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EXPERIMENT B3: CHEMICAL KINETICS
Learning Outcomes
Upon completion of this lab, the student will be able to:
- Measure the rate of a chemical reaction and determine the rate law.
- Evaluate the effect of temperature on the rate of a chemical reaction and determine the activation energy.
- Assess the role of a catalyst on the rate of a chemical reaction.
Introduction
Chemical kinetics is the study of speed of chemical reactions. The rate law is a mathematical expression that indicates the relationship between the concentrations of the reactants and the rate of the reaction. The rate of a reaction may be expressed in two forms: 1) the change in concentration of products over time or 2) the change in concentration of reactants over time. Consider the following chemical reaction: aA(aq) + bB(aq)! cC(aq) + dD(aq) Equation 1 In the above equation, a, b, c, and d are stoichiometric coefficients of the reactants and products. The rate of the reaction can be expressed as follows: € Reaction Rate = −
a
Δ[ A ]
Δ t
b
Δ[ B ]
Δ t
c
Δ[ C ]
Δ t
d
Δ[ D ]
Δ t Equation 2 In the expression for Reaction Rate, Δt is a unit of time during which the changes in concentrations of the reactants or products are measured. The unit for Reaction Rate is therefore molarity per time and is Ms-^1 or Mmin-^1. The rate law for this reaction is written as: Rate Law: Reaction Rate = € k × [ A ] x × [ B ] y Equation 3 In the Rate Law expression, “k” is referred to as the rate constant, a proportionality constant. “x” is referred to as the “order” of the reaction with respect to the reactant A and “y” is referred to as the “order” of the reaction with respect to the reactant B. The overall order of the reaction is given by x + y. Most chemical reactions are zero order, first order, or second order. The order of a chemical reaction and the order with respect to each individual reactant, can only be determined experimentally and is unrelated to the stoichiometric coefficients of the balanced chemical equation.
Therefore, for the reaction in Equation 1, “x” and “y” are unrelated to “a” and “b” and “x” and “y” can only be determined experimentally. If the values of “x” and/or “y” happen to be the same as “a” and/or “b”, it is purely a coincidence. The order of a reaction is always an integer or half-integer. Values of reaction order larger than three are almost impossible due to the extremely low probability collisions that would be necessary to result in such an order. If the experimental value of “x” in Equation 3 is 0, then it implies that the rate of the reaction is independent of the concentration of A. If the experimental value of “x” in Equation 3 is 1, then it implies that the rate of the reaction is proportional to the first power of the concentration of A. An important piece of information obtained from the Rate Law is the unit of the rate constant “k”. The unit of the rate constant depends on the order of the reaction. For instance, suppose for a particular reaction x = 1 and y = 2, the unit of “k” may be derived by dimensional analysis as follows: € Reaction Rate = k × [ A ]^1 × [B]^2 k = Reaction Rate [ A ] 1 × [ B ] 2 Unit of k = Ms −^1 M 1 M
2 =^ M^
− (^2) s − 1 Experimental Determination of Order of a Reaction In order to determine the order of a reaction with respect to each reagent, the reaction rate must be determined using various concentrations of the reactants, A and B. The experiment must be designed such that the rate of the reaction is measured by varying the concentration of A while keeping the concentration B a constant, and vice versa. Consider the following data: Experiment [A], M [B], M Rate, Ms-^1 1 0.1 0.1 0. 2 0.2 0.1 0. 3 0.1 0.2 0. Equation 3 gives the rate law expression: Reaction Rate = € k × [ A ] x × [ B ] y . Substituting the experimental data into the expression for the rate law results in the following three equations:
Equation 4 is rewritten as: 0.1 Ms-^1 = k[0.1]^1 [0.1]^2 Therefore k = 100 M-^2 s-^1 Equation 5 is rewritten as: 0.2 Ms-^1 = k[0.2]^1 [0.1]^2 Therefore k = 100 M-^2 s-^1 Equation 6 is rewritten as: 0.4 Ms-^1 = k[0.1]^1 [0.2]^2 Therefore k = 100 M-^2 s-^1 The average value of the rate constant k = 100 M-^2 s-^1 Therefore the Rate Law is: Reaction Rate = 100 M-^2 s-^1 [A]^1 [B]^2 Effect of Temperature on the Rate of a Reaction When reactions are conducted at elevated temperatures, the number of collisions of the reacting molecules increases due to the fact that the kinetic energy of the reacting molecules has increased. As a result, the rate of the reaction increases when the temperature is increased. The effect of temperature on the rate of a reaction is given by the Arrhenius equation. € k = Ae − Ea RT (^) Equation 7 In Equation 7: k is the rate constant A is the Arrhenius constant Ea is the activation energy of the reaction R is the universal gas constant (8.314 J/molK) T is the absolute temperature The activation energy, Ea, can be determined graphically by measuring the rate constant, k, and different temperatures. The mathematical manipulation of Equation 7 leading to the determination of the activation energy is shown below. € k = Ae − Ea RT ln k = ln A − Ea RT ln k = − Ea R
T
- ln A Therefore, a plot of the natural logarithm of the rate constant (ln k on the y-axis) vs. reciprocal of the absolute temperature (1/T) should yield a straight line whose
slope will be €
Ea R
. Since, R is the universal gas constant whose value is known (8.314 J/mol.K), the activation energy Ea can be calculated. Effect of Catalyst on the Rate of a Reaction Catalysts are chemical substances that increase the rate of a chemical reaction. A catalyst is not chemically modified as a result of the reaction and is recovered completely at the end of the reaction. Only a small amount of the catalyst is required for the rate enhancement. Metals and concentrated acids are examples of some common catalysts. The rate of a reaction depends on the activation energy of the reaction. The activation energy may be thought of as an energy barrier that reactants must cross prior to being transformed into products. Consider the exothermic transformation of reactant A to product B, in a single mechanistic step, whose energy diagram is shown in Figure 1. FIGURE 1 As shown in Figure 1, the transformation of A to B is a one-step mechanism that passes through a single transition state with an activation energy of Ea. The catalyst alters the reaction mechanism. The mechanism of the reaction in the presence of a catalyst, say X, will have more than a single step, but with lower activation energy. A possible reaction mechanism is shown below. Energy Reaction Progress Reactant, A Product, B Transition State Activation Energy, Ea
By providing an alternate path for the reaction, the catalyst, like high temperature, also increases the number of collisions of the reacting molecules. As a result the rate constant k, for the reaction is also altered in the presence of the catalyst. The effect of the catalyst on the rate constant of a reaction will be explored in this experiment.
Experimental Design
The kinetics of the reaction between iodide (I−) and peroxydisulfate (S 2 O 82 - ) will be explored in this experiment. The reaction is also referred to as the “Iodine Clock Reaction”. The chemical reaction between these two substances is shown in Equation 8 below. 3I−(aq) + S 2 O 82 - (aq) € ⇔ I 3 −(aq) + 2SO 42 - (aq) Equation 8 Both the reactants in Equation 8 are colorless aqueous solutions. In order to measure the rate of the reaction, further manipulations are necessary. A fixed amount of thiosulfate, S 2 O 32 -^ and starch are added to the reaction mixture. The thiosulfate, reacts with the triiodide (I 3 −) produced in Equation 8 according to the following chemical reaction. I 3 −(aq) + 2S 2 O 32 - (aq)! 3I−(aq) + S 4 O 62 - (aq) Equation 9 Once all the thiosulfate, S 2 O 32 - , is consumed, the chemical reaction shown in Equation 9 will stop. At this point, the triiodide, I 3 −^ will combine with the starch and form a complex that is dark blue/purple in color. I 3 −(aq) + starch! I 3 −-starch Equation 10 The rate of a reaction may be measured by two different methods:
- The amount of time it takes for a certain amount of reactant to be consumed.
- The amount of time it takes for a certain amount of product to be formed. In this experiment, the amount of time taken for a fixed amount of one of the products, I 3 −, to form is measured indirectly. Due to the fact that a constant amount of thiosulfate, S 2 O 32 - , is added to all the reactions, each reaction is then carried out until a constant amount of I 3 −^ is formed. As soon as this amount of I 3 −^ is formed, all the thiosulfate, S 2 O 32 - , is consumed (as per Equation 9), and the I 3 −^ formed thereafter forms a complex with the starch (as per Equation 10) and gives the reaction mixture the dark blue/purple color. Therefore, by measuring the time it takes for the appearance of the dark blue/purple color, the time it takes for a fixed amount of I 3 −^ to form is being measured. The molarity of the thiosulfate and the volume of thiosulfate used in this experiment are respectively, 0.012 M and 0.20 mL. The total volume of the reaction mixture is 1.9 mL. According to Equation 9, 1 mole of I 3 −^ reacts with 2 moles of S 2 O 32 -.
Reagents and Supplies
0.2% starch, 0.012 M Na 2 S 2 O 3 , 0.20 M KI, 0.20 M KNO 3 , 0.20 M (NH 4 ) 2 S 2 O 8 , 0.20 M (NH 4 ) 2 SO 4 , 0.0020 M Cu(NO 3 ) 2 Hot plate, thermometer, stopwatch (See posted Material Safety Data Sheets) NOTE: Even though the reaction mixture only consists of 0.2% starch, 0.012 M Na 2 S 2 O 3 , 0.20 M KI, and 0.20 M (NH 4 ) 2 S 2 O 8 , the following two reagents- 0.20 M (NH 4 ) 2 SO 4 and 0.20 M KNO 3 are also needed in order to maintain a constant ionic strength in all the reaction mixtures.
Procedure
PART 1: DETERMINATION OF ORDER OF THE REACTION WITH RESPECT TO I−
- Obtain four test tubes and label them as 1A, 2A, 3A, and 4A.
- Obtain four test tubes and label them as 1B, 2B, 3B, and 4B.
- Add the following volumes of the indicated reagents to test tubes labeled 1A, 2A, 3A, and 4A (All volumes are in mL). Reaction Test Tube
Starch
0.012 M
Na 2 S 2 O 3
0.20 M
KI
0.20 M
KNO 3
1 1A 0.10 0.20 0.80 0.
2 2A 0.10 0.20 0.40 0.
3 3A 0.10 0.20 0.20 0.
4 4A 0.10 0.20 0.10 0.
- Add the following volumes of the indicated reagents to test tubes labeled 1B, 2B, 3B, and 4B (All volumes are in mL). Reaction Test Tube
0.20 M
(NH) 2 S 2 O 8
0.20 M
(NH 4 ) 2 SO 4
1 1B 0.40 0.
2 2B 0.40 0.
3 3B 0.40 0.
4 4B 0.40 0.
- Prepare to start the stopwatch to record the time of the reaction. Press “start” as soon as the reagents have been combined as described below.
- Combine contents of test tube 1A and 1B. Start the stopwatch. Thoroughly mix the contents of the combined solution by transferring it back and forth between the two test tubes.
- When the dark blue/purple color forms, press the “stop” button on the stopwatch and the record the time in seconds in the data table.
- Repeat steps 6 and 7 with each of the other pairs of test tubes (combine 2A with 2B, 3A with 3B, and 4A with 4B).
- Empty the reaction mixtures into a large beaker labeled as “waste”. Empty the contents of the waste into a waste disposal container provided by the instructor.
- Perform a second trial of each reaction ensuring that the times are within 10% of each other.
PART 3: DETERMINATION OF ACTIVATION ENERGY OF THE REACTION
- Obtain four test tubes and label them as 9A, 10A, 11A, and 12A.
- Obtain four test tubes and label them as 9B, 10B, 11B, and 12B.
- Add the following volumes of the indicated reagents to test tubes labeled 9A, 10A, 11A, and 12A (All volumes are in mL). Reaction Test Tube 0.2 % Starch
0.012 M
Na 2 S 2 O 3
0.20 M
KI
0.20 M
KNO 3
Approximate temperature 9 9 A 0.10 0.20 0.40 0.40 10 °C 10 10 A 0.10 0.20 0.40 0.40 20 °C 11 11 A 0.10 0.20 0.40 0.40 30 °C 12 12 A 0.10 0.20 0.40 0.40 40 °C
- Add the following volumes of the indicated reagents to test tubes labeled 9B, 10B, 11B, and 12B (All volumes are in mL). Reaction Test Tube 0.20 M (NH) 2 S 2 O 8
0.20 M
(NH 4 ) 2 SO 4
Approximate temperature 9 9 B 0.20 0.60 10 °C 10 10 B 0.20 0.60 20 °C 11 11 B 0.20 0.60 30 °C 12 12 B 0.20 0.60 40 °C
- Prepare an ice bath to a temperature of around 10°C.
- Prepare to start the stopwatch to record the time of the reaction. Press “start” as soon as the reagents have been combined as described below.
- Combine contents of test tube 9A and 9B. Start the stopwatch. Insert the reaction mixture into the ice bath and constantly mix the contents.
- When the dark blue/purple color forms, press the “stop” button on the stopwatch and the record the time in seconds in the data table. Record the temperature of the reaction mixture.
- Combine test tubes 10A and 10B at room temperature and record the time for the reaction and the temperature of the reaction mixture.
- Prepare a hot water bath to a temperature of around 30°C.
- Combine contents of test tube 11A and 11B. Start the stopwatch. Insert the reaction mixture into the hot water bath and constantly mix the contents.
- When the dark blue/purple color forms, press the “stop” button on the stopwatch and the record the time in seconds in the data table. Record the temperature of the reaction mixture.
- Prepare a hot water bath to a temperature of around 40°C.
- Combine contents of test tube 12A and 12B. Start the stopwatch. Insert the reaction mixture into the hot water bath and constantly mix the contents.
- When the dark blue/purple color forms, press the “stop” button on the stopwatch and the record the time in seconds in the data table. Record the temperature of the reaction mixture.
- Empty the reaction mixtures into a large beaker labeled as “waste”. Empty the contents of the waste into a waste disposal container provided by the instructor.
- Perform a second trial of each reaction ensuring that the times are within 10% of each other and the temperatures are the same.
Data Table
PART 1: DETERMINATION OF ORDER OF THE REACTION WITH RESPECT TO I−
Reaction Time in seconds (Trial 1) Time in seconds (Trial 2) 1 2 3 4 PART 2: DETERMINATION OF ORDER OF THE REACTION WITH RESPECT TO S 2 O 82 − Reaction Time in seconds (Trial 1) Time in seconds (Trial 2) 5 6 7 8
PART 3: DETERMINATION OF ACTIVATION ENERGY OF THE REACTION
Reaction Time in seconds (Trial 1) Temperature, °C (Trial 1) Time in seconds (Trial 2) Temperature, °C (Trial 2) 9 10 11 12 PART 4: STUDY THE EFFECT OF A CATALYST ON THE RATE OF THE REACTION Reaction Time in seconds (Trial 1) Time in seconds (Trial 2) 13 14 15 16
- Enter the information from steps 1 through 4 in the following table. Reaction [I−], M [S 2 O 82 - ], M Average time (s) Reaction Rate Ms-^1 1 0. 2 3 4
- Rate Law: Reaction Rate = k × [I−]x^ × [S 2 O 82 - ]y. Substitute the information for Reactions 1-4 from above table into the rate law expression and solve for “x”. Reaction 1 Reaction 2 Reaction 3 Reaction 4
The order of the reaction with respect to I−^ =
PART 2: DETERMINATION OF ORDER OF THE REACTION WITH RESPECT TO S 2 O 82 −
- Calculate the molarity of the S 2 O 82 -^ in the reaction mixture for each reaction. Example: Reaction 5- M 1 = 0.20 M V 1 = 0.80 mL M 2 =? V 2 = 1.9 mL €
M 1 V 1 = M 2 V 2
M 2 =
0.20 M × 0.80 mL 1.9 mL
= 0.084 M
Reaction 6- Reaction 7- Reaction 8-
- Calculate the molarity of the I−^ in the reaction mixture for each reaction. Reactions 5- 8
- Calculate the average reaction time in seconds for each of the reactions.
- Calculate the average rate of reaction for each of the reactions. Reaction Rate = €
Δ[ I 3 −^ ]
Δ t
0.00063 M
Δ t
- Enter the information from steps 1 through 4 in the following table.
