Multiple Reactions. 授課教師：林佳璋. Definitions. There are four basic type of multiple reactions: series, parallel, complex, and independent. These types of multiple reactions can occur by themselves, in pair, or all together. When there is a combination of parallel and series reactions, they
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There are four basic type of multiple reactions: series, parallel, complex, and independent. These types of multiple reactions can occur by themselves, in pair, or all together. When there is a combination of parallel and series reactions, they
are often referred to as a complex reactions.
Parallel reactions (competing reaction)
The reactant is consumed by two different reaction pathways to form different products.
Series reactions (consecutive reaction)
The reactant forms an intermediate product, where reacts further to form another product.
formation of butadiene from ethanol
Involving a combination of both series and parallel reactions
cracking of crude oil to form gasoline
Occurring at the same time but neither the products nor reactants react with themselves or one another.
Of particular interest are reactants that are consumed in the formation of a desired product, D, and the formation of an undesired product, U, in a
competing or side reaction.
For a parallel reaction sequence
For a series reaction sequence
We want to minimize the formation of U and maximize the formation of D because the greater the amount of undesired product formed, the greater the cost of separating the undesired product U from
the desired product D (Figure 6-1).
As the cost of a reactor system increases in a attempt to minimize U, the cost of separating species U from D decreases.
Selectivity tells us how one product is favored over another when we have multiple reactions. We can quantify the formation of D with respect to U by
defining the selectivity and yield of the system.
The instantaneous selectivity of D with respective to U is the ratio of the rate formation of D to the rate of formation of U.
for a batch reactor
Example 6-1 when we have multiple reactions. We can quantify the formation of D with respect to U by
Develop a relationship between SD/U and for a CSTR.
Consider the instantaneous selectivity for the two parallel reactions just discussed:
For a CSTR the overall and instantaneous selectivities are equal.
Carrying out a similar analysis of the series reaction
instantaneous yield when we have multiple reactions. We can quantify the formation of D with respect to U by
~Instantaneous yield and overall yield are identical
for a CSTR.
~From an economic standpoint, the overall selectivities
and yields are important in determining profits.
However, the rate-based selectivities give insights in
choosing reactors and reaction schemes that will help
maximize the profit.
~There often is a conflict between selectivity and
conversion (yield) because you want to a lot of your
desired product (D) and at the same time minimize
the undesired product (U). However, in many
instances, the greater conversion you achieve, not
only do you make more D, but you also form more U.
for a batch system
for a flow system
Parallel reactions when we have multiple reactions. We can quantify the formation of D with respect to U by
rate of disappearance of A is
instantaneous selectivity is
To make this ratio as large as possible, we want to carry out the reaction in a manner that will keep the concentration of reactant A as high as possible during the reaction.
A batch or plug flow reactor should
be used in this case.
CSTR should not be chosen under these
If the reaction is carried out in the gas phase, we should run it without inerts and at high pressures to
keep CA high.
If the reaction is in the liquid phase, the use of diluents
should be kept to a minimum.
Case 2: when we have multiple reactions. We can quantify the formation of D with respect to U by 2>1
For this ratio rD/rU to be high, the concentration of A should be as low as possible
This low concentration may be accomplished by diluting the feed with inerts and running the reactor at low concentrations of A.
A CSTR should be used because the concentrations of reactants are
maintained at a low level.
A recycle reactor in which the product stream acts as a diluent could be used to maintain the entering concentrations of A at a low
Because the activation energies of the two reactions in case 1 and 2 are not given, it cannot be determined whether the reaction should be run at high or
The sensitivity of the instantaneous selectivity to temperature can be determined from the ratio of the specific reaction rates,
A is the frequency factor
E is the activation energy
The reaction system should be carried out at a low temperature to maximize SD/U.
The reaction system should be operated at the highest possible temperature to maximize SD/U.
Example 6-2 1 and 2 are not given, it cannot be determined whether the reaction should be run at high or
Reactant A decomposes by three simultaneous reactions to form three products, one that is desired, B, and two that are undesired, X and Y. These gas-phase reactions,
along with the appropriate rate laws, are called the Trambouze reactions.
The specific reaction rate are given at 300 K and the activation energies for reactions (1), (2), and (3) are E1=10000 kcal/mol, E2=15000 kcal/mol, and E3=20000 kcal/mol. How and under what conditions (e.g., reactor type(s), temperature, concentrations) should the reaction be carried out to maximize the selectivity of B for an entering
concentration of A of 0.4M and a volumetric flow rate of 2.0 dm3/s.
maximum 1 and 2 are not given, it cannot be determined whether the reaction should be run at high or
The selectivity with respect to B is
Because the concentration changes down the length of a PFR, we cannot operate at this maximum. As a result, a CSTR would be used and operated at this maximum.
The corresponding selectivity at CA* is
The mole balance on a CSTR for this liquid-phase reaction (v=v0) is
Run at as high a temperature as possible with existing equipment and watch out for other side reactions that
might occur at high temperatures.
Case 1: If
Run at low temperatures but not so low that a significant conversion is not achieved.
Case 2: If
For the activation energies given above
The selectivity for this combination of activation energies is independent of temperature!
If greater than 72% conversion of A is required, then the CSTR operated with a reactor concentration of 0.112 mol/dm3 should be followed by a PFR because the concentration and hence selectivity will decrease continuously from CA* as we move down the PFR to
an exit concentration CAf. Hence the system
would give the highest selectivity while forming more of the desired product B, beyond
what was formed at CA* in a CSTR.
The exit concentrations of X, Y, and B can be found from the CSTR mole balances
If 90% conversion were required then the exit concentration would be CAf=(1-0.9)/(0.4 mol/dm3)=0.04 mol/dm3.
The PFR mole balances for this liquid-phase reaction (v=v0) are
At τ=0, the values of the entering concentrations to the PFR are the exit concentrations from the CSTR.
maximizing S would be CD/U
The two reactions with recycle shown in (i) and (j) can be used for highly exothermic reactions. Here the recycle stream is cooled and returned to the reactor to dilute and cool the inlet stream thereby avoiding hot spots and runaway reactions. The PFR with recycle is used for gas-phase reactions, and the CSTR is used for liquid-phase reactions.
The last two reactors, (k) and (I), are used for thermodynamically limited reactions where the equilibrium lies far to the left (reactant
and one of the products must be removed (e.g., C) for the reaction to continue to completion.
The membrane reactor (k) is used for thermodynamically limited gas-phase reactions, while reactive distillation (I) is used for liquid-phase reactions when one of the products has a higher volatility (e.g., C) than
the other species in the reactor.
Example 6-3 would be C
For the parallel reactions
Consider all possible combinations of reaction orders and select the reaction scheme that will maximize SD/U.
Case I:1>2, 1>2
To maximize the ratio rD/rU, maintain the concentration of both A and B as high as possible. To do this, use
~A tubular reactor (Figure 6.3 (b))
~A batch reactor (Figure 6.3 (c))
~High pressures (if gas phase), and reduce inerts
Case II: would be C1>2, 1<2
To make SD/U as large as possible, we want to make the concentration of A high and
the concentration of B low. To achieve this result, use
~A semibatch reactor in which B is fed slowly into a large amount of A (Figure 6.3(d))
~A membrane reactor or a tubular reactor with side streams of B continually fed to
the reactor (Figure 6.3(f))
~A series of small CSTRs with A fed only to the first reactor and small amounts of B
fed to each reactor. In this way B is mostly consumed before the CSTR exit stream
flows into the next reactor (Figure 6.3(h))
Case III: would be C1<2, 1<2
To make SD/U as large as possible, the reaction should be carried out at low concentrations of A and B. To achieve this result, use
~A CSTR (Figure 6.3(a))
~A tubular reactor in which there is a large recycle ratio (Figure 6.3(i))
~A feed diluted with inerts
~Low pressure (if gas pressure)
Case IV: would be C1<2, 1>2
To make SD/U as large as possible, run the reaction at high concentration of B and low concentration of A. To achieve this result, use
~A semibatch reactor with A slowly fed to a large amount of B (Figure 6.3(e))
~A membrane reactor or a tubular reactor with side streams of A (Figure 6.3(g))
~A series of small CSTRs with fresh A fed to each reactor (Figure 6-3(h))
For series of consecutive reactions, the most important variable is time: space-time for a flow reactor and real time for a batch reactor. To illustrate the
important of the time factor, we consider the sequence
B is the desired product
~If the first reaction is slow and the second reaction is fast, it will be extremely
difficult to produce species B.
~If the first reaction (formation of B) is fast and the reaction to form C is slow,
a large yield of B can be achieved.
~However, if the reaction is allowed to proceed for a long time in a batch
reactor, or if the tubular flow reactor is too long, the desired product B will be
converted to the undesired product C.
Example 6-4 would be C
The oxidation of ethanol to form acetaldehyde is carried out on a catalyst of 4 wt % Cu-2 wt % Cr on Al2O3. Unfortunately, acetaldehyde is also oxidized on this catalyst to form carbon dioxide. The reaction is carried out in a threefold excess of oxygen and in dilute concentration (ca. 0.1% ethanol, 1% O2, and 98.9% N2). Consequently, the volume change with the reaction can be neglected. Determine the concentration of
acetaldehyde as a function of space-time,
The reactions are irreversible and first order in ethanol and acetaldehyde, respectively.
O2 in excess
Mole balance on A: would be C
Mole balance on B:
Mole balance on C: would be C
The yield has been defined as would be C
shown as a function of conversion in Figure E6-4.2
Another technique is often used to follow the progress for two reactions in series. The concentrations of A, B, and C are plotted as a singular point at different space time (e.g., 1’, 2’) on a triangular diagram (see Figure 6.-4). The vertices correspond to
pure A, B, and C.
For (k1/k2)>>1, a large quantity of B can be obtained.
For (k1/k2)<<1, very little B can be obtained.
Algorithm for Solution of Complex Reactions would be C
~In complex reactions systems consisting of combination of parallel and
series reactions, the availability of software packing (ODE solvers) makes
it much easier to solve problems using moles Nj or molar flow rates Fj
rather than conversion.
~For liquid systems, concentration is usually the preferred variable used in
the mole balance equations. The resulting coupled differential mole
balance equations can be easily solved using an ODE solver.
~For gas systems, the molar flow rates are usually the preferred variable in
the mole balance equation.
Mole Balances would be C
Table 6-1 gives the forms of the mole balance equation we shall use for complex reactions where rA and rB are the net rates of formation of A and B.
We find the net rate of reaction for each species in terms of the concentration of the reacting species in order to combine them with their respective mole
Net Rates of Reaction of the concentration of the reacting species in order to combine them with their respective mole
Consider the following reactions
Example 6-5 of the concentration of the reacting species in order to combine them with their respective mole
Consider the following set of reactions:
Write the rate law for each species in each reaction and then write the net rates of formation of NO, O2, and N2.
Net Rates of Reaction
Relative Rates of Reaction of the concentration of the reacting species in order to combine them with their respective mole
Net rate NO of the concentration of the reacting species in order to combine them with their respective mole
Net rate N2
Net rate O2
Stoichiometry: Concentration of the concentration of the reacting species in order to combine them with their respective mole
for liquid-phase reactions, v=v0
for gas-phase reactions
where fn represents the functional dependence on concentration of the net rate of formation
Multiple Reactions in a PFR/PBR of the concentration of the reacting species in order to combine them with their respective mole
We now insert rate laws written in terms of molar flow rates into the mole balances. After performing this operation for each species, we arrive at a coupled set of first-order ordinary differential equations to be solved for the molar flow rates as a
function of reactor volume.
In liquid-phase reactions, incorporating and solving for total molar flow rate is not necessary
at each step along the solution pathway because there is no volume change with reaction.
for gas-phase reactions
Combing mole balance, rate laws, and stoichiometry
solved simultaneously with a numerical package or by writing an ODE solver.
Example 6-6 of the concentration of the reacting species in order to combine them with their respective mole
Consider again the reaction in Example 6-5. Write the mole balances on a PFR in terms of molar flow rates for each species.
total molar flow rate of all the gases is
mole balance on NO of the concentration of the reacting species in order to combine them with their respective mole
mole balance on NH3
mole balance on H2O
mole balance on N of the concentration of the reacting species in order to combine them with their respective mole 2
mole balance on O2
mole balance on NO2
ODE solver algorithm is shown in Table E6-6.1 of the concentration of the reacting species in order to combine them with their respective mole
Table 6-2 shows the equation for species j and reaction i that are to be combined when we have q
reactions and n species.
Example 6-7 that are to be combined when we have q
The production of m-xylene by the hydrodealkylation of mesitylene over a Houdry Detrol
catalyst involves the following reactions:
m-xylene can also undergo hydrodealkylation to form toluene:
The second reaction is undesirable, because m-xylene sells for a higher price than toluene ($1.32/lb vs. $0.30/lb). Thus we see that there is a significant incentive to maximize the
production of m-xylene.
The hydrodealkylation of mesitylene is to be carried out isothermally at 1500R and 35 atm in a packed-bed reactor in which the feed is 66.7 mol% hydrogen and 33.3 mol% mesitylene. The volumetric feed rate is 476 ft3/h and the reactor volume (i.e., V=W/b) is
The rate laws for reactions 1 and 2 are, respectively that are to be combined when we have q
where the subscripts are: M=mesitylene, X=m-xylene, T=toluene, Me=methane, and
At 1500R, the specific reaction rates are
The bulk density of the catalyst has been included in the specific reaction rate (i.e, k1=k1’b).
Plot the concentration of hydrogen, mesitylene, and xylene as a function of space time. Calculate the space time where the production of xylene is a maximum (i.e., opt)
Solution that are to be combined when we have q
2.Rate laws and net rates:
3.Stoichiometry: that are to be combined when we have q
4.Combining: that are to be combined when we have q
If we know CM, CH, and CX, then CMe, and CT can be calculated from the reaction
These equations can be solved analytically. that are to be combined when we have q
Multiple Reactions in a CSTR that are to be combined when we have q
~If there is no volume change with reaction, we use concentration, Cj, as variables.
~If the reactions are gas phase and there is volume change, we use molar flow rates,
Fj, as variables.
The total molar flow for n species is
Example 6-8 that are to be combined when we have q
For the multiple reactions and conditions described in Example 6-7, calculate the conversion of hydrogen and mesitylene along with the exiting concentrations of mesitylene, hydrogen,
and xylene in a CSTR.
The moles of hydrogen consumed in reaction 1 are equal to the moles of mesitylene
consumed. Therefore, the conversion of hydrogen in reaction 1 is
The conversion of hydrogen in reaction 2 is just the overall conversion minus the
conversion in reaction 1; that is,
The yield of xylene from mesitylene based on molar flow rates exiting the CSTR for
The overall selectivity of xylene relative to toluene is
~In addition to using membrane reactors to remove a reaction product in order to
shift the equilibrium toward completion, we can use membrane reactor to
increase selectivity in multiple reactions.
~This increase can be achieved by injecting one of the reactants along the length of
~It is particularly effective in partial oxidation of hydrocarbons, chlorination,
ethoxylation, hydrogenation, nitration, and sulfunation reactions.
By keeping one of the reactants at a low concentration, we can enhance selectivity. By feeding a reactant through the sides of a membrane reactor, we can keep
its concentration low.
Example 6-9 Reactions
take place in the gas phase. The overall selectivities, , are to be compared for a membrane reactor (MR) and a conventional PFR. First, we use the instantaneous selectivity to determine
which species should be fed through the membrane
We see that to maximize SD/U we need to keep the concentration of A high and the concentration of B low; therefore, we feed B through the membrane. The molar flow rate of A entering the reactor is 4 mol/s and that of B entering through the membrane is 4 mol/s as
shown in Figure E6-9.1 For the PFR, B enters along with A.
The reactor volume is 50 dm3 and the entering total concentration is 0.8 mol/dm3. Plot the molar flow rates and the overall selectivity, , as a function of reactor volume for both the MR
Mole Balances for both the PFR and the MR
Net Rates and Rate Laws
Transport Law Reactions
The volumetric flow rate through the membrane is given by Darcy’s Law:
K is the membrane permeability (m/skPa) and Ps (kPa) and Pt (kPa) are the shell and tube side pressures, and At is the membrane surface area (m2).
The molar flow rate of B per unit volume of reactor is
no pressure drop
Fi vs. V
SD/U vs. V
We can easily modify the program, Table E6-9.1, for the PFR simply by setting RB equal to zero (RB=0) and the initial condition for B to be 4.0.
Notes that the enormous enhancement in selectivity the MR has over the PFR.
Example 6-10 simply by setting R
The following gas-phase reactions take place simultaneously on a metal oxide-supported catalyst:
Writing these equations in terms of symbols yields
Determine the concentrations as a function of position (i.e., volume) in a PFR.
Addition information: Feed rate = 10 dm3/min; volume of reactor = 10 dm3; and CA0 =
CB0 = 1.0 mol/dm3, CT0 = 2.0 mol/dm3
Solution simply by setting R
no pressure drop
Notes that there is a maximum in the concentration of NO (i.e., C) at approximately 1.5 dm3.
Closure (i.e., C) at approximately 1.5 dm
~After completing this section, the student should be able to describe the
different types of multiple reactions (series, parallel, complex, and independent)
and to select a reaction system that maximizes the selectivity.
~The student should be able to write down and use the algorithm for solving
CRE problems with multiple reactions.
~The student should also be able to point out the major differences in the CRE
algorithm for the multiple reactions from that for the single reactions, and then
discuss why care must be taken when writing the rate law and stoichiometric
steps to account for the rate laws for each reaction, the relative rates, and the
net rates of reaction.
~Finally, the student should feel a sense of accomplishment by knowing they
have now reached a level they can solve realistic CRE problems with complex
Homework-4 (i.e., C) at approximately 1.5 dm
In Int. J. Chem. Kinet., 35, 555 (2003), does the model overpredict the orthodiethyl benzene concentration? Please give a reason.