Department of Chemical Engineering Faculty of Engineering Universitas Indonesia 2013.

Slides:



Advertisements
Similar presentations
Lecture 20. Adsorption Phenomena
Advertisements

Topics to be covered in this module
CATALYSIS AND CATALYTIC REACTION MECHANISM PART 1
Chemical Equilibrium Advanced Higher Chemistry Unit 2b.
Reaction Rates and Equilibrium Chapter 17. Collision Theory or Model Molecules react by colliding with each other with enough energy and proper orientation.
{ Equilibrium Learning Goal: Students will understand that many factors influence chemical reactions and some reactions can achieve a state of dynamic.
Using and Controlling Reactions 1.  Most chemical reactions don’t go to completion.  Instead with the right conditions they will reach a balance between.
INTRODUCTION TO CATALYSIS –KINETICS OF CATALYTIC REACTIONS CH
Batch Stoichiometric Table SpeciesSymbolInitialChangeRemaining DD ________ ____________ CC B B A A InertI where and.
Enthalpy C 6 H 12 O 6 (s) + 6O 2 (g) --> 6CO 2 (g) + 6H 2 O(l) kJ 2C 57 H 110 O O 2 (g) --> 114 CO 2 (g) H 2 O(l) + 75,520 kJ The.
This continues our discussion of kinetics (Chapter 13) from the previous lecture. We will also start Chapter 14 in this lecture.
Semester 2 Review.
Adsorption. What is Adsorption? Adsorption is the transfer of a material from one liquid or gaseous state to a surface. The substance that is transferred.
Energy Changes in Chemical Reactions -- Chapter First Law of Thermodynamics (Conservation of energy)  E = q + w where, q = heat absorbed by system.
Daniel L. Reger Scott R. Goode David W. Ball Chapter 17 Chemical Thermodynamics.
Unit 4 Chemical Equilibrium. Completion of reactions Only dependant on limiting reactant? Do all reactions go to completion? Why or why not?
 There are several factors affect reaction rates ◦ Concentration ◦ Surface area (Particle size) ◦ Temperature ◦ Catalysts ◦ Inhibitors.
Chemical Kinetics Chapter 12. Chemical Kinetics The area of chemistry that concerns reaction rates.
Copyright©2004 by Houghton Mifflin Company. All rights reserved. 1 Introductory Chemistry: A Foundation FIFTH EDITION by Steven S. Zumdahl University of.
L17-1 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. Review: Unsteady State Nonisothermal.
Chemical Equilibrium Physical Equilibrium AND. Describe physical and chemical equilibrium Describe the conditions needed for equilibrium. Write the equilibrium.
Chapter 14 Chemical Kinetics (part 2). The Collision Model Goal: develop a model that explains why rates of reactions increase as concentration and temperature.
Chapter 14 Chemical Kinetics (part 2). The Collision Model Goal: develop a model that explains why rates of reactions increase as concentration and temperature.
Chapter 13.  Equilibrium is not static. It is a highly dynamic state.  Macro level reaction appears to have stopped  Molecular level frantic activity.
By Steven S. Zumdahl & Don J. DeCoste University of Illinois Introductory Chemistry: A Foundation, 6 th Ed. Introductory Chemistry, 6 th Ed. Basic Chemistry,
Kinetics, Thermodynamics and Equilibrium Regents Chemistry.
Equilibrium. Rates of reaction Different reactions happen at different speeds. There are ways to speed up or slow down a reaction. Changing the surface.
Chemical Equilibrium Reference: Chapter 9 Reactions Rates and Equilibrium.
Equilibrium. Reversible Reactions Able to proceed in both directions (forward and reverse) PE (kJ) Reaction coordinate Forward AE f Reverse AE r HH.
Chapter 16 Chemical Equilibrium.
Unit 61: Engineering Thermodynamics Lesson 4: Properties of Pure Substances.
Kinetics and Equilibrium. Kinetics Kinetics is the part of chemistry that examines the rates of chemical reactions. Collision theory is the concept of.
Collision Theory  Collision theory is a theory proposed independently by Max Trautz in 1916 and William Lewis in 1918, that qualitatively explains how.
Chapter 5 Rates of Chemical Reaction. 5-1 Rates and Mechanisms of Chemical Reactions 5-2 Theories of Reaction Rate 5-3 Reaction Rates and Concentrations.
Energy Changes in Chemical Reactions -- Chapter First Law of Thermodynamics (Conservation of energy)  E = q + w where, q = heat absorbed by system.
Kinetics, Thermodynamics and Equilibrium Regents Chemistry.
Section 4 Rates of Change. Objectives Describe the factors affecting reaction rates. Describe the factors affecting reaction rates. Explain the effect.
Heterogeneous Catalysis: Kinetics in Porous Catalyst Particles
Reactor analysis (Mass balances, Flow models, Reactors)
Unit 5: Solutions, Kinetics and Equilibrium.
Kinetics and Thermodynamics
ADSORPTION The removal of dissolved substances from solution using adsorbents such as activated carbon.
Chemical Equilibrium.
School for PhD June 8-12, 2015 Luigi PASQUA
Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of chemical reactions and the design of the reactors in which.
ANSWER SCHEME ASSIGNMENT 4
Chapter 17 Equilibrium.
Topic 7- Equilibrium.
Steps in a Heterogeneous Catalytic Reaction
Reaction Mechanism The reaction mechanism is the series of elementary steps by which a chemical reaction occurs. The sum of the elementary steps must give.
The rate and extent of chemical change
Catalysis and Heterogeneous Catalysis
Chapter 7.4 – Reaction Rates
Chapter 16 Equilibrium.
Signs of reactions and Rates of reactions
Flashcards for Unit 12.
Typical Reaction: reactants  products
Chapter 15 Chemical Equilibrium
Chemical Equilibrium.
Le Chatelier’s Principle
Expressions of the equilibrium constant K
Le Chatelier’s Principle
What does equilibrium mean?.
Reaction rates & equilibria
The reversibility of reactions
Chemical Equilibrium Equal but Opposite Reaction Rates
Unit 11: Kinetics and Equilibrium
Equilibrium Rate of Forward Reaction = Rate of Reverse Reaction
Catalysis Catalyst: A substance that speeds up a reaction without being consumed Enzyme: A large molecule (usually a protein) that catalyzes biological.
Chemical Equilibrium Chemistry, The Central Science, 10th edition
Presentation transcript:

Department of Chemical Engineering Faculty of Engineering Universitas Indonesia 2013

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

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)

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

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

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

 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

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)?

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:

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

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

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

 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