TUGAS KULIAH
TEKNIK REAKSI KIMIA I
Dosen : Prof. Dr. Ir. H. M. Rachimoellah, Dipl. EST
JURUSAN TEKNIK KIMIA
FAKULTAS TEKNOLOGI INDUSTRI
INSTITUT TEKNOLOGI SEPULUH NOPEMBER
SURABAYA
2013
Nama : Aristia Anggraeni S.
Pure A (1kmol/hr 1 atm) is fed to plugflow reactor where it reacts reversibly and isothermally at 1219 with elementary kinetics.
A 2R k1 = 200 hr -1, Kp = 1 atm
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
V
*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=
∫
dXAFa0 = 1000 mol/lt
Replacing at values gives
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
(P=1atm)∆ n=Kc
(
RT
P=1atm
)
=Ky(
π P=1atm)
∆ n
Kp = thermodynamics equilibrium constant
KC=
CRrCSs
CnR
Ky=yR
r y S s
yna
For our reaction A 2R
1 0 total = 1 1-Z 2Z total = 1 + Z
Mol fraction 11−Z
+Z
2X
1+Z
KC=Kr
π = Kp P1=
1atm
(1atm)1=1
yR2 yA=
(2Z/1+Z)2 (1−Z/1+Z)=
4Z2
1−Z2
From these two Z = 0,45
Bilangan Z = fraction XAe