Lecture 4 Chemical Reaction Engineering CRE is the

Lecture 4 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 4 – Tuesday �Block 1 �Mole Balances �Size CSTRs and PFRs given –r. A=f(X) �Block 2 �Rate Laws �Reaction Orders �Arrhenius Equation �Block 3 �Stoichiometry �Stoichiometric Table �Definitions of Concentration �Calculate the Equilibrium Conversion, Xe 2

Review Lecture 2 Reactor Mole Balances Summary in terms of conversion, X Reactor Differential Algebraic Integral X Batch t CSTR PFR X PBR 3 W

Review Lecture 2 Levenspiel Plots 4

Review Lecture 2 PFR 5

Review Lecture 2 Reactors in Series Only valid if there are no side streams 6

Review Lecture 2 Reactors in Series 7

Review Lecture 2 Two steps to get Step 1: Rate Law Step 2: Stoichiometry Step 3: Combine to get 8

Review Lecture 3 Building Block 2: Rate Laws Power Law Model: A reactor follows an elementary rate law if the reaction orders just happens to agree with the stoichiometric coefficients for the reaction as written. e. g. If the above reaction follows an elementary rate law 9 2 nd order in A, 1 st order in B, overall third order

Review Lecture 3 Arrhenius Equation k E = Activation energy (cal/mol) R = Gas constant (cal/mol*K) T T = Temperature (K) A = Frequency factor (same units as rate constant k) (units of A, and k, depend on overall reaction order) 10

Review Lecture 3 Reaction Engineering Mole Balance Rate Laws Stoichiometry These topics build upon one another 11

Review Lecture 3 Algorithm How to find Step 1: Rate Law Step 2: Stoichiometry Step 3: Combine to get 12

Building Block 3: Stoichiometry We shall set up Stoichiometry Tables using species A as our basis of calculation in the following reaction. We will use the stoichiometric tables to express the concentration as a function of conversion. We will combine Ci = f(X) with the appropriate rate law to obtain -r. A = f(X). A is the limiting reactant. 13

Stoichiometry For every mole of A that reacts, b/a moles of B react. Therefore moles of B remaining: Let ΘB = NB 0/NA 0 Then: 14

Batch System - Stoichiometry Table Species Symbol Initial Change A A NA 0 -NA 0 X B B NB 0=NA 0ΘB -b/a. NA 0 X C C NC 0=NA 0ΘC D D ND 0=NA 0ΘD Inert I NI 0=NA 0ΘI 15 NA=NA 0(1 -X) NB=NA 0(ΘBb/a. X) +c/a. NA 0 X NC=NA 0(ΘC+c/a. X ) +d/a. NA 0 ND=NA 0(ΘD+d/a. X) X -----N =N Θ I FT 0 Where: Remaining A 0 I NT=NT 0+δNA 0 X and δ = change in total number of mol per mol A reacted

Stoichiometry Constant Volume Batch Note: If the reaction occurs in the liquid phase or if a gas phase reaction occurs in a rigid (e. g. steel) batch reactor Then etc. 16

Stoichiometry Constant Volume Batch Suppose Batch: Equimolar feed: Stoichiometric feed: 17

Stoichiometry Constant Volume Batch If , then Constant Volume Batch and we have 18

Batch Reactor - Example Calculate the equilibrium conversion for gas phase reaction, Xe. Consider the following elementary reaction with KC=20 dm 3/mol and CA 0=0. 2 mol/dm 3. Find Xe for both a batch reactor and a flow reactor. 19

Batch Reactor - Example Calculate Xe Step 1: Step 2: rate law: 20

Batch Reactor - Example Symbol Initial Change Remaining A NA 0 -NA 0 X NA 0(1 -X) B 0 ½ NA 0 X NA 0 X/2 Totals: NT 0=NA 0 NT=NA 0 -NA 0 X/2 @ equilibrium: -r. A=0 21

Batch Reactor - Example Solution: At equilibrium Stoichiometry: Constant Volume: Batch Species A B 22 Initial NA 0 0 NT 0=NA 0 Change -NA 0 X +NA 0 X/2 Remaining NA=NA 0(1 -X) NB=NA 0 X/2 NT=NA 0 -NA 0 X/2

Batch Reactor - Example 23

Flow System – Stoichiometry Table Species Symbol Reactor Feed Change Reactor Effluent A A FA 0 -FA 0 X FA=FA 0(1 -X) B B FB 0=FA 0ΘB b/a. FA 0 X FB=FA 0(ΘB-b/a. X) Where: 24

Flow System – Stoichiometry Table Species Symbol Reactor Feed Change C C FC 0=FA 0ΘC +c/a. FA 0 X FC=FA 0(ΘC+c/a. X) D D FD 0=FA 0ΘD +d/a. FA 0 X FD=FA 0(ΘD+d/a. X) Inert I FI 0=A 0ΘI ----- FT 0 Where: Concentration – Flow System 25 Reactor Effluent FI=FA 0ΘI FT=FT 0+δFA 0 X and

Flow System – Stoichiometry Table Species Symbol Reactor Feed Change Reactor Effluent A A FA 0 -FA 0 X FA=FA 0(1 -X) B B FB 0=FA 0ΘB -b/a. FA 0 X FB=FA 0(ΘB-b/a. X) C C FC 0=FA 0ΘC +c/a. FA 0 X FC=FA 0(ΘC+c/a. X) D D FD 0=FA 0ΘD +d/a. FA 0 X FD=FA 0(ΘD+d/a. X) Inert I FI 0=FA 0ΘI ----- FI=FA 0ΘI FT 0 Where: 26 Concentration – Flow System FT=FT 0+δFA 0 X and

Stoichiometry Concentration Flow System: Liquid Phase Flow System: Flow Liquid Phase etc. 27 We will consider CA and CB for gas phase reactions in the next lecture

Heat Effects Isothermal Design Stoichiometry Rate Laws Mole Balance 28

End of Lecture 4 29