WIS AP Chem Lab #12 Kinetics Archer

13 Pages·2012·0.76 MB·English

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 (H O) 63 drops
2
0.10 M Hydrochloric Acid (HCl) 78 drops
2% Starch 30 drops
0.0010 M Sodium Thiosulfate (Na S O ) 30 drops
2 2 3
0.040 M Potassium Bromate (KBrO ) 78 drops
3
0.1 M Copper(II) Nitrate (Cu(NO ) ) 3 drops
3 2
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, H O, HCl, Starch, Na S O , and KBrO
2 2 2 3 3
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 Na S O
2 2 3
into a well (experiment 1)
5.) Add 2 drops of KBrO
3
6.) Stir the well with a toothpick
7.) Record the time from when KBrO was added until the first sign of blue color appears
3
8.) Repeat step 4 to 7 in another well for experiment 2: 4 drops KI, 2 drops H O, 2 drops HCl, 1 drop
2
starch, 1 drop of Na S O and 2 drops KBrO
2 2 3 3
9.) Repeat step 8 for experiment 3: 6 drops KI, 2 drops HCl, 1 drop starch, 1 drop Na S O , and 2
2 2 3
drops KBrO
3
10.) Repeat step 8 for experiment 4: 2 drops KI, 2 drops H O, 2 drops HCl, 1 drop starch, 1 drop
2
Na S O , and 4 drops KBrO
2 2 3 3
11.) Repeat step 8 for experiment 5: 2 drops KI, 2 drops HCl, 1 drop starch, 1 drop Na S O , and 6
2 2 3
drops KBrO
3
12.) Repeat step 8 for experiment 6: 2 drops KI, 2 drops H O, 4 drops HCl, 1 drop starch, 1 drop
2
Na S O , and 2 drops KBrO
2 2 3 3
13.) Repeat step 8 for experiment 7: 2 drops KI, 6 drops HCl, 1 drop starch, 1 drop Na S O , and 2
2 2 3
drops KBrO
3
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 Na S O
2 2 3
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 KBrO pipet about half full with the labeled liquid
3
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 KBrO into each of the 3 wells
3
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 KBrO into the other unreacted wells
3
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 Na S O ,
2 2 3
and 1 drop of Cu(NO ) into a well
3 2
2.) Add 2 drops of KBrO to the well
3
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
BrO - had been used up.
3
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6I-(aq) + BrO -(aq) + 6H+(aq)  3I (aq) + Br-(aq) + 3H O(l)
3 2 2
I (aq) + 2S O -(aq)  2I-(aq) + S O 2-(aq)
2 2 3 4 6

Na S O , KBrO ,
Experiment KI, 0.010 M Distilled HCl, 0.10 M Starch, 2 % 2 2 3 3
0.0010 M 0.040 M
Number (Drops) H O (Drops) (Drops) (Drops)
2 (Drops) (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 Concentration of Concentration of Concentration of Concentration of
Number KI (M) HCl (M) Na S O (M) KBrO (M)
2 2 3 3
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
Na S O (Experiment 1-7): 0.0010 × (1 ÷ 12) = 8.33 × 10-5 M Na S O
2 2 3 2 2 3
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KBrO (Experiment 1-3, 6-7): 0.040 × (2 ÷ 12) = 6.67 × 10-3 M KBrO
3 3
KBrO Experiment 4: 0.040 × (4 ÷ 12) = 0.0133 M KBrO
3 3
KBrO Experiment 5: 0.040 × (6 ÷ 12) = 0.0200 M KBrO
3 3
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
Mass of beaker plus 5 drops of water (g) 29.6302 29.5806 28.8205
Experiment
Mass of first 5 drops of water (g) 0.1285 0.0991 0.1322
1
Average mass of 1 drop of water (g) 0.02570 0.01982 0.02644
Mass of beaker plus 10 drops of water (g) 29.7543 29.7034 28.9387
Experiment
Mass of second 5 drops of water (g) 0.1241 0.1228 0.1182
2
Average mass of 1 drop of water (g) 0.02482 0.02456 0.02364
Mass of beaker plus 15 drops of water (g) 29.8813 29.8321 29.0655
Experiment
Mass of third 5 drops of water (g) 0.1270 0.1287 0.1268
3
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
Archer G11
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 S O 2- (M) 0.0010
2 3
Moles of S O 2- ions (moles) 2.5 × 10-8
2 3
Moles of BrO - reacted (moles) 4.1 × 10-9
3
Change in BrO - concentration (M) 1.4 × 10-5
3
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 BrO - (moles) 2.1 × 10-6
3
Rate Order
Rate Order of [I-] Rate Order of [BrO -] Rate Order of [H+]
3
1 1 2.5
Rate Law Expression Rate = k[I-][H+]2.5 [BrO -]
3
Rate Constant Data
Experiment 1 2 3 4 5 6 7
Value of
rate 2.0 × 102 2.0 × 102 290 180 240 270 220
constant,
Average Value of Rate Constant (M-3.5s-1) 228

Moles of S O 2- = (Molarity of S O 2-) × (Volume of 1 drop)
2 3 2 3
0.0010 × (2.461 × 10-5) = 2.46 × 10-8 moles S O -
2 3
Moles of BrO - reacted = (Moles of S O 2-) × (Mole ratio)
3 2 3
(2.46 × 10-8) × (1 ÷ 6) = 4.1 × 10-9 moles BrO -
3
Change in BrO - concentration = (Moles of BrO - reacted) ÷ (Volume of 12 drops)
3 3
(4.1 × 10-9) ÷ [12 × (2.461 × 10-5)] = 1.39 × 10-5 M BrO -
3
Rate = (Change in BrO - concentration) ÷ (Average time)
3
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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 BrO - = (Initial Volume) × (Initial Molarity)
3
0.040 × [2 × (2.683 × 10-5)] = 2.14 × 10-6 moles BrO -
3
Doubling the [I-]also caused the reaction rate to double; the rate order for [I-] is 1

Doubling the [BrO -] also caused the reaction rate to double; the rate order for [BrO -] is 1
3 3
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 KBrO )]
3
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
Archer G11
Determination of Activation Energy
Approximate Measured Measured Measured Time of Reaction (sec)
Temperature, Temperature Temperature Temperature-1,
Average
°C (°C) (K) (K-1) Trial 1 Trial 2
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
Measured Measured Rate of Rate
Average
Experiment Temperatur Temperatur Reaction Constant, k ln(k)
Time (s)
e (K) e-1 (K-1) (M/s) (M-3.5s-1)
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

1/T VS ln(k)
6.5
6.3
6.1
5.9
5.7
)
k 5.5
(
n 1/T VS ln(k)
l
5.3
Line of Best Fit
5.1
y = -2467.8x + 13.979
4.9
4.7
4.5
0.0031 0.0032 0.0033 0.0034 0.0035 0.0036 0.0037
1/T (K-1)

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
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Rate of Reaction = (Change in BrO - concentration) ÷ (Average Time)
3
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 KBrO )]
3
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
Reaction Time, seconds
Conditions
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,