Tugas 2 Kinrek.docx

TUGAS II KINETIKA REAKSI Program Studi Teknik Kimia

Diajukan oleh: FARHAH AYU FAJRIN 17/422463/PTK/ RENY OKTAVIANTI 17/422467/PTK/

PROGRAM PASCA SARJANA JURUSAN TEKNIK KIMIA FAKULTAS TEKNIK UNIVERSITAS GADJAH MADA YOGYAKARTA 2018

NH2 solution FL=250 mol/s.m2 10%NaOH CB= 2736 mol/m3 P=1 atm T= 40-45 C 1

3

2 Cl2 gas Fg= 100 mol/s.m2 2,36%Cl2

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 𝐹𝑔 𝐹𝑙 (𝑃𝐴3 − 𝑃𝐴1) = (𝐶𝐵1 − 𝐶𝐵3) 𝐴𝑐𝑠 𝑏 × 𝐶𝑡

(𝑃𝐴3 − 𝑃𝐴1) =

(𝑃𝐴3 − 𝑃𝐴1) =

(

𝐹𝑙 )𝜋 𝐴𝑐𝑠

𝐹𝑔 (𝐴𝑐𝑠) × 𝑏 × 𝐶𝑡

(𝐶𝐵1 − 𝐶𝐵3)

250 (273,6 − 𝐶𝐵3) 100 × 2 × 2736

PA3 = 0.125236 – 4.5687 x 10-4 CB3 *PA3=PA2 CB=

1 4.5687 x 10−4

(0.125236 – 0.00236) = 268.9517 mol/m3

*Pada bagian bawah, kAga x PA = 133 kla x CB =

mol mol x 0,0236atm = 3,1388 ℎ𝑟. 𝑚3. 𝑎𝑡𝑚 hr. m3

45 mol x 268,9517 mol/m3 = 12.102,8265 hr hr. m3

*Pada bagian atas, kAga x PA = kla x CB =

133 mol mol x 2,36 x 10−4 atm = 0,0314 hr . m3 . atm hr. m3

45 mol x 273,6mol/m3 = 12.312 hr hr. m3

Didapatkan bahwa kAgPA < klaCB, maka : -ra’’’ = kAga x PA = 133 PA

*Ketinggian tower Fg PA2 dPA Vr = ∫ π PA1 (−rA′′)a Fg PA2 dPA h x Acs = ∫ π PA1 133 PA x 1 h = h =

Fg

x

Acs x π 100 1

h = 100x

1 133

1

x

133 1

133

PA2 PA1

0.0236 2.36 x 10−4

x4,605

= 3,4583 m = 3,46 m

ln

ln