To determine the rate law for a chemical reaction among hydrogen peroxide, iodide and acid, specifically by observing how changing each of the concentrations of H2O2, I- and H+ affects the rate of reaction.
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Name: Tonny, Chan Kar Yu, Student ID: 10297729 Date of Experiment: 3rd December 2004, Group: B1 Title: Experiment 3 - Chemical Kinetics Objective: 1. To determine the rate law for a chemical reaction among hydrogen peroxide, iodide and acid, specifically by observing how changing each of the concentrations of H2O2, I- and H+ affects the rate of reaction. 2. To examine the effects of temperature and a catalyst on the rate of reaction. Chemicals and Apparatus: 0.8M hydrogen peroxide containing 0.001M sulphuric acid, 0.05M sodium thiosulphate(Na2S2O3), 0.05M potassium iodide, solid potassium iodide, 2.0M sulphuric acid, 0.05 acetic acid & sodium acetate buffer, 0.30M acetic acid, 0.02 Mo(VI) catalyst (containing 1.76g/L of ammonium hyptamolybdate ([NH4]6Mo7O244H2O)), 0.1% of freshly prepared starch solution, water bath at 40, electronic balance, 5.00cm3 pipette, 25cm3 pipette, 50.0cm3 burette, white tile, 125cm3 conical flask, 250cm3 conical flask, 250cm3 beaker, 10cm3 measuring cylinder, glass rod, 50cm3 measuring cylinder, thermometer and stop watch. Background: A description of the sequence of steps by which a chemical reaction takes place is called a reaction mechanism. Many studies go into trying to determine possible mechanisms. The rates at which reactions take place provide important insights. The rate law for a chemical reaction is a quantitative expression involving constants related to the nature of the chemical reaction and the concentrations of reactants. In order for reactants to react, they must come in contact with one another (or at least come nearby). The probability of a collision is related to a function of the concentration. So, for the general rate law: Rate=k [A]a[B]b[C]c [A], [B] and [C] represent concentrations of reactants and catalysts, and a, b, and c represent exponents that may or may not be related to the coefficients of the corresponding balanced chemical equation. The quantities in brackets are read as concentration and are raised to an appropriate power. Multiplied together with the constant (k), they give the rate of the reaction. The numerical values of a, b, and c must be determined by experimentation.
0.5763 = (0.4999) y In 0.5763 = y In 0.4999 y = 1.15 The order of reaction with respect to iodide was 1. 1 (c) the order of reaction with respect to H+ Reaction Mixture Time /s Acetic acid concentration Sodium acetate concentration Calculated H+ concentration / M A 118 0.01 0.01 1.8 x 10-5 D 198 0.1 0.01 1.8 x 10-4 Concentration of acetic acid in A = [(0.05)(30/1000)] / (150/1000) = 0.01 M Concentration of sodium acetate in A = [0.05 x (30/1000)] / (150/1000) = 0.01 M Ka =1.8 x 10-5 = [H+] (0.01) / 0.01 [H+] = 1.8 x 10-5 M Concentration of acetic acid in D = [0.05 x (30/1000) + 0.3 x (45/1000)] / (150/1000) = 0.1 M Concentration of sodium acetate in D = [0.05 x (30/1000)] / (150/1000) = 0.01 M 1.8 x 10-5 = [H+] (0.01) / (0.1) [H+] = 1.8 x 10-4 M The reaction order: x = ln (?t2/?t1) / ln ([H2O2]1/ [H2O2]2) = ln (118/198) / ln (1.8 x 10-5/1.8 x 10-4) =0.22 The order of reaction with respect to H+ was 0. 2 (a)(b) From my measurements, the reaction orders with respect to each of the three reactants H2O2, I- and H+ were 1, 1 and 0. So the rate expression for the uncatalyzed reaction was Rate = k [H2O2] 1 [I-] 1 And the total reaction order in the reaction was two. 2 (c) The rate law is inconsistent with the idea that the mechanism of the reaction was H2O (aq) + 2 I-(aq) + 2 H+ (aq) --> I2 (aq) + 2 H2O (l) , the rate law only included the rate determining steps of hydrogen peroxide and iodide. 2 (d) the first term in this rate law results from a rate determining nucleophilic attacked of the iodide anion upon the peroxide: I- + H-O-O-H --> H-O-I + H-O- [slow; rate determining] The second step involved the formation of the iodine dimer and is a very fast step: I-+ H-O-I + --> I2 + OH- [fast] The HOI (aq)
then reacted with I- to form I2 as above. Iodine was only slightly soluble in water and it was thought that they predominant aqueous species was I3-. Because water made a better leaving group than OH-; therefore, an acid-catalyzed mechanism was favored over the uncatalyzed mechanism, so 0.001 M H2SO4 was contained by hydrogen peroxide. If it was not for the catalyst, the reaction would occur so slowly that practical product formation would not be feasible. From a molecular point of view, the catalyst provides a surface for the reaction molecules to position themselves with one another so that when they do collide they will do so much more effectively. Conclusion: Iodine Clock Reaction was produced by a solution of hydrogen peroxide and sulphuric acid was added to a solution containing iodide ion, thiosulfate ion, and starch. After some time, the dark blue colour of a triiodide-starch complex was observed. The influences of initial reactant concentrations on the reaction were examined, as were the influence of volume and temperature. In the part I of the experiment, standardization of H2O2 solution, there would be a positive error, which was, the value of trial 2 was larger than that of trial 1, if the glassware were not rinsed clear. The stirring was required to mix the solution thoroughly, and we should stir the solution in each set of reaction mixtures in the same amplitude to ensure the stirring effect for all reaction mixtures were the same. Hazards: Hydrogen peroxide is corrosive to the eyes and the skin. The vapour irritates the respiratory tract. Ingestion of hydrogen peroxide may produce oxygen bubbles in the blood, resulting in shock. Sulphuric acid is corrosive and it is very corrosive to the eyes the skin and the respiratory tract. Inhalation of an aerosol of sulphuric acid may cause lung oedema. ?? ?? ?? ?? Tonny, Chan Kar Yu, Student ID: 10297729 College Chemistry I (249-05), Group: B1 - 1 -
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