LEMBAR PERHITUNGAN
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LEMBAR PERHITUNGAN Reaksi : CH 3 COOH + C 2 H 5 OH ↔ C 2 H 5 COOCH 3 + H 2 O A B C D t (menit) Volume Titran (ml) Ca Xa = 1-(Ca/Cao) 0 42.3 4.03 0,021 13 23.4 2.14 0,480 26 21.2 1.92 0,534 39 18.7 1.67 0,594 52 16.9 1.49 0,638 Cao= ρxVxkadarx 1000 BMx ml Cao= 1,039 x 72,02 x 0,99 x 1000 60 x 300 = 4,12 mol/L Cbo= ρxVxkadarx 1000 BMx ml Cbo= 0,807 x 226,25 x 0,96 x 1000 46 x 300 = 12,70 mol/L Ca= ( VxN ) NaoH−( VsampelxN.H 2 SO 4 ) Vsampel ¿ ( 0,5 xV )−( 5 x 0,2) 5
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Transcript of LEMBAR PERHITUNGAN
LEMBAR PERHITUNGAN
Reaksi : CH3COOH + C2H5OH ↔ C2H5COOCH3 + H2O
A B C D
t (menit)Volume Titran
(ml) Ca Xa = 1-(Ca/Cao)
0 42.3 4.030,021
13 23.4 2.140,480
26 21.2 1.920,534
39 18.7 1.670,594
52 16.9 1.490,638
Cao= ρ xV x kadar x 1000BM x ml
Cao=1,039 x72,02 x0,99 x100060 x300
= 4,12 mol/L
Cbo= ρ xV x kadar x 1000BM x ml
Cbo=0,807 x 226,25 x 0,96 x100046 x300
= 12,70 mol/L
Ca=(VxN ) NaoH−(Vsampel x N .H 2SO 4)
Vsampel
¿(0,5 xV )−(5 x0,2)
5
= (0,1 x V.NaOH) – 0,2
Xa= 1- Ca
Ca0
Ca = CaO (1-Xa) = 4,12 (1-Xa)
Cb = CbO – CaO.Xa = 12,70 – 4,12 Xa
Cc = Cao.Xa = 4,12 Xa
Cd = Ca0.Xa = 4,12 Xa
CboCao
= 12,704,12
= 3,08
-r a = -dCadt
= k1(Ca x Cb - Cc x CdK )
Cao x dXadt
= k1(Ca x Cb - Cc x CdK )
Cao x dXaodt
= k1(Cao(1-Xa ) x (Cbo - Cao x Xa ) - (Cao x Xa )(Cao x Xa )K )
Cao x dXadt
= k 1(Cao(1-Xa) x Cao(CboCao
- Xa) - Cao
2 x Xa
2
K )Cao x
dXadt
= k1(Cao(1-Xa ) x Cao(CboCao
- Xa) - Xa2
K )Cao x
dXadt
= k1 x Cao2 ((1-Xa ) x (CboCao
- Xa ) - Xa2
K )dXadt
= k1 x Cao ((1-Xa ) x ( M - Xa ) - Xa2
K )
Pada saat kesetimbangan
k = Cc x CdCa x Cb
= (Cao . Xa)(Cao .Xa )
Cbo(1-Xa )(Cbo-(Cao . Xa))
Mencari konstanta kesetimbangan
K = (Xae )2
(1-Xae )(M - Xae ) =
(0 ,638 )2
(1−0 ,638)(3 , 08−0 ,638 ) = 0,46
Mencari nilai Xa
dXadt
= k1 x Cao ((1-Xa ) x ( M - Xa ) - Xa2
K )dXadt
= k1 x 4,12 ((1-Xa ) x (3 ,08 - Xa ) - Xa2
0 , 46 )dXadt
= k1 x (−4 ,82 )( Xa2+ 3,49Xa -2,63 )
−14 , 82
∫0
XadXa
(Xa2+ 3,49Xa - 2,63 ) = k1 ∫
0
t
dt
Selesaikan ruas kiri dengan partial fraction
-0,21∫0
XadXa( Xa + 0,67 )(Xa - 4,16 )
A( Xa + 0,67)
+B(Xa - 4,16 )
= -0,21
A= 0,043B= - 0,043
Sehingga persamaan menjadi :
∫0
Xa0 ,043( Xa + 0,67)
−0 ,043(Xa - 4,16 )
dXa = k1. t
0,043 x ln [(Xa + 0,67 )(-4,16 )(Xa - 4,16 )(0,67 ) ] = k1 . t ≈ y = mx
t (menit) Xa Y XY X2
0 0,021 -0.010 0 013 0,480 -0.094 -1.216 16926 0,534 -0.109 -2.836 67639 0,594 -0.130 -5.057 152152 0,638 -0.147 -7.635 2704
Dengan kalkulator didapatkan
k1 = m =
n∑ xy−∑ x∑ y
n∑ x2− (∑ x )2 = 5,86 x 10-4 mol/menit
K= k1/k2
sehingga k2 =
5 ,86 x 10−4
0 ,46 = 12,74 x 10-4 mol/menit
