TUGAS KULIAH TEKNIK REAKSI KIMIA I
TUGAS KULIAH
TEKNIK REAKSI KIMIA I
Dosen : Prof. Dr. Ir. H. M. Rachimoellah, Dipl. EST
Nama : Aristia Anggraeni S.
NRP : 2313 105 007
JURUSAN TEKNIK KIMIA
FAKULTAS TEKNOLOGI INDUSTRI
INSTITUT TEKNOLOGI SEPULUH NOPEMBER
SURABAYA
2013
Pure A (1kmol/hr 1 atm) is fed to plugflow reactor where it reacts reversibly and isothermally at
1219 with elementary kinetics.
1
A
k1 = 200 hr -1, Kp = 1 atm
2R
2
Find :
a.) The size of reactor neede for 40% conversion
b.) The equilibrium conversion
Solution
a.) Volume of plug flow reaction the reaction is elementary. Hence first order forward, second order
reverse
-rA = k1 CA – k2 CR2
Thus the plug flow performance equation becomes
0,4
d X A 0,4 dX A
dX
V
=∫
=∫
=∫ A
2
F A0 0 -rA
N
N
0 k1 CA - k 2 C R
k1 A - k2 A
V
V
2
( ) ( )
*We must write these in terms of conversion. Since this
is not easy to see straight off. Let us go back to
definition of concentration.
V= FA0 =∫
V= FA0 =∫
dX A
[
N (1- X A )
N (2 X A )
k 1 ao
- k 2 A0
V0 (1+ εA XA )
V 0 (1+ ε A X A )
]
2
dX A
[
N -(1- XA )
X
k 1 ao
- 4k 2 C2A0 A
V0 (1+ εA XA )
1+ ε A XA
2
]
Now evaluate all the terms in the above performance expression
k1 = 200 hr -1
C A 0=
k2 conc =
PA0
=
RT
(
1 atm
¿ . atm
0,08206
1219 K
mol K
k 2 conc k 1 conc
=
Kc
Kp
RT
)
=
= 0,01 mol/lt
200 hr
¿ . atm
(¿¿−1)(100
)
mol
=20000 ¿
1 atm
mol . hr
¿
Fa0 = 1000 mol/lt
A
2R
1
0
1
2
2
2
ε A=
2−1
=1
1
Replacing at values gives
0,4
10
V=
3
∫
0
0,4
¿ 500 ∫
0
dX A
( 200 ) ( 0,0 1 )
(
1- XA
X
-4(2000)(0,01)2 A
1+ XA
1+ X A
)
(
2
)
2
(1+ X A ) dX A
1+5 X A
2
= (500) (9,8)
= 4900 liter = 4,9 m3
Solve graphically
XA
0
0,1
0,2
0,3
0,34
0,38
0,4
(1+XA)2/(1-5XA2)
1
1,27
1,80
3,07
4,25
6,85
9,80
10.00
9.00
8.00
7.00
6.00
5.00 A)2
(1+X
4.00 A2
1-5X
3.00
2.00
1.00
0.00
0
0.05
0.1
0.15
0.2
Xa
V = (500)( 1) = 500 liter = 0,5 m3 ... (a)
0.25
0.3
0.35
0.4
b.) Equilibrium conversion
The quickest way to find XAe is to recognize that occurs where (forward rate) = (backward
rate) ... or where ... k1 CA = k2 CR2
From the last writing of the performance equation this occurs where :
1 – 5XA2 = 0 ...or where ... XAe = 0,45 ... (b)
Alternatively we can conservatively retreat into formal thermodynamics. Here for the general
ideal gas reaction n hz = Rr + Ss
Δn = r + s - Q, we have
r=
Kp
RT
π
=K c
=K y
∆n
P=1 atm
P=1 atm
( P=1 atm)
(
) (
Kp = thermodynamics equilibrium constant
r
K C=
s
CR CS
R
Cn
r
s
y y
K y= R a S
yn
For our reaction
A
2R
1
0
1-Z
2Z total
total
Mol fraction
K C=
=1
=1+Z
1−Z
1+ Z
2X
1+ Z
Kr K p
1atm
= 1=
=1
π P (1 atm)1
y R2 ( 2 Z /1+Z )2
4 Z2
=
=
y A ( 1−Z /1+ Z ) 1−Z 2
From these two
Z = 0,45
Bilangan Z = fraction XAe
So, XAe = 0,45
∆n
)
TEKNIK REAKSI KIMIA I
Dosen : Prof. Dr. Ir. H. M. Rachimoellah, Dipl. EST
Nama : Aristia Anggraeni S.
NRP : 2313 105 007
JURUSAN TEKNIK KIMIA
FAKULTAS TEKNOLOGI INDUSTRI
INSTITUT TEKNOLOGI SEPULUH NOPEMBER
SURABAYA
2013
Pure A (1kmol/hr 1 atm) is fed to plugflow reactor where it reacts reversibly and isothermally at
1219 with elementary kinetics.
1
A
k1 = 200 hr -1, Kp = 1 atm
2R
2
Find :
a.) The size of reactor neede for 40% conversion
b.) The equilibrium conversion
Solution
a.) Volume of plug flow reaction the reaction is elementary. Hence first order forward, second order
reverse
-rA = k1 CA – k2 CR2
Thus the plug flow performance equation becomes
0,4
d X A 0,4 dX A
dX
V
=∫
=∫
=∫ A
2
F A0 0 -rA
N
N
0 k1 CA - k 2 C R
k1 A - k2 A
V
V
2
( ) ( )
*We must write these in terms of conversion. Since this
is not easy to see straight off. Let us go back to
definition of concentration.
V= FA0 =∫
V= FA0 =∫
dX A
[
N (1- X A )
N (2 X A )
k 1 ao
- k 2 A0
V0 (1+ εA XA )
V 0 (1+ ε A X A )
]
2
dX A
[
N -(1- XA )
X
k 1 ao
- 4k 2 C2A0 A
V0 (1+ εA XA )
1+ ε A XA
2
]
Now evaluate all the terms in the above performance expression
k1 = 200 hr -1
C A 0=
k2 conc =
PA0
=
RT
(
1 atm
¿ . atm
0,08206
1219 K
mol K
k 2 conc k 1 conc
=
Kc
Kp
RT
)
=
= 0,01 mol/lt
200 hr
¿ . atm
(¿¿−1)(100
)
mol
=20000 ¿
1 atm
mol . hr
¿
Fa0 = 1000 mol/lt
A
2R
1
0
1
2
2
2
ε A=
2−1
=1
1
Replacing at values gives
0,4
10
V=
3
∫
0
0,4
¿ 500 ∫
0
dX A
( 200 ) ( 0,0 1 )
(
1- XA
X
-4(2000)(0,01)2 A
1+ XA
1+ X A
)
(
2
)
2
(1+ X A ) dX A
1+5 X A
2
= (500) (9,8)
= 4900 liter = 4,9 m3
Solve graphically
XA
0
0,1
0,2
0,3
0,34
0,38
0,4
(1+XA)2/(1-5XA2)
1
1,27
1,80
3,07
4,25
6,85
9,80
10.00
9.00
8.00
7.00
6.00
5.00 A)2
(1+X
4.00 A2
1-5X
3.00
2.00
1.00
0.00
0
0.05
0.1
0.15
0.2
Xa
V = (500)( 1) = 500 liter = 0,5 m3 ... (a)
0.25
0.3
0.35
0.4
b.) Equilibrium conversion
The quickest way to find XAe is to recognize that occurs where (forward rate) = (backward
rate) ... or where ... k1 CA = k2 CR2
From the last writing of the performance equation this occurs where :
1 – 5XA2 = 0 ...or where ... XAe = 0,45 ... (b)
Alternatively we can conservatively retreat into formal thermodynamics. Here for the general
ideal gas reaction n hz = Rr + Ss
Δn = r + s - Q, we have
r=
Kp
RT
π
=K c
=K y
∆n
P=1 atm
P=1 atm
( P=1 atm)
(
) (
Kp = thermodynamics equilibrium constant
r
K C=
s
CR CS
R
Cn
r
s
y y
K y= R a S
yn
For our reaction
A
2R
1
0
1-Z
2Z total
total
Mol fraction
K C=
=1
=1+Z
1−Z
1+ Z
2X
1+ Z
Kr K p
1atm
= 1=
=1
π P (1 atm)1
y R2 ( 2 Z /1+Z )2
4 Z2
=
=
y A ( 1−Z /1+ Z ) 1−Z 2
From these two
Z = 0,45
Bilangan Z = fraction XAe
So, XAe = 0,45
∆n
)