CPB 415: Chemical Kinetics and Reactor DesignProblem Solutions

KaRE TExT Units 13 to 24

Problem 1. (Unit 14) Suppose the following mechanism (equations 1 and 2) is true for the overall reactionA+B+2DR+S

A+BX+R (1)

2D+XS (2)

If step 2 is the rate-determining step, derive a rate expression for the overall reaction that does not includethe concentration of reactive intermediates.

Solution. First, we can visually inspect the proposed mechanism for this case and determine that steps 1and 2 are mathematically independent and that the stoichiometric number of each step is 1 (b1= b2=1).

Since step 2 is the rate-determining step, the overall rate is equal to the forward rate of step 2 dividedby the stoichiometric number of step 2

roverall=r2,forward

b2= k2rd[D]

2[X]

where note that the stoichiometric number has been absorbed into the rate coefficient (for this particularcase b2=1, however this procedure is adopted in general). The concentration of X appears in the equation.This is not acceptable because X is a reactive intermediate.

To eliminate its concentration from the rate expression, all steps other than the rate-determining step areat quasi-equilibrium (for this example this is just step 1, but in general this would be all steps other thanthe rate-determining step). For step 1

K1,eq=[X][R]

[A][B]

which can be solved to obtain an expression for X in terms of constants and concentrations of stable species

[X] =K1,eq[A][B]

[R]

This can then be substituted into the rate expression, eliminating the concentration of X from it andproducing the desired result

roverall=k2rdK1,eq[A][B][D]

2

[R]

1

Problem 2. (Unit 14) Derive and simplify the rate expression for the following reaction mechanism if step3 is the rate determining step. How would the rate expression change if step 2 was rate determining?

Overall reaction: 2A+2B2Y+2ZMechanism:

A+BY+S (1)

B+ST+Z (2)

A+TY+Z (3)

Solution. To develop good habits, we will first determine if there is a linear combination of the mechanisticsteps that sums to give the macroscopically observed reaction. A necessary condition for there to be a linearcombination of mechanistic steps that sums to give the overall reaction is

i,overall=

s=all steps

bsi,s (4)

It will be easiest for me to set up the system of equations in a table (since we will ultimately want to constructa matrix).

i i,overall i,1 i,2 i,3

A 2 1 0 1

B 2 1 1 0

Y 2 1 0 1

S 0 1 1 0

T 0 0 1 1

Z 2 0 1 1

Now lets set up the resulting matrix in Octave or Matlab and computing the rank (see problem_2.m). Doingso we find that the rank of the matix is three.

We find that three (of the six total) equations resulting from applying equation 4 to each species aremathematically independent. Perfect! Since we have three independent equations and we have three stepsin the mechanism, we can solve for unique values of b1, b2, and b3. (This seems to be a confusing point forsome, so see page 6 of the Unit 11 Informational Reading for more details. Also, note that this is differentthan Probelm 12 of Problem set 1. For that case we had four independent equations resulting from equation4 but only three mechanistic steps. As a result, we could not solve for a unique set of bs, and the mechanismwas therefore not valid; it violated rule 1 that a mechanism must obey.) At this point we could solve for b1,b2, and b3, but it is not necessary.

Since step 3 is rate-determining, then the overall rate, r, is equal to the rate of the rate-determining step.The rate expression can be written for this step since it is elementary, and since it is rate-determining, theterm corresponding to the reverse rate can be dropped, leaving

r= k3CACT

This rate expression is not acceptable because it contains the concentration of T, and T is a reactiveintermediate. In fact, both S and T are reactive intermediates because they appear in the mechanism but notin the overall, macroscopically observed reaction. To eliminate their concentrations from the rate expression,we first assume that all steps other than the rate-determining step are quasi-equilibrated, and write thecorresponding equilibrium expressions:

K1=CYCSCACB

K2=CTCZCBCS

2

We can solve these expressions for the concentrations of S and T

CS=K1CACB

CY

CT =K2CBCS

CZ=

K1K2CACB2

CYCZ

Where now in our expression for CT we substituted in our expression for CS to eliminate if from theexpression. We can now plug in our final expression for CT to get our final, acceptable rate expression

r= k3CAK1K2CACB

2

CYCZ=

k3K1K2CA2CB

2

CYCZ

If step 2 was rate-determining, the overall rate would equal the forward rate of step 2. Step 1 would stillbe quasi-equilibrated, so our expression for K1 (which we solved for CS) would still be valid. Writing theforward rate of step 2

r= k2CBCS

We can then plug in our expression for CS to obtain an acceptable rate expression

r=k2K1CACB

2

CY

3

Problem 3. (Unit 15 and 16) Suppose that enzyme E catalyzes the conversion of substrate S to productP. The overall reaction is given by

SP

and is described by the following mechanism, with step 2 being irreversible.

E+SE-S (1)

E-SE-P (2)

E-PE+P (3)

Derive a Michaelis-Menton type of rate expression for this system.

Since no rate-determining step has been specified, we begin by choosing a reactant or product of theoverall reaction and generating an expression for the rate of the overall reaction with respect to the chosenspecies. Here we will choose the product, P, but the exact same answer could be obtained if we chose thesubstrate, S. In general, the rate of reaction j with respect to species i may be written as:

ri,j=

s=all steps

i,s

{

ks

(

m=all reactants

[m]m,s)

(

1

m=all species [m]m,s

Keq,s

)}

=

s=all steps

i,s

ks

(

m=all reactants

[m]m,s)

ks

Keq,s

(

m=all products

[m]m,s)

=

s=all steps

i,s

ks,f

(

m=all reactants

[m]m,s)

ks,r

(

m=all products

[m]m,s)

Applying to P, where note that the stoichiometric coefficient of P in steps 1 and 2 is 0 and the stoichiometriccoefficient in step 3 is 1, we obtain

rP,overall= k3,f[E-P] k3,r[E][P]

where note that the brackets just correspond to a concentration. This rate expression is not acceptablebecause it contains the concentration of a reactive intermediate (the enzyme-product complex). To eliminatethis reactive intermediate concentration, the Bodenstein steady state approximation (BSS, below) can beapplied to each reactive intermediate, RI.

0=

s=all steps

RI,s

ks,f

(

m=all reactants

[m]m,s)

ks,r

(

m=all products

[m]m,s)

In the present problem, there are three reactive intermediates, E, E-S, and E-P, so BSS can be written threetimes, once for each of the reactive intermediates. However, because E is an enzyme and is therefore con-served, we realize that only two of the resulting equations would be mathematically independent. Therefore,we only apply the BSS approximation twice, once for E-S and once for E-P. Doing so leads to

0= k1,f[E][S] k1,r[E-S] k2,f[E-S]

0= k2,f[E-S] k3,f [E-P] + k3,r[E][P]

The additional relationship we need is the conservation of catalyst or enzyme

E0=[E] + [E-S] + [E-P]

These three equations can now be solved to obtain expressions for the concentrations of the reactive inter-mediates in terms of constants and the concentrations of stable species. Here you may use a computeralgebra system to help solve the system of equations. Options available are the (free) open-source softwareMaxima, or MuPAD which comes installed with Matlab. Here I have used MuPAD (see problem_3.mn).MuPAD returns a set of seven solutions, each dependent on seven different conditions. Only the sixth solutionis possible (k3,f + Pk3,r 0 and k2,fk3,f + k3,fk1,r + Pk2,fk3,r + Pk1,rk3,r + Sk1,fk2,f + Sk1,fk3,f 0). Anequivalent solution is obtained with Maxima, which only returns the one solution (see problem_3.wxm).

4

So solving using MuPad or Maxima we have

[E] =E0k3,f(k1,r+ k2,f)

k3.fk1,r+ k3,fk2,f +(k3,fk1,f + k1,fk2,f)[S] + (k3,rk1,r+ k3,rk2,f)[P]

[E-P] =E0(k1,fk2,f[S] + (k3,rk1,r+ k3,rk2,f)[P])

k3,fk1,r+ k3,fk2,f +(k3,fk1,f + k1,fk2,f)[S] + (k3,rk1,r+ k3,rk2,f)[P]

[E-S] =E0k1,fk3,f[S]

k3,fk1,r+ k3,fk2,f +(k3,fk1,f + k1,fk2,f)[S] + (k3,rk1,r+ k3,rk2,f)[P]

Our expressions for [E] and [E-P] may be substituted into rP,overall to obtain our final expression

rP,overall=E0k1,fk2,fk3,f[S]

k3,fk1,f + k3,fk2,f +(k3,fk1,f + k1,fk2,f)[S] + (k3,rk1,r+ k3,rk2,r)[P]

We find that this takes the form of the general Michaelis-Menton equation

rP,overall=Vmax[S]

Km+ [S] +KP[P]

5

Problem 4. (Unit 15 and 16) Suppose that the enzyme E catalyzes the conversion of substrate S to productP (equation 1), but the enzyme is inhibited by inhibitors I and J according to the mechanism given in steps 2through 5. If reaction 4 is irreversible, use the Bodenstein steady state approximation to derive an acceptableexpression for the overall rate of reaction.

SP (1)

E+SES (2)

ES+ IESI (3)

ESE+P (4)

E+JEJ (5)

Solution. As in problem 2, to develop good habits, we will first determine if there is a linear combinationof the mechanistic steps that sums to give the macroscopically observed reaction. A necessary condition forthere to be a linear combination of mechanistic steps that sums to give the overall reaction is

i,overall=

s=all steps

bsi,s (6)

It will be easiest for me to set up the system of equations in a table (since we will ultimately want to constructa matrix).

i i,overall i,2 i,3 i,4 i,5S 1 1 0 0 0

P 1 0 0 1 0

E 0 1 0 1 1

ES 0 1 1 1 0

I 0 0 1 0 0

ESI 0 0 1 0 0

J 0 0 0 0 1

EJ 0 0 0 0 1

Now lets set up the resulting matrix in Octave or Matlab and compute the rank (see problem_4.m). Doingso we determine that the rank of the matix is five.

Hmm, five of the equations are mathematically independent, and there are only four mechanistic steps.Therefore, this would suggest that we can not solve for a unique set of bs. Did Prof Paluch throw us a curveball? No! Remember for homogeneous and enzymatic catalysis reactions, we have a conservation of enzymecriteria. Therefore, not all of the enzyme concentrations are independent of each other. Specifically, assumingthe initial amount of enzyme (or catalyst) is known, then one of the enzyme concentrations can be writtenin terms of the initial amount minus the concentration of all other enzyme forms. The proposed mechanismis therefore mathematically valid.

The overall rate of the apparent, macroscopically observed reaction j may be found with respect to speciesi as

ri,j=

s=all steps

i,srs

Thus, applying this to the product, P, leads to

rP,1=

s

P,srs=(0)r2+(0)r3+(1)r4+(0)r5

rP,1= r4

6

Because each step in a reaction mechanism must be an elementary reaction, the rate expressions for themechanistic steps are known, as given below for step 4, where note that the brackets denote either the partialpressure or the concentration of the species

rP,1= r4= k4[ES]

(

1[E][P]

K4[ES]

)

Since we are told that step 4 is irreversible, this may be reduced to

rP,1= k4[ES]

This rate expression is not acceptable because it contains the concentration of a reactive intermediate, namelythe enzyme-substrate complex, ES. To eliminate the concentration of reactive intermediates from the rateexpression, the Bodenstein steady-state (BSS) approximation can be applied to each reactive intermediate. Inthis case the reactive intermediates are E, ES, ESI, and EJ. However, if the BSS is applied to all four reactiveintermediates, the resulting equations will not be independent. Hence, one of the equations is replaced witha statement of the conservation of total catalyst (or in this case enzyme)

E0=[E] + [ES] + [ESI] + [EJ]

Applying BSS to E, ESI, and EJ leads to the following three equations

0= rE,1=(1)r2+(0)r3+(1)r4+(1)r5=k2[E][S] +k2K2

[ES] + k4[ES] k5[E][J] +k5K5

[EJ]

0= rESI,1=(0)r2+(1)r3+(0)r4+(0)r5= k3[ES][I]k3K3

[ESI]

0= rEJ,1=(0)r2+(0)r3+(0)r4+(1)r5= k5[E][J]k5K5

[EJ]

The conservation of enzyme and the three BSS equations can be used to solve for E, ES, ESI, and EJ as follows

[E] =E0

1+

(

k2[S]k2K2

+ k4

)

+

(

k2K3[S][I]k2K2

+ k4

)

+K5[J]

[ES] =

(

k2E0[S]

k2

K2+ k4

)

1+

(

k2[S]k2K2

+ k4

)

+

(

k2K3[S][I]k2K2

+ k4

)

+K5[J]

[ESI] =

(

k2K3E0[S][I]

k2

K2+ k4

)

1+

(

k2[S]k2

K2+ k4

)

+

(

k2K3[S][I]k2

K2+ k4

)

+K5[J]

[EJ] =K5E0[J]

1+

(

k2[S]k2K2

+ k4

)

+

(

k2K3[S][I]k2K2

+ k4

)

+K5[J]

We can plug in our equation for [ES] into the rate expression to obtain our final result

rP,1=

(

k2k4E0[S]

k2

K2+ k4

)

1+

(

k2[S]k2K2

+ k4

)

+

(

k2K3[S][I]k2K2

+ k4

)

+K5[J]

7

Problem 5. (Unit 17 and 18) The water gas shift reaction might take place according to the mechanismgiven in equations 1 and 2. If step 2 is irreversible, find an expression for the rate of water-gas shift in termsof the partial pressures of the stable species (and rate and equilibrium constants).

Overall, macroscopically observed reaction:

CO+H2OCO2+H2

Proposed mechanism:

CO+O-CO2+ (1)

H2O+H2+O- (2)

Solution. This problem is solved following the same procedures/steps as with the enzymatic reactionproblems encountered in the last two proplems. Species not appearing in the overall, macroscopic reaction(or vacant sites and adsorbed species) are considered reactive intermediates and must not appear in thefinal overall reaction. The only difference is that rather than having a conservation of catalysis or enzymeconstraint, now we have a conservation of surface sites. Also, the concentrations in our rate expression willchange slightly, as we will encounter shortly.

The overall rate of the apparent, macroscopically observed reaction is found as

ri,j=

s=all steps

i,srs

Recall that this is the rate of the overall reaction j with respect to species i. How is this related to thegeneral rate of reaction? We can use any species to generate an expression for the reaction rate. I will chooseH2. This choice was motivated by the fact that H2 only appears in step 2 which is irreversible, and it is aproduct, so its stoichiometric coefficient will be positive. Writing the reaction rate with respect to H2

rH2,overall=

s

H2,1rs=(0)r1+(1)r2= r2

Because each step in a reaction mechanism must be an elementary reaction, the form of the rate expressionsfor each mechanistic step is known and given by the usual equation

ri,s= ks

(

i=all reactants

[i]i,s)

(

1

i=all species [i]i,s

Ks

)

ri,s= ks

i=all reactants

[i]i,s ksKs

i=all products

[i]i,s

Before applying this expression to obtain an equation for r2. It is useful to pause and mention the concen-trations of the species in our mechanism. The species in the macroscopic equation are not a problem. Herewe will use partial pressures for their concentrations since they are gas phase species. For the vacant sitesand adsorbed species, they are written in terms of their surface concentration Ci,surf. For each species thismay be written as

Ci,surf=Csitesi

where Csites is the concentration of active sites (which is assumed to be constant) and i is the fractionalcoverage of species i. Since Csites is a constant, it is usually absorbed into the rate coefficient, ks.

Writing the expression for rH2,overall= r2 where we are told that step 2 is irreversible

rH2,overall= r2= k2PH2Ov

(

1PH2O

K2PH2Ov

)

= k2PH2Ov

8

where the subscript v stands for vacant. This rate expression is not acceptable because it contains theconcentration of a reactive intermediate, namely the vacant surface site. Since we are not told that any stepsare rate-determining, we will eliminate the concentrations of reactive intermediates from the rate expressionusing the Bodenstein steady-state approximation (BSSA). The reactive intermediates for this problem areadsorbed oxygen (O-) and the vacant surface sites (). However, if the BSSA is applied to both reactiveintermediates, the resulting equations will not be mathematically independent. Hence, one of the equationsis replaced with the conservation of surface sites

v+

i

i=1

v+ O=1

where the summation is over all adsorbed species (i). BSSA can be either applied to the vacant sites oradsorbed oxygen. Neither appears to be more advantageous than the other since they both appear in steps1 and 2. Lets apply BSSA to the vacant sites, where recall that step 2 is irreversible.

0= r,i=

s

,srs=(1)r1+(1)r2

0= k1PCOOk1K1

PCO2v k2PH2Ov

We can solve our BSSA equation and our conservation of surface sites equation for v and O. Since the twoequations are linear, this is not too difficult. A strategy I might apply would be to multiply the conservationof surface sites equation by k1PCO and then add it to the BSSA equation to eliminate O. You can thensolve for v, which may then be substituted back into the conservation of surface sites expression to obtainO. Doing so, we find

v=1

1+1

K1

PCO2PCO

+k2

k1

PH2O

PCO

O=

1

K1

PCO2PCO

+k2

k1

PH2O

PCO

1+1

K1

PCO2PCO

+k2

k1

PH2O

PCO

Finally, plugging back into our original rate expression

rH2,1=k2PH2O

1+1

K1

PCO2PCO

+k2

k1

PH2O

PCO

=k2PH2OPCO

PCO+1

K1PCO2+

k2

k1PH2O

9

Problem 6. (Unit 17 and 18) Suppose that the gas phase conversion of A to B is heterogeneously catalyzedand the mechanism is given by reactions 1 through 3. In addition, suppose that gas phase species Z adsorbsreversibly on the catalyst as given in equation 4. Derive an acceptable expression for the rate of generationof B taking place in the presence of Z. Your expression should not include any surface coverages.

A+A- (1)

A-B- (2)

B-B+ (3)

Z+Z- (4)

Solution. It is convenient to first write the rate expression for each of the mechanistic steps. Since they aremechanistic steps, they must be elementary, and consequently the rate expressions are given in equations 5through 8.

r1= k1,fPAv k1,rA (5)

r2= k2,fA k2,rB (6)

r3= k3,fB k3,rPBv (7)

r4= k4,fPZv k4,rZ (8)

We are asked to generate an acceptable expression for the rate of generation of B. Since gas phase speciesB only appears in reaction 3, the rate of generation of B will be equal to r3.

roverall,B= r3

Looking at that equation, it can be seen to contain fractional coverages, and consequently it is not acceptableas written. The Bodenstein steady state approximation is used to find expressions for the surface coveragein terms of the species partial pressures and other constants. The Bodenstein steady state approximationsays that the net rate of generation of reactive intermediates (surface species in this case) is equal to zero.Applying the steady state approximation to surface adsorbed A, B, and Z leads to equations 9 through 11

0= r1 r2 (9)

0= r2 r3 (10)

0= r4 (11)

We could also write the steady state approximation for the vacant site, but wed find that it isnt mathemat-ically independent. Instead, we use a statement of the conservation of catalytic sites as given in equation 12

1= v+ A+ B+ Z (12)

Our next step is to solve equations 9 to 12 for A, B, Z, and v. This problem requires alot of elbow grease towork out, so lets use a computer algebra system package to help us out. For some reason MuPAD struggledwith this one, so I only used Maxima (see problem_6.wxm).

Let us define the following constants:

kv= k1,rk2,rk4,r+ k1,rk3,fk4,r+ k2,fk3,fk4,r

kA= k1,fk2,fk4,r+ k1,fk2,rk4,r+ k1,fk3,fk4,r

kB= k1,rk3,rk4,r+ k2,fk3,rk4,r+ k2,rk3,rk4,r

kZ= k1,rk2,rk4,f + k1,rk3,fk4,f + k2,fk3,fk4,f

10

So that:

A=(k1,fk2,rk4,r+ k1,fk3,fk4,r)PA+(k2,rk3,rk4,r)PB

kv+ kAPA+ kBPB+ kZPZ

B=(k1,fk2,fk4,r)PA+(k1,rk3,rk4,r+ k2,fk3,rk4,r)PB

kv+ kAPA+ kBPB+ kZPZ

Z=k4,f(k1,rk3,f + k2,fk3,f + k1,rk2,r)PZ

kv+ kAPA+ kBPB+ kZPZ

v=k4,r(k1,rk3,f + k2,fk3,f + k1,rk2,r)

kv+ kAPA+ kBPB+ kZPZ

Plugging into equation 7 gives the desired rate expression for the generation of B

roverall,B= r3=k1,fk2,fk3,fk4,rPA k1,rk2,rk3,rk4,rPB

kv+ kAPA+ kBPB+ kZPZ

11

Problem 7. (Unit 22 to 24) A continuous stirred laboratory reactor with a volume of 450 mL is to be usedfor kinetic studies. The following experiment was performed to assess whether the reactor can be modeledas an ideal CSTR. A solution of 1 M salt water was fed to the reactor at a steady state of 75 mL per min.After steady state has been reached, the salt was removed from the feed so that pure water was being fed,still at a rate of 75 mL per min. The concentration of salt in the water leaving the reactor was measured atvarious times after removing the salt from the feed. The data are given below; can the reactor be modeledas an ideal CSTR?

time (min) concentration (M)

0 1

0.25 0.96

0.5 0.91

0.75 0.89

1 0.86

2 0.73

3 0.63

4 0.49

5 0.44

6 0.37

8 0.25

10 0.21

12 0.13

14 0.10

16 0.07

20 0.04

30 0.00

50 0.00

100 0.00

Solution. The experimental age function is calculated as

F () =wtw0wf w0

Since this is a dilute aqueous solution (and then pure water), assume that the density of the system isconstant. For this case the age function is equivalent to

F ()=CtC0Cf C0

where C0 = 1 M and Cf = 0 M. Taking from the first column above (since the step change occurred att=0) and Ct from the second column, this equation can be used to calculate F () for each value of . Theaccompanying Excel spreadsheet (problem_7.xls) shows how to do this.

Then, to determine if the reactor may be modeled as an ideal CSTR, we can compute the age functionfor an ideal CSTR as

F ()= 1 exp

(

t

)

where t is the average residence time computed as

t =fluidVCSTR

M=

VCSTR

V0 =

450mL

75mL/min=6min

since we are assuming the density is constant. Finally, to asses the ideality of the laboratory reactor, onewould plot the experimental F and the ideal CSTR F vs. . If they are equal (to within experimental error)then the lab reactor can be modeled as an ideal CSTR.

12

As seen in the accompanying Excel spreadsheet, the experimental age function and the age function ofan ideal CSTR are in very close agreement, suggesting that it may be possible to model the reactor as anideal CSTR. This problem may also be solved using Matlab (see the problem_7 folder).

13

Problem 8. (Unit 22 to 24) Suppose you need to measure the age function for a 12 L reactor in yourlaboratory. To generate the necessary data, a steady flow of water is established at a rate of 6 L/min. Astopwatch is started just as 24 g of a tracer is instantly added to the reactor inlet. The concentration oftracer in the outlet is then measured as a function of time. The resulting data are given in the table below.Plot the age function for the reactor as F () vs. .

Time (min) Concentration (g/L)

0.2 0.04

0.4 1.97

0.6 1.76

0.8 1.47

1.0 1.36

1.2 1.26

1.4 1.08

1.6 0.91

1.8 0.88

2.0 0.70

2.2 0.56

2.4 0.64

2.6 0.50

2.8 0.48

3.0 0.45

5.0 0.14

7.5 0.03

10 0.05

Solution. The problem describes the application of an impulse stimuli to the reactor, consequently, the agefunction is found as

F (t t0) =F () =M

t0

t

(wout(t)w0)dt

mtot

The times given in the table represent the time elapsed since the application of the stimulus (t0 = 0 min),and consequently they are equal to . The total mass used in the impulse is given as mtot= 24 g, and beforeand after the impulse, there was no tracer in the feed, so w0= 0. The fluid flowing in the reactor is water,and it volumetric flow rate is given, so its mass flow rate can be found through use of the density of water as

M = V=

(

6L

min

)

(

1000g

L

)

= 6000g

min

The data table gives mass concentrations in the outlet; these are related to the mass fraction in the outletaccording to

wout(t) =Cout(t)

=

Cout(t)

1000

L

g

After substitution of these known quantities and relationships, our equation for F () simplifies as

F ()=6000

0

t(

Cout(t)

1000

L

g 0

)

dt

24min 1=

(

0.25L

gmin

)

0

t

Cout(t)dt

Thus, for each entry in the table, we can readily calculate the value of the age function. Notice that theunits of the integral will cancel out the units preceding the integral giving a dimensionless age function, asexpected.

Since we do not have a functional form for Cout, but instead a set of discrete values, the integration willbe performed numerically, for example using the trapezoid rule. In the present case, we will use Excel. Pleasesee the accompanying Excel spreadsheet for further details (problem_8.xls).

14

Problem 9. (Unit 22 to 24) A 5 gallon bucket of red dye at a concentration of 5 M was dumped into theinlet of a pilot plant reactor. The reactor volume is 95 L and the liquid feed flows at a rate of 0.65 gal/min.At the same time, a spectrophotometer began analyzing the outlet stream from the reactor. The signal fromthe spectrophotometer is directly proportional to the concentration of the dye. The table below reports theratio of the instantaneous spectrophotometer signal to that prior to the addition of the dye. On the basis ofthe data that were obtained, do you believe that the reactor can be accurately modeled as a plug flow reactor?

time (min) signal

1 0.01

10 0.01

20 0.05

30 0.02

35 0.09

36 0.03

37 15.40

38 9.41

39 5.72

40 3.60

41 2.10

42 1.25

43 0.81

44 0.46

45 0.30

50 0.04

55 0.01

60 0.03

Solution. The situation described corresponds to the application of an impulse stimulus to the reactor. Theage function is found from the following equation where molar units are used and it has been noted that theconcentration of dye prior to the impulse was zero.

F (t t0)=F () =M

t0

t

(wout(t)w0)dt

mtot=

V

t0

t

(Cout(t)C0)dt

ntot=

V

t0

t

Cout(t)dt

ntot

The volumetric flow rate is given as 0.65 gal/min which corresponds to 2.46 L/min. The total number ofmoles in the impulse can be calculated as follows:

ntot=VimpulseCimpulse

ntot=(5 gal)(5M)

which after unit conversion gives ntot= 94.6mol.The concentration was not measured, however, but only a signal that is proportional to it. That is,

if denotes the proportionality between the signal and the concentration and I denotes the signal, theconcentration is related to the signal as follows:

Cout(t) =I(t)

The proportionality constant can be determined by noting that

ntot= V

0

Cout(t)dt= V

0

I(t)dt=V

0

I(t)dt

=ntot

V

0

I(t)dt

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Alternatively, one can rearrange the equation for the age function and eliminate the proportionality constantas follows:

F ()=V

t0

t

Cout(t)dt

ntot=

V

t0

t

I(t)dt

V

0

Cout(t)dt

=V

t0

t

I(t)dt

V

0

I(t)dt

=

t0

t

I(t)dt

0

I(t)dt

The integral can be evaluated numerically using the trapezoid rule, provided the data arent too noisy.Otherwise, a higher order scheme is necessary. For this problem, the noise level is small so the trapezoid willbe used. To check this, plot the signal vs. time and look for noise.

The evaluation of the integral of the signal vs. time is shown in the accompanying Excel spreadsheet(problem_9.xls) where it is found that the value of the integral is 40.5 min. For our experiment, taking 60

minutes to be the infinite time limit, having already calculated V and ntot we can calculate which is found toequal 0.95 mol/min. With this information, the age function can be computed at each sample point as follows

F () =V

t0

t

I(t)dt

ntot

This is also calculated in the Excel spreadsheet using the trapezoid rule at each sample time. These valuescan also be calculated using Matlab, as shown in the accompanying .m file.

The age function for an ideal PFR is zero for times less than the average residence time and 1 for timesgreater than the average residence time. Here, the average residence time is found as follows:

t=V

V=

95L

2.46L/min= 38.6min

The two age functions are plotted in the Excel spreadsheet and using Matlab (see the problem_8 folderfor Matlab files). The laboratory reactor deviates significantly from the ideal behavior. It would be a goodidea to examine the laboratory reactor to see if a cause for the deviation can be identified, and perhaps tryoperating at a higher flow rate to see if that makes the behavior more ideal.

Note, most real PFRs will show behavior similar to that of this problem. Here the deviation from idealbehavior does not appear to be severe, and the reactor probably could be modeled as an ideal PFR. It wouldbe useful to measure the stimulus experimentally because it is doubtful that the stimulus was a pure stepchange. Some of the deviation from ideal behavior seen in the plot may actually have been introduced bythe stimulus.

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Problem 10. (Unit 22 to 24) Suppose you have an 18 gal stirred tank in your lab that you want to useto perform kinetics experiments. You establish a steady flow of water through the reactor at a rate of 5gal/min. You then start a stopwatch, at just the same time that you start continuously adding dye to theinlet at a concentration of 2 oz/gal. You then proceed to measure the concentration of tracer in the outletas a function of time. The resulting data are given in the table below. On the basis of the age function ofthe stirred tank, do you believe it can be modeled as an ideal CSTR?

Time (min) Concentration (oz/gal)

0.0 0.00

1.0 0.24

1.5 0.53

2.0 0.77

2.5 0.98

3.0 1.04

4.0 1.34

5.0 1.47

6.0 1.58

7.0 1.62

8.0 1.67

9.0 1.89

10.0 1.93

12.0 1.97

14.0 1.86

16.0 2.06

18.0 2.02

20.0 1.92

Solution. The problem statement describes the application of a step change stimulus, so the age functionwill be computed as

F (t t0)=F () =wtw0wf w0

Where we wish to compare to the age function expected for an ideal CSTR

F ()= 1 exp

{

t

}

We will begin with the analysis of the laboratory reactor data. The times given in the table are the elapsedtimes since the application of the step change at t0 = 0min . Hence, these sampling times t

are also equalto . Before the step change was applied, there wasnt any tracer in the outlet, so w0 = 0. Additionally, ifwe wait long enough for the reactor to reach steady state, the outlet mass concentration of tracer will equalthe inlet mass concentration of tracer, 2 oz/gal. Since the fluid is water, this can be converted to a massfraction of tracer as

wf =Cf

=2 oz/gal

134 oz/gal= 0.015

Similarly, the weight fraction at any sampling time can be related to the mass concentration at that samplingtime as

wt=Ct

=

Ct

134 oz/gal= 0.0075Ct gal/oz

Substitution of these quantities and relationships into our age function gives

F ()=0.0075Ct gal/oz

0.015= 0.5Ct gal/oz

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which can be used to calculate F () for each of the sampling times given in the data table.The resulting data can be plotting in the form of F () vs. . If the stirred tank behaves like an ideal

CSTR, we experimental data should agree well with the model for an ideal CSTR. In order to plot the idealCSTR model, the average residence time must be determined as

t =V

V=

18 gal

5 gal/min= 3.6min

and then the age function can be computed as

F ()= 1 exp

{

3.6min

}

While there is some scatter in the data, it appears that the laboratory reactor can be modeled as anideal CSTR since the data points fall close to the expected line and the scatter is random, not systematic.Furthermore, if we construct a parity plot of the experimental RTD versus that of an ideal CSTR, we obtaina value of R2= 0.975, supporting our conclusion. Please see the accompanying Excel spreadsheet for details(problem_10.xls).

NOTE, this problem could have been solved exactly like 7 where we assumed the density was constant.Here I convert to weight fractions just for comparision.

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