Chemical Engineering - Industrial Organic Chemistry

Exercise 6 - Text

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INDUSTRIAL ORGANIC CHEMISTRY – EXERCISE 6 Methanol synthesis: development of the rate equations Methanol can be synthetizing upon a Cu/ZnO/Al 2O 3 catalyst using a mixture of carbon monoxide, carbon dioxide and hydrogen. Generally speaking, the kinetic mechanism can be based on two different reactions: the methanol synthesis from CO 2 (1) and the reverse water gas shift reaction (2). CO 6 + 3H 6 → CH 7OH + H 6O (1) CO 6 + H 6 → CO + H 6O (2) The following reaction scheme comprises several elementary steps which have to be considered for the synthesis of methanol and for the reverse water gas shift reactions. Vanden Bussche and Froment, Journal of Catalysis 161, 1–10 (1996) It is required to derive the reaction rates for the two reactions according to a Langmuir–Hinshelwood- Hougen-Watson (LLHW) model. The following assumptions can be employed: 1. Hydrogen and carbon dioxide adsorb dissociatively 2. All the active sites are equivalent 3. O*, H 2O* and H* are the only adsorbate present on the surface in non-negligible amount 4. The rate determining step of the methanol synthesis is the hydrogenation of the formate (HCO 2**) 5. The rate determining step of the reverse water gas shift is the dissociative adsorption of CO 2 A set of kinetic measurements obtained in an annular laboratory reactor are available for the determination of the kinetic constants. It is asked to evaluate the kinetic constants of the methanol synthesis and of the reverse water gas shift, given following assumptions: 1. The laboratory reactor is isothermal and isobaric 2. Negligible internal and external transport limitations 3. The adsorption constants are taken from literature Reactor data: Catalyst weight [g] 13.5 Packed bed density [g/cm 3] 1.322 Catalytic bed length [mm] 40.0 Reactor OD [mm] 19.0 Reactor ID [mm] 6.00 Adsorption constants (T [K], R [J/mol/K]): � ������ ������������ 0.499 exp l 17197 ������������ p ������ ������������������ 6.62 ∙10 ?55 exp l 124119 ������������ p ������ ������������������ ������ ������������ ������������ ������������ 3453.38 Equilibrium constants: Methanol synthesis: �������� (������) = 10 @/,22 � ?54 .9=6 A Reverse water gas shift: �������� (������) = 10 @.,3/ � ?6 .46= A Run # H 2 in CO 2 in N2 in TOT FLOW IN [Nl/h] TOT IN [mol/h] P [bar] T avg [°C] H 2 out [mol/h] CO 2 out [mol/h] CH 3OH out [mol/h] CO out [mol/h] H 2O out [mol/h] 1a 71.60% 23.90% 4.50% 114 5.09 70 235 2.857 0.904 0.258 0.036 0.271 1b 71.60% 23.90% 4.60% 114 5.09 70 235 2.809 0.885 0.207 0.042 0.420 1c 71.60% 23.90% 4.60% 114 5.09 70 235 2.897 0.912 0.253 0.041 0.241 2a 68.20% 22.70% 9.10% 114 5.09 70 235 2.697 0.856 0.207 0.039 0.360 2b 68.20% 22.70% 9.10% 114 5.09 70 235 2.686 0.868 0.279 0.043 0.227 2c 68.20% 22.70% 9.10% 114 5.09 70 235 2.673 0.877 0.230 0.037 0.339 3a 68.20% 22.70% 9.10% 114 5.09 50 235 2.863 0.894 0.272 0.039 0.063 3b 68.20% 22.70% 9.10% 114 5.09 50 235 2.888 0.906 0.207 0.039 0.169 3c 68.20% 22.70% 9.10% 114 5.09 50 235 2.850 0.887 0.239 0.039 0.143 4a 71.70% 23.90% 4.40% 57 2.54 50 235 1.449 0.463 0.118 0.030 0.139 4b 71.70% 23.90% 4.40% 57 2.54 50 235 1.381 0.450 0.056 0.033 0.331 4c 71.70% 23.90% 4.40% 57 2.54 50 235 1.440 0.454 0.089 0.030 0.207 5a 71.80% 23.90% 4.30% 28 1.25 50 235 0.701 0.222 0.078 0.024 0.041 5b 71.80% 23.90% 4.30% 28 1.25 50 235 0.682 0.216 0.057 0.023 0.102 5c 71.80% 23.90% 4.30% 28 1.25 50 235 0.714 0.225 0.023 0.030 0.138 6a 68.20% 22.70% 9.10% 550 24.55 90 235 14.368 4.671 0.805 0.062 0.767 6b 68.20% 22.70% 9.10% 550 24.55 90 235 14.870 4.821 0.833 0.064 0.210 6c 68.20% 22.70% 9.10% 550 24.55 90 235 14.368 4.615 0.794 0.061 0.790 7a 68.20% 22.70% 9.10% 550 24.55 70 235 14.652 4.894 0.875 0.060 0.344 7b 68.20% 22.70% 9.10% 550 24.55 70 235 14.652 4.771 0.824 0.050 0.446 7c 68.20% 22.70% 9.10% 550 24.55 70 235 14.820 4.827 0.901 0.057 0.124 Kinetic measurements: