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Chemical Reactor Design Chemical Reactor Design Youn-Woo Lee School of Chemical and Biological Engineering Seoul National University 155-741, 599 Gwanangro, Gwanak-gu, Seoul, Korea ywlee@snu. ac.kr http://sfpl. snu. ac. kr 第7章 化學反應裝置設計 Chemical Reactor Design Reaction Mechanisms , Pathways, Bioreactions and Bioreactors Seoul National University Objective Discuss the pseudo-steady-state-hypothesis and explain how it can be used to solve reaction engineering problems Write reaction pathways for complex reactions. Explain what an enzyme is and how it acts as a catalyst. Describe Michealis-Menten enzyme kinetics and rate law along with its temperature dependence. Discss h ow t o di stin gui sh th e diff er ent t ypes of enz yme inhibition. Discuss stages of cell growth and rate laws used to describe growth. Write material balances on cells, substrates, and products in bioreactors to size chemostats and plot concentration-time trajectoriesin batch reactors. Describe how physiologically-based pharmocokinetic models can be used to model alcohol metabolism. Seoul National University 7.1 Active Intermediates and Nonelementary Rate law • Elementary rate law - the reaction order of each species is identical with the stoichiometric coefficient of that species for the reaction as written n rA kC A • Non-elementary rate law - no direct correspondence between reaction order and stoichiometry 3 / 2 CH CHO CH + CO r kC 3 4 CH 3CHO CH 3CHO k C C 3/ 2 1 H 2 Br2 H2+Br+ Br2 2HBr rHBr CHBr k2CBr Seoul National University Non-elementary Reaction Non-elementary rate laws involve a number of elementary reactions and at least one active intermediate. An active intermediate is a high-energy molecule that reacts virtually as fast as it is formed. As a result, it is present in very small concentrations. Active intermediates (e.g., A*) can be formed by collision or interaction with other molecules. AMA + M → A* + M Here the activation occurs when translational kinetic energy is transferred into energy stored in internal degrees of freedom, particularly vibrational degrees of freedom. Seoul National University 7.1.1 Pseudo-Steady-State Hypothesis (PSSH) Because a reactive intermediate reacts virtually as fast as it is formed, the net rate of formation of an active intermediate (e.g., A*) is zero, i.e., rA* 0 This condition is also referred to as the Pseudo-Steady-State Hypothesis (PSSH). If the active intermediate (e.g., A*) appears in n reactions, then n rA* riA* 0 i1 Seoul National University Pseudo-Steady-State-Hypothesis (PSSH) Let’s consider the gas-phase decomposition of azomethane, AZO, to give ethane and nitrogen. (CH3)2N2 C2H6 + N2 azomethane (AZO) ethane nitrogen Experimental observation shows that: High concentration (> 1atm) r ~ C C 2 H 6 AZO Low concentration (< 50 mmHg) r ~ C 2 C 2 H 6 AZO How to expppglain this first and second order depending on the concentration of AZO? New proposed mechanism that consisting of three elementary reactions. Seoul National University Proposed mechanism k1 rxn 1: (CH3)2N2 + (CH3)2N2 (CH3)2N2 + [(CH3)2N2]* k2 rxn 2: [(CH3)2N2]* + (CH3)2N2 (CH3)2N2 + (CH3)2N2 k3 rxn 3: [(CH3)2N2]* C2H6 + N2 2 r1, AZO * k 1C AZO r2 , AZO * k 2 C AZO * C AZO r3 , AZO * k 3 C AZO * 3 2 rAZO * ri , AZO * k 1C AZO k 2 C AZO C AZO * k 3 C AZO * i 1 The concentration of the active intermediate, AZO*, is very difficult to measure, because it is highly reactive and very short-lived (~10-9 second). We need to express CAZO* in terms of CAZO. Seoul National University Pseudo-Steady-State Hypothesis (PSSH) 1. Life time of active intermediate ~0 2. Active intermediate is ppyresent only in low concentrations rate of formation of the active intermediate= rate of disappearance Net rate of formation of the active intermediate = 0 2 rAZO * 0 k 1C AZO k 2 C AZO C AZO * k 3 C AZO * (7-7) k C 2 C 1 AZO AZO * (7-8) k 2 C AZO k 3 k k C 2 r k C 1 3 AZO C 2 H 6 3 AZO * (7-9) k 2 C AZO k 3 Seoul National University k k C 2 r 1 3 AZO N 2 Low AZO Conc. k 3 k 2 C AZO High AZO Conc. k 2 C AZO k 3 k 2 C AZO k 3 k k r k C 2 r 1 3 C kC N 2 1 AZO N 2 AZO AZO k 2 @ low concentration @ high concentration apparent second order apparent first order It is normal procedure to reduce the additive constant in the denominator to 1 k k C 2 k C 2 r 1 3 AZO r 1 AZO N 2 N 2 k 3 k 2 C AZO 1 k C AZO additive constant Seoul National University 7.1.2 Searching for a Mechanism In many instances the rate data are correlated before a mechanism is found. Seoul National University Rules of Thumb for Development of a Mechanism 1. Species having the concentration appearing in the denominator of the rate law probably collide with the active intermediate, e.g., A+A*A + A* [co llisi on prod uct s] 2. If a constant appears in the denominator, one of the reaction steps is probably the spontaneous decomposition of the active intermediate , e. g., A* [decomposition products] 3. Species having the concentration appearing in the numerator of the rate law probably produce the active intermediate in one of the reaction step, ege.g., [reactant] A* + [other products] Seoul National University Rules of Thumb for Development of a Mechanism Upon application of Table 7-1 to azomethane example just discussed, we see the following from rate equation (7-12) k C 2 r 1 AZO C 2 H 2 1 k C AZO 1. Theactive idiintermediate, AZO*,collides wihith azomethane, AZO [Reaction 2], resulting in the concentration of AZO in the ditdenominator. 2. AZO* decomposes spontaneously [Reaction 3], resulting in a constant in the denominator of the rate expression. 3. The appearance of AZO in the numerator suggests that the active intermediate AZO* is formed from AZO. Referring to [Reaction 1], we see that this case is indeed true. Seoul National University Rules of Thumb for Development of a Mechanism k1 (CH3)2N2 + (CH3)2N2 (CH3)2N2 + [(CH3)2N2]* 3. Concentration of AZO in numerator: k C 2 AZO* is formed from AZO r 1 AZO N 2 1 k C AZO k2 [(CH3)2N2]* + (CH3)2N2 (CH3)2N2 + (CH3)2N2 1. CttifAZOConcentration of AZO k3 in denominator: [(CH3)2N2]`* C2H6 + N2 active intermediate AZO* collid e w ith AZO 2. A constant in denominator: AZO* decomposes spontaneously Seoul National University Steps to deduce a rate law 1. Assume an activated intermediate(()s) 2. Postulate a mechanism, utilizing the rate law obtained from experimental data, if possible. 3. Model each reaction in the mechanism sequence as an elementary reaction. 4. After writing rate laws for the rate of formation of desired product, write the rate laws for each of the active intermediates. 5. Use the PSSH 6. Eliminate the concentration of the intermediate species in the rate laws by solving the simultaneous equations developed in step 4 and 5. 7. If the derived rate law does not agree with experimental observation, assume a new mechanism and/or intermediates and go to step 3. A strong background in organic and organic chemistry is helpful in predicting the activated intermediates for the reaction under consideration. Seoul National University 7.1.3 Steps in a Chain Reaction A chain reaction consists of following sequence: Initiation Propagation Termination 1. Initiation : formation of an active intermediate 2. Propagation or chain transfer : interaction of an active intermediate with the reactant or product to produce another active intermediate 3. Termination : deactivation of the active intermediate Seoul National University Example 7-2 PSSH Applied to Thermal Cracking of Ethane The thermal decomposition of ethane to ethylene, methane, butane, and hdhydrogen is blibelieve dtd to procee did in thfllthe follow ing sequence: Initiation k1 (1) C2H6 2CH3 -r1, C2H6=k1[C2H6] Propagation k2 (2) CH3 + C2H6 CH4 + C2H5 -r2, C2H6=k2[CH3][C2H6] k3 (3) C2H5 C2H4 + H -r3, C2H4=k3[C2H5] k (4) H + C2H6 4 C2H5 +H2 -r4, C2H6=k4[H][C2H6] Termination k 5 2 (5) 2C2H5 C4H10 -r5, C2H5 =k5[C2H5] (a) Use the PSSH to derive a rate law for the rate of formation of ethylene (b) Compare the PSSH solution in Part (a) to that obtained by solving the complete set of ODE mole balances. Seoul National University Solution (a) Developing the Rate Law rate of formation of ethylene: r k [C H ] (1) 3,C2H4 3 2 5 for the active intermediates: CH3•, C2H5•, H• the net rates of rxn are r r r r r 0 C2H5 2,C2H5 3,C2H5 4,C2H5 5,C2H5 (2) r2,C H r3,C H r4,C H r5,C H 0 PSSH 2 6 2 4 2 6 2 5 r r r 0 0 (3) H 3,C2H 4 4,C2H6 r 2r r 0 (4) CH3 1,C2H6 2,C2H6 S()Substituting the rate laws into (4) 2k1[C2 H6 ] k2[CH3 ][C2 H6 ] 0 (5) 2k1 [CH3 ] (6) k2 (2)+(3) yields r r 0 2,C2H6 5,C2H5 2 (7) k2 [CH3 ][C2 H 6 ]-k5[C2 H5 ] 0 Seoul National University Solving for [C2H5•] gives us 1/ 2 1/ 2 k2 2k1k2 [C2H5 ] [CH3][C2H 6 ] [C2H 6 ] k5 k2k5 1/ 2 2k1 [C2H 6 ] (8) k5 (8) (1) 1/ 2 2k r k [C H ] k 1 [C H ]1/ 2 (9) C2H4 3 2 5 3 2 6 k5 r k [C H ] k [CH ][C H ] k [H][C H ] (10) Rate of C2H6 1 2 6 2 3 2 6 4 2 6 disappearance of ethane k3[C2 H5 ] k4[H][C2 H 6 ] 0 (3) 1/ 2 k 2k 3 1 1/ 2 [H] [C2 H 6 ] (11) k4 k5 1/ 2 2k r 3k [C H ] k 1 [C H ]1/ 2 (12) C2H6 1 2 6 3 2 6 k5 Seoul National University For a constant-volume batch reactor, 1/ 2 d[C H ] 2k 2 6 r 3k [C H ] k 1 [C H ]1/ 2 Combined C2H6 1 2 6 3 2 6 Mole balances dt k5 & rate laws 1/ 2 d[C H ] 2k 2 4 r k 1 [C H ]1/ 2 C2H 4 3 2 6 dt k5 0100.10 @1000K ] 3 0.08 -3 -1 CC2H4 k1=1.5x10 s m 6 3 dd 0.06 k2=232.3x 10 dm /l/mol•s 4 -1 k3=5.71x10 s 8 3 [mol/ 0.04 k =9.53x10 dm /mol•s 4 ii 9 3 C CC H k5=3.98x10 dm /mol•s 0.02 2 6 3 0000.00 C0, C2H6=0.1mol/dm 0 3 6 9 12 15 t [sec]
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