For a First-Order Reaction, the Following Rate Coefficients Were Found

CHE 304 (Spring 2010) ______

LAST NAM, FIRST

Problem set #2

For a first-order reaction, the following rate coefficients were found:

Temperature, °C / k, h-1
38.5 / 0.044
53.1 / 0.301
77.9 / 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

(d) State which system will give the highest overall conversion and which system will give the lowest overall conversion.

Solution

(a) X1 = 0.382, X2 = 0.618

(b) X1 = 0.5, X2 = 0.634

PFR:

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

(3)1 The exothermic reaction

A ® B + C

was carried out adiabatically and the following data recorded:

X

/ 0 / 0.2 / 0.4 / 0.5 / 0.6 / 0.8 / 0.9
- rA mol/L×min / 10 / 16.67 / 50 / 50 / 50 / 12.5 / 9.09

The entering molar flow rate of A was 300 mol/min.

(a) What are the PFR and CSTR volumes necessary to achieve 40% conversion?

(b) Over what range of conversions would the CSTR and PFR reactor volumes be identical?

(d) What conversion can be achieved if a 7.2 L PFR is followed in series by a 2.4 L CSTR?

(e) What conversion can be achieved if a 2.4 L CSTRR 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.

Solution

CSTR: V = = 2.4 liters

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 = = VCSTR

(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)[2] 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
Total / NT0 = NA0 + NB0 + ND0 / NT = NT0 + NA0X

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

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
Total / CT0 = CA0 + CB0 + CD0 / CT = CT0 + CA0X

Making a mole balance for species A yields

= rAV

Since NA = NA0(1 - X), we have

NA0 = - rAV

Substituting the rate of reaction gives

CA0 = kCA2CB CD-1

CA0 = kCA02(1 - X)2 CA0 CA0-1

= kCA0 (1 - X)2(4 - 2X)(1 + X)-1

= 2kCA0dt (E-1)

Using partial fraction we obtain

= + +

Equation (E-1) can then be integrated

2kCA0t = 3ln(1 - X) + - 3ln(2 - X) - 2 + 3ln(2)

2kCA0t = - 3ln + - 2 + 3ln(2)

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

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

k(T) = koexp

k(350) = koexp

= exp = = 0.1153

Taking the natural log of both sides yields

= -2.1602

For E = 100,000 J/mol and R = 8.314 J/mol×oK

= 1.796´10-4 Þ = 3.0367´10-3

T = 329.3oK

(5) The elementary gas-phase reaction[3]

(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
Total / FT0 = FA0 / FT = FA0(1 + 2X)

For the CSTR: V = , where - rA = kCA

The total concentration at any point, CT, and at the entrance, CT0, to the reactor are given by

CT = = and CT0 = =

Assuming negligible changes in the compressibility factor, Z, we have

Q = Q0

For isothermal system with no pressure drop Q = Q0, therefore

CA = = = CA0

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.

k2 = k1exp = 10-4exp = 0.0443 min-1

The initial di-tert-butyl peroxide (A) concentration at 10 atm and 127oC is given by

CA0 = = = 0.305 mol/L

The volume of the CSTR is then

V = = = = 16,312 L

For the PFR: V = FA0

Since CA = CA0, we have

V = dX =

V = V = [- 2X - 3´ln(1 - X)]

V = [- 2´0.95 - 3´ln(1 - .95)] = 2,098 L

For the batch reactor, the reaction time to achieve 95% conversion is given by

t = NA0

If we assume constant volume V = V0 then the pressure will increase for isothermal system and CA = = = CA0(1 - X), we have

t = CA0 = dX = [- ln(1 - X)]

t = [- ln(1 - .95)] = 67.6 min

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 = = = 3,935 L

Calculate the equilibrium conversion

For the reaction

A Û B + 2C

We have - rA = kfCA - kbCBCC2

At equilibrium - rA = kfCA - kbCBCC2 = 0 Þ KC = = = 0.025

= 0.025

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 =

- rA = kfCA - kbCBCC2 = kf(CA - CBCC2/KC)

- rA = kfCA0 - CA0(2CA0)2

- rA = kf

- rA = 0.0443

- rA = 5.0351´10-4 mol/L

The CSTR volume is then

V = = = 3,864 L

For the PFR: V = FA0 where - rA = kf

We can use the Matlab function quad to numerical integrate FA0. The Matlab function pfrvol is written with f = as follows:

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 .

> 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?

Initial composition / Volume (cm3)
P(mmHg) / yA / yB / yC / Initial / Final
500 / 0.5 / 0.5 / 0 / 200 / 150
600 / 0.333 / 0.667 / 0 / 300 / 233
600 / 0 / 0 / 1.0 / 200 / 293

Solution

K = 4.5914