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

The Rate Law for Chemical chemical reaction Among Hydrogen Peroxide, Iodide, and AcidTo determine the govern right for a chemic reaction among hydrogen peroxide, iodide and acid, specifically by notice how changing distributively of the preoccupationsExperiment 3 Chemical KineticsObjectives1. To determine the rate law of nature for a chemic 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, two important questions may be asked about a chemicalreaction(1)How far do the reactants move to yield products, and (2) howfast is the reaction? How far? is a question of chemical equilibriumwhich is the realm of chemical thermodynamics. How fast? is therealm of chemical kinetics, the composition of this experiment.In this experiment we will study the rate of oxidation of iodide i onby hydrogen peroxide which proceeds according to the followingreactionH2O2 (aq) + 2 I-(aq) + 2H+(aq) I2(aq) + 2H2O(l)By varying the concentrations of each of the three reactants (H2O2, I-and H+), we will be able to determine the cast of the reaction with discover to each reactant and the rate law of the reaction, which is ofthe formRate = 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 method is utilise to obtain the reaction order with respect to I-(mixtures A & C) and H+ (mixtures A & D).ProceduresPart I) Standardization of H2O2 Solution1. A stand, a buret hug and a white tile were collected toconstruct a titration set-up.2. A burette was rinsed with deionized water and then with 0.05 MNa2S2O3 solution.3. The stopcock of the burette was unopen and the sodiumthiosulphate solution was pour into it until t he liquid level was nearthe zero in mark. The stopcock of the burette was opened to allow thetitrant to fill up the extent and then the liquid level was adjusted nearzero.4. The initial burette reading was recorded in Table 1.5. 1.00 cm3 of the 0.8 M H2O2 solution was pipetted into a clean125 ... ...te of a reaction byproviding an alternative piece of ground for the reaction, usually with apathway of lower activation energy than that of the uncatalyzedreaction. at that place are some improvements in this experiment.First, hydrogen peroxide is unstable, it decomposes to water andtype O by time. Therefore do the titration as quick as possible.2H2O2(aq) 2H2O(I) + O2(g)Second, the concentration of iodine gain 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 human error, the problem can be minimized byperforming the titration by the same person. So, th e reading can beinterpreted by the same person and the color change can be observed by thesame person.ConclusionIn the experiment, the reaction was tack together 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 positive catalyst is added to thereaction.