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Examples for heterogeneous reaction

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Presentation on theme: "Examples for heterogeneous reaction"— Presentation transcript:

1 Examples for heterogeneous reaction

2 Model Discrimination e
H2 + C2H4  C2H6 Is carried over molybdenum catalyst run Rxn rate PE PEA PH 1 2 3.13 3 5.12 5 4 3.82 4.2 6 2.4 0.5 7 3.8 8 2.2 9 0.75

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10 Tut 12

11 P10-4(a,c) t-butyl alcohol (TBA) produced by liquid phase hydration (W) of isobutene (I) over Amberlyst-15 catalyst. The system is multiphase mixture of hydrocarbon, water and solid catalysts.

12 solution

13 (a) Surface reaction limited

14 (C) Eley Rideal Kinetics

15 P10-9 Methyl-ethyl ketone (MEK) produced from dehydrogenation of butan-2-ol (Bu) over zinc oxide catalyst Bu  MEK +H2 (A  B+C) The following data giving the reaction rate for MEK obtained from differential reactor at 490C P(Bu) 2 0.1 0.5 1 P(MEK) 5 P(H2) 10 r’(MEK) 0.044 0.04 0.069 0.06 0.043 0.059

16 P10-9 Suggest a rate law consist with the exp. Data
P(Bu) 2 0.1 0.5 1 P(MEK) 5 P(H2) 10 r’(MEK) 0.044 0.04 0.069 0.06 0.043 0.059 Any increase in Bu parial pressure slightly increases the reaction rate

17 P10-9 Suggest a rate law consist with the exp. Data
P(Bu) 2 0.1 0.5 1 P(MEK) 5 P(H2) 10 r’(MEK) 0.044 0.04 0.069 0.06 0.043 0.059 MEK partial pressure has little effect on the reaction rate

18 P10-9 Suggest a rate law consist with the exp. Data
P(Bu) 2 0.1 0.5 1 P(MEK) 5 P(H2) 10 r’(MEK) 0.044 0.04 0.069 0.06 0.043 0.059 The partial pressure of H2 has no effect on the reaction rate

19 For H2 and MEK showing little effect we may propose a reaction rate
P(Bu) 2 0.1 0.5 1 r’(MEK) 0.044 0.04 0.069 0.06 0.043 0.059 r’(MEK) P(Bu)

20 P10-9 (b) Suggest a reaction mechanism and a rate limiting step consistent with the rate law. One possible mechanism Assume step 2 is the limiting step and the others are steady state

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23 (c) Plot conversion up to 90% and reaction rate as function of catalyst weight for an entering molar flow rate of pure Bu of 10 mol /min and an entering pressure P0=10atm, Wmax = 23kg

24 (PBu/r’MEK)0.5 PBu

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28 Now consider the change in pressure

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31 Cyclohexanol  Cyclohexene+water
P10-11 Cyclohexene is produced by passing cyclohexanol over catalyst Cyclohexanol  Cyclohexene+water It is suspected that the reaction may involve dual-site mechanism but not known for certain. It is believed that the adsorption equilibrium constant for cyclohexanol is around 1 and it is roughly around One or Two orders of magnitude greater than the adsorption equilibrium constants for other compounds. Suggest a rate law and mechanism consistent with the data given Determine the constants needed for the rate law Why do you think estimates of the rate law parameters were given.

32 Run Reaction rate *105 P (cyclohexanol) P (cyclohexene) P (H2O) 1 3.3 2 1.05 5 3 0.565 10 4 1.826 1.49 6 1.36 7 1.08 8 0.862 9 1.37

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35 (b) To evaluate the rate constants use Polymath with non-linear equation regression k = KA= 4.76 KB= 0.259 KC=0.424

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37 P10-14

38 P10-14

39 Low PTTIP: 2ND order Rxn Since 1>>KP2TTIP/PPI High PTTIP: 1<<KP2TTIP/PPI 0-order Rxn High temperature KI very small such that P2TTIPKPKI<<PPI

40 P10-20(a)

41 P10-20(b)


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