laporan percobaan laju reaksi
DESCRIPTION
Laporan percobaan laju reaksi mata kuliah kimia Universitas Hasanuddin, Makassar 2013.TRANSCRIPT
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LAPORAN PRAKTIKUM
KIMIA DASAR
PERCOBAAN VI
LAJU REAKSI
NAMA : FEBRIANTO PAAIS
NIM : H21113015
HARI/TANGGAL PERCOBAAN : JUMAT, 15 NOVEMBER 2013
GOLONGAN / KELOMPOK : H3 / II
ASISTEN : ALFIAH ALIF
LABORATORIUM KIMIA DASAR
JURUSAN KIMIA
FAKULTAS MATEMATIKA DAN ILMU PENGETAHUAN ALAM
UNIVERSITAS HASANUDDIN
MAKASSAR
2013
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BAB IV
HASIL DAN PEMBAHASAN
4.1 Hasil Pengamatan
A. Pengaruh Konsentrasi Na2S2O3
Reaksi berlangsung pada suhu 27 .
KonsentrasiNa2S2O3 (M)
KonsentrasiH2SO4 (M)
Waktu(detik)
1/waktu (detik)
0,1 0,1 29 0,0068
0,08 0,1 37 0,0123
0,06 0,1 44 0,0227
0,04 0,1 81 0,0270
0,02 0,1 147 0,0345
B. Pengaruh Konsentrasi H2SO4
Reaksi berlangsung pada suhu 27 .
KonsentrasiH2SO4 (M)
KonsentrasiNa2S2O3 (M)
Waktu(detik)
1/waktu (detik)
0,1 0,1 29 0,0345
0,08 0,1 30 0,0333
0,06 0,1 34 0,0294
0,04 0,1 37 0,0270
0,02 0,1 52 0,0192
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C. Pengaruh Suhu
KonsentrasiH2SO4 (M)
KonsentrasiNa2S2O3 (M)
Suhu
( )Waktu (detik)
0,1 0,1 12 52
0,1 0,1 27 28
0,1 0,1 74 13
4.2 Reaksi
Na2S2O3 + H2SO4 Na2SO4 + H2S2O3
4.2 Perhitungan
4.3.1 Pengenceran
A. Pengenceran Na2S2O3
Mn.Vn =Ml.Vl
Mn =
1. M1 =
= 0,1 M
2. M2 =
= 0,08 M
3. M3 =
= 0,06 M
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4. M4 =
= 0,04 M
5. M5 =
= 0,02 M
B. Pengenceran H2SO4
1. M1 =
= 0,1 M
2. M2 =
= 0,08 M
3. M3 =
= 0,06 M
4. M4 =
= 0,04 M
5. M5 =
= 0,02 M
4.3.2 Grafik
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A. Pengaruh Konsentrasi Na2S2O3
Vn=
1. [Na2S2O3]awal = 0.1 M
[Na2S2O3]akhir = [Na2S2O3] x
= 0,1 M x
= 0,05 M
D [Na2S2O3] = [Na2S2O3]akhir - [Na2S2O3]awal
= 0,05 M – 0,1 M
= -0,05 M
Vn=
V1=
=
= 1.7241 x 10-3 m/det
2. [Na2S2O3]awal = 0.08 M
[Na2S2O3]akhir = [Na2S2O3] x
= 0,08 M x
= 0,04 M
-d [Na2S2O3]
Vawal
Vakhir
-d [Na2S2O3]n
Vawal
Vakhir
-d [Na2S2O3]n
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d [Na2S2O3] = [Na2S2O3]akhir - [Na2S2O3]awal
= 0,04 M – 0,08 M
= -0,04 M
V2=
=
= 1.0812 x 10-3 m/det
3. [Na2S2O3]awal = 0.06 M
[Na2S2O3]akhir = [Na2S2O3] x
= 0,06 M x
= 0,03 M
d [Na2S2O3] = [Na2S2O3]akhir - [Na2S2O3]awal
= 0,03 M – 0,06 M
= -0,03 M
V3=
=
= 6,8181 x 10-4 m/det
4. [Na2S2O3]awal = 0.04 M
-d [Na2S2O3]
Vawal
Vakhir
-d [Na2S2O3]
Vawal
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[Na2S2O3]akhir = [Na2S2O3] x
= 0,04 M x
= 0,02 M
d [Na2S2O3] = [Na2S2O3]akhir - [Na2S2O3]awal
= 0,02 M – 0,04 M
= -0,02 M
V4=
=
= 2,4691 x 10-4 m/det
5. [Na2S2O3]awal = 0.02 M
[Na2S2O3]akhir = [Na2S2O3] x
= 0,02 M x
= 0,01 M
d [Na2S2O3] = [Na2S2O3]akhir - [Na2S2O3]awal
= 0,01 M – 0,02 M
= -0,01 M
V5=
=
Vakhir
-d [Na2S2O3]
Vawal
Vakhir
-d [Na2S2O3]
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= 6,8027 x 10-5 m/det
No.[Na2S2O3] (M) V (m/det)
Log [Na2S2O3]
Log v
0,1 1,7241 x 10-3 -1 -2,7634
0,08 1,0812 x 10-3 -1.1 -2,9661
0,06 6,8181 x 10-4 -1.2 -3,1663
0,04 2,4691 x 10-4 -1.4 -3,6074
0,02 6,8027 x 10-5 -1.7 -4,1673
Slope = 1.9954
Intercept = -0.7763
Log Ka = Intercept
Ka = Inv log (-0.7763)
= 0.1674
[Na2S2O3]
V
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vA = Ka x [Na2S2O3]m
1,7241 x 10-3 = 0,1674 x (0,1)m
0,1m = 0,0102
m = 1,9913
=
=2,0056
α = 63,4989O
1.
a) Log V1 = log Ka1 + m x log [Na2S2O3]1
Log (1,7241 x 10-3) = log Ka1 + 2,0056 x log (0.1)
Log Ka1 = -2,7634 + 2,0056
= -0,7578
Ka1 = 0,1747
b) Log V2 = log Ka2 + m x log [Na2S2O3]2
Log (1,0812 x 10-3) = log Ka2 + 2,0056 x log (0.08)
Log Ka2 = -2,9661 + 2.1999
= -0,7662
Ka2 = 0,1713
c) Log V3 = log Ka3 + m x log [Na2S2O3]3
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Log (6,8181 x 10-4) = log Ka3 + 2,0056 x log (0.06)
Log Ka3 = -3.1663 + 2,4505
= -0,7158
Ka3 = 0,1924
d) Log V4 = log Ka4 + m x log [Na2S2O3]4
Log (2,4691 x 10-4) = log Ka4 + 2,0056 x log (0.04)
Log Ka4 = -3,6075 + 2,8037
= -0.8038
Ka4 = 0,1571
e) Log V5 = log Ka5 + m x log [Na2S2O3]5
Log (7.0921 x 10-5) = log Ka5 + 2,0056 x log (0.02)
Log Ka5 = -4,1492 + 3,4074
= -0,7418
Ka5 = 0,1812
2.a. Va1 = Ka1 x [Na2S2O3]1m
= 0,1747 x (0.1) 1,9797
= 1.8306 x 10-3 m/det
b. Va2 = Ka2 x [Na2S2O3]2m
= 0,1713 x (0.08) 1,9797
= 1.1540 x 10-3 m/det
c. Va3 = Ka3 x [Na2S2O3]3m
= 0,1924 x (0.06) 1,9797
= 7,3335 x 10-4 m/det
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d. Va4 = Ka4 x [Na2S2O3]4m
= 0,1571 x (0.04) 1,9797
= 2,6833 x 10-4 m/det
e. Va5 = Ka5 x [Na2S2O3]5m
= 0,1812 x (0.02) 1,9797
= 7,8471 x 10-5 m/det
B. Pengaruh Konsentrasi H2SO4
1. [H2SO4] awal = 0.1 M
[H2SO4]akhir = [H2SO4] x
= 0,1 M x
= 0,05 M
d[H2SO4] = [H2SO4]akhir-[H2SO4]awal
= (0.05 x
= -0,075 M
V1 =
=
= 2,5862 x 10-3 m/det
2. [H2SO4]awal = 0.08 M
[H2SO4]akhir = [H2SO4] awal x
Vawal
Vakhir
Vawal
Vakhir
d [H2SO4]
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= 0,08 M x
= 0,04 M
d [H2SO4]2 = [H2SO4]akhir - [H2SO4]awal
= 0,04 M – 0,08 M
= -0,04 M
V2=
=
= 1.3333 x 10-3 m/det
3. [H2SO4]awal = 0.06 M
[H2SO4]akhir = [H2SO4] x
= 0,06 M x
= 0,03 M
d [H2SO4] = [H2SO4]akhir - [H2SO4]awal
= 0,03 M – 0,06 M
= -0,03 M
V3=
=
-d [H2SO4]2
Vawal
Vakhir
-d [H2SO4]
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= 8,8235 x 10-4 m/det
4. [H2SO4]awal = 0.04 M
[H2SO4]akhir = [H2SO4] x
= 0,04 M x
= 0,02 M
d [H2SO4] = [H2SO4]akhir - [H2SO4]awal
= 0,02 M – 0,04 M
= -0,02 M
V4=
=
= 5,4054 x 10-4 m/det
5. [H2SO4]awal = 0.02 M
[H2SO4]akhir = [H2SO4] x
= 0,02 M x
= 0,01 M
d [H2SO4] = [H2SO4]akhir - [H2SO4]awal
= 0,01 M – 0,02 M
= -0,01 M
Vawal
Vakhir
-d [H2SO4]
Vawal
Vakhir
-d [H2SO4]
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V5=
=
= 1.9231 x 10-4 m/det
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No.[H2SO4] (M) v (m/det)
Log [H2SO4]
Log v
1 0,1 2,5862 x 10-3 -1 -2,5873
2 0,08 1.3333 x 10-3 -1.1 -2,8751
3 0,06 8,8235 x 10-4 -1.2 -3,0543
4 0,04 5,4054 x 10-4 -1.4 -3,2672
5 0,02 1.9231 x 10-4 -1.7 -3,7160
Slope = 1,5145
Intercept = -1,1614
Log Ka = Intercept
Ka = Inv log (-1,1614)
= 0.0689
vA = Ka x [H2SO4]m
2,5862 x 10-3 = 0.0689 x (0,1)m
0,1m = 0,0375
m = 1,4256
=
[Na2S2O3]
V
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= 1,6124
α = 58,1908O
1.
a. Log V1 = log Ka1 + m x log [H2SO4]1
-2,5873 = log Ka1 + 1,6124 x (-1)
Log Ka1 = -2,5873 + 1,6124
= -0,9749
Ka1 = 0,1059
b. Log V2 = log Ka2 + m x log [H2SO4]2
-2,8751 = log Ka2 + 1,6124 x (-1.1)
Log Ka2 = -2,8751 + 1,7736
= -1,1015
Ka2 = 0,0792
c. Log V3 = log Ka3 + m x log [H2SO4]3
-3,0543 = log Ka3 + 1,6124 x (-1.2)
Log Ka3 = -3,0543 + 1,9349
= -1,1195
Ka3 = 0,0760
d. Log V4 = log Ka4 + m x log [H2SO4]4
-3,2672 = log Ka4 + 1,6124 x (-1.4)
Log Ka4 = -3,2672 + 2,2574
= -1,0098
Ka4 = 0,0977
e. Log V5 = log Ka5 + m x log [H2SO4]5
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-3,7160 = log Ka5 + 1,6124 x (-1.7)
Log Ka5 = -3,7160 + 2,7411
= -0,9749
Ka5 = 0,1059
2.a. Va1 = Ka1 x [H2SO4]1m
= 0,1059 x (0.1) 1,6124
= 2,5852 x 10-3 m/det
b. Va2 = Ka2 x [H2SO4]2m
= 0,0792 x (0.08) 1,6124
= 1.3492 x 10-3 m/det
c. Va3 = Ka3 x [H2SO4]3m
= 0,0760 x (0.06) 1,6124
= 8,1415 x 10-4 m/det
d. Va4 = Ka4 x [H2SO4]4m
= 0,0977 x (0.04) 1,6124
= 5,4432 x 10-4 m/det
e. Va5 = Ka5 x [H2SO4]5m
= 0,1059 x (0.02) 1,6124
= 1,9296 x 10-4 m/det
C. Pengaruh Konsentrasi Suhu
[H2SO4]awal = 0,1 M
[H2SO4]akhir = 0,1 M x
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= 0,05 M
d[H2SO4] = [H2SO4]akhir-[H2SO4]awal
= 0,05 – 0,1
= -0,05 M
Vn =
V1 =
= 9,6154 x 10-4 m/det
V2 =
= 1,7857 x 10-3 m/det
V3 =
= 3,8461 x 10-3 m/det
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No.v (m/det) t (°C) ln v 1/t
1 9,6154 x 10-4 12 -6,9470 0,8333
2 1,7857 x 10-3 27 -6,3279 0,0370
3 3,8461 x 10-3 74 -5,5607 0,0135
Slope = -0,5693
Intercept = -3,2800
Log Ka = Intercept
Ka = Inv log (-3,2800)
= 5,2481 x 10-4
tg α =
=
= -0,5913
= -30.5958°
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1.a. tg α=
= 0,5913
Ea = tg α x R
= -0,5913 x (0,7178)1/2
= -0,5010
2.a. ln A1 = ln V1 +
= -6,9470 +
= -7,0077
A1 = 9,0489 x 10-4
b. ln A2 = ln V2+
= -6,3279+
= -6,3836
A2 = 1.6890 x 10-4
c. ln A3 = ln V3+
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= -5,5607+
= -5,5126
A3 = 4,0356 x 10-3
4.4 Pembahasan
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Daftar Pustaka
Sastrohamidjojo, Hardjono, 2010, Kimia Dasar, Penerbit GMUP : Jakarta.
Sunarya, Yayan, 2011, Kimia Dasar 1, Penerbit Yrama Widya : Jakarta.
Petrucci, 1999, Kimia Dasar Jilid 2 Terjemahan Dari : 1st General Chemistry,
Penerbit Erlangga : Jakarta.
Utami, 2009, Kimia 2 : Untuk SMA/MA Kelas XI, Program Ilmu Alam,
Pusat Perbukuan Departemen Pendidikan Nasional : Jakarta
Chang, Raymond. 2008. Kimia Dasar Jilid 1 Edisi Kedelapan. Jakarta :
Penerbit Erlangga