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

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

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

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

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

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

[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]

= 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

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

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

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]

= 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]

= 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]

V5=

=

= 1.9231 x 10-4 m/det

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

= 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

-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

= 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

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°

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+

= -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