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

)