Archer G11

Partner: Joon 12 January 2012

Kinetics of a Reaction

Purpose: The purpose of this experiment is to determine the rate constant and activation energy of a

reaction and to verify the effect of catalyst on the reaction. The significance of this lab is that the

methods of slowing down the decomposition of food could be determined. Slowing down the

decomposition of food is important because delivering food takes time. If the decomposition of the food

was not slowed down, the food would decompose before they reach the destination and become

inedible.

Hypothesis: The hypothesis is that the rate constant can be calculated from the initial rate and

concentration of reactants while the activation energy can be calculated from graph of rate and

temperature and catalyst will increase the rate of reaction by lowering the activation energy. By

changing the concentrations of the initial concentrations, the initial rate of reaction would change and

the order of reaction could be determined and made into rate law. The rate law could be use with the

initial rate to determine the rate constant. The activation energy could be found by altering the

temperature in which the reaction takes place. The graph can be made from reaction rate at different

temperature and the activation energy can be found from the graph. The effect of catalyst could be

verified by comparing the reaction with and without catalyst at the same condition.

Materials:

Materials Quantity

0.010 M Potassium Iodide (KI) 78 drops

Distilled water (H2O) 63 drops

0.10 M Hydrochloric Acid (HCl) 78 drops

2% Starch 30 drops

0.0010 M Sodium Thiosulfate (Na2S2O3) 30 drops

0.040 M Potassium Bromate (KBrO3) 78 drops

0.1 M Copper(II) Nitrate (Cu(NO3)2) 3 drops

Tap water 45 drops

15-mL Beral-type pipet 7 pipets

Pliers 1 pliers

Labeling tape 1 roll

Permanent marker 1 marker

50-mL beaker 7 beakers

0.0001-g precision balance 1 balance

Thermometer 1 thermometer

96-wells reaction strip 1 strip

Timer 1 timer

Toothpick 30 toothpicks

Water bath 1 bath

Ice 1 pack

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Procedures:

Part 1.Find the Volume of One Drop of Solution

1.) Use pliers to make a capillary tip for a Beral-type pipet

2.) Mass a beaker

3.) Add 5 drops of tap water into the beaker on the balance using the capillary tip Beral-type pipet

4.) Record the mass

5.) Repeat step 3 and 4 for 10 more drops

6.) Repeat step 2 to 5 for 2 more trials

Part 2. Determine the Reaction Rate and Calculate the Rate Law

Rate Order of KI

1.) Use a plier to make capillary tip for 6 Beral-type pipet

2.) Label the pipet as KI, H2O, HCl, Starch, Na2S2O3, and KBrO3

3.) Fill the pipet about 10 mL of the labeled liquid

4.) Add 2 drops of KI, 4 drops of distilled water, 2 drops of HCl, 1 drop of starch, 1 drop of Na2S2O3

into a well (experiment 1)

5.) Add 2 drops of KBrO3

6.) Stir the well with a toothpick

7.) Record the time from when KBrO3 was added until the first sign of blue color appears

8.) Repeat step 4 to 7 in another well for experiment 2: 4 drops KI, 2 drops H2O, 2 drops HCl, 1 drop

starch, 1 drop of Na2S2O3 and 2 drops KBrO3

9.) Repeat step 8 for experiment 3: 6 drops KI, 2 drops HCl, 1 drop starch, 1 drop Na2S2O3, and 2

drops KBrO3

10.) Repeat step 8 for experiment 4: 2 drops KI, 2 drops H2O, 2 drops HCl, 1 drop starch, 1 drop

Na2S2O3, and 4 drops KBrO3

11.) Repeat step 8 for experiment 5: 2 drops KI, 2 drops HCl, 1 drop starch, 1 drop Na2S2O3, and 6

drops KBrO3

12.) Repeat step 8 for experiment 6: 2 drops KI, 2 drops H2O, 4 drops HCl, 1 drop starch, 1 drop

Na2S2O3, and 2 drops KBrO3

13.) Repeat step 8 for experiment 7: 2 drops KI, 6 drops HCl, 1 drop starch, 1 drop Na2S2O3, and 2

drops KBrO3

14.) Repeat step 3 to 13 for 2 more trials

Part 3. Determine the Activation Energy

1.) Add 2 drops of KI, 4 drops of distilled water, 2 drops of HCl, 1 drop of starch, 1 drop of Na2S2O3

into a well

2.) Repeat step 1 for 5 more wells

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3.) Heat a water bath up to about 40°C

4.) Fill the KBrO3 pipet about half full with the labeled liquid

5.) Dip the reaction strip and the bulb part of the pipet into the warm water

6.) Wait for 5 minutes

7.) Add 2 drops of KBrO3 into each of the 3 wells

8.) Stir with a tooth pick

9.) Record the time until the blue color appears

10.) Record the temperature

11.) Put a pack of ice into a metal tray

12.) Add some water into the tray

13.) Measure the temperature

14.) Dip the reaction strip and the bulb part of the pipet into the cold water

15.) Wait for 5 minutes

16.) Add 2 drops of KBrO3 into the other unreacted wells

17.) Stir the wells with toothpicks

18.) Record the time until the blue color appears

Part 4. Observe the Effect of a Catalyst on the Rate

1.) Add 2 drops of KI, 3 drops of distilled water, 2 drops of HCl, 1 drop of starch, 1 drop of Na2S2O3,

and 1 drop of Cu(NO3)2 into a well

2.) Add 2 drops of KBrO3 to the well

3.) Record the time needed for the blue color to appear

4.) Repeat step 1 to 3 for 2 more trials

Results: The appearance of the blue color was so sudden. The blue color appeared with or without

stirring the well in which the reaction took place. The unstirred well, the blue color appeared faster than

with stirred well. However, the blue color appeared as a thin line. On the other hand, when the well is

stirred, the blue color appeared all over the solution, and the solution quickly turned blue. Unstirred, the

blue color floated in the middle of the solution leaving the rest of the solution in the color of the

uninvolved ions. For the stirred well, all of the solution turned blue thoroughly. Color change signify that

BrO3- had been used up.

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6I-(aq) + BrO3-(aq) + 6H+(aq) 3I2(aq) + Br-(aq) + 3H2O(l)

I2(aq) + 2S2O3-(aq) 2I-(aq) + S4O6

2-(aq)

Experiment Number

KI, 0.010 M (Drops)

Distilled H2O (Drops)

HCl, 0.10 M (Drops)

Starch, 2 % (Drops)

Na2S2O3, 0.0010 M (Drops)

KBrO3, 0.040 M (Drops)

1 2 4 2 1 1 2

2 4 2 2 1 1 2

3 6 0 2 1 1 2

4 2 2 2 1 1 4

5 2 0 2 1 1 6

6 2 2 4 1 1 2

7 2 0 6 1 1 2

Initial Concentration Table

Experiment Number

Concentration of KI (M)

Concentration of HCl (M)

Concentration of Na2S2O3 (M)

Concentration of KBrO3 (M)

1 1.7 × 10-3 0.017 8.3 × 10-5 6.7 × 10-3

2 3.3 × 10-3 0.017 8.3 × 10-5 6.7 × 10-3

3 5.0 × 10-3 0.017 8.3 × 10-5 6.7 × 10-3

4 1.7 × 10-3 0.017 8.3 × 10-5 0.013

5 1.7 × 10-3 0.017 8.3 × 10-5 0.020

6 1.7 × 10-3 0.033 8.3 × 10-5 6.7 × 10-3

7 1.7 × 10-3 0.050 8.3 × 10-5 6.7 × 10-3

Initial concentration = (Molarity of the reactant) × (Number of drops ÷ 12)

KI (Experiment 1, 4-7): 0.010 × (2 ÷ 12) = 1.67 × 10-3 M KI

KI Experiment 2: 0.010 × (4 ÷ 12) = 3.33 × 10-3 M KI

KI Experiment 3: 0.010 × (6 ÷ 12) = 5.00 × 10-3 M KI

HCl (Experiment 1-5): 0.10 × (2 ÷ 12) = 0.0167 M HCl

HCl Experiment 6: 0.10 × (4 ÷ 12) = 0.0333 M HCl

HCl Experiment 7: 0.10 × (6 ÷ 12) = 0.0500 M HCl

Na2S2O3 (Experiment 1-7): 0.0010 × (1 ÷ 12) = 8.33 × 10-5 M Na2S2O3

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KBrO3 (Experiment 1-3, 6-7): 0.040 × (2 ÷ 12) = 6.67 × 10-3 M KBrO3

KBrO3 Experiment 4: 0.040 × (4 ÷ 12) = 0.0133 M KBrO3

KBrO3 Experiment 5: 0.040 × (6 ÷ 12) = 0.0200 M KBrO3

Find the Volume of One Drop of Solution

Density of water (g/mL) 1.00

Trial 1 Trial 2 Trial 3

Mass of empty beaker (g) 29.5017 29.4815 28.6883

Experiment 1

Mass of beaker plus 5 drops of water (g) 29.6302 29.5806 28.8205

Mass of first 5 drops of water (g) 0.1285 0.0991 0.1322

Average mass of 1 drop of water (g) 0.02570 0.01982 0.02644

Experiment 2

Mass of beaker plus 10 drops of water (g) 29.7543 29.7034 28.9387

Mass of second 5 drops of water (g) 0.1241 0.1228 0.1182

Average mass of 1 drop of water (g) 0.02482 0.02456 0.02364

Experiment 3

Mass of beaker plus 15 drops of water (g) 29.8813 29.8321 29.0655

Mass of third 5 drops of water (g) 0.1270 0.1287 0.1268

Average mass of 1 drop of water (g) 0.02540 0.02574 0.02536

Average mass of 1 drop of water (Experiment 1-3) (g) 0.02531 0.02337 0.02515

Average mass of 1 drop of water (Trials 1-3) (g) 0.02461

Average volume of 1 drop of water (L) 2.46 × 10-5

Mass of first 5 drops of water = (Mass of beaker plus 5 drops of water) – (Mass of empty beaker)

Trial 1: 29.6302 – 29.5017 = 0.1285 g

Trial 2: 29.5806 – 29.4815 = 0.0991 g

Trial 3: 28.8205 – 28.6883 = 0.1322 g

Average mass of 1 drop of water experiment 1 = (Mass of first 5 drop of water) ÷ 5

Trial 1: 0.1285 ÷ 5 = 0.02570 g

Trial 2: 0.0991 ÷ 5 = 0.01982 g

Trial 3: 0.1322 ÷ 5 = 0.02644 g

Mass of second 5 drops of water = (Mass of beaker plus 10 drops of water) – (Mass of empty beaker)

Trial 1: 29.7543 – 29.5017 = 0.1241 g

Trial 2: 29.7034 – 29.4815 = 0.1228 g

Trial 3: 28.9387 – 28.6883 = 0.1182 g

Average mass of 1 drop of water experiment 2 = (Mass of second 5 drop of water) ÷ 5

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Trial 1: 0.1241 ÷ 5 = 0.02482 g

Trial 2: 0.1228 ÷ 5 = 0.02456 g

Trial 3: 0.1182 ÷ 5 = 0.02364 g

Mass of third 5 drops of water = (Mass of beaker plus 15 drops of water) – (Mass of empty beaker)

Trial 1: 29.8813 – 29.5017 = 0.1270 g

Trial 2: 29.8321 – 29.4815 = 0.1287 g

Trial 3: 29.0655 – 28.6883 = 0.1268 g

Average mass of 1 drop of water experiment 3 = (Mass of third 5 drop of water) ÷ 5

Trial 1: 0.1270 ÷ 5 = 0.02540 g

Trial 2: 0.1287 ÷ 5 = 0.02574 g

Trial 3: 0.1268 ÷ 5 = 0.02536 g

Average mass of 1 drop of water = Σ(Average mass of 1 drop of water experiment 1, 2, 3) ÷ 3

Trial 1: (0.02570 + 0.02482 + 0.02540) ÷ 3 = 0.02531 g

Trial 2: (0.01982 + 0.02456 + 0.02574) ÷ 3 = 0.02337 g

Trial 3: (0.02644 + 0.02364 + 0.02536) ÷ 3 = 0.02515 g

Average mass of 1 drop of water = Σ(Average mass of 1 drop of water trial 1, 2, 3) ÷ 3

(0.02531 + 0.02337 + 0.02515) ÷ 3 = 0.02461 g

Average volume of 1 drop of water = (Average mass of 1 drop of water) × (Density of water)

0.02461 × 1.00 = 0.02461 mL

Determine the Reaction Rate and Calculate the Rate Law

Time, seconds

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Experiment No. Trial 1 Trial 2 Trial 3 Average Temp. °C

1 166 202 148 172 24

2 70 100 85 85 24

3 46 49 31 36 40.5 24

4 90 101 103 98 24

5 47 47 49 47.67 24

6 23 24 21 22.67 24

7 9 9 13 10.33 24

Reaction Rate Calculation

Molarity of S2O32- (M) 0.0010

Moles of S2O32- ions (moles) 2.5 × 10-8

Moles of BrO3- reacted (moles) 4.1 × 10-9

Change in BrO3- concentration (M) 1.4 × 10-5

Experiment Reaction Rate (M/s)

Experiment 1 8.1 × 10-8

Experiment 2 1.6 × 10-7

Experiment 3 3.4 × 10-7

Experiment 4 1.4 × 10-7

Experiment 5 2.9 × 10-7

Experiment 6 6.1 × 10-7

Experiment 7 1.3 × 10-6

Moles of BrO3- (moles) 2.1 × 10-6

Rate Order

Rate Order of [I-] Rate Order of [BrO3-] Rate Order of [H+]

1 1 2.5

Rate Law Expression Rate = k[I-][H+]2.5 [BrO3-]

Rate Constant Data

Experiment 1 2 3 4 5 6 7

Value of rate

constant, 2.0 × 102 2.0 × 102 290 180 240 270 220

Average Value of Rate Constant (M-3.5s-1) 228

Moles of S2O32- = (Molarity of S2O3

2-) × (Volume of 1 drop)

0.0010 × (2.461 × 10-5) = 2.46 × 10-8 moles S2O3-

Moles of BrO3- reacted = (Moles of S2O3

2-) × (Mole ratio)

(2.46 × 10-8) × (1 ÷ 6) = 4.1 × 10-9 moles BrO3-

Change in BrO3- concentration = (Moles of BrO3

- reacted) ÷ (Volume of 12 drops)

(4.1 × 10-9) ÷ [12 × (2.461 × 10-5)] = 1.39 × 10-5 M BrO3-

Rate = (Change in BrO3- concentration) ÷ (Average time)

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Experiment 1: (1.39 × 10-5) ÷ 172 = 8.08 × 10-8 M/s

Experiment 2: (1.39 × 10-5) ÷ 85 = 1.64 × 10-7 M/s

Experiment 3: (1.39 × 10-5) ÷ 40.5 = 3.43 × 10-7 M/s

Experiment 4: (1.39 × 10-5) ÷ 98 = 1.42 × 10-7 M/s

Experiment 5: (1.39 × 10-5) ÷ 47.67 = 2.92 × 10-7 M/s

Experiment 6: (1.39 × 10-5) ÷ 22.67 = 6.13 × 10-7 M/s

Experiment 7: (1.39 × 10-5) ÷ 10.33 = 1.34 × 10-6 M/s

Moles of BrO3- = (Initial Volume) × (Initial Molarity)

0.040 × [2 × (2.683 × 10-5)] = 2.14 × 10-6 moles BrO3-

Doubling the [I-] also caused the reaction rate to double; the rate order for [I-] is 1

Doubling the [BrO3-] also caused the reaction rate to double; the rate order for [BrO3

-] is 1

Tripling the rate of [H+] caused the reaction rate to increase by 16 folds; the rate order for [H+] is 2.5

Value of rate constant = (Reaction rate) ÷ [(Initial concentration of KI) × (Initial concentration of HCl)2.5 ×

(Initial concentration of KBrO3)]

Experiment 1: (8.08 × 10-8) ÷ [(1.67 × 10-3) × (0.0167)2.5 × (6.67 × 10-3)] = 201 M-3.5s-1

Experiment 2: (1.64 × 10-7) ÷ [(3.33 × 10-3) × (0.0167)2.5 × (6.67 × 10-3)] = 204 M-3.5s-1

Experiment 3: (3.43 × 10-7) ÷ [(5.00 × 10-3) × (0.0167)2.5 × (6.67 × 10-3)] = 285 M-3.5s-1

Experiment 4: (1.42 × 10-7) ÷ [(1.67 × 10-3) × (0.0167)2.5 × (0.0133)] = 177 M-3.5s-1

Experiment 5: (2.92 × 10-7) ÷ [(1.67 × 10-3) × (0.0167)2.5 × (0.0200)] = 243 M-3.5s-1

Experiment 6: (6.13 × 10-7) ÷ [(1.67 × 10-3) × (0.0333)2.5 × (6.67 × 10-3)] = 272 M-3.5s-1

Experiment 7: (1.34 × 10-6) ÷ [(1.67 × 10-3) × (0.0500)2.5 × (6.67 × 10-3)] = 215 M-3.5s-1

Average value of rate constant = Σ(Value of rate constant) ÷ 7

(201 + 204 + 285 + 177 + 243 + 272 + 215) ÷ 7 = 228 M-3.5s-1

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Determination of Activation Energy

Approximate Temperature,

°C

Measured Temperature

(°C)

Measured Temperature

(K)

Measured Temperature-1,

(K-1)

Time of Reaction (sec)

Trial 1 Trial 2 Average

Time

0 0 273.15 3.66 × 10-3 228 195 212

20 24 297.15 3.37 × 10-3 166 202 148 172

40 41 314.15 3.18 × 10-3 62 59 60.5

Activation Energy Calculation

Experiment Measured

Temperature (K)

Measured Temperatur

e-1 (K-1)

Average Time (s)

Rate of Reaction

(M/s)

Rate Constant, k

(M-3.5s-1) ln(k)

1 273 3.66 × 10-3 212 6.6 × 10-8 160 5.09

2 297 3.37 × 10-3 172 8.1 × 10-8 2.0 × 102 5.30

3 314 3.18 × 10-3 60.5 2.3 × 10-7 570 6.35

Slope -2467.8

Activation Energy (kJ) 20.52

Measured Temperature = (Measured Temperature in °C) + 273

Experiment 1: 0 + 273 = 273 K

Experiment 2: 24 + 273 = 297 K

Experiment 3: 41 + 273 = 314 K

y = -2467.8x + 13.979

4.5

4.7

4.9

5.1

5.3

5.5

5.7

5.9

6.1

6.3

6.5

0.0031 0.0032 0.0033 0.0034 0.0035 0.0036 0.0037

ln(k

)

1/T (K-1)

1/T VS ln(k)

1/T VS ln(k)

Line of Best Fit

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Rate of Reaction = (Change in BrO3- concentration) ÷ (Average Time)

Experiment 1: (1.39 × 10-5) ÷ 212 = 6.56 × 10-8 M/s

Experiment 2: (1.39 × 10-5) ÷ 172 = 8.08 × 10-8 M/s

Experiment 3: (1.39 × 10-5) ÷ 60.5 = 2.30 × 10-7 M/s

Rate Constant = (Rate of Reaction) ÷ [(Initial concentration of KI) × (Initial concentration of HCl)2.5 ×

(Initial concentration of KBrO3)]

Experiment 1: (6.56 × 10-8) ÷ [(1.67 × 10-3) × (0.0167)2.5 × (6.67 × 10-3)] = 163 M-3.5s-1

Experiment 2: (8.08 × 10-8) ÷ [(1.67 × 10-3) × (0.0167)2.5 × (6.67 × 10-3)] = 201 M-3.5s-1

Experiment 3: (2.30 × 10-7) ÷ [(1.67 × 10-3) × (0.0167)2.5 × (6.67 × 10-3)] = 573 M-3.5s-1

Activation Energy = -[(Slope) × (Gas Constant)]

-[-2467.8 × (8.314 × 10-3)] = 20.52 kJ

Observe the Effect of a Catalyst on the Rate

Conditions Reaction Time, seconds

Trial 1 Trial 2 Trial 3 Average

Uncatalyzed Reaction 166 202 148 172

Catalyzed Reaction 56 62 61 59.67

Analysis: The hypothesis cannot be verified. Some errors have occurred during the process of the

experiment, altering the results to some degree. The reaction rates change as concentrations of

reactants change because increasing concentration increase the number of reactant molecules.

According to the collision theory, rate of reaction depends on the frequency, orientation, and energy.

Thus, increasing the concentration of reactants would also increase the amount of molecules which

results in higher chance of reactant molecules colliding and higher frequency. In order to determine the

rate law, the order of each reactant must be determine. To determine an order of a reactant, one must

see how the change in initial concentration affects the reaction rate. In order to do that, experiment

must be conduct in which the concentration of the key reactant, or the reactant will have its order

determined, varies while keeping other conditions, such as the concentration of other reactants and the

temperature, constant. By comparing the two data, the relationship between the key reactant’s

concentration and the reaction rate can be seen, allowing for the order of each reactant to be

determined. After all orders are determined, the rate law can be found. Reaction rate also increases as

temperature increases. This is because, at a certain temperature, the average kinetic energy of

molecules is the same. However, it is diverse in that some molecules have high energy while some have

low energy. Molecules with enough energy can react when collide with other molecules. Thus,

increasing the temperature would cause an increase the fraction of molecules that has enough energy

to react and so causes an increase in frequency and reaction rate. To determine the activation energy,

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the graph of ln(k) against 1/T must be made. After making the graph, calculate the line of best fit. The

slope of line of best fit is equal to negative activation energy over gas constant. Thus, by multiplying the

slope of the line of best fit with negative gas constant, activation energy could be determined. Reaction

rate is rate of reaction whereas specific rate constant is the rate constant at a specific temperature.

Reaction rate is affected by concentrations of reactants, temperature, etc. while rate constant is only

affected by the temperature. Catalyst helps increase the reaction rate. Activation energy was probably

lowered by the catalyst, allowing the more molecules to be able to react, and thus increase the

frequency and reaction rate. Two of the calculated orders were close to the reactant. However, the

other order was right in the middle of integer two and three. The check of the order gave the same

value for two of the calculated order, but not for the third. The points on the graph were slightly bent.

The data could be improved by using more solutions in the same ratio. This would decrease the chance

of error due to an inappropriate amount reactant. Also, a little too much or too little of some reactant

would not have much effect on the data if the experiment were to be done in a bigger scale.

Conclusion: The hypothesis could not be confirmed through this experiment. The results gotten from

the experiment showed that there were errors during the course of the experiment. Thus, these results

could not be use to prove that the hypothesis is correct. Some likely error that could have altered the

results was that the quantity of the solution was measured incorrectly. Because of the smallness in

scale, little change in the quantity of solution could cause a significant change in the data. For example,

adding one drop too much in an experiment of this scale could alter the data by a considerable amount.

It might have caused too much of the solution to be added into the overall solution, causing the

concentration of the reactant (if that solution was the reagent solution) to increase and the reaction

rate to increase. This could have caused the order of that reactant to increase. The reaction order was

used often for the later calculations. Thus, failure in measuring the solutions could alter almost all the

data. This is possibly what had happened during the course of this experiment. Another error that could

have occurred during the course of the experiment was that the viscosity of each liquid is different. The

liquid with a higher viscosity would accumulate more volume in one drop. Thus, if some of the reactant

has higher viscosity than other reactants, there will be more of that reactant which could increase the

reaction rate. If reaction rate increases while the volume was thought to be the same, it could cause an

increase to the reaction order calculated from the reaction rate. As noted earlier, the reaction order was

used to calculate many other data. Thus, this error could alter the results significantly. Some ways to

prevent these errors are to conduct the experiment in a larger scale. This would allow for some

inaccuracy in measurement. For example, if the experiment was conduct as 1 mL in place of one drop, a

single drop error would only cause one-fifth of error that would have been caused in the drops scale

(one drop is about one-fifth of a milliliter). The viscosity problems could be solved by this method as well

because the little change in volume due to the viscosity would cause a negligible effect on the data if the

experiment were to be conducted at a bigger scale.

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