Experiment_5__-_rates_of_reaction

Experiment_5__-_rates_of_reaction - 5 Rate of Reaction...

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5 Rate of Reaction Introduction This experiment will allow you to study the effects of concentration, temperature, and catalysts on a reaction rate. The reaction whose rate you will study is the oxidation of iodide ion by the strong oxidizing agent peroxydisulfate ion (S 2 O 8 2– ). 2 I (aq) + S 2 O 8 2– (aq) → I 2 (aq) + 2 SO 4 2– (aq) (SLOW) This reaction is slow enough to be convenient to study, unlike most common reactions involving ions in water which occur almost instantaneously. Measuring a reaction rate entails measuring some unique property of one of the products or reactants. In this experiment, the unique property you will look for is the color produced by the I 2 formed in the reaction. This color is intensified by allowing the I 2 to react with starch to form a deep blue complex. Starch is therefore included in the reaction mixture as in indicator for I 2 . It reacts only with the product I 2 and not with the reactant I (iodide ion). The starch-iodine color is so intense that it becomes visible when only an extremely low concentration of I 2 has been produced. The intensity of the color increases as the concentration of I 2 increases, but to accurately measure the change in intensity of the color as I 2 builds up would require the use of a spectrophotometer. A simpler method is to resort to a sort of chemical “trick.” An additional reactant is introduced that will remove the first 2.5 × 10 -5 moles of I 2 formed in the reaction. The additional reactant is thiosulfate ion, S 2 O 3 2- , which reacts with I 2 by the very fast reaction below: I 2 (aq) + 2 S 2 O 3 2– (aq) → 2 I (aq) + S 4 O 6 2– (aq) (FAST) Only a small amount of thiosulfate is added: 5.0 × 10 –5 moles, or enough to react with 2.5 × 10 –5 moles of I 2 . Once the reaction has produced more than this amount of I 2 , no more thiosulfate remains to react with it, and the starch-iodine color appears. No color appears until the reaction has produced 2.5 × 10 –5 moles of I 2 , because the I 2 initially formed immediately reacts with thiosulfate. Chemistry 1B Experiment 5 1
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Once all the thiosulfate has been used up, the color appears very suddenly and sharply. Its appearance can be used as a signal to measure the rate of the reaction. The time from mixing the reactants to the appearance of color can easily be measured. This length of time is the time required by the slow reaction to produce 2.5 × 10 –5 moles of I 2 . As the experiment will show, the time required varies with the concentrations of the reactants, with temperature, and with the presence of catalysts. The time is also a measure of the rate of the reaction . Reaction rate is formally defined as the change of concentration of reaction products per unit of time—in other words, at what rate, in moles per liter per second, does the concentration of the product increase as the reaction proceeds. In this experiment, the reaction will be performed in a total volume of 50 mL. Therefore, producing 2.5
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This note was uploaded on 09/04/2011 for the course CHEM 1B taught by Professor Fossum during the Spring '10 term at Laney College.

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Experiment_5__-_rates_of_reaction - 5 Rate of Reaction...

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