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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 k2 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 2n1SO3 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-+1proudcts 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. 3.60 102 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 DH 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 # TS # Ink# = G # / RT K # exp( G # / RT ) exp [H # TS # ) / 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 ( n1) 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 ( n1) ( 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