, the s tandard free energy change (∆Go ) is related to the

equilibrium constant (K) by the following equation,


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The equilibrium constant Kp2at temperature T2 can be calculated, if the equilibrium constant Kp1 at

temperature T1 is known provided the heat of the reaction (∆H) is known.

3.What is meant by VantHoff reaction isotherm? Derive the expression for a reaction isotherm of

general reaction aA +bB cC + dD.

(May-2005,Dec-12,Jan 2014,Dec 2014, May 2015)

Van‟t Hoff isotherm gives a quantitative relationship between the free energy change ( ) and

equilibrium constant (K). it can be derived as follows.


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4. Derive the Claussius- Clapeyron equation. Discuss its applications.

(Dec 2012,Dec 2014)

Consider a system consisting of only 1 mole of substance in two phases A and B. The free energies of

the substance in two phases A and B be GA and GB

Let the temperature and pressure of the system be T and P respectively.

The system is in equilibrium, so there is no change in free energy

GA = GB

If the temperature of the system rised to T + dT and Pressure becomes P + dP

GA + d GA = GB + d GB

G = H – TS

G= E + PV –TS

Diffentiating the above equation.

dG = dE + PdV + Vdp-TdS –SdT

dG = VdP-SdT

dGA = VAdP –SA dT ------------ (a)

dGB = VBdP –SB dT ------------(b)

Where VAand VB are the molar volume of phases A and B respectively.

SA and SB are molar entropies.

Since GA = GB

dGA = dGB

Substituting in equation (a) and (b)

VAdP –SA dT = VBdP –SB dT

(or)

This is Clapeyron equation.

The above equation was modified by Claussius and called Claussius –Clapeyron equation.

Let us consider the following equilibrium

Solid ↔ vapour Vv ›› Vs

Liquid↔ vapour Vv ›› Vl

PV = RT ; V= RT /P

P


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Integrating between the limits. P1 and P2 corresponding to T1 to T2

∫

∫

*

+.

This is the form of Claussius –Clapeyron equation.

5. Derive all the four Maxwell’s relations. (May 2014, May 2015)

Using the above fundamental equations any thermodynamic relationships can be derived.

Note: E, S, H,A,G,T,P,V are all state functions .

(

(

)

(

(

)

(

) (

)

(

)


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(

)

(

)

(

)

(

) (

)

(

)

(

)

(

)

(

)

(

) = (

)

(

)

(

)


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(

)

(

)

From equation 7 and 8 …

(

) (

) The above equations A, B, C and C are

Maxwell relationships.

6. Compare the reversible process with the irreversible process.(May 2014)

U

N

I

T

-

I

I

Reversible Process Irreversible process

Driving force and opposing force differ

by small amount.

Driving force and opposing force differ

by a large amount

It is a slow process It is a rapid process

The work obtained is more The work obtained is less

It is am imaginary process It is a real process

It consists of many steps It has only two steps i.e initial and final

It occurs in both the directions It occurs in only one direction

It can be reversed by changing

thermodynamic variables It can be reversed.


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UNIT-III

PHOTOCHEMISTRY AND SPECTROSCOPY

PART A

1. State Beer - Lamberts law. (A.U. Jan 2016)

This 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 the radiation „dI‟ with the thickness of

the absorbing solution „dx‟ is proportional to the intensity of the incident radiation I as well as the

concentration of the solution C.

It is mathematically represented as

- dI/dx = kIC, where k = molar absorption co–efficient.

2. Mention the essential condition for a molecule to be IR active. (A.U. Jan 2016)

As vibration to be active in IR, the dipole moment of the molecule must change.

3. What are the type of electronic transitions possible in ethylene (CH2=CH2) molecule?

(A.U. Dec 2015)

π→π* transition.

4. State and explain Grotthus- Draper Law. (A.U. Dec 2015)

Grotthus - Draper Law states that only the light which is absorbed by a substance can bring about a

photochemical change.

5. State Stark – Einstein law of photochemistry. (A.U. May 2015)

Stark – Einstein law of photochemical equivalence states that, in a primary photochemical process

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

6. Mention the possible electronic transition that can occur in organic molecules.

(A.U. May 2015)

σ→σ*: n→σ*: n→π*: π→π*

7. What is photochemistry? (A.U. Dec 2014)

Photochemistry is the branch of chemistry deals mainly the chemical applications brought about by

photons (light).

8. What is the purpose of a IR spectrometer? (A.U. Dec 2014)

Identification of functional groups present in the sample by examining the absorption or emission of

IR radiations is the purpose of IR spectrometer.


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9. What is phosphorescence? (A.U. Jun 2014)

The emission of radiation due to the transition from the triplet excited state T1 to the groups state S0

is called is phosphorescence (T1→S0). This transition is slow and forbidden transition.

10. What is meant by wave number? (A.U. Jun 2014)

It is the number of waves per centimeter.

11. What is meant by absorption of radiation? (A.U. Jan 2014)

Absorption of radiation is the process of excitation of molecules from the ground state to excited state.

12. What is Chemiluminescence? (A.U. Jan 2014)

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

known as Chemiluminescence.

13. 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 large extent.

It is not applied to suspensions.

Deviation may occur, if the solution contains impurities.

Deviation also occurs if the solution undergoes polymerization or dissociation.

14. Define Quantum yield or Quantum efficiency.

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

or products formed per quantum of light absorbed”

Ф = Number of molecules formed (or) reacted in a given time / Number of quanta(photons)

absorbed in the same time.

15. Define Photo-sensitization.

The foreign substance, which absorbs the radiation and transfers the absorbed energy to the

reactants, is called a photosensitizer.

This process is called photosensitized reaction (or) photosensitization.

Examples:-

Atomic Photosensitizers – Mercury, cadmium, Zinc.

Molecular Photosensitizers – Benzophenone, Sulphur dioxide.

16. What is quenching?

When the foreign substance in its excited state collides with another substance it gets converted into

some other product due to transfer of its energy to the colliding substance. This process is known as

quenching.


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PART B

1. Derive the expression for the Beer-lambert’s law state its disadvantages. (June-09, Dec-12,

Jan-14, Dec-14, & May-15)

According to this law, “when a beam of beam of monochromatic radiation is passed through a

solution of an absorbing substance, the rate of decrease of intensity of radiation „dI‟ with thickness of

the absorbing solution „dX‟ is proportional to the intensity of incident radiation „I‟ as well as the

concentration of the solution „C‟.”

It is mathematically represented as

-dI/dX = kIC

∫dI/I = -∫kC dX

lnI/I0 = - kCX

log I/I0 = k/2.303 CX

(or)

A = €Cx

€ = k/2.303 (molar absorptivity coefficient)

log I0/I= A (absorbance or Optical density)

Thus, the absorbance (A) is directly proportional to molar concentration (C) and thickness (or) path

length (X).

Limitations (or) Disadvantages of Beer-Lambert’s law

Beer – Lambert‟s law is not obeyed if the radiation used is not monochromatic.

It is applicable only for dilute solutions.

It is not applied to suspensions.

Deviation may occur, if the solution contains impurities.

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

2. What is Chemiluminescence? Bring out the mechanism of Chemiluminescence.

Chemiluminescence is the process of emission of visible light at ordinary temperature as a result of

chemical reaction. Thus, it is the reverse of a photochemical reaction. As the emission occurs at a

ordinary temperature, the emitted radiation is also known as cold light.


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Explanation

In a Chemiluminescence reaction, the energy released during the chemical reaction makes the

product molecule electronically excited. The excited molecule then emits radiation as it returns to the

ground state.

Examples of Chemiluminescence

1. Grignard reagent produces greenish blue light, on oxidation by air.

2. Bioluminescence : Emission of cold light by fire flies (glow worm) due to the aerial oxidation of

luciferon (a protein) in the presence of enzyme (luciferase)

Mechanism of Chemiluminescence:

Mechanism of chemiluminscence can be explained by considering electron transfer Anion – Cation

reaction.

Example:

Interaction between the aromatic anions (Ar-) and (Ar

+)

Ar- + Ar

+ Ar +

1Ar*

1Ar* Ar+hv

The aromatic anion (Ar-) contains two paired electrons in the bonding molecular orbital (BMO) and

one unpaired electron in the antibonding molecular orbital (ABMO). The aromatic Cation (Ar+)

contains only one electron in the bonding molecular orbital (BMO) and the ABMO is empty

When the electron is transferred from the ABMO of the anion (Ar-), to thee ABMO of the singlet

Cation( Ar+) excited state

1Ar* is formed. The excited state can be deactivated by the emission of

photon hv.


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Ar- Ar

Ar+

Ar*

A* Ar + hv

3. Write Short note on Photosensitization.

The foreign substance, which absorbs the radiation and transfers the absorbed energy to the reactants,

is called a photosensitizer.

Mechanism of Photosensitization

The mechanism of photosensitization and quenching can be explained by considering a general

Donor (D), Acceptor (A) system.

In a donor-acceptor system, only the donor D (ie., Sensitizer) absorbs the incident photon and gets

excited from ground stage (S0) to singlet state (S1). Then the donor, via inter system crossing (ISC),

gives the triplet excited state (T1 or D

3). The triplet excited state of the donor is higher than the triplet

state of the acceptor (A). This triplet excited state of the donor then collides with the acceptor

produces the triplet excited state of the acceptor (3A) and returns to the ground state (S0).

Products (quenching)


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

photosensitization.

However, if the products are resulted directly from the excited state of the donor (3D), then A is

called the quencher and the process is called quenching.

4. Explain the Photo physical processes with help of jablonski diagram. (or) With a jablonski

diagram, explain Radiative and non-radiative pathways for an electronic transition

Jablonski diagram for various photo physical processes

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

general types of transitions.

1. Non-radiative transition:

These transitions do not involve the emission of any radiations. This involves the following two

transitions

(a) Internal conversion(IC)

These transition involve the return of the activated molecule from the higher excited states to the first

excited states, ie,

S3S1 (or) T3T1

S2S1 (or) T3T1

The energy of activated molecule is given out in the form of heat through molecular collision. It occurs

in less than about 10-11

second.


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(b). Inter system crossing (ISC)

These transitions involve the return of the activated molecule from the status of different spins. ie,

between singlet and triplet states.

S2T2

S1T1

These transition are forbidden, occurs relatively at slow rates.

2. Radiative transitions:

These transitions involve the return of activated molecules from the singlet excited state S1 and

triplet excited state T1 to the ground state S0.These transitions are accompanied by the emission of

radiations. Thus, Radiative transitions involve the following two radiations.

(a). Fluorescence

The emission of radiation due to the transition from singlet excited state, S1 to ground state S0 .This

transition is allowed transition and occurs in about 10-8

second.

S1S0

(b). Phosphorescence

The emission of radiation due to the transition from the triplet excited state T1 to the ground state

S0.This transition is slow and forbidden transition.

T1S0

5.What are the types of photo physical process? Explain the Fluorescence and

Phosphorescence.(Dec-12,June-13, Jan-14 & May-14, May-15)

Fluorescence

When a beam of light is allowed to fall on substance, it gets excited and emits radiation with in short

time (10-8

sec). Emission stops as soon as incident radiation is cut off. This process is called

fluorescence.

The substance, which shows fluorescence, is called fluorescent substance.

Types of fluorescence

1. Resonance fluorescence

If the excited atom emits radiation of the same frequency, the process is known as resonance

fluorescence.

1. Sensitized fluorescence

If the molecule is excited, due to the transfer of part of excitation energy from the foreign substance,

it emits the radiation of lower frequency, the process is known as sensitized fluorescence.


33

Molecules have even number of electrons in the ground state (So) and are paired. When it absorbs

light, one of the paired electrons moves to the higher energy states (excited states) (S1, S2, S3…..etc.).

From the excited state the molecules returns to the ground state by the following process.

Mechanism of Fluorescence

(a) From the excited state, the molecules return to the first excited state (S1) through Internal

conversion (IC)

S3 S1

S2 S1

(b) From the S1 state the molecules returns to the ground state (So) by emitting radiation, called

fluorescence.

S1 So

Phosphorescence

When a beam of light is allowed to fall on a substance it gets excited and emits radiation, for some

time. Emission continuous for some time even after the incident radiation is cut off. This process is

called phosphorescence.

T1 So

Mechanism of Phosphorescence:

When a molecule absorbs

light, one of its paired electrons

moves from the ground state (So) to

the higher energy states (excites

states ) (S1, S2, S3…..etc.). From the


34

excited state the molecules return to the ground state by the following process.

(a) The molecule crosses from the singlet excited state to the corresponding triplet excited state

through Inter System Crossing (ISC)

S3 T3

S2 T1

S1 T1

(b) From the triplet excited state, the molecule returns to the first triplet excited state through

Internal Conversion (IC).

T3 T1

T2 T1

(c) From the T1 state, the molecule returns to the ground state (So) by emitting radiation, called

phosphorescence.

T1 S0

6. Explain the determination of quantum yield.

Experimental determination of number of molecule reacted

The number of molecule reacted in a given time can be calculated by the usual analytical

technique, used in chemical kinetics

Measurement of rate of reaction

The rate of reaction is measured by the usual methods. Small quantities of the samples are pipetted

out from the reaction mixture from time to time and the concentration of the reactants are continuously

measured by the usual volumetric methods (or) the change in some physical property such as

refractive index (or) absorption (or) optical rotation.


35

Experimental determination of amount of photons absorbed

A photochemical reaction occurs by the absorption of photon of the light by the reactant molecule.

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

Radiation emitted from a source of light (L)(S sunlight, tungsten filament, mercury vapour lamp) is

passed through the lens, which produces parallel beams. The parallel beams are then passed through a

filter (or) monochromator „B‟ which yields a beam of the desired (one) wavelength only. The light

from the monochromator is allowed to enter into the reaction cell „C‟ immersed in a thermostat,

containing the reaction mixture. The part of the light that is not absorbed fall on a detector „X‟ which

measure the intensity of radiation.

Among the so many detectors, the most frequently employed is the chemical actinometer.

The chemical actinometer

A chemical actinometer is a device used to measure the amount of radiation absorbed by the system

in a photochemical reaction. Using chemical actinometer, the rate of a chemical reaction can be

measured easily.

Uranyl oxlate actinometer

Uranyl oxlate actinometer is commonly used chemical actiometer. It consists of 0.05 M oxalic acid

and 0.01 M uranyl sulphate in water.

When it is exposed to radiation, oxalic acid undergoes decomposition to give CO2, CO and H2O.

UO22+

+ h (UO22+

)*

(UO22+

)* + (COOH)2 UO22+

+ CO2 + CO+H2O

The residual concentration of oxalic acid can be found out by titrating with standard KMnO4.

The amount of oxalic acid consumed is a measure of the intensity of radiation.

Calculation of the amount of radiation absorbed

First, the empty cell (or) the cell filled with solvent is exposed = Total incident light

to radiation and reducing is noted

Next, the cell is filled with the reactants and again the reading is noted = Residual energy

Total energy (photon) absorbed by the reacting mixture = Total incident energy - Total Residual

energy.


36

7. Explain the principle and Instrumentation of UV-Visible spectroscopy. (June-13, Jan-14,

June-14, Dec-14 & May-15)

Principle:

UV- Visible spectra arises from the transition of valence electrons within a molecule or ion

from a lower electronic energy level (ground state E0) to higher electronic energy level (excited state

E1). This transition occurs due to the absorption of UV (wavelength 100 – 400nm) or Visible (400 –

750nm) region of the electronic spectrum by a molecule or ion.

The actual amount of energy required depends on the difference in energy between the ground state

and the excited state of the electrons.

Components:

The various components of a visible UV spectrometer are as follows,

Radiation Source

In UV-Visible spectrometers, the most commonly used radiation sources are hydrogen (or)

deuterium lamps.

Requirements of a radiation source

It must be stable and supply continuous radiation

It must be of sufficient intensity

Monochromators

The monochromators is used to disperse the radiation according to the wavelength. The essential

elements of a monochromators are an entrance slit, a dispersing element and an exit slit. The

dispersing element may be a prism or grating or a filter.

Cells (sample cell and reference cell)

The cells, containing samples or references for analysis, should fulfill the following conditions.

They must be uniform in construction

The material of construction should be inert to solvents.

Detectors

There are three common types of detectors used in UV-visible spectrophotometers. They are barrier

layer cell, Photomultiplier tube and Photocell. The detector converts the radiation, falling on which,

into current. The current is directly proportional to the concentration of the solution.

Recording System

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

recorder pen.

Working of UV-Visible

spectrometer:

The radiation from the source is

allowed to pass through the


37

monochromator unit. The monochromator allow a narrow range of wavelength to pass through an exit

slit. The beam of radiation coming out of the monochromators is split into two equal beams. One half

of the beam is directed to pass through a transparent cell containing a solution of the compound to be

analyzed. The other half is directed to pass through an identical cell that contains only the solvent.

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

If the compound absorbs light at a particular wavelength then intensity of the sample beam (I) will be

less than that of the reference beam (Io). The instrument gives an output graph which is a plot of

wavelength (vs) absorbance of the light. This graph is known as an absorption spectrum.

8. Explain the principle and Instrumentation of Infra-red Spectroscopy with neat block

diagram.(June-13,Jan-14, June-14,Dec-15 & June-16)

Principle

IR spectra are produced by the absorption of energy by a molecule in the infra-red region and the

transitions occur between vibrational levels. So, IR spectroscopy is also known as vibrational

spectroscopy.

Instrumentation:

Components of IR Spectrometer

Radiation source

The main sources of IR radiations are:

1. Nichorme Wire

2. Nernst glower, which is filament containing oxides of Zr, Th, Ce held together with a binder.

When they are heated electrically at 1200 to 20000C they glow and produce IR radiation.

• Monochromator

It allows the light of the required wavelength to pass through, but absorbs the light of other

wavelength.

• Sample Cell

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

• Detector

IR detectors generally convert thermal radiant energy into electrical energy. There are so many

detectors, of which the following are important.

1. Photoconductivity cell

2. Thermocouple

3. Pyroelectric detectors

Recorder

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

recorder pen.


38

Working of Infra-red Spectrophotometer

The radiation emitted by the source is split into two identical beams having equal intensity. One of

the beams passes through the sample and the other through the reference sample. The half-beam

travelling through the sample cell, containing the sample becomes less intense

. When the two half beams (one coming from the reference and the other from the sample) recombine,

they produce an oscillating signal, which is measured by the detector. The signal from the detector is

passed to the recording unit and recorded.


39

Unit – IV

PHASE RULE AND ALLOYS

PART A

1. What is condensed (or) reduced phase rule? (A.U. Jan 2014, 2016)

Solid-liquid equilibrium of an alloy has practically no gaseous phase and the effect of pressure is

negligible. Therefore, experiments are conducted under atmospheric pressure. Thus the system in

which only the solid and liquid phase is considered and the gas phase is ignored is called a condensed

system. Since the pressure is kept constant, the phase rule becomes

F‟=C-P+1

This equation is called reduced or condensed phase rule

2. What is meant by component? Give suitable example.(A.U. Dec 2015, Jan 2016)

Component is defined as, „the smallest number of independently variable constituents, by means of

which the composition of each phase can be expressed in the form of a chemical equation‟.

Consider a water system consisting of three phases.

Ice(s) Water (1) Vapour(g)

The chemical composition of all the three phases is H2O, but is in different physical form hence the

number of component is one.

3. What is bronze? Why is it superior to steel?(A.U. Dec 2015)

Bronze is a copper alloy containing copper and tin.

They possess,

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

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

iii) Non-oxidizing, corrosion resistance and water resistance property.

4. What is hardening of steel? Mention its purpose. (A.U. May 2015)

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.

Purpose:

i)It is increase its resistance to wear, ability to cut other metals and strength, but steel becomes extra

brittle.

ii)It increase abrasion-resistance, so that it can be used for making cutting tools.

5. What is the degree of freedom at eutectic point in Lead-Silver system? (A.U. Dec 2014) F

1 = C – P + 1

C = 2, P = 3

Degree of freedom is zero i.e., F1=0


40

6. Mention any two significance of alloy making. (A.U. Dec 2014)

1. To increase the hardness of the metal.

2. To lower the melting points of the metal

3. To resist the corrosion of the metal.

4. To modify chemical activity of the metal.

5. To get good casting of metal.

7. What is triple point? (A.U.Jun 2014)

It is the point at which three phases namely solid, liquid and vapour are is at equilibrium.

8. Write down the composition of Nichorme. (A.U.Dec 2014)

Nichorme is an alloy of nickel and chromium.

Its composition is

Metal Percentage

Nickel

Chromium

Iron

Manganese

60%

12%

26%

2%

9. What are alloys? (A.U.Jan 2014)

An alloy is defined as “homogeneous solid solution of two or more different elements, one of which

is at least essentially a metal”.

10. State phase rule and explain the terms involved. (A.U.Dec 2005, Jun 2007)

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

magnetic forces but are influenced only by pressure, temperature and concentration, then the number

of degree of freedom (F) of the system is related to number of components (C) and number of phases

(P) by the following phase rule equation.

F = C – P + 2

11. Define phase (p) with suitable example (A.U. Dec 2009)

Phase is defined as, “any homogeneous physically distinct and mechanically separable portion of a

system which is separated from other parts of the system by definite bounders”.

Consider a water system consisting of three phases.

Ice(s) Water (1) Vapour (g)

Each phase is physically distinct and homogeneous and there are definite boundaries between phases.

So this forms three phases.

12. What is meant by degree of freedom (F)? (A.U.Dec 2011)

Degree of freedom is defined as, “The minimum number of independed variable factors such as

temperature, pressure and concentration, which must be fixed in order to define the system

completely”.


41

13. How many phases and components are present in the following

system? (A.U.Jun 2010, Dec 2012)

CaCO3 (s) CaO(s) + CO2(g)

It consists of two solid phases and one gaseous phase.

P=3; C=2;

F= C-P +2 = 2–3+2=1.

14. What is eutectic point? (A.U.Jun 2006)

It is the point at which two solid and one liquid phase are in equilibrium.

15. Mention the merits of phase rule. (or) Mention the applications of phase rule.

(i) It is applicable to both physical and chemical equilibria;

(ii) It is convenient method of classifying the equilibrium states in terms of phases, components and

degree of freedom.

(iii) It helps in deciding whether the given number of substances remains in equilibrium or not.

PART B

1.State the phase rule and explain the terms involved in it with examples.(May-09,June-06,Jan-

14 & june-14).

Definition of Phase Rule:

The number of degrees of freedom (F) of the system is related to the number of components (C),

and of phases (P), by the phase rule equation:

F = C – P + 2

Phase

It may be defined as “any homogeneous, physically distinct, and mechanically separable portion of a

system, having a definite boundary”.

Examples:

A system containing liquid water and water vapour,

H2O (l) H2O (g)

Immiscible liquids constitute separate phases, e.g., water and chloroform, P = 2.

Every solid substance constitutes a separate phase

CaCO3(s) CaO(s) + CO2 (g)

P = 3 (two solid and one gas phase)

Component

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

which the composition of each phase can be expressed in the form of a chemical equation”.


42

Example:

(a) Ice water (1) water vapour (g)

The composition of each phase is H2O. Hence, C = 1

Degree of freedom

It is the minimum number of independent variables such as concentration, pressure and temperature

to be fixed in order to define the system completely.

(i) For the equilibrium, water (l) water vapour (g)

Applying the phase rule, F = C – P + 2; since C = 1, and P = 2, F = 1

Either the temperature or the pressure need to be stated for defining the system completely.

Hence, the degree of freedom is one or system is univariate (or) mono variant.

2. Draw a neat one component water system and explain in detail. (or) Explain the various stable

and unstable equilibria in ice-water vapour system and state the effect of decreasing pressure on

each of the equilibria.

(May-2009, May-08, June-13, Jan-14,Dec-14,Dec-2015 & June-16)

One–component system – water system:

The phase diagram of water system is the study of behavior of three different phases of water with

temperature and pressure. It is a one-component system. The phase diagram of the water system

contains:

The curves OA, OB, OC

The triple point O, critical point C.

The areas AOC, AOB, BOC.

Curve OA (Vapourization curve):

It represents the equilibrium between water and vapour. At any point on the curve the following

equilibrium will exist.

water (1) water vapour (g)

This equilibrium will extend up to the critical temperature (374o C). Beyond the critical

temperature the equilibrium will disappear and only water vapour will exist.

Curve OB (Sublimation Curve of Ice):

(a) Ice water vapour (g)

Along the curves OA, OB, and OC, since C = 1, and P = 2, applying the phase rule;

F = C – P + 2 = 1 – 2 + 2 = 1 F = 1 (univariant)

The curves are univariant because, for a given value of pressure, the temperature is automatically fixed

on a curve, and vice versa.


43

Curve OC (Melting point curve of ice):

(a) Ice water (1)

It represents the equilibrium between ice and water. At any point on the curve, the above

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

point of ice decreases with increase of pressure.

The triple point is invariant because, variation of any one of the variables, temperature or

pressure, causes the disappearance of one of the three phases.

Triple point „O‟

Ice Water Water vapour

The curves OA, OB, and OC, meet at the triple point „O‟ where all the three phases, liquid water,

ice, and vapour, are in equilibrium. This occurs at 0.0075ºC and 4.58mm Hg pressure. Since C = 1 and

P = 3,

F = C – P + 2 = 1 – 3 + 2 = 0

F = 0 (invariant)

It represents the equilibrium between ice and water vapour. At any point on the curve, the above

equilibrium will exist. The equilibrium (line OB) will extend up to the absolute zero (-273oC), where

no vapour can be present and only ice will exist.

Metastable equilibrium, OB‟:

The curve, OB‟, is called vapour pressure curve of the super cooled water or metastable

equilibrium where the following equilibrium will exist.

Super cooled water vapour


44

Areas AOC, BOC & AOB:

Areas AOC, BOC & AOB, represent the phases, water, vapour, and ice, respectively. In all the

three areas, since C = 1, and P = 1,F = 1 – 1 + 2 = 2

F = 2 (Bivariant)

Sometimes water can be cooled below 0o C without the formation of ice is called super

cooled water. Super cooled water is unstable; it can be converted in to solid by slight

disturbance. It has higher vapour pressure than ice at a particular temperature. In order to define

the system at any point in the areas, it is essential to specify both pressure and temperature.

3. Discuss the Phase diagram of the silver-lead system and explain the eutectic mixture,

characteristics and uses. (or) State the condensed phase rule and discuss its application to Ag-Pb

system. (Dec-11,Jan-14,May-15, Dec-2015 & June-16)

Systems in which the pressure is kept constant, or those in which the vapour phase is not considered,

are known as condensed or reduced systems. For such systems, the phase rule becomes

F’ = C – P + 1 Reduced phase rule

The system is considered as reduced system.

F’ = C – P + 1,.

1. Curves:

Curve AO: Freezing point curve of Ag.

It shows the effect on freezing point of Ag on addition of lead in small quantities. A is the melting

point of pure Ag (961oC), where exists as solid silver is in equilibrium with liquid Ag

(C = 1).


45

The curve OA indicates that the melting point of Ag falls gradually as the addition of Pb increases,

till the lowest point O (303oC) is reached. At point O, the solution gets saturated with respect to lead.

Along the curve AO, solid Ag and liquid melt are in equilibrium

Ag(s) Ag-Pb melt (l)

(P = 2, C = 2), and hence, according to reduced phase rule equation: F‟ = 2 – 2 + 1 = 1 (monovariant).

Thus, either composition or temperature is needed to define the system along the curve AO.

Curve BO: freezing point curve of Pb.

BO represents the effect on freezing point of Pb on gradual addition of small amounts of Ag to it.

Point B is the melting point of pure lead (327oC). Along BO, the melting point gradually falls on the

addition of Ag, till lowest point O is reached. At O, the solution gets saturated with respect to Ag. As

seen above, along BO, F‟ = 1.

Eutectic Point

The curve AO and BO meet at O, which is called the eutectic point. Here three phases, solid Ag,

solid Pb, and liquid melt, are in equilibrium. Applying the reduced phase rule, (P – 3; C = 2),

F‟ = C – P + 1 = 2 – 3 + 1 = 0 (invariant)

Both variables temperature (303oC) and composition (97.4% Pb, 2.6% Ag: eutectic

composition) are fixed. If the temperature is raised above the eutectic temperature, the solid phases Ag

and Pb.

Disappear. Lowest melting point for the Pb-Ag system is the eutectic temperature.

Area AOB:

This region represents the single phase system, the solution of molten Ag and Pb

(P = 1; C = 2). Applying the reduced phase rule;

F‟ = C – P + 1 = 2 – 1 + 1 = 2 (Bivariant)

To define a system in this area, both temperature and composition are to be specified.

Extraction of silver from Argentiferrous leads (Pattinson”s process)

-Desilverisation of Argentiferrous lead is called Pattinson‟s process.

The Argentiferrous lead contains small amount of silver (0.1% of Ag). It is melted at a temperature

above the melting point of pure lead (327oC). Let the point p represent molten lead on the diagram. It

is then allowed to cool, the temperature of the melt falls along the dashed line pq. When the point q on

the curve BO is reached, solid pure lead begins to separate.


46

On further cooling, more of pure lead separates along the curve qO until the eutectic point O is

reached. Lead is continuously removed by means of ladles and the percentage of silver in the melt goes

on increasing up to 2.6%. This process is called Desilverisation of lead or Pattinson‟s process.

Uses of eutectic compounds:

Since eutectic mixtures are low-melting, they find use in safety devices, e.g., as plugs in fire

sprinklers, and pressure cookers.

Pb-Sn solders used in soldering electrical and electronic components.

Wood‟s metal fusion temperature 70ºC.

1. Draw a neat Zinc - Magnesium system and explain in detail.(or) with neat phase diagram explain

the salient features of Zn - Mg alloy system. (Dec-11,Jan-14,Dec-14)

This system is a two component system with a compound formation with congruent melting

point.

The Zn-Mg system consists of two simple eutectic diagrams joined together at point C. The

diagram to the left represents the eutectic system AE1 C, while that on the right the other eutectic

system BE2C.

a) Left side of the phase diagram:

Curve AE1:

The curve AE1AE1 is freezing point curve of Zn. Point A is the melting point of pure Zn (4200C).

the curve AE1 shows the melting point depression of Zn by the successive addition of Mg. Along

this curve AE1, solid Zn and the melt are in equilibrium.

Solid Zn Melt (l)


47

Point E1:

The point E1 is the eutectic point, where, three phases (solid Zn, solid MgZn2 and their melt) are

equilibrium. The temperature at this point is 3800C.

Solid Zn + Solid MgZn2 Melt (l)

b) Right side of the phase diagram:

Curve CE2:

The curve CE2 is freezing point curve of Mg. Point C is the melting point of pure Mg (6500C). The

curve CE2 shows the melting point depression of Mg by the successive addition of Zn. Along this curve

CE2, solid Mg and the melt are in equilibrium.

Solid Mg Melt (l)

Point Solid Zn Melt (l)

Point Solid Zn Melt (l)

Point E2 (Eutectic Point)

The point E2 is the eutectic point, where, three phases (solid Mg, Solid MgZn2, and their melt) are in

equilibrium.

Solid Mg + Solid MgZn2 Melt (l)

Curve E1BE2

The curve E1BE2 is freezing point curve of MgZn2, along the curve, solid MgZn2 and the melt are in

equilibrium.

Solid MgZn2 Melt (l)

Point B

The point „B‟ is the melting point of the compound MgZn2. The temperature at the point is 5900C.

Here the solid has the same composition as the liquid. So MgZn2 said to possesses congruent melting

point. The composition of MgZn2 is 33.3% and Zn is 67.7%.

Areas

1) Below the line AE

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

2) Below the line CE2

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

3) Below the point E1 and E2


48

The area the below the point E1 and E2 consists of solid Zn + solid MgZn2 and solid MgZn2 + solid

Mg respectively.

4) Below the line E1BE2

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

5) Above the line AE1BE2C

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

Applications of congruent system:

Chemical affinity between two metals can be studies

Formation of inter-metallic compounds or double compounds can be studied

The composition of inter-metallic compounds or double compounds at different temperatures can be

studied.

Thermal stability of such compounds can be studied.

2. Write note on heat treatment of steel. (or) Discuss any four heat treatment of steel in detail.

(May-03,Dec-05, Dec-09,,June-06, June-09, Jan-14, Dec-15 &June-16)

1. Annealing

Annealing is a softening process of steel. The steel is heated to higher temperature for sufficient time,

followed by slow cooling in a furnace

Purpose of annealing:

To remove the internal stress

To increase the match inability of steel

To increase ductility and toughness of steel

Types of annealing process:

(a) Low temperature annealing:

It involves heating of steel to a temperature below the lower critical temperature followed by slow

cooling.

(b) High temperature annealing:

It involves heating of a steel to a temperature about 30-50oc above the higher critical temperature

and holding it for sufficient time and to cool slowly.

2. Hardening (or) Quenching:

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

cooling it by immersing in oil or water or some other fluids.

Purpose of hardening:

It increases the hardness, abrasion resistance, wear resistance, ability to cut other materials.

3. Tempering:

Tempering is the process of heating the already hardened steel to a temperature lower than its own

hardening temperature and cool slowly to room temperature.


49

Purpose of Tempering:

To remove internal stress.

To increase ductility and toughness.

To retain the strength and hardness.

4. Normalizing:

Normalizing involves the process of heating steel to a definite temperature at about

30-50oC above its higher critical temperature and allowing it cool gradually in still air.

Purpose of normalizing:

To recover the homogeneity.

To remove internal stress.

To increases the toughness

5. Carburizing

Carburizing is the process of increasing carbon content at the surface of low carbon steel.

Purpose of carburizing:

To increase the hardness of the steel surface.

6. Nitriding:

Nitriding is the process of heating the metal alloy in presence of ammonia at a temperature to about

5500c.The atomic nitrogen combines with the iron on surface to form Iron nitride to form a hard

alloy.

Purpose of Nitriding:

To get super hard surface.

7. Cyaniding:

The steel article is immersed in the molten salts of sodium cyanide (or) potassium cyanide with very

small amount of Na2CO3 and NaCl at about 8700c followed by quenching in oil or water.

Purpose of Cyaniding:

To get harder the surface of medium carbon alloy.


50

UNIT V

NANOCHEMISTRY

PART A

1. Mention the difference between a nano rod and a nanowire. (A.U.May 2015,Jan 2016)

Nanowire Nano rod

Aspect ratio is greater than 20 Aspect ratio is less than 20

Conductivity is less Conductivity is more

Thin Thick

Greater flexibility Hard

2. What are nanomaterials? (A.U.Dec 2015)

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

dimension.

3. Write any two important applications of gold nano particles in medicine. (A.U.Dec 2015).

Gold coated nano shells convert light into heat, enabling the destruction of tumours.

Gold Nano shells are used for blood immuno assay.

4. Write any two applications of carbon nanotubes. (A.U.May 2015)

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

It is also used as light weight shielding materials for protecting electronics equipments.

CNTs are used effectively inside the body for drug delivery.

It is used in composites, ICs.

5. What is meant by Nanochemistry? (A.U.Dec 2014)

Nano- chemistry is the branch of nano-science, which deals with the chemical applications of

nanomaterials. It also includes the study of synthesis and characterisation of nanomaterials.

6. What is Nano rod? (A.U.Dec 2014)

Nanorod is two dimensional cylindrical solid material having an aspect ratio i.e., length to width

ratio less than 20.

7. What are nanowires? (A.U.Jun 2014)

Nanowire is two dimensional cylindrical solid material having an aspect ratio ie., length to

width ratio greater than 20.Diameter of the nanowire ranges from 10-100 nm.

8. What are carbon nanotubes? (A.U.Jan 2014)

Carbon nanotube is a tubular form of carbon with 1-3 nm diameters and a length of few nm to

microns.


51

9. Distinguish between bulk materials and nanomaterials. (A.U.Jun 2014)

Nano-particles Bulk particles

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

2. Collection of few

molecules

Collection of infinite number of

molecules

3. Surface area is more Surface area is less

4. Strength , hardness are more Strength , hardness are less

10. What is laser ablation? (A.U.Jan 2014)

It is a technique, which involves vaporization of target material by passing an intense pulsed

laser beam at a higher temperature.

11. What is CVD?

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

material with other gases, to produce a non-volatile solid that deposits automatically on a suitably

placed substrate.

12. What is the basic principle involved in solvothermal synthesis of nano-materials.

Solvothermal synthesis involves the use of solvent under high temperature (between 100oC to

1000oC) and moderate to high pressure (1atm to 10,000 atm) that facilitate the interaction of precursors

during synthesis.

13. What is Nanoclusters? (A.U.Dec 2012) Nano clusters are fine aggregates of atoms or molecules. The size of which ranges from 0.1 to 10

nm. Of all the nano materials, Nano clusters are the smallest sized nano materials because of their

close packing arrangement of atoms.

14. How are nano-particles prepared by precipitation techniques. (A.U.Jun 2013)

Nano-particles are synthesized by the precipitation reaction between the reactants in presence of

water soluble inorganic stabilizing agents.

15. Mention some characteristics of nanomaterial. (A.U.Jan 2010)

1. Nanomaterial‟s are very strong and withstand extreme strain and tension.

2. It possesses very good electrical properties and thermal conductivity.


52

PART B

1. Write short notes on i) Carbon Nano tubes ii) Nano Rods iii) Nano wire (AU, Apr/May, 2015)

1. Carbon nanotubes

Carbon Nano-tubes are allotropes of carbon with a nanostructure having a length-to-diameter ratio

greater than 1,000,000. When graphite sheets are rolled into a cylinder, their edges joined and form

carbon nano-tubes i.e., carbon nano-tubes are extended tubes of rolled graphite sheets.

TYPES OF CARBON NANO-TUBES

Depending upon the way in which graphite sheets are rolled, two types of CNTs are formed.

Single-walled Nano-tubes (SWNTs)

SWNTs consist of one tube of graphite. It is one-atom thick having of 2 nm and a length of 100 µm.

They exhibit important electrical properties. It is an excellent conductor.

Multi-walled Nano-Tubes (MWNTs)

MWNTs consist of multiple layers of graphite rolled in on themselves to form a tube shape. It exhibits

both metallic and semi-conducting properties. It is used for storing fuels such as hydrogen and

methane.

2. Nano rods

Nano rod is two dimensional cylindrical solid material having an aspect ratio.., length to width ratio

less than 20.

Examples: Zinc Oxide, Cadmium sulphide, Gallium nitride Nano rods.

Synthesis: Nano rods are produced by direct chemical synthesis.

Properties

1. Nano rods are two-dimensional materials,

2. It exhibits optical and electrical properties.

Applications of Nano rods.

1. Nano rods find application in display technologies.

2. It is also used in the manufacturing of micro mechanical systems, etc.,

3. Nano rods have used as cancer therapeutics.


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3. Nano wires

Nanowire is two dimensional cylindrical solid materials having an aspect ratio it, length to width ratio

greater than 20. Diameter of the nanowire ranges from 10 – 100 nm.

Example:

Metallic nanowires – Au, Ni, Pt

Nanowire of semiconductors - Si, GaN

Synthesis of nanowires:

Nanowire can be prepared by Template-assisted synthesis and Vapour liquid solid method.

Properties of nanowires:

1. Nanowire is two-dimensional material.

2. Conductivity of a nanowire is less than that of the corresponding bulk materials.

3. Silicon nanowires show strong photoluminescence characteristics.

Uses of Nanowires:

1. Nanowires are used for enhancing mechanical properties of composites.

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

3. It is also used to link tiny components into very small circuits.

2. Explain chemical vapour deposition technique of synthesis of Nano particles. (AU, Nov/Dec,

2014, AU, Apr/May, 2015, )

This process involves conversion of gaseous molecules into solid nanomaterial in the form of tubes,

wires or thin films. First the solid materials are converted into gaseous molecules and then deposited

as nanomaterial.

Example: CNT preparation

The CVD reactor consists of a higher temperature vacuum

furnace maintained at inert atmosphere. The solid substrate

containing catalyst like nickel, cobalt, iron supported on

substrate materials like, silica, quarts is kept inside the

furnace. The hydrocarbons such as ethylene, acetylene and

nitrogen cylinders are connected to the furnace carbon


54

atoms, produced by the decomposition at 10000C, condense on the cooler surface of the catalyst.

CVD Reactor: The CVD reactors are of generally two types

i. Hot - wall CVD: are usually tabular in form, and heating is accomplished by surrounding the reactor

with resistance elements.

ii. Cold-wall CVD: substrates are directly heated inductively by

graphite subsectors, while camper walls are air or water-

cooled.

Advantages:

Nanomaterial‟s, produced by this method, are highly pure.

It is economical.

Nanomaterial‟s, produced by this method, are defect free.

3. Explain laser ablation. (Or) How to synthesis Nano

materials by Ablation method? (AU, Nov/Dec, 2014)

LASER ABLATION (removing material from a solid using laser beam)

The method involves vapourisation of target

material containing small amount of catalyst (nickel

or cobalt) by passing an intense pulsed laser beam at

a higher temperature to about 1200C in a quartz tube

reactor. Simultaneously, an inert gas such as argon,

helium is allowed to pass into the reactor to sweep

the evaporated particles from the furnace to the

colder collector.

Uses:

i).Nano tube having a diameter of 10 to 20 nm and 100µm can be produced by this method.

ii).Ceramic particles and coating can be produced.

Advantages of laser ablation:

It is very easy to operate.

The amount of heat required is less.

It is eco-friendly method because no solvent is used.


55

This process is economical.

4.How are carbon Nano tubes synthesized? Explain in detail.(or) Describe the preparation of

any two methods of carbon nanotubes. (AU, Nov/Dec, 2014 & AU, Jan, 2014)

(i) Pyrolysis

Carbon Nano tubes are synthesized by the pyrolysis of hydro carbon such as acetylene at about 7000C

in the presence of Fe-Silica or Fe-graphite catalyst under inert condition.

(ii) Laser Evaporation (removing material from a solid using laser beam)

The method involves vapourisation of target

material containing small amount of catalyst (nickel

or cobalt) by passing an intense pulsed laser beam at

a higher temperature to about 1200C in a quartz tube

reactor. Simultaneously, an inert gas such as argon,

helium is allowed to pass into the reactor to sweep

the evaporated particles from the furnace to the

colder collector.

(iii) Carbon Arc Method

It is carried by applying direct current (60-100 A and 20-25 V) arc between graphite electrodes of 10-

20 µm diameters

(iv)Chemical Vapour Deposition

It is a process of chemically reacting, a volatile compound of

a material with other gases to produce a non-volatile solid

that deposited automatically on a suitable placed substrate.

Chemical vapor deposition (CVD) methods have been

successful in making carbon fiber, filament and Nano-tube

materials since more than 10–20 years ago.


56

5. Explain any six applications of Nano materials in various fields. (AU, May/jun, 2014)

1. 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 possesses 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.

2. 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 tumour.

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.


57

4. Gold Nano shells for blood Immune assay:

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

5. Gold Nano shells in Imaging:

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

6. Targeted drug delivery:

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

3. ELECTRONICS:

i) Quantum wires have electrical conductivity

ii) Nano radios are produced by using CNT MOSFET (Metal Oxide Semiconductor Field Effect

Transistor) is used for Amplifying and switching electronic signals.

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

4. BIOMATERIALS:

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

b. It is used as material for joint replacement.

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

implants etc.,

6. Explain about: a) Nano Clusture b) Nano wire (AU, Jan , 2014)

(a)Nano Cluster

Nano clusters are fine aggregates of atoms or molecules. The size of which ranges from 0.1 to

10nm.Of all the nanomaterial, Nano clusters are the smallest sized nano materials because of their

close packing arrangement of atoms

Example: CdS, ZnO, etc.,

All the atoms, in Nano cluster are bound by forces like metallic, covalent, ionic, hydrogen bond or

Vander Waals forces of attraction. Clusters of certain critical size are more stable than others. Nano

clusters containing of up to a couple of hundred atoms, but larger aggregates containing 103 or more

atoms, are called nanoparticles.

Magic number:

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


58

Production of Nano clusters

Nano clusters can be produced from atomic or molecular constituents or from the bulk material

either by bottom up process or top down process

Atomic clusters or molecular clusters are formed by the nucleation of atoms or molecules n

respectively.

Properties of Nano clusters

The reactivity of Nano clusters are decreased due to their decrease in size.

The melting point of Nano clusters are lower than the bulk material due to high surface to

volume ratio.

Application of Nano cluster

Nano cluster are used as catalyst in many reactions.

It is used in Nano based chemical sensors.

It is also used as a light emitting diode in quantum computers.

b) Nanowire

Nanowire with example

Nanowire is two dimensional cylindrical solid material having an aspect ratio length to width ratio

greater than 20. Diameter of the nanowire range from 10-100 nm

Synthesis of nanowires

Template assisted synthesis

Template assisted synthesis of nanowire is simple way to fabricate nanostructures. These 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.

Properties of nanowires

Nanowires are two-dimensional material

Conductivity of a nanowire is less than that of the corresponding bulk material.

Use of nanowire

Nanowire is used for enhancing mechanical properties of composites.

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


59

7. Distinguish molecules, Nano particles and bulk materials. (or) Compare the properties of

molecules, nanoparticles and bulk materials. (or) Distinguish molecules, nanoparticles and bulk

materials. (A.U. May 2015)(A.U Jan 2014)

S.No Properties Molecules Nano Particles Bulk Materials

1. Size of the

particle

Size is much

smaller

Size is larger than

molecules but

smaller than bulk

materials.

Size is much

larger than the

molecules and

nanoparticle

2.

Magnitude of

constituting

particles

Few

Angstroms(A0)

(10-10

m)

Angstrom to

nanometer 10-10

to

10-9

m

Microns to

millimeter

3.

Number of

constituting

particles

Two atoms for

molecules.

2 to several

thousands. Infinite

4. Electronic

structure confined confined continuous

5. Geometric

structure

Well defined

structure and

predictable.

Well defined

structure and

predictable.

Crystal structure

decides

Example NaCl, HCl (NaCl)n Gold bar &

Silver bar

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