Ap Chemistry, Kinetics Lab Report 1or27

Kinetics of a Reaction

Sebasthian Santiago #12-173 AP Chemistry Prof. Martinez Partners: Yanira Rodríguez Viviana Berdecía Alberto Mulero

Purpose The main purpose of this experiment is to utilize a microscale technique to determine the total rate law, activation energy, and rate constant for the oxidation of iodide ions by bromate ions in the presence of acid: 6I-(aq) + BrO3-(aq) + 6H+ (aq)  3I2 (aq) + Br-(aq) + 3H2O (l) Other purposes include:  

To calculate and compare the rate constant the reaction at different temperatures To observe the effect of a catalyst on reaction rate

Hypothesis If the concentration of a reactant is increased, the rate of the reaction will increase. However, an increase in concentration of one reactant will not increase the rate to the same extent as an equal increase of concentration of another reactant. I believe that a change is the concentration of hydrochloric acid will affect the rate the most. Also, if the temperature in which a reaction takes place is increased, the rate of the reaction will increase.

Materials Chemicals:       

Copper (II) Nitrate Solution, Cu(NO3)2, 0.10 M, 5 mL Hydrochloric acid solution, HCl, .10 M, 5mL Potassium iodide solution, KI, 0.010 M, 5mL Potassium bromate solution, KBrO3, 0.040 M, 5mL Sodium thiosulfate solution, Na2S2O3, 0.0010 M, 5mL Starch solution, 2%, 5mL Water, distilled, deionized

Equipment:            

Balance, 0.001-g or 0.0001-g precision Beaker, 10-mL or 50-mL Beral-type pipet with microtip, 7 Cassette tape case Cotton swabs for cleaning reaction strip wells Label tape, for pipets Marking pen Reaction strip, 12-well Thermometer, 0°C – 100°C Timer, seconds Toothpicks for stirring Trough for hot and cold water baths, shared

Data Data Table 1. Find the Volume of One Drop of Solution Mas of empty beaker (a) Mass of beaker plus 5 drops of water (b) Trial 1 Mass of first 5 drops of water (b) – (a) Average mass of 1 drop of water Mass of beaker plus 10 drops of water (c) Trial 2 Mass of second 5 drops of water (c) – (b) Average mass of 1 drop of water Mass of beaker plus 15 drops of water (d) Trial 3 Mass of third 5 drops of water (d) – (c) Average mass of 1 drop of water Average mass of 1 drop of water (Trials 1 – 3)

66.819 g 67.036 g .217 g .043 g 71.236 g .0200 g .040 g 67.436 g .200 g .040 g .041 g

Data Table 2. Determine the Reaction Rate and Calculate the Rate Law

Experiment No. 1 2 3 4 5 6 7

Trial 1 165 84 66 77 56 42 21

Time, seconds Trial 2 Trial 3 172 86 58 96 63 39 17

Average

167 84 51 83 67 40 18

168 85 58 85 62 40 19

Temp. °C

Reaction Rate M/s

22 22 22 22 22 22 22

Data Table 3. Determine the Activation Energy

Approximate Temperature, °C 0 20 40

Measured Temperature, °C

Measured Temperature, K

2 22 41

275.15 295.15 314.15

Measured Temperature-1, K-1 3.63 x 10-3 3.39 x 10-3 3.18 x 10-3

Time of Reaction, seconds Trial Trial Average 1 2 Time 270 165 65

277 172 73

Data Table 4. Observe the Effect of a Catalyst on the Rate

Uncatalyzed Reaction Catalyzed Reaction

Reaction Time, seconds 160 57

274 169 69

Data Table 5.Reaction Rate of Each Experiment

Experiment 1 Experiment 2 Experiment 3 Experiment 4 Experiment 5 Experiment 6 Experiment 7

Reaction Rate, M/s 8.3 x 10-8 1.6 x 10-7 2.4 x 10-7 1.6 x 10-7 2.3 x 10-7 3.5 x 10-7 7.4 x 10-7

Data Table 6.Initial Concentration of Reactants

Experiment 1 Experiment 2 Experiment 3 Experiment 4 Experiment 5 Experiment 6 Experiment 7

Initial Concentrations, Moles/Liter [I-] [BrO3-] -3 1.7 x 10 6.7 x 10-3 -3 3.4 x 10 6.7 x 10-3 -3 5.0 x 10 6.7 x 10-3 1.7 x 10-3 1.3 x 10-2 -3 1.7 x 10 2.0 x 10-2 -3 1.7 x 10 6.7 x 10-3 -3 1.7 x 10 6.7 x 10-3

[H+] 1.7 x 10-2 1.7 x 10-2 1.7 x 10-2 1.7 x 10-2 1.7 x 10-2 3.3 x 10-2 5.0 x 10-2

Data Table 7.Rate Constants Experiment Value of k, M-3s-1

1 25

2 25

3 25

4 25

5 23

6 28

7 26

Data Table 8.Calculate the Activation Energy Measured Temperature, K Trial 1 Trial 2 Trial 3

275.15 295.15 314.15

Measured Temperature-1, K-1 3.63 x 10-3 3.39 x 10-3 3.18 x 10-3

Average Time, s 274 169 69

Rate of Reaction, M/s 5.1 x 10-8 8.3 x 10-7 2.0 x 10-7

Rate Constant, k, M-3s-1

ln k

15 25 61

2.7 3.2 4.1

Post-Lab Calculations and Analysis Part 1. Calculate the Volume of One Drop of Solution Assume the density of water to be 1.00 g/mL Volume of one drop=

=

= 4.1 x 10-5 L

Part 2A. Calculate the Rate Calculate the number of moles of S2O32- present in one drop: moles S2O32- ions= Volume of 1 drop moles S2O32- ions= 4.1 x 10-5 L moles S2O32- ions=4.1 x 10-8 moles Calculate the moles of BrO3- which react as all of the S2O32- ion is used up: moles BrO3- reacted= moles S2O3-2 moles BrO3-- reacted= 4.1 x 10-8 moles S2O3-2 moles BrO3- reacted= 6.8 x 10-9 moles Calculate the value of –Δ[BrO3-] The value of –Δ[BrO3-] in all reactions, since all experiments have a total volume of 12 drops, is: –Δ[BrO3-]= –Δ[BrO3-]= –Δ[BrO3-]= –Δ[BrO3-]= 1.4 x 10-5 moles/L –Δ[BrO3-]=1.4 x 10-5 M Calculate the rate of reaction in each experiment and enter the results into the following table. Use the average time for each experiment. The rate of each reaction can be found by dividing –Δ[BrO3-] by the number of seconds for the reaction take place. Rate=

–

Experiment 1: Rate=

–

Rate= Rate=8.3 x 10-8 M/s See Data Table 5 for the reaction rate of all other experiments.

Part 2B. Calculate Initial Concentrations Vconcentrated

Mconcentrated = Vdilute

Mdilute = (Vconcentrated

Mdilute

Mconcentrated)/ Vdilute

Since V ∝ number of drops, Vconcentrated= number of drops of reactant & Vdilute=12 drops [I-]=

= 0.0017 M

See Data Table 6 for the initial concentration of all other reactants. Part 2C.Calculate the Order of Each Reactant 1. Iodide ion: Experiment 1 vs. Experiment 2 Rate= k[I-]x[BrO3-]y[H+]z Experiment 1: Rate1 8.3 x 10-8 M/s= k[1.7 x 10-3 M]x[6.7 x 10-3 M]y[1.7 x 10-2 M]z Experiment 2: Rate2 1.6 x 10-7 M/s= k[3.4 x 10-3 M]x[6.7 x 10-3 M]y[1.7 x 10-2 M]z =

= =

.52=.50x log (.52)=x log(.50) x=

= .94

Experiment 1 vs. Experiment 3 Comparison of experiment 1 and 3 gives us x=.97 x=1.0

2. Bromate Ion Experiment 1 vs. Experiment 4 Rate= k[I-]x[BrO3-]y[H+]z Experiment 1: Rate1 8.3 x 10-8 M/s= k[1.7 x 10-3 M]x[6.7 x 10-3 M]y[1.7 x 10-2 M]z Experiment 4: Rate2 1.6 x 10-7 M/s= k[1.7 x 10-3 M]x[1.3 x 10-2 M]y[1.7 x 10-2 M]z =

= =

.52=.52y log (.52)=y log(.52) y=

=1

Experiment 1 vs. 5 Comparison of experiment 1 and 5 gives us y=.95 y=1.0 3. Hydron ion Experiment 1 vs. Experiment 6 Experiment 1: Rate1 8.3 x 10-8 M/s= k[1.7 x 10-3 M]x[6.7 x 10-3 M]y[1.7 x 10-2 M]z Experiment 6: Rate2 3.5 x 10-7 M/s= k[1.7 x 10-3 M]x[6.7 x 10-3 M]y[3.3 x 10-2 M]z =

= =

.21=.52z log (.24)=z log(.52) z=

= 2.2

Experiment 1 vs. 7 Comparison of experiment 1 and 7 gives us z=2.0 z=2.0

4. Write the experimentally determined rate law. Rate= k[I-][BrO3-][H+]2 Part 2D. Find the rate constant Experiment 1: Rate= k[I-][BrO3-][H+]2 k= k= k=25 M-3s-1 See Data Table 7 for the values of k for experiments 2 to 7. Average value of k= 25M-3s-1 Part 3.Calculate the Activation Energy, Ea 1. Using your data from Data Table 3, calculate the rate, the rate constant, and the natural logarithm of the rate constant. Trial 1: Rate=

–

Rate=

= 5.1 x 10-8 M/s

k= k= k= 15 M-3s-1 ln k= ln (15 M-3s-1)= 2.7 See Data Table 8 for the calculations of Trial 2 and 3.

2. Graph the data with natural logarithm of the rate constant, ln k, on the vertical axis versus 1/T on the horizontal axis. Draw the straight line that is closest to most point and determine the slope of the line, which is equal to –Ea/R.

ln k

ln k vs 1/T 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 0.00310

0.00320

0.00330

0.00340

1/T (1/K)

3. Calculate the activation energy. m= -3100 K= Ea=25773 J/mol

26 kJ/mol

0.00350

0.00360

0.00370

Post-Lab Questions 1. Why does reaction rate change as concentrations of the reactants change? According to collision theory, the rate of a chemical reaction is directly proportional to the frequency of collisions between reactant molecules. When the concentration of the reactants is increased, the frequency of the collision also increases, and thus the reaction occurs more rapidly. Furthermore, as the concentration of the reactants is lowered, the frequency of the collisions decreases and the reaction occurs more slowly. Note, however, that if a reactant does not appear in the rate law, i.e. has a reaction order of zero, changing its concentration will not affect the rate of a reaction. 2. Explain the general procedure used to find the rate law. First, the initial concentrations of each reactant and the initial rate are determined. Then, several experiments in which the concentration of an individual reactant is varied are performed, and the rate of each reaction is calculated. Afterwards, the rate of each reaction and the concentration of each reactant are compared to the initial rate and concentration, and the order of each reactant is determined. Finally, after the order of each reactant has been determined, the rate and concentrations of the reactants are inserted in the rate law and the rate constant is calculated. 3. Why does reaction rate change as temperature changes? The rate of a reaction changes as temperature changes because of two reasons. As said before, collisions theory states that the rate of a chemical reaction is directly proportional to the frequency of collisions. As the temperature changes, the frequency of collisions also changes. If the temperature increases, the molecules begin to move at faster speeds and the frequency of collisions increase. Thus, the rate of the reaction increases. The same is also true for the opposite, as the temperature is lowered, so is the reaction rate. However this is not the main reason why the rate of a reaction changes as temperature changes. Collision theory also states that the rate of a chemical reaction increases as the fraction of molecules with a kinetic energy greater than the activation energy increases. The fraction of molecules, f, with a kinetic energy greater than the activation energy increases is equal to e-Ea/RT, where e is a mathematical constant, Ea is the activation energy, R is the universal gas constant, and T is the absolute temperature. As you can see, f changes exponentially when T changes, and when T increases, f also increases. This is what predominantly s for the change in reaction rate when there is a change in temperature.

4. Explain the general procedure used to determine the activation energy. To determine the activation energy of a reaction, one uses Arrhenius equation: ln k=

+ ln A

indirectly. The Arrhenius equation correspond to the equation y=mx + b, with y=ln k, x=1/T, b= ln A, and m=-Ea/R. One must first run a reaction on at least two different temperatures. Then, the rate constant at each temperature is determined. Afterwards, a graph is plotted, with the natural logarithm of each rate constant on the vertical axis and the inverse of the temperature (1/T) belonging to each rate constant on the horizontal axis. When the graph has been plotted, a straight line that covers most of the points (the best line fit) is drawn. The slope of this line is calculated. Finally, since the slope is equal to -Ea/R, one multiplies it by –R (-8.314 J K-1 mol-1). This value is equal to the activation energy of a reaction. 5. Differentiate between reaction rate and specific rate. The reaction rate is the negative of the change in concentration of a reactant per unit time. The rate of a reaction is not constant during the course of a chemical reaction due to its dependence on the concentration of the reactants. The rate constant, k, is a proportionality constant in the relationship between rate and concentrations. It does not vary during the course of a reaction, but it varies with temperature. The rate constant allows us to calculate the rate of reaction if the concentration of the reactants and the rate law are known. 6. Comment on the effect of the catalyst. Predict how the activation energy changes when a catalyst is added to the reaction. When the catalyst was added, the time of reaction decreased from 160 seconds to 57 seconds. This means that the catalyst increased the rate of a reaction. A catalyst does this by providing an alternate pathway in the reaction mechanism that requires lower activation energy. Providing a pathway that requires a lower activation energy means that the fraction of molecules with energy greater than Ea increases, therefore the rate of the reaction also increases. Thus, I predict that the activation energy of a reaction is lowered when a catalyst is added. 7. Make a general statement about the consistency of the data as shown by calculating the orders of reactants, and by the graphical analysis which leads to activation energy. Were the calculated orders close to integers? Did the check of the order give the same value for the order? Were the points on the graph close to a straight line? The data were reasonably consistent. The calculated orders were close to integers and the check of the order gave reasonably close values. The points in the graph were scattered, probably due to the difficulty of maintaining a constant temperature in the water bath, but they showed a linear relationship between the natural logarithm of the rate constant and the inverse of temperature.

8. Write out the “two-point” form of the Arrhenius equation which relates rate constants, temperatures and activation energy. ln = Where: k1 and k2= rate constants at absolute temperatures T1 and T2 Ea=activation energy R=Universal gas constant, equal to 8.314 J/(K mol) 9. How could you improve data? I would improve data by performing this experiment in a macroscale instead of a microscale because it would reduce the effect and influence of small measurement errors. Another thing I would do to improve data is to have better control in the water bath used to determine the activation energy of the reaction, and use more precise equipment.

Observations                

When the reaction reached completion, the solution turned blue. Instead of tap water, deionized water was used. The reaction well had a clear color. In order to minimize error, the pipets were held vertically and almost no air bubbles were introduced to them. The reaction that was catalyzed by copper nitrate occurred 103 seconds faster than the uncatalyzed reaction. After the reaction had reached completion and some minutes had ed, the reaction well turned yellow. All reactants, except copper nitrate, had a clear color. The reactant that affected the reaction rate that most was hydrochloric Acid The slowest reaction in part 2 was experiment number 1. The copper nitrate solution had a light blue color. The fastest reaction in part 2 was experiment number 7, which contained the greatest amount of hydrochloric acid. The temperature of the reaction well did not vary much during the course of the reaction. Each experiment was performed 3 times to assure accuracy and precision. In part 3, the rate and rate constant of the reaction was measured in 3 different temperatures: 2°C, 22° C, and 41°C. As the temperature of the water bath increased, so did the reaction rate. Some sources of error include: measurement error, introduction of air bubbles into pipets, not holding the pipets vertically, and lack of control in the temperature of the water bath.

Conclusion The rate of a chemical reaction, or reaction rate, is defined as the change in concentration of a reactant per unit time. Reaction rate depends on various factors, such as the concentration of reactants, presence of a catalyst, and temperature. In this experiment, the dependence of rate in each of these factors is observed. The reaction studied in this experiment is the oxidation of iodide ions by bromate ions in the presence of acid: 6I-(aq) + BrO3-(aq) + 6H+ (aq)  3I2 (aq) + Br-(aq) + 3H2O (l) The first part of the experiment involved the determination of the rate law. The rate law is an equation that relates the rate of a reaction to the concentration of reactants raised to various powers. A reaction’s rate law shows the reaction rate’s dependence on the concentration of each reactant. In order to determine the rate law, the reaction was performed several times with the concentration of an individual reactant varying in each trial. This procedure allowed the calculation of the order of each reactant and to calculate the rate constant. After the reaction was performed several times, the rate and concentration of reactants in each trail were compared. Of the 7 trials, the fastest one was the reaction that contained the highest concentration of hydrochloric acid, with a reaction rate of 7.4 x 10-7 M/s. The slowest reaction was the one that contained the lowest concentration of each reactant, with a reaction rate of 8.3 x 10-8 M/s. The order of each reactant was then determined. Iodide was found to have a reaction order of 1, meaning that if its concentration was doubled, the reaction rate was doubled. Bromate was also found to have a reaction order of 1. Finally, hydron was found to have a reaction order of 2, meaning that if its concentration were doubled, the reaction rate would quadruple. This means that the rate law of the reaction is: Rate= k[I-][BrO3-][H+]2 Also, it means that a change in the concentration of hydrochloric acid will affect the rate of the reaction by the biggest extent and that the reaction is fourth order overall. Since the rate law was now known, values of rate and concentration were plugged into the equation, and the rate constant, k, was determined. The average value of k on the 7 trials was equal to 25 M-3s1. Now that the rate law was known, another kinetic property of the reaction, activation energy, could be calculated. The activation energy of a reaction is the minimum energy required for two molecules to react. It can be calculated by using the Arrhenius equation: ln k=

+ ln A

To determine the activation energy, the rate constant of the reaction was measured at 3 different temperatures. Then, a graph was made with the natural logarithm of the rate constant on the vertical axis and the inverse of temperature in the horizontal axis. A regression line was drawn using computer software, and its slope was determined to be equal to -3100 K. Since the Arrhenius equation corresponds to the equation y=mx + b, the slope of the regression line will be equal to -Ea/R where Ea is the activation energy and R is the universal gas constant. Thus, the activation energy of the reaction was equal to -2.6 kJ/mol. Another thing that was observed was the effect of a catalyst on the rate of the reaction. When the catalyst was added, the reaction time lowered from

160 seconds to 157 seconds. Catalyst increase the rate of a chemical reaction by providing an alternate reaction pathway that requires a lower activation energy. According to the data, my hypothesis was correct. Hydrochloric acid was the reactant that affects the rate of the reaction the most. Also, as the concentration of the reactants, of the reactants increased, so did the rates. Furthermore, in part 3, the reaction that had the highest rate was the one performed at the highest temperature. There was minimal error in the data, which had consistency and reproducibility. Studying the kinetics of a reaction has various applications. It allows chemists and engineers to obtain the maximum yield of a desire product, such fertilizers, in a safe and economical manner. Knowing the rate law of a reaction allows us to know which reactants affect the rate the most and which affect it the least, which can be useful when trying to get a large quantity of a desired product. Finally, studying the kinetics and other properties of a reaction allows us to perform it in the most optimal environment.