004 Initial 2 X 10 4 example essay topic

Beaker 1 Beaker 2 Combo. 04 M Fe+3.15 M HNO 3 H 2 O. 04 M KI. 00364.00338. 00315 K Rate Constant (m-2's-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 = [S 2 O 3-2] initial / time (s) [S 2 O 3-2] initial = . 004 m x 10.00 mL / 100.00 mL = . 0004 [S 2 O 3-2] initial Rate initial for Combo #1, Combo #2, Combo #3: 1. (. 0004) / 98.34's = 2.03 x 10-6 2. (.

0004) / 44.75's = 4.47 x 10-6 3. (. 0004) / 35.19's = 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... 002 [I-] initial 3... 006 [I-] initial Rate initial = [S 2 O 3-2] initial / time (s) [S 2 O 3-2] initial = . 0004 [S 2 O 3-2] initial Rate initial for Combo #1, Combo #4, Combo #5: 1. (.

0004) / 98.34's = 2.03 x 10-6 4. (. 0004) / 336.82's = 5.94 x 10-6 5. (. 0004) / 49.88's = 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... 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] 1 [. 0004) / 5.22's Rate initial = 3.83 x 10-5 3.83 x 10-5 = K [. 004] 1 [. 004] 2 K = 598 M-2's-1 COLD: Rate initial = (. 0004) / 2043.47's Rate initial = 9.79 x 10-8 9.79 x 10-8 = K [. 004] 1 [.

004] 2 K = 1.53 M-2's-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 HNO 3 KI S 2 O 3-2 Starch Volume needed 150 mL 150 mL 100 mL 100 mL 50 mL Beaker 1 Beaker 2 Combo. 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 0 C. When the solution turned blue, the time was recorded. Finally, combination #1 was run at 45 0 C 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).