Date post: | 28-Feb-2018 |
Category: |
Documents |
Upload: | shukra-raj-regmi |
View: | 221 times |
Download: | 1 times |
of 201
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
1/201
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
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
2/201
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
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
3/201
Comprehensive Solution of Physical Chemistry..../2
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
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
4/201
Comprehensive Solution of Physical Chemistry..../3
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.
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
5/201
Comprehensive Solution of Physical Chemistry..../4
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.
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
6/201
Comprehensive Solution of Physical Chemistry..../5
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)]
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
7/201
Comprehensive Solution of Physical Chemistry..../6
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
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
8/201
Comprehensive Solution of Physical Chemistry..../7
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
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
9/201
Comprehensive Solution of Physical Chemistry..../8
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.
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
10/201
Comprehensive Solution of Physical Chemistry..../9
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)
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
11/201
Comprehensive Solution of Physical Chemistry..../10
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)
From equation (7) and (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.
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
12/201
Comprehensive Solution of Physical Chemistry..../11
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)
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
13/201
Comprehensive Solution of Physical Chemistry..../12
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)
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
14/201
Comprehensive Solution of Physical Chemistry..../13
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
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
15/201
Comprehensive Solution of Physical Chemistry..../14
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)
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
16/201
Comprehensive Solution of Physical Chemistry..../15
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)
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
17/201
Comprehensive Solution of Physical Chemistry..../16
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
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
18/201
Comprehensive Solution of Physical Chemistry..../17
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
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
19/201
Comprehensive Solution of Physical Chemistry..../18
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
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
20/201
Comprehensive Solution of Physical Chemistry..../19
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,
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
21/201
Comprehensive Solution of Physical Chemistry..../20
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
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
22/201
Comprehensive Solution of Physical Chemistry..../21
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.
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
23/201
Comprehensive Solution of Physical Chemistry..../22
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
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
24/201
Comprehensive Solution of Physical Chemistry..../23
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
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
25/201
Comprehensive Solution of Physical Chemistry..../24
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)
From equation (1) and (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.
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
26/201
Comprehensive Solution of Physical Chemistry..../25
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.
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
27/201
Comprehensive Solution of Physical Chemistry..../26
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]
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
28/201
Comprehensive Solution of Physical Chemistry..../27
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
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
29/201
Comprehensive Solution of Physical Chemistry..../28
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.
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
30/201
Comprehensive Solution of Physical Chemistry..../29
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
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
31/201
Comprehensive Solution of Physical Chemistry..../30
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
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
32/201
Comprehensive Solution of Physical Chemistry..../31
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.
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
33/201
Comprehensive Solution of Physical Chemistry..../32
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.
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
34/201
Comprehensive Solution of Physical Chemistry..../33
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.
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
35/201
Comprehensive Solution of Physical Chemistry..../34
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
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
36/201
Comprehensive Solution of Physical Chemistry..../35
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
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
37/201
Comprehensive Solution of Physical Chemistry..../36
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++
+
+ +
+ +
+
+
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
38/201
Comprehensive Solution of Physical Chemistry..../37
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.
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
39/201
Comprehensive Solution of Physical Chemistry..../38
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
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
40/201
Comprehensive Solution of Physical Chemistry..../39
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
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
41/201
Comprehensive Solution of Physical Chemistry..../40
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.
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
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
42/201
Comprehensive Solution of Physical Chemistry..../41
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
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
43/201
Comprehensive Solution of Physical Chemistry..../42
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
longer a step growth polymerization proceed higher the average molar
mass of the product.
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.
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
44/201
Comprehensive Solution of Physical Chemistry..../43
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
From equation (1) and (3)
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.
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
45/201
Comprehensive Solution of Physical Chemistry..../44
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.
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
46/201
Comprehensive Solution of Physical Chemistry..../45
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
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
47/201
Comprehensive Solution of Physical Chemistry..../46
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)
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
48/201
Comprehensive Solution of Physical Chemistry..../47
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.
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
49/201
Comprehensive Solution of Physical Chemistry..../48
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
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
50/201
Comprehensive Solution of Physical Chemistry..../49
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.
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
51/201
Comprehensive Solution of Physical Chemistry..../50
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|
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
52/201
Comprehensive Solution of Physical Chemistry..../51
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 (
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
53/201
Comprehensive Solution of Physical Chemistry..../52
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
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
54/201
Comprehensive Solution of Physical Chemistry..../53
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.
7/25/2019 Comprehensive Physical Chemistry Solution for Master Degree Level Student of Tribhuwan University Written by S
55/201
Comprehensive Solution of Physical Chemistry..../54
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=