ANNA UNIVERSITY QUESTION PAPER DEC-2014
ANSWER KEY
PART-A
1. What are copolymers?
It is the joint polymerization in which two or more different monomers combine to give a
polymer. High molecular weight compounds obtained by copolymerization are called
copolymers.
Example:
nCH2 == CH –– CH == CH2 + n CH2 == CH
Butadiene
Styrene
Copolymerization
–(– CH2 –– CH == CH – CH2–– CH2–– CH –)–
n
Polybutadiene-co-Styrene
(Styrene-butadiene rubber, SBR)
2) What are thermoplastics?
Thermoplastics are prepared by addition polymerization. They can be softened on
heating and hardened on cooling. They are soft and less brittle.
They are generally linear polymers. A weak Vander Waals force is present
between the two adjacent layers. They are soluble in organic solvents.
Examples: PE, PVC, PP, PS, nylons, Teflon, etc.
3) What is the significance of decrease in free energy?
The decreases free energy of process at constant temperature and pressure is equal to
the useful work obtainable from the system.
4) State Second law thermodynamics?
The second law of thermodynamics has been formulated to explain the spontaneity(feasibility) of
physical and chemical process. This law introduces two new thermodynamic functions entropy and
free energy to explain the spontaneity (feasibility) of process.
CLAUSIUS STATEMENT :It is impossible to construct a machine which transfer heat
from a lower temperature to a higher temperature.ds= dq/T.
5) What is photochemistry?
Photochemistry is the study of chemical reactions resulting from the exposure of light
radiations. The radiations of visible and ultraviolet portion spectrum ranging from 2,000 8,000 Å. Wavelength are mainly of concern in photochemical reaction.
6) What is the purpose of a IR spectrometer?
It is an important tool in the structural elucidation of organic compounds.
The region of electromagnetic radiation in between the visible and microwave
region is called infrared region. IR spectra is produced by the absorption of energy by
a molecule in the infrared region and the transitions occur between vibrational levels.
7) What is degree of freedom at eutectic point in Lead- silver system?
Eutectic point
Solid Pb + Solid Ag Melt
According to reduced phase rule equation
F' = C – P + 1 ; F' = 2 – 3 + 1 ; F' = 0
The system is non-variant.
At eutectic point or eutectic temperature and its corresponding composition,
97.4% Pb +2.6% Ag, is called eutectic composition. Below this point the eutectic
compound and the metal solidify.
8) Mention any two significance of alloy making.
(i) To lower the melting points of the metal
(ii) To increase the hardness of the metal
(iii) To modify chemical activity of the metal
(iv) To resist the corrosion of the metal
(v) To modify the colour of the metal
(vi) To get good casting of metal
9) What is meant by nanochemistry
Nanochemistry is a branch of nanoscience, deals with the chemical
applications of nanomaterials in nanotechnology.Nanochemistry involves the study of
the synthesis and characterisation of materials of nanoscale size.
Nanochemistry is a relatively new branch of chemistry concerned with the
unique properties associated with assemblies of atoms or molecules of nanoscale
(~1-100 nm), so the size of nanoparticles lies somewhere between individual atoms or
molecules (the ‘building blocks’) and larger assemblies of bulk material which.
10) What are nanorods?
Nanostructures shaped like long sticks or dowels with a diameter in the
nanoscale but having a length that is very much longer. Standard aspect ratios (length
divided by width) are 3-5nm.
PART-B
11. (A) i) Discuss the mechanism of condensation polymerization in detail. (8)
It is a reaction between simple polar groups containing monomer with the formation of
polymer and elimination of small molecules like H2O, HCl, etc., (A carboxylic acid
monomer and an amine monomer can join in an amide linkage).
1.Hexamethylene diamine and adipic acid condense to form a polymer, Nylon 6:6
(polymide)
2.Polymerisation of ω – amino caproic acid to form nylon- 6.
n H2N
(CH2) COOH
ω – amino caproic acid
[CO ( CH2 )5
NH]n
nylon- 6
In some cases condensation polymerisation takes place without the elimation of
small molecules like H2O, HCl, etc., but by just the opening of cyclic compounds.
ii) Explain any four important properties of polymers. (8)
1. Glass transition temperature
Rubber is an elastic substance at room temperature. But when it is closed to -79oC it
becomes a glossy brittle solid, which when struck, crumbles to a powder.
Glossy brittle solid
Rubber.
Thus for every polymer there exists a temperature below which it is a glossy brittle solid
and above which it is a soft elastic substance. This temperature is known as glass transition
temperature.
2. 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.
a)Isotactic polymer:
If the functional groups are arranged on the same side of the main chain, the polymer is
Isotactic polymer.EX:
b) syndiotactic polymer :
If functional groups are arranged in an alternating fashion, the polymer is called
syndiotactic polymer.
c) Atatic polymer:
If the functional groubs are arranged randomly, the polymer is called Atactic polymer.
3)Number- Average molecular mass:
Number Average molecular mass is the mass obtained by dividing the mass of the
polymer material with the total number of molecules present. Mathematically it is defined
as
4) Weight- average molecular mass:
Weight- average molecular mass not only depend on the number of particles, but also on
the molecular size. So, in averaging process molecular weight of each individual species is
multiplied by the mass and not by the number. Mathematically it is defined as
OR
B) i) Describe the emulsion polymerization technique. Give two examples (8)
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. Then initiator is added.The whole content 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 deemulsifier like 3% solution of Al2 (SO4)3
Monomer + Initiator + Surfactant
Polymer
Examples: Polyvinyl acetate, PVC are prepared by this method
II) Write down the preparation, propertie and uses of Nylon 6:6 (8)
Nylon 6:6 is manufactured by solution polymerization by condensing Hexamethylene
diamine and adipic acid in toluene solvent at higher temperature in an inert atmosphere.
Properties of Nylon 6:6
1.
2.
3.
4.
It is a horny translucent material.
Its melting point is high (264oC).
Nylon 6:6 is a less soft and stiff material when compared to nylon 6:6.
It is insoluble in common organic solvents, but soluble in formic acid and
cresol.
5. Both fibre and plastic have high tenstile strength and dimensional stability.
6. It shows good impact strength due to the large number of flexible groups.
7. As I/D ratio of the polymer filament is very high and there is high
intermolecular attraction between the molecules due to H-bonds between the
carbonyl and amide –NH- groups it can act as a very good fibres.
8. Both nylon fibre and plastic are resistant to abrasion.
9. It does not absorb water and can be dried easily.
Applications of nylon 6:6
As a fibre
1. The majority of the woven fibres are used in the manufacture of tyre cards.
2. It is for making socks, dress materials and ropes.
3. It is blended with wool to make wool more resistant to abrasion.
As a plastic
1. It is used as an engineering plastic.
2. It is used to make various parts due to its high tenstile strength, good impact
strength , good dimensional stability and its resistance to abrasion.
3. It is used in ball bearing, mounting, electrical connections, etc.,
4. It is also used to make wheels, bearings etc., which can run without use of
lubricants.
12 ) A) i) Derive Gibbs- Helmholtz equation
(8)
12 A) i) ∆G for areaction at 300 K is -16 k cal, ∆H for the reaction is -10k cal. What is the
entropy (∆s) of the reaction? What will be ∆G at 300 K?
(8)
B) i) Derive clausis- clapeyron equation and mention its applications.
(8)
Consider a system containing only 1 mole of a substance existing in 2
phases
A and B. The free energies of the substance in 2 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, there is no change in free energy.
i.e.,
GA = GB
.....(1)
If the temperature of the system be raised to T+dT, the pressure becomes
P + dP and the free energies become GA + dGA and GB + dGB respectively.
Then the equation (1) becomes,
GA + dGA = GB+ dGB
.....(2)
We know that,
G = H – TS
H = E + PV
(Gibbs free energy)
(Enthalpy)
G = E + PV – TS
dG = dE + PdV + VdP – TdS – SdT
But,
dE = dq – dw
.....(3)
.....(4)
d
q
=
dE + dw dq =
dE + PdV
We know that for reversible equation,
[ dw = PdV]
.....(5)
dq
= dS
T
dq = TdS
.....(6)
Substituting equation (6) in equation (5)
TdS = dE + PdV
dE = TdS – PdV
.....(7)
Substituting equation (7) in (4)
dG = TdS – PdV + PdV + VdP – TdS –SdT
dG = VdP – SdT
.....(8)
Here, the work done is due to volume change only, so equation (8) may be
applied to phase A as well as B
dGA
= VA dP – SA dT
.....(9)
dGB
= VB dP – SB dT
.....(10)
Where,
VA and VB are the molar volumes of phases A and B
respectively. SA and SB are their molar entropies.
Since,
GA
= GB
Hence from the 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)
(SB
(VB
SA )
VA )
.....(14)
=
dP
dT
.....(15)
SB – SA is denoted
S . SB – SA represents the change in entropy when
as
1 mole of the substance passes from the initial phase A to the final phase B.
Hence, equation (15) becomes
(VB
dT
S
dP
VA )
.....(16)
We know that entropy change
S =
S at constant T is
q
T
.....(17)
Substituting equation (17) in (16)
dP
dT
)
q
T (VB
VA
.....(18)
This is the Clausius-Clapeyron equation.
Applications
Calculation of molar elevation constant.
Calculation of boiling point or freezing point.
Determination of latent heat of vapourization.
Calculation of vapour pressure at another temperature.
B) i) Derive Vant Hoff’s isotherm equation. (8)
Standard Gibbs energy of a reaction
G o is the difference between standard
Gibbs energies of products and reactants in their standard states at the given
temperature.
Standard state means, the most stable physical form at the said conditions and
of high purity.
i.e., activity = 1.
Van’t Hoff has derived a relationship between the activities of the species
concerned in the reaction and G o of the reaction.
Consider a reversible reaction.
A
C
Gibbs free energy of reactant A is given by
GA = G oA + RT ln [A]
where
[A]
activity of A
G oA
Standard Gibbs energy of A
Similarly, we can write,
GB = G oB
RT ln [B]
GC G o
C
RT ln [C]
=
o
D
GD
RT ln [D]
= G
=
Gproducts – Greactants
=
[GC + GD] – [GA + GB]
G of the reaction
G = {G C
[B]}
RT ln [C]

= GC
G D
= GC
G
D
G =
G
DG
RT ln [C] [D]
ln [D]} A {G

GA
RT ln [C] [D] A G
B

GB
G
RT ln
[C][D]
[A] [B]
of a reaction at any instant.
[ G
= Standard Gibb’s free energy of the reaction.
.....(1)
ln [A]
B
G
RT ln [A] [B]
RT ln [A] [B]
.....(2)
ln
[[A], [B], [C] and [D] are activities of A, B, C and D at the given instant]
At equilibrium,
G
0
0
G
=
RT ln
[C]eq [D]eq
[A]eq [B]eq
Here the concentration refer to concentrations at equilibrium.
Keq = Equilibrium constant, can be simply referred as k.
G
= RT ln keq
G = –RT ln keq
.....(3)
k
[C]eq
[D]eq eq
[A]eq [B]eq
substituting this value in equation (2)
[C] [D]
=
–RT
ln
k
+
RT
ln
G
[A] [B]
.....(4)
This equation (4) is known as Van’t Hoff isotherm. The equation
G
T ln k
gives the relation between standard free energy and equilibrium constant.
G is standard free energy change when [A] = [B] = [C] = [D] = 1
VAN’T HOFF EQUATION
This equation tells the variation of equilibrium constant with temperature. This
is obtained by combining Van’t Hoff isotherm and Gibbs Helmholtz equation.
G
T
ln k
G
ln k =
.....(1)
R
T
CHEMICAL THERMODYNAMICS
2.41
Differentiately with respect to temperature
1
ln k
( G
/ T)
=
T
R
T
1
H
=
T2
R
(
From Gibbs Helmholtz equation)
ln k
H
=
T
RT 2
This is known as Van’t Hoff equation. On integration between the limits T1
and T2 and k1 and k2.
k2
T2
d ln k
H
T
2
RT

=
T1
k1
k
ln 2
k1
H
1
1
=
R
T2
T1
H
T2
T1
=
R
T1T2
.....(3)
This is another form of Van’t Hoff equation (for a reaction carried out at
constant pressure).
13 A) i) State and explain lambert-Beer law (8)
It states that when a beam of monochromatic radiation passes
through a homogeneous absorbing medium, the rate of decrease of intensity of
radiation dI with the thickness of the absorbing medium dx is proportional to the
intensity of incident radiation I.
Mathematically it is expressed as
dI
dx kI
.....(1)
Where k = absorption coefficient or proportionality constant
On integrating the equation (1) between limits I = I0
at x = x, we get
I
dI
x kdx
I
Io
x
ln
I
Io
I/IO = e-kx
I= IO. e-kx
.....(2)
This equation is known as Lambert’s
law
kx
at x = 0 and I = I
Beer’s law (or) Beer Lambert’s law
This law states that when a beam of monochromatic radiation passes
through a solution of an absorbing substance, the rate of decrease of intensity of
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
Mathematically, it is expressed as
dI
dx
kI c
.....(3)
Where k = Molar absorption coefficient.
On integrating the equation (3) between limits I = I0 at x = 0 and I = I at
x = x, we get
I
Io
ln
ln
dI
I
x
kCdx
=
0
I
Io
I
Io
=
kCx
=
– kCx
=
kCx
By taking natural logarithm
2.303 log
I
Io
log
Where,
= k / 2.303
I
Io
=
A
=
k
Cx
2.303
Cx
.....(4)
= molar absorptivity (or) molar extinction coefficient.
log I 0/I = A= Absorbance (or) Optical
density. The equation (4) is called BeerLambert’s law.
This equation is known as Beer’s law or Beer-Lambert’s law.
Thus, the absorbance A is directly proportional to the molar concentration
C and path length (x).
Application of Beer-Lambert’s law
Determination of unknown concentration
First absorbance ‘As’ of a standard solution of known concentration ‘Cs’ is
measured, then according to Beer-Lambert’s law.
AS
Cs x
=
AS
CS
x
....(5)
Now, we measure the absorbance Au, of a solution of unknown concentration
Cu, Since
and x are constants. We have
Au
Au
Cu
=
Cu x
Au
Cu
Cs
x
x
From equations (5) and (6)
As
=
Au
Cu
.....(6)
Cu
=
Au
Cs
As
.....(7)
Since the values of Au and As are experimentally determined values and Cs is
known, the value of unknown concentration Cu can be calculated from the
equation (7).
Limitations of Beer-Lambert’s law
If the solution contains impurities, deviation may occur.
If the solution undergoes polymerization, deviation also occurs
It is applicable for dilute solutions
It is not obeyed if the radiation used is not monochromatic.
It is not applied to suspensions.
ii) Explain photo sensitation. Mention its applications in photography.
The foreign substance (called sensitizer) 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:
(i) Molecular photosensitizers: Benzophenone, sulphurdioxide
(ii) Atomic photosensitizers: Cadmium, Mercury, Zinc, etc.
Mechanism of photosensitization
Consider a general Donor (D) – Acceptor (A) system in which D (the
sensitizer) absorbs the incident photon. The triplet state of D is higher in energy
than that of A (the reactant).
1
(i) On absorption of the photon, the singlet excited state of donor (i.e., D)
1
3
is produced. D in-turn gives the triplet excited state D via inter-system
crossing (ISC).
3
(ii) The triplet excited state D then collides with A (i.e., acceptor or
3
reactant), thereby producing the triplet excited state of acceptor, A and
the ground state of donor (So).
3
(iii) If A gives the desired reaction products, the mechanism is known as
photosensitization. However, if the desired reaction products results from
3
D state, then A is known as quencher and the process is called
quenching. Thus, the reactions depicting photosensitization and quenching
may be described as below:
D
1D
3D
1D
ISC
3D
D 3A
A
3A
products (photosensitization)
3D
products (Quenching)
11
D
(or S 1)
A
ISC
2A
2D
(or T1)
(or T1)
S0
S0
Donor, D
(Sensitizer)
Acceptor, A
(Reactant)
The triplet excited state of the
1
sensitizer or donor ( D) must be
higher in energy level than the triplet
excited state of the reactant or acceptor
3
( A) so that enough energy is
available to raise the reactant
molecule to its triplet excited
state.
Examples for photosensitized
reactions:
1. Dissociation of H2
molecule.
2. Photosynthesis in plants.
B) i) Explain
1.quantum efficiency and
QUANTUM EFFICIENCY( )
The number of moles of the substance undergoing photochemical change per
Einstein of radiation absorbed is known as quantum efficiency.
Alternatively, it is the number of molecules of the substance undergoing
photochemical change per quantum of radiation absorbed.
Thus,
No. of molecules reacting in a given time
No. of quanta of light absorbed in the same time
If we measure the rate of formation of product, then quantum yield can be
written as,
d[Pr oduct] / dt
Number of quanta absorbed
d[Pr oduct] /
dt
I abs
Where,
Iabs = Number of quanta absorbed.
Low or high quantum yield
6
The quantum efficiency varies from zero to 10 . If a reaction obeys the
Einstein law, one molecule is decomposed per photon, the quantum yield
1.
(i) Low quantum yield
When the number of molecules decomposed is less than one per photon, the
quantum yield
and the reaction has a low quantum yield.
1
(ii) High quantum yield
When two or more molecules are decomposed per photon, the quantum yield
and the reaction has a high quantum yield.
1
Conditions for high and low quantum yield
All the reactant molecules should be initially in the same energy state and
hence equally reactive.
Molecules in the activated state should be largely unstable and decompose
to form the products.
The reactivity of the molecules should be temperature independent.
Causes (or) Reasons for low quantum yield
Excited molecules may get deactivated before they form products.
Excited molecules may lose their energy by collisions with non-excited
molecules.
Molecules may receive insufficient energy thereby reaction does not occur.
Reversion of the primary photochemical reaction.
Recombination of fragments (obtain by dissociation) will give low
quantum yield.
Example:
(i) Dimerization of anthracene to dianthracene
2C14 H 10
h
C 28H 20
Quantum yield = 2, Actually it is found to be 0.5.
Because, the above reaction is reversible
2C14 H 10
h
C H
therm 28 20
al
Causes (or) Reason for high quantum yield
Absorption of radiation produces atoms or free radicals, which initiate a series
of chain reactions.
Formation of an intermediate product, which acts as a catalyst.
If the reaction is exothermic the heat evolved may activate other molecules,
without absorption of additional quanta of radiation.
The active molecules produced by primary absorption may collide with other
molecules and activate them which inturn activate other reacting molecules.
Example:
Decomposition of HI
In the primary reaction, HI molecule absorbs a photon and dissociated to produce
are H and I.
HI
h
H + HI
I+I
2HI + h
reaction
H
I
H2 + I
.....(1) Primary process
.....(2) Secondary process
I2
H2 + I2
.....(3) Overall
In the secondary reaction, step(1) H react with HI to produce H2 and I. This I
react with step(1) I to produce I2.
T he o ve r all r eact io n sho ws t hat t he t wo HI a r e de co mpo sed fo r
one photon ( h ). so the quantum yield ( ) =2.
DETERMINATION OF QUANTUM EFFICIENCY (OR) QUANTUM
YIELD
Quantum yield,
of a photochemical reaction is expressed as
Number of molecules reacted (or) formed
(or)
Number of photons (or) Quanta of radiation absorbed
Number of moles reacted (or) formed
Number of Einsteins of radiation absorbed
We can calculate the quantum yield from the determination of the following:
(i) Determination of the number of molecules reacted in a given time.
(ii) Determination of the amount of photons absorbed in the same time.
2. chemiluminescence
CHEMILUMINESCENCE
If light is emitted at ordinary temperature, as a result of chemical reaction
the process is called as chemiluminescence. It is the reverse of a photochemical reaction.
Explanation:
In a chemiluminiscence 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:
(i) When pyragallol is oxidised by hydrogen, peroxide, chemiluminescence
is produced.
(ii) Oxidation of decaying wood and containing certain bacteria.
(iii) Grignard reagent produces greenish blue light on oxidation by air.
Mechanism of chemiluminescence
It can be explained by considering anion-cation reactions.
Example:
–
+
Interaction between the aromatic anions (Ar ) and cations (Ar ). (Fig.3.5)
AMBO
+
+
BMO
Ar –
Ar +
Ar *
Ar
Ar
1Ar*
Ar
1 Ar*
Ar
h
Ar
–
The aromatic anion (Ar ) contains one unpaired electron in the antibonding
molecular orbital (AMBO) and two paired electrons in the bonding molecular
+
orbital (BMO). The AMBO of the aromatic cation Ar
is empty. When the
electron
–
+
is transferred from the AMBO of the anion (Ar ) to the AMBO of the cation (Ar ).
1 *
So, the singlet excited state Ar is formed. The excited state can be deactivated by
the emission of photon h
.
ii) Discuss the principle and instrumentation of UV- Visible spectrometer.
Instrumentation
I. Components
The most important components of UV-Visible spectrometer are
(i)
Radiation source
(ii)
Monochromators
(iii) Cells
(iv)
Detectors
(v)
Recorder
(i) Radiation source
Hydrogen or deuterium lamps are commonly used as radiation source.
Requirements of a radiation source : It must be stable and supply continuous
radiation, sufficient intensity.
Beam
splitte
r
Recorder
Detector
Monochromat
or
Sourc
e
sampl
e
referenc
e
Fig. 3.13 Block diagram of uv-visible spectrophotometer
(ii) Monochromator:
Selects appropriate wavelength of light. The important elements of a
monochromator are
an entrance slit
a dispersing element
an exit slit
The dispersing element may be a prism or a grating or a filter.
(iii) Cells
It is made out of quartz or fused silica. The cells must contains the following
characteristics
They must be uniform in construction.
The material of cell should be inert to solvents.
They must transmit the light of the wavelength used.
(iv) Detectors
The most commonly used detectors are
(a) Barrier layer cell
(b) Photomultiplier tube
(c) Photocell
These detectors converts light radiations into electrical signals.
(v)
Recorder
The signal from the detector is finally received by the recorder.
II. Working of UV-Visible spectrometer
The radiation from the source is allowed to pass through the monochromator unit.
The monochromator allows a narrow range of wavelength to pass through an exit slit.
The beam of radiation coming out of the monochromator 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 another half is directed to pass through
an identical cell which contains only the solvent.
The instrument is designed in such way that it can compare the intensities of these
two beams.
If the sample absorbs light at a particular wavelength, then intensity of the sample
beam (I) will be less than that of the reference beam (I0).
The instruments give output graph, which is a plot of wavelength Vs. absorbance
of the light. The graph obtained is known as absorption spectrum.
14) A) i) Draw neat one component water system and explain
The water system
Water exists in three possible phase 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 phase diagram for the water
system is shown in Fig.4.1.
This phase diagram contains curves, areas and triple point.
(i) Curve OA
The curve OA is called vapourisation curve, it represents the equilibrium
between water and vapour, at any 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 upto the critical temperature
o
(374 C). Beyond the critical temperature the equilibrium will disappear only water
vapour will exist.
C
Critical
pressure
218 atm
Solid ice
A
Melting curve
Liquid water
Pressure
Vapour
pressure
curve
4.58 mm
Metasta
ble
equilibr
ium
B’
Sublima
tion
Curve
O
Water vapour
Tri
ple
poi
nt
Criti
cal
Temperat
ure
M.
P.
B
273oC
374o
0.0075o
Temperat
ure
Fig. 4.1 Phase diagram of water system
(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
o
This equilibrium (line OB) will extend upto the absolute zero (–273 C),
where no vapour can be present and only ice will exit.
(iii) Curve OC
The curve OC is called melting point curve of ice, it represents the
equilibrium between ice and water. At any point on the curve the following
equilibrium will exist.
Ice
Vapour
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 and 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(s)
Water(l)
Vapour(g)
The degree of freedom of the system is zero. Ie., nonvariant. This is predicted
by the phase rule.
F = C–P+2 ; F=1–3+2 ; F = 0.
o
Temperature and pressure at the point ‘O’ are 0.0075 C 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
o
Sometimes water can be cooled below 0 C without the formation of ice, this
water is called super-cooled water. Super cooled water is unstable and it can be
converted into solid by “seeding” or by slight disturbance.
(vi) Areas
Area AOC, BOC, AOB represents water, ice and vapour respectively. In order
to define the system at any point in the areas, it is essential to specify both temperature
and pressure. The degree of freedom of the system is two i.e. Bivariant.
This is predicted by the phase rule.
F = C–P+2 ; F=1–1+2 ; F = 2
ii) Draw neat Zinc- Magnesium alloy system and explain.
Zinc-Magnesium Alloy system
Zn-Mg binary alloy system is an example for the formation of compound
with congruent melting point.
A compound is said to possess congruent melting point, if it melts exactly
at a constant temperature into a liquid having the same composition as that of
solid.
The phase diagram of Zn-Mg binary alloy system consist of two sides
(Fig.4.3).
(i) Left side consists of Zn and MgZn2 system.
(ii) Right side consists of MgZn2 and Mg
system.
Left side
(a) Curve AE1
The curve AE1 is freezing point curve of Zn. Point A indicates the melting
point of pure Zn. 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
(b) Point E1 (Eutectic
point)
Point E1 is the eutectic point, where three phases are in equilibrium.
o
The temperature at this point is 380 C.
Solid Zn + Solid MgZn2
melt
Right side
(c) Curve CE2
It is the freezing point curve of Mg. Point C is the melting point of pure
Mg. 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
(d) Point E2 (Eutectic point)
Point E2 is the eutectic point, where three phases are in equilibrium. The
o
temperature at this point is 347 C.
solid Mg + solid MgZn2
Melt
(e) Curve E1BE2
It is the freezing point curve of MgZn2. Along the curve, solid MgZn2
and the melt are in equilibrium.
solid MgZn2
melt
(f) Point B
It is the melting point of the compound MgZn2. Here the solid has the same
composition as the liquid MgZn2 is said to possesses congruent melting point.
(g) Areas
Below the
line AE1
Below the
line CE2
Below the
line
E1BE2
Below the
point E1
and E2
Above the
line
AE1BE2C
It consists of
solid Zn and
the solution.
It consists of
solid Mg and
the solution.
It consists of
solid MgZn2
and the
solution.
It consists of
solid Zn +
solid MgZn2
and
solid MgZn2
+ solid Mg
respectively.
It consists of
only liquid
phase.
B) i) Write a note on :
1. Annealing and
Annealing:
Annealing means softening. This is done by heating the metal to a certain
high temperature, followed by very slow cooling in a planned manner in a furnace.
Purpose :
It increases the machinability.
It also removes the imprisoned gases and internal stresses.
Types of annealing
Annealing can be done in two ways.
Low temperature annealing (or) Process annealing
High temperature annealing (or) Full annealing
(a) Low temperature annealing (or) Process annealing
It involves in heating steel to a temperature below the lower critical
temperature followed by slow cooling.
Purpose :
It reduces hardness
It improves machinability by relieving the internal stresses or internal
strains.
It increases ductility and shock resistance.
(b) High temperature annealing (or) Full annealing
o
It involves in heating steel to a temperature about 30 to 50 C above the
higher critical temperature and holding it, at that temperature for sufficient time
to allow the internal charges to take place and then, cooling gradually to room
temperature.
The approximate annealing temperatures of various grades of carbon
steel are
o
Mild steel = 840 – 870 C
o
Medium – Carbon steel = 780 – 840 C
o
High carbon steel = 760 – 780 C
Purpose
It makes the steel softer, together with an appreciable increase in its
toughness.
It increases the ductility and machinability.
2. Nitriding
Nitriding
It is a process of getting super-hard surface. In this operation, the metal alloy
o
is heated in presence of ammonia at a temperature of about 550 C. The nitrogen
combines with the surface constituents of the alloy to form extremely hard nitrides.
This process is quite time-consuming.
Purpose : To get super-hard surface
ii) Discuss the composition, properties and uses of brass and bronze.
Brass (Copper alloy)
Brass is the most important copper alloy. It containing mainly copper
and zinc.
They possess
Lower melting points than Cu and Zn.
Water resistance property
Greater strength than pure Cu.
Good corrosion resistance property.
Greater durability than pure Cu.
Greater machinability than pure Cu.
Important brasses their properties and uses
Some important brasses and their composition, properties and uses are
summarized in the following Table-4.2.
Type of
Brasses
Composition
Brass
Cu = 6090% Zn =
40–10%
Main
Brasse
s (i)
Commercia
l Brass (or)
Guilding
metal (or)
French
gold
(ii)Dutch
- metal
or Low
brass
Cu =
90% Zn =
10%
Cu =
80% Zn =
20%
Table 4.2
Properties
Uses
(i) They possess
greater strength,
durability and
machinability than
pure copper.
(ii) They possess
lower melting point
than Cu and Zn
(iii) They are
good corrosion
resistant against
water
(i) It is golden in
colour
(ii) It is stronger
and harder than
pure Cu.
(i) It is golden in
colour
(ii) It i s suitable
for all drawing and
forming operations
Forgings,
hardwares, rivets,
screws,
costumes,
jewellery,
etc.
Cheap jewellery,
musical
instruments, tubes,
battery caps,
name plates,
flexible hoses,
etc.
PHASE RULE & ALLOYS
(iii)
Cartridg
e brass
(or)
Spinnin
g brass
Special
brasses
(i) German
Silver
(ii)Admiral
y brass or
Tobin
bronze
Cu =
70% Zn =
30%
It contains
metals other
than Cu and
Zn Cu = 25–
50% Zn = 10
–35% Sn = 5–
35%
Cu=59–62%
Sn = 0.5–
1.5% Zn =
rest
4.31
(i) It is soft, ductile,
harder and
stronger than Cu.
(ii) It can be cold
deformed like
extraction
Condenser tubes,
sheet fabrication,
household
articles, etc.
Utensils,
(i) It possesses good
strength and
tablewares, bolts,
corrosion– resistance
screws, cutlery,
to salt water (ii) It is
ornaments,
ductile, malleable and
corrosion- resistant
looks
implements,
like silver.
decorative articles,
It possesses high
It is used in
etc.
corrosion and abrasion - propellers
resistances
and marine
works.
4.19.2 Bronze (Copper alloy)
Bronze is also a copper alloy. It containing copper and tin.
They possess
Non-oxidizing property.
Lower melting point than steel.
Corrosion resistance property
water resistance property
Be t t er hea t and elect r ical co nduc t ing pr o per t y t han mo st o f
the steels.
Important Bronze their properties and uses
Some important bronzes and their compositions, properties and uses are
summarized in the following Table 4.3.
Table 4.3
Type of
Bronzes
Composition
Bronze
Cu = 80–
95% Sn =
20–5%
They are tough,
strong and corrosion
resistant.
Cu = 89–92%
Sn = 11 – 8%
It is soft, ductile
and durable.
Pumps, valves,
wires, flanges,
utensils, coins,
statues, etc.
(ii)Gun
Metal
Cu = 85%
Zn = 4%
Sn = 8%
Pb = 3%
It is hard, tough.
strong to resist
the force of explosion
Foundary works,
heavy-load bearings,
water fittings,
marline pumps,
parts of highpressure steam
plants, etc.
(iii)
Aluminum
bronze
Cu = 90–93%
Al = 10-7%
(i) It is quite strong
and readily fusible.
(ii) It gives good
castings
(iii) It is resistant to
corrosion, even at
high temperatures
(iv) It possess good
Bushes,
casting operations,
bearings,
jewellery,
utensils,
photoframes,
coins, etc.
Main
Bronzes
(i) Coinage
Bronze
(or)
Properties
Uses
abrasion-resistance
(v) It is golden yellow
in colour.
(iv) High
Sn = 10–13%
phosphorous P = 0.4 to 1%
bronze
Cu = rest
(i) It is hard, brittle
and abrasion, resistant
(ii) It possess low
co-efficient of friction.
Bearings, gears,
taps, springs,
bushes, turbine
blades, fuses, etc.
(v) Nickel
bronzes
(i) It is better corrosion
resistant than copper.
(ii) It is hard and
higher tensile strength
than copper.
For rolling
purposes,
valves,
unhardened shafts
and general
purpose semi-hard
Cu = 90%
Ni = 9%
Fe = 1%
bearings.
15 A i) Explain the chemical vapour deposition with a neat diagram.
Chemical vapour deposition
Chemical Vapor Synthesis (CVS) is a modified Chemical Vapor Deposition
(CVD) method where the process parameters are adjusted to form nanoparticles
instead of film. The entire range of reaction regimes and co rrespo nding
microstructures (epitaxial, polycrystalline, columnar, granular films and aerogel
coatings as well as nanopowders) are shown in the below.
Both in CVD and CVS, precursors are metalorganics, carbonyls, hydrides,
chlorides and other volatile compounds in gaseous, liquid or solid state. The major
limitation of the CVS process is the availability of appropriate precursor materials.
The energy for the conversion of the reactants into nanoparticles is supplied in
hot wall (external furnace), flame (reaction enthalpy), plasma (microwave or radio
frequency) and laser (photolysis or pyrolysis) reactors.
The most important process parameters determining the quality and usability
of the nanopowders are the total pressure (typical range from 100 to 100000 Pa),
the precursor material (decomposition kinetics and ligands determining the impurity
level), the partial pressure of the precursor (determining the production rate and
particle size), the temperature or power of the energy source, the carrier gas (mass
flow determining the residence time) and the reactor geometry.
The nanoparticles are extracted from the aerosol by means of filters,
thermophoretic collectors, electrostatic precipitators or scrubbing in a liquid.
Modifications of the precursor delivery system and the reaction zone allow the
synthesis of pure oxides, doped oxides, coated nanoparticles, functionalized
nanoparticles and granular films.
ii) Describe the preparation of any two methods of carbon nanotubes.
Synthesis of carbon nanotubes
Carbon nanotubes can be synthesized by any one of the following methods.
(i)
(ii)
(iii)
(iv)
Pyrolysis of hydrocarbons
Laser evaporation
Carbon arc method
Chemical vapour deposition
(i) 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
o
(1200 C) 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 copper collector, on which they condense as carbon nanotube.
(ii) Chemical vapour deposition
It involves decomposition of vapour of hydrocarbons such as methane,
o
acetylene, ethylenes, etc. at high temperatures (1100 C) in presence of metal
nanoparticle catalyst like nickel, cobalt, iron supported on MgO or Al2 O3 .
Carbon atoms produced by the decomposition condense on a cooler surface of the
catalyst.
Chemical vapour deposition
Chemical Vapor Synthesis (CVS) is a modified Chemical Vapor Deposition
(CVD) method where the process parameters are adjusted to form nanoparticles
instead of film. The entire range of reaction regimes and co rrespo nding
microstructures (epitaxial, polycrystalline, columnar, granular films and aerogel
coatings as well as nanopowders) are shown in the below.
Both in CVD and CVS, precursors are metalorganics, carbonyls, hydrides,
chlorides and other volatile compounds in gaseous, liquid or solid state. The major
limitation of the CVS process is the availability of appropriate precursor materials.
The energy for the conversion of the reactants into nanoparticles is supplied in
hot wall (external furnace), flame (reaction enthalpy), plasma (microwave or radio
frequency) and laser (photolysis or pyrolysis) reactors.
The most important process parameters determining the quality and usability
of the nanopowders are the total pressure (typical range from 100 to 100000 Pa),
the precursor material (decomposition kinetics and ligands determining the impurity
level), the partial pressure of the precursor (determining the production rate and
particle size), the temperature or power of the energy source, the carrier gas (mass
flow determining the residence time) and the reactor geometry.
Fig. 5.9
The nanoparticles are extracted from the aerosol by means of filters,
thermophoretic collectors, electrostatic precipitators or scrubbing in a liquid.
Modifications of the precursor delivery system and the reaction zone allow the
synthesis of pure oxides, doped oxides, coated nanoparticles, functionalized
nanoparticles and granular films.
5.9.7 Laser Ablation
LASiS is the acronym for Laser Ablation Synthesis in Solution and it is a
commonly used method for obtaining colloidal solution of nanoparticles in a variety
of solvents.
In the LASiS method, nanoparticles are produced during the condensation
of a plasma plume formed by the laser ablation of a bulk metal plate dipped in a
liquid solution. LASiS is usually considered a top–down physical approach.
In the past years, laser ablation synthesis in solution (LASiS) emerged as a
reliable alternative to traditional chemical reduction methods for obtaining noble
metal nanoparticles (NMNp).
LASiS is a ‘‘green’’ technique for the synthesis of stable NMNp in water or in
organic solvents, which does not need stabilizing molecules or other chemicals. So
obtained NMNp are highly available for further functionalization or can be used wherever
unprotected metal nanoparticles are desired.
B) i)Explain laser ablation
(8)
Laser Ablation
LASiS is the acronym for Laser Ablation Synthesis in Solution and it is a
commonly used method for obtaining colloidal solution of nanoparticles in a variety
of solvents.
In the LASiS method, nanoparticles are produced during the condensation
of a plasma plume formed by the laser ablation of a bulk metal plate dipped in a
liquid solution. LASiS is usually considered a top–down physical approach.
In the past years, laser ablation synthesis in solution (LASiS) emerged as a
reliable alternative to traditional chemical reduction methods for obtaining noble
metal nanoparticles (NMNp).
LASiS is a ‘‘green’’ technique for the synthesis of stable NMNp in water or in
organic solvents, which does not need stabilizing molecules or other chemicals. So
obtained NMNp are highly available for further functionalization or can be used wherever
unprotected metal nanoparticles are desired.
ii) Discuss any four applications in the nanotechnology
(i) Nanoparticles applications in medicine
(8)
Nanoparticles are used for early diagnosis of infectious disease. The
nanoparticles attach to molecules in the blood stream indicating the start
of an infection. When the sample is scanned for Raman scattering the
nanoparticles enhance the Raman signal, allowing detection of the
molecules indicating an infectious disease at a very early stage.
Researchers are developing a nanoparticle intended to make very early
detection of cancer tumors easier. When the nanoparticles attach to a
cancer tumors the nanoparticles release “biomarkers”, molecules called
peptides. The idea is that since each nanoparticle carries several peptides
a high concentration of these biomarkers will occur even at very early
stages of cancer, allowing early detection of the disease.
Quantum Dots (crystalline nanoparticles) that identify the location of
cancer cells in the body.
Nanoparticles that deliver chemotherapy drugs directly to cancer cells.
old nanoparticles that allow heat from infrared lasers to be targeted on
cancer tumors.
Porous silica nanoparticles used to deliver chemotherapy drugs to cancer
cells.
A nanoparticle cream that releases nitric oxide gas to fight staph infections.
Nanoparticles called micelles used to carry and deliver a chemotherapy drug
called paclitaxel to bladder cancer cells.
(ii) Nanoparticles applications in manufacturing and Materials
Silicate nanoparticles can be used to provide a barrier to gasses (for example
oxygen), or moisture in a plastic film used for packaging. This could slow
down the process of spoiling or drying out in food.
Zinc oxide nanoparticles can be dispersed in industrial coatings to protect
wood, plastic, and textiles from exposure to UV rays.
Silicon dioxide crystalline nanoparticles can be used to fill gaps between
carbon fibers, thereby strengthening tennis racquets.
Silver nanoparticles in fabric are used to kill bacteria, making clothing
odour-resistant.
(iii)
Nanoparticles applications in Energy and Electronics
Silicon nanoparticles coating anodes of lithium-ion batteries can increase
battery power and reduce recharge time.
Semiconductor nanoparticles are being applied in a low temperature
printing process that enables the manufacture of low cost solar cells.
A layer of closely spaced palladium nanoparticles is being used in a
hydrogen sensor. When hydrogen is absorbed, the palladium nanoparticles
swell, causing shorts between nanoparticles. These shorts lower the
resistance of the palladium layer.
(iv) Nanoparticles applications in Environment
Researchers are using go ld nano par ticles embedded in a poro us
manganese oxide as a room temperature catalyst to breakdown volatile
organic pollutants in air.
Iron nanoparticles are being used to clean up carbon tetrachloride
pollution in ground water.
Iron oxide nanoparticles are being used to clean arsenic from water wells.