Lecture 5 Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of chemical reactions and the design of the reactors.

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Presentation transcript:

Lecture 5 Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of chemical reactions and the design of the reactors in which they take place.

Lecture 5 Block 1: Mole Balances Block 2: Rate Laws Chapter 4 Lecture 5 Block 1: Mole Balances Block 2: Rate Laws Block 3: Stoichiometry Stoichiometric Table: Flow Definitions of Concentration: Flow Gas Phase Volumetric Flow Rate Calculate the Equilibrium Conversion Xe

Reactor Mole Balances Summary Review Lecture 2 Reactor Mole Balances Summary in terms of conversion, X Reactor Differential Algebraic Integral CSTR PFR Batch X t PBR W

Algorithm How to find Review Lecture 3 Step 1: Rate Law Step 2: Stoichiometry Step 3: Combine to get

Reaction Engineering These topics build upon one another Review Lecture 3 Reaction Engineering Mole Balance Rate Laws Stoichiometry These topics build upon one another

Flow System Stoichiometric Table Chapter 4 Species Symbol Reactor Feed Change Reactor Effluent A FA0 -FA0X FA=FA0(1-X) B FB0=FA0ΘB -b/aFA0X FB=FA0(ΘB-b/aX) C FC0=FA0ΘC +c/aFA0X FC=FA0(ΘC+c/aX) D FD0=FA0ΘD +d/aFA0X FD=FA0(ΘD+d/aX) Inert I FI0=FA0ΘI ---------- FI=FA0ΘI FT0 FT=FT0+δFA0X Where: and Concentration – Flow System

Stoichiometry Liquid Systems Concentration Flow System: Chapter 4 Stoichiometry Concentration Flow System: Liquid Phase Flow System: Liquid Systems Flow Liquid Phase etc.

Liquid Systems If the rate of reaction were then we would have Chapter 4 Liquid Systems If the rate of reaction were then we would have This gives us

Stoichiometry for Gas Phase Flow Systems Chapter 4 Stoichiometry for Gas Phase Flow Systems Combining the compressibility factor equation of state with Z = Z0 Stoichiometry: We obtain:

Stoichiometry for Gas Phase Flow Systems Chapter 4 Stoichiometry for Gas Phase Flow Systems Since , Using the same method,

Stoichiometry for Gas Phase Flow Systems Chapter 4 Stoichiometry for Gas Phase Flow Systems The total molar flow rate is: Substituting FT gives: Where

For Gas Phase Flow Systems Chapter 4 For Gas Phase Flow Systems Concentration Flow System: Gas Phase Flow System:

For Gas Phase Flow Systems Chapter 4 For Gas Phase Flow Systems If –rA=kCACB This gives us FA0/-rA X

For Gas Phase Flow Systems Chapter 4 For Gas Phase Flow Systems

Chapter 4 Example: Calculating the equilibrium conversion (Xef) for gas phase reaction in a flow reactor Consider the following elementary reaction where KC=20 dm3/mol and CA0=0.2 mol/dm3. Calculate Equilibrium Conversion or both a batch reactor (Xeb) and a flow reactor (Xef).

Chapter 4 Gas Flow Example (Xef) Solution: Rate Law:

Gas Flow Example (Xef) Species Fed Change Remaining A FA0 -FA0X Chapter 4 Gas Flow Example (Xef) Species Fed Change Remaining A FA0 -FA0X FA=FA0(1-X) B +FA0X/2 FB=FA0X/2 FT0=FA0 FT=FA0-FA0X/2

Gas Flow Example (Xef) Gas isobaric P=P0 A FA0 -FA0X FA=FA0(1-X) B Chapter 4 Gas Flow Example (Xef) A FA0 -FA0X FA=FA0(1-X) B FA0X/2 FB=FA0X/2 Stoichiometry: Gas isothermal T=T0 Gas isobaric P=P0

Gas Flow Example (Xef) Pure A  yA0=1, CA0=yA0P0/RT0, CA0=P0/RT0 Chapter 4 Gas Flow Example (Xef) Pure A  yA0=1, CA0=yA0P0/RT0, CA0=P0/RT0 At equilibrium: -rA=0

Chapter 4 Gas Flow Example (Xef) Flow: Recall Batch:

Chapter 4

Chapter 4

Chapter 4

End of Lecture 5