Chapt er 6 MULTIPLE REACTIONS REACTIONS If there are multiple rxns, use concentrations not conversions. intermediate
Consecutive rxns
1. Series Reactions 2. Parallel Reactions
3. Complex Reactions: Series and Parallel
4. Independent
None of the products or reactants are common.
These might occur in combination or by themselves.
Desired & Undesired Rxns We want to get the desired product, D and avoid the formation of undesired one, U. k D In the parallel rxn A ⎯ ⎯→ ⎯ →D
A ⎯ U → U k
In series
D U ⎯→ ⎯ → D ⎯ ⎯→ ⎯ →U A ⎯
k
k
D Cost
A
Reactor
Total Cost
Reactor Cost
Seperator Seperator Cost
U
Low
Efficiency
High
1
Selectivity and Yield Selectivity tells us how one product is favored over another when multiple rxns take place
Exit flow rate For batch
~
S DU
=
N D N U
tend
Reaction yield = ratio of the rxn rate of a given product to the rxn rate of key reactant A.
Determine the instantaneous selectivity, S D/U, for the liquid phase reactions: 2
A + B ⎯ ⎯→ D
r D = k 1C A C B
A + B ⎯ ⎯→ U 1
r U 1 = k 2 C A C B
A + B ⎯ ⎯→ U 2
r U 2 = k 3 C A C B
3
Sketch the selectivity as a function of the concentration of A. Is there an optimum and if so what is it? SD/U1U 2 =
Find the optimum concentration
2
r D r U1 + r U 2
dS dC A
=
k 1C A C B k 2C A C B + k 3C A C B 3
= 0 = k 1 [k 2 + k 3C A CA = *
*2
=
k 1C A k 2 + k 3C A
2
]− k C [2k C ] *
1
A
*
3
A
k 2 k 3
2
Parallel Rxns : For the following competing rxns D A ⎯ D ⎯→
(desired)
U ⎯→ A ⎯ U
(undesired )
k k
r D = k D C A
α 1
r U = k U C A
α 2
Rate of disappeara nce of A - r A = r D + r U = k D C A
α 1
+ k U C A
α 2
where α 1 and α 2 are positive rxn orders.
S D/U =
r D r U
=
k D k U
⋅ CA
α 1 −α 2
Maximizing the the Desired Desired Product Product for for one one Reactant Reactant Case1: If α 1 > α 2 ⇒ SD/U = Case2: If α 1 < α 2 ⇒ SD/U =
k D k U k D
⋅ CA
⋅
α 1 −α 2
k U C A
1 α 2 −α 1
SD/U
as C A
SD/U
as C A
In Case 1: If gas rxn, we should run it without any inerts and at high pressures to keep C A high. If liquid phase rxn, the diluents should be minimum. CSTR should not be used, batch or plug flow should be used. In Case 2: Dilute the feed with inerts. CSTR should be used, because C A decreases rapidly.
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Effect of Temperature on Selectivity It can be determined from SD/U ~
k D k U
=
AD AU
⋅ e -([E
D - E U ]/RT)
where A : freq factor,
E : activation energy SD/U
Rxn should be operated at the highest possible T to max. SD/U
ED > EU ⇒ T
SD/U
ED < EU ⇒ T
2. Series Reactions
This series reaction could also be written as Reaction (1) Reaction (2)
Species A:
Species B:
4
Using the integrating factor, i.f.:
at t = 0, C B = 0
When should you stop the reaction to obtain the maximum amount of B? Let's see
Then
And
5
The Integrating Factor (1) If you only had an expression of the form
(2)
things would be much easier, then you could integrate with respect to z and find y(z). (3) then you could manipulate equation (3): (4) where the term in brackets is the left hand side of equation (1). You need du/dz = u f(z). Recall (5)
If you define
and f(z) = dq/dz (i.e.
, then
(6) This satisfies the condition that du/dz=u f(z). Therefore, and substituting into equation (4), (7)
where the term in brackets is the left hand side of equation (1). CONCLUSION: If your problem is of the form
you can multiply both sides of the equation by the
6
(which you should be able to evaluate, since you know f(z)), to yield
or, substituting from equation (7)
Derivation
f(t)=k 2, so
From equation (12), the solution is then
The constant can be obtained from the intitial condition that at t=0, CB=0;
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3. Algorithm for Multiple Reactions
Reactor Selection & Operating Conditions Conditions:: Consider two simultaneous rxns k 1 ⎯→ A + B ⎯ D
r D = k 1C A 1 C B
k 2 ⎯→ A + B ⎯ U
r U = k 2 C A 2 C B
S D/U =
r D r U
=
k 1 k 2
α
β 1
α
CA
(α 1 −α 2 )
CB
β 2
( β 1 − β 2 )
We usually want to max SD/U. So the reactors should be designed to max SD/U B B
∆H>>0
A A
B
B
B D
A A
∆H>>0
C
A D
Thermodinamically limited rxns A + B <-> C + D
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Algo ut io n o f Comp lex Reactio Al go ri th m for Sol Solut React io ns : In combinations of parallel and series rxns, ODE solver packages make life easier. •
Write the rxns
•
Write the mole balances
•
Net rate laws
•
Stoichiometry
•
Combine & Solve
4. Applications of Algo rithm (1) (2)
NOTE: The specific reaction rate k1A is defined with respect to species A. NOTE: The specific reaction rate k2C is defined with respect to species C.
Case 1: PFR Mole Balances A : B:
dF A dV dF B dV
C:
dF C
D :
dF D
dV dV
Rate Laws
= r A = r B = r C = r D
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r A = r 1A + r 2A
Species A
r 1A = -k 1A C A C B
For reaction (1):
2
r 1B = 2r 1A r 1C = -r 1A r 1A = -k 1A C A C B - r 2A
For reaction (2):
2
=
2
- r 2C 3
r 2A = −
2
r 2D = −
1
3 3
2
( -r 2C ) = −
3
3
2
3
2
k 2C C C C A
r 2C 2
Species A
r A = − k 1A C A C B −
Species B
r B = r 1B = − 2 k 1A C A C B
Species C
r C = k 1A C A C B − k 2C C C C A
Species D
r D = r 2D = −
2
3
2
r 2C 3
k 2C C C C A 2 3
=
1 3
3
2
k 2C C C C A
2
Writing the net rates Stoichiome try
liquid : v = v0 FA = CAv0
Combine Species A
dC A dV
=
r A v0 2 3
− k 1A CA CB − k 2CCC CA 2
dCA dV Species B
dC B dV dCB dV
Species C
dCC dV dCC dV
Species D
dC D dV dCD dV
= = =
3
v0
Evaluate solution using Polymath of MatLab
r B v0
k1A = 0.5
− 2k 1A C AC B
2
k2C = 2.0
v0
=
v0 =5.0
r C
@ t=0; V=0, C A0=4, CB0=4, CC0=0, CD0=0
v0 k 1A C A C B − k 2C C C C A 2
= =
3
2
Vf = 5 dm3
v0
r D v0 3
=
2
1 k 2CC C C A 3
2
v0
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Case 2: CSTR
Species A
V=
(FA0 - FA )
Species B
V=
(FB0 - FB )
Species C
V=
(FC0 - FC )
Species D
V=
(FD0 - FD )
- r A
- r B
=
v 0 (C A0 - C A ) 2 2 3 2 k 1A C A C B + k 2C CC C A 3
=
v 0 (C B0 - C B ) 2k 1AC A C B
- r C - r D
=
FC
=
FD
=
r C r D
=
2
v 0C C k 1A C A C B − k 2C CC C A 2
3
2
v 0C D 1 3
3
k 2C CC C A
2
We will specify V, C A0, CB0 along with the specific reaction rates k ij. This formulation leaves us with four equations and four unknowns (C A, CB, CC and CD)
⎡ ⎣
2
f(C A ) = - V ⋅ ⎢ k 1A C A C B +
[
2
3
2
⎤ ⎦
k 2C C C C A ⎥ + v 0 (C A0 - C A ) 3
]
f(C B ) = - V ⋅ 2 k 1A C A C B + v 0 (C B0 - C B )
[
2
]
f(C C ) = V ⋅ k 1A C A C B − k 2C C C C A − v 0 C C 2
3
2
⎡1 3 2⎤ f(C D ) = V ⋅ ⎢ k 2C C C C A ⎥ − v 0 C D ⎣3 ⎦
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Case 3: Semibatch Liquid Phase Species A Species B Species C Species D
dC A dt dC B dt dC C dt dC C dt
= − k 1A C A C B − 2
=
v0 [C B0
− CB ]
V
2 3
k 2C C C C A − 3
2
− 2 k 1A C A C B
= k 1A C A C B − k 2C C C C A − 2
=
1 3
3
k 2C C C C A − 3
2
2
v0 C A
V
2
v0 C C
FB0
V
v0 C D
V
A
V = V0 +v 0 t
Parameters
k 1A = 0.5 v 0 = 1.2
dm 6 mol2 hr
dm3 hr
Initial conditions
k 2C = 2 C B0 = 4 CAi = 4
dm 2 mol4 hr mol dm
3
V0 = 2dm 3
C Bi = 0
CCi = 0 C Di = 0
t f = 8 hr
PolyMath Solutions
12
THE ALGORITHM Mole balances
Rate Laws
Relative rates Rxn (1) A + 2B ->C
Rxn (2) 2A + 3C ->D
Net rates Species A Species B Species C Species D
2
Species A
r A = − k 1A C A C B −
Species B
r B = r 1B = − 2 k 1A C A C B
Species C
r C = k 1A C A C B − k 2C C C C A
Species D
r D = r 2D = −
Stoichiometry
2
3
3
k 2C C C C A 2
2
CA =
r 2C 3
FT0 FA0 P T0
⋅
⋅
⋅
v 0 FT0 P0 T
3
=
1 3
2
3
2
k 2C C C C A
= CT0 ⋅
2
FA0 P T0
⋅
⋅
FT0 P0 T
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Combine
∆P = 0, P = P0 , T =T0 Ci =CT0
Species A
Species B
Species C
Species D
dFA dV dFB dV dFC dV
Fi FT 2
= −k 1A C A C B − k 2C CC C A 2
3
2
3
= −2k 1A CT0
3
⎛ FA ⎞⎛ FB ⎞ ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ ⎝ FT0 ⎠⎝ FT0 ⎠
2
2
= k 1A CT0
3
⎛ FA ⎞⎛ FB ⎞ 1 ⎛ F ⎞ ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ − k 2CC T0 5 ⎜⎜ C ⎟⎟ ⎝ FT0 ⎠⎝ FT0 ⎠ 3 ⎝ FT0 ⎠
⎛ F ⎞ = k 2CC T0 ⎜⎜ C ⎟⎟ dV 3 ⎝ FT0 ⎠ FT = FD +FC + FB +FA dFD
1
5
3
⎛ FA ⎞ ⎜⎜ ⎟⎟ F ⎝ T0 ⎠
k 2C = 1.3
3
⎛ FA ⎞ ⎜⎜ ⎟⎟ ⎝ FT0 ⎠
2
2
Parameters
k 1A = 0.05,
Initial Cond.
V = 0, FA = 10, FB = 20, FC = 0, FD = 0, Vf = 200
For aa CSTR CSTR
CT0 = 0.8
Species A
V=
(FA0 - FA )
Species B
V=
(FB0 - FB )
Species C
V=
(FC0 - FC )
Species D
V=
(FD0 - FD )
Total :
FT = FD +FC + FB +FA
- r A - r B - r C - r D
Five equations five unknowns X=
FA0 - FA FA0
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ConcentrationConcentration-Time Trajectories 1. The following concentration-time trajectories were observed in a batch reactor
Which of the following reaction pathways best describes the data
2. Sketch the concentration-time trajectory for the reaction
Solution: Part 1 Choice B is the answer. Choices A and C are incorrect because they show species B eventually consumed, which is clearly not the case
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Solution: Part 2
(1) B virtually consumed so no more D can be produced in reaction 2. (2) Rates of Consumption of A and Ba are virtually the same. (3) Rate of consumption of B greater than that of A owing to Reaction 2
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