Fast, Faster, Fastest: Exploring Chemical Catalysis

Transcription

1 Fast, Faster, Fastest: Exploring Chemical Catalysis Used with the permission of Dr. Russell Larsen, Department of Chemistry, University of Iowa with special thanks to the Fall Semester CHM 33 FS02 Teaching Assistants of the Mizzou Chemistry Department, most notably John Brockman and Jon Fitzsimmons OVERVIEW Topics include: heterogeneous and homogeneous catalysis biocatalysis and inhibition Reaction rate determined by measuring rate of oxygen gas formation INTRODUCTION The formation of products from reactants is the essence of chemistry. Thermodynamics can be used to identify reactions that are spontaneous (favorable), as well as the equilibrium state of the reaction. However, on a practical level the rate with which reactants are formed is as important as the overall thermodynamic favorability. If the reaction is too slow, products may never form. If the reaction is too fast, the reaction can destroy the system (for example, an explosion). There are several strategies for implementing control over the rate of a reaction: the concentration of the reactants and products on which the rate depends can be varied, the temperature at which the reaction takes place can be adjusted, a catalyst can be used to speed up a reaction, or an inhibitor can be used to slow down the reaction. In this experiment, you will investigate the influence of several types of catalysts and an inhibitor on the rate of a chemical reaction. The reaction that you will focus on is the decomposition of hydrogen peroxide, H 2 O 2. The reaction can be written as: 2 H 2 O 2 (aq) O 2 (g) + 2 H 2 O (l) For reactants to become products they must first overcome a potential energy barrier. This barrier is called the activation energy. In this reaction, the barrier might be imagined to be a state in which the oxygen-hydrogen bonds of one hydrogen peroxide molecule are being stretched (broken) as oxygen-hydrogen bonds on an adjacent molecule are being formed. The course of the reaction might be pictured as: In this diagram, the difference in energy between the products and reactants is related to the thermodynamic.

2 spontaneity of the reaction, and the rate of the reaction depends on the magnitude of the activation barrier between the reactants and the top of the activation barrier. The species with the highest energy is called the transition state. Transition state species have a very fleeting existence (i.e sec). Since both products and reactants lie lower on the potential energy curve, transition state species quickly decay to form either product or reactants. As indicated in the diagram, the decomposition of hydrogen peroxide reaction is spontaneous ( G = 118. kj/mol) but is slow under ambient conditions due to the large activation energy (E a = 76 kj/mol). To traverse the path from reactants to products, molecules must have enough thermal energy to scale the activation barrier. At any given temperature (other than absolute zero), molecules have a distribution of energy. That is to say, some molecules have a higher instantaneous energy than other molecules. The average amount of thermal energy that molecules possess at any given temperature is on the order of RT (at room temperature, RT=8.3 J mol 1 K 1 * 298 K = 2.5 KJ/mol). The fraction of the molecules that have enough energy to overcome a potential energy barrier of magnitude E a depends exponentially on the ratio of the activation to the thermal energy. Mathematically, this takes the form of the Arrhenius equation (1), k Ea / RT Ae (1) where k is the rate constant of the reaction, A is a constant that depends on how frequently the reactants are found in the proper configuration to react, and the exponential term expresses the fraction of the molecules that have enough thermal energy to overcome the activation barrier. Given this mathematical form, two strategies for increasing the rate of the reaction become apparent. All other things being equal, (1) the rate of the reaction will increase exponentially with increasing temperature, and (2) the rate of the reaction will increase exponentially with decreasing activation energy. The first strategy is straightforward to implement by simply heating the reaction system. The second method, decreasing the activation energy, can be achieved by adding a catalyst to the system. Catalysts A catalyst is a substance that can increase the rate of a reaction while not being consumed. Catalysts usually function by breaking the reaction up into a sequence of steps with an overall reduction in the activation energy and thus overall enhancement of the rate. A reaction diagram for a catalyst is given below. A catalyst does not appear in the balanced chemical reaction, but it does influence the rate of the reaction. It is useful to define three types of catalysts: (1) Homogeneous catalysts are chemicals that are in the same phase as the reactants; (2) Heterogeneous catalysts are chemicals that are in a different phase from the reactants; and (3)

3 Enzymes are polypeptides (proteins) capable of catalyzing reactions. You will be studying examples from each of these three classes of catalysts. In the presence of a catalyst, the reaction rate of hydrogen peroxide decomposition can be enhanced substantially. First, consider the homogeneous catalytic decomposition of hydrogen peroxide by iodide (I ). The elementary steps and relative rates are given by, Step 1. H 2 O 2 (aq) + I (aq) H 2 O (l) + IO (aq) Step 2. H 2 O 2 (aq) + IO (aq) H 2 O (l) + O 2 (g) + I (aq) overall 2 H 2 O 2 (aq) 2 H 2 O (l) + O 2 (g) SLOW FAST (What would be the expected rate law for this mechanism? answer: rate = k[h 2 O 2 ][ I ]). Catalysis by iodide is able to promote the speedy formation of oxygen and water largely due to the reduction in the activation energy to about 57 kj/mol (compared to 76 kj/mol for the uncatalyzed reaction). Heterogeneous catalysis is also possible for the hydrogen peroxide decomposition reaction. Manganese(IV) oxide (MnO 2 ) is a black solid that is insoluble in water. Since the solid is insoluble, any enhancement of the rate of decomposition must occur through a heterogeneous pathway. Because the reaction takes place at the solid/liquid interface, one of the problems that must be addressed experimentally is that molecules must be continuously transported to the interfacial region. For instance, if a layer of O 2 gas forms on the solid, new hydrogen peroxide molecules, which are in the solution phase, cannot reach the active surface in order to react. Continuous stirring is one simple solution that works in this case. One of the most significant advantages of heterogeneous catalysts is that since they are present as a separate phase they can be separated from the reaction mixture and recycled. In addition, you will investigate biocatalysis. Hydrogen peroxide is a strong oxidant, and therefore can cause serious molecular damage to cells (thus its use as a disinfectant). It is known to readily react with unsaturated fatty acids in cell membranes causing damage to the membrane structure. To protect the cell against oxidative damage, the cell produces enzymes that can accelerate the decomposition of hydrogen peroxide into the less harmful molecules of oxygen and water. One enzyme that has evolved to serve this function is catalase. Catalase is an iron-heme containing enzyme (you will be studying two other heme proteins later this semester, myoglobin and hemoglobin). A heme is a large cyclic molecule with a pocket in the center that is capable of binding metal ions. Catalase is a tetramer (made up of four parts), with a total molecular weight of about 250,000 g/mol. Each of the tetrameric subunits contains one active site composed of a heme-bound iron(iii) surrounded by protein. This active site is the location where H 2 O 2 binds and decomposes to oxygen and water. Structures of catalase from several sources have been obtained, and one structure of catalase from bovine liver obtain by I. Fita and M.G. Rossmann is presented as ribbon representation below (Heme units shown in black, helix dark gray, and sheet medium gray.). Although simple iron salts (e.g. FeCl 3 ) possess some catalytic activity for the decomposition of hydrogen peroxide, further enhancement is achieved by the binding of the iron within the heme of the enzyme. The activation energy of the catalyzed reaction is only about 8.8 kj/mol! Catalase is a very efficient enzyme, and in the cell, it is able to decompose hydrogen peroxide nearly as fast as hydrogen peroxide can reach the enzyme. All three components of the enzyme, the iron (III) ion, the heme, and the protein are essential to proper catalytic function of catalase. Because all three components are vital to catalytic function, the disruption of any one of the components can inhibit catalysis. An inhibitor is a chemical species that slows the rate of the reaction. In the case of catalase, copper (II) ions can function as an inhibitor. Copper (II) is thought to compete with the iron for the heme binding site. If a copper (II)

4 ion is bound in the heme pocket, the enzyme does not function as a catalyst, and therefore the rate decreases. EXPERIMENTAL METHOD Since the decomposition of hydrogen peroxide produces oxygen gas as one of the products, the rate of gas formation can be used to measure the rate of the reaction. You will measure the volume of gas produced versus time. To do this you will need to set up a reaction vessel and attach a tube to a buret as shown on the next page. 1) Fill a 600 ml beaker 2/3 of the way with water. 2) Fill a buret with water. 3) Insert the tubing into the mouth of the buret, then using a thumb or finger, seal off the open end of the buret and invert the buret into the beaker. Attach the buret to a clamp. 4) Allow air into the buret so that the volume of water in the buret reads at 50 ml. This will make it very easy to measure 10 ml of gas produced. Explanation: the gas produced from the decomposition of H 2 O 2 will displace the water in the tube as it rises in the buret. Gas will be formed in the reaction flask and flow through the rubber tube. The change in the volume of gas produced can be measured by monitoring the displaced volume in ml in the 50 ml buret tube. The buret is filled with water, so the oxygen produced in the reaction flask will displace the water in the filled inverted buret. From volume of gas produced per second, the rate of the reaction can be obtained. Using the ideal gas law, PV=nRT, the volume per second data can be converted to moles of O 2 per second. However, a small correction for water vapor must first be made. Since the oxygen gas passes through water, the gas evolved is not pure oxygen; the gas also contains water vapor. Assuming that the total pressure is atmospheric pressure (760 torr) and the temperature of the room is 25 C, the vapor pressure of water is 24 torr. Using Dalton s law of partial pressures, this means that oxygen gas is present at 736 torr ( Torr). Once the data are transformed into units of (moles O 2 /sec) the stoichiometry of the reaction and volume of the solution in the reaction flask, can then be used to find the rate of decomposition of H 2 O 2 in units of mol/(l sec). These are the units that should be used when reporting your final results. As a source of catalase, you will be using a potato (or other vegetable) extract. Within eukaryotic cells, hydrogen peroxide using and destroying enzymes are kept within special membrane compartments called peroxisomes to protect the cell from the harmful effects of peroxides. By breaking the cells open, these enzymes can be released into solution. A catalase containing extract will be formed by blending vegetables in a blender, and separating the catalase containing solution from the solid components. EXPERIMENTAL PROCEDURE For each run, measure 25 ml of hydrogen peroxide solution into a 125 ml erlenmyer flask, and begin to swirl. Start each kinetic run by pouring the appropriate catalytic solution/solid into the flask of hydrogen peroxide solution (this marks time = 0). Monitor how long it takes to produce 10.0 ml of gas. You will want to determine the reaction rate for each of the following eight mixtures:

5 Kinetic Run Volume of H 2 O 2 (in ml) Volume of Distilled Water (ml) Amount of Catalyst Amount of Copper(II) inhibitor Time to form 10.0 ml of gas (in seconds)* H 2 O 2 solution 25.0 ml 5.0 ml 0.00 ml 0.00 ml This run will not produce results. Use this null as part of your discussion. 1 M KI 25.0 ml 0.0 ml 5.00 ml 0.00 ml MnO 2 (use weigh paper) 25.0 ml 5.0 ml 0.50 g 0.00 ml Catalase 25.0 ml 0.0 ml 5.0 ml 0.00 ml Catalase / with 1M Cu ml 0.0 ml 5.0 ml ~10 drops *If the rate is too slow, record the time to produce a smaller volume and then extrapolate to the time that it would take to form 10 ml (assuming a constant rate). The Catalase solution will be provided. Test the solution for catalytic activity as described above, and report the time that the reaction mixture took to form 10.0 ml of gas on the chalkboard. Record, report, and compare your results to those of your classmates. For the final part of this experiment, you should choose one variable (such as concentration, temperature, surface area, etc.), formulate a hypothesis and design an experiment to test the validity of your hypothesis. An appropriate hypothesis should take at least 3 new kinetic runs to either support or refute the hypothesis. For example, you might formulate the hypothesis that since proteins are known to denature when heated that if the catalase solution was boiled before using, the solution would no longer decompose hydrogen peroxide. Each student will need her/his own hypothesis, so be creative. A group of students may work together and pool data, if the scope of the problem is sufficient to warrant collaboration. You should present your results in an appropriate manner (e.g. graph, table, etc.). You should include some mention of real world application of commerically available enzyme products (e.g. products that ulitize enzymes to carry out some process). References 1. L.R. Summerline, C. L. Borgford, and J. B. Ealy, Chemical Demonstrations: a Sourcebook for Teachers, vol. 2, second edition, American Chemical Society: Washington D.C., A. L. Lehninger, Principles of Biochemistry, Worth Publishers, Inc., New York, I. Fita, M.G. Rossmann, PROC.NAT.ACAD.SCI.USA, Vol. 82, p. 1604, J. Teggins and C. Mahaffy, J. Chem. Educ., Vol. 74, p. 566, D.R. Kimbrough, M.A. Magoun, R. Langfur, J. Chem. Educ., Vol.74, p210, 1997