Chemical Reaction
Engineering
Lecture 6
Elementary reactions
•
•
Kinetics of chemical reactions determined by the elementary reaction steps.
Molecularity of an elementary reaction is the number of molecules coming
together to react in one reaction step (e.g. uni-molecular, bimolecular,
termolecular)
Uni-molecular: first order in the reactant
A ⎯⎯
→P
d [ A]
= − k[ A]
dt
Bimolecular: first order in the reactant
A + B ⎯⎯
→P
d [ A]
= − k[ A][ B]
dt
Proportional to collision rate
H + Br2 ⎯⎯
→ HBr + Br
! rate of disappearance of individual components can be
calculated as:
1 dni d ξ
v=
=
ν i dt
dt
Equilibrium constant
• Let’s consider a reaction of diphenyl formation
k1
⎯⎯
→ C12 H10 + H 2
2C6 H 6 ←⎯⎯
k −1
• rate of change for benzene
d [C6 H 6 ]
= rB = −2k1CB 2 + 2k−1CD CH 2
dt
• noticing that the equilibrium constant: K C = k1 k−1
⎡ 2 CD CH 2 ⎤
d [C6 H 6 ]
= rB = −2k1 ⎢CB +
⎥
dt
KC ⎦
⎣
! At equilibrium rb=0, so
in terms of
concentration
CD CH 2
⎡ 2 CD CH 2 ⎤
⎢C B +
⎥ = 0; K C =
2
K
C
C
B
⎣
⎦
Reaction rate
• Reaction rate is not a constant, it depends
– on concentration of the reagents
– on the temperature
– on the total pressure in the reactions
involving gas phase
– ionic strength and solvent in liquid state
– presence of a catalyst
Le Chatelier Principle
Dependence on the concentraion can be calculated knowing reaction
mechanism, as before
⎡ 2 CD CH 2 ⎤
d [C6 H 6 ]
= rB = −2k1 ⎢CB +
⎥
dt
K
C
⎣
⎦
Temperature dependence:
Arrhenius equation
k (T ) = Ae − E / RT
Reaction rate
• Below, we will consider only the concentration and
temperature dependence of the reaction rate
Dependence on the concentration can be calculated knowing reaction
mechanism, as before
C C ⎤
⎡
d [C6 H 6 ]
D H2
= rB = −2k1 ⎢CB 2 +
⎥
dt
K
C
⎣
⎦
Temperature dependence:
Arrhenius equation
k (T ) = Ae − E / RT
Reaction rate
• From the collision theory, only molecules
with the energy higher than the activation
energy can react:
• The usual “rule of thumb that the
reaction rate doubles with every
10ºC is not always true:
k (T1 ) k (T2 ) = e
⎛1 1⎞
− E / R⋅⎜ − ⎟
⎝ T1 T2 ⎠
Back to the reactors
aA + bB ⎯⎯
→ cC + dD
• if we are interested in species A we can define A as the basis
of calculation
b
c
d
A + B ⎯⎯
→ C+ D
a
a
a
XA =
• conversion:
Moles of A reacted
Moles of A fed
• number of moles A left after conversion:
N A = N A0 − N A0 X = N A0 (1 − X )
N A N A0 (1 − X )
CA =
=
V
V
• number of moles B left after conversion:
N B = N B 0 − N A0
b
X
a
N B N B 0 − ( b a ) N A0 X N A0 ( Θ B − ( b a ) X )
CB =
=
=
V
V
V
Batch reactors
• For every component in the reactor we can write after
conversion X is achieved:
d
Flow reactors
• Equations for flow reactors are the same with number
of moles N changed for flow rate F [mol/s].
Flow reactors
• For a flow system a concentration at any point can be
obtained from molar flow rate F and volumetric flow
rate
FA moles/time moles
CA =
=
=
v
liters/time
liter
• For reaction in liquids, the volume change is
negligible (if no phase change occurred):
FA
CA =
= C A0 (1 − X )
v
b ⎞
⎛
CB = C A0 ⎜ Θ B − X ⎟
a ⎠
⎝
Flow reactor with variable flow rate
• In a gas phase reaction a molar flow rate might
change as the reaction progresses
⎯⎯
→ 2 NH 3
N 2 + 3H 2 ←⎯
⎯
4 moles
2 moles
• In a gas phase reaction a molar flow rate might
change as the reaction progresses
total concentration: CT =
total volume rate:
v = v0
Cj =
FT
P
=
v ZRT
at the entrance:
CT 0 =
FT 0
P0
=
v
Z 0 RT0
FT P0 Z T
FT 0 P Z 0 T0
Fj
v
= Fj
Fj P Z 0 T0
⎛ FT P0 Z T ⎞
⎜ v0
⎟ = CT 0
FT P0 Z T
⎝ FT 0 P Z 0 T0 ⎠
Flow reactor with variable flow rate
• In a gas phase reaction a molar flow rate might
change as the reaction progresses
P0 T
P T
= v0 (1 + ε X ) 0
P T0
P T0
FT P0 Z T
v = v0
FT 0 P Z 0 T0
v = v0 (1 + y A0δ X )
FT = FT + FA0 ⋅ δ ⋅ X
FA0 ( Θ j + ν j X )
Cj =
P T
v0 (1 + ε X ) 0
P T0
Example
• For reaction below calculate
– equilibrium conversion in a constant batch reactor;
– equilibrium conversion in a flow reactor;
– assuming the reaction is elementary, express the rate of
the reaction
– plot Levenspil plot and determine CSTR volume for 80%
conversion
• assume the feed is pure N2O4 at 340K and 202.6kPa.
KC=0.1mol/l; kA=0.5 min-1.
⎯⎯
→ 2 NO2
N 2O4 ←⎯
⎯
Example
1. Batch reactor
CBe 2
4C A0 2 X e 2
4C A0 X e 2
KC =
=
=
1− X e
C Ae C A0 (1 − X e )
C A0 =
y A0 P0
= 0.071mol/dm3 .
RT0
X e = 0.44
Example
2. Flow reactor
FA FA0 (1 − X ) C A0 (1 − X )
=
=
CA =
1+ ε X
v v0 (1 + ε X )
CB =
C A0 X
1+ ε X
CBe 2
4C A0 2 X e 2
4C A0 X e 2
=
=
KC =
C Ae C A0 (1 − X e ) (1 − X e )(1 + ε X e )
C A0 =
y A0 P0
= 0.071mol/dm3 .
RT0
X e = 0.51
Example
3. Rate laws
⎡
CB 2 ⎤
− rA = k A ⎢C A −
⎥
K
C ⎦
⎣
–
Constant volume
–
Flow
–
Levenspiel plot
⎡
4C A0 2 X 2 ⎤
− rA = k A ⎢C A0 (1 − X ) −
⎥
K
C
⎣
⎦
⎡ C A0 (1 − X )
4C A0 2 X 2 ⎤
− rA = k A ⎢
−
2⎥
K C (1 + ε X ) ⎥⎦
⎢⎣ 1 + ε X
Example
4. CSTR volume for X=0.4, feed of 3 mol/min
⎡
4C A0 2 X 2 ⎤
– Constant volume −rA = k A ⎢C A0 (1 − X ) −
⎥
⎣
− rA | X =0.4 = 7 ⋅10−4
FA0 X
V=
= 1714 dm3
−rA | X
–
Flow
KC
⎦
Example: Hippo’s digestion
•
A hippo's stomach consists of three distinct chambers - the parietal blind
sac, the stomach, and the glandular stomach which connects to the
intestines.
CSTR
Fermentation reaction
PFR
Hippo’s digestion
• Stomach (CSTR)
A + M ⎯⎯
→P+ M
• Assuming Monod kinetics in the stomach:
kC ACM
−r =
CA + KM
C A = C A0 (1 − X ) ; CB = C A0 (Θ B + YM X )
kC A0 2 (1 − X )(YM + Θ M X )
−rAM 1 =
=
C A0 (1 − X ) + K M
b + cX − eX 2 −1
1 − aX
=
=
;
1 − aX
rAM 1 b + cX − eX 2
Hippo’s digestion
• Intestine (PFR)
A + E ⎯⎯
→ EA ⎯⎯
→P+ E
• Assuming Michealis-Menten kinetics:
kC A
−r =
CA + KM
C A = C A0 (1 − X ) ;
−1 1 + b(1 − X )
=
rAM 2
a(1 − X )
Hippo’s stomach
• Maximum conversion rate in the stomach
• Maximum conversion rate
in the intestine:
• CSTR is more efficient for
autocatalytic reaction; PFR
is more efficient for enzyme
catalyzed reaction
Hippo’s stomach
• for CSTR
• for PFR
• studies indicate that X2=45%; 75% of it happens in CSTR
Hippo’s stomach
• Our starting point:
X0=0; X1=0.34 X2=0.45;
• Fitting the data
V1=0.45m3
• flow rate 40 kg/day, with density of grass 365kg/m3,
volumetric rate 0.13 m3/day
Questions
•
•
•
•
A hippo is full grown in 4.5 years. A baby hippo follows its mother around
and eats one half of what she eats. However, the rate of digestion in the
autocatalytic CSTR stomach, -rAM1, is the same as the mother hippo as is
the conversion. Assuming that in the growing stages the hippo stomach
volume is proportianal to its age. How old is the baby hippo?
What is the conversion in the intestine (PFR) for the baby hippo diiscussed
in Question (1) assuming that Vmax is one half that of the mother?
The hippo has picked up a river fungus and now the effective volume of the
CSTR stomach compartment is only 0.2 m3. The hippo needs 30%
conversion to survive. Will the hippo survive?
The hippo had to have surgery to remove a blockage. Unfortunately, the
surgeon, Dr. No, accidentally reversed the CSTR and PFR during the
operation. Oops!! What will be the conversion with the new digestive
arrangement? Can the hippo survive?