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Klautzsch 1 Mikey Klautzsch Chemistry Lab 262-020 L. Santiago Delgado

Rate Laws: The Iodine Clock Reaction Date of Experiment: 02/11/14 and 02/18/14 Date of Submission: 02/25/14 Abstract: The purpose of the experiment was to determine the rate law, rate law constant, activation energy, and frequency factor for a reaction by measuring the concentrations of Potassium Iodine (KI), Hydrochloric Acid (HCl), Potassium Bromate (KBrO3), and Sodium Thiosulfate (Na2S2O3) in an ionic solution. For Part One of the Rate Laws Lab, 12 trails were done at a constant temperature with different amounts of each substance (except for the first 3 which were control tests) and 4 drops of indicator to indicate when the reaction was finished. The time for the reaction to occur was measured and used to determine the rate law for the reaction by determining the orders with respect to each substance (I- BrO3-H+). The rate law was found to be: Rate = k [I-][BrO3-][H+]2. This was then used to determine the average rate law constant (k), for part one, which was found to be 43.4 M-3s-1. The standard deviation for the rate constant was 6.98. For Part Two of the Rate Laws Lab, five trials were done with the same amount of each substance, including the indicator, during every trial but different temperatures (20, 30, 40, 50, and 60 ℃ ). This data was then used to find the activation energy and frequency factor of the reaction. By using the Arrhenius Plot −Ea 1 + ln ( A ) , it was found that the activation energy for the Equation, R T ln ¿ reaction was 49.5 kJ/mole and the frequency factor for the reaction was found to be 3.4 × 1010 M-3s-1. (k )=¿

( )

Theory: The rate of a reaction is measured by finding out how long it takes for a specific amount of substance to be completely used up to make a specific amount of product in a reaction.2 For the reaction aA + bB  cC + dD the rate would be as so: −1 Δ[ A ] −1 Δ [B] +1 Δ[C ] +1 Δ[ D ] = = = Rate = d Δt c Δt b Δt a Δt The reaction rate is defined as the negative of the change in concentration of the reactant per change in time because the reactant concentrations decrease. Since the product concentration increases as the reaction proceeds, the change in concentration of a product is positive.1 Once the rate of a reaction is found (in M/s), the rate law can then be determined, which was the purpose of the Iodine Clock Reaction, Part 1. The rate law is an expression that shows the relationship between the concentration of the

Klautzsch 2 reactants and the rate of the reaction.1, 2 For the above reaction, aA + bB  cC + dD, the rate law would be: Rate = k [A]x[B]y Where k is the rate constant (at a constant temperature), and x and y are the order of the reaction with respect to the concentration of A and B. 2 These values must be experimentally determined and have no correlation to the stoichiometric coefficients. These values are determined by measuring the initial rate of reaction when the concentration of one substance is held constant and changing the concentration of the other.1,2 Once the orders of the reaction have been found, one can then determine the value of the rate constant (k), which should be the same number (or very close to it) for every trial.1,2 The reaction between Iodide and Bromate occurs in three elementary steps, with the 3rd step being the slow one. The slow step is the rate-determining step. 1 During this 2rd step all the thiosulfate is used up and the first traces of iodine are produced which causes a blue color to appear because of the reaction with the starch indicator. 2 It is important to note that the stoichiometric coefficients during the elementary steps are used to determine the rate at which thiosulfate is consumed: I −2 ∆[ ¿¿ 2] −∆ [ S 2 O 3 ] = 3 ∆t 6∆t Rate=¿ Rate Law = k [I-]x [BrO3-]y [ H+]z Part one of the experiment was done at a constant temperature, but what would happen if the concentrations of the reactants were kept the same and the temperature was changed? The rate of a reaction is depends on temperature. The higher the temperature, the more kinetic energy the system will have, thus resulting in more particles surpassing the activation energy for the reaction which speeds up the rate of the reaction.2 Arrhenius’s Equation shows the relationship between rate and temperature: −E a

k = A e RT Taking the logarithms of this equation produces the Arrhenius Plot Equation, which resembles the equation for a straight line (y=mx+b): −E a 1 ln k = + ln A R T

()

Procedure: All procedural information for Rate Law part 1 is described in the CH 262 Rate Laws Part 1 Laboratory Manual, pages 3-4. All procedural information for Rate Law part 2 is described in the CH 262 Rate Laws Part 2 Laboratory Manual, pages 1-2. Materials: Two 125 mL flasks, 9 pipets (four 5 mL and five 10 mL), One stopwatch, 68

Klautzsch 3 drops of starch, 180 mL of .01 M KI, 170 mL of .00025 M Na2S2O3, 145 mL of .1 HCl, 185 mL H2O, 180 mL of .04 M KBrO3, Hot Plate, 160 mL beaker, Water bath and extra water. Data: Table 1: Summary of Data Collected for Each Reaction With Different Amounts:

Trial 1a 1b 1c 2 3 4 5 6 7 8 9 10

Beaker A Beaker A Flask B Flask B Flask C 0.010 M 0.00025 M 0.04 M Temp ( ℃ ) KI Na2S2O3 0.10 M HCl H20 KBrO3 Time (s) 10 10 10 10 10 32.1 22.5 10 10 10 10 10 31.7 22.5 10 10 10 10 10 31.6 22.5 5 10 10 15 10 66.4 22.5 15 10 10 5 10 20.6 22.5 20 10 10 0 10 10.7 22.5 10 10 10 15 5 61.8 22.5 10 10 10 5 15 21.9 22.5 10 10 10 0 20 16.3 22.5 10 10 5 15 10 91.4 22.5 10 10 15 5 10 14.8 22.5 10 10 20 0 10 8 22.5

Table 2: Summary of Data Collected for Each Reaction at Different Temperatures:

Beaker A Trail 1 2 3 4 5 Total mL

.01 M KI 10 10 10 10 10 40

Beaker A Flask B Flask B .00025 M .10 M Na2S2O3 HCl H2O 10 5 15 10 5 15 10 5 15 10 5 15 10 5 15 40

20

60

Flask C .04 M KBrO3 Time (s) Temp ( ℃ ) 10 91.40 22.5 10 47.80 30 10 28.90 40 10 16.50 50 10 8.70 60 40

Klautzsch 4

Calculations: Sample Calculations for: 1.) Total Volume of Solution: To find the total volume of the solution, simply add the amounts of every substance together. mL KI + mL Na2S2O3 + mL HCl + mL H2O + mL KBrO3 Total Volume: 10 mL KI + 10 mL Na2S2O3 + 10 mL HCl + 10 mL H2O + 10 mL KBrO3 = 50 mL 2.) Initial Concentration of a Substance: The initial concentrations of each substance are to be calculated. To do this simply set the concentrations of the substance and the amount of the substance equal to the total volume of the solution (calculation 1) and solve for the new concentration. M1V1 = M2V2 (.01 M KI)(10 mL) = (initial concentration (M2))(50mL) Initial Concentration of KI = M2 = .002 M KI 3.) Rate: To calculate the rate of a reaction, look at the stoichiometric coefficients of the substances: 6I- + BrO3- + 6H+  3I2 + Br- + 3 H2O which breaks down into the fast reaction: I2 + 2S2O32-  2I- + S2O62- and the slow reaction: I2 + starch  dark blue compound. It is important to note that the rate is not equal to time, but it is a unit of the change in concentration per unit time (M/s). 2−¿ S 2 O 3¿ Rate = ¿ ∆¿ ¿ = [.00005M S2O3-]/6(32.1 s) Rate = 2.60 × 10-7 M/s 4.) Natural Log of Rate: ln (rate) = ln (2.60 × 10-7 M/s) = -15.16 (unit-less) 5.) Rate Constant (k): To calculate the rate constant, rearrange the rate law, rate=k[I-][BrO3-][H+]2, to solve for k. The order of the reaction with respect to each substance must be determined first (refer to figures 1-3).

Klautzsch 5 −¿¿ I ¿ −¿ ¿ Br O3 2Rate Constant for S2O3 = k = ¿ H + ¿¿ ¿ Rate ¿ 2 .02 M ¿ ¿ k = [ .002 M ] [ .008 M ] ¿ 40.6 M-3s-1 2.60 × 10−7 M /s ¿ 2

6.) 1/Temperature (1/Kelvin): simply divide one by the temperature in kelvin. 1 1 for TrialOne= =.0034 K −1 T 295.5 K 7.) Temperature in Kelvin (K): Add 273 to the temperature in Celsius. K = ℃ + 273 K = 22.5 + 273 = 295.5 K 8.) Finding the Frequency Factor (A): To calculate the frequency factor, create a graph of the ln (k) versus 1/Temperature. Use the Arrhenius Plot Equation Arrhenius Plot Equation (seen below) to calculate the value of A. (refer to figure 4 for value of ln(A). −Ea 1 (k )=¿ + ln ( A ) R T ln ¿ y = m x + b ln (A) = b = y intercept (refer to figure 4)

( )

ln (A) = 24.2 A = e = 3.37 × 1010 (M-3s-1) 9.) Finding the Activation Energy (Ea): To calculate the activation energy (Ea) create a graph of the ln(k) versus 1/Temperature. Use the Arrhenius Plot Equation (seen below) to calculate the value of Ea. This is done by multiplying 24.2

the slope of the line (figure 4) by -8.31

kPa∗ L . mole K

Klautzsch 6 −Ea 1 + ln ( A ) R T ln ¿ = m x + b

(k )=¿ y −E a R

= Slope of Line = m (refer to figure 4)

−E a R Ea =5954 10.)

( )

= -5954

× 8.31

kPa∗ L mole K

R = 8.31

kPa∗ L mole K

= 49500 J/mol =49.5 kJ/mol

Natural Log of Rate Constant (k): for Trial 1 on Table 6 ln (rate constant) = ln (56.98) ln (k) = 4.04 (Unit-less)

11.) Reaction Orders: Refer to figures 1, 2 and 3. The slopes of the lines rounded to the nearest whole number are the reaction orders.

Results: Table 3: Results for Initial Concentration Day One: Total Volume Initial Initial Initial Trail (mL) Initial [KI] [Na2S2O3] [HCl] [KBrO3] 1a 50 0.002 0.00005 0.02 0.008 1b 50 0.002 0.00005 0.02 0.008 1c 50 0.002 0.00005 0.02 0.008 2 50 0.001 0.00005 0.02 0.008 3 50 0.003 0.00005 0.02 0.008 4 50 0.004 0.00005 0.02 0.008 5 50 0.002 0.00005 0.02 0.004 6 50 0.002 0.00005 0.02 0.012 7 50 0.002 0.00005 0.02 0.016 8 50 0.002 0.00005 0.01 0.008 9 50 0.002 0.00005 0.03 0.008 10 50 0.002 0.00005 0.04 0.008 *Note: See sample calculation numbers 1, and 2 for calculations.

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Table 4: Results for Initial Concentrations and Rate Day Two: Total Initial Initial Trail Volume [Na2S2O3] Initial [KI] [KBrO3] Initial [HCl] 1 50 0.00005 0.002 0.008 0.01 2 50 0.00005 0.002 0.008 0.01 3 50 0.00005 0.002 0.008 0.01 4 50 0.00005 0.002 0.008 0.01 5 50 0.00005 0.002 0.008 0.01 *Note: See sample calculation numbers 1, 2, and 3 for calculations.

Rate (M/s) 9.12E-08 1.74E-07 2.88E-07 5.05E-07 9.58E-07

Table 5: Results for the Rate, Natural Log of Rate, and Rate Constant (k): Trial Rate (M/S) ln(Rate) k (M-3s-1) 1a 2.60E-07 -15.2 40.6 1b 2.63E-07 -15.2 41.1 1c 2.64E-07 -15.1 41.2 2 1.26E-07 -15.9 39.2 3 4.05E-07 -14.7 42.1 4 7.79E-07 -14.1 60.8 5 1.35E-07 -15.8 42.1 6 3.81E-07 -14.8 39.6 7 5.11E-07 -14.5 39.9 8 9.12E-08 -16.2 57.0 9 5.63E-07 -14.4 39.1 10 1.04E-06 -13.8 40.7 1 (average) 2.62E-07 -15.2 40.9 Average: 43.4 STD: 6.98 *Note: See sample calculations numbers 3, 4 and 5 for calculations. Table 6: Results for Temperature, Rate, Rate Constant (k), Natural Log of k, and 1/T Trail Temp (K) Time (s) Rate (M/s) k (M-3s-1) ln k 1/T (1/K)

Klautzsch 8 1 295.5 91.4 9.12E-08 56.98 4.04 2 303 47.8 1.74E-07 108.96 4.69 3 313 28.9 2.88E-07 180.22 5.19 4 323 16.5 5.05E-07 315.66 5.75 5 333 8.7 9.58E-07 598.66 6.39 *Note: See sample calculations numbers 3, 5, 6, 7, and 10 for calculations.

0.00338 0.00330 0.00319 0.00310 0.00300

Table 7: Natural Log of Initial Concentrations of I-, BrO3-, and H+: Trial ln[I-] ln[BrO3-] ln[H+] 1 -6.215 -4.828 -3.912 2 -6.908 3 -5.809 4 -5.521 5 -5.521 6 -4.423 7 -4.135 8 -4.605 9 -3.507 10 -3.219

ln(rate) vs. ln[I-]

ln(rate)

-13.0 -7.000 -6.800 -6.600 -6.400 -6.200 -6.000 -5.800 -5.600 -5.400 -13.5 -14.0

f(x) = 1.26 x − 7.28 R² = 0.97

-14.5 -15.0 -15.5 -16.0 -16.5

ln[I-] Figure 1: Natural Log of the Rate Versus the Natural Log of Concentration of I-: Reaction Order= Slope of the Line = 1

Klautzsch 9

ln(rate) vs. ln[BrO3-]

ln (rate)

-13.5 -5.600 -5.400 -5.200 -5.000 -4.800 -4.600 -4.400 -4.200 -4.000 -14.0

f(x) = 0.96 x − 10.54 R² = 1

-14.5 -15.0 -15.5 -16.0

ln [I-] Figure 2: Natural Log of the Rate Versus the Natural Log of Concentration of BrO3-: Reaction Order= Slope of the Line = 1

ln(rate) vs. ln[H+] -12.5 -4.800 -4.600 -4.400 -4.200 -4.000 -3.800 -3.600 -3.400 -3.200 -3.000 -13.0

ln (rate)

-13.5

f(x) = 1.74 x − 8.25 R² = 0.99

-14.0 -14.5 -15.0 -15.5 -16.0 -16.5

ln [I-] Figure 3: Natural Log of the Rate Versus the Natural Log of Concentration of H+: Reaction Order= Slope of the Line = 2

Klautzsch 10

Arrhenius Plot ln k vs. 1/T 7.00 6.00

ln k

5.00 4.00

f(x) = − 5953.86 x + 24.24 R² = 0.99

3.00 2.00 1.00 0.00 0.00300 0.00305 0.00310 0.00315 0.00320 0.00325 0.00330 0.00335 0.00340 0.00345

1/T (1/K) Figure 4: Natural Log of the Rate Constant (k) Verse 1/Temperature (1/Kelvin): Activation Energy (Ea)= 49.5 kJ/mol (see sample calculation number 9) Frequency Factor (A)= 3.37 × 1010 M-3s-1 (see sample calculation number 8) Conclusion: The goal of the Rate Laws: Iodine Clock Reaction: Part 1 and 2, was to determine the rate law, rate law constant, activation energy, and frequency factor for the specific reaction of Potassium Iodine (KI), Hydrochloric Acid (HCl), Potassium Bromate (KBrO3), and Sodium Thiosulfate (Na2S2O3) in an aqueous solution, by measuring the concentrations of the reactants. For Part One of the Rate Laws Lab, 12 trails were done at a constant temperature with different amounts of each substance, although, the first three trials were all done with the same amounts of each substance and used as a control test. The average rate for the first reaction was 2.62 × 10-7 M/s. Refer to Table 5 for the rate values. Once all the rates were found, the order of each reactant were found to be one for Iodide, one for Bromate, and two for Hydrogen, thus meaning the over all order for the reaction was four. The rate law for the reaction was fond to be: Rate = k [I-] [BrO3-] [ H+]2 Finally, after finding the rates for the different reactions at a constant temperature, the initial concentration of each substance, and the order of reaction with respect to each substance, the rate constant (k) was found by solving for k. This value was an average between all the trails and found to be 43.4 M-3s-1 with the standard deviation being 6.98. By finding the Rate Law, the orders of the reaction with respect to each substance, and the average rate constant, the objectives for Part One of the lab were met.

Klautzsch 11

For Part Two of the Rate Laws Lab, five trials were done with a constant amount of Potassium Iodine, (10 mL KI), Hydrochloric Acid (5 mL HCl), Potassium Bromate (10 mL KBrO3), Water (15 mL H2O), and Sodium Thiosulfate (10 mL Na2S2O3). Every trial was done at different temperatures, 20, 30, 40, 50, and 60 ℃ . It was found that for every ten degrees the temperature increased, the time it took for the reaction to take placed was divided in half (refer to Table 6). This data was used to find the rates of the reaction, which was then used to find the different values of the rate constant (k) at different temperatures (refer to table six for 5 rate constant results). By using the plot form of Arrhenius’s Equation (which resembles the y=mx+b): −E a 1 ln k = + ln A R T −E a The activation energy (Ea) and frequency factor (A) could now be found. Since R is equal to the slope of a line (m), and ln(A) is equal to the y-intercept (b), a plot showing the ln(k) (y axis) verse 1/T (x axis) was made (refer to Figure 4). The equation for this line was found to be y = -5953.9x + 24.242. The activation energy was then kPa∗ L resulting with found (see calculation 9) by multiplying -5953.9 by -8.31 mole K 49500 Joules/mole or 49.5 kilojoules/mole as compared to the literature value for the activation energy was found to be 51.8 kJ/mole.1 The frequency factor (A) was found (see calculation 8) by using the value of 24.242, the y intercept, and putting it as the exponent of e, which resulted in a frequency factor of 3.37 × 1010 M-3s-1. The objectives of part two were met once these two values were found.

()

To improve the experiment more trials for each temperature and each amount of substance could be done. For the first trails (1a, 1b, and 1c table 1) the times were all within 1 second of each other. The data that was found for the k constant had a standard deviation of 6.98, which is relatively large. This was large because during trials number four and eight (table 5) the values found for the rate constant (k) were 57 and 60 M-3s-1. These values were way over our average and any other of the trails, which explains the large standard deviation. To improve the experiment trails four and eight should be re-done which could produce better data.

References:

Klautzsch 12 1. Chemical Kinetics in Chemistry: A Molecular Approach 3rd ed.; Harbour: Nivaldo J. Trio, Upper Saddle River, NJ, United States, 2011; pp 596-633. 2. Burand, Michael, W. Ch 262 Laboratory Manual Rate Laws, The Iodine Clock Reactions, Part 1 and Part 2; Oregon State University; Winter 2014, p1-5, 1-3....