Chem 162 - Lab 2 - Page 1 of 11 Name and ID Alice Meng-Ju...

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Page 1 of 11 Grading: 60 pts. Name and ID: Alice Meng-Ju Hsieh Lab Partner: Steven H. Quiz Section: AA Experiment 2: Chemical Kinetics Part I: A Clock Reaction Part II: Crystal Violet-Hydroxide Reaction By signing below, you certify that you have not falsified data and that you have not plagiarized any part of this lab report. Signature: Purpose and Method Part I: Clock Reaction (3 pts) The purpose of Part I is to determine the order of reaction with respect to each reagent, the rate constant at room temperature and zero degree celsius, and the activation energy by using the initial rates method. To accomplish this, we will combine the solutions we prepared in two test tubes and measure the amount of time it takes for a color of light blue to appear. Then we can determine the order of the reaction based on the time it takes and the concentrations of each reagent used. Some of the formulas that will be needed for this part of the lab are as follows: ln(k2/k1) = -(Ae/R)(1/T2-1/T1) where k is the rate constant and k2/k1 is proportional to rate2/rate1, Ae is the activation energy, R is the ideal gas constant and T is the temperature in Kelvins Rate = k[BrO3-]b[I-]i[H3O+]h k = Rate/[BrO3-]b[I-]i[H3O+]h where b, i, and h represents the order of reactants. Part II: Crystal Violet-Hydroxide Reaction (3 pts) Part II of the experiment is to dermine the order of the reaction with respect to CV+ and OH-, as well as the rate constnat (k) for the reaction at room temperature. We will accomplish this by performing the reaction CV+ + OH- --> CVOH. From the absorbance values measured and several graphs, the order of the reaction can then be determined. We will graph about three graphs: [A] VS. t, 1/[A] vs. T graph, and ln[A] vs. t graph. We will use the following formula to determine the molar absorptivity: Concentration = (Absorbance)/(Molar Absorptivity)
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Page 2 of 11 Stock Solutions Soln. Conc. Units 0 M 0.010 M 0.040 M HCl 0.100 M Experiment Test Tube #1     Test Tube #2 Temp. Run  Celsius Time Rate # ml ml ml ml ml C sec. M/s 1 0.75 0.50 0.25 0.50 0.50 24.1 212.38 7.9E-08 2** 0.50 0.50 0.50 0.50 0.50 25.3 62.58 1.3E-06 3 0.25 0.50 0.75 0.50 0.50 23.5 39.18 1.4E-06 4 0.00 0.50 1.00 0.50 0.50 23.5 37.34 1.1E-06 5 0.75 0.50 0.50 0.25 0.50 24.6 188 4.4E-08 6** 0.50 0.50 0.50 0.50 0.50 23.8 67.09 1.2E-06 7 0.25 0.50 0.50 0.75 0.50 25.6 51.11 1.6E-06 8 0.00 0.50 0.50 1.00 0.50 23 39.87 2.1E-06 9 0.75 0.50 0.50 0.50 0.25 23.9 232.5 3.6E-08 10** 0.50 0.50 0.50 0.50 0.50 24.0 80.09 1.0E-06 11 0.25 0.50 0.50 0.50 0.75 23.2 40.97 2.0E-06 12 0.00 0.50 0.50 0.50 1.00 24.6 29.41 2.8E-06 13 0.50 0.50 0.50 0.50 0.50 0 230 3.6E-08 14* 0.75 0.50 0.50 0.50 0.25 24.4 36 2.3E-06 15* 0.75 0.50 0.50 0.50 0.25 0 64 1.3E-07 * Measurements made with a drop of 1.0 mM Ammonium Molybdate ** Repeated measurements for calculating k and Ea A drop of 1% Starch indicator solution is included in each run. S 2 O 3 2- I - BrO 3 - H 2 O S 2 O 3 2- I - BrO 3 - H + Data, Calculations and Graphs Please provide a sample rate calculation here (NEATLY handwritten is fine): (2 pts) M1V1 = M2V2 0.50mL S2O3 2- x 0.0005M x 1L/1000mL = M2 x 0.0025L M2 = 1.0E-4 M Initial rate of reaction = 1/6 x d[S2O3 2-]/dt = 1/6 x 1.0E-4M / 212.38s = 7.85E-8M/s
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Page 3 of 11 Experiment Temp
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This note was uploaded on 10/27/2010 for the course CHEM 162 taught by Professor N. during the Summer '08 term at University of Washington.

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Chem 162 - Lab 2 - Page 1 of 11 Name and ID Alice Meng-Ju...

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