Tabel Hasil Perhitungan Dekomposisi H2O2 dengan Katalis FeCl3 ( T1 = suhu ruang)

t (menit) T1 ( K) A (mmol) C (ml)D (M) E (mol) F (mol) [G] (M)

-log [G] Ln Slope K1 Ln x K1 1/T1A x C 2 x D A - E F/ 50 ml

0 301 8 0 0 0 8 0.16 0.79588002 0.795 0 0 0 0.00332231 301 8 0 0 0 8 0.16 0.79588002 0.795 0 0 0 0.00332232 301 8 0 0 0 8 0.16 0.79588002 0.795 0 0 0 0.00332233 301 8 0 0 0 8 0.16 0.79588002 0.795 0 0 0 0.00332234 301 8 0 0 0 8 0.16 0.79588002 0.795 0 0 0 0.00332235 301 8 0 0 0 8 0.16 0.79588002 0.795 0 0 0 0.00332236 301 8 0 0 0 8 0.16 0.79588002 0.795 0 0 0 0.00332237 301 8 0 0 0 8 0.16 0.79588002 0.795 0 0 0 0.00332238 301 8 0 0 0 8 0.16 0.79588002 0.795 0 0 0 0.00332239 301 8 0 0 0 8 0.16 0.79588002 0.795 0 0 0 0.0033223

10 301 8 0 0 0 8 0.16 0.79588002 0.795 0 0 0 0.0033223Tabel Hasil Perhitungan Dekomposisi H2O2 dengan Katalis FeCl3 ( T2 = 30 C )

t (menit) T2 ( K) A (mmol) C (ml)D (M) E (mol) F (mol) [G] (M)

-log [G] Ln Slope K2 Ln x K2 1/T2A x C 2 x D A - E F/ 50 ml

0 303 8 0 0 0 8 0.16 0.79588002 0.795 0 0 0 0.00330031 305 8 0 0 0 8 0.16 0.79588002 0.795 0 0 0 0.00327872 306 8 0 0 0 8 0.16 0.79588002 0.795 0 0 0 0.0032683 309 8 0 0 0 8 0.16 0.79588002 0.795 0 0 0 0.00323624 311 8 0 0 0 8 0.16 0.79588002 0.795 0 0 0 0.00321545 313 8 0 0 0 8 0.16 0.79588002 0.795 0 0 0 0.00319496 314 8 0 0 0 8 0.16 0.79588002 0.795 0 0 0 0.00318477 314.5 8 0 0 0 8 0.16 0.79588002 0.795 0 0 0 0.00317978 315 8 0 0 0 8 0.16 0.79588002 0.795 0 0 0 0.00317469 315 8 0 0 0 8 0.16 0.79588002 0.795 0 0 0 0.0031746

10 315 8 0 0 0 8 0.16 0.79588002 0.795 0 0 0 0.0031746

Keterangan :

[H2O2] = 0,2 M [FeCl3] = 0,005 M A = mol H2O2 D = Gas O2 terbentuk Volume = 40 ml Volume = 10 ml B = mol FeCl3 E = n H2O2 bereaksi (mol)mol H2O2 = 0,2 M x 40 ml = 8 mmol mol FeCl3 = 0,005 M x 10 ml = 10 mmol F = n H2O2 sisa (mol)C = ΔV (ml)

ΔV = perbedaan volume buret t = waktu (menit) T = Suhu ( K) G = Konsentrasi H2O2 sisa (M)

0 1 2 3 4 5 6 7 8 9 100

0.020.040.060.08

0.10.120.140.160.18

I. Grafik [H2O2] vs. waktu (t) (T1=suhu ruang)

(t) menit

[H2O2]

0 1 2 3 4 5 6 7 8 9 100

0.020.040.060.08

0.10.120.140.160.18

IV. Grafik [H202] vs waktu(t) (T2=303 K)

(t) menit

[H202]

0 1 2 3 4 5 6 7 8 9 100

0.10.20.30.40.50.60.70.80.9

f(x) = 0.79588002

II. Grafik - log [H2O2] vs waktu (t) (T1=suhu ruang)

(t) menit

-log [H2O2]

0 1 2 3 4 5 6 7 8 9 100

0.10.20.30.40.50.60.70.80.9

f(x) = 0.79588002

V. Grafik - log [H2O2] vs waktu (t) (T2=303 K)

(t) menit

-log [H202]

00.10.20.30.40.50.60.70.80.9

1

III. Grafik Ln.K1 vs 1/T (T1=suhu ruang)

1/T1

Ln.K1

00.10.20.30.40.50.60.70.80.9

1

VI. Grafik Ln.K2 vs 1/T2 (T2=303 K )

1/T 2

Ln.K2

0 1 2 3 4 5 6 7 8 9 100

0.020.040.060.08

0.10.120.140.160.18

IV. Grafik [H202] vs waktu(t) (T2=303 K)

(t) menit

[H202]

0 1 2 3 4 5 6 7 8 9 100

0.10.20.30.40.50.60.70.80.9

f(x) = 0.79588002

V. Grafik - log [H2O2] vs waktu (t) (T2=303 K)

(t) menit

-log [H202]

00.10.20.30.40.50.60.70.80.9

1

VI. Grafik Ln.K2 vs 1/T2 (T2=303 K )

1/T 2

Ln.K2