SUMMER-16 EXAMINATION Model Answer

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1 (ISO/IEC Certified) SUMMER-6 EXAMINATION Subject code :(756) Page of Important Instructions to examiners: ) The answers should be examined by key words and not as word-to-word as given in the model answer scheme. ) The model answer and the answer written by candidate may vary but the examiner may try to assess the understanding level of the candidate. 3) The language errors such as grammatical, spelling errors should not be given more Importance (Not applicable for subject English and Communication Skills. ) While assessing figures, examiner may give credit for principal components indicated in the figure. The figures drawn by candidate and model answer may vary. The examiner may give credit for any equivalent figure drawn. 5) Credits may be given step wise for numerical problems. In some cases, the assumed constant values may vary and there may be some difference in the candidate s answers and model answer. 6) In case of some questions credit may be given by judgement on part of examiner of relevant answer based on candidate s understanding. 7) For programming language papers, credit may be given to any other program based on equivalent concept.

2 (ISO/IEC Certified) SUMMER-6 EXAMINATION Subject code :(756) Page of Q No. Answer marks Total marks A Attempt any THREE Aa Ab Chain reaction: In Chain reaction, the intermediate is formed in the first reaction called chain initiation step. It then combines with the reactant to form product and more intermediates in the chain propagation step. Occasionally these intermediates are destroyed in chain termination step. Non chain reaction: R I* Initiation step I* + R I* + P Propagation step I* P Termination step In non-chain reaction, the intermediate is formed in the first reaction and then disappears as it reacts further to give products. R I* I P Relation between conversion and equilibrium constant for second order reversible reaction Reaction is A + B R +S Let mole of A and mole of B is present initially. x A mole of A reacts at equilibrium P is the total pressure. When mole of A reacts, mole of B also reacts. Then mole of R and mole of S is produced Moles initially present A B R S Total - - moles

3 (ISO/IEC Certified) SUMMER-6 EXAMINATION Subject code :(756) Page 3 of Moles reacted / x A x A x A x A produced at equilibrium Moles present at - x A - x A x A x A equilibrium Mole fraction - x A / - x A / x A / x A / Partial pressure P(- x A ) / P(- x A ) / P x A / P x A / K p = K p (- x A ) = x A K p (-x A + x A ) = x A = K p - K p x A + K p x A = x A ( ) ( K p - ) x A K p x A + K p = 0 x A = ± ( ) ( ) A- C A = 0.5 mol / l -r A =.7 * 0-3 mol / l min c C A = 0.05 mol / l -r A = 0.3 * 0-3 mol / l min -r A = KC A n Taking log, log -r A = log K +n log C A Log ( -r A / -r A ) = n log ( C A / C A ) Log (.7 * 0-3 / 0.3 * 0-3 ) = n log( 0.5 / 0.05 ) n = Order of the reaction is

4 (ISO/IEC Certified) SUMMER-6 EXAMINATION Subject code :(756) Page of Relation between C A and X A i)constant volume System C A =C A0 (-X A ) ii)variable volume system B Attempt any ONE 6 Ad Ba Integrated form of rate expression for zero order reaction Rate equation is -r A = - In terms of concentration = C A0 = k 6 - = k Integrating between appropriate limits dca= k dt.5 C A0 - C A = kt for t In terms of fractional conversion C A0 = k Integrating between appropriate limits.5

5 (ISO/IEC Certified) SUMMER-6 EXAMINATION Subject code :(756) Page 5 of C d = k dt C A0 X A = kt for t B- Theories of reaction rate constant: 6 b. Arrhenius theory. Collision theory 3. Transition state theory Transition State Collision Theory Arrhenius Theory theory 5 i) Prediction of reaction Prediction of reaction Prediction of Experiment rates is more rates is less accurate reaction rates is accurate with with experiment. more accurate experiment. with experiment ii) Temperature Temperature Temperature Mathematical Dependency is Dependency is Dependency is

6 (ISO/IEC Certified) SUMMER-6 EXAMINATION Subject code :(756) Page 6 of equation k T. e k T. e k. e iii) Activated The theory views The theory views that No activated complex that the formation of the decomposition of complex is activated complex is activated complex is formed very rapid& very rapid & formation decomposition of of activated complex is activated complex is slow. slow. Attempt any TWO 6 -a Derivation for temperature dependency of rate constant from collision theory The collision rate of molecules in a gas is found from the kinetic theory of gases. For the bimolecular collision of like molecules A, 8 Z AA = A n A Where = A C A...(i) = number of collisions of A with A / sec.cm 3 = diameter of molecule, cm M = (molecular weight) / N, mass of a molecule, gm N = Avogadro s number C A = concentration of A, mol / liter n A = NC A / 0 3, number of molecules of A / cm 3 k = Boltzmann constant For the bimolecular collision of unlike molecules in a mixture of A and B, kinetic theory gives Z AB = {( A+ B) /} n A n B 8πkT +

7 (ISO/IEC Certified) SUMMER-6 EXAMINATION Subject code :(756) Page 7 of = {( A+ B) /} 8πkT + C A C B...(ii) If every collision between reactant molecules results in the transformation of reactants into product, these expressions give the rate of bimolecular reactions. The actual rate is much lower than that predicted and this indicates that only a small fraction of all collisions result in reaction. This suggests that only those collisions that involve energies in excess of a given minimum energy E lead to reaction. From the Maxwell distribution law of molecular energies the fraction of all bimolecular collisions that involves energies in excess of this minimum energy is given approximately by e -E / RT where E Thus the rate of reaction is given by RT. 3 -r A = - = k C A C B.(iii) = ( collision rate, mole / liter.sec)* (fraction of = Z AB / RT e-e collision involving energies in excess of E ) = {( A+ B) /} 8πkT + e -E / RT C A C B (iv) Comparing iii and iv, -b Given: X A =0.70,t = 3 min k T / -E / RT e For batch reactor, in which first order chemical reaction is taking place 8 -r A = k.c A -ln( - X A ) = kt -ln(-0.70) = k * 3 k = min - For Plug flow reactor ( constant density system): = =

8 (ISO/IEC Certified) SUMMER-6 EXAMINATION Subject code :(756) Page 8 of τ = C..( ) dx τ = k. ( X ) Space time τ = (-ln(-x A)) = ( ln( 0.7)) = 3 min.. Space velocity = s = / τ = /3 = min - For Mixed flow reactor : Space timeτ = (0.7) τ = C. X ( r ) τ = C. X C. X = r kc ( X ) = 5. min. Space velocity = s = / τ = /5. = min - -c Packed bed(fixed bed) reactor 8 Explanation:Fixed bed reactors normally consist of one or more tubes packed with catalyst particles and are generally operated in vertical position. The catalyst particles may be of a variety of sizes and shapes granular, pelleted, cylinder, spheres etc. Wires of metals are made into screens in the case of expensive catalysts. Chances of hot spot formation are there in fixed bed reactor. In order to prolong the time between regeneration and shut down, the reactor tube may be made longer than required. The

9 (ISO/IEC Certified) SUMMER-6 EXAMINATION Subject code :(756) Page 9 of problem of regeneration of catalyst is a serious issue in this type of reactor. Fluidized bed reactor with regenerator: Explanation: Consider a bed of fine particles and gases passing upward through them. On increasing the flow rate, a point is reached when the solids become suspended in the gas stream. In this state the bed is said to be fluidized and the particular inlet gas velocity is called minimum fluidizing velocity. If the gas velocity is much larger than this minimum, the bed takes the appearance of a violently boiling liquid with solids in vigorous motion and large bubbles rising rapidly through the bed. In this case we have the bubbling fluidized bed. Industrial reactors particularly for solid catalyzed gas phase reactions often operate as bubbling beds with gas velocities equal to 9 to 30 times minimum fluidizing velocity. 3 Attempt any FOUR 6 3-a C NH3 = 0.05 mol / l C H =.5 mol / l C N =. mol / l T = 000K N + 3 H NH 3

10 (ISO/IEC Certified) SUMMER-6 EXAMINATION Subject code :(756) Page 0 of K c = = 0.05 / (.5 3 *.) =.97 * 0-3 (g mol/ l ) - Kp=Kc(RT) Δn Δn = - = - Kp =.97 * 0-3 * ( *000) - =. * 0-7 atm - 3-b Definition of Half life : Itisthetime requiredtoreducetheconcentrationofthereactantto halfofitsoriginalvalue. Relation between half life and rate constant for first order reaction: Characteristics. Half-life of first order reaction is independent of initial concentration of the reactant 3-c Definition: i)fugacity: It is the fictitious (imaginary) pressure for real gases ii) Chemical potential: Itisdefined asthechangeinfree energyofasubstancewhenonemoleofitis addedtosuchalarge quantityofthesystemthatthere is noappreciablechangeinoverallcompositionofthesystematconstant temperature&pressure. iii) chemical equilibrium constant: It is the ratio of rate constants of forward and backward reaction. Or It is the ratio of product of the concentration of the products to the product of concentration of the reactants. iv) Gibb s free energy: Itistheenergyactuallyavailabletodousefulwork.It predicts thefeasibilityandequilibrium

11 (ISO/IEC Certified) SUMMER-6 EXAMINATION Subject code :(756) Page of conditionsforchemicalreactionsat constanttemperatureandpressure. 3-d Types of intermediates in non- chain reaction:.freeradicals.freeatomsorlargerfragments of stable molecules which contain one or more unpaired electrons are called free radicals.the unpaired electrons designated by a dot in the chemical symbol for the substance. Eg.CH C H.Ions andpolarsubstances. Electrically charged atoms, mark each moleculesorfragmentsofmoleculessuchasna +,OH -,NH + arecalledions.theymayact asintermediatesinreaction. 3.Molecules:ConsidertheconsecutivereactionA R S Thisisamultiple reaction.iftheproductrishighlyreactive,itsconcentration inthereactionmixturecanbecometoosmall tomeasure.insuchasituation,r isnotobservedand canbeconsideredtobea reactiveintermediate..transitioncomplexes.the collisionbetweenreactantmoleculesresultina widedistributionofenergiesamongtheindividual molecules.thiscanresultin strainedbonds,unstableformofmoleculesorunstableassociationofmolecules whichcantheneitherdecomposetogiveproductsorbyfurther collisionreturn tomolecules inthenormalstate.suchunstableformsarecalledtransition complexes. 3-e Steps involved in solid catalyzed gas phase reactions. Diffusion of the reactant from bulk fluid phase to external surface of catalyst. Diffusion of reactant tfrom pour mouth into catalyst pores 3. Adsorption of reactant into catalyst surface. Chemical reaction to form product 5. Deadsorption of product 6. Diffusion of deadsorped product from interior of catalyst pores to surface 7. Diffusion of product to the bulk of fluid phase

12 (ISO/IEC Certified) SUMMER-6 EXAMINATION Subject code :(756) Page of Attempt any THREE Derivation of integrated rate expression for irreversible second order reactiona Products: A product In terms of conversion -r A = -d CA /dt = kc A C A =C A0 (-X A ) d CA = -C A0.d XA Aa Ab - d CA = C A0. d XA C A0 (d XA /dt)=k. C A0 (-X A ) On integrating / C A0 { X A /(-X A )}= kt Catalyst deactivation categories. Deactivation by Fouling. Chemical Deactivation - Poisoning 3. Deactivation by Thermal Degradation and Sintering Description: Deactivation by Fouling Generally fouling covers all phenomena where the surface is covered with adeposit. The most widely known form of fouling of catalysts is coke formation. Most of coke formation arises as a result of carbonaceous residues covering the active sites of a heterogeneous catalyst surface, subsequently decreasing the active surface area of the catalyst. In addition the deposition of rust and scale from elsewhere in the catalytic system is not uncommon. Coke-forming processes also involve chemisorption of different kinds of carbons or condensed hydrocarbons that may act as catalyst poisons causing the chemical deactivation of the catalyst. Eg. The fouling of zeolite catalysts occurs in the form of coke molecules limiting the access of the reactant hexane to the active sites of catalyst.

13 (ISO/IEC Certified) SUMMER-6 EXAMINATION Subject code :(756) Page 3 of Due consideration should be given to any other method of catalyst deactivation CA0 =0. mol/l t / = 5 min CA0 = 0.mol/lt / = 568 min t = () [( ) ] ( t / / t / ) = ( C A0 / C A0 ) -n Taking log and rearranging n =-(log(t / / t / ) / log(c A0 / C A0 )) n = -( log (5/5 68)/( log (0./0.)) n= +.0 =.0 Order of the reaction is Derivation of Kp = K c (RT) Δn Consider the reaction aa +bb +. K c = (C R r. C S s ) / (C A a.c B b ) For ideal gas C i = p i / RT rr + ss. Therefore K c = {( p R / RT) r. ( p S / RT) s } / { ( p A / RT) a. ( p B / RT) b } = (p r R.p s S ) / (p a A. p b (r + s + (a+b= )) B ). ( / RT) R is the ideal gas constant T the absolute temperature in K P is the pressure in atm iek c = K p. ( / RT) Δn where K p = (p R r. p S s ) / (p A a. p B b ) Δn = (r + s + -(a + b +.))is the difference in the number of moles of product and reactant n Or K p = K c (RT) B Attempt any one 6 Ac Ad Ba Rate of disappearance of A is -dc A / dt = ( ) / = 0.05 mol / l hr Rate of disappearance of B is -dc B / dt = ( ) / = 0.05 mol / l hr Since rate of disappearance of A is high, the reaction A R proceeds at a faster rate 6

14 (ISO/IEC Certified) SUMMER-6 EXAMINATION Subject code :(756) Page of Bb Rate of formation of R is dc R / dt = (0.05-0) / = 0.05 mol / l hr Rate of formation of S is dc S / dt = (0.05-0) / = 0.05 mol / l hr Derivation for entropy change of an ideal gas From first law of thermodynamics, du = dq dw dq = du + dw..(i) But dq = TdS.(ii) dw = PdV (iii) For ideal gas du= nc v dt (iv) Substituting ii,iii and iv in i TdS = n C v dt + PdV ds = + For ideal gas P = Substituting ds = Integrating between limits + ds = nc v + nr S = nc v ln + nrln (v) For isothermal operation, S = nrln V V For ideal gas PV = RT = Therefore = (vi) Also Cp Cv = R..(vii) Substituting vi and vii in v S = n(c p R)ln + nrln

15 (ISO/IEC Certified) SUMMER-6 EXAMINATION Subject code :(756) Page 5 of S = nc p ln + nrln For isothermal operation, S = nrln P P For constant pressure processnc p ln 5 Attempt any TWO 6

16 (ISO/IEC Certified) SUMMER-6 EXAMINATION Subject code :(756) Page 6 of 5-a Derivation of integrated rate expression for first order reversible reaction A B For first order unimolecular reaction: A B The net rate of disappearance of A 8 ethe rate of formation of B r r = k C k C = k C k C dc dt = dc dt = k C k C Let initial mole ratio of B to A to be M Concentration of A at time t is Therefore on differentiating Concentration of B at time t is C = C ( X ) dc = C dx C = C + C X M = C C dc dt = C dx dt = k C ( X ) k (C + C X ) The reaction is composed of two elementary reaction = C From eq. I, we have = C [k ( X ) k (M + X )]eq. I At equilibrium,net rate is zero,therefore k ( X ) = k (M + X ) K = = = ( ) ( ) eq.iii dx dt = [k ( X ) k (M + X )] = 0 and X = X (fractional conversion) 0 = C [k ( X ) k (M + X )] eq.ii

17 (ISO/IEC Certified) SUMMER-6 EXAMINATION Subject code :(756) Page 7 of Substituting the value of k from eq.iii in above equation dx dt = k ( X ) k ( X ) (M + X ) (M + X ) dx dt = k ( X )(M + X ) ( X )(M + X ) (M + X ) dx dt = k M + X MX X X M X + MX + X X (M + X ) Collecting the terms,we get dx dt = k [X (M + ) X (M + )] (M + X ) dx dt = k (M + ) (M + X ) (X X ) dx = k (M + ) dt X X (M + X ) ln = () ( ) k.t ln = ln = Eq. V is the desired integrated rate equation for reversible first order reaction. Graphical representation A plot of () ( ) k.t eq. IV ln X X vs time gives a straight line with slope equal to X eq.v (M + ) (M + X ) k.

18 (ISO/IEC Certified) SUMMER-6 EXAMINATION Subject code :(756) Page 8 of 5-b Derivation of performance equation for PFR In PFR, the composition of the fluid varies from point to point along the flow path. So the material balance for a reaction component must be made for a differential element of volume dv. 8 Taking material balance of reactant A over the reactor, Input of A to reactor = Output of A from reactor + Disappearance of A due to chemical reaction + Accumulation of A within the reactor. For plug flowreactor, the last term is zero. Input of A to reactor = Output of A from reactor + Disappearance of A due to chemical reaction.()

19 (ISO/IEC Certified) SUMMER-6 EXAMINATION Subject code :(756) Page 9 of Let, F A0 = Molar feed rate to the reactor. C A0 = Molar concentration of A in stream entering the reactor ( moles/volume) X A = Fractional conversion of A F A = F A0 ( - X A ) = F A0 - F A0. X A df A = - F A0. d X A For the differential volume dv Input of A in moles/time = F A Output of A from reactor in moles/time = F A + df A Disappearance of A due to chemical reaction in moles/ time = (-r A ) dv Substituting all the terms in the material balance equation () F A = F A + d F A + (-r A ).dv - d F A = (-r A ) dv F A0. d X A = (-r A ) dv dv F = dx r This is the equation accounting for A in the differential section of the reactor of volume dv. For the whole reactor, the equation must be integrated. dv F A0 r A = dx A x A = 0 = Where X A and (-r A ) are evaluated at the exit conditions For constant volume system : = x A = = C A 0 C Ao Graphical representation:

20 (ISO/IEC Certified) SUMMER-6 EXAMINATION Subject code :(756) Page 0 of 5-c Comparison of MFR and PFR 8 ) For any given duty & for all positive reaction orders,size of MFR is always larger than marks that of PFR. The volume ratio Vm / Vp increases with reaction order. each ) For low values of conversion,reactor size is only slightly affected by type of flow. As for conversion approaches zero, volume / size ratio of these reactors approaches unity. The size any ratio increases very rapidly at high values of conversion. four 3) Design of reactor is affected by density variation during reaction. Expansion(density points decrease) during reaction increases the size ratio.as density decreases, the effectiveness of CSTR wrt PFR. Density increase during reaction has the opposite effect. ) The performance equation for MFR is τ. C = C V = X ( + ε. X ) F k( X ) For PFR τ. C = C... V F = ( + εaxa) K ( X ). dx 5) For a given space time, conversion in a PFR is higher than in MFR. 6) It is possible to operate MFR under isothermal conditions whereas with PFR it is difficult.

21 (ISO/IEC Certified) SUMMER-6 EXAMINATION Subject code :(756) Page of 7) MFR have long residence time compared to PFR. 8) MFR is not suitable for high pressure reactionswhereas PFR is suitable. 6 Attempt any FOUR 6 6-a Space time:it is the time necessary to process one reactor volume of feed at specified conditions. Unit: (units of time) second,minute,hour Space velocity: It is the number of reactor volumes of feed at specified conditions which can be treated in unit time Unit: (reciprocal of time) second -,minute -,hour - 6-b Temperature increase is not desirable for exothermic reaction. Van t Hoff equation is = For exothermic reaction, H is negative. When temperature is increased, dt is positive; the overall equation is negative, which means ln K is negative. When ln K is negative, the value of K is low which denotes lower concentration of products. Therefore temperature increase is not desirable for exothermic reaction. 6-c Significance of activation energy. Activation energy is the minimum energy that the reactants must acquire before reaction takes place to give the product. Activation energy can be considered as a potential energy barrier. Only those molecules that have sufficient energy to reach the top of the barrier and form activated complex react. Smaller the activation energy (low barrier), greater will be the number of activated molecules and faster will be the rate of reaction. It will help to find out the nature of the reaction (endothermic or exothermic)it will give the difference between activated complex and energy of reactant. 3

22 (ISO/IEC Certified) SUMMER-6 EXAMINATION Subject code :(756) Page of 6-d Differential method of analysis of data )Assume a mechanism and from it obtain a rate equation of the form r = dc dt = kf(c) ) From experiment obtain concentration-time data and plot them. 3) Draw a smooth curve through this data. ) Determine the slope of this curve at suitably selected concentration values.these slopes( ) are the rates of reaction at these composition. 5) Evaluate f(c) for each composition. 6) Plot vs f(c)for each composition.if we get a straight line through origin,the rate equation is consistent with the data.if not,the another rate equation should be tested.

23 (ISO/IEC Certified) SUMMER-6 EXAMINATION Subject code :(756) Page 3 of 6-e Method of feeding when PFR s are connected in parallel Consider PFRs are connected as follows For the PFRs connected in parallel,the whole system can be treated as a single plug flow reactor of volume equal to total volume of the individual units if the feed is distributed in such a manner that fluid stream which meet have the same composition.thus for reactors in parallel,v/f must be the same for each parallel line. Let us consider from above figure that volume of reactor in branch A is V & of reactor in branch B is V. For branch A For branch B = = () () For the PFRs in parallel,the ratio V/F must be identical if the conversion is to be the same in each branch. X= x =x 3 V F0 = V Fo F0 = F0 + F0 Therefore Feed flow to first branch should be multiple of. V

24 (ISO/IEC Certified) SUMMER-6 EXAMINATION Subject code :(756) Page of For example: If V = 80 lit & V = 0 lit = = = Therefore /3 rd of feed must be fed to branch A.