Post on 23-Feb-2016
description
Group 5
Desti AndaniShinta Leonita
Wisnu Wardana
Department of Chemical EngineeringFaculty of EngineeringUniversitas Indonesia
2013
HOMOGENEOUS HETEROGENEOUS
Form Soluble metal complexes, usually mononuclear
Metals, usually supported, or metal oxides
Active site Well-defined, discrete molecules Poorly definedPhase Liquid Gas/SolidTemperature Low (<250C) High (250-500oC)Activity High VariableSelectivity High VariableDiffusion Facile Can be very importantHeat transfer Facile Can be problematicProduct separation Generally problematic FacileCatalyst recycle expensive simpleCatalyst modification Easy DifficultReaction mechanism Reasonably well understood Not obvious
Sensitivity to deactivation Low High
Major Differences between Homogeneous and Heterogeneous Catalyst
Major Differences between Homogeneous and Heterogeneous Reaction
HOMOGENEOUS REACTION HETEROGENEOUS REACTION
Definitionall 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 rate No Yes
Example 3H2(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)
Major Differences between Catalyst and Biocatalyst
CATALYST BIOCATALYST
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 terms Inorganic catalyst Organic catalystReaction rate Typically slower Several times faster
SpecificityThey are not specific and therefore end up producing residues with errors
Biocatalyst are highly specific producing large amount of good residues
Conditions High temperature Mild conditions, physiological pH and temperature
Example vanadium oxide amylase, lipase
Catalytic Reaction Steps Connected with Mass Transfer
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 surface4. Reaction on the surface of the catalyst (e.g., AB)5. Desorption of the products (e.g., B) from the surface6. Diffusion of the products from the interior of the
pellet to the pore mouth at the external surface7. Mass transfer of the products from the external
pellet surface to the bulk fluid
Calculate dCA/dt for Reversible Reaction
The reaction given below:
The solution is:
(Pseudo Equilibrium)
Calculate dCA/dt for Irreversible Reaction The reaction given below:
The solution is:
Apparatus for the volumetric method
Sensitive beam-type balance used for the gravimetric method
Equipment arrangement for the dynamic method
EXERCISE 1 FOR CHAPTER 5Dinitrogen 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)?
P/P0
Volume adsorbed (cm3/g)Sample 1 Sample 2
0.02 23.0 0.150.03 25.0 0.230.04 26.5 0.320.05 27.7 0.380.10 31.7 0.560.15 34.2 0.650.20 36.1 0.730.25 37.6 0.810.30 39.1 0.89
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:
P/P0 P (kPa)
Sample 1 Sample 2Volume
adsorbed (cm3/g)
Volume adsorbed (cm3/g)
0.02 2.026 99.274 23.0 0.000887 0.15 0.1360.03 3.039 98.261 25.0 0.00124 0.23 0.1340.04 4.052 97.248 26.5 0.00157 0.32 0.1300.05 5.065 96.235 27.7 0.0019 0.38 0.1390.10 10.13 91.17 31.7 0.0035 0.56 0.1980.15 15.195 86.105 34.2 0.00515 0.65 0.2710.20 20.26 81.04 36.1 0.00693 0.73 0.3420.25 25.325 75.975 37.6 0.00887 0.81 0.4120.30 30.39 70.91 39.1 0.011 0.89 0.482
1is
C
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 10-8 m3/g, then using Eq. 2 specific area of solid for Sample 1 is 7.45 x 10-3 m2/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/P0 to get slope (s) & intercept (i) that 1/(s + i) is equal to VM. From graphic, VM = 28.49 cm3/g = 2.849 x 10-8 m3/g then using Eq. 2 specific area of solid for Sample 1 is 7.66 x 10-3 m2/g
0 5 10 15 20 25 30 3520
25
30
35
40
45
f(x) = 0.542291298392219 x + 24.2527906976744
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
0 0.05 0.1 0.15 0.2 0.25 0.3 0.350
0.002
0.004
0.006
0.008
0.01
0.012
f(x) = 0.0352543742355373 x + 9.14694933469218E-05
0 5 10 15 20 25 30 350
0.10.20.30.40.50.60.70.80.9
1f(x) = 0.0244304659580492 x + 0.210968992248062
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 10-10 m3/g, then using Eq. 2 specific area of solid for Sample 1 is 1.95 x 10-4 m2/g
BET constant for Sample 2:Using Eq. 3, then BET constant of solid for Sample 2 is = 16
0 0.05 0.1 0.15 0.2 0.25 0.3 0.350
0.1
0.2
0.3
0.4
0.5
0.6
f(x) = 1.29029120200058 x + 0.0859710019141397
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
Thank You