comprehensive physical chemistry solution for master degree level student of tribhuwan university...

7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S

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COMPREHENSIVE

SOLUTION OF PHYSICAL

CHEMISTRY

For master degree level students of Tribhuwan

University, Kathmandu, Nepal

2013SHUKRA RAJ REGMI

Central Department of ChemistryTribhuwan University Kathmandu Nepal


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Comprehensive Solution of Physical Chemistry..../1

Acknowledgement

It is an immense pleasure for me to present thiscomprehensive solution of Physical chemistry for master degree

level chemistry students of Tribhuwan University, with an aim to

cover up completely the recent developments of syllabus for the

first part examination of physical chemistry of TU Nepal. This

book focused to provide detail solution with effective method to

defense examination properly. Due to the lack of reference

materials, burdensome and complex mathematical problem,

pattern less question and many more difficulties makes the

physical chemistry very complicated and severe. To avoid such

problem I have introduced a non-profitable, error free guideline

for curious intellectual chemistry circle of our University. I hope

that this book will be a fruitful product for our students. I have

cordial appreciation and sincere gratitude to our respected Prof.

Dr. Kedar Nath Ghimire, Prof. Dr. Megh Raj Pokhrel, Prof. Dr.

Rameshor Adhikari, and Prof. Dr. Deba Bahadur Khadka fortheir tremendous instruction, support and encouragement in

writing this book.

Shukra Raj Regmi

CDC, TU, 2013

[email protected]


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Content

TU Question Solution

2057 3-38

2058 39-75

2059 ........... 76-98

2060 99-120

2061 121-130

2062 131-149

2063 ... 150-158

2064 159-172

2065 173-181

2066 182-190

2067 191-201


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TRIBHUVAN UNIVERSITY

Institute of Science and Technology

2057Master Level/I Year/Science Full Marks: 100Chemistry (Chem. 512) Pass Marks: 40

Physical Chemistry Time: 4 hours

Candidates are required to give their answers in their own words as far as

practicable.

Attempt any FOUR questions from Group A and any EIGHT

questions from Group B.Group A

Comprehensive Question [415=60]

1. Why quenching method cannot be used to study fast

reaction? Describe advantage and limitations of continuous and

stopped flow techniques use to study fast reaction.

Show that the reaction A B + C, First Order Forward (rate

constant K1) and second order reverse (rate constant K2) relaxes

exponentially for small displacement from equilibrium. Find anexpression for the relaxation time in terms of K1and K2.

When a chemical reaction has been allowed to proceed for a

certain time, and the composition is analyzed in leisure by trapping the

reaction intermediates is called quenching method. Generally

quenching of chemical reaction is made by the rapidly cooling,

complete consumption of one reactant, neutralizing of acid base

medium, altering the pH and by adding large volume of solvent on

reaction mixture, which are only possible for the reactions that areslow enough to monitor. Nevertheless most of the fast chemical

reactions are proceed within the rage from femtosecond to

nanoseconds. Hence, the time taken that to mix the reactants or to

bring them to a specified in comparison to the half life of the reaction,

an appreciable error will be introduced,hence the initial time cannot

be determined accurately. This is the time that takes to make a

measurement of concentration may be significant compared to the half

life. The time taken to quench the mixture takes the little reactionalready proceed which cannot be predicted exactly without error,

therefore quenching method cannot be used to study fast reaction.


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a. Advantage and Limitation of Continuous Flow Techniques

The continuous flow method was developed by Hatridge and

Roughton. In this method, two reagents are placed in a separate

container and driven through a special mixing chamber into an

observation tube partial reaction occurs and a steady state is set up. At

various points along the observation tube the composition of reaction

progress is determined by optical, thermal or spectroscopic method.

By measuring the reaction mixture at various distances, a

concentration vs. time curve is obtained from which the various

information of reaction rate can be determined. The advantages of this

method are as follows:

1. After partial reaction occurs at the mixing chamber steady state isset up, to which along the various point of observation tube the

composition of reaction progress is determined without any

difficulties.

2. Different reaction times can be selected by varying the flow ratealong the outlet tube.

3. The spectroscopic finger prints are not needed in order tomeasure the concentration of reactants and products.

Even of this significant analysis for the rate can be measured in

different time interval by different physical as well as chemical

method it has some limitation. They are:

1. Large number of reactants is necessary.

2. It is difficult to maintain the constant speed.

3. It is difficult to observe through spectrometer at small timeintervals.

b. Advantage and Limitation of Stopped Flow Method

In this method the reagent are mixed very quickly in a small

chamber fitted with a syringe instead of outlet tube. The flow ceaseswhen the plunger of the syringe reaches a stop and the reaction

continues in the mixed solution. Observation, commonly using

spectroscopic techniques such as UV absorption circular dichroism

and fluorescence emission are made on the sample as a function of

time. The method is more superior than continuous flow method

which has following advantages.

1. The technique allows for the study of reactions that occur on the

millisecond to second time scale.


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2. Small amount of reagent is sufficient to analyze the overall

reaction progress.

3. It is appropriate to study of small samples for many biochemical

reactions.

4. It is widely used to study the kinetics of protein folding andenzyme action.

Some limitation of stopped flow methods are:

1. It is less sensitive than continuous flow method.

2. Fluorescence, electrical conductivity and optical rotation are

convenient properties to measure such reaction but difficult to

handle.

3. High cost, and unappropriated for most of the fast reaction.

Given the reaction,

A

1

2

K

KB + C

Consider,

a0 b0 c0 initial concentration

a b c equilibrium concentration at certain

time

ae be ce final equilibrium

So the rate of reaction is,

1 2dx

K [A] K [B][C]dt

At equilibrium, dx

0dt

1 e 2 e e

K a K b c 0 ___________________ (1)

Let the displacement of the species at timetbe x.From the new equilibrium then,a = ae+x, b=bex, c=cex

So the rate of relaxation at timetis

dx

dt= 1 e 2 e eK (a x) K (b x)(c x)

= 2

1 e 2 e e 1 2 e 2 e 2K a K b c (K K b K c )x K x ____ (2)

Since x is very small, K2x2 0, and K1aeK2bece= 0 [From equation (1)]


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Then,

1 2 e 2 e

dxK K b K c x

dt ____________ (3)

or, 1 2 e 2 edx

K K b K c dtx

On integration,

1 2 e 2 e

ln x K K b K c t c __________ (4)

Applying boundary condition when t = 0, x = x0, 0ln x c

0

1 2 e e

xln K K b c t

x _____________ (5)

Equation (5) shows that rate of relaxation depends upon two rate

constants then

1

2

KK

K

When 0x

ln 1x

, t =

**and, 2 e eK K (b c ) 1

or,

2 e e

1

K K (b c ) _________________ (6)

The rates of relaxation process are defined by a relaxation time ( )which is the time at which 0ln x / x equal to 1.

Here 0x

ln 1x

, 0x

ex

, 0

x 1

x e

1

0.3679 36.79%2.718

i.e. at the relaxation time the system has gone 36.79% onthe way to the

new equilibrium position.From equation (5)

** 0

1 2 e e

xln K K b c t

x

or, 1 2 e e0

xln K K b c t

x

**or, 1 2 e e

K K b c t

0

xe

x


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or,

t

0x x e __________________________ (7)

Equation (7) shows that for small displacement from equilibrium. This

reaction relaxes exponentially and equation (6) gives the expression

for the relaxation time.2. Deduce the Born expression for the enthalpy of ion-solvent

interaction.

Born expression suggests a simple thought process for

calculating the free energy change ( ES

G ) for ion solvent interaction

i.e. the work of transferring an ion from vacuum into the solvent by

using a thermodynamic cycle. The basic idea behind which is the law

of conservation of energy.

ESG = Work of discharging in vacuum + Work of charging insolvent

Born model has yielded the free energy change resulting from the

transfer of ion from a vacuum to a solvent. The important derives from

the fact that systems in nature try to attain a state of minimum free

energy. The Born expression provides the information that the ion

exist more stably in the solvent than in vacuum with considering

ES

G negative. It also shows that all ions would be involved in ion

solvent interaction than be left in vacuum. The prediction for smallerthe value of ri(rigid ionic sphere of radius) and large the E (electric

force medium) then greater will be the magnitude of the free energy

change in negative direction. The Born expression also interlinks the

electrostatics with chemistry and reduced very complicated situation to

a simple one by the choice of a simple mode.

Before finding out experimental testing of Born theory it is

preferable to recover from the theoretical expression. For ES

G , the

enthalpy and the entropy changes associates with ion-solventinteractions, because it is the heat of ion solvent interactions, rather

than the free energy, which is obtained directly from the

experimentally measured heat changes to occur when ionic solids are

dissolved in a solvent.

From thermodynamics

G = HTS= E + PVTS where H=E+PV

Differentiate on both side,

or, dG = dE + PdV + VdPTdSSdT


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or, dG = TdSPdV + PdV + VdPTdSSdTor, dG = VdPSdT ___________________ (1)At constant pressure,

P

GS

T

_______________________ (2)

or, I S

P

d GS

T

___________________ (3)

or, I ST S

P

d GS

dT

3. Describe the free electron gas model of metal. Derive the

expression for electrical conductivity of metal on the basis of this

model. Point out the achievements of this model? Where did it

failed?

The density of silver is 10.5103kgm3. The atomic weight of

silver is 107.9. Assuming that each silver atom provides one

conduction electron, calculate the electron density. The

conductivity of silver at 200C is 6.8107ohm1m1. Calculate also

the mobility of electron in silver. [e=1.61019C]

Free Electron Gas Model of MetalAs early as 1900, Drude regarded a metal as a Lattice with

electrons moving through it in much the same way as the molecules of

ideal gas in container. The idea was refined by Lorentz in 1923. He

applied kinetic theory of gas and Maxwell Boltzmann distribution law

to the electron gas. So this theory is also known as Drude-Lorentz free

electron gas theory. The basic assumptions of this theory are:

1. Metal comprised a lattice of rigid spheres (positive ions),embedded in a gas of free valence electrons which could move in

the interstices.

2. There exist an electrostatic attraction between positive ions(kernels) and the electron gas cloud.

3. The collisions between electrons are elastic. Between collisionsthe electrons move in a straight line and are influenced by the

heavy positive ions.

4. The collisions are instantaneous events which abruptly alter the

velocity of the electrons.5. The electrons lose all the extra energy gain from the applied field

up to the collision.


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6. The average distance an electron travels between collisions (themean free path) is of order of distance apart of the fixed ions (the

lattice separations).

7. In absence of an electric field the electrons move in all possibledirections randomly and thereby the average thermal velocity of

all the electrons is zero. Thus, there is current flow in absence of

electric field. When an electric field is applied electrons move in

a potential gradient and produce a current in the metal which is

proportional to an applied voltage and explains the origin of

Ohm's law.

8. The motions of electrons obey Newton's law of motion.

9. The mutual repulsion between electrons is neglected.

10. The potential field due to positive ions is uniform everywhere.Expression for Electrical Conductivity of Metals

In an electric field, electrons lend to drift move in one direction

than in any other. Then the average increase in velocity of the

electrons, which is simply the drift velocity, dV is

dV a ________________ (1)

where, a = acceleration of electrons between collisions

= relaxation time or mean free time, the average timebetween scattering event.

From Newton's second law

F = am

F e E

am m

_________________ (2)

where, E = electric field

e = electronic chargem = mass of electron

From equation (1) and (2),

de E

Vm

__________________ (3)

If we multiply equation (3) by total charge per unit volume, ne, where

n is electron per cm3in the gas, we have

2

d ne Ene Vm

_______________ (4)


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The quantity on the left side is simply current per unit area or current

density J.

2ne

J Em

_________________ (5)

Equation (5) is the Ohm's law expressed in terms of current densityand electric field. Ohm's law in terms of current density and electric

field is,

J E ______________________ (6)

From equation (5) and (6)

2ne

m _____________________ (7)

This is required expression for electrical conductivity of metal.The electron mobility, is defined as the average drift velocity perunit electric field.

= dV

E

= e E 1

.m E

e

m

_____________________ (8)


ne ______________________ (9)

Equation (9) allows us to define electrical conductivity in terms of

electron density and electron mobility.

Achievements of Free Electron Gas Model

The classical free electron theory is highly successful inexplaining many properties of metals.

1. The high electrical and thermal conductivity of metals can beexplained by the consideration of free electron gas, which is free

to move anywhere inside the metallic lattice.

2. When electric field is applied electrons move in a potentialgradient and produce a current in the metal which is proportional

to an applied voltage. This explains the origin of Ohm's law.

3. It correctly predicted the magnitude of resistivity.


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4. The semi-quantitative agreement of Widemann-Franz law isanother success of this theory.

5. This theory successfully explains the lusture and opacity ofmetals.

Failure of Free Electron Gas ModelAlthough classical free electron theory successfully explains

some properties of metals, it fails to explain

1. Mean free path

The general expression for resistivity of metal according to

classical free electron gas theory is given by

2

m

ne _________________ (1)

The resistivity of copper at 200C is 1.69108ohmm and densityof free electrons, n = 8.51028. Thus

=2

m

ne

=

31

228 19 8

9.1 10

8.5 10 1.6 10 1.69 10

= 2.47 1014

sec.

But,

C

_______________________ (3)

or, = C = 2.471014 1.154 105

= 2.85nm

The experimentally observed value for is nearly ten times the

above value. Classical theory thus could not explain the large

variations in values.2. Molar heat capacity of electron gas

According to law of equipartition of energy, every free electron

has an average kinetic energy B3

K T2

, so that in 1kmol of metal, in

which there are N atoms and, therefore, N free electrons, assuming

that each atom contributes one valence electron to the electron gas, the

total energy of the electron is given by

B3V Nk T2

__________________ (4)


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The energy for 1 mole of electron is given by

B

3V k T

2 ____________________ (5)

We know,

BRkN

Differentiating with respect to T at constant volume,

V elC = BV

V 3 3R k N

T 2 2

_____ (6)

= 23 263

1.38 10 6.02 102

V elC is called the 'electron specific heat', when heat is suppliedto the material, the free electrons also absorb part of heat. That is,

molar specific heat = 12.5KJ/Kmol/K.

The electronic specific heat due to free electrons is about

hundred times greater than the experimentally predicted value. Since

the heat capacity of solid due to atomic vibration is 3R, free electrons

should make significant contribution to the total specific heat of metal.

We know, however, that at least at high temperature, Dulong and Petit

law (atomic weight specific heat = 6.4) holds good and total specificheat of a solid is given by 3R, this means that free electrons do not

contribute significantly to the heat capacity of metal. We can,

therefore, conclude that the law of equipartition and hence classical

Maxwell Boltzmann statistics cannot be applied to the free electrons in

a metal.

3. Wiedemann Franz law

Statement: At not too low temperatures the ratio of thermal

conductivity to the electrical conductivity is directly proportional tothe absolute temperature, with the value of the constant of

proportionality independent of the particular metal. For it to support

the picture of an electron gas.22 2 2

B B

2

k Tn / 3m k KT

3 ene / m

___ (7)

The Lorentz number L is defined as

KL

T ______________________ (8)


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and according to equation (7) should have value22

8Bk

L 1.12 103 e

wattohm/K2

For copper at 200C, the electrical resistivity and thermal conductivity

are 1.72108m and 386Wm1K1.

Now,

L =

K

T

= 7

386

293 5.81 10

= 2.26 108WK2

This value of Lorentz number does not agree with the valuecalculated from the classical formula given in equation (8). Thus the

classical assumption that all the free electrons of a metal participate in

thermal conduction is not correct.

4. The observed electronic contribution to magnetic susceptibility isabout 1% of the predicted one.

5. This theory fails to explain ferromagnetism, superconductivity,photoelectric effect, Compton Effect and black body radiation.

6. According to this theory, is proportional to T. Butexperimentally it was found that is proportional to T.

7. In classical theory, the collision time C is considered to be

equal to relaxation time . This is only possible when we consider the

collisions are perfectly elastic and the moving directions is perfectly

random i.e., after each collision the electron had no memory of what

went before. However it is possible that the scattering is very weak in

which case such as assumption is unrealistic.

4. How are different thermodynamic parameters related with

partition function? Describe the significance of partition function.

The canonical ensemble partition function, Z is defined as

Z = iE

expkT

______________ (1)

Where,Ei= Energy of a system


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k = Boltzmann constant

T = Temperature on absolute scale

The partition function is related with different thermodynamic

parameters as:

1. Energy and partition functionLet us consider an ensemble to assemblies N. The average

energy of the system is given by

i i

i

N EE

N

______________ (2)

But, Boltzmann distribution law for such ensemble is written as

ii 0

EN N exp

kT

i0 i

i0

EN E exp

kTEE

N expkT

__________ (3)

Now, on differentiating equation (1) with respect to T at constant V,

we get

ii2

V

EZ 1 E expT kTkT

or, 2iiV

E ZE exp kT

kT T

From equation (3),

2

V

ZkT

TE

Z

or, 2

V

lnZE kT

T

_____________ (4)

2. Heat capacity CVand partition function

We have,

VV

EC

T

________________ (5)


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2

V

V V

ln ZC kT

T T

=2

2

2

V V

ln Z ln Zk 2T T

T T

This can also be written as

2

V 2 2

V

k ln ZC

1T

T

____________ (6)

(See illustration 1 below)

3. Third law of thermodynamics and partition function

From second law of thermodynamics, we have

VTdS C dT

or,V

dTds C

T _________________ (7)

Now, entry change for finite process is given by

T T

V00 0

CdS S S dTT

or,

stnd

T2

00

12 function

function

1 ln ZS S . kT dT

T T T

___ (8)

Integrating by parts we have

T2 2

0 20V V

1 lnZ 1 lnZS S kT kT dT

T T TT

(See illustration 2 below)

T

00

V V

lnZ lnZS S kT k dT

T T

= T TT 0V

lnZkT k lnZ

T

0V

lnZ ZS S kT klnZ klnT T 0

___ (9)


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On comparing temperature independent terms on both sides of

equation (9) we get,

0

ZS k ln 0

T

But, from third law of thermodynamics,at T = 0, S = 0

Z

k ln 0T 0

Thus, from equation (9)

V

lnZS kT k lnZ

T

ES k ln ZT

______________ (10)

4. Helmholtz free energy and partition function:

Helmholtz free energy, A is given by

A = ETSSubstituting the expression for E and S we have

2

V V

ln Z ln ZA kT T kT k ln Z

T T

A kT lnZ _________________ (11)

5. Enthalpy and partition function

We have,

H E PV

Also,T

AP

V

Substituting value of A, we get

T

lnZP kT

V

2

V T

lnZ lnZH kT kT V

T V

V T

lnZ lnZH kT T V

T V

_____ (12)

6. Gibb's free energy and partition function


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Gibb's free energy, G is given by

G H Ts E PV TS A PV

Substituting the expressions for A and P, we have

T

lnZG kT lnZ kT V

V

T

lnZG kT V ln Z

V

______ (13)

7. Chemical potential ( ) and partition function

The chemical potential for any component in a mixture is

given by

j

i T,Vi i iT ,V,n T ,V

A lnZkT lnZ kTn n n

____ (14)

Where ni= no. of moles of ithtype of species.

Also, we have, for ideal gas.NN!Z z

For ithtype of molecules, we havenin !i

iZ z i i i i iln Z n ln z n ln n n (Stirling approximation)

Differentiating with respect to niat constant T and V, we get

i ii i

i i iT,V T,V

ln z nlnZn ln z ln n 1

n n n

For ideal gas,

i

i T ,V

ln z 0n

i

i iT ,V

zlnZln

n n

Now from equation (14)

ii

i

zkT ln

n

___________ (15)

Significance of partition function


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The partition function summarizes in a convenient mathematical

form as how the energy of a system of molecules is partitioned among

the molecules. It has following significances:

1. It is a pure dimensionless quantity

2. It provides a bridge to link the microscopic properties ofindividual molecules such as their energy levels, moments of

inertia etc. with the macroscopic properties like entropy, heat

capacity etc. of a system of a large no. of molecules.

3. Its value depends on the molar mass, the temperature, the molar

volume, the inter nuclear distances, the molecular motion and the

intermolecular forces.

4. From Boltzmann distribution law

0 ii i

0

N EN g expg kT

_________ (16)

Since,iN N

0 0ii

0 0

N NEN g exp z

g kT g

0

0

N N

g z

Now, from equation (16)

iii

Eg exp

N kT

N z

Thus,

00 gN 1

N z z

, if g0= 1

Since, 0N

N is the mole fraction of the molecules occupying the

ground state, hence the partition function is equal to the

reciprocal of the mole fraction of the molecules in the ground

state.

Or,0

Nz

N


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5. Since,0

N1

N , hence z is always either equal to 1 (at absolute

zero N0=N, and only one energy level is accessible) or greater

than one. At higher temperature z is much larger than unity

because only a fewer molecule occupy the ground state ofenergy.

6. It describes the mode of distribution of molecules in the various

energy levels.

7. The partition function applies to a system of any physical state

and is calculated from the data obtained from quantum

mechanics and spectroscopy.

Illustration 1:

21V VF V V

1lnZ lnZ 1lnZ T .T Td T T

2

1VT V

ln Z ln ZT

d T

or,

2 Q2T2 V

2 1 11 V T TT VV V

lnlnZ lnZ T

TT

= 1 12 2T 2 T

2

V VV V

ln Z T ln Z. T

T T TTT

=2

2

2 2 2V

ln Z 2T ln Z 1. T

T T T T

2 23 4

2 21V VT V

lnZ lnZ lnZ2T T

T T

2 22

2 2 21V VT V

k lnZ lnZ lnZk T

TT T

Illustration 2:

We have,


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duu.vdx u vdx . vdx dx

dx

Thus,

T2

0V

uv

1 ln Z

kT dTT T T

= 1T T T2 2

0 0V V

1 ln Z ln ZkT dT . kT dT dT

T T T T T T

=T

2 2

20V V

1 ln Z 1 ln ZkT .kT dT

T T TT

5. Discuss vibrational rotational spectra of diatomic molecules.

What are P, Q and R branches of rotational spectra? Explain why

CO2molecule is microwave inactive but IR active.

When sufficiently excited, a molecule can vibrate as well as

rotate. The vibratory motion of the nuclei of a diatomic molecule can

be represented as the vibration of a sample harmonic oscillator, i.e. an

oscillator in which the restoring force proportional to the displacement

in accordance with Hook's law. For such as oscillator wave mechanics

shows that the vibrational energy EV is related to the fundamentalvibrational frequency

0 by the relation

V 0

1E v h

2

_________ (1)

Here v is the vibrational quantum number, which may take on the

values v=0, 1, 2 .... etc. Equation (1) reveals that such an oscillator

retains in its lowest vibrational level v=0 the energy of the oscillator,

cannot be removed from the molecule even by cooling below 00

K.According to Born-Oppenheimer approximation, a diatomic

molecule can execute rotations and vibrations quite independently.

Under these circumstances the energy levels of a diatomic molecule

are given by

v J v v' J J'E E E E E E E

or, 2

0 0 2

1 1 hE h v h v' h J J 1 J' J' 1

2 2 8 I

Since, J J J' 1


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2

0 2

hE h v v' h J

4 I

_____ (2)

And hence

0 2

h

v J4 I

_____________ (3) [Wherev v v'

]

Since, EVis very much larger than EJ, even at high temperatures

only vibrational states corresponding to v=0 and v=1 are excited.

The selection rules for vibrational rotational spectra of diatomic

molecules is v 1, 2 etc and J 1 . Thus v=0 v=1 transitions

fall into two categories.

i) P Branch:

Here

J 1 (i.e. JJ1)

P 0 2h

J 1 J J J 14 I

or, P 0 2h

J4 I

_____________ (4)

ii) R Branch:

Here

J 1 (i.e. JJ+1)

R 0 2h

J 1 J 2 J J 14 I

or, R 0 2h

J 14 I

__________ (5)

There are no lines at 0 (the Q branch) because transitionsfor which J 0 are forbidden.

The spacing between the lines in both the P and R branch is

2

h

4 I

Hence, by measuring the frequencies of these lines, the moment

of inertia of the molecule and the length of chemical bond can be

calculated.


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For a molecule to be microwave active and to give rotational

spectrum, the dipole moment of molecule should change during

rotation. CO2molecule is linear molecule with no net dipole moment

and there will be no change in dipole moment during rotation hence no

interaction with radiation. Thus, CO2molecule is microwave inactive.

For a molecule to be IR active and to given vibrational spectrum,

the vibration of molecule should be such that it asymmetric and can

change the dipole moment during vibration (i.e. vibration can interactwith radiation). CO2molecule is a linear molecule with no net dipole

J = 4

J = 3

J = 2

J = 1

J = 0

J = 4

J = 3

J = 2

J = 1

J = 0

J=+1J=1

v = 1

v = 0

P branch R branch

Q Branch

Fig.: Vibrational-rotational spectrum

of diatomic molecules


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moment but during the vibration its dipole moment changes and hence

CO2molecule is IR active.

6. What are partial molar quantities? Derive Gibbs Duhem

equation and describe its physical significance.

The partial molar quantity of any component i is defined as therate of change of property with change in the no. of moles of the

component when the temperature, pressure and no. of moles of other

components of the system are constant i.e. for any extensive property

Y it is denoted by Yi,m(or iY ) and given by

j

i,mi T,P,n

YY

n

In other words, partial molar quantity is the increase in the totalproperty when one mole of that component is added to large excess of

another substance at constant T and P. This may also be interpreted as

the contribution of a component to the total property of the system.

Partial molar quantities Symbol Defining equation

Partial molar volumeVi,m

ji T,V,n

V n

Partial molar energy Ei,m ji T,V,nE n

Partial molar enthalpy Hi,m j

i T,V,nH n

Partial molar entropy Si,m j

i T,V,nS n

Partial molar free energy Gi,m i j

i T,V,nG n

Partial molar work function Ai,m ji T,V,nA n Derivation of Gibbs Duhem equation

For multicomponent system, Gibb's free energy G is the function

of T, P and amount (ni) of various species. For a binary system

1 2G f T ,P,n ,n

1 2 1 2

1

1 2

P,n ,n T ,n ,n 1 T,P,n

G G GdG dT dn dn

T P n


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or,

1 2 1 2

1,m 1 2,m 2P,n ,n T,n ,n

G GdG dT dP G dn G dn

T P

___ (1)

where, j

i,m i

i T,P,n

GG j i

n

i = Chemical potential of component in the system.

For a closed system dn1= 0, dn2= 0 and equation (1) gives

i

i i

nP,n T,n

G GdG dT dP

T P

____________ (2)

But, we have

dG SdT VdP _____________ (3)

Comparing coefficients of equation (2) and (3), we have

iP,n

GS

T

;

iT ,n

GV

P

_____ (4)


i 1 2 2dG SdT VdP dn dn ____ (5)

At constant T and P, dT =0 and dP = 0

1 1 2 2dG dn dn ____________ (6)

On integration for a constant composition, we get

1 1 2 2G n n _______________ (7)

or, 1 1 1 1 2 2 2 2dG dn n d dn n d ___ (8)

Since, G is a state function equation (5) and equation (6), On equating

give2

i i

i 1SdT VdP n d 0

________ (9)

Equation (9) is Gibbs Duhem equation.

Physical significance of Gibbs Duhem Equation

1. Gibbs Duhem equation suggests that the partial molar quantities

cannot change independently of each other as for isothermal and

isobaric process 1 1 2 2n d n d . If 1 increases for some reason

2,mY must decrease accordingly.


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2. The molar quantities are not the function of no. of moles but the

partial molar quantities are functions of no. of moles of

component.

3. Gibbs Duhem equation is valid for all compositions. Howeveri

change with composition of solution and hence they must bemeasured separately for each composition.

4. Partial molar quantities are intensive (intrinsic) properties. Thus

Gibbs Duhem equation relates extensive properties with

intensive properties.

Group B

Short Answer Questions [85=40]

7. Starting from Michaelis Menten equation2 0 0

K [K] [S]V=

Km+[S]

Derive Eadie-Hofstee equation and describe haw will you evaluate

the Michaelis constant and maximum velocity of enzymolysis

From Eadie-Hofstee plot.

Here, 2 0 0K [K] [S]

V

Km [S]

2 0 0

Km [S]1

V K [E] [S]

or,0

2 0 2 0

1 Km 1 1. Since[S] [S]

V K [E] [S] K [E]

Which is Lineweaver- Burk equation

or,

1 km 1 1

.V Vmax [S] Vmax

or,1 km

Vmax VV [S]

This equation is the Eadie-Hofstee equation. A graph can be plot V/[S]

Versus V which shows the slope and intercept. From where we can

calculate maximum velocity of enzymolysis, which has the advantage

that it tends to spread but the points to a greater extent than the Line

weaver Burk plot.


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8. Show that in step growth polymerization the molar mass of

the product increases with reaction time.

It is define as the average no. of monomer residues per polymer

molecules this is the ration of the initial concentration of end groups,

at the time of interest. It is represented by or n

No Co Co 1

N C Co(1 P) 1 P

1

1 P

____________(I)

Consider a hydroxyl acid HO M COOH , we can consider the

formation of polyester from such monomer and measure its progress in

terms of the concentration of the -COOH groups in the sample (which

is denoted 'A') for these groups gradually disappear the reaction canoccur between molecules containing any number of monomer unit

chains of many different length can grow in the reaction mixture.

In the absence of catalyst, over all order of polymerization is

second order.

or, d[A]

K[OH][A]dt

or, 2d[A] K[A]d t

Since[A] [OH]

or, 2

d[A]Kdt

[A]_________(I)

On, Integrating 1

Kt const[A]

Boundary condition t=o, [A]=[A]0

0

1 1

Kt[A] [A]


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0

0

[A][A] Kt

1 [A]__________(II)

The extent of polymerization p= 0

0

[ A] [ A]

[A]

=

0

0

[A]

1 [A] Kt

Now, for degree of polymerization n = 0[A]

[A]

=

0

0 0

[A]

[A] / 1 [A] Kt

0n 1 [A] Kt_________(III)

The average length grows linearly with time, Therefore, the

longer a step growth polymerization proceed higher the average molar

mass of the product.

Let P be the probability that a molecular group e.g., [-COOH] has

formed link with another group e.g., [-OH] then at starting point P=0,

and at the end P=1. The average no. of monomers linked together in apolymer that has been formed by stepwise Polymerization increases as

links are formed i.e. the average no. of monomers linked together

increases as the probability that a monomer has not formed a link

decrease this probability is (1P), so we can conclude that1

or n1 P

a) if P=0, n =1, indicating no polymerization.

b) if P=1, n = , indicating the requirement of P 1. to obtain

polymer with high molar mass.

c) As P grows with time, also grows with time

n or =1+Kt[A]

So number average molar mass increases with time as

Mn= n Mmonomer=(1

kt [A])Mmonomer.

9. What is meant by degree of polymerization? Calculate the

number average degree of polymerization of an equimolar


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mixture of hexamethylene diamine and adipic acid for the extent

of reaction 0.995.

10. How electrical double layer is formed? Briefly describe the

structural models for the interfacial region.

Electro chemical measurement involves heterogeneous system because

an electrode can only donate or accept electrons from a species that is

present in a layer of solution that is immediately adjacent to the

electrode. This layer may have a composition that differs significantly

from that of the bulk of the solution.

Consider the structure of the solution immediately adjacent to an

electrode when a positive potential is first applied to that electrode.

Immediately after impressing the potential, there will be a momentary

surge of current which rapidly decays to zero if no reactive species ispresent at the surface of the electrode. This current is a charging

current that creates an excess of negative charge at the surface of the

two electrons. As a consequence of ionic mobility, however the layers

of solution immediately adjacent to the electrodes acquire on opposing

charge. The surface of the metal electrode is shown as having an

excess of positive charge as a consequence of an applied positive

potential. The charged solution layer consists of two parts

1. Compact inner layer (d0 to d1), in which the potential decreaseslinearly with distance from the electrode surface.

Fig. (a)

Electrode

+++++++++

20

300A

o

d0

d1

d2

d

Bulk ofsolution

d0

d1

d2

0

Fig. (b)

d

Fig.: Electrical double layer formed at electrodesurface as a result of an applied potential.


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2. A diffuse layer (d1 to d2), in which the decrease is exponential.

This assemblage of charge at the electrode surface and in the solution

adjacent to the surface is termed an electrical double layer.

The structural model for interfacial region arose from the work of

Helmholtz and Perrin. They thought that the charge on the metal

would draw out from the randomly dispersed ions in solution a

counter-layer of a charge of an opposite sign. In this way the

electrified interface will consist of two sheets of charge, one on the

electrode and other in the solution as shown in figure, the term double

layer is arise. The charge densities on the two sheets are equal in

magnitude but opposite in sign, exactly as in a parallel -plate

capacitor. The drop in potential between these two layer of charge is,

then a linear one

11. Explain the terms diffusion current and half wave potential

with their importance in polarographic analysis.

The current which is proportional to the concentration of

depolarizer or active species and helps to appreciate the quantitative

analysis of polarography is called diffusion current.

The limiting current is the sum of the diffusion current and

migration current, in case other interfering factors are not present. But

in real practice migration current is eliminated and hence the diffusion

current becomes the limiting current.

Different current can be explained by the Ilkovic equation.2 11

3 62Iav 607n m t c

Where = Diffusion coefficient

t= time

c=concentration of the solution

n=no. of electrons involved


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Fig: A typical polarogram showing half wave potential and diffusion current

Half-wave potential E1/2

The potential at which at which the current is equal to one half of the

limiting current is called half wave potential.It is closely related to the

standard potential of the half reaction but not identical to that

constant.the current of current-voltage curve is cathodic current

resulting from reduction and is consider to be +magnitude, the height

of the curve i.e wave height is called diffusion current and potential

corresponding to the half of the diffusion current and potential

corresponding to the half of the diffusion current is called half wave

potential.The potential does not depend on the concentration of thereacting materials but it is characteristic of the nature of reacting

materials.The half wave potential is useful for the identification of the

component of solution by measureing the half wave potential of

different component present in solution.the potential at a polarised

electrode obeys the Nernst equation and the concentration of electro

active species is directly related to the current this allows one to relate

the potential and the current to each other via the equation.

1 2RT Id IE E lnnF I


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Which at 250c takes the form

1 2

0.0591 Id IE E log

n I

Where I is the measured current at any point (minus the residual

current)This equation sometimes known as the equation of thepolarographic wave, clearly shows that for a reversible reaction when

I Id 2 the measured potential is 1 2E . A graph of E versus

log Id I I should produce a straight line with slope -0.0059I/n,

thus enabling the number of electrons taking part in the reaction to be

determined. It also show why reactions involving more than one

electron given sharper polar graphic wave than single - electron

processes. Notice that the equation predicts1 2

E is concentration

independent, however several criteria must be satisfied if the equation

is to be useful firstly since it is derived from the Nernst equation it will

only apply to reversible professes. Also it is necessary to correct both I

and Id for residual currents and the correct the measured potential for

any IR drop in the cell. The supporting electrolyte in paleographic

experiments raises the overall conductance of the solution thus

reducing R. Although the half wave potential is independent of

concentration in the given cell it does depend on the exact nature of

the reacting species and in most cases it will not have exactly the same

value as the standard electrode potential of the species E . The most

common situation is where a free ion in solution is reduced to a metal,

which dissolves in mercury to form an amalgam. This is a reversible

reaction, but the product is thermodynamically stabilized when it

forms the amalgam so that E1/2 is not equal toE .

Few application of half wave potential1)plot of i/(id-i) Vs E applied for a polarographic wave provide

polarographic scan which is a way to determine the potential at which

an ion or molecule will reduced . Hence polarography is useful pilot

tchnique for stablishing the condition for a sucessful macroscale

controlled potential electrolysing using a mercury cathod.

2)Half wave potential for the series of compound reflect electronic

structure in a systematic manner to provide useful data on substituent

effect that correlate nicely with other molecular properties.

3)It is applicable to analysis include the determination of the smallqantity of metal ion like Cu,Zn,Fe,Cd etc.


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4)In favourable case ,two or more component can be determined with

a single polarogram hence E1/2 value must sufficiently different for

the individual wave to be distinguished.

12 How can you determine the concentration of an ion with

ion selective electrode? What is its limitation?

An electrode that generates a potential in response to the

presence of a solution of specific ions is called an ion selective

electrode. There not only collect the electrons but also take part in half

cell ran (While other noble metal electrodes does not take part in

electrochemical reaction but acts as collector and donor of electrons

only) e.g.

In rod responds to Zn++ion-so large no. of metals can acts as ion

selective electrode for their own ions. A glass electrode can be

responsive to Na+,K+and NH4+ions by being dropped with Al2O3and

B203.

An ion selective electrode consist of a membrane that response more

or less relatively to a single ionic species .It is in compound contact

with solution of the ion to be determinate one side and usually with a

solution having a fixed ionic activity on other side which is in contact

with a suitable reference electrode.An electrode pair may be dipped

into the solution of the substance to be determined and concentration

of the solution can be obtained from the potential observed.there are

mainlytwo types of electrode.

1)Non-crystalline membrane glass electrode.

2)Crystalline membranes.

Non-crystalline glass electrode are useful for pH measurement silicateglass for H+,Na+ and for cations other than protons.While crystalline

membrane glass electrode useful for single crystal LaF3 for F-, and

polycrystalline or mixed crystal Ag2s for s2- and Ag+.

Limitation of Ion-selective electrode:-

1. They measure activity rather than concentration of an ion.

2. Interference from ionic strength effect is possible.

3. On complication or prolongation interference can occur.

4. They are rather selective but not always specific.5. They are PHlimited.


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6. Response is not very good for dilute solutions.

7. Periodic calibration is a must to get good results.

8. Impurities in solutions cause serious errors.

14. Derive thermodynamically an expression relating the value ofmolal freezing point constant of a solvent with latent heat of

fusion.

The depression in freezing point of 1kg solvent by the addition of

1 mole of solute to it is called molal freezing point depression

constant.

We have,

Depression of freezing point

2f

f 2f

RT

T XH _______

For a dilute solution

2 2 2 2 12

1 2 1 1 1

n n n m MX

n n n W M 1000

____________ (2)

Where,

m2= molality of solution

W1= mass of solvent in gmM1= molar mass of solvent

From equation (1) and (2)2

f 2 1f

f

RT m MT

H 1000

=2

f 12

f

RT Mm

1000 H

or, f f 2T k m ________________ (3)

Where,2

f 1f

f

RT Mk

1000 H

= molal freezing point constant

fH = latent heat of fusion.


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15. From the following energy levels calculate the partition

function for system at 1000K.

Energy /102J 0.53 0.75 1.27 1.95 2.09 3.5

Degeneracy 2 1 3 3 1 1

What fractions of the molecules are in lowest energy level?Molecular partition function is given by

i iz g exp E kT Where,

gi= Degeneracy

k = Boltzmann constant = 1.381023JK1

T = 1000K

Thus,

20 231z 2exp 0.53 10 1.38 10 1000 1.362

20 232z 1exp 0.75 10 1.38 10 1000 0.58

20 233z 3exp 1.27 10 1.38 10 1000 1.195

20 234z 3exp 1.95 10 1.38 10 1000 0.73

20 234z 1exp 2.09 10 1.38 10 1000 0.22

20 235z 1exp 3.5 10 1.38 10 1000 0.08

z = z1+ z2+ z3+ z4+ z5

= 1.362 + 0.58 + 1.195 + 0.73 + 0.22 + 0.08

z = 4.16

Now,

Fraction of molecules of ground energy level = 1z

z

=1.362

24%4.16


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16. Assume that the energy of two particles in the field of each

other is given by

r 9a b

Ur r

Where a and b constants, show that in a stable configuration theenergy of attraction is 9 times the energy of repulsion.

Let us consider the attractive energy between particles in the

stable configuration is a0

a

r and the repulsive energy is r9

b

r .

9 8

a 0 0

r 0

r ara.

r b b

_____________ (1)

But,

r 9

a bU

r r and

0

r

2 100 0r r

dU a 9b0

dr r r

or,2 10

0 0

a 9b

r r

Thus,1

80

9br r

a

_____________ (2)

From equation (1) and (2),1

.88a

r

a 9b9

b a

a r9 _________________ (3)

Hence, energy of attraction is 9 times the energy of repulsion.

17. What are point defects? Explain with a neat sketch how the

cation impurity of higher valency creates cation vacancy in the

crystal lattice?

Any deviation from a chemically pure, stoichiometric, 'perfect'crystal is called an imperfection or defect. The localized defective

regions of the crystal that are confined to a volume which is of atomic


37/201


dimensions are called point defects. Foreign atoms, vacant lattice sites

and extra or missing electrons are examples of imperfections

belonging to this class.

At absolute zero, crystals tend to have a perfectly ordered

arrangement. As the temperature increases, the amount of thermal

vibration of ions in their lattice sites increases, and if the vibration of a

particular ion becomes large enough, it may jump out of its lattice site.

This constitutes a defect. The higher the temperature, the greater the

chance that lattice sites may be unoccupied. Since the no. of defects

depends on the temperature, they are sometimes called thermodynamic

defects.

Cation vacancies are present in alkali halides containing in

additions the divalent (or higher valent) elements. If crystal of KCl is

grown with controlled amounts of CaCl2, the density varies as if a K+

lattice vacancy were formed for each Ca2+ion in the crystal. The Ca2+

enters the lattice in a normal K+site and the two Clions enter two Cl

sites in KCl crystal. Demand of charge neutrality results in a vacant

metal ion site.

18. What is meant by chemical shift? Why tetramethylsilane is

used as reference in NMR spectroscopy?

The frequency at which proton absorbs depends on the magnetic

field which that proton feels and this effective field strength is not

exactly the same as the applied field strength. The effective field

strength at each proton depends on the environment of that proton e.g.

electron density at the proton and presence of other near by protons.

Compared to naked proton, a shielded proton requires a higher

applied field strength and a deshielded proton requires a lower applied

Fig.: Production of lattice energy vacancy bythe solution of CaCl

2in KCl: to ensure electrical

neutrality a positive ion vacancy is introduced intothe lattice with each divalent cation Ca++.

+ Ca++

+

+ +

+ +

+

+


38/201


field strength to provide the particular effective field strength at which

absorption occurs. Shielding thus shifts the absorption upfield, and

deshielding shifts the absorption downfield. Such a shift in the

position of NMR absorptions arising from shielding and deshielding

by electrons circulating about the proton itself, are called chemical

shift.

Tetramethylsilane, (CH3)4Si(TMS) is used as a reference in NMR

spectroscopy due to following reasons:

1. Tetramethylsilane (TMS) has 12 H-atoms and therefore, a very

small amount of it gives a relatively large signal.

2. Because all hydrogen atoms are equivalent they give single

signal.

3. Since Si is less electronegative than carbon, the protons of TMSare in regions of high electron density. They are as a result high

shielded, and the signal from TMS occurs in a region of the

spectrum where few other H-atoms absorb as a result, most NMR

signals appear in the downfield direction from TMS.

4. TMS is relatively inert.

5. It is volatile (B.pt. 270C), after the spectrum has been determined

the TMS can be removed from the sample easily by evaporation.


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TRIBHUVAN UNIVERSITY

Institute of Science and Technology

2058Master Level/I Year/Science Full Marks: 100Chemistry (Chem. 512) Pass Marks: 40

Physical Chemistry Time: 4 hours

Candidates are required to give their answers in their own words as far as

practicable.

Attempt any FOUR questions from Group A and any EIGHT

questions from Group B.

Group AComprehensive Question [415=60]

1. Distinguish between step growth and chain polymerization

reaction with respect to their mechanism and molecular weights of

the Polymers. Discuss the kinetics of step growth polymerization

and show that the degree of Polymerization is linear function of

time. It takes 10 min. to achieve a degree or Polymerization of 100

units. How long will take to achieve a degree of polymerization of

200 units?Step Growth Polymerization:

Step growth polymerization involves the condensation of

monomer units and the elimination of small molecules usually water.

The repeating unit of condensation polymer has not same composition

to that of monomer. Thus the molar mass is not the integral multiple of

the monomer. The molar mass of polymer builds up slowly and the

average molar mass of the polymer builds up slowly and the average

molar mass of the polymer grows with time the various steps may beMonomer + Monomer Dimer

Dimer +Dimer Tetramer

Monomer + Dimer Trimer and so on.

Some examples are

1. Formation of polyamide e.g. nylon - 66.

2nH o2 2 2 2

2 2 4

nH N (CH ) NH nHooC (NH )6 CooH

H[NH (CH )6 NHCo(CH ) CooH


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Nylon polyamide

2. Formation of poly. ethylene terphthalate

2 2nHocH CH OH nHO c

Kinetics of Step Groth polymerization

From above mechanistic scheme, it is clear that in step growth

polymerization any two monomers in the reaction mixture can link

together at any time and growth of the polymer is not confined to

chains that are already forming. The longer a stepwise polymerization

proceeds, the higher the average molar mass of the product. Consider a

hydroxyl acid HO M COOH , we can consider the formation of

polyester from such monomer and measure its progress in terms of the

concentration of the -COOH groups in the sample (which is denoted'A') for these groups gradually disappear the reaction can occur

between molecules containing any number of monomer unit chains of

many different length can grow in the reaction mixture.

In the absence of catalyst, over all order of polymerization is

second order.

or, d[A]

K[OH][A]dt

or, 2d[A]

K[A]dt

Since[A] [OH]

or, 2

d[A]Kdt

[A]_________(I)

On, Integrating 1

Kt const[A]

Boundary condition t=o, [A]=[A]0

0

1 1Kt

[A] [A]

0

0

[A][A] Kt

1 [A]__________(II)

The extent of polymerization p= 0

0

[A] [A]

[A]

=

0

0

[A]

1 [A] Kt


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Now, for degree of polymerization n = 0[A]

[A]

=

0

0 0

[A]

[A] / 1 [A] Kt

0n 1 [A] Kt_________(III)

The average length grows linearly with time, Therefore, the



Chain polymerization:-

In chain polymerization an activated monomer attacks another

monomer. links to it then that unit attacks another monomers and soon. The monomer is used polymers are formed readily and only the

yield not the average molar mass, of the polymer is increased by

allowing long reaction time.Chain polymerization occurs by addition

of monomers to a growing polymer, often by a racial chain process. It

results in the rapid growth of an individual polymer chain for each

activated monomer. Examples include the addition polymerization of

ethene, methyl methacrylate and styrene as in-

0 02 2 2 2CH CHX CH CHX CH CHXCH CHX

There are three basic types of reaction step in a chain polymerization

process

a. Initiation

Where I is the initiator, R the radical I forms, and M1 is a

monomer radical. We have shown a ieaction in which a radical is

produced, but in some polymerizations the initial step leads to the

formation of an ionic chain carrier. The rate determining step is theformation of the radicals R by homolysis of the initiator, so the rate of

initiation equal to the V1given above.

0 0I R R

0

01M R M

b. Propagation

If we assume that the rate of propagation is in dependent of chain size

for sufficiently large chains, then we can use only the equation givenabove to describe the propagation process. Consequently for


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sufficiently large chains, the rate of propagation is equal to the over all

rate of polymerization.

Because this chain of reaction propagates quickly, the rate at

which the total concentration of radicals grows in equal to the rate of

the rate determining initiation step. It follows that

.

Pr oduction

d Mzfki I

dt

o0

1 2M M M

00

2 3M M M

00

n 1 nM M M

c. Termination

0 0

n m nM M M Mmutualter mination

o o

n m n mM M M M disproportionnation

00

n nM M M M chain transfer

In mutual termination two growing radical chain combine. In

termination by disproporationation a hydrogen atom transfers fromone chain to another corresponding to the oxidation of the donor and

the reduction of acceptor. In chain transfer, a new chain initiates at the

expense of the one currently growing.

Here we suppose that only mutual termination occurs. If we

assume that the rate of termination is independent of the length of the

chain the rate law of termination is2

.Vt kt M __________(1)

and the rate of change of radical concentration by this process is.

2.

depletion

d M2kt M

dt

__________(2)

The steady state approximation gives

.

2.

d M2fki I 2kt M 0

dt

___________(3)

The steady state concentration of radical chains is therefore


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1 2

1 2. fkiM I

kt

__________________(4)

Because the rate of propagation of the chains is the negative of the rate

at which the monomer is consumed, we can write Vp d M dt Vp= kp[.M][M]

Vp= kp

1 2fki 1 2

I Mkt

__________(5)

This rate is also the rate of polymerization,

Assumptions

i. Radical concentration is determined by initiation and termination

steps

ii. Propagation step neither increases nor decreases the radical

concentration.

iii. The constant does not depend on the size of the radical.

From above mechanism, an activated monomer attacks another

monomer and links to it that unit attacks another monomer and so on.

The slower the initiation of the chain the higher will be the average

molar mass of the polymer.The average length grows linearly with time, Therefore, the



2. Discuss the Michaelis-Menten Mechanism of enzyme-

catalyzed action show that the enzyme catalyzed reaction is first

order and zero order with respect to substrate, s, at low and high

concentration of substrate respectively. Describe line weaver and

Bulk plot to evaluate Michaelis constant and maximum velocity of

the reaction.

The enzyme catalyzed conversion of substrate at 250C has

Michaelis constant 0.035 mol L1. The rate of the reaction is

1.15103 molL1S1when the substrate concentration is 0.110 mol

L1 . What is the maximum velocity of this enzymolysis?

L. Michaelis lies and M.L Menten (1913) proposed the

mechanism of enzyme catalyzed reaction which involves equilibrium

among the enzyme, the substrate and the complex such that a complex

is formed to give either the product or the substrate.


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1 2K K

E S ES P E

The rate of product formation according to this mechanism is

2

d[P]K [ES]

dt _____________ (1)

Similarly the rate of formation of ES complex is

1d[ES]

K E Sdt

K-1 [ES]K2[ES]

At steady stated[ES]

dt= 0

K1[E] [S]K-1[ES]K2[ES] =0_____ (2)we have, [E]0 = [E] +[ES]

because the substrate is typically in large excess relative to enzyme,

the free substrate concentration is approximately equal to the initial

substrate concentration and so, we can write [S] [S]0. So

[ES] =

1

1 2

K

(k k )[E] [S]

=[E][S]

km _________________ (3)

Where km= 1 2

1

K K

Kis Michaelis- Menten constant


2K

Km[E] [S] ________________ (4)

and [E]0= [E] +[ES]

[E]0 = [E] +[E][ES]

Km

or, E =

0E

[S]Km

Km

_________________ (5)

From equation (4) and (5 )

02

0

[E]K

Km [S]Km

Km

[S]0 Since [S] [S]0.


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or,

2 0 0

0

K [E] [S]

Km [S] ________________ (6)

Equation(6) is the Michaelis - Menten equation.

Case-I

Enzyme catalyzed reaction is zero order with respect to substrate

When[S]0 is very high

[S]0 Km

Then, Km + [S] [s]

So, equation (6) becomes

2 0 00

K [E] [S]

[S]

= K2[E]0 __________________ (7)

Case-II

Enzyme catalyzed reaction is 1storder with respect to substance

when [S]0 Km the rate is proportional to the [S]0

In this case, [S]0 + Km Km

2 0 0K [E] [S]

Km________ (8)

i.e. 0[S]

Here the relation is first order with

respect to substrate.

Line weaver and Bulk plot for

evaluate Michaelis constant andmaximum velocity of the reaction -

We can write the Michaelis Menten

equation

max

0

V

1 Km

[S]

_________________ (9)

By rearranging this expression in to a form this is a men able to data

analysis by linear regulation

Vmax

Substrate concentration [S]

Fig.: The variation of the rate of an enzyme-catalyzed reaction with substrate concentration.

The approach to a maximum rate Vmax. For large[S] is explained by the Michaelis-Menten mechanism.


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max max 0

1 1 Km 1

V V [S] ___________ (10)

From Figure,

Km =slope

intercept

A line weaver - Burk plot is a

plot of

1 against

0

1

[S] and

according to equation 10 it should

yield a straight line with slope of

max

Km

Vand a y intercept at

max

1

Vand

x intercept at1

Km. The value of Kb

is then calculated from y- intercept.

The final conclusion can we draw

from line weave - Burk plot is the

eqn 10 concurrent the required all

parameters that occur in the Michaelis - Menten Mechanism.

Given

Michaelis constant (Km) = 0.035 molL1

Rate of reaction ( ) = 1.15 10-3mol L1S1

Substrate concentration [S] = 0.110 mol L1

Maximum velocity ( max) = ?

we have,

max =

Km1

[S]

=

30.035 1 1.15 100.110

= 1.515103mol L1S1

Slope =K

M/ vmax

1/ KM

1/ [S] 00

1/ v

Fig.: A Lineweaver-Burk plot for thanalysis of an enzyme-catalyzedreaction that proceeds by a Michaelis-Menten mechanism and thsignificance of the intercepts and thslope.

1/ vmax


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3. Give an introductory account of various models proposed for

the structure of electrical double layer. Justify the reason for

proposing more than one model.

a. The Helmholtz Perrine model-

This is the simplest model which treats the double layer like aparallel plate condenser with a molecular distance apart. When a metal

electrode is brought in contact with solution, the charge on the metal

will drew out from the randomly dispersed ions in solution a counter

layer of charge of opposite sign. The electrified interface consists

therefore of two sheets of change one on the electrode and the other in

the solution hence the term double layer. The charge densities on the

two sheets are equal in magnitude but opposite in sign, exactly as in

the parallel plate capacitor.

This is the non structural, thermodynamic thinking and has

yielded information on the gross quantities present at the double layer.

But the relations are in terms of changes of the various experimental

conditions and quantities. The language is that of differentials.

Consider the Lippmann eqn i.e. the relation between charges of

surface tension and changes of cell potential.

v

=qm _________________ (1)

Experimentally, vs. V curve is parabolic can this be explained on

thermodynamic frame work.

Integrating eqn1

= qm dv ______________ (2)But is qm a function of v, we can't integrate without knowing this.

Since there are structural questions we can't solve thermodynamically,

this was done by Helmholtz and Perrine. The potential difference 'V'across the condenser is

v =q4 d

____________________ (3)

Where ddistance between plates

Dielectric constant based on the parallel plate model of thedouble layer.

dv = 4 d

dq M ______________ (4)


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Now, we can integrate lippmann eqnas

= m Mu

q dq

or, =4

1

2

q2m ____________ (5)

when, qm= 0, thenmax

=constant, So

=max

4

2

mq

2

or, =max

4 d

1

2V2 ____________ (6)

This is the equation of the parabola symmetrical aboutmax

Helmholtz-

Perrin model would be quite satisfactory for electrocapillary curveswhich are perfect parabola.

The double layer in Troble

Electrocapillary curve shows a marked deviation from perfect parabola

and show a marked sensitivity to the nature of the anion present in the

electrolyte.The asymmetry of electrocapillary curves has important

implications for the capacities of the double layers.For a typical

electrocapillary curve on differentiation ,a charge density (qm)

Vs.electrode potential(V) plot yield not a straight line but two,straightlines meting at q=0,Now differential capacity C is given-

C = dq/dV

So a plot of C Vs.V can be obtain by differentiation of the qm Vs.V

curve .It turns that when we starts from an asymmetric electrocapillary

curve the interface displays a differential capacity,which is not

constant with electrode potential.

A parallel plate condenser has a specific (i.e per unit area of

plate)which is given by rearranging equation (4) as

dq/dv =/4d = cIf E and d are taken constant this model predict a constant capacity

,but this is not the case .So it is appear that an electrified interface does

not behave like a simple double layer so this modelis too native an

approach.

Some drawback of the model

1)Thickness of the layer is small.2)It can not account for Dern effect or streaming potential.


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3)It can not explain the variation of differential capacitance with

dielectric constant.

b. The GouyChapman model

According to the Gouy chapman ,diffuse charge modeldifferential capacity of an electrical interface should not be a constant,rather than it could shown as inverted.GouyChapman thought toliberate the ions from a parallel sheet. Once they become free, the

behavior of ions in the vicinity of the electrode is affected by the

electric force arising from the electrode charge and by thermal Jostling

equilibrium between electric and thermal forces is attained and thus

also a time average ionic distribution. The technique of analysis of the

differ double layer proceeds along the same lines as in the theory of

the long range ion-ion interaction - considering whole electrode as

central ion i.e. the charged electrode will be enveloped by charge

cloud (consisting of an excess of ions of opposite charges). This ionic

atmosphere represents a falling off, with distance laming parallel to

the electrode and at increasing distance out into the solution. The

potential too will decay with distance, asymptotically setting down to

a constant value (zero) in the bulk of solution.

The corresponding field or gradient of potential at a distance x

from the electrode according to the diffuse- Charge model of Gouyand Chapman is given by the expression

0 0 x

0

8kTc zedysinh

dx 2kT

______(i)

Where,

C0= Concentration of the ithspecies in bulk solution

= dielectric constant of solution

= permittivity of free spacex

= outer potential difference x from electrodeThis equation (i) spells outs the relation between the electric field and

the potential at any distance x from the electrode.

The total diffuse charge in solution qdand now potential varies with

distance is given by Gauss's law is:-

0

dq

dx

___________(2)

to determine the total diffuse- charge density qd, the Gaussian boxshould extend from a place very close to the electrode x=0, then to a


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place deep inside the solution x Wherex

0 . Hence with these

conditions the total diffuse charge density scattered in the solution

under the interplay of thermal and electrical forces qd is given from

equation (1) and (2) is

1 2 0 00 zeqd 2 2 CokT) sin h 2kT ______(3)

Where0

is the potential at x=o to the bulk of the solution where the

potential is taken as zero.

The variation of potential with distance can be obtained from the

integration of equation (1) assuming 0 0 x(Ze )/(2kT) Ze /2kT

(Ze0x

)/(2kT)

In

1 22 2

0x

0

2Coz e x cons t an tkT

_____(4)

On integrating the equation (4) by taking the boundary x 0 ,

x 0 therefore

kT

x 0 e ______(5)

From above equation the potential decays exponentially as the

distance from the electrode increases further as the solution

concentration Co increases K increases and x falls more and moresharply this potential distance relation is an important and simple

result of Gouy Chapman model. It forms a valuable basis for thinking

about the interaction of the diffuse charges around what are called

colloidal particles.


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The diffuse layer is formed because as the distance from the

interface increases, the intensity of field gradually decreases and thecounter ions of double layer scatter. So a diffuse layer is formed but at

higher concentration the compact layer will be formed.

The C vs. V curve is inverted parabola. Hence according to thesimple diffuse change theory the differential capacity of an electrified

interface should not be a constant. This is of course a welcome result

++

+ +

Solution

OHP

Excess chargedensity of metal qm

Fig.: The excess charge density on the OHP is smallerin magnitude than the charge on the metal. Theremaining charge is distributed in the solution.

Excess charge densityon OHP = |qOHP|


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over HelmholtzPerrin model.Although the weakness of the parallelplate model has been the details of such dependence The main fact is

that the experimental capacity potential curves are just not the inverted

parabola, which the GouyChapman diffuse model predicts. In verydilute solution (


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Where,

C = total differential capacity of interface

CH= HelmholtzPerrin capacityCG = GouyChapman capacitySo, Helmholtz and Gouy regions do store ions and are

consecutive in a direction normal to the electrode and the total

potential difference is across the whole interface.

Linear variation

x =a x = 0

Fig.: The potential variation according to thin model

Fig.: A layer of ions stuck to the electrode

and the remainder scattered in cloud fashion

x =a x = 0

Scatterd ion

Two region of charge separation

+

+++++++++++ a

Fig.: The corresponding total differentialcapacity C is given by Helmholtz and GC model

CH 1C

= 1CH

1CG

+ CG

STERN MODEL


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The potential difference across the electrode solution interface can

given as

4.What are the forces contributing to the bonding of an ionic

crystal? Obtain an expression for lattice energy of an ionic crystal.

The repulsion potential exponent (n) is related to

compressibility as4

0

2

18rn=1+

kAe

Calculate the value of repulsive potential exponent and lattice

energy of NaCl in Joule/mol having given r0=2.81A0, k=3.310

12cm2/dyne,e=4.81010esu, A=1.747, N =6.0221023

The interacting force which might contribute to the bonding of an

ionic crystal are:i) Attractive force due to coulomb interaction between positive and

negative ions.

ii) Attraction due to polarization of individual ions in the field of

other ions.

iii) Short range repulsive force due to overlapping of electron

density of the ions. The force exists when the ions come closer

than the closet distance from coulomb attraction.

iv) Van der Waals interactionbetween the ions.

For two ions of charges Z1e and Z2e separate by a distance r, the

attractive energy is2

1 2

0

Z Z e

4 r

and therefore this energy is2

0

e

4 r

,

2

0

4e

4 r

,2

0

9e

4 r

if both the atoms are respectively monovalent, divalent

and trivalent. For the whole crystal, the coulomb potential energy may

be written as2

1 2

0

AZ Z e

4 r

. The minus sign shows that the net coulomb

energy is attractive. The constant A is known as Madelung constant.


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To prevent the lattice from collapsing, there must also be

repulsive forces between the ions. These repulsive forces become

more noticeable when the electron shells of neighbouring ions begin to

overlap and they increase strongly in this region with decreasing

values of r. This is repulsive force arise from the interaction of the

electron cloud surrounding an atom. Born showed that this repulsive

energy is equal to nB

rwhere n is called repulsive exponent.

The total energy of one ion due to the presence of all other ions is

given by

21 2

r n0

AZ Z e BU

4 r r

________ (1)

For univalent alkali halides, Z1= Z2= 1; and

2

r n

0

Ae BU

4 r r

___________ (2)

The total energy per kilomole of the crystal is

2

r n0

B AeU N

4 rr

_________ (3)

The potential energy will be minimum at the equilibrium spacing r0

0

2r

n 1 2

0 0 0r r

dU nB AeN 0

dr r 4 r

__ (4)

Repulsive energy, U2=b

rn

a

rmb

rn+U2=

comprehensive physical chemistry solution for master degree level student of tribhuwan university...

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