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