1 / 17

Group 5 Desti Andani Shinta Leonita Wisnu Wardana

Group 5 Desti Andani Shinta Leonita Wisnu Wardana. Department of Chemical Engineering Faculty of Engineering Universitas Indonesia 2013. Major Differences between Homogeneous and Heterogeneous Catalyst. Major Differences between Homogeneous and Heterogeneous Reaction.

bjorn
Download Presentation

Group 5 Desti Andani Shinta Leonita Wisnu Wardana

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Group 5DestiAndaniShintaLeonitaWisnuWardana Department of Chemical Engineering Faculty of Engineering Universitas Indonesia 2013

  2. Major Differences between Homogeneous and Heterogeneous Catalyst

  3. Major Differences between Homogeneous and Heterogeneous Reaction

  4. Major Differences between Catalyst and Biocatalyst

  5. Catalytic Reaction Steps Connected with Mass Transfer Steps in a Catalytic Reaction: • Mass transfer (diffusion) of the reactant(s) (e.g., species A) from the bulk fluid to the external surface of the catalyst pellet • Diffusion of the reactant from the pore mouth through the catalyst pores to the immediate vicinity of the internal catalytic surface • Adsorption of reactant A onto the catalyst surface • Reaction on the surface of the catalyst (e.g., AB) • Desorption of the products (e.g., B) from the surface • Diffusion of the products from the interior of the pellet to the pore mouth at the external surface • Mass transfer of the products from the external pellet surface to the bulk fluid

  6. Calculate dCA/dt for Reversible Reaction • The reaction given below: • The solution is: (Pseudo Equilibrium)

  7. Calculate dCA/dt for Irreversible Reaction • The reaction given below: • The solution is:

  8. Apparatus for the volumetric method • Sensitive beam-type balance used for the gravimetric method • Equipment arrangement for the dynamic method

  9. Exercise 1 for chapter 5 Dinitrogen adsorption data: (a)Calculate the BET surface area per gram of solid for Sample 1 using the full BET equation and the one-point BET equation. Are the values the same? What is the BET constant? (b)Calculate the BET surface area per gram of solid for Sample 2 using the full BET equation and the one-point BET equation. Are the values the same? What is the BET constant and how does it compare to the value obtained in (a)?

  10. Normal boiling point of dinitrogen is 77 K and the saturated vapour pressure P0 = 1.05 bar = 101.3 kPa. Assuming mass of each sample is 1 gram. Table modification for the answer: Equation needed:

  11. BET surface area for Sample 1 using the one-point BET equation: Plotting V against P to get the ‘Point B’ as VM VM = 27.7 cm3/g = 2.77 x 10-8m3/g, then using Eq. 2 specific area of solid for Sample 1 is 7.45 x 10-3m2/g BET surface area for Sample 1 using the full BET equation: From Eq. 1, plotting data in the form P/[V(P0-P)]against P/P0to get slope (s) & intercept (i) that 1/(s + i) is equal to VM. From graphic, VM = 28.49 cm3/g = 2.849 x 10-8m3/g then using Eq. 2 specific area of solid for Sample 1 is 7.66 x 10-3 m2/g

  12. Graphic full BET method of solid for Sample 1: BET constant for Sample 1: Using Eq. 3, then BET constant of solid for Sample 1 is = 389.889 BET surface area for Sample 2 using the one-point BET equation: Plotting V against P to get the ‘Point B’ as VM

  13. VM = 0.38 cm3/g = 3.8 x 10-10 m3/g, then using Eq. 2 specific area of solid for Sample 1 is 1.02 x 10-4 m2/g BET surface area for Sample 2 using the full BET equation: From Eq. 1, plotting data in the form P/[V(P0-P)]against P/P0 to get slope (s) & intercept (i) that 1/(s + i) is equal to VM. VM= 0.72 cm3/g = 7.2 x 10-10m3/g, then using Eq. 2 specific area of solid for Sample 1 is 1.95 x 10-4m2/g BET constant for Sample 2: Using Eq. 3, then BET constant of solid for Sample 2 is = 16

  14. Difference value of surface area using one-point BET eq. and full BET eq. : Discrepancy value between those method illustrated the dangers in relying on the estimation of a single point either by inspection (point B method) therefore point B is not particularly well defined and the BET full method more empirical. • Comparison of BET constant between Sample 1 and 2 : Comparison of the BET constant obtained from Sample 1 & 2 indicated its depends on the difference on volume adsorbed of each sample that showed by slope and intercept of line that used to calculate the layer of adsorbed gas quantity

  15. Thank You

More Related