Reaction Kinetics: The Iodination of Acetone I

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Transcription:

1 Reaction Kinetics: The Iodination of Acetone I David Bartholomew Evan Beachnaw Steven Erickson Kevin Ramirez Karina Ramirez

2 Abstract The purpose of this experiment was to measure the rate constant k, for the reaction of acetone and iodine at room temperature, which resulted in 4.47x10-5 M -1 s -1. K is the rate constant for the reaction and does not depend on the concentration of the reactants. The rate constant is only affected by changes in temperature, which remain constant throughout the experiment at room temperature. The values of m, n and p were also calculated and were found to be 1, 0, and 1, respectively. These values represent the reaction rate orders with respect to acetone (m), iodine (n) and hydrogen ion (p). The results agree with the established notion that the order with respect to the concentration of iodine is zero order. Introduction Kinetics is a subfield that studies the rate and mechanism for a chemical reaction. The mechanisms are the species involved, and how many of each species are involved in a reaction. In chemistry, it is often the case that a certain product is needs to be produced in great quantities for whatever reason. The reaction or method used to make this product needs to have maximum efficiency by knowing which components accelerate the reaction, which ones slow it down and which ones do not affect it at all. The study of the rates of a reaction is called kinetics and concerns itself with reaction rates, which are defined as the change in concentration of the reactants and products over the time interval where the change occurs (3). The rate depends on several factors, which are the concentration of the reactants and products, partial pressure of the reactants and products, the temperature of the reaction, the order or molecularity of the species, area available for the reaction, and the presence or absence of a catalyst or retardant (6). This experiment will analyze the kinetics of the reaction between acetone and iodine because it has a rate that is easily measured at room temperature, and the color allows for a visual assessment of concentration. The time that it takes for the color to dissapear, is the time it takes for the iodine to be consumed by the reaction. Another important characteristic is that the order of the reaction with respect to Iodine is zero, so the rate does not depend on the concentration of iodine. The Iodine will be the limiting reagent in the presence of excess acetone and the hydrogen ion catalyst. CH 3 COCH 3 (aq) + I 2 CH 3 COCH 2 I (aq) + H + (aq) + I - (aq) The average reaction rate of a reactant, measured as a change in molar concentration, is given by the following equation: The concentration of the reactants is negative because it decreases over time, and gives the rate a positive value. On the other hand, the products increase over time and the rate is calculated with the

3 following equation: [ ] Since the rate for this reaction depends on the concentration of hydrogen ion and the concentrations of the two reactants, the empirical rate law of this reaction is then determined by: Rate = [I 2] / t = k [acetone] m [I 2 ] n [H + ] p The brackets indicate the concentration of the specified substance, k is the rate constant, and m, n and p represent the orders of the reactions with respect to acetone, iodine and hydrogen ion, respectively. Cato Maximilian Guldberg and Peter Waage first established the Law of Mass Action (a mathematical formula that attempts to predict and explain chemicals in equilibrium) in 1864. The two scientists built off Claude Louis Berthollet s concepts of reversing chemical reactions. When van t Hoff independently discovered this theory, it prompted Guldberg and Waage to elaborate on the conclusions of the experiments and results the two worked on previously. Around 1879 in Germany, is when the two extrapolated these findings [5]. A + B A' + B' This was the first equation developed by Guldberg and Waage [5]. The above equation is the beginning of showing that when a substance reaches a state of equilibrium that the rate of the forward process equals that of the reverse. This normally happens while inside a closed system [1, 4]. An example includes a reaction that causes a thermal equilibrium. This is when the loss of heat equals the heat gained. Equilibriums may also be mechanical and chemical in nature [4]. All equilibrium situations can be mathematically predicted using the law of mass action [4]. All chemical reactions have the potential to be reversible. When the reaction is completed, the products and reactants continue to revert back and forth to one another [1, 3, 4]. Below is a rendition of what the dynamic reactions are: Below shows the Forward rate equal to the Reverse rate. This is when the substance is at equilibrium and the rate of each reaction is equal to itself [1, 3]. kf [A] A [B] B = [C] C [D] D

4 At equilibrium, there is a customary ratio of products relative to reactants. The equilibrium constant is below: kf [A] A [B] B = [C] C [D] D [ ] eq = Equilibrium concentrations in molarity [1, 3] Formalism Arrhenius equation for rate constant Ea = the activation energy in kj/mol R = gas law constant (8.314/K*mol) T = temperature in Kelvin (K) A = frequency factor Le Châtlier s Principal Experimental Use the following concentrations to create the kinetic reactions. Iodine was the final component in each trial as it was the element, which started the reaction. Run 4 M Acetone 1 M HCl Water 0.005 M I 2 1 2 ml 2 ml 4 ml 2 ml 2 4 ml 2 ml 2 ml 2 ml 3 2 ml 4 ml 2 ml 2 ml 4 2 ml 2 ml 2 ml 4 ml Table 1. Volumes of reactants mixed in each of the different runs of the experiment. 1. Gather four beakers and label them appropriately Acetone, HCl, Water, and Iodine. 2. Add approximately 35 ml of each component accordingly and add parafilm to cover and prevent evaporation. 3. Designate each solution a graduated cylinder and disposable pipette. 4. Setup 3 large test tubes in a beaker and then label them Trial 1, 2, and 3.

5 5. Each run consisted of three tubes of the same chemicals in an assembly line arrangement (view the table above), adding the Iodine last and then immediately starting the timer. For example, run 1 should be placed in tubes 1-3 with 2 ml of 4M Acetone, 2 ml of 1 M of HCl, 4 ml Water, and finally the 2 ml of 0.005 M of I 2 simultaneously starting a stopwatch. 6. When the solution enters equilibrium, it turns clear. 7. Finally, these values contain the data desired. 8. Repeat steps four through seven. Results and Discussion Data Table Run Elapsed Time Average Time [I₂] (M) (seconds) (seconds) Rate (M/sec) [ ] 1 150 158.001M 6.33x10 ⁶ M/sec 160 165 2 78 70.001M 1.43x10 ⁵M/sec 68 65 3 72 72.001M 1.39x10 ⁵ M/sec 71 72 4 298 310.002M 6.45x10 ⁶ M/sec 280 351 From the rate information we can determine the orders with respect to acetone (m), iodine (n) and hydrogen ion (p) by varying the amounts of reactants and measuring the effect on the rate. Once the orders of the reaction are known we will be able to calculate the rate constant, k. The concentration of the reactants is determined by the following equation, using as an example the concentration of acetone: [Acetone]= (4.0 mole Acetone / 1000 ml) Vol Acetone (ml) / (0.010 L) The volume of acetone for each run is taken from table 1, which indicates the volumes of reactants mixed in each trial. A similar table is organized once all the concentration values are obtained. Run # [I 2 ] (M) [Acetone] (M) [H + ] (M) 1 0.001 0.8 0.2 2 0.001 1.6 0.2 3 0.001 0.8 0.4

6 4 0.002 0.8 0.2 To solve for m, the runs where the concentration for acetone is constant but the others remain the same. Therefore: m= rate 2 / rate 1 (1.43x10-5 M/s) / (6.33x10-6 M/s) = (1.6 M) / (0.8 M) = 2 m ln(2.259) = m ln(2) m = ln(2.259) / ln(2)= 1.18 1 The order rate for acetone equals 1, which remains constant with the coefficient in the reaction equation. To find the value of k, it is possible to use any run and have similar results. In this case we will use run #2 to calculate this value by using the following formula: k = rate / [acetone] m [I 2 ] n [H + ] p k = 1.43x10-5 / (1.6 M) 1 (0.001 M) 0 (0.2 M) 1 = 4.47x10-5 M -1 s -1 The k values for the other runs are as follows: Run #1: k = 3.96x10-5 M -1 s -1 Run #3: k = 4.34x10-5 M -1 s -1 Run #4: k = 4.03x10-5 M -1 s -1 Conclusion References 1. Dr. Eddie W. Ong, Private Communications. 2. N. J. Tro and R. S. Boikess, Chemistry: A Molecular Approach, 2 nd Ed., (Pearson Education, New Jersey, 2011, pp. 618, 621, 623, 799. 3. Phoenix College General Chemistry II Lab Manual, (Premium Source Publishing, Phoenix, 2006), p.47-50. 4. Sullivan, Dan M. "Equilibrium." Chemistry: Foundations and Applications. Ed. J. J. Lagowski. Vol. 2. New York: Macmillan Reference USA, 2004. 63-65. Gale Virtual Reference Library. Web. 8 Dec. 2011.

7 http://ezproxy.pc.maricopa.edu:2084/ps/i.do?id=gale%7ccx3400900179&v=2.1&u=mcc_phoe &it=r&p=gvrl&sw=w 5. Wikipedia contributors. Law of mass action [Internet]. Wikipedia, The Free Encyclopedia; 2011 Nov 27, 23:33 UTC [cited 2011 Dec 9]. Available from: http://en.wikipedia.org/w/index.php?title=law_of_mass_action&oldid=462808458. 6. Dr. Eddie W. Ong. Chemistry II Lab Notes.