The Rate Law for Chemical Reaction Among Hydrogen Peroxide, Iodide, and

The Rate Law for Chemical Reaction Among Hydrogen Peroxide, Iodide, and erosiveTo watch the estimate law for a chemical response among enthalpy peroxide, iodide and acid, specifically by observing how changing sepa assessly of the concentrationsExperiment 3 Chemical KineticsObjectives1. To determine the rate law for a chemical reaction amonghydrogen peroxide, iodide and acid, specifically by observing howchanging each of the concentrations of H2O2, and H+ affects the rateof reaction.2. To observe the effects of temperature and catalyst on the rateof reaction.IntroductionGenerally, deuce important questions may be asked about a chemicalreaction(1)How far do the reactants interact to yield products, and (2) how debased is the reaction? How far? is a question of chemical equilibriumwhich is the area of chemical thermodynamics. How fast? is therealm of chemical kinetics, the subject of this try out.In this experiment we will study the rate of oxidation of iodide ionby hydrogen p eroxide which proceeds according to the followingreactionH2O2 (aq) + 2 I-(aq) + 2H+(aq) I2(aq) + 2H2O(l)By change the concentrations of each of the three reactants (H2O2, I-and H+), we will be able to determine the direct of the reaction withrespect to each reactant and the rate law of the reaction, which is ofthe moldRate = k H2O2xI-yH+zBy knowing the reaction times (t) and the concentrations of H2O2 oftwo separate reaction mixtures (mixtures A & B), the reaction order ofH2O2, x, can be calculated.x = log(t2/ t1) / log ( H2O21/H2O22 )The same(p) method is used to obtain the reaction order with respect to I-(mixtures A & C) and H+ (mixtures A & D).ProceduresPart I) normalization of H2O2 Solution1. A stand, a buret clamp and a snow-white tile were collected toconstruct a titration set-up.2. A burette was rinsed with deionized pissing and then with 0.05 MNa2S2O3 consequence.3. The stopcock of the burette was closed and the sodiumthiosulphate solution was pour into it until the silver-tongued level was nearthe zero in mark. The stopcock of the burette was opened to allow thetitrant to fill up the tip and then the liquid level was adjusted nearzero.4. The initial burette reading was put down in Table 1.5. 1.00 cm3 of the 0.8 M H2O2 solution was pipetted into a clean one hundred twenty-five ... ...te of a reaction byproviding an alternative pathway for the reaction, usually with apathway of lower activation energy than that of the uncatalyzedreaction.There are more or less improvements in this experiment.First, hydrogen peroxide is unstable, it decomposes to water andoxygen by time. so do the titration as quick as possible.2H2O2(aq) 2H2O(I) + O2(g)Second, the concentration of iodine annex is due to the iodide canbe oxidized by oxygen which promoted by acids. Therefore do thetitration as quick as possible.4I-(aq) + O2(g) + 4H+(aq) 2I2(aq) + 2H2O(aq)Third, as for the man error, the problem can be minimized byperforming the titration by the same perso n. So, the reading can betaken by the same person and the color change can be observed by thesame person.ConclusionIn the experiment, the reaction was found to be zero order respect to(H+), it is first order respect to iodide, (I-) , it is first orderrespect to hydrogen peroxide, (H2O2). Hence the rate law is Rate = kH2O2I-.The rate of reaction is increase when the temperature is increase andthe rate is increase when a validatory catalyst is added to thereaction.