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# To determine the rate law for a chemical reaction among hydrogen peroxide, iodide and acid, specifically by observing how changing each of the concentrations

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Introduction

Experiment 3 - Chemical Kinetics Objectives 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, and H+ affects the rate of reaction. 2. To observe the effects of temperature and catalyst on the rate of reaction. Introduction Generally, two important questions may be asked about a chemical reaction: (1)How far do the reactants interact to yield products, and (2) how fast is the reaction? "How far?" is a question of chemical equilibrium which is the realm of chemical thermodynamics. "How fast?" is the realm of chemical kinetics, the subject of this experiment. In this experiment we will study the rate of oxidation of iodide ion by hydrogen peroxide which proceeds according to the following reaction: H2O2 (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 order of the reaction with respect to each reactant and the rate law of the reaction, which is of the form: Rate = k [H2O2]x[I-]y[H+]z By knowing the reaction times (?t) and the concentrations of H2O2 of two separate reaction mixtures (mixtures A & B), the reaction order of H2O2, x, can be calculated. x = log(?t2/ ?t1) / log ( [H2O2]1/[H2O2]2 ) The same method is used to obtain the reaction order with respect to I- (mixtures A & C) and H+ (mixtures A & D). Procedures Part I) Standardization of H2O2 Solution 1. A stand, a burette clamp and a white tile were collected to construct a titration set-up. 2. A burette was rinsed with deionized water and then with 0.05 M Na2S2O3 solution. 3. The stopcock of the burette was closed and the sodium thiosulphate solution was pour into it until the liquid level was near the zero mark. ...read more.

Middle

dm3 / [(30+30+45+25+5+5+10) /1000 dm3] = 0.01 mol dm-3 Since Ka = [H+] [CH3COONa] / [CH3COOH] 1.8x10-5 = [H+] (0.01) / (0.1) [H+] = 1.8x10-4mol dm-3 Reaction Mixture Time / s Acetic acid concentration/ M Sodium acetate concentration/ M Calculated H+ Concentration/ M A 172 0.01 0.01 1.8x10-5 D 169 0.1 0.01 1.8x10-4 The order of reaction respect to iodide: Since x = log(?t2/ ?t1) / log([H+]1/[H+]2) x = log(169/172)/log[(1.8x10-5)/( 1.8x10-4)] x = 0.0076 = 0 It is zero order reaction respect to H+. 2(a) From your measurements, what are the reaction orders with respect to each of the three reactants H2O2, I- and H+? Ans: It is zero order respect to H+, first order respect to iodide, I- , and first order respect to hydrogen peroxide, H2O2. (b) What is the rate expression (rate law) for the uncatalyzed reaction? What is the total reaction order? Ans: The rate law is: Rate = k[H2O2][I-] where k is rate constant Therefore it is second order respect to the reaction. (c) Is the rate law consistent with the idea that the mechanism of the reaction is H2O2 (aq) + 2 I-(aq) + 2H+(aq)--> I2 (aq) + 2H2O(l) (all in one step)? Explain why or why not. Ans: If the mechanism involves only one step only, then it means that this step is the rate determining step. The rate law is Rate = k[H2O2][I-]. The rate determining step should include only one hydrogen peroxide and one iodide molecule only, but the above reaction involve one hydrogen peroxide , two iodide and two hydrogen ion molecule. Also, the hydrogen ion should not be present in the rate determining step as it is zero order and the rate law does not include it. Actually, according to collision theory, it is not favour that 3 molecules collides together and proceed a reaction. Collision of two or one molecule is more favourable. ...read more.

Conclusion

(iii). Since the color change from iodine to iodide cannot be used to accurate the end point (the change in color brown to yellow to colourless is difficult to observe), starch is used as the indicator. Starch + iodine --> blue complex (iv). The rate equation for a reaction may help to deduce the mechanism of the reaction. The order of a reaction is an important implication of the rate determining step. From the order, we can deduce the mechanism and the rate determining step. The orders are experimentally found, they are not necessarily equal to the coefficient of the reactant in the equation unless the reaction is a single reaction. (v). A catalyst is a substance which alters the rate of a particular chemical reaction, but itself is chemically unchanged at the end of the reaction. A catalyst can change the rate of a reaction by providing an alternative pathway for the reaction, usually with a pathway of lower activation energy than that of the uncatalyzed reaction. There are some improvements in this experiment. First, hydrogen peroxide is unstable, it decomposes to water and oxygen by time. Therefore do the titration as quick as possible. 2H2O2(aq) --> 2H2O(I) + O2(g) Second, the concentration of iodine increase is due to the iodide can be oxidized by oxygen which promoted by acids. Therefore do the titration 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 by performing the titration by the same person. So, the reading can be taken by the same person and the color change can be observed by the same person. Conclusion In the experiment, the reaction was found to be zero order respect to (H+), it is first order respect to iodide, (I-) , it is first order respect to hydrogen peroxide, (H2O2). Hence the rate law is Rate = k[H2O2][I-]. The rate of reaction is increase when the temperature is increase and the rate is increase when a positive catalyst is added to the reaction. ?? ?? ?? ?? 1 ...read more.

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