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A. Lampiran Perhitungan Diketahui : -
V awal NaOH
= 20 mL
-
N NaOH
= 0,02 N
-
V awal CH3COOC2H5
= 25 mL
-
N CH3COOC2H5
= 0,02 N
-
V HCl
= 10 mL
-
N HCl
= 0,02 N
-
V NaOH + CH3COOC2H5
= 45 mL
-
V campuran yang diambil
= 5 ml
Ditanya
: Orde reaksi dan konstanta laju reaksi?
Jawab
:
-
V NaOH =
𝑉 𝑎𝑤𝑎𝑙 𝑁𝑎𝑂𝐻 𝑉 𝑡𝑜𝑡𝑎𝑙
x V camp. yang diambil
20 𝑚𝐿
= 45 𝑚𝐿 x 5 mL = 2,22 mL -
V CH3COOC2H5 =
𝑉 𝑎𝑤𝑎𝑙 CH3COOC2H5 𝑉 𝑡𝑜𝑡𝑎𝑙
x V camp. yang diambil
25 𝑚𝐿
= 45 𝑚𝐿 x 5 mL = 2,78 mL -
Molek NaOH awal = V NaOH x N NaOH = 2,22 mL x 0,02 N = 0,0444 mmol
-
Molek CH3COOC2H5 awal = V CH3COOC2H5 x N CH3COOC2H5 = 2,78 mL x 0,02 N = 0,056 mmol
-
Molek HCl awal = V HCl x N HCl
= 10 mL x 0,02 N = 0,2 mmol CH3COOC2H5(aq) + NaOH(aq) m:
0,056 mmol
CH3COONa(aq) +
C2H5OH(aq)
0,044 mmol
r:
x
x
x
x
s:
(0,056-x) mmol
(0,044-x)
x mmol
x mmol
mmol
NaOH(aq) +
HCl(aq)
(0,044-x)
0,2
mmol
mmol
r:
y
y
y
y
s:
-
(0,2-y)
y mmol
y mmol
m:
NaCl(aq) + H2O(l)
mmol
Range volume NaOH titrasi secara teori
Pada saat t = 0
-
Mol HCl yang bereaksi = mol NaOH awal = 0,044 mmol
-
Mol HCl sisa = mol HCl awal – mol HCl yang bereaksi = 0,2 mmol – 0,044 mmol = 0,156 mmol
-
Mol NaOH titrasi = mol HCl sisa = 0,0156 mmol
-
V NaOH titrasi
= =
𝑀𝑜𝑙 𝑁𝑎𝑂𝐻 𝑡𝑖𝑡𝑟𝑎𝑠𝑖 𝑁 𝑁𝑎𝑂𝐻 0,156 𝑚𝑚𝑜𝑙 0,02 𝑁
= 7,8 mL
Pada saat t = ∞
-
Mol HCl yang bereaksi = 0 (karena NaOH habis bereaksi)
-
Mol HCl sisa = mol HCl awal – mol HCl yang bereaksi = 0,2 mmol – 0 mmol = 0,2 mmol
-
Mol NaOH titrasi = mol HCl sisa = 0,0156 mmol
-
V NaOH titrasi
= =
𝑀𝑜𝑙 𝑁𝑎𝑂𝐻 𝑡𝑖𝑡𝑟𝑎𝑠𝑖 𝑁 𝑁𝑎𝑂𝐻 0,2 𝑚𝑚𝑜𝑙 0,02 𝑁
= 10 mL HCl sisa 1. t = 3 menit Molek HCl sisa = Molek NaOH = V NaOH titrasi x N NaOH = 7,8 mL x 0,02 N = 0,156 mmol 2. t = 8 menit Molek HCl sisa = Molek NaOH = V NaOH titrasi x N NaOH = 8 mL x 0,02 N = 0,16 mmol 3. t = 15 menit Molek HCl sisa = Molek NaOH = V NaOH titrasi x N NaOH = 8,1 mL x 0,02 N = 0,162 mmol 4. t = 25 menit
Molek HCl sisa = Molek NaOH = V NaOH titrasi x N NaOH = 8,3 mL x 0,02 N = 0,166 mmol 5. t = 40 menit Molek HCl sisa = Molek NaOH = V NaOH titrasi x N NaOH = 8,4 mL x 0,02 N = 0,168 mmol 6. t = 65 menit Molek HCl sisa = Molek NaOH = V NaOH titrasi x N NaOH = 8,5 mL x 0,02 N = 0,17 mmol 7. t = 2880 menit Molek HCl sisa = Molek NaOH = V NaOH titrasi x N NaOH = 8,7 mL x 0,02 N = 0,174 mmol Mol NaOH sisa (b-x) 1. t = 3 menit Molek NaOH sisa
= Molek HCl – Molek HCl sisa = 0,2 mmol – 0,156 mmol = 0,044 mmol
2. t = 8 menit Molek NaOH sisa
= Molek HCl – Molek HCl sisa = 0,2 mmol – 0,160 mmol = 0,040 mmol
3. t = 15 menit Molek NaOH sisa
= Molek HCl – Molek HCl sisa = 0,2 mmol – 0,162 mmol
= 0,038 mmol 4. t = 25 menit Molek NaOH sisa
= Molek HCl – Molek HCl sisa = 0,2 mmol – 0,166 mmol = 0,034 mmol
5. t = 40 menit Molek NaOH sisa
= Molek HCl – Molek HCl sisa = 0,2 mmol – 0,168 mmol = 0,032 mmol
6. t = 65 menit Molek NaOH sisa
= Molek HCl – Molek HCl sisa = 0,2 mmol – 0,170 mmol = 0,030 mmol
7. t = 2880 menit Molek NaOH sisa
= Molek HCl – Molek HCl sisa = 0,2 mmol – 0,174 mmol = 0,026 mmol
Mol HCl yang bereaksi (x) 1. t = 3 menit Mol HCl yg bereaksi = Molek NaOH awal – Molek NaOH sisa = 0,0444 mmol – 0,044 mmol = 0,0004 mmol 2. t = 8 menit Mol HCl yg bereaksi = Molek NaOH awal – Molek NaOH sisa = 0,0444 mmol – 0,040 mmol = 0,00044 mmol 3. t = 15 menit Mol HCl yg bereaksi = Molek NaOH awal – Molek NaOH sisa = 0,0444 mmol – 0,038 mmol = 0,0064 mmol 4. t = 25 menit
Mol HCl yg bereaksi = Molek NaOH awal – Molek NaOH sisa = 0,0444 mmol – 0,034 mmol = 0,0104 mmol 5. t = 40 menit Mol HCl yg bereaksi = Molek NaOH awal – Molek NaOH sisa = 0,0444 mmol – 0,032 mmol = 0,0124 mmol 6. t = 65 menit Mol HCl yg bereaksi = Molek NaOH awal – Molek NaOH sisa = 0,0444 mmol – 0,030 mmol = 0,0144 mmol 7. t = 2880 menit Mol HCl yg bereaksi = Molek NaOH awal – Molek NaOH sisa = 0,0444 mmol – 0,026 mmol = 0,0184 mmol CH3COOC2H5 sisa (a-x) 1. t = 3 menit Mol CH3COOC2H5
= Molek CH3COOC2H5 – Molek HCl yang bereaksi = 0,056 mmol – 0,0004 mmol = 0,0556 mmol
2. t = 8 menit Mol CH3COOC2H5
= Molek CH3COOC2H5 – Molek HCl yang bereaksi = 0,056 mmol – 0,0044 mmol = 0,0516 mmol
3. t = 15 menit Mol CH3COOC2H5
= Molek CH3COOC2H5 – Molek HCl yang bereaksi = 0,056 mmol – 0,0064 mmol = 0,0496 mmol
4. t = 25 menit Mol CH3COOC2H5
= Molek CH3COOC2H5 – Molek HCl yang bereaksi = 0,056 mmol – 0,0104 mmol
= 0,0456 mmol 5. t = 40 menit = Molek CH3COOC2H5 – Molek HCl yang bereaksi
Mol CH3COOC2H5
= 0,056 mmol – 0,0124 mmol = 0,0436 mmol 6. t = 65 menit = Molek CH3COOC2H5 – Molek HCl yang bereaksi
Mol CH3COOC2H5
= 0,056 mmol – 0,0144 mmol = 0,0416 mmol 7. t = 2880 menit = Molek CH3COOC2H5 – Molek HCl yang bereaksi
Mol CH3COOC2H5
= 0,056 mmol – 0,0184 mmol = 0,0376 mmol Mencari Nilai k 1. t = 3 menit k1
1
1
𝑏(𝑎−𝑥)
= 𝑡 𝑥 (𝑎−𝑏) 𝑥 ln 𝑎(𝑏−𝑥) 1
1
= 180 𝑠 𝑥 (0,056 𝑚𝑚𝑜𝑙−0,0444 𝑚𝑚𝑜𝑙) 𝑥 ln 1
0,0444 𝑚𝑚𝑜𝑙 (0,0556 𝑚𝑚𝑜𝑙) 0,056 𝑚𝑚𝑜𝑙 (0,044 𝑚𝑚𝑜𝑙)
1
= 180 𝑠 𝑥 0,0116 𝑚𝑚𝑜𝑙 𝑥 ln 1,0018831169 0,0018813461
= 2,088 𝑠 .
𝑚𝑚𝑜𝑙
= 0,0009 s . mmol 2. t = 8 menit k2
1
1
𝑏(𝑎−𝑥)
= 𝑡 𝑥 (𝑎−𝑏) 𝑥 ln 𝑎(𝑏−𝑥) 1
1
= 480 𝑠 𝑥 (0,056 𝑚𝑚𝑜𝑙−0,0444 𝑚𝑚𝑜𝑙) 𝑥 ln 1
1
0,0444 𝑚𝑚𝑜𝑙 (0,0516 𝑚𝑚𝑜𝑙) 0,056 𝑚𝑚𝑜𝑙 (0,040 𝑚𝑚𝑜𝑙)
= 480 𝑠 𝑥 0,0116 𝑚𝑚𝑜𝑙 𝑥 ln 1,0227857143 0,0225299971
= 5,568 𝑠 .
𝑚𝑚𝑜𝑙
= 0,0040 s . mmol 3. t = 15 menit
k3
1
1
𝑏(𝑎−𝑥)
= 𝑡 𝑥 (𝑎−𝑏) 𝑥 ln 𝑎(𝑏−𝑥) 1
1
= 900 𝑠 𝑥 (0,056 𝑚𝑚𝑜𝑙−0,0444 𝑚𝑚𝑜𝑙) 𝑥 ln 1
0,0444 𝑚𝑚𝑜𝑙 (0,0496 𝑚𝑚𝑜𝑙) 0,056 𝑚𝑚𝑜𝑙 (0,038 𝑚𝑚𝑜𝑙)
1
= 900 𝑠 𝑥 0,0116 𝑚𝑚𝑜𝑙 𝑥 ln 1,034887218 0,0342924527
= 10,44 𝑠 .
𝑚𝑚𝑜𝑙
= 0,0033 s . mmol
4. t = 25 menit k4
1
1
𝑏(𝑎−𝑥)
= 𝑡 𝑥 (𝑎−𝑏) 𝑥 ln 𝑎(𝑏−𝑥) 1
1
= 1500 𝑠 𝑥 (0,056 𝑚𝑚𝑜𝑙−0,0444 𝑚𝑚𝑜𝑙) 𝑥 ln 1
0,0444 𝑚𝑚𝑜𝑙 (0,0456 𝑚𝑚𝑜𝑙) 0,056 𝑚𝑚𝑜𝑙 (0,034 𝑚𝑚𝑜𝑙)
1
= 1500 𝑠 𝑥 0,0116 𝑚𝑚𝑜𝑙 𝑥 ln 1, 0633613445 0,0614349706
= 17,4 𝑠 .
𝑚𝑚𝑜𝑙
= 0,0035 s . mmol 5. t = 40 menit k5
1
1
𝑏(𝑎−𝑥)
= 𝑡 𝑥 (𝑎−𝑏) 𝑥 ln 𝑎(𝑏−𝑥) 1
1
= 2400 𝑠 𝑥 (0,056 𝑚𝑚𝑜𝑙−0,0444 𝑚𝑚𝑜𝑙) 𝑥 ln 1
0,0444 𝑚𝑚𝑜𝑙 (0,0436 𝑚𝑚𝑜𝑙) 0,056 𝑚𝑚𝑜𝑙 (0,032 𝑚𝑚𝑜𝑙)
1
= 2400 𝑠 𝑥 0,0116 𝑚𝑚𝑜𝑙 𝑥 ln 1,0802678571 0,0772090262
= 27,84 𝑠 .
𝑚𝑚𝑜𝑙
= 0,0028 s . mmol 6. t = 65 menit k6
1
1
𝑏(𝑎−𝑥)
= 𝑡 𝑥 (𝑎−𝑏) 𝑥 ln 𝑎(𝑏−𝑥) 1
1
= 3900 𝑠 𝑥 (0,056 𝑚𝑚𝑜𝑙−0,0444 𝑚𝑚𝑜𝑙) 𝑥 ln 1
1
0,0444 𝑚𝑚𝑜𝑙 (0,0416 𝑚𝑚𝑜𝑙) 0,056 𝑚𝑚𝑜𝑙 (0,030 𝑚𝑚𝑜𝑙)
= 3900 𝑠 𝑥 0,0116 𝑚𝑚𝑜𝑙 𝑥 ln 1,0994285714 0,0947905643
= 45,24 𝑠 .
𝑚𝑚𝑜𝑙
= 0,0021 s . mmol 7. t = 2880 menit
k7
1
1
𝑏(𝑎−𝑥)
= 𝑡 𝑥 (𝑎−𝑏) 𝑥 ln 𝑎(𝑏−𝑥) 1
1
= 172800 𝑠 𝑥 (0,056 𝑚𝑚𝑜𝑙−0,0444 𝑚𝑚𝑜𝑙) 𝑥 ln 1
1
0,0444 𝑚𝑚𝑜𝑙 (0,0376 𝑚𝑚𝑜𝑙) 0,056 𝑚𝑚𝑜𝑙 (0,026 𝑚𝑚𝑜𝑙)
= 172800 𝑠 𝑥 0,0116 𝑚𝑚𝑜𝑙 𝑥 ln 1,1465934066 0,1367952911
= 2004,48 𝑠 .
𝑚𝑚𝑜𝑙
= 0,00007 s . mmol
V awal NaOH
= 20 mL
-
N NaOH
= 0,02 N
-
V awal CH3COOC2H5
= 25 mL
-
N CH3COOC2H5
= 0,02 N
-
V HCl
= 10 mL
-
N HCl
= 0,02 N
-
V NaOH + CH3COOC2H5
= 45 mL
-
V campuran yang diambil
= 5 ml
Ditanya
: Orde reaksi dan konstanta laju reaksi?
Jawab
:
-
V NaOH =
𝑉 𝑎𝑤𝑎𝑙 𝑁𝑎𝑂𝐻 𝑉 𝑡𝑜𝑡𝑎𝑙
x V camp. yang diambil
20 𝑚𝐿
= 45 𝑚𝐿 x 5 mL = 2,22 mL -
V CH3COOC2H5 =
𝑉 𝑎𝑤𝑎𝑙 CH3COOC2H5 𝑉 𝑡𝑜𝑡𝑎𝑙
x V camp. yang diambil
25 𝑚𝐿
= 45 𝑚𝐿 x 5 mL = 2,78 mL -
Molek NaOH awal = V NaOH x N NaOH = 2,22 mL x 0,02 N = 0,0444 mmol
-
Molek CH3COOC2H5 awal = V CH3COOC2H5 x N CH3COOC2H5 = 2,78 mL x 0,02 N = 0,056 mmol
-
Molek HCl awal = V HCl x N HCl
= 10 mL x 0,02 N = 0,2 mmol CH3COOC2H5(aq) + NaOH(aq) m:
0,056 mmol
CH3COONa(aq) +
C2H5OH(aq)
0,044 mmol
r:
x
x
x
x
s:
(0,056-x) mmol
(0,044-x)
x mmol
x mmol
mmol
NaOH(aq) +
HCl(aq)
(0,044-x)
0,2
mmol
mmol
r:
y
y
y
y
s:
-
(0,2-y)
y mmol
y mmol
m:
NaCl(aq) + H2O(l)
mmol
Range volume NaOH titrasi secara teori
Pada saat t = 0
-
Mol HCl yang bereaksi = mol NaOH awal = 0,044 mmol
-
Mol HCl sisa = mol HCl awal – mol HCl yang bereaksi = 0,2 mmol – 0,044 mmol = 0,156 mmol
-
Mol NaOH titrasi = mol HCl sisa = 0,0156 mmol
-
V NaOH titrasi
= =
𝑀𝑜𝑙 𝑁𝑎𝑂𝐻 𝑡𝑖𝑡𝑟𝑎𝑠𝑖 𝑁 𝑁𝑎𝑂𝐻 0,156 𝑚𝑚𝑜𝑙 0,02 𝑁
= 7,8 mL
Pada saat t = ∞
-
Mol HCl yang bereaksi = 0 (karena NaOH habis bereaksi)
-
Mol HCl sisa = mol HCl awal – mol HCl yang bereaksi = 0,2 mmol – 0 mmol = 0,2 mmol
-
Mol NaOH titrasi = mol HCl sisa = 0,0156 mmol
-
V NaOH titrasi
= =
𝑀𝑜𝑙 𝑁𝑎𝑂𝐻 𝑡𝑖𝑡𝑟𝑎𝑠𝑖 𝑁 𝑁𝑎𝑂𝐻 0,2 𝑚𝑚𝑜𝑙 0,02 𝑁
= 10 mL HCl sisa 1. t = 3 menit Molek HCl sisa = Molek NaOH = V NaOH titrasi x N NaOH = 7,8 mL x 0,02 N = 0,156 mmol 2. t = 8 menit Molek HCl sisa = Molek NaOH = V NaOH titrasi x N NaOH = 8 mL x 0,02 N = 0,16 mmol 3. t = 15 menit Molek HCl sisa = Molek NaOH = V NaOH titrasi x N NaOH = 8,1 mL x 0,02 N = 0,162 mmol 4. t = 25 menit
Molek HCl sisa = Molek NaOH = V NaOH titrasi x N NaOH = 8,3 mL x 0,02 N = 0,166 mmol 5. t = 40 menit Molek HCl sisa = Molek NaOH = V NaOH titrasi x N NaOH = 8,4 mL x 0,02 N = 0,168 mmol 6. t = 65 menit Molek HCl sisa = Molek NaOH = V NaOH titrasi x N NaOH = 8,5 mL x 0,02 N = 0,17 mmol 7. t = 2880 menit Molek HCl sisa = Molek NaOH = V NaOH titrasi x N NaOH = 8,7 mL x 0,02 N = 0,174 mmol Mol NaOH sisa (b-x) 1. t = 3 menit Molek NaOH sisa
= Molek HCl – Molek HCl sisa = 0,2 mmol – 0,156 mmol = 0,044 mmol
2. t = 8 menit Molek NaOH sisa
= Molek HCl – Molek HCl sisa = 0,2 mmol – 0,160 mmol = 0,040 mmol
3. t = 15 menit Molek NaOH sisa
= Molek HCl – Molek HCl sisa = 0,2 mmol – 0,162 mmol
= 0,038 mmol 4. t = 25 menit Molek NaOH sisa
= Molek HCl – Molek HCl sisa = 0,2 mmol – 0,166 mmol = 0,034 mmol
5. t = 40 menit Molek NaOH sisa
= Molek HCl – Molek HCl sisa = 0,2 mmol – 0,168 mmol = 0,032 mmol
6. t = 65 menit Molek NaOH sisa
= Molek HCl – Molek HCl sisa = 0,2 mmol – 0,170 mmol = 0,030 mmol
7. t = 2880 menit Molek NaOH sisa
= Molek HCl – Molek HCl sisa = 0,2 mmol – 0,174 mmol = 0,026 mmol
Mol HCl yang bereaksi (x) 1. t = 3 menit Mol HCl yg bereaksi = Molek NaOH awal – Molek NaOH sisa = 0,0444 mmol – 0,044 mmol = 0,0004 mmol 2. t = 8 menit Mol HCl yg bereaksi = Molek NaOH awal – Molek NaOH sisa = 0,0444 mmol – 0,040 mmol = 0,00044 mmol 3. t = 15 menit Mol HCl yg bereaksi = Molek NaOH awal – Molek NaOH sisa = 0,0444 mmol – 0,038 mmol = 0,0064 mmol 4. t = 25 menit
Mol HCl yg bereaksi = Molek NaOH awal – Molek NaOH sisa = 0,0444 mmol – 0,034 mmol = 0,0104 mmol 5. t = 40 menit Mol HCl yg bereaksi = Molek NaOH awal – Molek NaOH sisa = 0,0444 mmol – 0,032 mmol = 0,0124 mmol 6. t = 65 menit Mol HCl yg bereaksi = Molek NaOH awal – Molek NaOH sisa = 0,0444 mmol – 0,030 mmol = 0,0144 mmol 7. t = 2880 menit Mol HCl yg bereaksi = Molek NaOH awal – Molek NaOH sisa = 0,0444 mmol – 0,026 mmol = 0,0184 mmol CH3COOC2H5 sisa (a-x) 1. t = 3 menit Mol CH3COOC2H5
= Molek CH3COOC2H5 – Molek HCl yang bereaksi = 0,056 mmol – 0,0004 mmol = 0,0556 mmol
2. t = 8 menit Mol CH3COOC2H5
= Molek CH3COOC2H5 – Molek HCl yang bereaksi = 0,056 mmol – 0,0044 mmol = 0,0516 mmol
3. t = 15 menit Mol CH3COOC2H5
= Molek CH3COOC2H5 – Molek HCl yang bereaksi = 0,056 mmol – 0,0064 mmol = 0,0496 mmol
4. t = 25 menit Mol CH3COOC2H5
= Molek CH3COOC2H5 – Molek HCl yang bereaksi = 0,056 mmol – 0,0104 mmol
= 0,0456 mmol 5. t = 40 menit = Molek CH3COOC2H5 – Molek HCl yang bereaksi
Mol CH3COOC2H5
= 0,056 mmol – 0,0124 mmol = 0,0436 mmol 6. t = 65 menit = Molek CH3COOC2H5 – Molek HCl yang bereaksi
Mol CH3COOC2H5
= 0,056 mmol – 0,0144 mmol = 0,0416 mmol 7. t = 2880 menit = Molek CH3COOC2H5 – Molek HCl yang bereaksi
Mol CH3COOC2H5
= 0,056 mmol – 0,0184 mmol = 0,0376 mmol Mencari Nilai k 1. t = 3 menit k1
1
1
𝑏(𝑎−𝑥)
= 𝑡 𝑥 (𝑎−𝑏) 𝑥 ln 𝑎(𝑏−𝑥) 1
1
= 180 𝑠 𝑥 (0,056 𝑚𝑚𝑜𝑙−0,0444 𝑚𝑚𝑜𝑙) 𝑥 ln 1
0,0444 𝑚𝑚𝑜𝑙 (0,0556 𝑚𝑚𝑜𝑙) 0,056 𝑚𝑚𝑜𝑙 (0,044 𝑚𝑚𝑜𝑙)
1
= 180 𝑠 𝑥 0,0116 𝑚𝑚𝑜𝑙 𝑥 ln 1,0018831169 0,0018813461
= 2,088 𝑠 .
𝑚𝑚𝑜𝑙
= 0,0009 s . mmol 2. t = 8 menit k2
1
1
𝑏(𝑎−𝑥)
= 𝑡 𝑥 (𝑎−𝑏) 𝑥 ln 𝑎(𝑏−𝑥) 1
1
= 480 𝑠 𝑥 (0,056 𝑚𝑚𝑜𝑙−0,0444 𝑚𝑚𝑜𝑙) 𝑥 ln 1
1
0,0444 𝑚𝑚𝑜𝑙 (0,0516 𝑚𝑚𝑜𝑙) 0,056 𝑚𝑚𝑜𝑙 (0,040 𝑚𝑚𝑜𝑙)
= 480 𝑠 𝑥 0,0116 𝑚𝑚𝑜𝑙 𝑥 ln 1,0227857143 0,0225299971
= 5,568 𝑠 .
𝑚𝑚𝑜𝑙
= 0,0040 s . mmol 3. t = 15 menit
k3
1
1
𝑏(𝑎−𝑥)
= 𝑡 𝑥 (𝑎−𝑏) 𝑥 ln 𝑎(𝑏−𝑥) 1
1
= 900 𝑠 𝑥 (0,056 𝑚𝑚𝑜𝑙−0,0444 𝑚𝑚𝑜𝑙) 𝑥 ln 1
0,0444 𝑚𝑚𝑜𝑙 (0,0496 𝑚𝑚𝑜𝑙) 0,056 𝑚𝑚𝑜𝑙 (0,038 𝑚𝑚𝑜𝑙)
1
= 900 𝑠 𝑥 0,0116 𝑚𝑚𝑜𝑙 𝑥 ln 1,034887218 0,0342924527
= 10,44 𝑠 .
𝑚𝑚𝑜𝑙
= 0,0033 s . mmol
4. t = 25 menit k4
1
1
𝑏(𝑎−𝑥)
= 𝑡 𝑥 (𝑎−𝑏) 𝑥 ln 𝑎(𝑏−𝑥) 1
1
= 1500 𝑠 𝑥 (0,056 𝑚𝑚𝑜𝑙−0,0444 𝑚𝑚𝑜𝑙) 𝑥 ln 1
0,0444 𝑚𝑚𝑜𝑙 (0,0456 𝑚𝑚𝑜𝑙) 0,056 𝑚𝑚𝑜𝑙 (0,034 𝑚𝑚𝑜𝑙)
1
= 1500 𝑠 𝑥 0,0116 𝑚𝑚𝑜𝑙 𝑥 ln 1, 0633613445 0,0614349706
= 17,4 𝑠 .
𝑚𝑚𝑜𝑙
= 0,0035 s . mmol 5. t = 40 menit k5
1
1
𝑏(𝑎−𝑥)
= 𝑡 𝑥 (𝑎−𝑏) 𝑥 ln 𝑎(𝑏−𝑥) 1
1
= 2400 𝑠 𝑥 (0,056 𝑚𝑚𝑜𝑙−0,0444 𝑚𝑚𝑜𝑙) 𝑥 ln 1
0,0444 𝑚𝑚𝑜𝑙 (0,0436 𝑚𝑚𝑜𝑙) 0,056 𝑚𝑚𝑜𝑙 (0,032 𝑚𝑚𝑜𝑙)
1
= 2400 𝑠 𝑥 0,0116 𝑚𝑚𝑜𝑙 𝑥 ln 1,0802678571 0,0772090262
= 27,84 𝑠 .
𝑚𝑚𝑜𝑙
= 0,0028 s . mmol 6. t = 65 menit k6
1
1
𝑏(𝑎−𝑥)
= 𝑡 𝑥 (𝑎−𝑏) 𝑥 ln 𝑎(𝑏−𝑥) 1
1
= 3900 𝑠 𝑥 (0,056 𝑚𝑚𝑜𝑙−0,0444 𝑚𝑚𝑜𝑙) 𝑥 ln 1
1
0,0444 𝑚𝑚𝑜𝑙 (0,0416 𝑚𝑚𝑜𝑙) 0,056 𝑚𝑚𝑜𝑙 (0,030 𝑚𝑚𝑜𝑙)
= 3900 𝑠 𝑥 0,0116 𝑚𝑚𝑜𝑙 𝑥 ln 1,0994285714 0,0947905643
= 45,24 𝑠 .
𝑚𝑚𝑜𝑙
= 0,0021 s . mmol 7. t = 2880 menit
k7
1
1
𝑏(𝑎−𝑥)
= 𝑡 𝑥 (𝑎−𝑏) 𝑥 ln 𝑎(𝑏−𝑥) 1
1
= 172800 𝑠 𝑥 (0,056 𝑚𝑚𝑜𝑙−0,0444 𝑚𝑚𝑜𝑙) 𝑥 ln 1
1
0,0444 𝑚𝑚𝑜𝑙 (0,0376 𝑚𝑚𝑜𝑙) 0,056 𝑚𝑚𝑜𝑙 (0,026 𝑚𝑚𝑜𝑙)
= 172800 𝑠 𝑥 0,0116 𝑚𝑚𝑜𝑙 𝑥 ln 1,1465934066 0,1367952911
= 2004,48 𝑠 .
𝑚𝑚𝑜𝑙
= 0,00007 s . mmol
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