5-1 Factors Affecting Reaction Rates Lab Report Dr. Z Edits
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Factors Affecting Reaction Rates
Kristen Brown
11/25/2023
Data
Activity 1
Data Table 1: Calibration Trial
Na
2
S
2
O
3
(drops)
Reaction time (sec)
1
8
28
2
7
23
3
6
19
Question 1:
How many drops will be used in the remaining experiments? I will be using 6 drops in the remaining experiments. Activity 2
Data Table 2a. Effects of KI (I
-
) Concentration Trial
KI
(drops)
HCl
(drops)
Starch
(drops)
H
2
O
(drops)
Na
2
S
2
O
3
(drops)
H
2
O
2
(mL)
Time
1
(sec)
Time
2
(sec)
Average
time
(T
avg
)
(sec)
Average
Rate
(1/T
avg
)
1
8
2
4
0
6
0.4
19.2
18.85
0.053
2
8
2
4
0
6
0.4
18.5
3
6
2
4
2
6
0.4
26.36
26.04
0.038
Trial
KI
(drops)
HCl
(drops)
Starch
(drops)
H
2
O
(drops)
Na
2
S
2
O
3
(drops)
H
2
O
2
(mL)
Time
1
(sec)
Time
2
(sec)
Average
time
(T
avg
)
(sec)
Average
Rate
(1/T
avg
)
4
6
2
4
2
6
0.4
25.72
5
4
2
4
4
6
0.4
32.77
32.48
0.031
6
4
2
4
4
6
0.4
32.18
7
2
2
4
6
6
0.4
72.89
72.32
0.014
8
2
2
4
6
6
0.4
71.75
Data Table 2b. Effects of H
2
O
2
Concentration Trial
KI
(drops)
HCl
(drops)
Starch
(drops)
H
2
O
(mL)
Na
2
S
2
O
3
(drops)
H
2
O
2
(mL)
Time
1
(sec)
Time
2
(sec)
Average
time
(T
avg
) (sec)
Average
Rate
(1/T
avg
)
1
8
2
4
0
6
0.4
19.48
19.19
0.052
2
8
2
4
0
6
0.4
18.89
3
8
2
4
0.1
6
0.3
27.10
26.87
0.037
4
8
2
4
0.1
6
0.3
26.63
5
8
2
4
0.2
6
0.2
43.37
43.16
0.023
6
8
2
4
0.2
6
0.2
42.94
7
8
2
4
0.3
6
0.1
99.28
98.97
0.010
8
8
2
4
0.3
0.1
98.66
©2016 2 Carolina Biological Supply Company
Data Table 2c. Effects of HCl (H
+
) Concentration
Trial
KI
(drops)
HCl
(%)
Starch
(drops)
H
2
O
(drops)
Na
2
S
2
O
3
(drops)
H
2
O
2
(mL)
Time
1
(sec)
Time
2
(sec)
Average
time
(T
avg
) (sec)
Average
Rate
(1/T
avg
)
1
8
100%
4
0
0.4
26.69
26.37
0.038
2
8
100%
4
0
0.4
26.05
3
8
75%
4
0
0.4
21.18
21.03
0.048
4
8
75%
4
0
0.4
20.88
5
8
50%
4
0
0.4
24.09
23.86
0.042
6
8
50%
4
0
0.4
23.62
7
8
25%
4
0
0.4
22.90
22.59
0.044
8
8
25%
4
0
0.4
22.29
Activity 3
Data Table 3. Temperature Effects
Water bath trial
Temperature of the water bath
(°C)
Reaction time (sec)
Cold water
5
04.07
Room-temperature water
25
01.01
Hot water
45
00.39
Question 2:
Explain how each of these treatments affected the reaction rate. Describe the effect at a molecular level.
a. Concentration – The concentration was kept the same for all temperature baths. The basic 8-2-4-6 + 0.4mL with no addition of water inside the test tubes was used.
b. Temperature – Each temperature bath rapidly changed color, almost immediately, other than the cold water bath. The hot water bath was immediate, the room temp water took a ©2016 2 Carolina Biological Supply Company
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second longer than the hot water bath, and the cold water bath took a couple seconds of swirling before it finally changed color. Either way, getting the reactants up to the same temperature as their baths drastically increased the reaction time compared to not getting the reactants up to temp. Activity 4
Data Table 4. Catalyst Trial
Water (drops)
Diluted CuSO
4
Solution (drops)
Reaction time (sec)
1
4
0
14.68
2
3
1
7.68
3
2
2
6.16
4
1
3
1.49
5
0
4
1.26
Question 3:
How did the addition of copper(II) sulfate affect the reaction rate? Did the amount of catalyst affect the reaction rate? After adding in the diluted copper(ii), trial 2 immediately started to turn a light purple, and the following trials with more drops of copper(ii) turned a milky purple. Trial 1 remained clear until the hydrogen peroxide was added since it had no copper(ii) added. The reactions that had the diluted copper(ii) turned purple almost immediately. I had to swirl trial 2 and trial 3, but trial 4 and 5 was immediate. Activity 5: Data Table 5. Orders of Reactants in the Rate Law
Determine the actual Rate Law: R = k[I
–
]
m
[H
2
O
2
]
n
[H
+
]
p
Calculated Reaction order (X) (e.g. (
concentration
1
concentration
2
)
X
=
rate
1
rate
2
)
You must show your full calculation work for credit. Reactant
Concentration
Average
Rate
(1/T
avg
)
Calculated Reaction
order (X)
Show work! Average
Reaction
order
Reaction
Order
(integer)
KI (I
)
8 drops
0.053
=
2.7088114
8
m = 3
(8/6)
x = 0.053/0.038
Ln(1.33333333) = Ln(1.39473684)
KI (I
)
6 drops
0.038
©2016 2 Carolina Biological Supply Company
Reactant
Concentration
Average
Rate
(1/T
avg
)
Calculated Reaction
order (X)
Show work! Average
Reaction
order
Reaction
Order
(integer)
0.33270575/
0.28768207
= 1.15650499
(6/4)
x = 0.038/0.031
Ln(1.5) = ln(1.22580645)
0.20359896/
= 0.40546511
= 0.50213682
KI (I
)
4 drops
0.031
(4/2)
x = 0.031/0.014
Ln(2) = ln(2.21428571)
0.79492987/
0.69314718
= 1.14684138
KI (I
)
2 drops
0.014
You must show your full calculation work for credit. Reactant
Concentration
Average
Rate
(1/T
avg
)
Calculated Reaction
order (X)
Show work! Average
Reaction
order
Reaction
Order
(integer)
H
2
O
2
0.4 mL
0.052
=
1.1857219
1
n = 1
(0.4/0.3)
x = 0.052/0.037
Ln(1.33333333) = Ln(1.40540541)
0.34032581/
0.28768207
= 1.18299277
H
2
O
2
0.3 mL
0.037
©2016 2 Carolina Biological Supply Company
Reactant
Concentration
Average
Rate
(1/T
avg
)
Calculated Reaction
order (X)
Show work! Average
Reaction
order
Reaction
Order
(integer)
(0.3/0.2)
x = 0.037/0.023
Ln(1.5) = ln(1.60869565)
0.4754237/
0.40546511
= 1.1725391
H
2
O
2
0.2 mL
0.023
(0.2/0.1)
x = 0.023/0.010
Ln(2) = ln(2.3)
0.83290912/
0.69314718
= 1.20163386
H
2
O
2
0.1 mL
0.010
You must show your full calculation work for credit. Reactant
Concentration
Average
Rate
(1/T
avg
)
Calculated Reaction
order (X)
Show work! Average
Reaction
order
Reaction
Order
(integer)
HCl (H
+
)
100%
0.038
-
0.1832814
7
p = 0
(1.0/0.75)
x = 0.038/0.048
Ln(1.33333333) = Ln(0.79166667)
-0.23361485/
0.28768207
= -0.81205913
HCl (H
+
)
75%
0.048
(0.75/0.50)
x = 0.048/0.042
Ln(1.5) = ln(1.14285714)
0.13353139/
0.40546511
= 0.32932893
HCl (H
+
)
50%
0.042
(0.50/0.25)
x = ©2016 2 Carolina Biological Supply Company
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Reactant
Concentration
Average
Rate
(1/T
avg
)
Calculated Reaction
order (X)
Show work! Average
Reaction
order
Reaction
Order
(integer)
0.042/0.044
Ln(2) = ln(0.95454545)
-0.04652002/
0.69314718
= -0.0671142
HCl (H
+
)
25%
0.044
Rate Law
third order with respect to I-, first order with respect to H
2
O
2
, and zero order with respect to H+
Overall Order of the Reaction
3 + 1 + 0 = 4
Question 4:
Use the rate-law expression you
determined for Data Table 5 to answer the following questions. Answer using doubled, tripled, quadrupled, halved, quartered, no change, etc. A)
If the concentration of I
–
was doubled, how would that affect the reaction rate?
If the reaction is first order with respect to
I-, then doubling the concentration of
I-
would double the reaction rate. If the reaction is second order with respect to
I-, then doubling the concentration of
I-
would quadruple the reaction rate. Zero order with respect to I- would have no change to the rate.
B)
If the concentration of H
2
O
2
was halved, how would that affect the reaction rate?
If the reaction is first order with respect to
H
2
O
2
, then halving the concentration of
H
2
O
2
would halve the reaction rate. If the reaction is second order with respect to
H
2
O
2
, then halving the concentration of
H
2
O
2
would reduce the reaction rate to a quarter of its original rate. Zero order with respect to H
2
O
2 would have no change to the rate.
©2016 2 Carolina Biological Supply Company
C)
If the concentrations of I
–
and H
2
O
2
were both doubled, how would that affect the reaction rate?
If the reaction is first order with respect to both
I-
and
H
2
O
2
, then doubling the concentrations of both would quadruple the reaction rate. This is because the rate is proportional to the concentration in a first-order reaction, so if you double the concentration, you double the rate. If you double the concentration of two first-order reactants, you quadruple the rate.
If the reaction is second order with respect to
I-
and first order with respect to
H2O2, then doubling the concentrations of both would increase the reaction rate by a factor of 8. This is because in a second-order reaction, the rate is proportional to the square of the concentration, so if you double the concentration, you quadruple the rate. If you double the concentration of a second-order reactant and a first-order reactant, you
increase the rate by a factor of 8.
D)
If the concentration of H
+
(HCl) was doubled, how would that affect the reaction rate?
If the reaction is first order with respect to
H+, then doubling the concentration of
H+
would double the reaction rate. If the reaction is second order with respect to
H+, then doubling the concentration of
H+
would quadruple the reaction rate. Zero order with respect to H+ would have no change to the rate.
©2016 2 Carolina Biological Supply Company
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Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY