L17-1
Review: Unsteady State
Nonisothermal Reactor Design
Q
Goal: develop EB for unsteady state reactor
An open system (for example, CSTR)
Fin
Fout
Hin
Hout
W
dEsys
dt

Q

Rate of
rate of heat
accumulation
flow from
=
of energy in
surroundings
system
to system
dEsys
dt
W
n

 FE
i i
i1
in

Rate of work
Rate of energy
done by
added to
+
system on
system by
surroundings
mass flow in
 0 steady state
dEsys
dt
n
 FE
i i
i1
out
Rate of energy
leaving system
by mass flow
out
 0 unsteady state
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L17-2
Review: Simplified EB for Well-Mixed
Reactors
n
n
Q  WS   FH
i i
i1
n
  FH
i i
in
i1
out
dHi n dNi d
  Ni
  Hi
 PV 
dt
dt
dt
i1
i1
Total Energy Balance for unsteady state
Constant PV variation
Made following
dNi
substitutions & dHi  Cpi dT
d
 Fi0  F i  irA V
PV   0
dt
dt
solved for dT/dt: dt
dt
n
Energy balance for
unsteady state reactor
with phase change:
Q  WS   Fi0 Hi  Hi0   H RX  T  rA V 
i1
n
 NiCpi

dT
dt
i1
n
Q  WS   Fi0Cpi  T  Ti0   H RX  T rA V 
Energy balance for
dT
i1

unsteady state reactor
n
dt
 NiCpi
without phase change:
i1
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Review: Unsteady State EB,
Liquid-Phase Reactions
L17-3
n
Q  WS   Fi0Cpi  T  Ti0   H RX  T rA V 
i1
n
 NiCpi

dT
dt
i1
For liquid-phase reactions, often Cp = SiCpi is so small it can be neglected
When Cp can be neglected, then:
n
 NiCpi  NA0Cps where Cps  SiCpi is the heat capacity of the solution
i1
If the feed is well-mixed, it is convenient to use:
SFi0Cpi  FA0Cps
Plug these equations
& Ti0 = T0 into the EB:
Q  WS  FA 0 Cps  T  Ti0   H RX  T  rA V  dT

NA0 Cps
dt
This equation for the EB is simultaneously solved with the mass
balance (design eq) for unsteady state, nonisothermal reactor design
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Review: Nonisothermal Batch
Reactor Design
L17-4
n
Q  WS   Fi0Cpi  T  Ti0   H RX  T rA V 
i1
0
n
 NiCpi

dT
dt
i1
No flow, so:

Q  WS  H RX  T  rA V 
n
 NiCpi
i1
Put the energy balance N  N     X  where   Ni0 &  C  C
i
A0
i
i A
i
i pi
P
NA 0
in terms of conversion:

Q  WS  H RX  T  rA V 
n

NA0   iCpi  Cp X A 
i1


dT
dt
Solve with the batch reactor design
dX A
N
equation using an ODE solver (Polymath): A0 dt  rA V
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Review: Adiabatic Nonisothermal
Batch Reactor Design
L17-5
0 0
For negligible
Q  WS  H RX  T rA V  dT Substitute:  iCpi  Cps


stirring work:
n
dt Rearrange


NA0   iCpi  Cp XA 
i1

XA
T
dX A
dT
 
 
X 0 CpS  Cp X A
T H RX  TR   Cp  T  TR 
A0
0
CpS  T  T0 
CpS  T  T0 
Integrate &  X 
 XA 
A
solve for XA:
HRX  T 
 H RX  TR   Cp  T  TR 

Solve for T:

 H RX  T0   X A
 H RX  T0   X A


T  T0  
 T  T0  n
Cps  X A Cp
 iCpi  X A Cp
Solve with the batch reactor design
equation using an ODE solver (Polymath)
i1
XA
dX A
0 rA V
t  NA0 
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L17 Basic Catalysis & Reaction
Mechanisms
L17-6
• Though we have discussed the use of catalyst in a PBR, we have not
discussed the process of catalysis itself
• An understanding of catalysis, the mechanisms, and catalytic reactor design
are the subject of the next few lectures
• Catalyst properties
• Steps involved in a catalytic reaction
• Development of a rate law using steps in catalytic reaction
• Different types of catalyst mechanisms
• Design of catalytic reactors
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L17-7
Catalysts & Catalysis
• ~1/3 of the GNP of materials produced in the US involve a catalytic process
• A Catalyst is a substance that speeds up the rate of reaction but is not
changed by the reaction
• A catalyst lowers the energy barrier by promoting a different molecular
pathway (mechanism) for the reaction
• Homogeneous catalysis: catalyst is in solution with at least 1 reactant
• Heterogeneous catalysis: more than 1 phase, usually solid and fluid or
solid and gas is present. Reaction occurs at solid/liquid or gas interface.
Many catalyst are porous (high surface area)
catalyst
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Steps in a Heterogeneous Catalytic
Reaction
7. Diffusion of B
from external
surface to the bulk
fluid (external
diffusion)
1. Mass transfer
of A to surface
2. Diffusion of A
from pore mouth to
internal catalytic
surface
L17-8
3. Adsorption of
A onto catalytic
surface
6. Diffusion of B
from pellet interior
to pore mouth
5. Desorption of product
B from surface
4. Reaction on surface
Ch 10 assumes steps 1,2,6 & 7 are fast, so only steps 3, 4, and 5 need to be considered
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L17-9
Adsorption Step
The adsorption of A (gas phase) on an active site S is represented by:
A
A
A(g) + S ⇌ A·S
I
-S-S-S-S-S-SS: open (vacant) surface site
A·S: A bound to a surface site
Rate of adsorption = rate of attachment – rate of detachment
rAD  k APACv  k ACAS Conc of sites occupied by A
partial pressure of A
Molar conc of vacant sites on surface
• Rate is proportional to # of collisions with surface, which is a function of PA
• Rate is proportional to # of vacant (active) sites, Cv, on the surface
• Active site: site on surface that can form a strong bond with adsorbed species
k
In terms of the adsorption equilibrium constant KA where K A  A
kA


kA

r

k
P
C

C
rAD  k APACv  k ACAS
AD
A A v
A S 
k


A


C
 rAD  k A  PA Cv  A S 
KA 

Equation I
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L17-10
Site Balance
Ct: Total number of active sites per unit mass of catalyst divided by
Avogadro’s # (mol/g cat)
Cv: Number of vacant sites per unit mass of catalyst divided by Avogadro’s #
Vacant
active site
Active site
occupied by A
A
Active site
occupied by B
B
Surface
Cv is not measurable, but the total number of sites Ct can be measured
In the absence of catalyst deactivation, assume the total number of
active sites remains constant:
Site balance:
Ct = Cv + CA·S + CB·S
We will use the site balance equation to put Cv in terms of measurable species
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L17-11
Langmuir Isotherm Adsorption
Adsorption of carbon monoxide onto a surface: CO + S ⇌ CO·S

CCOS 
kA

r

k
P
C

KA 
AD
A  CO v

rAD  k APCOCv  k  ACCOS
K
kA

A 
Determine the concentration of CO adsorbed on the surface at equilibrium
At equilibrium, rAD = 0:
Rearrange &
solve for CCO·S


C
rAD  0  k A  PCOCv  COS 
KA 

CCOS

 PCOCv
KA
 CCOS  K APCOCv
Put Cv in terms of Ct using the site balance; only CO can absorb on the surface:
Ct  Cv  CCOS
 Ct  CCOS  Cv Insert into eq for CCO·S from rxn rate
CCOS  K APCOCv  CCOS  K APCO  Ct  CCOS  Solve for CCO·S
 CCOS  K APCOCt  K APCOCCOS  CCOS  K APCOCCOS  K APCOCt
K P C
 CCOS  A CO t Concentration of CO adsorbed on
1  K APCO surface vs PCO→ Langmuir Isotherm
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L17-12
Surface Reaction Step
After the molecule is adsorbed onto the surface, it can react by a few different
mechanisms
1. Singe site mechanism: Only the site to which the reactant is absorbed is
involved in the reaction

kS
CBS 
A
B
where
K

r

k
C

A·S ⇌ B·S
S
S
S  A S

k S
K
I
I

S 
⇌
-S-SEquation IIa
2. Dual site mechanism: Adsorbed reactant interacts with another vacant
site to form the product
A·S + S ⇌ S + B·S
A
B

CBSCv 
rS  k S CASCv 
I
I
 Equation IIb
⇌
K

S 
-S-S-S
-S-S-S3. Eley-Rideal mechanism: reaction between adsorbed reactant and a
molecule in the gas phase A·S + B(g) ⇌ C·S
B
C
A

CCS 
r

k
C
P

I
S
S  A S B
 Equation IIc
I
⇌
K

S 
-S-S-S-S-S-S
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L17-13
Desorption Step
Products are desorbed into the gas phase
C·S ⇌ C + S
C
C
I ⇌
-S-S-S-S-S-S-

PCCv 
rD,C  kD CCS 

K
D,C 

Equation III
where KD,C 
kD
k D
Note that the desorption of C is the reverse of the adsorption of C
rD,C  rAD,C
Also the desorption equilibrium constant KD,C is the reciprocal of the
adsorption equilibrium constant KC
1
KD,C 
KC
Substituting 1/KC for KD,C in the rate equation for product desorption gives:
rD,C  kD CCS  KCPCCv 
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L17-14
Derive a Rate Law for Catalytic Rxn
• Postulate catalytic mechanism, and then derive the rate law for that
mechanism
• Assume pseudo-steady state hypothesis (rate of adsorption = rate of
surface reaction = rate of desorption)
• No accumulation of species on the surface or near interface
• Each species adsorbed on the surface is a reactive intermediate
• Net rate of formation of species i adsorbed on the surface is 0, ri·S=0
• One step is usually rate-limiting
• If the rate-limiting step could be sped up, the entire rxn would be faster
• Although reactions involve all 7 steps, (for chapter 10) only adsorption,
surface reaction, or desorption will be rate limiting
• The surface reaction step is rate limiting ~70% of the time!
• Steps to derive the rate law
• Select among types of adsorption, surface reaction, and desorption
• Write rate laws for each individual step, assuming all are reversible
• Postulate which step is rate limiting
• Use non-rate-limiting steps to eliminate the surface concentration
terms that cannot be measured
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Consider A ⇌ B and assume the following mechanism is correct:

CAS 
1. Adsorption: A(g) + S ⇌ A·S rAD  k A  PA Cv 

K

A 
L17-15

CBSCv 
r

k
C
C

2. Surface reaction: A·S + S ⇌ S + B·S S
S  A S v
K S 


PBCv 
3. Desorption: B·S ⇌ B + S rD  kD CBS  K 

D 
We need to select one of these 3 reactions as the rate limiting step, then
derive the corresponding rate equation, and see if this rate eq matches
experimental data. Which step is the most logical to start with?
a) Adsorption
b) Surface reaction
c) Desorption
d) None of the above
e) Any of these would be “logical” - they all have equal probability of being
the rate limiting step
The surface reaction step as is rate limiting ~70% of the time
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Consider A ⇌ B and assume the following mechanism is correct:

CAS 
1. Adsorption: A(g) + S ⇌ A·S rAD  k A  PA Cv 

K

A 
L17-16

CBSCv 
r

k
C
C

2. Surface reaction: A·S + S ⇌ S + B·S S
S  A S v
K S 


PBCv 
3. Desorption: B·S ⇌ B + S rD  kD CBS  K 

D 
Derive the rate equation for when the surface reaction is rate limiting (true
~70% of the time)

C C 
r 'A  rS  k S CASCv  BS v 
KS 

1. CA·S, Cv, and CB·S need to be expressed in terms of measurable quantities
• For surface reaction-limited mechanisms, kS is small and kA and kD are
relatively large
• Therefore rAD/kA and rD/kD are very small, and can be approximated as
equal to zero
2. Use this relationship to eliminate CA·S and CB·S from their respective rate
equations and the site balance to eliminate CV
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Consider A ⇌ B and assume the following mechanism is correct:
3. Desorption
1. Adsorption
2. Surface reaction


C
rAD  k A  PA Cv  AS 
KA 


CBSCv 
rS  k S CASCv 

K

S 
Derive the rate equation for when
the surface reaction is rate limiting
L17-17

PC 
rD  kD CBS  B v 
KD 


CBSCv 
r 'A  rS  k S CASCv 

K

S 
Use rAD/kA =0 and rD/kD =0 to eliminate CA·S and CB·S from their respective
rate equations and the site balance to eliminate CV
Use rAD/kA =0 & rAD
equation to solve for CA·S:
CAS
rAD

 0  PA Cv 
kA
KA

Use rD/kD =0 & rD equation
to solve for CB·S:
r
PC
 D  0  CBS  B v
kD
KD

CAS 
rAD  k A PA Cv 

K

D 
CAS
 PA Cv
KA
 CAS  K APACv

PC 
rD  kD CBS  B v 
KD 

PC
 B v  CBS
KD
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Consider A ⇌ B and assume the following mechanism is correct:
3. Desorption
1. Adsorption
2. Surface reaction
L17-18



C
PC 

C C 
rAD  k A  PA Cv  AS  rS  k S CASCv  BS v 
rD  kD CBS  B v 
KA 
KD 
KS 



Derive the rate equation for when the surface reaction is rate limiting

CBSCv 
r 'A  rS  k S CASCv 

K

S 
PBCv
 CBS
rAD/kA =0 & rD/kD =0
CAS  K APACv
KD
Use site balance to solve for CV:
Ct  Cv  CAS  CBS
 Cv  Ct  CAS  CBS Make substitutions for CA·S & CB·S, solve for Cv
PC
PC
 Cv  Ct  K APA Cv  B v  Cv  K APA Cv  B v  Ct
KD
KD

PB 
 Cv  1  K APA 
 Ct

KD 

 Cv 
Ct
1  K APA 
PB
KD
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Consider A ⇌ B and assume the following mechanism is correct:
3. Desorption
1. Adsorption
2. Surface reaction


C
rAD  k A  PA Cv  AS 
KA 


CBSCv 
rS  k S CASCv 

K

S 
L17-19

PC 
rD  kD CBS  B v 
KD 

Derive the rate equation for when the surface reaction is rate limiting
Ct
PBCv
 CBS
Cv 
CAS  K APACv
KD
1  K APA  PB KD

CBSCv 
r 'A  rS  k S CASCv 
Substitute in CA·S, CB·S, &Cv

KS 

2
2



 
Ct
Ct
PB 

 r 'A  rS  k S K APA 





 1  K APA  PB KD  K SKD  1  K APA  PB KD  
2

 
Ct
PB 
 r 'A  rS  k S 
K
P

  A A K K 
1

K
P

P
K

A A
B D 
S D
This is the rate equation in terms of measurable species and rate
constants for the mechanism given in the problem statement
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Evaluating a Catalytic Reaction
Mechanism
L17-20
• Collect experimental data from test reactor
• See if rate law is consistent with data
• If not, then try other surface mechanism (i.e., dual-site
adsorption or Eley-Rideal) or choose a different rate-limiting
step (adsorption or desorption)
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Consider A ⇌ B and assume the following mechanism is correct:
3. Desorption
1. Adsorption
2. Surface reaction


C
rAD  k A  PA Cv  AS 
KA 


CBSCv 
rS  k S CASCv 

K

S 
L17-21

PC 
rD  kD CBS  B v 
KD 



C
Now derive the rate equation for
r ' A  rAD  k A  PA Cv  AS 
when adsorption is rate limiting:
KA 

Conc of vacant and occupied sites must be eliminated from the rate equation
If adsorption is rate limiting, kS>>kAD, so rS/kS can be approximated as 0. Then:

CBS Cv 
rS  k S CASCv 

K

S 

CBS Cv
 CAS Cv
KS

rS
C C
 0  CAS Cv  BS v
kS
KS

CBS
 CAS
KS
Need to put CB·S in
measureable terms
If adsorption is rate limiting, kD>>kAD, so rD/kD can be approximated as 0. Then:

P C 
rD  kD CBS  B v 
KD 


rD
P C
 0  CBS  B v
kD
KD

PB Cv
 CBS
KD
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Consider A ⇌ B and assume the following mechanism is correct:
3. Desorption
1. Adsorption
2. Surface reaction
L17-22


C
rAD  k A  PA Cv  AS 
KA 


PBCv 

CBSCv 
r

k
C

rS  k S CASCv 
D
D  BS


K
K

D 

S 


C
Now derive the rate equation for
r ' A  rAD  k A  PA Cv  A S 
when adsorption is rate limiting:
KA 

Conc of vacant and occupied sites must be eliminated from the rate eq
CBS
 CAS
KS
PB Cv
 CBS Solve for Cv using the site balance equation
KD
Make substitutions  C  C  CBS  PB Cv
Ct  Cv  CAS  CBS
t
v
KS
KD
for CA·S & CB·S
Ct

PB
PB  
PB Cv PB Cv
 Cv
 Ct  Cv  1 

 Ct  Cv 

PB
PB

K
K
K
K SK D
KD
1



S D
D
K SKD KD
Substitute
PB Ct
PB Ct
Substitute
Cv into the CBS 
CAS 

PB
PB  CB·S into

PB
PB 
expression
KD  1 

K SK D  1 

 CA·S:

K
K
K
for CB·S:
K
K
K

S D
D

S D
D
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Consider A ⇌ B and assume the following mechanism is correct:
3. Desorption
1. Adsorption
2. Surface reaction
L17-23



C
PC 

C C 
rAD  k A  PA Cv  AS  rS  k S CASCv  BS v 
rD  kD CBS  B v 
KA 
KD 
KS 



Now derive the rate equation for

C A S 

r
'

r

k
P
C

A
AD
A A v

when adsorption is rate limiting:
K

A 
PB Ct
PB Ct
Ct
CAS 
CBS 
 Cv
P
P




P
P
P
P
K SKD  1  B  B 
KD  1  B  B  1  B  B
K SKD KD
 K SKD KD 
 K SKD KD 




Use these eqs
P
C
P
C
A t
B t



r

k

to replace CA·S
AD
A
PB
PB


PB
PB  
1


& Cv in rAD:
K A K SK D  1 

 K K

K
K
K
K
S
D
D

S D
D 


k 


P
P
Factor  r  k C 
A
B


AD
A t
PB
PB

out Ct:

PB
PB  
1


K A K SKD  1 

 K K

K
K
K
K
S
D
D

S D
D 

Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L17-24
The gas phase hydromethylation of toluene: C6H5CH3 + H2 → C6H6 + CH4 is
to be carried out in a PBR. Plot the conversion and the partial pressures of
toluene, hydrogen and benzene as a function of catalyst weight.
FA0 = 50 mol toluene/min, P0 = 40 atm, T= 913K, a= 9.8 x 10-5 kg-1, Feed is
30% toluene (species tol), 45% hydrogen (species H) and 25% inerts (I)
Rate law:rT' 
kPHPtol
k= 0.00087 mol/atm2·kg cat·min
1  K tolPtol  KBPB KB = 1.39 atm-1 Ktol= 1.038 atm-1
'
kPHPtol
dXtol rtol
'

r

Rate
law:

T
Mole balance:
1  K tolPtol  KBPB
dW
FA0
 1  Xtol  P 
 1  Xtol  P   T0 
Stoichiometry:Ctol  Ctol,0 
    CT  CT0  1   X  P 

tol  0 
 1   Xtol  P0   T 
1
Need concentrations in terms of pressure
Ptol,0
Ptol

P

C
RT

 Ctol
 Ctol,0
Ptol Vtol  ntolRT
tol
tol
RT
RT
Total P
Ptol Ptol,0  1  Xtol  P 
 1  Xtol 
P



   P  Ptol,0 
 y where y=  
RT
RT  1   Xtol  P0  Total P tol
 1   Xtol 
 P0 
at inlet
Ftol,0 0.3

 0.3    0.3  0   0
  y tol,0   1  1  1  1  0 y Tol,0 
FT0
1
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L17-25
The gas phase hydromethylation of toluene: C6H5CH3 + H2 → C6H6 + CH4 is
to be carried out in a PBR. Plot the conversion and the partial pressures of
toluene, hydrogen and benzene as a function of catalyst weight.
FA0 = 50 mol toluene/min, P0 = 40 atm, T= 913K, a= 9.8 x 10-5 kg-1, Feed is
30% toluene (species tol), 45% hydrogen (species H) and 25% inerts (I)
Rate
law:rT'
2·kg cat·min
kPHPtol
k=
0.00087
mol/atm

1  K tolPtol  KBPB KB = 1.39 atm-1 Ktol= 1.038 atm-1
'
dXtol rtol
kPHPtol

Rate law:rT' 
Mole balance:
dW
FA0
1  K tolPtol  KBPB
Stoichiometry:
Ptol,0  y tol,0P0
P
 Ptol  Ptol,0 1  Xtol  y where y=  
 P0 
 Ptol,0  0.3  40atm  12atm
 0
y tol,0  0.3
 0
Use the Ergun equation to evaluate y:
Isothermal rxn & =0, so: y  1  a W 1 2  Ptol  Ptol,0 1  Xtol  1  a W 
12
Need an equation for PB & PH
PB  Ptol,0  Xtol  y


PH2  Ptol,0 H2  Xtol y
H2 
FH2
Ftol
0.45

 1.5
0.3
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L17-26
The gas phase hydromethylation of toluene: C6H5CH3 + H2 → C6H6 + CH4 is
to be carried out in a PBR. Plot the conversion and the partial pressures of
toluene, hydrogen and benzene as a function of catalyst weight.
FA0 = 50 mol toluene/min, P0 = 40 atm, T= 913K, a= 9.8 x 10-5 kg-1, Feed is
30% toluene (species tol), 45% hydrogen (species H) and 25% inerts (I)
k= 0.00087 mol/atm2·kg cat·min
kPHPtol

Rate
-1 K = 1.038 atm-1
K
=
1.39
atm
B
tol
1  K tolPtol  KBPB
'
 0
dXtol rtol
kPHPtol
'
y tol,0  0.3

Rate law:rT 
Mole balance:
dW
FA0
1  K tolPtol  KBPB
 0
law:rT'
Stoichiometry:
Ptol  Ptol,0 1  Xtol  1  a W 
12
PB  Ptol,0  Xtol  y
Ptol,0  12atm
PH2  Ptol,0 1.5  Xtol  y y  1  a W 
12
 
Finally, calculate the kg P
1atm
12
5 1


1

9.8

10
kg W

1

a
W


cat where P0 = 1atm:
40atm
P0

 0.000625  1  9.8  10 5 kg1W


12
 Wfinal  10197.7kg
Plug equation in boxes into Polymath to solve
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L17-27
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L17-28
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L17-29
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.