Tutorial Chemical Reaction Engineering:
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1 Dipl.-Ing. ndreas Jöre Tutorial hemical Reaction Engineering: 5. Kinetics I Institute of Process Engineering, G25-27, andreas.joere@ovgu.de 22-May-4
2 Equilibrium: Information about what is possible but no time eq information c c t Kinetics answer the question how fast a reaction reaches its equilibrium state c inetics equilibrium X eq equilibrium inetics feasible region t T Kinetic information is needed for reactor design and optimization RE: Kinetics I 22-May-4 2
3 Kinetic description requires nowledge about the reaction rate r Definition: reaction rate r vol dni dn rmass V m R i For systems with constant volume (liquid phase reactions) cat i i c r vol i t i r vol dn n with ci V V R i i i i i R t esides this definition there is also a inetic approach Depends on the system, catalyst, chemical mechanisms, Examples: power law, Langmuir-Hinshelwood, Eley-Rideal, Mars-vanKrevelen, Michaelis-Menten / Monod, RE: Kinetics I 22-May-4 3
4 In this tutorial: power law for a liquid phase with constant volume ni r T c j j i i Order of component i (= ν i for elementary reactions) Σn i = overall order of the reaction Example: r T c c overall order of reaction = 2.7 Tas 5.: Series reaction (e.q. selective oxidation sequence) Power law approach 2 Questions:.7 Integrated inetic expression for all three components c t c, c t, c t oncentration-time curves for different parameter sets RE: Kinetics I 22-May-4 4
5 Kinetic approach: power law, first order for all reactants r r c c 2 2 Using the approach for the mass balances r c r r2 c 2c r2 2c Different solution strategies for the coupled ODE system possible Linear system! Laplace Transformation! Holds for a batch system with constant volume i R j i, j r j RE: Kinetics I 22-May-4 5
6 Laplace transformation st L f t F s f t e i st L F s e Fsds with s 2πi i for t f t for t Reduces ODE to E E are easy to solve ODE System hard Solution in t Transformations and inverse transformations can be found in tables! lgebraic equations time domain frequency domain easy Solution in s RE: Kinetics I 22-May-4 6
7 Laplace tables RE: Kinetics I 22-May-4 7
8 Some simple rules for LPT d f t f t F s pplication to our system yields From () follows f t sf s s s c s s s 2 s s c 2 s s s c s c s 2 3 c s s s 2 s s s 2 s c * 2 s s c 2 s s s 2 2c s s s 2 3 * RE: Kinetics I 22-May-4 8
9 Inverse transformation generates the time functions! RE: Kinetics I 22-May-4 9
10 Important transformations from the tables Solution in time domain c t c e t t t 2 e e ct c 2 c t c e t t e e at s a at bt e e a b s a s b at bt be ae ab a b s s a s b. min.3 2 RE: Kinetics I 22-May-4
11 Important transformations from the tables Solution in time domain c t c e t t t 2 e e ct c 2 c t c e t t e e at s a at bt e e a b s a s b at bt be ae ab a b s s a s b min.3 2 RE: Kinetics I 22-May-4
12 Important transformations from the tables Solution in time domain c t c e t t t 2 e e ct c 2 c t c e t t e e at s a at bt e e a b s a s b at bt be ae ab a b s s a s b 2 min.3 2 RE: Kinetics I 22-May-4 2
13 Important transformations from the tables Solution in time domain c t c e t t t 2 e e ct c 2 c t c e t t e e at s a at bt e e a b s a s b at bt be ae ab a b s s a s b min.3 2 RE: Kinetics I 22-May-4 3
14 Important transformations from the tables Solution in time domain c t c e t t t 2 e e ct c 2 c t c e t t e e at s a at bt e e a b s a s b at bt be ae ab a b s s a s b min.3 2 RE: Kinetics I 22-May-4 4
15 If an intermediate is consumed very fast r c r r c c 2 2 odenstein steady state approximation r c 2 2 Simplified solution c t c e c t c t 2 t c t c e t 2 min.3 2 RE: Kinetics I 22-May-4 5
16 If an intermediate is consumed very fast r c r r c c 2 2 odenstein steady state approximation r c 2 2 Simplified solution c t c e c t c t 2 t c t c e t 5 min.3 2 RE: Kinetics I 22-May-4 6
17 If an intermediate is consumed very fast r c r r c c 2 2 odenstein steady state approximation r c 2 2 Simplified solution c t c e c t c t 2 t c t c e t min.3 2 RE: Kinetics I 22-May-4 7
18 If an intermediate is consumed very fast r c r r c c 2 2 odenstein steady state approximation r c 2 2 Simplified solution c t c e c t c t 2 t c t c e t min.3 2 RE: Kinetics I 22-May-4 8
19 Tas 5.2: Equilibrium reaction with constant volume D Question: Derivation of an integrated inetic expression Stoichiometric mixture at t = c t mol/l, c t mol/l c t mol/l, c t mol/l Using the extend of reaction yields c c c c D D Only description of is necessary! D c t c t c t c t c c c t D RE: Kinetics I 22-May-4 9
20 Power law approach and balance for r r c c c c D r r c c c c c c c 2 D Introducing conversion d X X t c c c t 2 2 c X c X c 2 2 c X c X dx 2 2 c 2X X X K K 2 RE: Kinetics I 22-May-4 2
21 ontinued d X c 2X X X K c 2X X Solve ODE with separation of variables Integration K d X 2K c 2X X K ˆ X t d X c X 2 ˆ ˆ K t 2X X K dˆ t RE: Kinetics I 22-May-4 2
22 From integration tables (e.g. ronstein) we find the standard integral X X d Xˆ 2aXˆ b ln with 4ac b ˆ 2 ax bxˆ c 2aXˆ b hec, if Δ < and evaluate conversion integral 4ac b K K X ˆ X d X K K K ln K 2 2 X ˆ ˆ K K 2 X K X X K K K 2 4 X 2 Include this intermediate result in the ODE and integrate over t RE: Kinetics I 22-May-4 22
23 ontinued X X K K K K 2c exp t K K K K K Solve for X X t K K 2c With the conversion 2c exp t K K exp t K K X t c c c t RE: Kinetics I 22-May-4 23
24 ontinued The final result c t c 2c exp t K K exp t K K K K 2c / min / min. / min. / min RE: Kinetics I 22-May-4 24
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