L4 Rate laws and stoichiometry.pptx

L4-1
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
Ideal CSTR
Design Eq
with XA:
Review: Design Eq & Conversion
D
a
d
C
a
c
B
a
b
A 



fed
A
moles
reacted
A
moles
XA 
BATCH
SYSTEM: A
0
A
j
0
j
j X
N
N
N 

  











j
A
0
A
j
j
0
T
j
T X
N
N
N
N 
FLOW
SYSTEM: A
0
A
j
0
j
j X
F
F
F 

  











j
A
0
A
j
j
0
T
j
T X
F
F
F
F 
r
X
F
V
A
A
0
A


V
r
dt
dX
N A
A
0
A 

Ideal Batch Reactor
Design Eq with XA:



A
X
0 A
A
0
A
V
r
dX
N
t
A
A
0
A r
dV
dX
F 

Ideal SS PFR
Design Eq with XA:



A
X
0 A
A
0
A
r
dX
F
V
'
r
dW
dX
F A
A
0
A 

Ideal SS PBR
Design Eq with XA:



A
X
0 A
A
0
A
'
r
dX
F
W
j≡ stoichiometric coefficient;
positive for products, negative
for reactants

L4-2
Review: Sizing CSTRs
We can determine the volume of the CSTR required to achieve a specific
conversion if we know how the reaction rate rj depends on the conversion Xj
A
A
0
A
CSTR
A
A
0
A
CSTR X
r
F
V
r
X
F
V 












Ideal SS
CSTR
design eq.
Volume is
product of FA0/-rA
and XA
• Plot FA0/-rA vs XA (Levenspiel plot)
• VCSTR is the rectangle with a base of XA,exit and a height of FA0/-rA at XA,exit

FA 0
rA
X
Area = Volume of CSTR
X1
V 
FA 0
rA



X1
 X1

L4-3
FA 0
rA
Area = Volume of PFR
V  0
X1

FA 0
rA






dX
X1
Area = VPFR or Wcatalyst, PBR
dX
'
r
F
W
1
X
0 A
0
A
 









Review: Sizing PFRs & PBRs
We can determine the volume (catalyst weight) of a PFR (PBR) required to
achieve a specific Xj if we know how the reaction rate rj depends on Xj
A
exit
,
A
X
0 A
0
A
PFR
exit
,
A
X
0 A
A
0
A
PFR dX
r
F
V
r
dX
F
V  













Ideal PFR
design eq.
• Plot FA0/-rA vs XA (Experimentally determined numerical values)
• VPFR (WPBR) is the area under the curve FA0/-rA vs XA,exit
A
exit
,
A
X
0 A
0
A
PBR
exit
,
A
X
0 A
A
0
A
PBR dX
r
F
W
r
dX
F
W  













Ideal PBR
design eq.
dX
r
F
V
1
X
0 A
0
A
 










L4-4
Numerical Evaluation of Integrals (A.4)
Simpson’s one-third rule (3-point):
       
 
2
1
0
2
X
0
X
f
X
f
4
X
f
3
h
dx
x
f 



h
X
X
2
X
X
h 0
1
0
2 



Trapezoidal rule (2-point):
     
 
1
0
1
X
0
X
f
X
f
2
h
dx
x
f 


0
1 X
X
h 

Simpson’s three-eights rule (4-point):
         
 
3
2
1
0
3
X
0
X
f
X
f
3
X
f
3
X
f
h
8
3
dx
x
f 




3
X
X
h 0
3 

h
2
X
X
h
X
X 0
2
0
1 



Simpson’s five-point quadrature :
           
 
4
3
2
1
0
4
X
0
X
f
X
f
4
X
f
2
X
f
4
X
f
3
h
dx
x
f 





4
X
X
h 0
4 


L4-5
Review: Reactors in Series
2 CSTRs 2 PFRs
CSTR→PFR
VCSTR1 VPFR2
VPFR2
VCSTR1
VCSTR2
VPFR1
VPFR1
VCSTR2
VCSTR1 + VPFR2
≠
VPFR1 + CCSTR2
PFR→CSTR
A
A0
r
-
F
 


i j
CSTR
PFR
PFR V
V
V
If is monotonically
increasing then:
CSTR
i j
CSTR
PFR V
V
V
  


L4-6
L4: Rate Laws & Stoichiometry
• Reaction Rates (–rA )
1. Concentration
2. Temperature
3. Reversible reactions
• How to derive an equation for –rA [–rA = f(XA)]
1. Relate all rj to Cj
2. Relate all Cj to V or u
3. Relate V or u to XA
4. Put together
A
A
X
A
A0
0
dX
r
t N
V

 
A
A
A0
F X
V
r


A
X
A
A0
0 A
r
dX
V F

 
A
A
X
A
A0
0
dX
r
W
'
F

 

L4-7
Concentration and Temperature
• Molecular collision frequency  concentration
• Rate of reaction  concentration
   
A A A B
-r k T f C ,C ,...
  
 
Reaction rate is a function of temperature and concentration
CA : Concentration of A CB : Concentration of B
• As temperature increases, collision frequency increases
• Rate of reaction = f [( CA, CB, ……), (T)]
• At constant temperature : r = f(CA, CB, …….)
Specific rate of reaction, or rate constant,
for species A is a function of temperature

L4-8
Elementary Reactions & Rate Laws
• Dependence of reaction rate –rA on concentration of chemical species in the
reaction is experimentally determined
• Elementary reaction: involves 1 step (only)
• Stoichiometric coefficients in an elementary reaction are identical to the
powers in the rate law:
C
B
A 

 
 

B
A
A
A C
C
k
r 

Reaction order:
•  order with respect to A
•  order with respect to B
• Overall reaction order n = 
Zero order: -rA = kA k is in units mol/(volume∙time)
1st order: -rA = kACA k is in units time-1
2nd order: -rA = kACA
2 k is in units volume/(mol∙time)
3rd order: -rA = kACA
3 k is in units volume2/(mol2∙time)

L4-9
Examples:
• This reaction is not elementary, but under some conditions it
follows an elementary rate law
• Forward reaction is 2nd order with respect to NO and 1st order
with respect to O2 (3nd order overall)
Overall Stoichiometric Equations
• Overall equations describe the overall reaction stoichiometry
• Reaction order cannot be deduced from overall equations
Compare the above reaction with the nonelementary reaction
between CO and Cl2
2 2
2NO O 2NO
 
2
NO NO NO O2
r k C C
 
2 2
CO Cl COCl
 
3 2
CO CO Cl2
r kC C
 
Forward reaction is 1st order with respect to CO and 3/2 order with
respect to Cl2 (5/2 order overall)

L4-10

L4-11
Specific Rate Constant, kA
kA is strongly dependent on temperature
Where :
A = Pre-exponential factor or frequency factor (1/time)
E = Activation energy, J/mol or cal/mol
R = Gas constant, 8.314 J/mol K (or 1.987 cal/mol K)
T = Absolute temperature, K
Arrhenius Equation
  E RT
A
k T Ae

To determine activation energy E, run
the reaction at several temperatures,
and plot ln k vs 1/T. Slope is –E/R
Taking ln of
both sides:
E 1
lnk lnA
R T
 
   
 
1/T
ln k -E/R

L4-12
Reversible Reactions
kA
k A
aA b B c C d D

 
KC: concentration equilibrium constant (capital K)
a b a b
fA A A B fA A A B
r k C C r k C C
    
At equilibrium, the reaction rate is zero, rA=0
Rate of disappearance of A (forward rxn):
c d
bA A C D
r k C C


Rate of generation of A (reverse reaction):
A,net A fA bA
r r r r
  
a b c d
A A A B A C D
r 0 k C C k C C

   
c d
C D
A
C
a b
A A B
C C
k
K
k C C

  
Thermodynamic equilibrium relationship
RX
C C 1
1
H 1 1
K (T) K (T )exp
R T T
 
 

 
 
 
 
 
KC is temperature dependent
(no change in moles or CP):
HRX: heat of reaction
If KC is known for temperature T1, KC for temperature T can be calculated
a b c d
A A A B A C D
r k C C k C C

   
a b c d
A A B A C D
k C C k C C

 

L4-13
L4: Rate Laws & Stoichiometry
• Reaction Rates (–rA ) 
1. Concentration 
2. Temperature 
3. Reversible reactions 
• How to derive an equation for –rA [–rA = f(XA)]
2. Relate all Cj to V or u
3. Relate V or u to XA (Wednesday)
4. Put together (Wednesday)
A
A
X
A
A0
0
dX
r
t N
V

 
A
A
A0
F X
V
r


A
X
A
A0
0 A
r
dX
V F

 
A
A
X
A
A0
0
dX
r
W
'
F

 

L4-14
• rA as a function of Cj is given by the rate law
• The rate relative to other species (rj) is determined by stoichiometry
D
a
d
C
a
c
B
a
b
A 


 “A” is the limiting reagent
     
a
d
r
a
c
r
a
b
r
r D
C
B
A 



 rj is negative for reactants,
positive for products
In general:
j
A
j
r
r
 

j≡ stoichiometric coefficient
positive for products, negative for reactants
a
d
a
c
1
a
b
d
c
A
B 




 




L4-15
For the reaction the rate of O2
disappearance is 2 mol/dm3•s (-rO2= 2 mol/dm3•s).
What is the rate of formation of NO2?
2 2
2NO O 2NO
 
j
A
j
r
Hint: r
 

 
2
2
NO
O
r
r
2 1
   
2 2
O NO
2 r r
  
2 2
NO NO
3 3
mol mol
2 2 r 4 r
dm s dm s
 
   
 
 
 
rNO2 = 4 mol/dm3•s

L4-16
2a. Relate all Cj to V (Batch System)


B
A
A
A C
C
k
r 

Reaction rate is a function of Cj:
How is Cj related to V and XA? Batch:
j
j
N mol
C
V L
 
D
a
d
C
a
c
B
a
b
A 



 
B0 A0 A
B
B
b
N N X
N a
C
V V
 
  
 
 
 
C0 A0 A
C
C
c
N N X
N a
C
V V
 
  
 
 
A
A
N
C
V

 
A0 A0 A
A
N N X
C
V


Put NA in
terms of XA:
 
D0 A0 A
D
D
d
N N X
N a
C
V V
 
  
 
 
Do the same for
species B, C, and D:
Cj is in terms of XA and V. But what if V varies with XA? That’s step 3a!
A
0
A
j
0
j
j X
N
N
N 



L4-17
2a. Additional Variables Used in
Textbook
 
 
  
 
 
B0 A0 A
B
B
b
N X
N a
C
N
V V
Book uses
term Θi:
  
A0
0
0
i
A
i
0 i
C
C
N
N
So species Ni0 can be removed from the equation for Ci
 
 
 
  
 
 
 
 
A0
A0 A
A
B0
A
0
0
B
N
b
X
N
N N
1 a
C
N
V
Multiply numerator by NA0/NA0:
 

 
 
 
   



 
 
A
B
B
B
A
A
A0
0
B
V
b
X
X
C
b
a
C
a
N
C
D
a
d
C
a
c
B
a
b
A 




L4-18
T
0 0 0 T0 0
ZN RT
PV
P V Z N RT

3a. Relate V to XA (Batch System)
Volume is constant (V = V0) for:
• Most liquid phase reactions
• Gas phase reactions if moles reactants = moles products
       
2 2 2
CO g H O g CO g H g
 
If the volume varies with time, assume the equation of state for the gas phase:
At time t: PV = ZNTRT and at t=0: P0V0 = Z0NT0RT0
P: total pressure, atm Z: compressibility factor
NT: total moles T: temperature, K
R: ideal gas constant, 0.08206 dm3∙atm/mol∙K
d c b change in total # moles
where = 1
a a a Moles A reacted
 
 
    
 
 
Want V in terms of XA. First find and expression for V at time t:
NT at time t is:
0 T
0
0 0 T0
P N
T Z
V V
P T Z N
   
 
     
 
    
T j T0 A0 A
j
d c b
N N N 1 N X
a a a
 
     
  
 
T T0 A0 A
N N N X

  
What is
NT at t?

L4-19
  
T
A
T0
N
1 X
N

3a. Relate V to XA (continued)
T T0 A0 A
N N N X

 
d c b change in total # moles
where = 1
a a a Moles A reacted
 
 
    
 
 
0 T
0
0 0 T0
P N
T Z
V V
P T Z N
   
 
    
 
    
Can we use the eq. for NT above to
find an expression for NT/NT0?
A0
A0
T0
= =mole fraction of A ini
S t
ubstitut ially pr
e:
N
y
N
esent
 
A0
S e
ubsti xpans
tute ion
: r
y facto


T T
0
0
0
A
T 0
0 T0
P T Z
Plug : into
N N
V V
P T
1 X
N N
Z

   
 
 
   
 
    
 
 
  
 
    
 
   

0
0
0
A
0
P T Z
V V
P T
1
Z
X

 
T0
T
A
T0 T0
A0
T0
N
N
N
N
X
N N

  
T
A
T0
A0
N
1 y X
N


L4-20

 
T T0
A
T0
N N
X
N

What is the meaning of ε?
When conversion is
complete (XA=1):
Tf T0 A
T0
N N Change in total # moles at X 1
N total moles fed

 
 
The expansion factor,, is the fraction of change in V per mol A reacted
that is caused by a change in the total number of moles in the system
A0
A0
T0
N
d c b
expansion factor: y 1
a a a N
 
 
    
 
 
T
A
T0
N
1 X
N

 
If we put the following
equation in terms of ε:
  
T
A
T0
N
1 X
N


 
T T0
T0 A
N N
N X

 
  
 
   
 
   


0
0 A
0 0
P T Z
V V
Z
1
P T
X
 where

L4-21
4a. Put it all together (batch reactor)
Batch:
 
  
 
 

  
 
  

 
0
j j0 j
j
A0 A
0 A
0 0
V P T
N N N X
Z
V 1
P T
C
X
Z


   
   
   
 


j A0 0 0
A 0
0 A
j
j T Z
P
1 X P T Z
C
C
X
C 

For a given XA, we can calculate Cj and plug the Cj into –rA=kCj
n

j
j
C
N
V


j0 A A
j
j 0
C
N
V
N X

j j j A A
N N N X
 
 0 0
 
  
 

  
 
   


0
0 A
0 0
P T Z
V V 1 X
P T Z


0
0
i0
i
V
N
C
What about flow systems?

L4-22
2b. Relate all Cj to u (Flow System)
How is Cj related to uand Xj?
Flow:
j
j
F mol s mol
C
L s L
u
  


B
A
A
A C
C
k
r 

Reaction rate is a
function of Cj:
D
a
d
C
a
c
B
a
b
A 



 
B0 A0 A
B
B
b
F F X
F a
C
u u
 
  
 
 
 
C0 A0 A
C
C
c
F F X
F a
C
u u
 
  
 
 
A
A
F
C
u

 
A0 A0 A
A
F F X
C
u


Put FA in
terms of XA:
 
D0 A0 A
D
D
d
F F X
F a
C
u u
 
  
 
 
Do the same for
species B, C, and D:
We have Cj in terms of XA and u, but what if u varies with XA? That’s step 3b!
A
0
A
j
0
j
j X
F
F
F 



L4-23
3b. Relate u to XA (Flow System)
Start with the equation of
state for the gas phase:
T
T
N
P
C
ZRT V
  
What is CT0 at the
entrance of the reactor?
T0 0
T0
0 0 0
F P
C
Z RT
u
 
 
 
T
T0 0 0 0 0
F ZRT 1 P
F Z RT 1 P
u
u

T
PV ZN RT

Rearrange to put in terms
of CT, where CT = NT/V:
T
T
F
C
u

Can we relate
CT to u? T
1
F ZRT
P
u
 
 
 
 
0
T
0
T0 0 0
P
F Z T
F Z T P
u u
    
      
 
   
Rearrange to put
in terms of u:
Put in terms of u0: T0 0 0 0
0
1
F Z RT
P
u
 
 
 
 
Use these 2 equations to
put uin terms of known or
measurable quantities
T
F P
ZRT
u
 

L4-24
3b. Relate u to XA (continued)
 
T T0 A0 A
subst F F F X
: and simplify
itute in 
    
     
 
  

 
0
0
T0 0
T0 A0
0
A P
Z T
F Z T P
F F X

u u
When conversion is
complete (XA=1):
Tf T0 A
T0
N N Change in total # moles at X =1
N total moles fed


 
    
     
 
   
0
0
T0
T
0 0
P
Z T
F Z T P
F
u u

A0
substitute y
:  
 
   
     
 
  
A0
0
0 A
0 0
P
Z T
y
1 X
Z T P
u 
u
    
  
    
 
   
A 0
0
0
A
T 0
0
0
X P
Z T
1
Z T
F
F P

u u
 
 
 
   
 
 
A0 0
A0 A0 A0
A0 A0
T0 T0 T0 0 T0
Simplify wit
N V
F N F
y y
F N N V F
Because
:
h
u
u
 
   
      
 
  
0
0 A
0 0
P
Z T
1 X
Z T P
u u 

L4-25
4b. Put it all together (flow reactor)
Flow:
 
  
 

  
 


 

  
0
j j0 j
0 A
0 0
0 A
j
A
P T Z
1 X
P
C
F
T
F F X
Z
u
u 

   
   
   
 


j A0 0 0
A 0
0 A
j
j T Z
P
1 X P T Z
C
C
X
C 

n

j
j
C
F
u


j0 A A
j
j 0
C
F
V
F X

j j j A A
F F F X
 
 0 0
 
  
 

  
 
   


0
0 A
0 0
P T Z
1 X
P T Z
u u 

0
0
i0
i
F
C
u
This is the same equation as that for the batch reactor!

L4-26
4. Summary: Cj in terms of Xj
Batch:
 

 
  
 

  
 
   
j j0 j A0 A
j
0
0 A
0 0
N N N X
C
V P T Z
V 1 X
P T Z


j0 j A0 A 0 0
j
A 0
C C X T Z
P
C
1 X P T Z


    
     
   
 
j0
j0
0
N
C
V

Flow:
 

 
  
 
   
 
   
j j0 j A0 A
j
0
0 A
0 0
F F F X
C
P T Z
1 X
P T Z

u
u 
j0
j0
0
F
C
u

j0 j A0 A 0 0
j
A 0
C C X T Z
P
C
1 X P T Z


    
     
   
 
This is the same equation as that for the batch reactor!
n

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