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Lecture 5

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 - Thursday 1/20/2011. Block 1: Mole Balances Block 2: Rate Laws Block 3: Stoichiometry

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Lecture 5

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  1. 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.

  2. Lecture 5 - Thursday 1/20/2011 • Block 1: MoleBalances • Block 2: RateLaws • Block 3: Stoichiometry • Stoichiometric Table: Flow • Definitions of Concentration: Flow • Gas Phase Volumetric Flow rate • Calculate the Equilibrium Conversion Xe

  3. Reactor Mole Balance SummaryIn terms of conversion X Batch t X CSTR W PFR PBR

  4. How to find Algorithm Step 1: Rate Law Step 2: Stoichiometry Step 3: Combine to get

  5. ReactionEngineering Mole Balance Rate Laws Stoichiometry These topics build upon one another

  6. FlowSystem Stochiometric Table Where: and Concentration – Flow System

  7. Stoichiometry Concentration Flow System: LiquidPhase Flow System: Flow LiquidPhase Liquid Systems etc.

  8. Liquid Systems If the rate of reactionwere thenwewouldhave This gives us

  9. ForGas Phase Flow Systems • Stoichiometry Combining the compressibility factor equation of state with Z = Z0 Stoichiometry: Weobtain:

  10. Stoichiometry ForGas Phase Flow Systems

  11. Stoichiometry ForGas Phase Flow Systems The total molar flow rate is: Substituting FT gives:

  12. ForGas Phase Flow Systems Concentration Flow System: Gas Phase Flow System:

  13. ForGas Phase Flow Systems If –rA=kCACB This gives us FA0/-rA X

  14. ForGas Phase Flow Systems

  15. Example: Calculating the equilibrium conversion for gas phase reaction in a flow reactor, Xef Consider the following elementary reaction with KC=20 dm3/mol and CA0=0.2 mol/dm3. CalculateEquilibriumConversion or both a batch reactor (Xeb) and a flow reactor (Xef).

  16. Gas Flow Example Xef Rate law: Solution:

  17. Gas Flow Example Xef

  18. Gas Flow Example Xef Stoichiometry: Gas isothermal T=T0, isobaric P=P0

  19. Gas Flow Example Xef Pure A  yA0=1, CA0=yA0P0/RT0, CA0=P0/RT0 @ eq: -rA=0

  20. Gas Flow Example Xef Flow: Batch: Recall

  21. End of Lecture 5

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