ChE 101 – Chemical Reaction Engineering

Winter 2014

Homework #2

Solutions

1. Consider the following reversible elementary reaction:

A ↔2B

The rate coefficient of the forward reaction is kf = 25 hr-1, and the rate coefficient of the reverse

reaction is kb = 4M-1 hr-1. The aqueous phase reaction is carried out in a constant volume batch reactor.

The reactor is initially charged with 1 mole of pure A and 1 mole of pure B at 1 atm. The reactor volume

is 1 L.

(a) Write the rates of reaction for A and B.

(b) Write rA as a function of the conversion of A.

(c) What is the equilibrium conversion of A?

(d) Find the conversion of A as a function of time.

(e) How long does it take for the system to reach 90% of the equilibrium conversion?

(a) 𝑟𝐴 = 𝑘𝑏𝐶𝐵2 − 𝑘𝑓𝐶𝐴 = 4𝐶𝐵

2 − 25𝐶𝐴

𝑟𝑏 = 2𝑘𝑓𝐶𝐴 − 2𝑘𝑏𝐶𝐵2 = 50𝐶𝐴 − 8𝐶𝐵

2

(b) Let X be the fractional conversion of A, and make an ICM table.

Species Initial(mol) Change(mol) Final(mol)

A 1 -X 1-X

B 1 +2X 1+2X

Total 2 X 2+X

Given that the reactor volume is 1L, we know

𝐶𝐴 =𝑁𝐴

𝑉= 1 − 𝑋

𝐶𝐵 =𝑁𝐵

𝑉= 1 + 2𝑋

Thus the rate of change of A is given by:

𝑟𝐴 = 4(1 + 2𝑋)2 − 25(1 − 𝑋)

(c) Recall that 𝐾𝑒𝑞 = ∏ 𝑎𝑗

𝑣𝑗𝑆𝑗=1 = exp (−

∆𝐺𝑅𝑜

𝑅𝑇)Where aj is the activity of species j and vj is the

stoichiometric coefficient of species j. We also know that 𝐾𝑒𝑞 =𝑘𝑓

𝑘𝑏. When working with ideal

solutions, the activity is equal to the concentration of the species, thus:

𝐾𝑒𝑞 =𝑘𝑓

𝑘𝑏=

25

4

𝐾𝑒𝑞 = ∏ 𝑐𝑗

𝑣𝑗

𝑆

𝑗=1

=𝐶𝐵

2

𝐶𝐴=

(1 + 2𝑋𝑒𝑞)2

1 − 𝑋𝑒𝑞

Where Xeq is the equilibrium conversion of A. Equating and solving these two expersions, we find

that:

(1 + 2𝑋𝑒𝑞)2

1 − 𝑋𝑒𝑞=

25

4

𝑋𝑒𝑞 = 0.4375

(d) We need simply to integrate the rate expression from part b.

𝑟𝐴 =𝑑𝐶𝐴

𝑑𝑡=

𝑑(1 − 𝑋)

𝑑𝑡= −

𝑑𝑋

𝑑𝑡= 4(1 − 2𝑋)2 − 25(1 − 𝑋)

This can be solved by separating and integration. 𝑑𝑋

𝑑𝑡= −4(1 + 2𝑋)2 + 25(1 − 𝑋)

𝑑𝑋

𝑑𝑡= −16𝑋2 − 41𝑥 + 21

𝑑𝑋

𝑑𝑡= −(𝑋 + 3)(16𝑋 − 7)

𝑑𝑋

(𝑥 + 3)(16𝑋 − 7)= −𝑑𝑡

∫𝑑𝑋

(𝑥 + 3)(16𝑋 − 7)= − ∫ 𝑑𝑡

𝑡

𝑡=0

𝑋

𝑋=0

−1

55∫

𝑑𝑋

𝑥 + 3+

16

55∫

𝑑𝑋

16𝑋 + 7= −𝑡

𝑋

𝑋=0

𝑋

𝑋=0

1

55(ln|16𝑋 − 7| − ln|𝑋 + 3| − ln (

7

3)) = −𝑡

ln (7 − 16𝑋

𝑋 + 3) = −55𝑡 + ln (

7

3)

7 − 16𝑋

𝑋 + 3=

7

3exp (−55𝑡)

𝑋 =21(1 − exp(−55𝑡))

48 + 7 exp(−55𝑡)

Note: at the equilibrium conversion, 16X-7 =0. All conversions of interest will be below this value, thus

|16X-7| = 7-16X for this reaction. In addition, |X+3| = X+3 for all X of interest.

(e) We want to reach 90% of the equilibrium conversion found previously, thus X = 0.9X 0.4375 =

0.39375. Reorganizing the result from part d, we find:

𝑡90% = −1

55(ln(7 − 16 ∗ 0.39375) − ln(0.39375 + 3) − ln (

7

3)) = .0441 ℎ𝑟 = 2.65 min = 159 𝑠𝑒𝑐

2. In a batch reactor the following reaction takes place: A+2B 2C+ D

The reactor is initially charged with CAo and CBo and CBo/CAo = 3.0. Find the value of k

if CAo is 0.01 M and after 10 min of reaction the conversion of A is 0.5.

To find the rate constant, we first need to know the values of CA and CB in the reactor.

A + 2B 2C + D

Given our definition for conversion, and a simple stoichiometric argument, we know that

CA = CAo*(1-XA)

CB = CBo – 2*CAo*XA

By the law of mass action, the rate of change of A is given by :

𝑑𝐶𝐴

𝑑𝑡= 𝐶𝐴𝑜

−𝑑𝑋𝐴

𝑑𝑡= −𝑘𝐶𝐴𝐶𝐵

2 = −𝑘𝐶𝐴𝑜(1 − 𝑋𝐴)(𝐶𝐵𝑜 − 2𝐶𝐴𝑜𝑋𝐴)2

𝑑𝑋𝐴

𝑑𝑡= 𝑘(1 − 𝑋𝐴)(𝐶𝐵𝑜 − 2𝐶𝐴𝑜𝑋𝐴)2

∫𝑑𝑋𝐴

(1 − 𝑋𝐴)(𝐶𝐵𝑜 − 2𝐶𝐴𝑜𝑋𝐴)2= ∫ 𝑘𝑑𝑡

10

0

.5

0

Integrating gives

𝑘 = 1.2 ∗ 1021

𝑚𝑖𝑛 𝑀2

3. Consider the following gas phase reaction, which is carried out in a constant pressure batch

reactor:

A + B C

It is found that starting with 1 mole of A and 2 moles of B, the volume after 20 minutes

decreased by 20% from an initial volume of 1 L at a pressure of 1 atm. If the same reaction is to

be carried out in a constant volume reactor with an initial volume of 1L and an initial pressure of

1 atm, calculate the conversion after 20 minutes.

For a batch reactor at constant pressure, 𝑑𝑁𝑗

𝑑𝑡= 𝑉𝑣𝑗𝑟. From a TCF table, it can be found that:

𝑁𝐴 = 1 − 𝑋

𝑁𝐵 = 2 − 𝑋

𝑁𝐶 = 𝑋

∑ 𝑁 = 3 − 𝑋 = 3(1 −1

3𝑋)

Because all species are at constant pressure with an initial volume Vo, and assuming they are

ideal gasses,

𝑉 = 𝑉𝑜 (1 −1

3𝑋)

. 8𝑉𝑜 = 𝑉𝑜 (1 −1

3𝑋)

𝑋 = .6

By definition, we know that:

𝐶𝐴 =𝑁𝐴

𝑉=

1 − 𝑋

1 −13 𝑋

𝐶𝐵 =𝑁𝐵

𝑉=

2 − 𝑋

1 −13

𝑋

From the rate law above,

𝑑𝑁𝐴

𝑑𝑡= −𝑉𝑘𝐶𝐴𝐶𝐵 = −𝑘 (1 −

1

3𝑋) ∗

1 − 𝑋

1 −13 𝑋

∗2 − 𝑋

1 −13 𝑋

𝑑𝑁𝐴

𝑑𝑡= −𝑘 ∗

(1 − 𝑋)(2 − 𝑋)

1 −13 𝑋

𝑑𝑋

𝑑𝑡= 3𝑘 ∗

(1 − 𝑋)(2 − 𝑋)

3 − 𝑋

Separate and integrate.

∫(3 − 𝑋)𝑑𝑋

(1 − 𝑋)(2 − 𝑋)= 3𝑘 ∫ 𝑑𝑡

𝑡

𝑡=0

𝑋

𝑋=0

∫ (1

𝑋 − 2−

2

𝑋 − 1) 𝑑𝑋 = 3𝑘𝑡

𝑋

𝑋=0

ln(2 − 𝑋) − 2 ln(1 − 𝑋) − ln(2) = 3𝑘𝑡

𝑘 =1

3𝑡(ln (

2 − 𝑋

(1 − 𝑋)2) − ln(2))

Plugging in numbers,

𝑘 =1

3 ∗ 20𝑚𝑖𝑛(ln (

2 − 0.6

(1 − 0.6)2) − ln(2)) = 0.0246 𝑚𝑖𝑛−1𝑀−1

In a constant volume reactor, this simplifies to:

𝐶𝐴 =𝑁𝐴

𝑉= 1 − 𝑋

𝐶𝐵 =𝑁𝐵

𝑉= 2 − 𝑋

𝑑𝑋

𝑑𝑡= 𝑘 ∗ (1 − 𝑋)(2 − 𝑋)

Separate and integrate as before:

∫𝑑𝑋

(1 − 𝑋)(2 − 𝑋)= 𝑘 ∫ 𝑑𝑡

𝑡

𝑡=0

𝑋

𝑋=0

∫ (1

𝑋 − 2−

1

𝑋 − 1) 𝑑𝑋 = 𝑘𝑡

𝑋

𝑋=0

ln(2 − 𝑋) − ln(1 − 𝑋) − ln(2) = 𝑘𝑡

ln (2 − 𝑋

1 − 𝑋) = ln(2) + 𝑘𝑡

1 +1

1 − 𝑋= 2 exp(𝑘𝑡)

𝑋 = 1 −1

2 exp(𝑘𝑡) − 1

Plugging in numbers,

𝑋 = 1 −1

2 exp(0.246 ∗ 20) − 1= 0.560

4. An aqueous reaction gave the following data in a batch experiment:

t T(°C) X

10 50 0.5

20 50 0.666

10000 50 0.999

10 80 0.8

What residence time would be required for 90% conversion at 80°C in a pressurized CSTR

starting with the same initial composition?

Step 1: We need to look at the reaction data to determine the order of the reaction. If the

reaction were first order,

𝑟 =𝑑𝐶

𝑑𝑡= −𝐶𝑜

𝑑𝑋

𝑑𝑡= −𝑘𝐶 = −𝑘𝐶𝑜(1 − 𝑋)

𝑑𝑋

𝑑𝑡= 𝑘(1 − 𝑋)

ln(1 − 𝑋) = −𝑘𝑡

Solve for k in the case of each reaction.

@𝑡 = 10: 𝑘 = 0.0693

@𝑡 = 20: 𝑘 = 0.0548

@𝑡 = 10000: 𝑘 = 0.000691

This indicates that the reaction is not first order.

Next say the reaction is second order:

𝑟 =𝑑𝐶

𝑑𝑡 = −𝐶𝑜

𝑑𝑋

𝑑𝑡= −𝑘𝐶2 = −𝑘𝐶𝑜(1 − 𝑋)2

𝑑𝑋

𝑑𝑡= 𝑘𝐶𝑜(1 − 𝑋)2

𝑘𝐶𝑜𝑡 =1

1 − 𝑋− 1

𝑘𝐶𝑜 =1

𝑡∗ (

1

1 − 𝑋− 1)

Solve for kCo in each case again:

@𝑡 = 10: 𝑘𝐶𝑜 = 0.1

@𝑡 = 20: 𝑘𝐶𝑜 = 0.0997

@𝑡 = 10000: 𝑘𝐶𝑜 = 0.0999

As these are all in agreement, this must be a second order reactor. Using the data at 80°C,

kCo=0.4 will be used in the calculation of the residence time.

In a CSTR,

𝜏 =𝐶𝑜 − 𝐶

𝑟(𝐶)=

𝐶𝑜 − 𝐶

𝑘𝐶2=

𝑥𝐶𝑜

𝑘𝐶𝑜2(1 − 𝑋)2=

0.90

0.4(1 − 0.90)2= 225

5. Transient of Cell Culture growth in a CSTR.

While the traditional role of chemical engineers is heavily focused on the very simple chemical

processes and reactions of the petrol industry, the field continues to grow in to new and

developing fields. One such area is the use of biological organisms to facilitate the creation of

desirable products. This includes metabolites such as yeast’s production of ethanol, enzymes,

heavily used throughout the world both for research and medicinal purposes, and even whole-

cell catalytic systems that require the immobilization of living cells within a reactor system to

perform otherwise difficult to achieve catalytic conversions.

Imagine that as a chemical engineer, you’re asked to continuously produce these cells in a

bioreactor. Your client has told you that the cells require only sucrose as a limiting substrate,

and grow at a specific growth rate described by

𝜇 =𝜇𝑚𝑎𝑥𝑆

𝐾𝑚 + 𝑆

𝑔𝑐𝑒𝑙𝑙

𝑔𝑐𝑒𝑙𝑙𝑚𝑖𝑛

Where S is the sucrose concentration, 𝜇𝑚𝑎𝑥 = 0.05 1/min, and Km = 2 g/L. This form of a growth

rate is very similar to those found in enzyme kinetics, which will be covered later in the class. In

addition, you are told that the metabolic needs of the cells consume half of the sucrose for

energy during growth without providing additional cell mass (This is called a growth yield).

(a) Write down the differential equations governing the transient startup of this CSTR.

(b) The volume of the reactor is one liter. Given a feed sucrose concentration of 10 grams per

liter, an initial charge of .1 g/L cells in this solution, and flow rate of .01 L/min model the

transient concentrations of both Sucrose and Cells in the system. Approximately how long

does it take for this system to reach steady state?

(c) Use your simulation to evaluate a range of flow rates through the system. Briefly describe

how the behavior of the system changes with flow rate, and what considerations should be

taken into account when attempting to optimize this reactor.

(a) Use a simple mass balance: accumulation = in – out + generation. The feed contains no cells,

thus: 𝑑𝑋

𝑑𝑡= −𝑋𝑣 + 𝜇𝑋𝑉 = −𝑋𝑣 +

𝜇𝑚𝑎𝑥𝑆

𝐾𝑚 + 𝑆𝑋𝑉

For the sucrose substrate, keep in mind that twice the amount of substrate is consumed. 𝑑𝑆

𝑑𝑡= 𝑆𝑖𝑛𝑣 − 𝑆𝑣 − 2𝜇𝑋𝑉

= 𝑣(𝑆𝑖𝑛 − 𝑆) − 2𝜇𝑋𝑉

= 𝑣(𝑆𝑖𝑛 − 𝑆) − 2𝜇𝑚𝑎𝑥𝑆

𝐾𝑚 + 𝑆𝑋𝑉

(b) The following code was used to simulate this reactor:

%Define Adjustable Parameters

V=1; %L

mumax=.05; %1/min

Yield = 0.5;

Km = 2; %g/L

Sin=10; %g/L

v=.05; %L/min

tmax=250; %min

%Define initial conditions

Xo=0.1; %g/L

So=10; %g/L

ode_system = @(t,y) [-y(1)*v+mumax*y(2)/(Km+y(2))*y(1)*V; v*(Sin-y(2))-

(1/Yield)*mumax*y

(2)/(Km+y(2))*y(1)*V];

options = odeset('InitialStep', 1, 'Maxstep', 1);

[T,Y]=ode45(ode_system,[0,tmax],[Xo,So],options);

Cells=Y(:,1);

Sucrose=Y(:,2);

plot(T,Cells, T, Sucrose)

xlabel('Time(min)')

ylabel('Concentration (g/L)')legend('Cells', 'Sucrose')

(c) v= .02:

v=.03:

v=.04:

v=.05:

As the flow rate through the system increases, the overall conversion of glucose to cells

decreases. Due to the higher flow rate, however, a greater total number of cells are produced as

the flow rate increases. This is balanced by the greater feed costs due to wasted substrate. At

sufficiently high flow rates, the initial charge of cells are no longer capable of growing fast

enough to avoid being washed out of the reactor completely.