Chapter 6: Multiple Reactions CHE 324 Fall 2024 PDF

Chapter 6: Multiple Reactions Chapter 8 in Fifth Edition ChE 324: Kinetics and Reactor Design (A) Fall 2024 1  Introduction In many situations, more than one reaction takes place at the same time. These reactions can inf...

Chapter 6: Multiple Reactions Chapter 8 in Fifth Edition ChE 324: Kinetics and Reactor Design (A) Fall 2024 1  Introduction In many situations, more than one reaction takes place at the same time. These reactions can influence one another as well as influence what products are produced in what amounts. Multiple reactions are different reactions that can occur at the same time. The analysis of multiple reactions is very similar to single reactions, except for the fact that we cannot use conversions X at all. The reasons for that are: 1. Since we have more than one reaction  we have more than one limiting reactant. A limiting reactant for one reaction might be not be a limiting reactant for another. Remember: conversion is defined with respect to the limiting reactant. 2. Since we have more than one reaction, we will have more than conversion. 2  Classification of Multiple Reactions 1) Parallel or competing reactions k1 B A k2 C Desired product 2) Series reactions A k1 B k2 C Desired product 3) Independent reactions k1 k2 A B C D Crude oil cracking k k 4) Complex reactions 1 C+D A + B → 2 →E A + C  3 Concentration Requirements & Reactor Selection kD D How do concentration requirements play into A+B reactor selection? kU U CA0υ0 CAυ0 CB0υ0 CBυ0 PFR PFR (or PBR): concentration is high CSTR: concentration is at the inlet & progressively drops always at its lowest value to the outlet concentration (that at outlet) CBυ0 Semi-batch: concentration of Batch: concentration one reactant (A as shown) is is high at t=0 & high at t=0 & progressively CA(t) CB(t) progressively drops drops with increasing time, with increasing time CA whereas concentration of B can be kept low at all times 4 Review: Multiple Rxns & Selectivity 1) Parallel / competing rxns k1 k2 2) Series rxns A B C k1 B Desired product A k1 k2 3) Complex rxns A+B C+D A+C E k2 C instantaneous rate selectivity, SD/U instantaneous yield, YD rate of formation of D rD (at any point or time in reactor) = SD U = rate of formation of D r rate of formation of = U rU YD = D rate of consumption of A −rA  overall rate selectivity, SD U overall yield, Y  S = = N D Final moles of desired product D DU FD at NU Final moles of undesired product  flow YD = exit FA0 − FA FD Exit molar flow rate of desired product at S D=U = batch Y = N D FU Exit molar flow rate of undesired product D tfinal NA0 − NA Goal: maximize selectivity / yield to maximize production of desired product 5 Calculate the yield of forming B in a CSTR and PFR when the conversion of A is 90% and CA0 = 4 mol/L. The following reactions occur in the reactor: k mol kC A B B r= B k=B 2 A C rC = k CCA k C = 1 min−1 L ⋅ min What is the expression for the yield of B for a CSTR? CBυ0 CB Y = FB = (overall yield)  Y B = →  Y B B CA0υ0 − CAυ0 CA0 − CA FA0 − FA We know CA0 and CA when XA=0.9. How do we get CB? In - Out + Gen. = Accum. dN C υ FB0 − FB + rB V = B → −FB + rB V =0 → rB =B 0 0 dt V 0 CB mol CB mol → rB = → = rB 2 = →2 τ=CB τ L ⋅ min τ L ⋅ min Use the mole balance on A to find τ (at 90% conversion) In - Out + Gen. = Accum. dNA → CA0υ0 − CAυ0 = −rA V FA0 − FA + rA V = dt 0 V CA0 − CA CA0 − C A = −rA −rAτ → → C A0 − C A = τ = υ0 −rA 6 Calculate the yield of forming B in a CSTR and PFR when the conversion of A is 90% and CA0 = 4 mol/L. The following reactions occur in the reactor: kB mol kC A B r= B k=B 2 A C rC = k CCA k C = 1 min−1 L ⋅ min CB mol CA0 − CA −rA = rB + rC Y B = 2 τ = CB =τ What is –rA? CA0 − CA L ⋅ min −rA mol 1 Plug -rA back into → −rA= kB + k CCA → −rA 2 = + CA expression for τ L ⋅ min min CA0 − CA CA0 − CA CA0 = 4 mol/L, and at =τ = →τ −rA mol 1 XA=0.9, CA= 0.4 mol/L 2 + CA L ⋅ min min mol mol 4 − 0.4 τ = L L →τ = 1.5 min Residence time for XA = 0.9 mol 1  mol  2 +  0.4  L ⋅ min min  L  mol r τ 2 (1.5min ) Y B = B → Y B = L ⋅ min  = CA 0 − CA → YB 0.83 mol mol 4 − 0.4 L L 7 Calculate the yield of forming B in a CSTR and PFR when the conversion of A is 90% and CA0 = 4 mol/L. The following reactions occur in the reactor: kB mol kC −1 A B r= B k=B 2 A C rC = k C C A k C = 1 min L ⋅ min mol 1 → −rA= kB + k CCA = → −rA 2 + CA L ⋅ min min What is the expression for the yield of B for a PFR? CBυ0 CB Y = FB (overall yield) =  Y B = → Y B B CA0υ0 − CAυ0 CA0 − CA FA0 − FA Use the mass balance to get CB dFB dCBυ0 dCB dCB mol = rB → =rB → = rB → = 2 dV dV dτ d τ L ⋅ min CB mol τ mol mol → ∫ dCB = 2 L ⋅ min ∫ dτ → CB −= CB0 2 (τ − 0 ) → CB = 2 τ CB0 0 L ⋅ min L ⋅ min 0 Use the mole balance on A to find τ (at 90% conversion) dFA dC Aυ0 dC A dC A  mol 1  = rA → = rA → = rA → = −  2 + C A dV dV d τ d τ  L ⋅ min min  CA dC A  mol  1 dC A 1 τ → −2 = + CA  → ∫ = ∫ dτ dτ  L  min CA0 − ( 2mol L + C A ) min 0 8  Calculate the yield of forming B in a CSTR and PFR when the conversion of A is 90% and CA0 = 4 mol/L. The following reactions occur in the reactor: kB mol kC A B r= B k=B 2 A C rC = k CCA k C = 1 min−1 L ⋅ min mol 1 → −rA= kB + k CCA = → −rA 2 + C L ⋅ min min A CB mol Y B = C = 2 τ Use mole balance on A to find τ (at XA = 0.9) CA0 − CA B L ⋅ min  mol  CA −  2 + C A0  dC A 1 τ  L = 1 → ∫ mol = ∫ d τ → ln (τ − 0 ) CA0 −  2 + C  min 0  mol  min  A −  2 + C A  L   L   mol mol   2 + 4  CA0 = 4 mol/L → ln  L L  = 1 τ → 0.92 min = τ CA = 0.4 mol/L  mo l mol  min  2 + 0. 4   L L  mol Yield was better in C r τ 2 0.92min the CSTR, but the  YB = B = B → =  YB L ⋅ min →=  YB 0.51 residence time CA0 − CA CA0 − CA mol mol 4 − 0.4 was longer L L 9 Series (Consecutive) Reactions k1 k2 A D U Time is the key factor here!!! (desired) (undesired) Spacetime τ for a flow reactor Real time t for a batch reactor To maximize the production of D, use: Batch CSTRs in series or PFR/PBR or n and carefully select the time (batch) or spacetime (flow) 10 Concentrations in Series Reactions k1 k2 -rA = k1CA A B C rB,net = k1CA – k2CB If the above reaction takes place in a PFR: a) Find CA, CB, and CC b) Find the time (space time) to quench the reaction when the concentration of B will be at maximum c) Find the overall selectivity and yield at this quench time Additional information: 11 Concentrations in Series Reactions k1 k2 -rA = k1CA A B C rB,net = k1CA – k2CB How does CA depend on τ? dFA dC A = −k1C A → υ0 C A0e −k1τ −k1C A → C A = = dV dV How does CB depend on τ? dFB = dV k1C A − k 2CB → υ0 = dCB dV k1 C A0e −k1τ − k 2CB ( ) Substitute V υ0 =τ → = dCB dτ ( k1 CA0e−k1τ − k 2CB → dCB dτ ) k1 CA0 e−k1τ + k 2CB = ( ) Use integrating factor (reviewed → ( d CBek 2τ )= k C e( 2 1) → CB = k −k τ  e−k1τ − e−k 2τ  k1CA0   1 A0  k 2 − k1  on Compass) dτ   CC = CA0 − CA − CB 12 Reactions in Series: Cj & Yield B CA = CA0e−k1τ A C  e−k1τ − e−k 2τ  CB = k1CA0     k 2 − k1  CC = CA0 − CA − CB τopt The reactor V (for a given υ0) and τ that maximizes CB occurs when dCB/dt=0 dCB  k1CA0  =   dτ  k 2 − k1  −( k1e −k1τ + k 2 e −k 2τ ) = 0 1 k τ opt = ln 1 k1 − k 2 k 2 V = τ so Vopt = υ0τ opt υ0 13