LEMBAR PERHITUNGAN

Variabel 1
1. Konsentrasi reaktan awal
a. Casam asetat0 (CA0 )
ρ × V × kadar × 1000
CA0 =
BM × V total
0,996 × 42,07 × 0,96 × 1000
CA0 =
60,052 × 220
CA0 = 3,047
b. Cmetanol0 (CB0 )
ρ × V × kadar × 1000
CB0 =
BM × V total
0,784 × 171,019 × 0,96 × 1000
CB0 =
32,042 × 220
CB0 = 18,259
Cmetanol0
c. M = C
asam asetat0
18,259
M=
3,047
M = 5,993

2. Konversi setelah reaksi

a. CA
(V × N)NaOH − Vsampel × NHCl
CA =
Vsampel
(58,7 × 0,35) − (5 × 0,25)
CA (0) = = 3,039
5
(15,7 × 0,35) − (5 × 0,25)
CA (10) = = 0,849
5
(13 × 0,35) − (5 × 0,25)
CA (20) = = 0,66
5
(12 × 0,35) − (5 × 0,25)
CA (30) = = 0,59
5
(10,5 × 0,35) − (5 × 0,25)
CA (40) = = 0,485
5
b. XA
CA
XA = 1 −
C A0
3,039
XA (0) = 1 − = 0,0026
3,047
0,849
XA (10) = 1 − = 0,721
3,047
0,66
XA (20) = 1 − = 0,783
3,047
0,59
XA (30) = 1 − = 0,806
3,047
0,485
XA (40) = 1 − = 0,841
3,047

1
 t (menit) volume NaOH (ml) 𝐂𝐀 𝐗𝐀
0 58,7 3,108 0,0026
10 15,7 0,849 0,721
20 13 0,66 0,783
30 12 0,59 0,806
40 10,5 0,485 0,841

3. ∆H dan ∆G
CH3COOH + CH3OH ↔ CH3COOCH3 + H2O
∆G°f reaksi = ∆G°f produk - ∆G°f reaktan

Diketahui data ΔH°f standar (Smith dkk, 2001):

∆H°f 298 CH3COOH = - 484500 J/mol
∆H°f 298 CH3OH = - 238660 J/mol
∆H°f 298 CH3COOCH3 = - 445890 J/mol
∆H°f 298 H2O = - 285830 J/mol

∆H°f 298 = (∆H°f 298 CH3COOCH3 + ∆H°f 298 H2O) - (∆H°f 298 CH3COOH +
∆H°f 298 CH3OH)
= (- 445890 - 285830) - (- 484500 - 238660)
= - 8560 J/mol

Diketahui data ∆G° standar (Yaws, 1997):

∆G°f 298 CH3COOH = - 389900 J/mol

∆G°f 298 C2H5OH = - 166270 J/mol
∆G°f 298 CH3COOCH3 = - 324200 J/mol
∆G°f 298 H2O = - 237129 J/mol

Maka:
∆G°f 298 = (∆G°f 298 CH3COOCH3 + ∆G°f 298 H2O) - (∆G°f 298 CH3COOH +
∆G°f 298 CH3OH)
= (- 324200 - 237129) - (- 389900 - 166270)
= - 5159 J/mol
Dari persamaan Van’t Hoff:
∆G°f 298 = −RT ln K

2
 −∆G°298
ln K =
RT
J
− (−5159 )
ln K = mol
J
8,314 ∙ 298 K
mol K
K = 8,023

Suhu operasi = 40°C, maka harga K pada suhu operasi 40°C (313 K):
K 313 ∆H°298 1 1
ln =− ( − ′)
K 298 R T T
J
K 313 (−8560)
ln =− mol ( 1 − 1 ) K
K 298 J 313 298
8,314
mol K
K 330
ln = −0,1656
8,023
K 330 = 6,7987

CH3COOH + CH3OH ↔ CH3COOCH3 + H2O

M CA0 CB0 0 0
R CA0.XA CA0.XA CA0.XA CA0.XA
S CA0-CA0.XA CB0-CA0.XA CA0.XA CA0.XA
Pada saat kesetimbangan,
CC ∙ CD (CA0 XA )(CA0 XA )
QC = =
CA ∙ CB (CA0 (1 − XA )) ∙ (CB0 − CA0 XA )
(XAe )2
QC =
(1 − XAe ) ∙ (M − XAe )
(0,841)2
QC =
(1 − 0,841) ∙ (5,993 − 0,841)
QC = 0,863

4. Konstanta Laju Reaksi

CH3COOH + CH3OH ↔ CH3COOCH3 + H2O
(A) (B) (C) (D)
CA = CA0 − CA0 × XA
CB = CB0 − CA0 × XA
CC = CD = CA0 × XA
k1 k1
K 40 = → k2 =
k2 K
dCA
− = −rA = k1 CA CB − k 2 CC CD
dt

3
 CC CD
−rA = k1 (CA CB − )
𝐾
dXA CC CD
CA0 × = k1 (CA CB − )
dt 𝐾

dXA
CA0 × = k1 (CA0 (1 − XA ) (CB0 − CA0 × XA
dt

(CA0 × XA )(CA0 × XA )
− ))
𝐾

2
dXA CB (CA0 × XA )
CA0 × = k1 (CA0 (1 − XA )CA0 ( 0 − XA ) − )
dt CA0 𝐾

dXA 2 CB0 XA 2
CA0 × = k1 CA0 ((1 − XA ) × ( − XA ) − )
dt CA0 𝐾

dXA CB XA 2
= k1 CA0 ((1 − XA ) × ( 0 − XA ) − )
dt CA0 K

dXA XA 2
= k1 CA0 ((1 − XA ) × (M − XA ) − )
dt K

dXA XA 2
= k1 × 3,047 ((1 − XA ) × (5,993 − XA ) − )
dt 6,7987

dXA 1
= k1 × 3,047 (5,993 − 6,993XA + XA 2 − X 2)
dt 6,7987 A
dXA
= 3,047k1 (0,8529XA 2 − 6,993XA + 5,993)
dt
dXA
= 3,572k1 (XA 2 − 8,199X A + 7,0266)
dt
XA 𝑡
1 dXA
∫ = k 1 ∫ 𝑑𝑡
3,572 0 (XA 2 − 8,199XA + 7,0266) 0

Dengan menggunakan rumus, dapat ditemukan kedua akar A = 7,2267 dan

B = 0,9723, maka XA 2 − 7,2662XA + 6,0517 = (XA − 7,2267)(XA −
0,9723).
XA t
dXA
∫ = 3,572k1 ∫ dt
0 (XA 2 − 8,199XA + 7,0266) 0
XA t
dXA
∫ = 3,572k1 ∫ dt
0 (XA − 7,2267)(XA − 0,9723) 0
XA XA
A B
∫ dXA + ∫ dXA = 3,572k1 t
0 (XA − 7,2267) 0 (X A − 0,9723)

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 A B A(XA − 0,9723) + B(XA − 7,2267)
=
(XA − 7,2267) (XA − 0,9723) (XA − 0,9723)(XA − 7,2267)
A(XA − 0,9723) + B(XA − 7,2267) = 1
A+B = 0
7,2267A + 0,9723B = 1
0,9723A + 0,9723B = 0
Eliminasi untuk mendapatkan 6,2544A = 1, A = 0,1599, dan B = −0,1599.
XA XA
0,1599 −0,1599
∫ dXA + ∫ dXA = 3,572t
0 X A − 7,2267 0 X A − 0,9723

0,1599[ln(XA − 7,2267) − ln(XA − 0,9723)] = 3,572k1 t

(XA − 7,2267)(−0,9723)
k1 t = 0,0447ln [ ]
(XA − 0,9723)(−7,2267)
m.x = y

t CA0 XA y x.y x2 y2
0 3 0.0026 0.000104 0 0 1.0816E-08
10 3 0.721 0.05578 0.5578 100 0.00311141
20 3 0.783 0.068 1.36 400 0.004624
30 3 0.806 0.07365 2.2095 900 0.00542432
40 3 0.841 0.08396 3.3584 1600 0.00704928
∑ = 100 15.235 3.1536 0.281494 7.4857 3000 0.02020902

𝑛. ∑𝑥𝑦 + ∑𝑥. ∑𝑦
𝑘1 =
𝑛. ∑𝑥 2 − (∑𝑥)2
5.7,4857 + 100.0,281494
𝑘1 = = 0,01311
5.3000 − (100)2
𝑘1 0,01311
𝑘2 = = = 0,00193
𝐾 6,7987

Variabel 2

1. Konsentrasi reaktan awal

a. Casam asetat0 (CA0 )
ρ × V × kadar × 1000
CA0 =
BM × V total
0,996 × 42,34 × 0,96 × 1000
CA0 =
60,052 × 220
CA0 = 3,067
b. Cmetanol0 (CB0 )
ρ × V × kadar × 1000
CB0 =
BM × V total
0,784 × 172,128 × 0,96 × 1000
CB0 =
32,042 × 220
CB0 = 18,402

5
 Cmetanol0
c. M = C
asam asetat0
18,402
M=
3,067
M=6

2. Konversi setelah reaksi

a. CA
(V × N)NaOH − Vsampel × NHCl
CA =
Vsampel
(46 × 0,35) − (5 × 0,2)
CA (0) = = 3,02
5
(16,5 × 0,35) − (5 × 0,2)
CA (10) = = 0,955
5
(11 × 0,35) − (5 × 0,2)
CA (20) = = 0,57
5
(9,5 × 0,35) − (5 × 0,2)
CA (30) = = 0,465
5
(8,5 × 0,35) − (5 × 0,2)
CA (40) = = 0,396
5
b. XA
CA
XA = 1 −
C A0
3.02
XA (0) = 1 − = 0.015
3,067
0,955
XA (10) = 1 − = 0,688
3,067
0,57
XA (20) = 1 − = 0,814
3,067
0,465
XA (30) = 1 − = 0,848
3,067
0,396
XA (40) = 1 − = 0,871
3,067
t (menit) volume NaOH (ml) 𝐂𝐀 𝐗𝐀
0 46 5,19 0.015
10 16,5 0,955 0,688
20 11 0,57 0,814
30 9,5 0,465 0,848
40 8,5 0,396 0,871

3. ∆H dan ∆G
CH3COOH + CH3OH ↔ CH3COOCH3 + H2O
∆G°f reaksi = ∆G°f produk - ∆G°f reaktan

Diketahui data ΔH°f standar (Smith dkk, 2001):

∆H°f 298 CH3COOH = - 484500 J/mol
∆H°f 298 CH3OH = - 238660 J/mol

6
∆H°f 298 CH3COOCH3 = - 445890 J/mol
∆H°f 298 H2O = - 285830 J/mol

∆H°f 298 = (∆H°f 298 CH3COOCH3 + ∆H°f 298 H2O) - (∆H°f 298 CH3COOH +
∆H°f 298 CH3OH)
= (- 445890 - 285830) - (- 484500 - 238660)
= - 8560 J/mol

Diketahui data ∆G° standar (Yaws, 1997):

∆G°f 298 CH3COOH = - 389900 J/mol

∆G°f 298 C2H5OH = - 166270 J/mol
∆G°f 298 CH3COOCH3 = - 324200 J/mol
∆G°f 298 H2O = - 237129 J/mol

Maka:
∆G°f 298 = (∆G°f 298 CH3COOCH3 + ∆G°f 298 H2O) - (∆G°f 298 CH3COOH +
∆G°f 298 CH3OH)
= (- 324200 - 237129) - (- 389900 - 166270)
= - 5159 J/mol
Dari persamaan Van’t Hoff:
∆G°f 298 = −RT ln K
−∆G°298
ln K =
RT
J
− (−5159 )
ln K = mol
J
8,314 ∙ 298 K
mol K
K = 8,023

Suhu operasi = 40°C, maka harga K pada suhu operasi 40°C (313 K):
K 313 ∆H°298 1 1
ln =− ( − ′)
K 298 R T T
J
K 313 (−8560) 1 1
ln =− mol ( − )K
K 298 J 313 298
8,314
mol K
K 330
ln = −0,1656
8,023
K 330 = 6,7987

CH3COOH + CH3OH ↔ CH3COOCH3 + H2O

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 M CA0 CB0 0 0
R CA0.XA CA0.XA CA0.XA CA0.XA
S CA0-CA0.XA CB0-CA0.XA CA0.XA CA0.XA
Pada saat kesetimbangan,
CC ∙ CD (CA0 XA )(CA0 XA )
QC = =
CA ∙ CB (CA0 (1 − XA )) ∙ (CB0 − CA0 XA )
(XAe )2
QC =
(1 − XAe ) ∙ (M − XAe )
(0,871)2
QC =
(1 − 0,871) ∙ (6 − 0,871)
QC = 1,1466

4. Konstanta Laju Reaksi

CH3COOH + CH3OH ↔ CH3COOCH3 + H2O
(A) (B) (C) (D)
CA = CA0 − CA0 × XA
CB = CB0 − CA0 × XA
CC = CD = CA0 × XA
k1 k1
K 40 = → k2 =
k2 K
dCA
− = −rA = k1 CA CB − k 2 CC CD
dt
CC CD
−rA = k1 (CA CB − )
𝐾
dXA CC CD
CA0 × = k1 (CA CB − )
dt 𝐾

dXA
CA0 × = k1 (CA0 (1 − XA ) (CB0 − CA0 × XA
dt

(CA0 × XA )(CA0 × XA )
− ))
𝐾

2
dXA CB (CA0 × XA )
CA0 × = k1 (CA0 (1 − XA )CA0 ( 0 − XA ) − )
dt CA0 𝐾

dXA 2 CB0 XA 2
CA0 × )
= k1 CA0 ((1 − XA × ( − XA ) − )
dt CA0 𝐾

dXA CB0 XA 2
= k1 CA0 ((1 − XA ) × ( − XA ) − )
dt CA0 K

8
dXA XA 2
)
= k1 CA0 ((1 − XA × (M )
− XA − )
dt K

dXA XA 2
= k1 × 3,047 ((1 − XA ) × (6 − XA ) − )
dt 6,7987

dXA 1
= k1 × 3,047 (6 − 7XA + XA 2 − X 2)
dt 6,7987 A
dXA
= 3,047k1 (0,8529XA 2 − 7XA + 6)
dt
dXA
= 3,572k1 (XA 2 − 8,207X A + 7,0348)
dt
XA 𝑡
1 dXA
∫ = k1 ∫ 𝑑𝑡
3,572 0 (XA 2 − 8,207XA + 7,0348) 0

Dengan menggunakan rumus, dapat ditemukan kedua akar A = 7,2346 dan

B = 0,9724, maka XA 2 − 8,207XA + 7,0348 = (XA − 7,2346)(XA −
0,9724).
XA t
dXA
∫ = 3,572k1 ∫ dt
0 (XA 2 − 8,207XA + 7,0348) 0
XA t
dXA
∫ = 3,572k1 ∫ dt
0 (XA − 7,2346)(XA − 0,9724) 0
XA XA
A B
∫ dXA + ∫ dXA = 3,572k1 t
0 (XA − 7,2346) 0 (X A − 0,9724)

A B A(XA − 0,9724) + B(XA − 7,2346)

=
(XA − 7,2346) (XA − 0,9724) (XA − 0,9724)(XA − 7,2346)
A(XA − 0,9724) + B(XA − 7,2346) = 1
A+B = 0
7,2346A + 0,9724B = 1
0,9724A + 0,9724B = 0
Eliminasi untuk mendapatkan 6,2622A = 1, A = 0,1597, dan B = −0,1597.
XA XA
0,1597 −0,1597
∫ dXA + ∫ dXA = 3,572t
0 X A − 7,2346 0 X A − 0,9724

0,1597[ln(XA − 7,2346) − ln(XA − 0,9724)] = 3,572k1 t

(XA − 7,2346)(−0,9724)
k1 t = 0,0447ln [ ]
(XA − 0,9724)(−7,2346)
m.x = y

t CA0 XA y x.y x2 y2
0 3 0.015 0.00062 0 0 0.000000625
10 3 0.688 0.05048 0.5048 100 0.00254823

9
 20 3 0.814 0.07578 1.5156 400 0.00574261
30 3 0.848 0.08634 2.5902 900 0.0074546
40 3 0.871 0.09532 3.8128 1600 0.0090859
∑ = 100 15.335 3.236 0.30854 8.4234 3000 0.024831965

𝑛. ∑𝑥𝑦 + ∑𝑥. ∑𝑦
𝑘1 =
𝑛. ∑𝑥 2 − (∑𝑥)2
5.8,4234 + 100.0,30854
𝑘1 = = 0,0145942
5.3000 − (100)2
𝑘1 0,0145942
𝑘2 = = = 0,002147
𝐾 6,7987

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