Teknik Reaksi Kimia – Semester Semester 5.
UTS (Take Home) TEKNIK REAKSI KIMIA Andre Deri Deri ci NI M , 12210 1221000 001 1
1. Diketahui data kinetika dari suatu eksperimen sebagai berikut : t (min)
CA (mol/dm3)
0
0,08
50
0,118
100
0.0386
150
0,0356
200
0,0322
250
0,0955
300
0,0974
Dengan persamaan laju kinetika :
α
A
Tentukan Orde Reaksi pada persamaan laju reaksi tersebut! Jawaban :
Metode Differensial
] α
A
α
α
Mencari nilai
dengan metode finite difference :
Andre Derici, 12210001 12210001
1
Teknik Reaksi Kimia – Semester 5.
()() ()
( ) [ ]
( ) ( ) 0,000414 () ( ) ( ) () ( ) ( ) () ( ) ( ) () ( ) ( ) ()
[ ] () ()() [ ] () t(min)
CA (mol/dm3)
dCA/dt
(-)dCA/dt
ln CA
ln (-) dCA/dt
0
0,08
0,001934
(-)0,001934
-2,525728644
6,248164
50
0,118
-0,000414
0,000414
-2,137070655
-7,789644
100
0,0386
-0,000824
0,000824
-3,254503003
-7,10134
150
0,0356
-0,000064
0,000064
-3,335409641
-9,65662
200
0,0322
0,000639
(-) 0,000639
-3,435788826
7,355606
250
0,0995
0,000652
(-) 0,000652
-2,307597635
7,335465
300
0,0974
-0,000736
0,000736
-2,328929068
-7,21428
Andre Derici, 12210001
2
Teknik Reaksi Kimia – Semester 5.
10 y = 0.4875x - 0.2001
8 6 4 2
Series1
0 -4
-3
-2
-1
-2
0
Linear (Series1)
-4 -6 -8 -10 -12
Berdasarkan persamaan
Dan berdasarkan persamaan linear pada grafik y = 0,4875x + 0,2001 dapat disimpulkan : A= 0,2001 dan B= 0,4875 Average ln Ca = - 2,7607 Bx
= α ln Ca
0,4875 = α ln Ca α
= 0,4875/2,7607
α
= 0,1766
nilai K ln K = A ln k = 0,2001 k = 1,2215
Andre Derici, 12210001
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Teknik Reaksi Kimia – Semester 5.
2. Suatu persamaan reaksi enzimatic trigliserida:
Dengan nilai laju kinetik persamaan reaksi masing-masing:
Tuliskan persamaan laju kinetikanya dan bagaimana hasil pemodelannya (gunakan matlab)? Jawab : Mol awal Trigliserida (TG) = 1 mol Mol awal CH3COOCH3 = 6 mol Asumsi : waktu tinggal 1jam (3600detik)
Persamaan Model Kinetik
-(k’ )[TG][CH COOCH ] + (K’ )[RCOOCH ][DG] (k’ )[TG][CH COOCH ] + (K’ )[RCOOCH ][DG] – (K’ )[DG][CH COOCH ] + (K’ )[RCOOCH ][MG] (k’ )[DG][CH COOCH ] + (K’ )[RCOOCH ][MG] – (K’ )[MG][CH COOCH ] + (K’ )[RCOOCH ][TA] (k’ )[MG][CH COOCH ] + (K’ )[RCOOCH ][TA] 1
1
3
5
3
3
3
3
2
2
3
3
3
3
4
3
3
3
6
3
3
3
5
3
3
4
3
3
6
3
Andre Derici, 12210001
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Teknik Reaksi Kimia – Semester 5.
(k’ )[TG][ 1
CH3COOCH3]
+
(K’2)[RCOOCH3][DG]
+
(K’3)[DG][
CH3COOCH3] –
(K’4)[RCOOCH3][MG] + (K’5)[MG][ CH 3COOCH3] – (K’6)[RCOOCH3][TA]
(k’1)[TG][ CH3COOCH3] + (K’2)[RCOOCH3][DG] + (K’3)[DG][ CH3COOCH3] –
(K’4)[RCOOCH3][MG] + (K’5)[MG][ CH 3COOCH3] – (K’6)[RCOOCH3][TA]
Simulasi Matlab Script
%UTS Nomor 2 %-Andre Derici,12210001
function kineticUTS2 tspan = [0 3600]; x0 = [1;0;0;0;0;6]; ode45 (@f,tspan,x0);
function dydt = f(t,x) k1 = 0.0311; k2 = 0.0176; k3 = 0.1124; k4 = 0.1271; k5 = 0.1129; k6 = 0.0915; dTG = -k1*x(1)*x(6)+k2*x(5)*x(2); dDG = k1*x(1)*x(6)-k2*x(5)*x(2)-k3*x(2)*x(6)+k4*x(5)*x(3); dMG = k3*x(2)*x(6)-k4*x(5)*x(3)-k5*x(3)*x(6)+k6*x(5)*x(4); dTA = k5*x(3)*x(6)-k6*x(5)*x(4); dRCOOCH3 = k1*x(1)*x(6)-k2*x(5)*x(2)+k3*x(2)*x(6)k4*x(5)*x(3)+k5*x(3)*x(6)-k6*x(5)*x(4); dCH3COOCH3 = -k1*x(1)*x(6)+k2*x(5)*x(2)-k3*x(2)*x(6)+k4*x(5)*x(3)k5*x(3)*x(6)+k6*x(5)*x(4); dydt = [ dTG dDG dMG dTA dRCOOCH3 dCH3COOCH3 ];
Andre Derici, 12210001
5
Teknik Reaksi Kimia – Semester 5.
Hasil Simulasi Matlab
Andre Derici, 12210001
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