lembar perhitungan perbaikan
DESCRIPTION
lemperTRANSCRIPT
P4
LEMBAR PERHITUNGAN1. Perhitungan Reagen
NaOH 0,1 N
gr = 8 gr
Etil Asetat 0,2 N
V = 20,33 ml
HCl 0,075 N
V = 5 ml
2. Perhitungan Proses Batch
NaOH = 0,1 N
Tinggi tangki = 7 cm
Etil asetat = 0,1 N
t pengambilan = 2 menit
HCl = 0,075 N
V yang dititrasi = 5 ml
Konsentrasi NaOH sisa (Ca) = Reaksi: NaOH +CH3COOC2H5( CH3COONa + C2H5OH
A + B ( C + D
Persamaan kecepatan reaksi
dimana Ca=Cb
2
y = mx+c Variabel 1 Proses Batch Suhu Kamar
t(x)V HCl(ml)Ca-ln Cao/Ca (y1)1/Ca (y2)xy1xy2x2
03,40,051019,60784000
230,0450,12516322,222220,25032644,444444
430,0450,12516322,222220,50065388,8888916
630,0450,12516322,222220,750979133,333336
1212,40,1860,37548986,274511,501958266,666756
Trial Orde Reaksi Variabel 1 Grafik Trial Reaksi Orde 1
Grafik Trial Orde 2
Jadi, orde reaksi variabel 1 adalah orde 2, dengan x=t dan y= 1/Ca
m = k = 0,392157c = Variabel 2 Proses Batch Suhu 40Ct(x)V HCl(ml)Ca-ln Cao/Ca (y1)1/Ca (y2)xy1xy2x2
02,60,039025,64103000
22,40,0360,08004327,777780,16008555,555564
42,40,0360,08004327,777780,320171111,111116
62,40,0360,08004327,777780,480256166,666736
129,80,1470,240128108,97440,960512333,333356
Trial Orde Reaksi Variabel 2 Grafik Trial Reaksi Orde 1
Grafik Trial Reaksi Orde 2
Jadi, orde reaksi variabel 2 adalah orde 2, dengan x=t dan y= 1/Ca
m = k = 0,320513c = Variabel 3 Proses Batch Suhu 50Ct(x)V HCl(ml)Ca-ln Cao/Ca (y1)1/Ca (y2)xy1xy2x2
02,90,0435022,98851000
22,40,0360,18924227,777780,37848455,555564
42,40,0360,18924227,777780,756968111,111116
62,40,0360,18924227,777781,135452166,666736
1210,10,15150,567726106,32182,270904333,333356
Trial Orde Reaksi Variabel 3 Grafik Trial Reaksi Orde 1
Grafik Trial Reaksi Orde 2
Jadi, orde reaksi variabel 3 adalah orde 2, dengan x=t dan y= 1/Ca
m = k = 0,718391c = 3. Perhitungan Persamaan Kontinyu
Neraca massa total
input output = akumulasi
( Fo - 0 = (dV = Fo.dt
V = Fo.t ............(1) Neraca massa komponen
akumulasi = input output laju konsumsi konversi
= Fo.Cao 0 V.k.Ca2
.........(2)
Persamaan (1) dan (2) diselesaikan dengan orde 4
k1 = k2 = k3 = k4 = (Ca = Ca model = (Ca+ Ca model sebelumnya
(t = 2 menit Variabel 1 Proses Kontinyu Suhu Kamary = 0,392157x + 20,392k = 0,392157 Cao = 1/20,392=0,049t (menit)V HCl(ml)Cak1k2k3k4CaCa model
01,50,0225-0,00242453-0,00217-0,00219631-0,0019743-0,00220,0225
21,30,0195-0,000321090,0003164,40278E-05-0,000351347,9E-060,02031
41,30,0195-0,00107109-0,00031-0,0005308-0,00054638-0,00050,02032
61,30,0195-0,00132109-0,00065-0,0008099-0,00072046-0,00080,01977
Variabel 2 Proses Kontinyu Suhu 40Cy = 0,320513x + 26,28205k = 0,320513 Cao = 1/26,28205 = 0,038t (menit)V HCl(ml)Cak1k2k3k4CaCa model
01,60,024-0,002758580,000308-0,00310210,0010105-0,00120,024
21,50,0225-0,00167453-0,00069-0,0011211-0,0008784-0,0010,02278
41,50,0225-0,00204953-0,0012-0,0014573-0,0011349-0,00140,02175
61,50,0225-0,00217453-0,00146-0,0016334-0,0013019-0,00160,02033
Variabel 3 Proses Kontinyu Suhu 50C
y = 0,718391x + 24,42529k = 0,718391 Cao = 1/24,42529 = 0,0409t (min)V HCl(ml)Cak1k2k3k4CaCa model
01,60,024-0,00275860,000308-0,00310210,0010105-0,00120,0409
21,50,0225-0,0016745-0,00069-0,0011211-0,000878-0,0010,03968
41,30,0195-0,00069610,000183-7,46923E-05-0,000282-0,00010,03865
61,30,0195-0,0010711-0,00028-0,0004685-0,000492-0,00050,03852
81,30,0195-0,0012586-0,00057-0,0007058-0,000651-0,00070,03801