Tugas 2 Kinetika Reaksi.docx

TUGAS 2 KINETIKA REAKSI Program Studi Magister Teknik Kimia

Dosen Pengampu: MUHAMMAD MUFTI AZIS, S.T., M.Sc., Ph.D.

Dikerjakan oleh: HASAN RAHMAN MUHARRAM (17/422468/PTK/12047) MUHAMMAD ABDUL GHONY (17/422471/PTK/12050) MUKMIN SAPTO PAMUNGKAS (17/422472/PTK/12051)

DEPARTEMEN TEKNIK KIMIA FAKULTAS TEKNIK UNIVERSITAS GADJAH MADA YOGYAKARTA 2018

SOAL:

CAi = PAin / HA =

101,325 Pa = 28.95 mol / m3 3 3,500 Pa.m / mol

CB >> CA, so we can assume CB ≅ constant

√DA x k x CB √DA x k x CB MH = = kAla kAl a =

2 3 √1.4 x 10−9 m x 0.433 m x 300 mol 3 s mol.s m

0.025 s−1 m2 120 3 m

= 2.0469

DB x CBout x HA Ei = 1 + b x DA x PAi 300

= 1+

mol x 3,500 Pa .m3 / mol 3 m

1 x 101,325 Pa

Ei > 5 MH, so E = MH = 2.0469

= 11.3627

−ra′′′ =

1 1 HA HA + + kAga kAla x E k x CB x fl

PA

Using fig. 23.4:

Reaction zone approaches interface, then we assume PA = PAi, 1 so the resistance in gas film is negligible: kAga = 0

101,325 Pa 3,500 Pa . m3 / mol 3,500 Pa . m3 / mol 0 + + 0.025 s−1 x 2.0469 m3 mol 0.433 x 300 3 x 0.08 mol. s m -ra’’’ = 1.4764 mol/m3.s −ra′′′ =

Using equation from Plug Flow G/ Mixed Flow L – Mass Transfer + Reaction in Bubble. YAin

Vr = Fg

∫ YAout

dYA (−rA′′)a

YAin

Vr = vo x Cgo

dYA ∫ (−rA′′)a

YAout 3

m mol 0.6041 m = 0.0363 x 28.95 3 s m 3

1

dYA ∫ mol 1.4764 x1 YAout m3 . s YAout = 0.1513

Fraction of entering CO2 absorbed = dYA dYA = YA in - YA out = 0.8487

1 3 2

Step 1. Express the material balance and find CB2 in the exit stream PA1 = 0.0236 % x 1 atm = 2.36 x 10-4 atm PA2 = 2.36 % x 1 atm = 0.0236 atm CB1 = 10 % x 2,736 mol/m3 = 273.6 mol/m3 Using equation from Plug Flow G/ Plug Flow L – Mass Transfer + Reaction in a Countercurrent Tower, dilute systems (solution containing a relatively small quantity of solute as compared with the amount of solvent). (PA3 − PA1) =

Fl (Acs) π Fg (Acs) b x CT

(CB1 − CB3)

mol 2 x 1 atm s. m −4 (PA3 − 2.36 x 10 atm) = (273.6 mol/m3 − CB3) mol mol 100 x 2 x 2736 3 s. m2 m 250

PA3 = 0.125236 – 4.5687 x 10-4 CB3 At the bottom of the tower, PA3 = PA2, so 1

CB2 = 4.5687 x 10−4 (0.125236 – 0.00236) = 268.9517 mol/m3

Step 2. Find which of the many forms of rate equation should be used. At top, mol mol −4 x 2.36 x 10 atm = 0.0314 hr. m3 . atm hr. m3 mol mol −1 kla x CB = 45 hr x 273.6 3 = 12,312 m hr. m3 kAga x PA = 133

At bottom, mol mol kAga x PA = 133 x 0.0236 atm = 3.1388 hr. m3 . atm hr. m3 mol mol −1 kla x CB = 45 hr x 268.9517 3 = 12,102.8265 m hr. m3 At both ends of the tower kAgPA < klaCB; therefore, gas-phase resistance controls and we have a pseudo first-order reaction as given by Eq. 16 of Chapter 23 -ra’’’ = kAga x PA = 133 PA

Step 3. Determine the tower height. Fg PA2 dPA Vr = ∫ π PA1 (−rA′′)a Fg PA2 dPA h x Acs = ∫ π PA1 133 PA x 1 Fg 1 PA2 x ln Acs x π 133 PA1 100 1 0.0236 h = x ln 1 133 2.36 x 10−4 h = 3.46 m h =