Kinetics Formal Essay, Research Paper Beaker 1 Beaker 2 Combo .04 M Fe+3 .15 M HNO3 H2O .04 M KI .004 M S2O3-2 Starch H2O 1 10.00 20 20 10.00 10.00 5 25 2 20.00 10 20 10.00 10.00 5 25

Kinetics Formal Essay, Research Paper

Beaker 1 Beaker 2

Combo .04 M Fe+3 .15 M HNO3 H2O .04 M KI .004 M S2O3-2 Starch H2O

1 10.00 20 20 10.00 10.00 5 25

2 20.00 10 20 10.00 10.00 5 25

3 30.00 0 20 10.00 10.00 5 25

4 10.00 20 20 5.00 10.00 5 30

5 10.00 20 20 15.00 10.00 5 20

Data Table #1

Combo #1a Combo#1b Combo#1c Combo#2 Combo#3 Combo#4 Combo#5

TempInitial0C 40C 230C 450C 230C 230C 22.20C 21.90C

TempFinal0C 1.90C 230C 44.80C 230C 230C 22.20C 21.90C

TimeFinal 34:03.47 1:38.34 0:05.22 0:44.75 0:35.19 5:36.82 0:49.88

1/T(in 0K) .00364 .00338 .00315

K RateConstant(m-2s-1)

ln K .425 3.46 6.39

Data Table #2

Order with respect to Fe+3

Combo # Fe+3 initial Fe+3 adjusted Rate initial Log Fe+3 adjusted Log rate initial

1. .004 .0038 2.03 x 10-6 -2.4 -5.69

2. .008 .0078 4.47 x 10-6 -2.1 -5.35

3. .012 .0118 5.68 x 10-6 -1.9 -5.25

Calculations:

[Fe+3] initial:

1. .04 M Fe+3 x 10.00 mL/100.00 mL = .004 [Fe+3] initial

2. .04 M Fe+3 x 20.00 mL/100.00 mL = .008 [Fe+3] initial

3. .04 M Fe+3 x 30.00 mL/100.00 mL = .012 [Fe+3] initial

[Fe+3] adjusted:

1. .004 [Fe+3] initial -2 x 10-4 = .0038 [Fe+3] adjusted

2. .008 [Fe+3] initial -2 x 10-4 = .0078 [Fe+3] adjusted

3. .012 [Fe+3] initial -2 x 10-4 = .0118 [Fe+3] adjusted

Rate initial = ? [S2O3-2] initial / time (s)

[S2O3-2] initial = .004 m x 10.00 mL / 100.00 mL = .0004 [S2O3-2] initial

Rate initial for Combo #1, Combo #2, Combo #3:

1. ? (.0004) / 98.34s = 2.03 x 10-6

2. ? (.0004) / 44.75s = 4.47 x 10-6

3. ? (.0004) / 35.19s = 5.68 x 10-6

Data Table #3

Order with respect to I-

Combo # I- initial I- rate initial log I- initial Log rate initial

1. .004 2.03 x 10-6 -2.40 -5.69

4. .002 5.94 x 10-6 -2.70 -6.23

5. .006 4.01 x 10-6 -2.22 -5.40

Calculations:

[I-] initial:

1. .04 M KI x 10.00 mL/100.00 mL = .004 [I-] initial

2. .04 M KI x 10.00 mL/100.00 mL = .002 [I-] initial

3. .04 M KI x 10.00 mL/100.00 mL = .006 [I-] initial

Rate initial = ? [S2O3-2] initial / time (s)

[S2O3-2] initial = .004 m x 10.00 mL / 100.00 mL = .0004 [S2O3-2] initial

Rate initial for Combo #1, Combo #4, Combo #5:

1. ? (.0004) / 98.34s = 2.03 x 10-6

4. ? (.0004) / 336.82s = 5.94 x 10-6

5. ? (.0004) / 49.88s = 4.01 x 10-6

Data Table #2

Order with respect to Fe+3

Combo # Fe+3 initial Fe+3 adjusted Rate initial Log Fe+3 adjusted Log rate initial

1. .004 .0038 2.03 x 10-6 -2.4 -5.69

2. .008 .0078 4.47 x 10-6 -2.1 -5.35

3. .012 .0118 5.68 x 10-6 -1.9 -5.25

Data Table #3

Order with respect to I-

Combo # I- initial I- rate initial log I- initial Log rate initial

1. .004 2.03 x 10-6 -2.40 -5.69

4. .002 5.94 x 10-6 -2.70 -6.23

5. .006 4.01 x 10-6 -2.22 -5.40

A) Order of reaction with respect to [Fe+3] ? 1

B) Order of reaction with respect to [I-] ? 2

C) Rate law = K [Fe+3]1 [I-]2

2.03 x 10-6 = K [.004]1 [.004]2

K = 31.7 M-2s-1

D) ln K = ln 31.7 = 3.46

Rate initial for hot and cold:

HOT:

Rate initial = ? (.0004) / 5.22s

Rate initial = 3.83 x 10-5

3.83 x 10–5 = K [.004]1 [.004]2

K = 598 M-2s-1

COLD:

Rate initial = ? (.0004) / 2043.47s

Rate initial = 9.79 x 10-8

9.79 x 10-8 = K [.004]1 [.004]2

K = 1.53 M-2s-1

1/T ln K

Cold .00364 .425

Room .00338 3.46

Hot .00315 6.39

Ea

M = – Ea/R

M = -18001

-Ea = -18001 x .008314

-Ea = -149.7

Ea = 149.7 kJ

Kinetics Lab Formal

Introduction

Kinetics of chemical reactions is how fast a reaction occurs and determining how the presence of reactants affects reaction rates. In this experiment the rate of reaction for Fe+3 and I- is determined. Because the rate of chemical reactions relates directly to concentration of reactants, the rate law is used to find the rate constant, and calculated with specified temperatures.

Two catalyst reactants are used in the experiment, thiosulfate and starch, to dictate the time of reactions.

The order with respect to Fe+3 and I- is also determined by graphing the slope of the log rate initial as a function of the log (Fe+3) or (I-). The activation energy is also graphed with the rate constant as a function of the inverse of the temperature.

Procedure

The volumes of solutions were obtained and placed into two separate beakers as shown in the tables below.

Reagent Fe+3 HNO3 KI S2O3-2 Starch

Volume needed 150 mL 150 mL 100 mL 100 mL 50 mL

Beaker 1 Beaker 2

Combo .04 M Fe+3 .15 M HNO3 H2O .04 M KI .004 M S2O3-2 Starch H2O

1 10.00 20 20 10.00 10.00 5 25

2 20.00 10 20 10.00 10.00 5 25

3 30.00 0 20 10.00 10.00 5 25

4 10.00 20 20 5.00 10.00 5 30

5 10.00 20 20 15.00 10.00 5 20

Part A:

The two beakers were allowed to chill for 15 minutes. Their contents were then mixed and put back on ice. Combination #1 was run at room temperature on a separate trial. The temperature was recorded at 23 0C. When the solution turned blue, the time was recorded. Finally, combination #1 was run at 45 0C and the solution was monitored until it became blue.

Part B:

Combinations #2 – #5 were all run at room temperature. The temperature varied slightly for each combination (Data Table #1). Each solution was timed until it appeared blue (Data Table #1).