CHEM 122L General Chemistry Laboratory Revision 1.3 Determination of the Rate Constant for an Iodine Clock Reaction To learn about Integrated Rate Laws. To learn how to measure a Rate Constant. To learn about Clock Reactions. In this laboratory exercise, we will measure the Rate Constant k for the oxidation of Iodide (I - ) by Peroxysulfate (S 2 O 8 2- ). 2 I - (aq) + S 2 O 8 2- (aq) I2 (aq) + 2 SO 4 2- (aq) (Eq. 1) This reaction is referred to as a Clock Reaction because the progress of (Eq. 1) is monitored by a secondary Clocking reaction which consumes the product I 2 as it is produced and triggers a color change when the Clocking Reagent is itself completely consumed. For this purpose, we use Thiosulfate (S 2 O 3 2- ) as the Clocking Reagent to reduce the I 2 : 2 S 2 O 3 2- (aq) + I2 (aq) S 4 O 6 2- (aq) + 2 I - (aq) (Eq. 2) As long as any Thiosulfate Ion is present, none of the Iodine produced in (Eq. 1) remains; it is consumed as quickly as it is produced. (This, of course, requires that (Eq. 2) proceed very rapidly. This is in fact the case.) As soon as the Thiosulfate is used up, Iodine will begin to appear in the solution. The presence of Iodine is then dramatically detected by the formation of a blue Starch-Iodine molecular complex and the reaction is said to Clock at this point. (Actually, the starch molecules complex with I 3 - which is formed by a reaction between I 2 and I -.) We can set the Clock Point by simply adding differing amounts of Thiosulfate Ion to the solution. The Rate at which a chemical reaction proceeds is typically influenced by the amount of each reactant present and the temperature of the reaction vessel. And, typically, this relationship between the Reaction Rate and Reagent Concentration takes a simple form known as the Rate Law: Rate = k [A] n [B] m (Eq. 3) where A and B are generic reacting Species, k is a reaction specific proportionality constant known as the Rate Constant, and n and m are the Reaction Order. The Rate Law parameters k, n and m must be determined experimentally. Once the Rate Law has been established, the Rate of the chemical reaction can be determined. However, this is usually not the information which is sought. Consider how it would be received
P a g e 2 if while traveling over hill and dale to grandma s house, one of the kids asks, How long till we get there?, and the father answers, We are traveling at 60mph!. Similarly, we usually want to know how long it will be until the reaction is complete, or, alternatively, how much of a given reagent remains after a specified period of time, not how fast the reaction is proceeding at a given moment. (This is a bit of an over statement, but...) The Rate itself is not an answer to either of the above queries. In order to answer questions like How Long? or How Much Remains?, we must relate the Concentration of the reactants to the Time period over which the reaction has proceeded. This means we must Integrate the Rate Law. Because the Integration can be tricky, we will restrict ourselves to a few simple cases: Zeroth Order Rxns Rate = k (Eq. 4) First Order Rxns Rate = k [A] (Eq. 5) Second Order Rxns Rate = k [A] 2 (Eq. 6) We will illustrate the Integration for the Zeroth Order case. First we apply the definition of the Reaction Rate: Rate = - (Eq. 7) and equate this with the Rate Law: - = k (Eq. 8) This result is rearranged so all the Concentration variables are on the left-side of the equality and all Time variables on the right: d[a] = - k dt (Eq. 9) Integrating this from an Initial Time Point ([A] o at t=0) to an arbitrary Time Point ([A] at t): = - k (Eq. 10) gives us the desired relationship: [A] - [A] o = - k t (Eq. 11) allowing us to determine the Concentration of Species A, [A], at any given Time t. This relationship is typically inverted into a Linear form: [A] = [A] o - k t (Eq. 12)
P a g e 3 This is Linear because a plot of [A] versus t results in a straight line with Slope = - k and Intercept = [A] o. Integrating our other simple Rate Laws into their Linear form provides us with: Zeroth Order Rxns [A] = [A] o - k t (Eq. 13) First Order Rxns ln [A] = ln [A] o - k t (Eq. 14) Second Order Rxns = + k t (Eq. 15) A simple example is in order. Consider the Concentration data for the decomposition of Dinitrogen Pentoxide: 2 N 2 O 5 (g) 2 N 2 O 4 (g) + O 2 (g) Time (sec) [N 2 O 5 ] (M) 0 1.40 200 1.24 400 1.10 600 0.98 800 0.87 1000 0.77 1200 0.68 1400 0.60 1600 0.53 1800 0.47 2000 0.42 This reaction has been shown to be First Order in N 2 O 5 ; meaning the Rate Law can be written in Differential form as: or in Integral form as: Rate = k [N 2 O 5 ] ln [N 2 O 5 ] = ln [N 2 O 5 ] o - k t Thus, a plot of the Natural Log of the above Concentration data vs. Time should give us a straightline with a slope = k. This is in fact the case:
P a g e 4 0.4 0.2 0 ln(conc.) -0.2-0.4-0.6-0.8-1 y = -0.0006x + 0.339 0 500 1000 1500 2000 Time [sec] For this decomposition reaction, we have k = -slope = 0.0006 sec -1. This process can be applied in reverse. If we do not know the order of a given reaction, we can simply plot the data in all three Linear forms and see which results in a straight-line. This then gives us the Reaction Order. Of course, it is possible none of the simple forms gives us a straightline or, worse, the data has such a subtle curvature it becomes impossible to determine which graph gives us the best straight-line. We now have a method for determining the Rate Constant k of a given reaction. Provided we know the Reaction Order, we plot the Concentration data according to the appropriate equation (Eqs. 13 15) and determine the Slope of the Line. This Slope gives us k. For the present case, (Eq. 1), it has been determined the reaction kinetics proceeds according to a Mixed-Second Order Rate Law: Rate = k [I - ] [S 2 O 8 2- ] (Eq. 16) This presents us with our first problem; mixed-second order kinetics is not one of our simple integrated rate law forms. What to do? If we use a vast excess of Iodide, then its Concentration will barely change over the course of the reaction. This means that at any time during the reaction: [I - ] ~ [I - ] o (Eq. 17)
P a g e 5 Thus, we can re-write our Rate Law as: Rate = k [I - ] o [S 2 O 8 2- ] (Eq. 18) or, defining a new Rate Constant k as:: we can write the Rate Law as: k = k [I - ] o (Eq. 19) Rate = k [S 2 O 8 2- ] (Eq. 20) (Eq. 20) is said to be a Pseudo-First Order Rate Law with a Pseudo-First Order Rate Constant k. Keep in mind, it is k and not k we are after. However, if we can measure k at a known I - concentration, then we can use (Eq. 19) to determine k. And, we can measure k by obtaining [S 2 O 8 2- ] vs. Time data and plotting this according to the Linear form, (Eq. 14), for a First Order reaction: ln [S 2 O 8 2- ] = ln [S 2 O 8 2- ] o - k t (Eq. 21) We can obtain [S 2 O 8 2- ] vs. Time data fairly easily by varying the amount of Thiosulfate used in the Clocking Reaction, (Eq. 2). The amount of S 2 O 8 2- remaining at the Clocking Point is related to the amount of Thiosulfate used by the reaction stoichiometry of (Eqs. 1 and 2): At the Clocking Point, the number of moles of Peroxysulfate (S 2 O 8 2- ) which has reacted is exactly 1/2 the number of moles of Thiosulfate (S 2 O 3 2- ) orginally added. As an example, suppose we mix the following: 5mL of 1M KI 7mL of 0.2M Na 2 S 2 O 3 5mL of 0.15M K 2 S 2 O 8 1mL of Starch Sol'n
P a g e 6 18mL total Then, at the Clocking Point when the blue Iodine-Starch complex appears: # mmoles S 2 O 3 2- added = ( 0.2M ) x ( 0.007 L ) x ( 1000 mmole / 1 mole ) = 1.40 mmole # mmoles S 2 O 2-2- 8 consumed = 1.4 mmole S 2 O 3 x ( 1 mmole I 2 / 2 mmole S 2 O 2-3 ) x ( 1 mmole S 2 O 2-8 / 1 mmole I 2 ) = 0.70 mmole # mmole S 2 O 8 2- remaining = ( 0.15M ) x ( 0.005 L ) x ( 1000 mmole / 1 mole ) - 0.7mmole = 0.75 mmole - 0.70 mmole = 0.05 mmole [S 2 O 8 2- ] remaining = 0.05 mmole / 18 ml = 0.00278 ln ( [S 2 O 8 2- ] ) remaining = ln ( 0.00278 ) = -5.886 Thus, by arranging for the reaction to run under Pseudo-First Order conditions, and by varying the Clocking Reagent (Thiosulfate Ion) concentration, we can generate the needed data for a determination of the Rate Constant k for (Eq. 1).
P a g e 7 Pre-Lab Questions 1. The dimerization of Cyclopentadiene (C 5 H 6 ) in Benzene at 25.1 o C: 2 C 5 H 6 C 10 H 12 has been followed by A. Wasserman, J. Amer. Chem. Soc. (1936), 1028. His data yields: Time [min] [C 5 H 6 ] (M) 0 1.358 1600 1.177 2910 1.077 4650 0.977 9060 0.792 14,370 0.617 21,460 0.470 This reaction follows 2 nd Order kinetics: Rate = k [C 5 H 6 ] 2 Make an appropriate plot, using a software package like Excel, and determine the Rate Constant k. 2. The oxidation of H 2 S in chlorinated water proceeds via: Cl 2 (aq) + H 2 S(aq) S(s) + 2 HCl(aq) This is first order in each reactant. The second order rate constant at 28 o C is 3.5x10-2 M -1 sec -1. If we start with concentrations of each reactant at [Cl 2 ] o = 0.05M and [H 2 S] o = 5.0x10-5 M, what is the value of the Pseudo-First Order Rate Constant?
P a g e 8 Procedure 1. Obtain 4 small beakers and label them: Label KI Na 2 S 2 O 3 K 2 S 2 O 8 Starch Water Amount ~40 ml ~40 ml ~40 ml ~10 ml ~40 ml Obtain the indicated amount of each solution. 2. Place six 15 cm test tubes in a test tube rack. Label them 1-6. Add, by means of a Mohr pipet, the amounts of the reagents listed: ml Water ml Starch Tube # ml 1M KI ml of 0.2M Na 2 S 2 O 3 1 5 7.0 0 1 2 5 6.5 0.5 1 3 5 6.0 1.0 1 4 5 5.0 2.0 1 5 5 4.0 3.0 1 6 5 2.0 5.0 1 3. To tube #1, add 5.0 ml of 0.15 M Potassium Peroxysulfate (S 2 O 8 2- ) by pipet. Record the time to the nearest second when addition is begun. As soon as delivery is complete, stopper the test tube and INVERT the test tube at least 5 times. (If the system is not mixed well, only part of the solution will clock and you will have to repeat that run.) Place this in a large beaker filled with Room Temperature water. 4. Repeat this for test tubes #2 - #6. Start each reaction no more than one minute after the preceding one. 5. Record the exact time when the blue color appears; i.e., the Time at which the reaction Clocks. 6. Record the temperature of the Water bath.
P a g e 9 Data Analysis 1. Prepare a Table of the following form and fill-in the needed Concentration values. Tube # 1 ml Na 2 S 2 O 3 mmoles S 2 O 3 2- added mmoles S 2 O 8 2- consumed mmoles S 2 O 8 2- remaining [S 2 O 8 2- ] remaining ln [S 2 O 8 2- ] remaining elapsed time 2 3 4 5 6 2. Prepare a graph of ln [S 2 O 8 2- ] vs. Time using a software package such as Excel. Use the software package s Linear Least Squares analysis routine to determine the Slope of the line. 3. Determine the Pseudo-First Order Rate Constant k for the Peroxysulfate oxidation of Iodide. 4. Calculate the Rate Constant k for the Peroxysulfate oxidation of Iodide; see (Eq. 19). Note: You must provide appropriate Units of Measurement for k and k.
P a g e 10 Post Lab Questions 1. Why is it important that our test tubes be placed in a Water bath? Be thorough in your answer. 2. In theory, we could perform a single run for this experiment and use the equation: ln([s 2 O 8 2- ]) = ln([s 2 O 8 2- ] o ) - k' t to determine k', and hence k. Why have we gone to the trouble of performing this experiment several times, preparing a graph of the results, and determining the slope of the graph in order to determine k. Be specific. 3. The chemical reaction of (Eq. 1) is written in Net Ionic form. If Potassium Ion (K + ) is the only Spectator Ion present, write the Total Molecular Equation for this reaction. 4. Write the Lewis Structure for the Thiosulfate Ion (S 2 O 3 2- ) used in the Clocking reaction (Eq. 2). Hint: Thiosulfate is the thiol (Sulfur based) analog of Sulfate (SO 4 2- ). So, start by working out the Lewis Structure for the Sulfate Ion. 5. The Clocking reaction (Eq. 2) results in the Reduction of I 2 to I -. Show the Sulfur is oxidized by explicitly assigning Oxidation Sates to the S in each relevant species.