Concentrations of stock solutions: Name: __________________________________ Section: ___ CHE 112 - Kinetics A.) Determining the Rate Law Varying KIO3 [ ]0 [H3O+]0 [I-]0 [H3AsO3]0 Time (s) Average Time...

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Concentrations of stock solutions: Name: __________________________________ Section: ___ CHE 112 - Kinetics A.) Determining the Rate Law Varying KIO3 [ ]0 [H3O+]0 [I-]0 [H3AsO3]0 Time (s) Average Time (s) rate (M s-1) 0.0015 0.0015 Varying KI [ ]0 [H3O+]0 [I-]0 [H3AsO3]0 Time (s) Average Time (s) rate (M s-1) 0.0015 0.0015 Varying pH [ ]0 [H3O+]0 [I-]0 [H3AsO3]0 Time (s) Average Time (s) rate (M s-1) 0.0015 0.0015 *All concentrations in M Show sample calculations in the space below Write the rate law: Name: __________________________________ Section: ___ CHE 112 - Kinetics B.) Temperature Effects Initial concentrations used for each trial [ ]0 _____________ [H3O +]0 _____________ [I -]0 _____________ [H3AsO3]0 ___0.0015 M_____ Temp (oC) Temp (K) 1/T (K-1) Time (s) Rate (M s-1) K (units ?) ln(k) * Show sample calculations in the space below Name: __________________________________ Section: ___ CHE 112 - Kinetics Create a graph of ln(k) vs 1/T in Excel and paste it in the space below Calculate the Activation Energy Name: __________________________________ Section: ___ CHE 112 - Kinetics Questions 1.) Determine the mass of potassium iodate needed to prepare a 0.010 M solution with a volume of 50.00 mL. 2.) What is required for an effective collision to occur? 3.) Explain the relationship between concentration and reaction rate. 4.) When doe the color change occur in this experiment? a.) When all of the KIO3 reacts b.) When all of the KI reacts c.) When all of the H3AsO3 reacts d.) When all of the reactants have reacted 5.) What is the difference between an average rate and an instantaneous rate? What did you calculate in this lab? Name: __________________________________ Section: ___ CHE 112 - Kinetics 6.) What is the overall reaction order for the experiment? 7.) What are the units of the rate constant? 8.) Explain why reaction rate increases as temperature increases. Iodine Clock Kinetics Thomas M. Moffett Jr., SUNY Plattsburgh, 2020 In this experiment you will study the iodine clock reaction to quantitatively determine the rate law and the activation energy for the following reaction. 6 H3O + (aq) + IO3 − (aq) + 8 I- (aq) → 3 I3 − (aq) + 9 H2O (l) There are several different iodine clock reactions. In this reaction the iodate (IO3 −) is oxidized into triiodide (I3 −). The triiodide forms a dark blue complex with starch. The rate law for the reaction can be written as shown below. rate = k[H3O +]x[IO3 −]y[I-]z If the solutions were mixed as is, the triiodide concentration and color would gradually increase as the reaction progresses. Instead of this, we want a sharp change in color. To accomplish this, we can add a reducing agent (arsenious acid, H3AsO3) which will quickly reduce the triiodide to iodide (I-). H3AsO3 (aq) + I3 − (aq) + 5 H2O (l) → HAsO4 2− + 3 I- (aq) + 4 H3O + This reaction happens quickly, and keeps the triiodide from reacting with the starch until all of the arsenious acid is consumed. Once that happens the triiodide will quickly bind with the starch resulting in a fast color change from clear to dark blue. This means that the color change occurs when all of the reducing agent is gone, not when the iodate or iodide is completely consumed. To perform the experiment two solutions are mixed. In one solution potassium iodate, starch, arsenious acid, and a buffer to maintain hydronium concentration are mixed together. The other solution is potassium iodide. The solutions are then mixed together and the reaction is timed until there is a color change. The rate of reaction comes from the disappearance of the arsenious acid. rate = − ∆[H3AsO3] ∆t You will simulate this reaction using a computer program. In the experiment, you will vary the concentrations of the reactants and measure how long it takes for the reaction to occur. You will use this data to determine the reaction order of each chemical in the rate law. To do so, you will compare two reactions where only one chemical is changed, the change in rate is then due solely to the change in concentration of that chemical. Example: A (aq) + B (aq) → AB (aq) rate = k[A]x[B]y Experiment [A]0 (M) [B]0 (M) Rate (M s-1) 1 0.20 0.30 5.0 2 0.20 0.60 10 3 0.40 0.30 20 To find the reaction order for A, we need to find two reactions where only the concentration of A is varied. In this example, we would use experiments 1 and 3. We would then set up a ratio comparing concentration of A to rate. ( [A]3 [A]1 ) x = rate3 rate1 ( 0.40 0.20 ) x = 20 5 2x = 4 x = 2 the reaction is second order for A We would then use experiments 1 and 2 to determine the rate order of B. ( [B]2 [B]1 ) y = rate2 rate1 ( 0.60 0.30 ) y = 10 5 2y = 2 y = 1 the reaction is first order for B Now that both reaction orders are known, we can write the rate law. rate = k[A]2[B] By entering data from one of the experiments into the rate law, we can calculate the value of the rate constant. k = rate [A]2[B] = 5.0 mol L s (0.20 mol L ) 2 (0.30 mol L ) = 420 L2 mol2s Activation energy can be determined by measuring the reaction rate at various temperatures. The Arrhenius equation relates activation energy to the rate constant. RT/-Eae k = FIGURE 2 - MIXING FIGURE 1 - SOLUTION PREPARATION In the above equation, Ea is the activation energy, k is the rate constant, T is the temperature in Kelvin, and R is the gas constant (8.314 J/mol K). The equation can be rewritten in the form of a straight line by taking the natural log of both sides. ??(?) = ??(?) − ?? ? ( 1 ? ) By graphing ln(k) (y-axis) vs 1/T (x-axis) you can create linear plot. The slope of the line is equal to -Ea/R. In the second part of the experiment you will measure the reaction rate at various temperature while maintaining constant initial concentrations. Procedure: The experiment will be done using the simulation available at: http://web.mst.edu/~gbert/IClock/Clock.html The table on the left side of the computer window (figure 1) is used to prepare solutions. You can change the concentrations of the KIO3, H3AsO3, KI, and the pH of the buffer. The buffer will change the concentration of H3O + (listed as H+ on the website). You can also change the temperature. 1. Leave the temperate at 25 o C for all of part A. 2. Set up the reaction with the desired concentrations. 3. Click on mix the solutions, this will bring up a new screen (figure 2). 4. Click on start, this will mix the solutions and stat the timer. Click on stop timer when you see a color change. 5. To change the concentrations, click on prepare solutions. A. Determining the Rate Law 1.) Leave the temperature at 25 oC for all of part A. http://web.mst.edu/~gbert/IClock/Clock.html FIGURE 3 - INITIAL CONCENTRATIONS 2.) Run two trials where only the concentration of KIO3 is changed. Run the simulation, and record the initial concentrations and time for the reaction. You should run several simulations of each mixture (just click on Reset, and then start to rerun a sample.). You want to select a set of conditions that does not have a very short reaction time (1 or 2 seconds). Initial concentrations are listed in the window on the right side of the computer screen (Figure 3). The initial concentrations are half of what you selected in the solution prep because the solutions dilute each other when mixed. 3.) Run two trials where only the concentration of KI is varied. Record all of the relevant data. 4.) Run two trials where only the pH of the buffer is varied. Record all of the relevant data. B. Determining the Activation Energy 1.) In this part, you will not change concentrations, only temperature. 2.) Run the simulation at each temperature. Again, you should complete several trials each time. Data Analysis Part A • Determine the rate of reaction for the disappearance of H3AsO3 rate = − ∆[H3AsO3] ∆t * Δ[H3AsO3] = [H3AsO3]t - [H3AsO3]0 * [H3AsO3]t = 0 M * Δt = average reaction time • Determine the rate order for each reactant by comparing the rates of two reactions where only one component is varied. (round rate orders to the nearest whole number) ( [reactant]2 [reactant]1 ) x = rate2 rate1 • Write the rate law for the reaction: rate = k[H3O +]x[IO3 −]y[I-]z Part B • Calculate the rate of reaction for the disappearance of H3AsO3 • Use the rate law that you determined in part A to calculate the value of the rate constant at each temperature. • Take the natural log of the rate constant • Create a graph of ln(k) vs 1/T (temperature must be in Kelvin). • Calculate the activation energy (slope = -Ea/R) Concentrations of stock solutions: Name: __________________________________ Section: ___ CHE 112 - Kinetics A.) Determining the Rate Law Varying KIO3 [ ]0 [H3O+]0 [I-]0 [H3AsO3]0 Time (s) Average Time (s) rate (M s-1) 0.0015 0.0015 Varying KI [ ]0 [H3O+]0 [I-]0 [H3AsO3]0 Time (s) Average Time (s) rate (M s-1) 0.0015 0.0015 Varying pH [ ]0 [H3O+]0 [I-]0 [H3AsO3]0 Time (s) Average Time (s) rate (M s-1) 0.0015 0.0015 *All concentrations in
May 04, 2021
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