Lab 10 Factors Affecting Chemical Kinetics

Fundamentals of Chemistry Lab # 10 Factors Affecting Chemical Kinetics Objective The main objective of this experiment is to determine the order of a reaction with respect to one of the reactants. Some other sub­objectives include: knowing how to set up a rate law equation and knowing the different factors that affect the rate of a chemical reaction. Methods and Materials Old Nassau Reaction Step 1: IO3­ + 3HSO3­ ® I­ + 3SO42­ + 3H+ Step 2: 5I­ + 6H+ + IO3­ ® I­ + 3H2O + 3I2 Step 3: I2 + HSO3­ + H2O ® 2I­ + SO42­ + 3H+ Amount of HSO3­ in moles = .0020 * .02 = 4 * 10­5 mol of HSO3­ Amount of IO3­ in moles = .02 * .02 = 4 * 10­4 mol of IO3­ Amount of HSO3­ needed in reaction = 4*10­4 mol of IO3­ * (3 mol of HSO3­/1 mol of IO3­) = .0012 mol of HSO3­ Amount of IO3­ needed in reaction = 4*10­5 mol of HSO3­ * (1 mol of IO3­/3mol of HSO3­) = 1.33 * 10­5 mol of IO3­ Because all of the HSO3­ can react it is the limiting reactant and the IO3­ is in excess A solution of 20 mL of 0.0020 M sodium bisulfate solution was measured and poured into a beaker. Next 20 mL of 0.020 M potassium iodate solution was poured into a second beaker. The temperature of both solutions were adjusted until they were within 0.5 degrees C then recorded. Then the potassium iodate solution was added to the bisulfate solution. The time when the solutions are mixed was recorded and also the time was recorded when the solution produces a dark blue starch iodine complex. A second trial was done. The experiment was repeated. For this part of the experiment 20 mL of sodium bisulfate and 10 mL of iodate solution were used. Then 10 mL of water was added to the 10 mL of iodate solution to bring it to 20 mL. The concentration of iodate solution was calculated. Before the solutions were mixed, the temperatures of the solutions had to be the same as they were previously. The temperatures and volumes were recorded for the second attempt and the time it takes for the dark blue color to appear was calculated. This part was repeated. Now a third part was done. For this one 5 mL of iodate solution and 15 mL of water were used along with the original 20 mL of sodium bisulfate. As before the temperatures were recorded with the volumes and time of reaction. A second trial was done on this part. Peroxide Oxidation of Tartaric Acid Cl H2C4H4O6 (aq) + 5H2O (aq) ¾Co ¾¾ ® 8H2O (l) + 4CO2 (g) 2 A 30 mL of 0.22 M potassium sodium tartrate solution was mixed with a 10 mL of 6% hydrogen peroxide solution. The solution was heated to 55 °C After the solution reached the right temperature it was placed on a mat and changes were observed. Then 3 mL of 0.30 M cobalt chloride solution was added to the solution. The time and temperature was recorded at the time the cobalt chloride was added. This time is time zero. Changes were recorded as the new solution turns a pink color. When the pink color starts to appear the time and temperature were recorded. This time was expressed in seconds. The solutions were disposed of into the sink. Next a second 30 mL portion of sodium potassium tartrate, and a 10 mL portion of 6% hydrogen peroxide were mixed and placed into a second beaker. The solution was heated to 65 °C . When the temperature was reached 3 mL of cobalt chloride solution was added to the original solution. The temperature and time was recorded when the mixing started. Again the time of mixing is time zero. The amount of time it took for the solution to turn pink was recorded. Results and Discussion Rate for all Solution Concentrations in Iodine Clock Reaction Rate = change in [A] / change in Time Solution 1 Trial 1 Change in Potassium Iodate = ­.01 mol / L Change in Time = 1 / 775 sec Rate = (.01 mol / L) / (1 / 775 sec) Rate = 1.29 * 10­5 mol / L / sec Trial 2 Change in Potassium Iodate = ­.01 mol / L Change in Time = 1 / 1440 sec Rate = (.01 mol / L) / (1 / 1440 sec) Rate = 6.94 * 10­6 mol / L / sec Average Rate = 9.92 * 10­6 Solution 2 Trial 1 Change in Potassium Iodate = ­.015 mol / L Change in Time = 1 / 1525 Rate = (.015 mol / L) / (1 / 1525) Rate = 9.84 * 10­6 mol / L / sec Trial 2 Change in Potassium Iodate = ­.015 mol / L Change in Time = 1 / 1480 sec Rate = (.015 mol / L) / (1 / 1480 sec) Rate = 1.01 * 10­5 Average rate = 9.97 * 10­6 Solution 3 Trial 1 Change in Potassium Iodate = ­.0175 mol / L Change in Time = 3480 sec Rate = (.0175 mol / L) / (1 / 3480 sec) Rate = 5.03 * 10­6 mol / L / sec Trial 2 Change in Potassium Iodate = ­.0175 mol / L Change in Time = 2810 sec Rate = (.0175 mol / L) / (1 / 2810 sec) Rate = 6.23 * 10­6 mol / L / sec Average rate = 5.63 * 10­6 mol / L / sec Determination of Order of Reaction Solutions 1 and 3 9.92 * 10­6 / 5.63* 10­6 mol / L / sec = (.01)a / (.0025)a 1.76 mol / L / sec = (4)a a = log 1.76 / log 4 a = 0.5 Solutions 2 and 3 9.97 *10­6 / 5.63 * 10­6 mol / L sec = (.005)a / (.0025)a 1.77 = 2a a = log 1.77 / log 2 a = .8 average order value = 1 Peroxide Oxidation Trial 1 Trial 2 Initial Time 0 0 Final Time 47 sec 70 sec Initial Temp 56 C 57 C Final Temp 76 C 70 C Color change Brown ® green ® yellow ® pink As the cobalt was added to the solution the temperature immediately went up. The color also started to change. A green color was seen followed by a fizzing sound as the solutions reacted with each other. When the green color was seen it showed that the intermediate complex was present in the reaction. This color change and temperature increase allowed the reaction to progress quicker than normal. The reaction was definitely exothermic because heat was given off into the surroundings. Conclusion In this lab the average order of an Iodine Clock reaction was found to be 1 and simple rates were calculated from concentrations and time. Also the value of a catalyst was learned and how concentration temperature along with a catalyst affects the reaction.