OER Lab 5B- Iodine - Clock Reaction-2

CHEM 1412 Lab – 5 1 Kinetics Measuring the Rate of a Chemical Reaction Please Watch This Video: https://www.youtube.com/watch?v=TdXamAGRHe4 Objectives: List factors that affect reaction rate. ✓ Be able to plot data to determine rate of the reaction. ✓ Determine the rate order for each reactant and the rate constant for the reaction. ✓ Determine the overall order of the reaction. ✓ Write the rate law of the reaction. Reagents: 1. 0.01 M KI 2. 0001 M Na 2 S 2 O 3 3. 0.04 M KBrO 3 4. 0.10 M HCl 5. Distilled water 6. 1% starch Equipment and Materials: 10.0 mL graduated cylinders (5) or pipettes w/extractor. 250 mL Erlenmeyer flasks (2) 250 mL beakers (2) One 50mL beaker Disposable dropper Digital timer or stopwatch Label tape Safety: Take caution when handling all reagents, particularly iodine, potassium bromate, and hydrochloric acid. Wearing gloves is highly recommended. Rinse and wash any skin that comes in contact with the reagents. Waste Disposal: All reagents should be discarded in the appropriate waste container, particularly for halogenated waste. Consult your lab instructor and dispose as directed in your lab.

CHEM 1412 Lab – 5 2 Theoretical Background: Kinetics of a reaction tell us how quickly it is occurring or what the rate of the reaction is proceeding to completion. Rate is defined as; • the change in concentration of the reactants (a decline or - rate) per unit of time or • the change in concentration of the products (an increase or + rate) per unit of time. For a reaction of A ➔ B where A is the reactant and B is the product… Rate is typically a numerical value with units but can also be written as a rate expression as follows: • the decline of reactant (A) ➔ ???? ?? ? = − ∆[ ? ] / ∆ ? • the increase of product (B) ➔ ???? ?? ? = + ∆[ ? ] / ∆ ? When concentration versus time is plotted graphically, the slope of the graph is the rate of the reaction. Typically, units are M/s. Note that a declining rate or slope is denoted with a negative sign while a positive rate or slope is increasing. If a chemical equation contains coefficients, they will affect the rate expression as follows: x A + y B → z C Rate Expression: − 1∆ [ ? ]/ ? ∆ ? = − 1∆ [ ? ] / ? ∆ ? = + 1 ∆ [ ? ]/ ? ∆ ? Types of Rates: Instantaneous rate is calculated for one specific data point of time and corresponding concentration; gives a snapshot glimpse of an “instant” in a reaction. Average rate is calculated at a variety of time intervals (and corresponding concentrations) throughout the reaction and gives an overall impression of rate in the reaction. If the data below for the decomposition of hydrogen peroxide was graphed, both the average and instantaneous rates could be calculated from the slope depending on the data points selected. Note the table and figure calculated from the slope depending on the data points selected. Note the table and figure below.

CHEM 1412 Lab – 5 4 • Catalyst: the presence of a catalyst increases the rate of a reaction by decreasing the activation energy, the minimum amount of energy required for the reaction to proceed. Catalysts create the new pathways or mechanisms for collisions to occur so that they occur earlier at lower energy level s, thus speeding up the reaction. Think of them like a chemical “shortcut.” Enzymes are common biological catalysts, but they can occur in many forms and physical states. Rate Laws: In contrast to the rate expression, the rate law only focuses on the reactants (and not products as well). The coefficients of reactants are also arbitrary. A rate law states the relationship of the rate as proportional to • the rate constant, k • the concentrations of the reactants in molar units (M) as raised to the power of their reaction order or exponents. See the reaction below and its corresponding rate law. a A + b B → c C + d D ➔ ???? ??? = ? [ ? ] ? [ ? ] ? The rate law utilizes orders of reaction, which are the exponents of each reactant concentration. Orders of reaction reflect the exponential relationship that explains how the change in concentration of a reactant can affect the rate. Orders may be zero, first, or second and are mathematically determined from data points of concentration and rate. To do so, pick two data points, ensuring that one reactant amount changes but the other stays constant. That is the only way to determine how the rate is affected only by that reactant. Then determine by what exponential factor the reactant’s concentration cha nge affected the rate as a result. Consider the following reaction: S 2 O 8 2 – (aq) + 2I – (aq) → I 2 (aq) + 2SO 4 2 – (aq) peroxydisulfate ion iodide ion iodine sulfate ion EXP # [ S 2 O 8 2 – ] [ I – ] Rate 1 0.16 0.24 1.10x10 -4 2 0.16 0.48 4.40x10 -4 3 0.32 0.24 2.20x10 -4

CHEM 1412 Lab – 5 5 Sample Problem: Calculate the order for S 2 O 8 2 – and I – using the experimental data. Step 1 : Select two data points for S 2 O 8 2 – in which [ S 2 O 8 2 – ] is changing while [ I – ] remains constant. Only then can you attribute any changes in corresponding rates to be directly because of changing [ S 2 O 8 2 – ]. Here we will select Exp 1 and 3 . Note how [ S 2 O 8 2 – ] changes while [ I – ] remains constant. Step 2 : Write the rate laws for BOTH Exp 1 and 3 . Rate= k [ S 2 O 8 2 – ] x [ I – ] y Rate 1 : 1.10 x 10 -4 = k [0.16] x [0.24 ] y Rate 3 : 2.20 x 10 -4 = k [0.32] x [0.24 ] y Step 3 : Write the rate laws as ratio of each to determine proportions and thus order of S 2 O 8 2 ???? 3/ ???? 1= ? [ S 2 O 8 2 – ] ? [ I – ] ? / ? [ S 2 O 8 2 – ] ? [ I – ] ? (2.20 x 10 − 4 ) /(1.10 ? 10 − 4 ) = ? [0.32].[0.24] ? / ? [0.16].[0.24] ? Notice that k, [ I – ], and y cancel out leaving you with 2 = 2 ? ➔ ? = 1, order of S 2 O 8 2 – is 1 or first order . A theoretical interpretation concludes that as [ S 2 O 8 2 – ] doubles from 0.16M to 0.32M (and [ I – ] remains the same at 0.24M), the rate also doubles (2.20 x10 -4 vs 1.10 x10 -4 M/s). So, this one-to-one exponential relationship is indicative of a first order reaction. Repeat the same for [ I – ] using data from Exp 1 and 2 . ???? 2 / ???? 1 = ? [ S 2 O 8 2 – ] ? [ 𝐼 ] ? / ? [ S 2 O 8 2 – ] ? [ 𝐼 ] ? (4.40 x 10 − 4 ) /(1.10 ? 10 − 4 ) = ? [0.16] ? [0.48] ? / ? [0.16] ? [0.24] ? Notice that k, [ S 2 O 8 2 – ], and x cancel out leaving you with 4 = 2 ? ➔ y = 2, order of I – is 2 or second order . The theoretical interpretation notes that as [ I – ] doubles from 0.24M to 0.48M and [ S 2 O 8 2 – ] remains the same, the rate also quadruples (4.40 x10 -4 vs 1.10 x10 -4 M/s). This means that the effect was exponential to the power of two, indicative of a second order reaction. If there had been no correlation between a change in the reaction rate and a reactant concentration, such that rate was directly attributed to the k, this would indicate a zero-order reaction since any concentration value to the power of zero would be one. If Rate = k [A] 0 whereby [A] 0 = 1, then Rate= k. The overall order of the reaction would be the sum of each reactant order ➔ Overall order = x + y In the case of this reaction, the overall reaction order would be 3 . Once the orders of the reaction are determined, the rate constant, k, can be calculated using any data points because k is a “constant.” The rate law is simply rearranged to solve for k. ???? = ? [ S 2 O 8 2 – ] 1 [ I – ] 2 .

CHEM 1412 Lab – 5 7 The reaction, however, is quite slow so we will be running another reaction in tandem that is much faster and of course, related to our original reaction, because it will be consuming one of the reactants. This fast reaction is the one that we will time as our “clock reaction.” This is when a reducing agent called sodium thiosulfate (Na 2 S 2 O 3 ) enters the picture. 3 I 2 (aq) + 6 S 2 O 3 −2 (aq) ⟶ 6 I − (aq) + 3 S 4 O 2 −2 (aq) Notice that the thiosulfate ion is a strong reducer here and reduces molecular iodine back into its anion. When this occurs, the dark blue color no longer appears. When S 2 O 3 −2 is completely consumed, the solution color is dark blue. Therefore, we can use the appearance of the blue color complex to track the consumption of S 2 O 3 −2 ion. So how can we use all of this information to help us determine the rate of this reaction as well as the reaction orders for the three reactants and finally, the rate constant? Recall the rate can also be expressed as the negative or decline of any of the three reactants if the stoichiometry of the chemical equation is taken into consideration. However, since the observation of that is not practical due to the speed of the reaction, rather than calculate the rate of disappearance or consumption of BrO3-, we can instead calculate the rate of disappearance of the S 2 O 3 − 2 ion and then extrapolate that to the three reactants by looking at the stoichiometry of the balanced equation. ???? = = - ∆ [ ? 2 𝑂 3 2− ] / ∆ ?𝑖?? ( ? ) = - [ ? 2 𝑂 3 2− ] ? − [ ? 2 𝑂 3 2− ] 𝑖 /( ? ? − ? 𝑖 ) Relating this to the simplified chemical equations, it should be noted that for each bromate ion, reacted, 3 moles of molecular iodine (I 2 ) are formed. And for each 3 moles of molecular iodine, 6 moles of thiosulfate ion react (or a 2:1 in simplified form). Therefore, looking back at the rate expression, the rate of disappearance of the bromate ion is 1/6 that of the rate of the disappearance of the thiosulfate ion. 6 I − (aq) + 1BrO 3 − (aq) + 6 H+ (aq) ⟶ 3I 2 (aq) + Br − (aq) + 3 H 2 O (l) 3 I 2 (aq) + 6 S 2 O 3 − 2 (aq) ⟶ 6 I − (aq) + 3 S 4 O 2 − 2 (aq)  1 I 2 (aq) + 2 S 2 O 3 − 2 (aq) ⟶ 2 I − (aq) + 1 S 4 O 2 − 2 (aq) (simplified) Rate Expression: − (1/ ? ). ∆ [ ? ] / ∆ ? = − (1/ ? ) ∆ [ ? ] / ∆ ? Rate Expression BrO 3 − as related to S 2 O 3 − 2 : • Rate of BrO 3 − = − (Rate of S 2 O 3 − 2 ) ( − 1/3 ??? ??𝑂 3 − / ??? 𝐼 2 ). ( 1/2 ??? 𝐼 2 / ??? ? 2 𝑂 3 2− ) Thus, Rate [BrO 3 − ]= (1/6) Rate [S 2 O 3 −2 ]. You can do the same for the other reactants, I − and H + .

CHEM 1412 Lab – 5 8 When writing the rate law for this reaction, three reactants will be considered, each with its own reaction order, which can be determined mathematically from experimental data. Rate= k [I − ] x [BrO 3 − ]y [H + ] z To find to the initial concentrations of reagents, you will need to look at the table in the procedure that outlines the concentration and volume of each reagent used. You may utilize the M 1 V 1 = M 2 V 2 equation to help you determine the initial concentration or simply use the molarity and volume (equation below) to find the moles of each reagent and divide that by the total volume of the solution, which you will find to be 50mL or 0.050L. The latter method may be helpful in determining which mixtures to compare to determine the order for each reagent. The table in your report form should also help in organizing that data. Molarity reagent x Liters reagent used = mols reagent ? ?? ??????? 𝑖? ?????𝑖?? ?𝑖????? = ( ???? ??????? ) / ( 0.050 ? ?????𝑖?? ) Or use the same data and plug into M 1 V 1 = M 2 V 2 Again, we will use the S 2 O 3 −2 ion data to find the initial rate since we can calculate the initial concentration from the procedure, and we know the final concentration should always be zero (0) because the blue color only appears upon the full consumption of the thiosulfate ion and presence of I 2 forming a complex with the starch. From there, finding the rate of disappearance of the thiosulfate ion is simple, and the rates of the other reagents can be extrapolated.

CHEM 1412 Lab – 5 10 Procedure: Collect and prepare your glassware, ensuring it is clean prior to use. Take great caution in keeping track of your vessels so as not to cross contaminate your reagents. This can significantly interfere with your results. Label each vessel with the reagent name as you work, including your beakers and graduated cylinders. From the instructor’s lab bench or designated area, measure ~100m l of Na 2 S 2 O 3 , KI, KBrO 3 , and HCl in 4 clean 250mL beakers. Take these back to your work bench. You will use these as your “supply” for the experiment’s trials. You will also need a bottle of distilled water. You may premeasure ~5ml of starch in a small 50mL beaker or 10mL graduated cylinder if you wish or simply acquire the 3-5 drops as needed for each trial. You will be measuring reagents and making two solutions (Flask 1 and 2) for each reaction mixture and its trial(s). Label these as well. If you are working in a group, it may be prudent to assign tasks to each member to keep track of your steps. You will be conducting trials for four reaction mixtures in this experiment. These reaction mixtures will consist of varying the volume (and thus amount and initial concentration) of your reactants. This will of course influence the rate of the reaction. Refer to the table on the foll owing page for the “recipe” for each half of the reaction mixtures (i.e., Flask 1 and 2) which will tell you the amount of each reagent to use for each reaction mixture/trial. Measure and prepare Flask 1 and Flask 2 reagents, but do NOT add the two reaction mixtures. It is important to keep the two reaction mixtures separate until you are ready to time the reaction! Take note of the volumes and concentrations of each reagent used. You will need this information for your calculations. Also take note that that the final volume of all four trials is 50ml. Do not forget to add the starch indicator or you will not see the color change! FLASK 1 FLASK 2 Reaction Mixture 0.001 M Na 2 S 2 O 3 0.01 M KI H 2 O Starch 0.04 M KBrO 3 0.10 M HCl 1 10.0 ml 10.0 ml 10.0 ml 3-5 drops 10.0 ml 10.0 ml 2 10.0 ml 20.0 ml 0 ml 3-5 drops 10.0 ml 10.0 ml 3 10.0 ml 10.0 ml 0 ml 3-5 drops 20.0 ml 10.0 ml 4 10.0 ml 10.0 ml 0 ml 3-5 drops 10.0 ml 20.0 ml Note: 50ml total volume for all reaction mixtures Work on one trial at a time to avoid confusion and for safety. You will need to ensure that one member of your group is timing the reaction as soon as the flask contents are mixed . Familiarize yourself with the timer/stopwatch before starting. Once you have prepared Flask 1 and Flask 2 for the first reaction mixture’s trial 1, you may begin. You can also place a white piece of paper under flask 1 to better see the color change. When ready, add the solution in flask 2 to flask 1 and gently swirl . Time how long it takes for the solution to turn dark blue and record it. This may take anywhere from 1-3 minutes. Repeat this trial so that you can ensure your values are no greater than 10% apart and thus consistent. Note that you will need to convert the time values from minutes and seconds to seconds only. Dispose of all reagents as instructed by your instructor. You may wish to compare your times with other groups. Repeat these steps for remaining reaction mixtures 2-4 with one trial only if your times are like that of your classmates.

CHEM 1412 Lab – 5 11 Name : _______________________________ Lab Report: Timetable: Reaction Mixture Time in minutes (min) Time in seconds (sec) Average Time (sec) 1 Trial 1 Trial 2* 2 Trial 1 Trial 2* 3 Trial 1 Trial 2* 4 T rial 1 Trial 2* *If necessary Moles and Concentrations: Reaction Mixture S 2 O 3 − 2 BrO 3 − I − H + 1 Initial Moles of Reagent Initial Concentration (M) 2 Initial Moles of Reagent Initial Concentration (M) 3 Initial Moles of Reagent Initial Concentration (M) 4 Initial Moles of Reagent Initial Concentration (M)

CHEM 1412 Lab – 5 13 Show Orders Calculations: Reaction order for (x) I - : ______ Reaction order for (y) BrO 3 - : ______ Reaction order for (z) H + : _______ Show Rate Constant Calculations: Rate Constant Reaction Mixture 1: _____________ Rate Constant Reaction Mixture 2: _____________ Rate Constant Reaction Mixture 3: _____________ Rate Constant Reaction Mixture 4: _____________ Average Rate Constant (include units): ______________

CHEM 1412 Lab – 5 14 Post- laboratory Questions and Exercises Name _______________________________ Due after completing lab. Answer in the space provided. Date __________________ 1. List the four reagents in this experiment and explain their roles in the reaction. 2. How would the omission of starch from the reaction mixture affect the experiment? 3. What would be the rate constant for a first order reaction of 2H 2 O 2 (l) → 2H 2 O (l) + O 2 (g) if the concentration of hydrogen peroxide is 0.015M at a rate of 2.31 x 10-4 M/s? (Show calculation) 4. What is the overall order of the reaction with the rate law: Rate = k [I - ][S 2 O8 2 - ]? _______________ 5. Why does iodine clock reaction turn blue? 6. List three catalysts that can be used in reactions.