Download presentation

Presentation is loading. Please wait.

Published byBryanna Guttridge Modified about 1 year ago

1
Department of Chemical Engineering Faculty of Engineering Universitas Indonesia 2013

2
HOMOGENEOUSHETEROGENEOUS Form Soluble metal complexes, usually mononuclear Metals, usually supported, or metal oxides Active siteWell-defined, discrete moleculesPoorly defined PhaseLiquidGas/Solid TemperatureLow (<250C)High ( oC) ActivityHighVariable SelectivityHighVariable DiffusionFacileCan be very important Heat transferFacileCan be problematic Product separationGenerally problematicFacile Catalyst recycleexpensivesimple Catalyst modificationEasyDifficult Reaction mechanismReasonably well understoodNot obvious Sensitivity to deactivation LowHigh

3
HOMOGENEOUS REACTIONHETEROGENEOUS REACTION Definition all reactants are in the same phase more than one phase in reactants Equilibrium Constant Rate (K) Equal between forward and reverse reaction Difference between forward and reverse reaction Surface area affects the reaction rateNoYes Example3H2(g) + N2(g) --> 2NH3(g) Zn(s) + 2HCl(aq) --> H2(g) + ZnCl2(aq) Ag+(aq) + Cl-(aq) --> AgCl(s)C(s) + O2(g) --> CO2(g)

4
CATALYSTBIOCATALYST Definition catalysts are substances that increases or decrease the rate of a chemical reaction but remain unchanged may be broadly defined as the use of enzymes or whole cells to increase speed in which a reaction takes place but do not affects the thermodynamics of reaction Molecular weight low molecular weight compounds High molecular weight globular protein or whole cells Alternate termsInorganic catalystOrganic catalyst Reaction rateTypically slowerSeveral times faster Specificity They are not specific and therefore end up producing residues with errors Biocatalyst are highly specific producing large amount of good residues ConditionsHigh temperature Mild conditions, physiological pH and temperature Examplevanadium oxideamylase, lipase

5
Steps in a Catalytic Reaction: 1. Mass transfer (diffusion) of the reactant(s) (e.g., species A) from the bulk fluid to the external surface of the catalyst pellet 2. Diffusion of the reactant from the pore mouth through the catalyst pores to the immediate vicinity of the internal catalytic surface 3. Adsorption of reactant A onto the catalyst surface 4. Reaction on the surface of the catalyst (e.g., A B) 5. Desorption of the products (e.g., B) from the surface 6. Diffusion of the products from the interior of the pellet to the pore mouth at the external surface 7. Mass transfer of the products from the external pellet surface to the bulk fluid

6

7
The reaction given below: The solution is: (Pseudo Equilibrium)

8
The reaction given below: The solution is:

9

10
Apparatus for the volumetric method Sensitive beam-type balance used for the gravimetric method Equipment arrangement for the dynamic method

11
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)?

12
Normal boiling point of dinitrogen is 77 K and the saturated vapour pressure P 0 = 1.05 bar = kPa. Assuming mass of each sample is 1 gram. Table modification for the answer: Equation needed:

13
BET surface area for Sample 1 using the one-point BET equation: Plotting V against P to get the ‘Point B’ as V M V M = 27.7 cm 3 /g = 2.77 x m 3 /g, then using Eq. 2 specific area of solid for Sample 1 is 7.45 x m 2 /g BET surface area for Sample 1 using the full BET equation: From Eq. 1, plotting data in the form P/[V(P 0 -P)] against P/P 0 to get slope (s) & intercept (i) that 1/(s + i) is equal to V M. From graphic, V M = cm 3 /g = x m 3 /g then using Eq. 2 specific area of solid for Sample 1 is 7.66 x m 2 /g

14
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 = BET surface area for Sample 2 using the one-point BET equation: Plotting V against P to get the ‘Point B’ as V M

15
V M = 0.38 cm 3 /g = 3.8 x m 3 /g, then using Eq. 2 specific area of solid for Sample 1 is 1.02 x m 2 /g BET surface area for Sample 2 using the full BET equation: From Eq. 1, plotting data in the form P/[V(P 0 -P)] against P/P 0 to get slope (s) & intercept (i) that 1/(s + i) is equal to V M. V M = 0.72 cm 3 /g = 7.2 x m 3 /g, then using Eq. 2 specific area of solid for Sample 1 is 1.95 x m 2 /g BET constant for Sample 2: Using Eq. 3, then BET constant of solid for Sample 2 is = 16

16
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

17

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google