Determination of a Rate Law Essay

703 WordsJun 26, 20113 Pages
Determination of a Rate Law and Temperature Dependence of a Rate Constant By Marvin Coleman March 7, 2011 Abstract: From the shown calculations & graphical analysis, the experimentally determined rate law is rate = K[I-].969 [H2O2].991 and the experimentally determined activation energy is 59.50 kJ/mole. Introduction: The rate of a reaction varies at different temperatures and reactant concentrations. In this experiment, the orders and dependence of the rate constant of the products used are determined by the following chemical reaction: 2I- + H2O2-1 + 2H+ -> I2 + 2H2O-2 The general rate law for this reaction is: Rate = k [I-]x [H2O2-]y where k is the rate constant, x and y are the orders, and the rate is equal to…show more content…
I2 + 2S2O3-2 -> 2I- + S4O6-2 .0001 moles S2O3-2 (1 mole I2)/ (2 moles S4O6-2) = .00005 moles I2 g. (.00005 moles I2 / .04 L) / 450 sec = 2.78 x 10-6 M I2/ sec h. log rate = log (2.78 x 10-6) = -5.56 i. M1 V1 = M2 V2 40 ml x = (2 ml) (.3 M) x = (2 ml/ 4 ml) (.3 M) = .015 M [I-] j. log [I] = log [.015] = -1.82 k. M1V1 = M2V2 (40mL)x = (6 mL) (.1 M) x = (6 mL) (.1M)/ (40 mL) x = .015 M H2O2 l. log [H2O2] = log [.015] = -1.82 Results: Cara and I were able to find that as the temperature increased, the faster the reaction took place. We came to the conclusion that the experiment we were conducting was an endothermic reaction. We were also able to conclude the rate order of the hydrogen peroxide and the iodine, which were both first-order. Discussion: For the most part, the results turned out accurately. There were, however, several causes for miscalculations. First, our measurements were not as exact as they should have been for several of the experiments. Second, something went awry for trial 2, at least I believe, because our k value is extremely high. Maintaining the correct temperatures for the experiments needing 30 and 40 degrees was more difficult than expected, which is also likely to contribute in data not exactly comparable to the predicted results. Conclusion: The slope of the line for Iodide at room temperature was: y = 0.969x + 3.6104 The slope for hydrogen peroxide at room temperature was: y =