Chemical Kinetics Lab - Chris Stanton Lab Report Chemical...

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Chris Stanton 11/16/06 Lab Report: Chemical Kinetics I. Introduction The goal of this experiment is to determine the chemical rate law for the oxidation of iodide ion by persulfate ion: S 2 O 8 2- + 3I - 2SO 4 2- + I 3 - . Rate = k[S 2 O 8 2- ] m [I - ] n To determine this overall rate law, you first have to test experimental rate laws, and in this case, the iodine clock-initial rate method is used. This is composed of two reactions, the first being the reaction of starch with tri-iodide, I 3 - , to form a dark blue complex easily visible with the eye. The second reaction is the oxidation of thiosulfate ion by tri-iodide to form the iodide ion: 2S 2 O 3 2- + I 3 - S 4 O 6 2- + 3I - . Several runs are done of these reactions, data being recorded each time. By varying the initial iodide ion concentration in runs and then varying the concentration of persulfate ion, the exponents n and m can be calculated and the overall rate law can be determined. The second part to this experiment involves the addition of a copper catalyst. The catalytic effect of Cu 2+ on the persulfate-iodide reaction may be used to quantitatively determine [Cu 2+ ] in a solution. By using the same techniques in the first part of the experiment and varying the concentration of Cu 2+ . II. Prelaboratory Questions 1. Run [I - ] [S 2 O 8 2- ] [S 2 O 3 2- ] 1 0.2 M 0.2 M 0.01 M 2 0.1 M 0.2 M 0.01 M 3 0.05 M 0.2 M 0.01 M 4 0.025 M 0.2 M 0.01 M 5 0.2 M 0.1 M 0.01 M 6 0.2 M 0.05 M 0.01 M 7 0.2M 0.025 M 0.01 M 2. ∆[S 2 O 8 2- ] = -½ ∆[S 2 O 3 2- ] because the two equations (both listed in Introduction) that link these two ions to each other involves a coefficient of 2 in front of the S 2 O 3 2- ion. Since the entire reaction takes place with such a minute concentration of S 2 O 3 2- , the rate law can be assumed to be equal to -k[S 2 O 8 2- ] i m [I - ] i n , where the concentrations are actually the initial concentrations of each reactant. This is why ∆[S 2 O 8 2- ] = -½ ∆[S 2 O 3 2- ], considering that S 2 O 3 2- has a coefficient of 2 and the rate law has a negative sign in it. 3. Cu2+, or a specific catalyst in any reaction, speeds up the reaction by lowering the activation energy (∆G ± ) for the reaction. This allows the reaction to need less energy to move to equilibrium (get over the “hump” in a reaction coordinate diagram). Catalysts do not change the overall thermodynamics for the reaction and ∆G° is unchanged.
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III. Experimental Procedure: Laboratory Period I 1. Three solutions are needed before the experimental runs can begin. First, 100 mL each of 0.01 M Na 2 S 2 O 3 , 0.2 M KNO 3 , and 0.2 M (NH 4 ) 2 SO 4 are obtained. The 0.01 M Na 2 S 2 O 3 must be dissolved in deionized water and the calculation to do this is shown below. (0.01M)(1L/1000mL)(100mL) = X(1mol/248.12g) X = 0.248 g Na 2 S 2 O 3 needed for 100 mL of 0.01 M solution 2. The next solutions you need to make are 250 mL of 0.2 M KI and 250 mL of 0.2 M (NH 4 ) 2 S 2 O 8 . These also need to be dissolved in deionized water, so the calculation for each is shown below.
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