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M.Sc. DEGREE (CSS) EXAMINATION, JANUARY 2016
F 5071
Third Semester, Branch III: Chemistry
CH 3C 11 – CHEMICAL KINETICS, SURFACE CHEMISTRY AND
PHOTOCHEMISTRY
(2012 Admission onwards)
Time: 3 hrs
Max. Weight: 30
SCHEME OF VALUATON
Section A: Any 4 correct points – A, 3 points – B, 2 points – C, 1 point – D
Section B: Any 8 correct points – A, 6 points – B, 4 points – C, 2 points – D
Section C: Any 20 correct points – A, 15 points – B,10 points – C, 5 points – D
This scheme is only a guideline for valuation. Any other attempt which is
equally valid and correct, may be given due credit.
Section A
(Answer any ten questions. Each question carries a weight of 1)
1. Explain the significance of rate determining step in a multistep reaction
The overall rate of the reaction isdetermined by the rate determining step.
Consider the reaction in which two reactants A and B combine to form the
intermediate X which change into the product Y.
K1
K2
A + B ------------ X--------------- Y
K-1
Suppose K2>> K-1 i.e.As soon as X is formed, it is transformed into Y.
Always the slowest step isthe ratedetermining step. Hence the initial step is
the ratedetermining step. According to steady state approximation.
d[ X ]
0
dt
K1 [A] [B] - K-1 [X] - K2[X] = 0
K1 [A] [B] = {K-1 + K2} [X]
[X ] 
K1[ A][ B]
K 1  K 2
Rate of the reaction = K2 [X] =
K1 K 2 [ A][ B]
K 1  K 2
K-1 is ignored in comparison with K2
 Rate 
K1 K 2 [ A][ B]
 K1[ A][ B]
K2
But if rate constant for the formation of product K2<<K=1 then the
overall rate = K2 [X] becomes.
Rate =
K1 K 2 [ A][ B]  K1 K 2 
[ A][ B] since K2isvery small,
 
K 1  K 2
 K 1 
when compared to K-1. In this case, reaction 2 which is the conversion of the
intermediate to the product is the rate - determining step.
2)). Explain the principle of SEM
 A scanning electron microscope (SEM) is a type of electron microscope
that produces images of a sample by scanning it with a focused beam
of electrons.
 The electrons interact with atoms in the sample, producing various
signals that contain information about the sample's
surface topography and composition.
 The electron beam is generally scanned in a raster scan pattern, and the
beam's position is combined with the detected signal to produce an
image.
 SEM can achieve resolution better than 1 nanometre. Specimens can be
observed in high vacuum, in low vacuum, in wet conditions (in
environmental SEM), and at a wide range of cryogenic or elevated
temperatures.
 The most common SEM mode is detection of secondary electrons emitted
by atoms excited by the electron beam.
 The number of secondary electrons that can be detected depends, among
other things, on the angle at which beam meets surface of specimen i.e.
on specimen topography.
 By scanning the sample and collecting the secondary electrons that are
emitted using a special detector, an image displaying the topography of
the surface is created.
3.)
Distinguish between activated complex andtransition state.
The initial andfinal states of a reaction differ by the energy U 0 , the
standard internalenergy change in the overall reaction. Between the initial and
final states, the energy usually passes through a maximum which is known
as the
transition state and it corresponds to the top of the barrier. The
molecular species at this maximum are known as activated complexes.
Variation of energy between the initial and final states of a reaction.
4)
Write London equation for calculating activation energy.
Quantum - mechanical methods could be usedto calculate the energies of
reaction intermediates such as the activated complexes. In a diatomic molecule,
E  A  . The integral A is the columbine energy
which is almostequivalent
tothe classical energy of the system. The integral  is the exchange energy.
For
a
triatomic
species,

the
London
equation
is

E  A  B  C  1 / 2(   ) 2  | (  ) 2  (   ) 2 1 / 2
of we consider a general triatomic complete ABC, the energy of the diatomic
molecule B-C, with A removed is A   where A is the Columbic energy and
 is the exchange energy.
 is the
exchange energy for the diatomic
molecule A-C and  is the exchange energy for A-B.
5.) Explain Bioluminescence with example.
Bioluminescence is
the
production
and
emission
of light by
a
living organism. It is a form of chemiluminescence. Bioluminescence occurs
widely in marine vertebrates and invertebrates, as well as in some fungi,
microorganisms including some bacteria and terrestrial invertebrates such
as fireflies. In some animals, the light is produced by symbiotic organisms such
as Vibrio bacteria.
The principal chemical reaction in bioluminescence involves the light-emitting
pigment luciferin and the enzyme luciferase, assisted by other proteins such
as aequorin in some species. The enzyme catalyzes the oxidation of luciferin. In
some
species,
the
type
of
luciferin
requires cofactors such
as calcium or magnesium ions, and sometimes also the energy-carrying
molecule adenosine triphosphate (ATP). The best-known bioluminescent
creature is the firefly. In the firefly, the chemicals luciferase and luciferin
combine with adenosine triphosphate (ATP) to form the luciferase-luciferinATP complex. ATP is found in cells to provide energy for essential cellular
processes. The complex quickly goes from its excited state to a lower energy
state, emitting the difference in energy as visible light with a greenish glow.
All firefly larvae are bioluminescent, while some adult species do not have the
ability to glow. Firefly larvae are sometimes referred to as glow-worms, but true
glow-worms
are
similar
to
beetles,
and
are
also
bioluminescent.Bioluminescence has five functions that have been identified to
date: camouflage, attraction, repulsion, communication (between bacteria), and
illumination.
6) Define Zeta potential.
 Zeta potential is a measure of the magnitude of the electrostatic or charge
repulsion/attraction between particles.
 It is one of the fundamental parameters known to affect stability.
 It is the potential difference between the dispersion medium and the
stationary layer of fluid attached to the dispersed particle.
 The magnitude of the zeta potential indicates the degree of electrostatic
repulsion between adjacent, similarly charged particles in a dispersion.
 Colloids with high zeta potential (negative or positive) are electrically
stabilized while colloids with low zeta potentials tend to coagulate.
7) How does temp influences photochemical reaction
The primary process of light absorption in photochemical reaction is
independent of temp. Effect of temp depends up on the type and nature of
secondary process. If the secondary process involves the active atom or radical
produced in the primary process, its activation energy is very small and thus
very small temp. coefft. of the overall process is observed. If one or more of the
secondary process possess large activation energy, then the photochemical
reaction exhibits a larger value of temp. coefft. If the secondary process involve
a reversible reaction of appreciable energy of reaction and if the equilibrium
const appears in the final rate expression, then a large temp.coefft is expected.
If a photochemical reaction has a quantum efficiency of one and shows no temp
dependence, the reaction involves only primary process. The determination of
temp.coefft of a photochemical reaction helps in establishing the mechanism of
the reaction.
8) What is the condition under which BET isotherm approximates to Langmuir
adsorption isotherm?
We have the BET eqn, [P/V(Po-P)] = (1/VmC) + [(C-1) (P/Po)]/VmC
Multiplying both sides by Po, [PPo/V(Po-P)] = (Po/VmC) + [(C-1)
(PPo/Po)]/VmC
We have C = (K1/K) & K = 1/Po
Therefore K1 = C/Po
[PPo/V(Po-P)] = (1/VmK1) + [(C-1) P]/VmC
Assuming C ˃˃1 & Po ˃˃ P
So [P/V]= (1/VmK1) + [P]/Vm . This is Langmuir eqn.
Stern –Volmerequation given the ratio of the quantum yield for
9.
fluorescence in the absence and presence of a quencher.
A + h v  Ax
K1
Ax  A  hv
Ax k2 A
Ax + Q
K3
 A  Q1

= 1 + Ksv [Q]  Stern - Volmer eqn.
Q
 -Quantum yield for fluorescence in the absence of quencher
Q - quantumyield for fluorescencein the presence of quencher.
KSV - Sten - VolmerConst = KSV = k3T =
k3
k1  k 2
[Q] - Con. of the quencher.

Q
Slope = k3T = KSV
[Q]
10)
Explain with one example anionic surfactants.
 Surfactants- substances which get preferentially adsorbed at the air-water,
oil-water and solid-water interfaces, forming an oriented monolayer
wherein the hydrophilic groups point towards the aqueous phase and the
hydrocarbon chains point towards the air or towards the oil phase.
The surfactants can be cationic, anionic or non - iono genic.
Anionic
surfactants have a negatively charged head.
EgSodiumpalmitatale (C15H31CooNa), allylsulphonates Cn H 2n1SO3 M 
11). Write Gibb`s adsorption isotherm. Explain the term involved.
 Gibbs derived a thermodynamic relationship between the surface or
interfacial tension γ and the surface excess Γ (adsorption per unit area).
 At constant temperature, the Gibbs adsorption equation is dγ =
−∑(ni/A)dμI = −∑ΓidμI, where (ni/A) = Γi is the number of moles of
component I adsorbed per unit area and μI is the chemical potential of the
surfactant solution.
 For a single surfactant component, the Gibbs adsorption equation is
simply dγ/dlog C = −2.303 Γ RT, where C is the surfactant concentration,
R is the gas constant, and T is the absolute temperatures.
 The Gibbs adsorption equation allows one to determine the amount of
surfactant adsorption Γ (moles m−2) from a plot of log γ (the surface
tension at the air/water interface or interfacial tension at the liquid/liquid
interface) versus log C.
12)
Unimolecular gas phase reaction follows first order kinetics at high
pressure and 2nd order at law pressure. Why?
In unimoleculargas phase reactions, the reactant molecules acquire
activation energy by collision with one another.
formedwill decompose into products.
decompose immediately.
decomposition.
But the activatedmolecules do not
There is a time lag between activation and
The deactivation of the molecule takes place as a result
ofcollusion withless energetic molecules.
A + A - A + A* - activation
K
Ax + A 
2A - deactivation
2
The activated complex
K
A x 
Products - decomposition
3
According to steady state concentration,
d[ Ax ]
0
dt
K1 [A]2- K2[A] [Ax] - K3 [Ax] = 0
K1 [A]2 = {K2[A]+K3} [Ax]
[Ax] =
K1[ A]2
K 2 [ A]  K 3
The rate of formation of the products is
d [ P]
dt
 K3[ A x ] 
K1 K 3 [ A]2
K 2 [ A]  K 3
There are two opposing processes. The activated complex can undergo
either decompositionor deactivation. If the conditions are such that there are
more collisions during a duration of stable existence, then deactivation occur
during high pressure conditions. Hence K2 [A] >> K3, then
d [ p] K1 K 3 [ A]2 K1 K 3
Rate =


[ A]  K 1[ A]
dt
K 2 [ A]
K2
Thus the reaction follows first order kinetics at high pressure.
At low pressure, the rate of decomposition is greater than the rate of
deactivation K3 >>K2[A], then
d [ P] K1 K 3 [ A]2
Rate =

 K1 [ A]2 - The reaction follows second order kinetics.
dt
K3
13.
What is potential energy surface? Explain its significance.
In potential energy surface, potential energy is plotted against bond
distances and angles. For a diatomic molecule, we can contract a potentialenergy curve in a two-dimensional diagram by plotting energy against the
distance. If 3 atoms are involved, three parameters are required to describe
it and to plot energy against these three distances, a four dimensional
diagram is necessary. In order to construct such a diagram, one parameter has
to be fixed at a particular value.
The potential energy surfaces consist of a reactant valley and a product
valley which meet at a col or saddle point. The paths of steepest descent from
the col into the two valleys constitute a path called minimum energy path. The
actual reaction paths followed by individual reaction systems are not
exactly the same as this minimum energy path and it depend on the energy
states of the colliding molecules.
A section through the minimum energy path is known as a potential energy profile. The maximum point in this profile, at the col in the potential
energy surface has particular significance. It is a position of maximum energy
along the minimum - energy path and a position of minimum energy with
respect to motions at right angles to the path. Systems represented by a
small region round this point are known as activated complexes and their
state is called the transition state for the reaction.
Section B
1. (Answer any five questions. Each question carries a weight of 2) Explain
the principle of Surface Enhanced Raman Spectroscopy.
 Surface – enhanced Raman spectroscopy (SERS) is a surface-sensitive
technique that enhances Raman Scattering by molecules adsorbed on
rough metal surfaces or by nanostructures such as plasmonic-magnetic
silica nanotubes. The enhancement factor can be as much as 10 10 to 1011
which means the technique may detect single molecules. There are two
theories electromagnetic theory and chemical theory to explain the
mechanism of enhancement effect of SERS.
 The chemical theory applies only for species that have formed a chemical
bond with the surface, so it cannot explain the observed signal
enhancement in all cases, whereas the electromagnetic theory can apply
even in those cases where the specimen is physiorbedon the surface.
 According to electromagnetic theory the increase in intensity of the
Raman signal for adsorbates on particular surfaces occurs because of an
enhancement in the electric field provided the surface. When the incident
light strikes the surface, localized surface plasmons are excited. The field
enhancement is greatest when the Plasmon frequency is in resonance with
the radiation. In order for scattering to occur, the plasmon oscillations
must be perpendicular to the surface. If they are in plane with the surface,
no scattering will occur. It is because of this requirement that roughened
surface or arrangements of nanoparticles are typically employed in SERS.
 For many molecules, often those with a lone pair of electrons, in which
the molecules can bond to the surface, a different enhancement
mechanism that does not involve surface plasmons has been described.
This chemical mechanism involves charge transfer between the
chemisorbed species and the metal surface.
15). Describe the stabilizing action of surfactants
When a surfactant adsorbes on the interface, the interfacial tension between two
phases decreases. The reduced surface tension depends on the concentration of
the surfactant according to the Gibb's Isotherm.
According to Gibb's Isotherm
2 =
 C 2 dr
RT dc2
2
=
excess concentration of the solute per unit area of the
surface.
C2 - concentration of the solute
r -
surface energy
 2 is
 dr 
+ve if   is negative.
 dc2 
The limiting value of the decrease of surface tension with
  dr 

is called the surface activity.
 dc2  C 0
concentration,ie 
2
Soaps, dyestuffs, detergents belong to surfactants.
If there is a strong
interaction between the solvent - solvent molecular, the solute molecular are
pushed up to the surface from the bulk of the solution
2
is +ve. Any
substance which exhibits +ve deviation from Raoults law is expected to have a
positive value of 
Stabilization a action of surfactants can be explained via two mechanism.
(1) Steric stabilization (2) Electrostatic stabilization
Steric Stabilization:-
It arises from a physical barrier to contact and
coalescence.
For eg. High mol.wt polymers can adsorb on the surface of the dispersed phase
droplets and extend significantly into the continuous phase, providing a volume
restriction or a physical barrier for particle interaction. As polymer coated
particles approach, the polymers are forced into close proximity and repulsive
forces arise, keeping the particles apart from each other.
Clays can also
stabilize emulsion via steric mechanism.
Electrostatic stabilization is based on the mutual repulsive forces that are
generated when electricity charged surface approach each other. Ionic or
ionisable surfactants form a larged layer at the interface. For an oil in water
emulsion, this layer is neutralised by counter icons are termed a double layer.
If the counter ions are diffused, the dispersed droplets act as charged spheres
as they approach each other. If the repulsive forces are strong enough, the
droplets are repelled before they can make contact and coalesce, and the
emulsion is stable.
Electrostatic stabilization is significant in oil-water
interface - Where electric double layer thickness is much greater in water than
oil.
16.The decomposition of N2O5 takes place according to the following
mechanism. Derive the rate law.
N2O5= NO2 + NO3
NO2 + NO3  NO2 + O2 +NO
NO+ NO3 2NO2
N 2O2
NO2  NO3
- (1)
k2
NO3  NO2 
NO  NO2  O2
- (2)
k3
NO3  NO 
2NO2
- (3)
Rate of formation of O2 is
d [O2 ]
 k 2 [ NO2 ] [ NO3 ]
dt
- (4)
Applying the steady state approximation to NO and NO3, we get
d [ NO]
 O  k 2 [ NO3 ][ NO2 ]  k3 [ NO3 ][ NO]
dt
- (5)
d [ NO3 ]
 O  k1 [ N 2O5 ]  k  1[ NO2 ] [ NO3 ]  k 2 [ NO3 ][ NO2 ]  k3 [ NO3 ][ NO] - (6)
dt
From (5)
We get [ NO] 
k2
[ NO2 ]
k3
- (7)
From (6)
[ NO3 ] 
k1 [ N 2O5 ]
(k  1  2k 2 )[ NO2 ]
- (8)
(8) in (4)


k1 [ N 2O5 ]
d [O2 ]
 [ NO2 ]
 k 2 
dt
 (k  1  2k2 ) ( NO2 ) 
Also
d [O2 ]
k 2 k1

[ N 2O5 ]  k [ N 2O5 ]
dt
k  1  2k 2
- (9)
 d [ N 2O5 ]
 k1[ N 2O5 ]  k  1[ NO2 ] [ NO3 ]
dt
- (10)
(8) in (10)
we get
 d [ N 2O5 ] 2k1k 2 [ N 2O5 ]

dt
k  1  2k 2
- (11)
Either (9) or (11) can be considered on the rate law.
15.Explain primary and secondary kinetic salt effect
Influence of ionic strength of solutions on reaction rates - Salt effects
(a) Primary kinetic salt effect
The ionic strength of the medium influences the activity coefficient which
in turn will influence the rate and rate constant of a reaction. The influence of
ionic strength on the rate constant of a reaction by way of its influence on the
activity coefficient of the reactant molecules is called primary salt effect. The
ionic strength of the medium can be altered by adding electrolytes to the
solution. The electrolytes can be reactants, products or an added inert salt
whose ions do not interact with either the reactant or the product. This effect of
inert salts on the reaction rate is called primary kinetic salt effect.
Bronsted and Bjerrum provided an effective treatment of these reactions
in their activated rate theory applied to charged particles. Suppose the reaction
occurs between two ions - ion A with a charge of ZA and ion B with a charge of
ZB. Let the reaction proceeds through an activated complex (AB#)ZA + ZB
AZA + BzB  [AB(ZA+ZB)#]  Products
#
K =
a AB
( ZA  ZB ) #
a AZA aBZB
For ions, the equilibrium constant is expressed in terms of activities rather than
concentration.
#
K =
YAB
Y ZA
Y BzB
Z
( ZA  ZB ) #
[AB(ZA + ZB)#]
[AZA] [BZB]
[AB(ZA + ZB)#] = K# [AZA] [BZB]
YAZA YBZB
YAB ( ZA  ZB )#
According to the transition state theory,
Rate = [ABZA + ZB)#] (KT/h)
Rate = K# [AZA] [BZB]
YAZA YBZB
(KT/h)
YAB ( ZA  ZB )#
For a bimolecular reaction between A and B, the experimental rate of the
reaction is rate = K' [AZA] [BZB] where K' is the experimentally evaluated rate
constant.
K' [AZA] [BZB] = K#[AZA] [BZB]
YAZA YBZB
(KT/h)
YAB( ZA  ZB )#
YAZA YBZB
YAB ( ZA  ZB )#
K' = K# (KT/h)
K# (KT/h) = K0
K' = K0
YAZA YBZB
Y AB( ZA  ZB )#
This result is known as Bronsted - Bjerrum equation
Taking log of the above equation,
ZB
( ZA ZB )#
In K' = In K0 + In (Y ZA
)
A ) + In (Y B ) - In (Y AB
The activity coefficient of the ion depends upon the ionic strength of the
solution.
1 = 1/2
m z
2
i i
where mirepresent molarity and z1 is the charge of the ion i.
The Debye - Huckel Limiting Law relates the activity coefficient to the ionic
strength of the solution.
In Y1 = AZi2 (I)1/2
A is a constant
For an aqueous solution at 25oC, the constant A = 0.509.
Then In Yi = 0.509 Zi2 (I)1/2
In K' = In K0 = [0.509 ZA2 + 0.509 ZB2 - 0.509 (ZA+ZB)2] (I)1/2
= In K0 - [0.509 ZZ2 + 0.509 ZB2 - 0.509 ZA2 - 0.509 ZB2 - 1.018 ZAZB]
(I)1/2
In K' - In K0 + 1.018 ZAZB (I)1/2
The above equation indicates the rate constant of an ionic reaction in solution
should depend on the ionic strength. This is the kinetic salt effect.
In (K'/K0) = 1.018 ZAZB(I)1/2
K' increases or decreases depending upon the product of ZA and ZB. Suppose
one of the reactants is neutral, then ZA ZB = 0 and In K' = In K0. Then
irrespective of the variation of the ionic strength, the rate constant remains the
same which means the added electrolyte will not affect the rate constant. The
variation of observed rate constant depends upon the signs of ZAZB.
Consider the reactions
1. [Co(NH3)5Br]2+ Hg2+ Products ZAZB = +4
2. S2O82-+1proudcts ZAZB= +2
3. CH3COOC2H5 +OH  products ZAZB=0
4. [Co(NH3)5Br]2 + OH  products ZAZB = -2
(Graph)
Reactions between pairs of ions of like charge are usually accelerated by
increasing the ionic strength. But reactions between ions of unlike charges are
usually slowed down by increase in ionic strength. The rates of reaction
between to uncharged molecules or between of inert salts. These assumptions
have been confirmed by experimental results. A plot of In (K'/K 0) against (I)1/2
in a dilute solution will give a straight line with a slope equal to ZAZB.
Secondary salt effect
The rate of a reaction will change on altering the effective concentration
of the catalytic species. The change in the rate of the reaction as a result of the
change in the concentration of one of the reacting ions in the presence of a
foreign electrolyte that changes the ionic strength of the solution is known as
the secondary salt effect. It is independent of the primary salt effect and it will
change the degree of dissolution of the reacting electrolyte.
If the reactions are catalyzed by acids or bases that is, H  or OH  ions,
the addition of inert salt affects the concentration of H  or OH  ions. The rate
of reactions is proportional to the concentration of H  or OH  ions. Thus the
rate constant will depend on the ionic strength.
Consider the hydrolysis of cane sugar catalyzed by a weak acid.
C12 H 22O11  H 2O H   C6 H12O6  C6 H12O6
The rate of reaction Kexp  K[ H  ]
The catalyst comes from the dissociation of weak acid.
HA  H   A
K=
[ H  ] [ A ] Y [ H  ] Y [ A  ]

[ HA]
Y [ HA]
[H  ] 
K [ HA]
Y [ HA]
[ A] Y [ H  ] Y [ A ]
If the reaction is carried out at a fixed ratio of [HA]/ [A-] by making use of
buffer solution, then [ H  ]  K '
Kexp = KK '
Y [ HA]
Y [ H  ]Y [ A ]
Y [ HA]
Y [ HA]
 K0


Y [ H ]Y [ A ]
Y [ H  ] Y [ A ]
In Kexp = In K0 + In Y [HA] - In Y [H+] - In Y [A-]
Using the Debye Huckel Limiting Law we get,
In Kexp = In K0 + 1.018 (I) 1/2
Kexp increases with an increase in the ionic strength of the solution.
18)Biacetyl triplets have a quantum yield of 0.25 for phosphorescence and a
measured lifetime of the triplet state of 10 milliseconds. The phosphorescence
is quenched by a compound Q with a diffusion controlled rate of 10 10 I mol-1s-1.
What concentration of Q is required to reduce the phosphorescence yield to
half?
Quantum yield without
Quencher  0 = 0.25
Life time of excited state To = 10 x 10-3s
Rate of quenching kq = 1010 l mol-1 s-1
quantum yield with quencher -
0.25
2
[  Q]
O
Q
= 1+ Ksv [Q], where [Q] = con. of quencher
O
= 1+kq To (Q)
Q


1
[Q] =  O  1 
 Q
 kqTo
0.25 
= 
1 
 0.125
= 10-8
1
3 1
 10 l mol s 10 10 s
10
1 1
mol/l
========
The relaxation time for attaining equilibrium in A +B = C has been found to
the 100μs. The equilibrium constant of the reaction is 1010. Evaluate the rate
constants k1 and k-1.
For the reaction
A+B
relaxation time T =
C,
1
1
1


k1[ A] k1[ B] k 1[c]
At equilibrium, K1[A] = k1[B] = k-1[c]
T=
3
k 1
T = 100
100 = 3/k-1
k-1 = 3 x 10-2
Keq =
k1
, k1 = k-1 x Keq
k 1
k1 = 1010 x
3
= 3 x 108
100
=======
20)
In an experiment to measure quantum efficiency of a photochemical
reaction, the absorbing substance was exposed to 490nm light from a 100W
source for 45 minutes. The intensity of transmitted light was 40% of the
intensity of the incident light. As a result of irradiation, 0.344 mol of the
absorbing substance decomposed. Determine the quantum efficiency

ofmoleculesreacted
No.ofphotonsabsorbed
Pulse radiant energy = Radiant power x time
= 100 W x 45 x 60 S
= 270000 J
Amount of light absorbed = 60%
Absorbed energy (Q) =
270000  60
100

162 103 J
9
Vo. of photons =
Q 162 103  J  490 10 19 10m

hc
6.62 10 34 JS  3 108 m
20.
= 3.99 x 1023
No. of molecules in 0 -344mol = 0.344 x 6.02.3x1023

0.344  6.023 10 23

3.99 10 23
21) The intrinsic viscosity of a solution of poly isobutylene at 200c is 180 cm3
per gram. If [n] is related to the viscosity-average molar mass Mvisc by the
expression
[n] =3.6 x 10-2(Mvisc)0.64, calculate the molar mass of the polymer.
   3.60 102 visc
0.64
180 cm3g-1 = 3.60 x 10-2 [Mvisc]0.64
[Mvisc]0.64 =
180cm 3 g 1
 5.0 103
2
3.60 10
[Mvisc] = 6.0 x 105g/mol
Section C.
Answer any two questions. Each question carries a weight of 5
22 (a) Explain the principle of Lasers in the study of photochemical kinetics
Special features of laser beam of great value in photochemical kinetics are
1. It is highly monochromatic, covering only a narrow range of wavelengths.
2. With a suitable geometry for the laser medium,(eg.ruby laser),the beam can
be extremely narrow and coherent, the beam divergence can be of the order of a
mill radian.
3. High intensity.
Pulsed lasers-In pulsed lasers, radiation are emitted for a period of very fast
duration foreg.Iodine lasers can be used to study the rate of dissociation of
water. Pulsed lasers are important in studying very fast chemical reactions
Iodine laser-effective species,being iodine atom produced photolitically.They
are often produced in an excited state by the irradiation of C 3 H 7 I with UV
irradiation.
C 3 H 7I +hv - C 3 H 7 +I
The transition between the excited iodine atoms and ground state iodine atoms
corresponds to a wavelength of 1315 nm.Irradiaion of the system with such
radiation induces laser action,with the production of powerful coherent beam of
1315wavelength.Diferent technique such as Q-switched mode, free running
mode, and mode locked oscillating system produces pulses of short duration
b). Derive Michelis- Menten equation for an enzyme catalysed reaction.
Step I - Formation of the Enzyme -Substrate complex
K1

E+S

k 1
ES
(1)
Step - II Decomposition of the complex
k2
ES  P + E (Slow) - (2)
E - free enzyme
S - Substrate
ES - enzyme substrate - complex
Rate of the reaction in given by
r, k2 [ES] - (3)
Using steady state approximation for ES
d [ ES ]
 K1[ E ][ S ]  K 1[ ES ]  K 2 [ ES ]  0
dt
- (4)
The equilibrium between the free and the bound enzyme is given by the
enzyme conservation equation.
[E] 0 = [E] + [ES]
-(5)
Where [E]o is the total enzyme concentration
[E] = [E]0 - [ES]} [S] - k-1 [ES] - k2 [ES] = 0 -(7)
On sim
[ES] =
k1[ E 0]]S ]
k 1  k 2  k1[ S ]
-(8)
(8) in (3)
r
k1k 2 [ E 0][ S ]
k 1  k 2  k1[ S ]
- (9)
 k1
(9)
r=
k [ E 0][ S ]
k 2 [ E 0][ S ]
= 2
- (10)
(k 1  k 2 / k1  [ S ]
km  [ S ]
km is a new constant km =
k 1  k 2
k1
Eqn. (10) is known as the Michael is - Mentenequation
a) Explain the principle of Auger and ESCA
 Auger Electron Spectroscopy (AES) provides information about the
chemica composition of the outermost material comprising a solid surface
or interface.
 The principal advantages of AES over other surface analysis methods are
excellent spatial resolution (< 1 µm), surface sensitivity and detection of
light elements.

AES uses a primary electron beam to excite the sample surface.
 When an inner-shell electron is ejected from a sample atom by the
interaction with a primary electron, an electron from an outer shell fills
the vacancy.
 To compensate for the energy change from this transition, an Auger
electron or an x-ray is emitted.
 For light elements, the probability is greatest for the emission of an Auger
electron, which accounts for the light-element sensitivity for this
technique.
 The energy of the emitted Auger electron is characteristic of the element
from which it was emitted.
 Detection and energy analysis of the emitted Auger electrons produces a
spectrum of Auger electron energy versus the relative abundance of
electrons. Peaks in the spectrum may be used to identify the elemental
composition of the sample surface.
 Auger electrons have relatively low kinetic energy, which limits their
escape depth. Any Auger electrons emitted from an interaction below the
surface will lose energy through additional scattering reactions along its
path to the surface.
 Auger electrons emitted at a depth greater than about 2 - 3 nm will not
have sufficient energy to escape the surface and reach the detector. Thus,
the analysis volume for AES extends only to a depth of about 2 nm.
ESCA
o In photoelectron spectroscopy, UV radiation is used to excite the
valance electrons (UPES).
o However a higher energy is required to ionize inner electrons or
core electrons. Xray sources are used to effect such ionizations
and the technique is the called Xray photo electron spectroscopy
(XPES) or Electron Spectroscopy for Chemical Analysis (ESCA).
o In this case a X-ray beam is produced by electron bombardment of
a clean metal target, such as Al or Mg, resulting in the emission of
radiation at very specific energies, Eg: the Kα lines of Al occur at
1486.6eV.
o ESCA probes the binding energies of core electron in an atom or
molecule. Such electrons play only little part in chemical bonding,
but different chemical environment can produce small changes in
their binding energies.
o This is because the formation of a bond, changes the distribution of
electrons in the molecule and hence produces changes in the
effective nuclear charge and hence the binding energies.
o Consider the case of Carbon-Fluorine bond.
The highly
electronegative fluorine attracts electron and so reduces the
electron density around the carbon nucleus.
o The carbon is now less shielded and its effective nuclear charge is
increased. Thus the core carbon experiences an increased nuclear
forces and binding energies.
All these phenomena can be
effectively studied by ESCA or XPES
(b) Write a note on dimerisation of anthracene
Anthracene in benzene solutiondimerizes and simultaneously fluoresces when
irradiated with light of a certain frequency  . The frequency of fluorescence
is  1 . Where '   The most accepted mechanism is given below.
k1
A+ h 
 A x [A - Anthracene excitation] - (1)
K2
Ax+A  A2 [dimerisation] - (2)
k3
Ax  A + h  1 [fluorescence]
Applying S.S.a to Ax,
d [ Ax ]
dt
- (3)
d[ A x ]
 0 - (4)
dt
 k1 Ia  k 2 [ A x ] [ A]  k3 [ A x ]  0 - (5)
Ia - Intensity of light absorbed
[ Ax ] 
k1 Ia
- (6)
k 2 [ A]  k3
The rate of formation of the dimer A2 is given by
r
d [ A2 ]
 k2 [ Ax ]
dt
ie
r
[ A]
- (7)
k1k 2 Ia[ A]
- (8)
k 2 [ A]  k 3
Quantum yields for the formation of the dimer is given by
No.of moles of
No.of einsteens
d [ A2]
k k [ A]
dt

 1 2
Ia
k 2 [ A]  k 3
A2 
A2 
A2
formed
absorbed
- (9)
k1[ A]
-(10)
[ A]  k 3 / k 2
If [A] <<k3/k, it can be neglecteddenominater
A2 
k1k 2 [ A]
A2 increases with increase in the [A] concentration of the
k3
monomer.
If [A] >> k3/k2, A2  k1
is the quantum yield is independent of [A] & intensity of light absorbed.
a) Greenhouse effect
(a)
Green house effect.
It is the process by which radiation from a planet's atmosphere warms
the planet's surface to a temperature above what it would be without its
atmosphere.
In a greenhouse solar radiations pass through the transparent glass and
heat up the soil and the plants.
The warm soil and plants remit infrared
radiations. This mechanism keeps the energy of the sun trapped in the green
house. Similar phenomenon happens in the surface of the earth also. Sun emits
different types of wavelengths, consisting of UV, visible region and IR. Of
these harmful UV radiations are absorbed by ozone layer in the stratosphere
visible and IR radiation's pass through the atmosphere and reach the surface
of the earth. some gases in the atmosphere, CO2, H2O, Ozone, CH4,N2O,
CFCl3, CF2Cl2etc has the property of allowing visible light to pass through it
but absorbing the IR radiations reflected from the surface of the earth.
There radiations are re-emitted in all directions and some are redirected back
towards earth's surface and heat up the atmosphere. This heating of the earth
due to trapped radiations is called green house effect.
24 (b) Ely - Ridel Mechanism of heterogeneous catalysis. The sequence
of
events in a surfacecatalysed reaction comprises
(1)
Diffusion of reactants to the surface (fast)
(2)
Adsorption of the reactants on the surface.
(3)
Surface diffusion of reactants to active sites.
(4)
Reaction of the adsorbed species (often rate determining)
5)
Desorption of the reaction products.
6)
Diffusion of the products away from the surface.
Consider the exchange reaction between
H2
H2
 D2
D
I
 M
_H DH
I
M
 H
D  MH

M- Metal surface
Here reaction takes place between chemisorbed atoms and a colliding molecule
or a physically adsorbed molecule.
25)
(a) What are the assumptions in Transition state theory? Following the
theory derive an equation for rate constant.
(b) Compare Transition state theory with Collision theory.
Conventional Transition State Theory (CTST)
This theory represents an alternative approach to the collision theory to explain
the mechanism of a reaction. This theory was put forward by H. Eyring and M.
Polanyi (Eyring equation). The important postulates of this theory are
1.
The reacting molecules must form an activated complex before being
converted to products.
2.
There exists an equilibrium between the activated complex and the
reactants,
The main assumptions of CTST are
1.
Molecular systems that have surmounted the colin the direction of
products cannot turn back and form reactant molecules again.
2.
The energy distribution among the reactant molecules is in accordance
with the Maxwell - Boltzmann distribution. It is assumed that even when the
whole system is not at equilibrium, the concentration of those activated
complexes that are becoming products can also be calculated using equilibrium
theory.
3.
It is permissible to separate the motion of the system over the col from
the other motions associated with the activated complex.
4.
A chemical reaction can be satisfactory treated in terms of classical
motion over the barrier, quantum effects being ignored.
One example is the reaction between CO and NO2. The conversion of
reactants to products occurs through the formation of a transition state or
activated complex.
CO  NO [O  C ........... O  N  O] CO2  NO
Reactants
Products
H   54 Kcal mol 1
The activated complex is not stable and the atoms are linked together
with loose valence bonds. The reaction energy diagram presents the energies of
a reaction which makes the explanation of transition state theory easier. (Figure)
The bimolecular reaction between A and B can be represented as
A+B  AB*  Products
A.C. or T.S.
The reaction rate is equal to the concentration of the activated complex and the
frequency of the decomposition of activated complex.
Rate = [Activated complex] x frequency of decomposition of activated complex
---(1)
If [A], [B] and [AB]* represent the concentration of A,B and AB* at time t, then
equilibrium constant K# for the formation of the activated complex is
K#= [AB]*
[A] [B]
Concentration of the A.C. is [AB]*= K# [A] [B] -----(2)
In CTST, the situation at the col in the potential energy surface is considered.
Consider a dividing surface that passes through the col which is constructed
perpendicular to the minimum - energy path, which is the path of steepest
descent from the lowest point of the col into the reactant and product valleys.
CTST focuses attention entirely on the dividing surface that passes through the
col and is concerned with the rate at which systems pass through that dividing
surface.
Consider two parallel diving surfaces separated by a very small
distance  . Systems that are within this small distance are activated complexes
by definition; those to the left are reactants and those to the right are products.
Systems entering the region of thickness  from the left hand side must exit
into the product side and end up as products. Similarly those entering from the
product side must become reactants.
Frequency of the decomposition of activated complex
The activated complex must have one of its vibrational degrees of
freedom unstable. If Evib is the vibrational energy which causes the rupture of
the bond, then according to Planck's quantum theory.
Evib= hv
............(3)
This vibration is responsible for disrupting the complex into the product. The
frequency of such vibrations will be low and the average energy Evib will be of
the order of classical vibrational energy KT. Thus
Evib = KT
....... (4)
hv = KT of V = KT/h
.............(5)
-d[A]/dt = K#[A] [B] (KT/h)
.............. (6)
If K' is the rate constant of the reaction, the experimentally obtained rate law is
-d[A]/dt = K' [A] [B] ............(7)
Comparing equations (6) and (7)
K# [A] [B] (KT/h) = K' [A] [B]
K' = K# (KT/h)
...........(8)
K# cannot be measured experimentally.
The equilibrium constant K# can be expressed in thermodynamic terms. Thus
G #  RTInK #  H #  TS #
Ink# = G # / RT
K #  exp( G # / RT )  exp [H #  TS # ) / RT ]
= exp (H # / RT ) exp (S # ) / R
G #
..............(9)
- standard free energy H # - standard enthalpy S # - standard entropy
corresponding to a gram molecule.
K '  ( KT / h) exp (H # / RT ) exp (S # ) / R
............(10)
E #  H #  nRT where n is the change in the number of molecules in complex
formation.
E  RT  H #  nRT  H #  RT (n  1)
Then  H #   E  RT (n  1)
..........(4)
Consider the equation K '  ( KT / h) exp (H # / RT ) exp( S # / R)
Substituting for  H # in the above equation.
K '  ( KT / h) exp [ E  RT (n  1).RT ) exp( S # / R)
K '  ( KT / h) exp (S # / R) e  E / RT e ( n1)
or
K ' [e( n 1) ( KT / h) exp (S # / R] e E / RT e E / RT
.......... (5)
Compare this equation with the Arrhenius equation. Then we get the frequency
factor as
A  e( n 1) ( KT / h) exp( S # / R)
For a unimolecular reaction,  n = 0
Then K  e ( KT / h) exp (S # / R)e E / RT
For a bimolecular reaction n   1
K  e 2 ( KT / h) exp (S # / R)e  E / RT
For reactions in solution, whether unimolecular or bimolecular,  n = 0 or
negligible.
Comparison with Collision theory
In collision theory, the frequency factor A is proportional to Z.
K = P.Ze-E/RT
In transition state theory, the frequency factor A is a measure of entropy of
activation.
k '  [e ( n1) ( KT / h) exp (S # / R] e  E / RT
In K '   E / RT  InT  [S # / R  In k / h  (n  1)]
S # is
independent of temperature and InT is negligible in comparison to E/RT.
Then a plot of InK against 1/T would be linear.
Entropy of activation
On knowing the experimental values of the rate constant and the
activation energy, the value of entropy of activation can be calculated from
equation (5). The experimental value of S # gives an idea about the nature of
the transition state.
If the entropy of activation S # is positive, the activated complex is more
disordered when compared to the basic state. This occurs for unimolecular gas
reactions. If S # is negative, the activated complex is more ordered. As the
freedom of activity of reactants is lost when they form the activated complex,
the entropy of activation decreases.
Consider a first order gas phase unimolecular reaction in which the
entropy of activation is due to intramolecular atomic reorganization when
reactant molecules are transformed into an activated complex. Example is
C2 H 6  2CH 3
The pre-exponential factor A for the dissociation of ethane into methyl
radicals is high (A=167) and corresponds to S # 858K = 58-1mol-1. The activated
complex is a single molecular species, hence no gain in transitional entropy. On
considering the enthalpy of activation, the two methyl fragments are only
weakly bonded together at the transition state.
Hence some molecular
vibrations of the activated complex have low force constants corresponding to
shallow potential energy curves with closely spaced quantum energy levels.
For a general reaction, X 2  2 X