Chapter 1 CHEMICAL REACTION ENGINEERING.pdf

Chemical reaction engineer

1. Chemical Reaction Engineering I

2. Assessments
The mark distribution is subject to
change
40 % course work (CW) and 60%
Examination
•Test 1 25% of CW
•Mid Semester Exam 40% of CW
•Test 2 35% of CW

3. Content
•Introduction to CRE
•Isothermal ideal reactor systems
•Non-ideal flow and mixing
•Thermal effects

4. Reaction Kinetics
•Reaction kinetics deals with the study of the rates at which
chemical reactions occur (how fast the reaction occurs and
the variables that affect these rates.
•All chemical reactions take place at a finite rate depending
on the conditions, the most important of which are the
concentration of reactants and products, temperature,
and presence of a catalyst, promoter, or inhibitor
•Reaction Order -This is a number that relates the rate of
reaction with the concentration of the reacting
substances. The order gives information on how the
reactants are depleted (being used up) in a particular
reaction. It can only be found through experiments and
NOT by simply looking at the chemical reaction equation

5. The rate of reaction involves the change in the amount of one of the
components participating in the reaction per unit time. The rate is
proportional to concentration of the reactants and thus can be written
as



where k is the rate constant and n is called the reaction order, [A]
means concentration of A and can also be written as
C
A
Therefore the order of a reaction with respect to a reactant is the
power to which the concentration of the reactant is raised in the rate
law, and the overall order of reaction is the sum of the powers of the
concentrations of products involved in the rate law.
When: n=0 the reaction is zero order. It means that the reaction
rate is not dependent on the concentration of the particular species.

6. Reaction Kinetics
n=1 the reaction is first order. The reaction rate is
directly proportional to the concentration of the
species involved.
n=2 it is called second order. For a single species
reaction it means the rate is proportional to the
square of the concentration of the species.
n=n the reaction is nth order. For a single species
reaction it means the rate is proportional to the
nth power of the concentration of the species
.

7. Determination of reaction order from experimental data
Determine the rate order for the reaction between CO and
NO2 using the experimental information in Table 2.1







Experiment CO (initial mol/L) NO
2 (initial mol/L) Initial Rate (mol/L.h)
1 5×10
-4
0.35×10
-4
3.4×10
-8

2 5×10
-4
0.7×10
-4
6.8×10
-8

3 5×10
-4
0.175× 10
-4
1.7×10
-8

4 1.02×10
-3
0.350× 10
-4
6.8×10
-8

5 1.53×10
-3
0.350× 10
-4
10.2×10
-8

8. First consider experiments 1 and 2, [CO] is constant whilst [NO
2] is
doubled leading to doubling also of the initial rate of reaction
?????????????????????????????????????????? ???????????????????????? 2
?????????????????????????????????????????? ???????????????????????? 1
=
????????????22
????????????21
??????

6.8??????10
−8
3.4??????10
−8
=
0.7??????10
−4
0.35??????10
−4
??????

2=2
??????

Therefore
??????=1
Hence the rate is first order with respect to [NO
2].

9. Now consider experiments 1 and 4, they show that [NO
2] is constant whilst
[CO] is doubled also leads to doubling of rate. Hence performing the same
procedure above, the rate is also first order with respect to [CO]. Therefore the
rate equation becomes ????????????????????????=??????????????????????????????
2
Meaning that the order of reaction is first order with respect to NO
2 and CO
and the overall order of reaction is second order. Let us now look at graphical
determination of order of reactions

10. The reaction between rate of the following reaction is shown in Table 3.2. Determine the
order of reaction.PBA→+ .
A (mol/L) B (mol/L) Rate (mol/s)
Experiment 1 1 1 1
Experiment 2 1 2 2
Experiment 3 2 1 8

11. With respect to A let us consider experiments 1 and 3.
322
1
8
)
1
2
(
3
=⇒=⇒= n
nn

With respect to B let us consider experiments 1 and 2
122
1
2
)
1
2
(
1
=⇒=⇒= n
nn

The overall order is 3+1=4

12. The rate of reaction with regards to the reactant A is given as
dt
dC
r
A
A
=−

where rA is the reaction rate and CA is the concentration of A (as mentioned above)
This applies for constant volume reaction. We will use constant volume reaction to better
understand the concepts. Remember also
[]
n
A
Akr=

where n is the order of reaction and [A] is the concentration of A.
Combining the two equations and bearing in mind that the A is being consumed and hence
depleting, we end up having the following equation for the reaction rate at constant volume
[]
n
A
A
Akr
dt
dC
−=−=

We will now look at different orders of reaction and see how they look graphically.

13. Zero Order
[] kAk
dt
dC
A
−=−=
0

In order to find the relationship we then integrate between the following limits
t=0 to t=t and CA=CAO and CA, CAO is the initial concentration of A
ktCC
ktCC
dtkdCA
k
dt
dC
AOA
AOA
tC
C
A
A
AO
−=
−=−
⇒−=
−=
∫∫
0

Therefore a plot of concentration of A, CA vs. time is a straight line with the rate constant as
the slope and initial concentration of A, CAO as the intercept.

14. First Order
For first order reaction r=k CA, where A is a reactant. Therefore the rate is written as
ktCC
ktCC
dtkdC
C
kdt
C
dC
kC
dt
dC
AOA
AOA
t
A
C
C A
A
A
A
A
A
AO
−=
−=−
−=
−=
−=
∫∫
lnln
lnln
1
0

Therefore a plot on a graph of ????????????????????????????????????
???????????? ???????????????????????? ???????????? should be a straight line where the slope is the
reaction constant and the intercept is ln CAO. The plot of ln (CA/CAO) vs. t should also be a
straight line.

15. 2
nd
Order
The general form of a second order rate expression is r=-kCA
2
. The mathematical conversion
of this equation into a linear form yields
AOA
AOA
AOA
t
A
C
C A
A
A
A
A
C
kt
C
C
kt
C
kt
CC
dtkdC
C
kdt
C
dC
kC
dt
dC
A
AO
11
11
11
1
0
2
2
2
+=
−−=−
−=+−
−=
−=
−=
∫∫

Therefore a plot of
1
????????????????????????
???????????????????????? ???????????? vs time should be a straight line.

16. Nth order equation
If rate equation is given by r=-k CA
n
via mathematical manipulation the linear equation
becomes
11
11
0
1
1
11
1
1
)
11
(
1
1
1
−−
−−
+
+=
+
−=−
+
−
−=
−=
−=
∫∫
n
AO
n
A
n
AO
n
A
t
A
C
C
n
A
n
A
A
n
A
A
Cn
kt
Cn
kt
CCn
dtkdC
C
kdt
C
dC
kC
dt
dC
A
AO


A plot of
vst
C
n
A
1
1
−
should be s straight line.

17. Example
Activity 3.8
Determine if the following experimental conditions describe a second order equation with
respect to CA
Time (mins) CA (mol/L)
0 0.1300
4 0.0872
8 0.0650
12 0.0535
16 0.0453
20 0.0370
30 0.0281

18. Example
First calculate 1/ CA
Time (mins) CA 1/ CA
0 0.1300 7.7
4 0.0872 11.5
8 0.0650 15.4
12 0.0535 18.7
16 0.0453 22.1
20 0.0370 27.0
30 0.0281 35.6

19. Example
Then plot 1/ CA vs t


The experimental data fits second order. The rate equation
2
9329.0
ACr=
Rate constant k = 0.9329 L⋅min/mol
Initial concentration:
1
????????????????????????????????????
=7.7189 and ????????????
????????????????????????=0.1296(~0.130) mol/L
y = 0.9329x + 7.7189
R² = 0.9985
0
5
10
15
20
25
30
35
40
0 5 10 15 20 25 30 35
1/
C
A
t

20. Arrhenius Equation: The influence of temperature on reaction rate
Single reaction

This equation shows the dependence of the rate constant on temperature i.e. it is used to show
the effect of temperature on the reaction itself. The rate constant is temperature dependent
and is given by
RT
E
Ak
eAk
Aek
Aek
A
RT
E
RT
E
RT
E
A
A
A
−=
+=








=
=
−
−
−
lnln
lnlnln
lnln

where EA is the activation energy (kJ/mol), A is the pre exponential term (units for k are the
same as units for A), R is the gas constant (=8.314x10
-3
kJ/mol⋅K) and T is temperature in
Kelvins.

21. Activity 3.6
Calculate the activation energy of the following experimental data.

k (s−1) ×10
−8
4.53 59.3 197 613
T (◦C) 90 110 120 130

Solution
Determine the natural logarithm of rate constant and the inverse of temperature in
kelvins.

k (s−1) ×10
−8
4.53 59.3 197 613
T (◦C) 90 110 120 130
ln k −16.910 −14.338 −13.137 −12.002
T (K) 363 383 393 403
1/T × 10
−3
2.754 2.610 2.544 2.480



The activation energy is determined from the slope
kgJ
E
R
E
A
A
/149
929.17314.8
929.17
=
×=
−=−

y = -17.929x + 32.466
-18
-16
-14
-12
-10
-8
-6
-4
-2
0
2.45 2.5 2.55 2.6 2.65 2.7 2.75 2.8
ln k
1/T

22. The rate constant of a particular reaction at 273K is 7.78×10
-7
s
-1
and at 338 K, the rate
constant is 4.87×10
-3
s
-1
. Calculate the activation energy of the reaction
Solution:
1
7
3
338273
273
338
338273273
338
338273338273273
338
273338
273338
338
338
273
273
103
338
1
273
1
1078.7
1087.4
ln314.8
11
ln
11
ln
11
ln
lnlnlnln
lnln
lnln
−
−
−






−
×
×
==








−
=








−=×








−=−=








−−








−=−








−=








−=
kJmol
TT
k
k
R
E
TT
ER
k
k
TTR
E
RT
E
RT
E
k
k
RT
E
A
RT
E
Akk
RT
E
Ak
RT
E
Ak
A
A
AAA
AA
A
A

23. The activation energy for the reaction
62242
HCHHC →+
is 180.75 kJ/mol. Determine the effect on the reaction rate constant if the reaction
temperature is increased from 120°C to 500°C.
Solution
120
120
lnln
RT
E
Ak
A
−=

500
500lnln
RT
E
Ak
A
−=

1119.27
200
500
120
500
120
500
500120120
500
500120120500120
500
120500
120500
120
120
500
500
1043.6
19.27ln
773
1
393
1
314.8
180750
ln
11
ln
11
ln
lnlnlnln
lnln
lnln
×==
=














−=








−=








−=−=








−−








−=−








−=








−=
e
k
k
k
k
k
k
TTR
E
k
k
TTR
E
RT
E
RT
E
k
k
RT
E
A
RT
E
Akk
RT
E
Ak
RT
E
Ak
A
AAA
AA
A
A

24. Reversible reactions
Let us consider a simple reversible reaction given below
mBnA⇔
where the forward reaction constant is k1 and the reverse being k2

Concentration at initial point is A=a, B=0, at time, t: A=a-x, B=x
Rate of forward reaction =k1(a-x)
n
, Rate of reverse reaction =k2(x)
m

Net rate of reaction =
mn
xkxak
dt
dx
)()(
21−−=
At equilibrium, Net rate =0, and x=xe
Therefore
m
e
n
e
m
e
n
e
xxakk
xkxak
/)(
)()(
12
21−=
=−

Substituting for k2 in original equation, we have:
m
e
mn
e
n
xxxakxak
dt
dx
/)()(
11−−−=

25. If both forward and reverse reactions are first order then
( )
xx
x
at
x
k
t
x
ak
xx
dt
x
ak
dx
xx
xx
x
ak
dt
dx
e
ee
e
e
t
e
x
e
e
e
−
=
=−−
=
−
−=
∫∫
ln
ln
)(
1
)(
1
1
0
1
0
1

26. Find the values of kf and kb for reaction
BA⇔
if [A]0 = 0.30 mol dm
–3
[A]e = 0.172 mol dm
–3

[A]10 = 0.285 and [A]100 = 0.21 mol dm
-3




Solution:
00703.0
744.0
2744.0
2
1
0052.0
015.0172.0
172.0
ln
103.0
172.0
ln
1
1
1
==∴=
−
=
=
−×
=
−
=
−
k
k
x
xa
k
k
s
xx
x
at
x
k
e
e
e
ee

If the equilibrium concentration is not know the various ways to write the rate equation
mBnA⇔
If the reaction is first order then







+−==
=⇒+−=
K
C
Ck
dt
dC
r
k
k
KCkCkr
B
A
A
A
BAA
1
2
1
21

27. Alternatively
BA⇔
( )
( ) ( )
( ) ( ) tCk
kk
CkCkCk
kk
tCkCkCk
kk
CkCkCk
kk
dt
CkCkCk
dCA
dt
CkCkCk
dCA
CCkCk
dt
dC
r
CCCCkCkr
AOAAOA
AOAOAOAAOA
C
C
t
AAOA
AAOA
AAOA
A
A
AAOBBAA
A
AO
=
+
−






+−
+
=






+−
+
−






+−
+
=
+−
=
+−
−−==
−=⇒−=
∫ ∫
1
21
221
21
221
21
221
21
0221
221
21
21
ln
1
ln
1
ln
1
ln
1
)(
)(