ERT 316: REACTION ENGINEERING
CHAPTER 3
RATE LAWS &
STOICHIOMETRY
1
Lecturer: Miss Anis Atikah Ahmad
Email: anisatikah@unimap.edu.my
Tel: +604 976 3245
OUTLINE

PART 1: Rate Laws
Relative Rates of Reaction
 Reaction Order & Rate Law
 Reaction Rate Constant, k


PART 2: Stoichiometry
Batch System Stoichiometric Table
 Flow System Stoichiometric Table
 Calculation for Concentration in terms of
Conversion

1. RELATIVE RATES OF REACTION
aA  bB  cC  dD
rC rD
rA
rB

 
a b c
d
Reaction
Stoichiometry
EXAMPLE
2 NO  O2  2 NO2
rNO rO2 rNO2


 2 1
2
If NO2 formed at 4 mol/m3/s
(r NO = 4 mol/m3/s), what is
2
the rate of formation of NO?
1. RELATIVE RATES OF REACTION
2 NO  O2  2 NO2
If NO2 formed at 4 mol/m3/s (r NO = 4 mol/m3/s),
2
what is the rate of formation of NO?
rNO rO2 rNO2


 2 1
2
rNO rNO2

2
2
3
rNO 4mol / m / s

2
2
rNO
 4mol / m3 / s 
  4mol / m3 / s
 2
2


1. RELATIVE RATES OF REACTION
EXERCISE

The Reaction:
2 A  3B  5C
is carried out in a reactor. If at a particular point,
the rate of disappearance of A is 10 mol/dm3/s,
what are the rates of B and C?
1. RELATIVE RATES OF REACTION
2 A  3B  5C

The relative rates are
rA
rB rC


2 3 5
Given, the rate of disappearance of A, -rA, is 10mol/dm3/s
3/s
r
=
-10
mol/dm
A
 Thus, solving the rates of B & C;

rA rC

2 5
rA
rB

2 3
3
rB 
 10mol / dm3 / s
2

 15mol / dm3 / s


5
rC 
 10mol / dm3 / s
2
 25mol / dm 3 / s

2. REACTION ORDER & RATE LAW


Rate law is a kinetic
expression that gives the
relationship between
reaction rate, -rA, and
concentration.
The reaction rate (rate of disappearance) depends
on temperature and composition.
It can be written as the product of reaction rate
constant, kA and a function of concentrations
(activities) of the reactants involved in the
reaction:
 rA  k A T  fnCA , CB ...
2. REACTION ORDER & RATE LAW

Rate law is a kinetic
expression that gives the
relationship between
reaction rate, -rA, and
concentration.
For reaction in which the stoichiometric coefficient is 1
for ALL species:
1NaOH  1HCl  1NaCl  1H 2O
we shall delete the subscript on the specific reaction
rate, (e.g.; A in kA) to let
k  k NaOH  k HCl  k NaCl  k H 2O
2.1 POWER LAW MODELS & ELEMENTARY RATE LAWS

Power Law Model:


 rA  kCA CB
The rxn is 𝛂 order wrt reactant A
AND
The rxn is 𝛃 order wrt reactant B
The overall order of the reaction, n;
n   
2.1 POWER LAW MODELS & ELEMENTARY RATE LAWS

The unit of the specific reaction, k, will vary with the
order of reaction.
A  Products

Concentration 
k
1 n
Time
 rA  k A
k  mol / dm3  s
Second order (n=2)
 rA  k ACA
 rA  k AC A2
Third order (n=3)
 rA  k AC
k  s 1
k  dm3 / mol  s
2
3
k  dm / mol   s 1
Zero order (n=0)
First order (n=1)
3
A
2.1 POWER LAW MODELS & ELEMENTARY RATE LAWS

Elementary reaction: a chemical reaction in which one or
more of the chemical species react directly to form products
in a single reaction step and with a single transition
state.
Unimolecular reaction
A  Products
A  B  Products

Bimolecular reaction
Elementary rate law:
The rxn is said to follow the elementary rate law if the
stoichiometic coefficients are IDENTICAL to the reaction
order of each species.
2 NO  O2  2 NO2
 rNO  k NOC CO2
2
NO
Nonelementary
rxn
But follows the
elementary rate law!
EXAMPLES OF REACTION RATE LAWS
EXAMPLES OF REACTION RATE LAWS
EXAMPLES OF REACTION RATE LAWS
2.2 NON-ELEMENTARY RATE LAWS
Non-elementary rate laws: reactions that do not
follow simple rate laws (power rate laws).
 Example 1: Homogeneous Rxn

CO  Cl2  COCl2
Gas phase
synthesis of
phosgene
The kinetic rate law is:
 rCO  kCCOC
32
Cl 2
Rxn order: first order wrt to CO, three-halves
order wrt Cl2, five-halves order overall.
2.2 NON-ELEMENTARY RATE LAWS

Example 2: Heterogeneous Rxn
C6 H 5CH 3  H 2  C6 H 6  CH 4
Gas-solid catalyzed
rxn:
Hydrodemethylation
of toluene (T)
cat
T  H2  B  M
cat
The rate of disappearance of toluene per mass of
catalyst is:
r 
'
T
kPH 2 PT
1  K B PB  KT PT
In terms of partial
pressure rather than
concentrations
where KB & KT is the adsorption constants.
2.3 REVERSIBLE REACTIONS
aA  bB ⇌ cC  dD

For reversible rxn, all rate laws must reduce to the
thermodynamic relationship relating the reacting
species concentrations at equilibrium.
c
Ce
a
Ae
d
De
b
Be
C C
KC 
C C
Thermodynamic
Equilibrium
Relationship
2.3 REVERSIBLE REACTIONS
EXAMPLE: combination rxn of 2 mol of benzene to form 1 mol
H2 and 1 mol diphenyl.
kB
2C6 H 6 ⇌ C12 H10  H 2
k-B
symbolically;
2B
kB
⇌k D  H 2
-B
The rate of disappearance of benzene;
 rB, forward  kBC
2
B
OR
rB, forward  kBC
2
B
The reverse rxn btween diphenyl & hydrogen;
k-B
C12 H10  H 2 2 ⇌ C6 H 6
The rate of formation of benzene (in reverse direction);
rB ,reverse  k  B C D C H 2
2.3 REVERSIBLE REACTIONS
The net rate of formation of benzene is;
rB  rB ,net  rB , forward  rB ,reverse
 kBC  k BCDCH 2
2
B
Multiplying both sides by -1, we obtain the rate law of
disappearance of benzene, -rB
 rB  kBC  k BCDCH 2
2
B
 2 k B

 k B  CB 
CDCH 2 
kB


2.3 REVERSIBLE REACTIONS
 2 k B

 rB  k B  CB 
CDCH 2 
kB


Replacing the ratio of the reverse & forward rate law
constant by equilibrium constants;
 2 CDCH 2
 rB  k B  CB 
KC




where
kB
Concentration
 K C  equilibrium constant
kB
3. THE REACTION RATE CONSTANT
k A T   Ae
 E / RT
Arrhenius
equation
A= preexponential factor or frequency factor
E= activation energy, J/mol or cal/mol
R=gas constant = 8.314 J/mol-K = 1.987 cal/mol-K
T= absolute temperature, K
A
-no of collision
e  E / RT -probability that
the collision will
result in a reaction
3. THE REACTION RATE CONSTANT
k A T   Ae

 E / RT
Activation energy is a measure of the minimum
energy that the reacting molecules must have
in order for the reaction to occur (energy required to
reach transition state).
Transition state
Energy barier
e
Reactants
Products
k
- no of collision that
result in a rxn
A
-total no of collision
 E / RT
probability that
- the collision will
result in a rxn
3. THE REACTION RATE CONSTANT
k A T   Ae
 E / RT
Taking a natural logarithm;
E1
ln k A  ln A   
R T 

E ⬆, k ⬆, -r = ⬆
The larger the
activation energy,
the more
temperature
sensitive k and thus
the reaction rate.
4. BATCH SYSTEMS STOICHIOMETRIC TABLE

Purpose of developing stoichiometric table:

To determine the no of moles of each species
remaining at a conversion of X.
4. BATCH SYSTEMS STOICHIOMETRIC TABLE

Components of stoichiometric table:
Species
A
B
C
D
I
Totals
Initially
(mol)
Change
(mol)
refers to moles of
species reacted or
formed
Remaining
(mol)
4. BATCH SYSTEMS STOICHIOMETRIC TABLE
aA + bB  cC + dD

Recall from Chapter 2:

N A0  N A 
X
N A0
moles of A reacted
N A  N A0  N A0 X

Factorizing;
N A  N A0 1  X 
moles of A remaining
in the reactor at a
conversion
of X
4. BATCH SYSTEMS STOICHIOMETRIC TABLE
Moles B
reacted, NB
Moles C
formed, NC
Moles D
formed, ND

Moles B reacted
Moles A reacted
b
  N A0 X 
a
c
  N A0 X 
a
d
  N A0 X 
a
Moles A reacted
4. BATCH SYSTEMS STOICHIOMETRIC TABLE
b
moles B remaining  N
 N A0 X
B
0
in the system, NB
a
moles of B
initially in the
system
NC
ND
c
 N C 0  N A0 X
a
d
 N D 0  N A0 X
a
moles of B
reacted
moles of C
formed
moles of D
formed
4. BATCH SYSTEMS STOICHIOMETRIC TABLE
Species Initially
(mol)
A
N A0
B
C
D
I
Totals
N B0
NC 0
N D0
NI0
NT 0
Change
(mol)
Remaining (mol)
 N A0 X 
N A  N A0  N A0 X
b
 N A0 X 
a
c
N A0 X 
a
d
N A0 X 
a
b
N A0 X
a
c
N C  N C 0  N A0 X
a
d
N D  N D 0  N A0 X
a
NI  NI 0
-
N B  N B0 
d c b 
NT  NT 0      1 N A0 X
a a a 

4. BATCH SYSTEMS STOICHIOMETRIC TABLE

Total no of moles per mole of A reacted can be
calculated as:
d c b 
NT  NT 0      1 N A0 X
a a a 
 N T 0   N A0 X
where
Change in the total number of moles per
mole of A reacted
d c b
    1
a a a
4. BATCH SYSTEMS STOICHIOMETRIC TABLE
 rA  kCA
 rA  kCA
2
 rA  k AC
3
A
Can we express concentration of each species??
Species
A
B
C
D
I
Totals
Initially Change Remaining Concentration
4. BATCH SYSTEMS STOICHIOMETRIC TABLE

Concentration of each species in terms of
conversion can be expressed as:
Recall from
stoichiometric
table
N A N A0 1  X 
CA 

V
V
N B N B 0  b / a N A0 X
CB 

V
V
N C N C 0  c / a N A0 X
CC 

V
V
N D N D 0  d / a N A0 X
CD 

V
V
Remaining (mol)
A
B
C
D
N A  N A0  N A0 X
b
N A0 X
a
c
N C  N C 0  N A0 X
a
d
N D  N D 0  N A0 X
a
N B  N B0 
4. BATCH SYSTEMS STOICHIOMETRIC TABLE
N B 0  b / a N A0 X
CB 
V
N A0 N B 0 / N A0  b / a X 

V
N A0  B  b / a X 

V
N C 0  c / a N A0 X
CC 
V
N A0 N C 0 / N A0  c / a X 

V
N A0 C  c / a X 

V
4. BATCH SYSTEMS STOICHIOMETRIC TABLE
N D 0  d / a N A0 X
CD 
V
N A0 N D 0 / N A0  d / a X 

V
N A0  D  d / a X 

V
N i 0 Ci 0
yi 0
i 


N A0 C A0 y A0
4. BATCH SYSTEMS STOICHIOMETRIC TABLE
Species Initially
A
B
N A0
N B0
C
NC 0
D
N D0
I
NI0
Change
 N A0 X 

Remaining
N A  N A0  N A0 X
b
N A0 X  N B  N B 0  b N A0 X
a
a
c
N A0 X  N C  N C 0  c N A0 X
a
a
d
N A0 X  N D  N D 0  d N A0 X
a
a
-
NI  NI 0
Concentration
N A0 1  X 
CA 
V
N A0  B  b / a X 
V
N   c / a X 
CC  A0 C
V
CB 
N A0  D  d / a X 
CD 
V
C IO
 N B0 b 
b
N B  N B 0  N A0 X  N A0 
 X 
a
 N A0 a 
b 

 N A0   B  X 
a 

i 
N i 0 Ci 0
y

 i0
N A0 C A0 y A0
4. BATCH SYSTEMS STOICHIOMETRIC TABLE
Species Initially
A
B
N A0
N B0
C
NC 0
D
N D0
I
NI0
Change
 N A0 X 

Remaining
N A  N A0  N A0 X
Concentration
N A0 1  X 
CA 
V
b
N A0 X  N B  N A0   B  b X  CB  N A0  B  b / a X 
a 
a

V
c
N A0 X  NC  N A0  C  c X  CC  N A0 C  c / a X 
a
V
a 

d
N A0 X  N D  N A0   D  d X  CD  N A0  D  d / a X 
a
a 
V

C IO
NI  NI 0
-
4. BATCH SYSTEMS STOICHIOMETRIC TABLE
EXAMPLE
Given the saponification for the formation of soap
from aqueous caustic soda & glyceryl stearate is:
3NaOH aq   C17 H 35COO 3 H 5  3C17 H 35COONa  C3 H 5 OH 3
Letting X the conversion of sodium hydroxide, set up a
stoichiometric table expressing the concentration of
each species in terms of its initial concentration and the
conversion.
4. BATCH SYSTEMS STOICHIOMETRIC TABLE
EXAMPLE
We know that this is a liquid-phase reaction.
Therefore, V=V0
3NaOH aq   C17 H 35COO 3 H 5  3C17 H 35COONa  C3 H 5 OH 3
3A  B  3C  D
a  3 b 1 c  3 d 1
N A0 1  X  N A0 1  X 
CA 

 C A0 1  X 
V
V0
N A0  B  b / a X 
1 

CB 
 C A0  B  X 
V0
3 

4. BATCH SYSTEMS STOICHIOMETRIC TABLE
EXAMPLE
Species Initially
A
B
N A0
N B0
C
NC 0
D
N D0
I
NI0
Total
NT 0

Change
Remaining
 N A0 X
N A  N A0 1  X 
1
N A0 X  N B  N A0   B  1 X 
3 
3

N A0 X
NC  N A0 C  X 
1 
1

N

N



N A0 X  D A0  D 3 X 
3
NI  NI 0
-
0
NT  N T 0
Concentration
C A  C A0 1  X 
1 

C B  C A0   B  X 
3 

CC  C A0 C  X 
1 

C D  C A0   D  X 
3 

C IO
5. FLOW SYSTEMS STOICHIOMETRIC TABLE

Purpose of developing stoichiometric table:

To determine the effluent flow rate of each species at
a conversion of X.
5. FLOW SYSTEMS STOICHIOMETRIC TABLE

Components of stoichiometric table:
Species
A
B
C
D
I
Totals
Feed rate to
reactor
(mol/time)
Change within Effluent rate
the reactor
from reactor
(mol/time)
(mol/time)
5. FLOW SYSTEMS STOICHIOMETRIC TABLE
Species
A
B
C
D
I
Totals
Feed rate
to reactor
(mol/time)
FA0
Change
within the
reactor
(mol/time)
 FA0 X 
Effluent rate from
reactor (mol/time)
FA  FA0  FA0 X
Concentration
(mol/L)
CA 
FA0 1  X 

b 

FA0  B  b / a X 
FB 0   B FA0  b F X  FB  FA0   B  X 
A0
a  CB 

a
FC 0  C FA0
FD 0   D FA0
FI 0  i FA0
FT 0
c
FA0 X 
a

c 

FC  FA0   C  X  C  FA0 C  c / a X 
C
a 


d 

d
F

F


FA0 X  D A0  D a X  CD  FA0  D  d / a X 
a

F 
FI  FA0 I
C I  A0 I
FT  FT 0  FA0 X

QUIZ 5

Given a liquid phase reaction:
A+ 2B  C + D
The initial concentration of A and B are 1.8 kmol/m3
and 6.6 kmol/m3 respectively. Construct a
stoichiometric table for a flow system considering A as
the basis of calculation.
Fi 0
Ci 0
yi 0
i 


FA0 C A0 y A0
ANSWER FOR QUIZ 5
A+ 2B  C + D
Given:
C A0  1.8kmol / m3
CC 0  0kmol / m3
CBO  6.6kmol / m3
CDO  0kmol / m3
Since C & D are
products.
From stoichiometry, we know that,
a 1 b  2 c 1 d 1
Ci 0
i 
C A0
6.6
B 
 3.67
1.8
d c b
     1  1
a a a
0
C 
0
1 .8
0
D 
0
1.8
ANSWER FOR QUIZ 5
Species
A
B
Feed rate to
reactor
(mol/time)
Change
within the
reactor
(mol/time)
FA0
 FA0 X 
FB 0  FA0 B
Effluent rate from
reactor (mol/time)
FA  FA0 1  X 
 2FA0 X  FB  FA0  B  2 X 
C
FC 0  FA0C
FA0 X 
FC  FA0 X
D
FD 0  FA0 D
FA0 X 
FD  FA0 X
Totals
FT 0
FT  FT 0  FA0 X
ANSWER FOR QUIZ 5
Substituting the numerical values;
Species
Feed rate to
reactor
(mol/time)
Change
within the
reactor
(mol/time)
FA0
 FA0 X
FA  FA0 1  X 
FB 0  3.67 FA0
 2 FA0 X
FB  FA0 3.67  2 X 
C
FC 0  0
FA0 X
FC  FA0 X
D
FD 0  0
FA0 X
FD  FA0 X
A
B
Totals
FT 0
Effluent rate from
reactor (mol/time)
FT  FT 0  FA0 X
6. CONCENTRATION IN TERMS OF CONVERSION
1. For liquid phase:

Batch System:
V  V0
CA 
N A N A0 1  X 

V
V
CB 
N B N A0  B  b / a X  N A0  B  b / a X 
 C A0  B  b / a X 


V0
V
V
NC
N A0 C  c / a X   C A0 C  c / a X 
N A0 C  c / a X 
CC 


V0
V
V
ND
N A0  D  d / a X   N A0  D  d / a X   C   d / a X 
CD 
A0
D

V
V
0
V
6. CONCENTRATION IN TERMS OF CONVERSION
1. For liquid phase:

Flow System - 
CA 
FA
CB 
FB
CC 
FC
CD 
FD





 0
FA0 1  X 

FA0  B  b / a X 
FA0  B  b / a X 
 C A0  B  b / a X 


0

FA0 C  c / a X 
0
 C A0 C  c / a X 
FA0  D  d / a X  FA0  D  d / a X 



0
 C A0  D  d / a X 
FA0 C  c / a X 



6. CONCENTRATION IN TERMS OF CONVERSION
2. For gas phase:

Batch System
CA 
NA
V
Need to substitute V
from gas law equation
From equation of state;
At any time t,
PV  ZNT RT
(1)
At initial condition (t=0)
P0V0  Z 0 NT 0 RT0
(2)
T= temperature, K
P= total pressure, atm (1 atm= 101.3 kPa)
Z= compressibility factor
R= gas constant = 0.08206 dm3-atm/mol-K
6. CONCENTRATION IN TERMS OF CONVERSION
2. For gas phase:

Batch System
PV  ZNT RT
(1)
P0V0  Z 0 NT 0 RT0
(2)
Recall from stoichiometric table
Dividing (1) by (2);
 P0  T  Z  NT
V  V0    
 P  T0  Z 0  NT 0
Dividing (4) by NT0 ;
N A0
NT
 1
X
NT 0
NT 0
 1  y A0 X
NT  NT 0  N A0 X
(3)
(4)
6. CONCENTRATION IN TERMS OF CONVERSION
2. For gas phase:

Applies for both
batch and flow
systems
Batch System
NT
 1  y A0 X
NT 0

NT
 1  X
NT 0
Rearranging;
d c b
 NT
      1
a a a
 NT 0
 y A0
Will be substitute
in (3)
NT  NT 0

NT 0 X
At complete conversion (for irreversible rxn): X=1, NT=NTf

N Tf  N T 0
NT 0
6. CONCENTRATION IN TERMS OF CONVERSION
2. For gas phase:

Batch System
 P0  T  Z  NT
V  V0    
 P  T0  Z 0  NT 0
(3)
Substituting the expression for NT/NT0 in (3),
 P0  T  Z 
V  V0    1  X 
 P  T0  Z 0 
T
P 
V  V0  0 1  X 
T0
P
(5)
If the compressibility factor are not change
significantly during rxn, Z0⩳Z
6. CONCENTRATION IN TERMS OF CONVERSION
2. For gas phase:
Flow System

Cj 
Need to substitute υ
from gas law equation
Fj

From gas law, at any point in the reactor,
CT 
FT


P
ZRT
(1)
At the entrance of reactor;
CT 0
FT 0
P0


0 Z 0 RT0
(2)
Dividing (1) by (2)
 FT  P0  T 
  
  0 
 FT 0  P  T0 
(3)
6. CONCENTRATION IN TERMS OF CONVERSION
2. For gas phase:

Flow System
Recall from stoichiometric table
 FT  P0  T 
  
  0 
 FT 0  P  T0 
FT  FT 0  FA0X
Substituting for FT;
 FT 0  FA0X
  0 
FT 0

 P0  T 
  
 P  T0 
 FA0
 P0  T 
 0 1 
X   
 FT 0
 P  T0 
P0  T 
P0  T 
 0 1  y A0X     0 1  X   
P  T0 
P  T0 
(4)
6. CONCENTRATION IN TERMS OF CONVERSION
2. For gas phase:

Flow System
Cj 
Need to substitute υ
from gas law equation
Fj

P0  T 
  0 1  X   
P  T0 
Substituting υ & Fj;
Cj 
FA0  j  v j X 

0  1  x 

 C A0

P0 T 

P T0 
 v j X   P  T0
 
1  x   P0  T
j
(4)
F j  F j 0  j  v j X 
(5)
Stoichiometric
coefficient
(d/a, c/a, -b/a, -a)
6. CONCENTRATION IN TERMS OF CONVERSION
2. For gas phase:
aA + bB  cC + dD
Flow System
Concentration for each species:

FA0 1  X 
 1  X  T0  P
 C A0 
 
 1  x  T  P0
  0 1  X 
CA 
FA
CB 
FB
FA0  B  b / a X   C  B  b / a X  T0
A0

1  x  T

CC 
FC
FA0 C  c / a X   C C  c / a X  T0

A0
1  x  T

CD 
FD
FA0  D  d / a X   C  D  d / a X  T0  P 
A0

1  x  T  P0 







FI  FI 0  I  C A0 I T0  P 
CI 

1  x  T  P0 

P
 
 P0 
P
 
 P0 
P0
P
T 
 
 T0 
SUMMARY

Relative rate of reaction:
aA  bB  cC  dD
rA rB rC rD
  
a b c d

Power Law Model:


 rA  kCA CB
SUMMARY

Elementary rate law:
The rxn that in which its stoichiometic coefficients are IDENTICAL to the
reaction order of each species.

Non-elementary rate laws:
The reactions that do not follow simple rate laws (power rate laws) in
which its stoichiometic coefficients are NOT IDENTICAL to the reaction
order of each species.

Reversible reaction:
All rate laws must reduce to the thermodynamic relationship relating the
reacting species concentrations at equilibrium.
SUMMARY

Reaction Rate Constant, k
k A T   Ae E / RT
E ⬆, k ⬆, -r ⬆
The larger the activation energy,
the more sensitive k is, (towards
the change in temperature)
SUMMARY

Stoichiometric Table for Batch Systems
Species Initially
A
B
N A0
N B0
C
NC 0
D
N D0
I
NI0
Change
Remaining
 N A0 X 
N A  N A0  N A0 X

b
N A0 X  N B  N B 0  b N A0 X
a
a
c
N A0 X  N C  N C 0  c N A0 X
a
a
d
N A0 X  N D  N D 0  d N A0 X
a
a
-
NI  NI 0
SUMMARY

Stoichiometric Table for Flow Systems
Species
A
B
C
D
I
Totals
Feed rate to
reactor
(mol/time)
FA0
FB 0   B FA0
FC 0  C FA0
FD 0   D FA0
FI 0  i FA0
FT 0
Change within the
reactor (mol/time)
 FA0 X 
Effluent rate from
reactor (mol/time)
FA  FA0  FA0 X
b
 FA0 X 
a
c
FA0 X 
a
b 

FB  FA0   B  X 
a 

d
FA0 X 
a
-
d 

FD  FA0   D  X 
a 

c 

FC  FA0   C  X 
a 

FI  FA0 I
FT  FT 0  FA0 X
SUMMARY

Expression of V and υ in calculating the concentration
of each species:

Batch systems

Liquid phase:
V  V0

Gas phase:
T  P0 
V  V0 1  X   
T0  P 

Flow systems

Liquid phase:
  0

Gas phase:
T  P0 
  0 1  X   
T0  P 
QUIZ 6

Derive a concentration for each species for the
isothermal gas phase reaction below, neglecting
the pressure drop:
A+BC