CHE 304 (Spring 2010)
__________________
LAST NAM, FIRST
Problem set #2
For a first-order reaction, the following rate coefficients were found:
Temperature, °C
38.5
53.1
77.9
k, h-1
0.044
0.301
1.665
Determine the activation energy of the reaction in kJ/mol.
Solution
When these values are plotted in a diagram of ln k versus 1/T, with T in degrees Kelvin, a
straight line is obtained with slope –E/R, leading to an E value of 82.4 kJ/mol.
(2)1 There are two reactors of equal volume available for your use: one a CSTR, the other a PFR.
The reaction is second order ( rA = kCA2 = kCA02(1  X)2), irreversible, and is carried out
isothermally
AB
There are three ways you can arrange your system:
(a) Reactors in series: CSTR followed by PFR
(b) Reactors in series: PFR followed by CSTR
(c) Reactors in parallel with half the feed rate going to each reactor after which the exit
streams are mixed.
(d) State which system will give the highest overall conversion and which system will
give the lowest overall conversion.
Solution
C
A
0
C
A
1
C
A
2
(a) X1 = 0.382, X2 = 0.618
(b) X1 = 0.5, X2 = 0.634
(c)
CSTR:
PFR:
X1 = 0.5
X1 = 2/3
X1,ave = 0.5(0.5 + 2/3) = 0.5833
(d) State which system will give the highest overall conversion and which system will give the
lowest overall conversion.
System 2: highest conversion.
System 3: lowest conversion
1
Fogler, H. S., Elements of Chemical Reaction Engineering, Prentice Hall, 1999
(3)1 The exothermic reaction
AB+C
was carried out adiabatically and the following data recorded:
X
 rA mol/Lmin
0
10
0.2
16.67
0.4
50
0.5
50
0.6
50
0.8
12.5
0.9
9.09
The entering molar flow rate of A was 300 mol/min.
(a)
(b)
(c)
(d)
(e)
(f)
What are the PFR and CSTR volumes necessary to achieve 40% conversion?
Over what range of conversions would the CSTR and PFR reactor volumes be identical?
What conversion can be achieved in a 10.5 L CSTR?
What conversion can be achieved if a 7.2 L PFR is followed in series by a 2.4 L CSTR?
What conversion can be achieved if a 2.4 L CSTRR is followed in series by a 7.2 L PFR?
Plot the conversion and rate of reaction as a function of PFR reactor volume up to a
volume of 10 L.
Solution
CSTR:
V=
(300)(0.4)
= 2.4 liters
50
PFR:
V = 7.2 liters
(b) Over what range of conversions would the CSTR and PFR reactor volumes be identical?
For a feed stream that enters the reaction with a previous conversion of 0.40 and leaves at any
conversion up to 0.60, the volumes of the PFR and CSTR will be identical because the rate is
constant over this conversion range
VPFR = FA0 
0.6
0.4
F  0.6  0.4 
dX
= A0
= VCSTR
 rA
 rA
(c) Therefore 70% conversion can be achieved in a 10.5 L CSTR.
(d) 60% conversion can be achieved if a 7.2 L PFR is followed in series by a 2.4 L CSTR.
(d) What conversion can be achieved if a 2.4 L CSTR is followed in series by a 7.2 L PFR?
90% conversion can be achieved if a 2.4 L CSTR is followed in series by a 7.2 L PFR.
(f) Plot the conversion and rate of reaction as a function of PFR reactor volume up to a
volume of 10 L.
X=0.1:.1:.9;Ai=X;
ira=[.1 .08 .06 .04 .02 .02 .02 .05 .08 .11];
Area=0;
for i=1:9
Area=Area+.5*(ira(i)+ira(i+1))*.1;
Ai(i)=Area;
end
Vol=300*Ai;
figure(2)
plot(Vol,X);grid on
xlabel('V(liter)');ylabel('X')
rate=1.0./ira(2:10);
figure(3)
plot(Vol,rate);grid on
xlabel('V(liter)');ylabel('Reaction rate (mol/L*min)')
(4)1 Consider an ideal batch reactor with the irreversible homogeneous reaction
A + 2B  3C + D
This liquid phase reaction has the rate equation
− rA = kCA2CB CD-1
The reaction rate constant is k = 1.5 L/mol∙hr at 350oK. The activation energy of the reaction is
100 kJ/mol. The initial concentrations are: CA0 = 2.0 mol/L, CB0 = 4.0 mol/L, CC0 = 0 mol/L, and
CD0 = 1.0 mol/L. We will neglect any reaction that takes place while the initial charge is being
added to the reactor, and while the reactor and contents are being heated to reaction temperature.
(a) How much time is required for the concentration of A to reach 0.10 mol/L if the reactor
run isothermally at 350oK? What is the concentration of C at this time?
(b) The reactor will be run isothermally at 350oK. The concentration of A in the final product
must be less than 0.20 mol/L, and the molecular weight of C is 125. An average of 10
hours is required between the batches in order to empty and clean the reactor, and prepare
for the next batch. How large must the reactor be in order to produce 200,000 kg of C
annually (with 8000 hours per year of operation)?
(c) We want to produce 200,000 kg of C annually, with a final concentration of A of 0.20
mol/L or less. The only reactor available has a working volume of 1400 L. At what
temperature does the reactor have to be operated, if it is operated isothermally? Once
again, an average of 10 hours is required between batches to empty and clean the reactor,
and to prepare for the next batch.
Solution
(a) Time required for the concentration of A to reach 0.10 mol/L.
The fractional conversion of reactant A is defined as
X = (NA0 − NA)/ NA0
Therefore the number of moles of A at any time, t, is given by
NA = NA0(1 − X)
We now create a stoichiometric table for this batch system (A + 2B → 3C + D).
Species
Initial number of moles, t = 0
Number of moles at t = t
A
NA0
NA = NA0(1  X)
B
NB0
NB = NB0  2NA0X
C
0
NC = 3NA0X
D
ND0
ND = ND0 + NA0X
NT0 = NA0 + NB0 + ND0
NT = NT0 + NA0X
Total
For liquid phase reaction, we can assume the volume V is constant. Dividing each term in the
second and third columns of the above table by V yields
1
Roberts, G. W., Chemical Reactions and Chemical Reactors, Wiley, 2006
Species
Initial number of moles, t = 0
Number of moles at t = t
A
CA0
CA = CA0(1  X)
B
CB0
CB = CB0  2CA0X
C
0
CC = 3CA0X
D
CD0
CD = CD0 + CA0X
CT0 = CA0 + CB0 + CD0
CT = CT0 + CA0X
Total
Making a mole balance for species A yields
dN A
= rAV
dt
Since NA = NA0(1  X), we have
NA0
dX
=  rAV
dt
Substituting the rate of reaction gives
CA0
dX
= kCA2CB CD-1
dt
C

dX
C

CA0
= kCA02(1  X)2 CA0  B 0  2 X  CA0-1  D 0  X 


dt
 C A0

 C A0

1
dX
= kCA0 (1  X)2(4  2X)(1 + X)-1
dt
(1  X )dX
= 2kCA0dt
(1  X ) 2 (2  X )
Using partial fraction we obtain
(1  X )
3
2
3
=
+
+
2
2
2 X
(1  X ) (2  X ) 1  X
(1  X )
Equation (E-1) can then be integrated
2kCA0t = 3ln(1  X) +
2
 3ln(2  X)  2 + 3ln(2)
1 X
2
2 X 
2kCA0t =  3ln 
 2 + 3ln(2)
 +
 1 X  1 X
(E-1)
When CA = 0.1 mol/L, X = 0.9, we have
(2)(1.5)(2)t =  3ln(11) + 18 + 3ln(2) = 12.8858
t = 2.1476 hr
At this time the concentration of C is
CC = 3CA0X = 3(2)(0.9) = 5.4 mol/L
(b) Volume of the reactor to produce 200,000 kg of C annually for CA = 0.2 mol/L
When CA = 0.2 mol/L, X = 0.8, we have
(2)(1.5)(2)t =  3ln(6) + 8 + 3ln(2) = 4.7042
t = 0.784 hr
The total batch time is
ttot = 10 + 0.784 = 10.784 hr
The number of batches, nb, per year is
nb = 8000/10.784 = 742 batches/year
The concentration of C is
CC = 3CA0X = 3(2)(0.8) = 4.8 mol/L
The annual production of C is
742 (batches/year)4.8V (mol/batch) 0.125 (kg/mol) = 200,000 (kg/year)
The volume of the required reactor is
V = 449 L
(c) Operating temperature of a 1400 L reactor to produce 200,000 kg of C annually for CA = 0.2
mol/L
The annual production of C is
2.41400 (mol/batch)  0.125 (kg/mol)  8000 (hr/yr)/ttot(yr) = 200,000 (kg/year)
ttot = 16.8 hr
The reaction time is t = 16.8  10 = 6.8 hr. For CA = 0.2 mol/L, X = 0.8, we have
2kCA0t =  3ln(6) + 8 + 3ln(2) = 4.7042
Therefore k = 4.7042/(226.8) = 0.17295 L/molhr
The Arrhenius relationship can be used to calculate the required temperature
 E 
k(T) = koexp  

 RT 
E 

k(350) = koexp  

 350 R 
 E1
1   0.17295
k (T )
= exp    
= 0.1153
  =
1.5
k (350)
 R  T 350  
Taking the natural log of both sides yields

E1
1 
 
 = -2.1602
R  T 350 
For E = 100,000 J/mol and R = 8.314 J/moloK
1
1
1

= 1.79610-4 
= 3.036710-3
T 350
T
T = 329.3oK
(5) The elementary gas-phase reaction3
(CH3)3COOH(CH3)3  C2H6 + 2CH3COCH3
is carried out isothermally in a flow reactor with no pressure drop. The specific reaction rate at
50oC is 10-4 min-1 and the activation energy is 85 kJ/mol. Pure di-tert-butyl peroxide enters the
reactor at 10 atm and 127oC and a molar flow rate of 4.0 mol/min. Calculate the reactor volume
to achieve 90% conversion in a CSTR and a PFR.
If this reaction is to be carried out at 10 atm and 127oC in a batch mode with 95% conversion,
what reactor size would be required to process (4.0 mol/min  60 min/h  24 h/day) 3600 mol of
di-tert-butyl peroxide per day? You can assume a downtime of 6 hr between batches for cleaning
and feed preparation.
Assume that the reaction is reversible with equilibrium constant KC = 0.025 mol2/L6 and
calculate the equilibrium conversion and then repeat the calculation for the CSTR and the PFR to
achieve 95% of the equilibrium conversion.
Solution
Solution ----------------------------------------------------------------------------------------(CH3)3COOH(CH3)3  C2H6 + 2CH3COCH3
A
B
2C
We now create a stoichiometric table for this flow system.
Species
Feed rate to reactor
Effluent rate from reactor
A
FA0
FA = FA0(1  X)
B
0
FB = FA0X
C
0
FC = 2FA0X
FT0 = FA0
FT = FA0(1 + 2X)
Total
For the CSTR: V =
FA 0 X
, where  rA = kCA
 rA
The total concentration at any point, CT, and at the entrance, CT0, to the reactor are given by
CT =
F
P0
P
FT
=
and CT0 = T 0 =
ZRT
Z 0 RT0
Q0
Q
Assuming negligible changes in the compressibility factor, Z, we have
Q = Q0
3
FT P0 T
FT 0 P T0
Fogler, H. S., Elements of Chemical Reaction Engineering, Prentice Hall
For isothermal system with no pressure drop Q = Q0
CA =
FT
, therefore
FT 0
F (1  X )
FA
(1  X )
= A0
= CA0
Q0 (1  2 X )
Q
(1  2 X )
The reaction rate constant k at 127oC can be evaluated with the gas constant R = 8.314 J/moloK
= 0.08205 Latm/moloK.
E 1
1 
1 
 85,000  1
k2 = k1exp      = 10-4exp 


  = 0.0443 min-1
R
T
T
8
.
314
323
400



2 
  1
The initial di-tert-butyl peroxide (A) concentration at 10 atm and 127oC is given by
CA0 =
PA0
10
=
= 0.305 mol/L
RT
(0.08205)( 400)
The volume of the CSTR is then
V=
FA 0 X
F X (1  2 X )
(4.0)(0.95)(1 1.9)
= A0
=
= 16,312 L
kC A
kC A0 (1  X )
(0.0443)(0.305)(0.05)
For the PFR: V = FA0 
X
0
Since CA = CA0
dX
 rA
(1  X )
, we have
(1  2 X )
F
1 2X
X
dX = A0  2 X  3 ln( X  1) 0
1 X
kC A0
V=
FA0
kC A0

V=
FA0
kC A0
FA0
X 1 

  2 X  3 ln(  1 ) V = kC [ 2X  3ln(1  X)]
A0
V=
4.0
[ 20.95  3ln(1  .95)] = 2,098 L
(0.0443)(0.305)
X
0
For the batch reactor, the reaction time to achieve 95% conversion is given by
V
t = NA0 
0
dX
 rAV
If we assume constant volume V = V0 then the pressure will increase for isothermal system and
N (1  X )
N
CA = A = A 0
= CA0(1  X), we have
V
V0
V
t = CA0 
0
t=
1
dX
=
k
 rA

X
0
1
1
dX = [ ln(1  X)]
1 X
k
1
[ ln(1  .95)] = 67.6 min
0.0443
The total cycle time is then tc = 67.6 + 660 = 428 min. Therefore we have 2460/412  3 runs
per day. The initial moles of di-tert-butyl peroxide fed to the reactor is
NA0 = 3600/3 = 1200 moles
The batch reactor volume is finally
V=
N A0
1200
=
= 3,935 L
0.305
C A0
Calculate the equilibrium conversion
For the reaction
A  B + 2C
We have  rA = kfCA  kbCBCC2
At equilibrium  rA = kfCA 
kbCBCC2
= 0  KC =
kf
kb
C C
= B C
CA
2
= 0.025
( XC A0 )( 2 XC A0 ) 2
= 0.025
(1  2 X ) 2 C A0 (1  X )
With an initial concentration CA0 = 0.305 mol/L, we have
X340.3052 = 0.025(1 + 2X)2(1  X)
The above nonlinear equation can be solved with the Matlab function solve
solve('x^3*4*.305^2-.025*(1+2*x)^2*(1-x)')
The result is X = X eq = 0.512. For 95% of the equilibrium conversion X = 0.95X eq = 0.4864
V=
FA 0 X
 rA
 rA = kfCA  kbCBCC2 = kf(CA  CBCC2/KC)
k
X
(1  X )
X2
 f CA0
(2CA0)2
KC
(1  2 X )
(1  2 X )
(1  2 X ) 2
 rA = kfCA0
 rA = kf
C A0
(1  2 X )
2

1 4C A0 X 3 
(
1

X
)



K C (1  2 X ) 2 

0.305
 rA = 0.0443
1  2  0.4864

1 (4)(0.305)2 (0.4864)3 
(1  0.4864)  .025 (1  2  0.4864)2 


 rA = 5.035110-4 mol/L
The CSTR volume is then
V=
FA 0 X
(4.0)(0.4864)
=
= 3,864 L
5.0351 10 4
 rA
For the PFR: V = FA0 
X
0
2

1 4C A0 X 3 
C A0
dX
where  rA = kf
(1  X ) 

K C (1  2 X ) 2 
 rA
(1  2 X ) 
We can use the Matlab function quad to numerical integrate FA0 
X
0
pfrvol is written with f =
dX
. The Matlab function
 rA
FA0
as follows:
 rA
function f=pfrvol(x)
k=.0443;cao=0.305;Kc=.025;
ra=k*cao*(1-x-(4*cao^2*x.^3)./(1+2*x).^2/Kc)./(1+2*x);
f=4.0./ra;
The expression f must allow for the fact that X is a vector in the function. We now can use the
function quad to integrate

0.4864
0
fdX .
>> quad('pfrvol',0,0.4864)
ans =
5.6867e+002
Therefore the volume for the PFR is 569 L.
(6)4 Equilibrium with respect to the reaction
A(g) + B(g) = C(g)
will be studied by measuring the volume change accompany the reaction. The temperature and
pressure are held constant and the initial volume and the final volume of the reacting system are
recorded. Three tested were made and are summarized in the table. Has equilibrium been
established? If so what is the value of K?
P(mmHg)
500
600
600
yA
0.5
0.333
0
Initial composition
yB
yC
0.5
0
0.667
0
0
1.0
Volume (cm3)
Initial
Final
200
150
300
233
200
293
Solution
K = 4.5914
K’s are constant for the three cases: equilibrium has been established.
(7) Run Murder Mystery (http://www.engin.umich.edu/~cre/icm/cre.html)
You will find the program Murder Mystery in the CHE 304 distribution folder, then
Kinetics, then click on Murder.bat. Turn in the last page of the program with performance
number.
4
Kyle, B.G., Chemical and Process Thermodynamics, Prentice Hall, 1999