.1-5.35 3..012.01185.68 x 10-6-1.9-5.25 Combo #I- initialI- rate initiallog I- initialLog rate initial 1..0042.03 x 10-6-2.40-5.69 4..0025.94 x 10-6-2.70-6.23 5..0064.01 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 Rate initial for hot and cold: Rate initial = (.0004) / 5.22s Rate initial = 3.83 x 10-5 3.83x 10–5 = K [.004]1 [.004]2 Rate initial = (.0004) / 2043.47s Rate initial = 9.79 x 10-8 9.79 x 10-8 = K [.004]1 [.004]2 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.ProcedureThe volumes of solutions were obtained and placed into two separate beakers as shown in the tables below.ReagentFe+3HNO3KIS2O3-2StarchVolume needed150 mL150 mL100 mL100 mL50 mLBeaker 1Beaker 2Combo.04 M Fe+3.15 M HNO3H2O.04 M KI.004 M S2O3-2StarchH2O110.00202010.0010.00525220.00102010.0010.00525330.0002010.0010.00525410.0020205.0010.00530510.00202015.0010.00520Part 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 ...
