03 PreLab_Kinetics2-Erio_KEY (2)
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Subject
Chemistry
Date
Feb 20, 2024
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Pages
5
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Document ID:
104 Pre-Lab, Kinetics of Erioglaucine #2, KEY Date of last revision:
2/9/24 Page 1 of 5
Kinetics of Erioglaucine with Sodium Hypochlorite Part 2: Pre-Lab Activity ANSWER KEY
SAMPLE DATA
To familiarize you with the data processing you will complete for this lab, sample data has been created for a hypothetical reaction involving two reactants, A and B. Each line of the table below represents a single run of the experiment, and each run is comprised of many absorbance readings over a period of time. For the sake of space, the absorbance vs. time data sets are not provided here. Instead, processed data (similar to what you would read off the LabQuest output) is provided as the basis for these problems. Run [A] at t = 0 sec (M) [B] (M) Temp (°C) 1 7.24 x 10
-5
1.250 23.4 2 7.22 x 10
-5
1.250 23.3 3 3.61 x 10
-5
1.250 23.4 4 3.63 x 10
-5
1.250 23.4 5 7.25 x 10
-5
0.625 23.2 6 7.24 x 10
-5
0.625 23.5 7 7.23 x 10
-5
1.250 13.5 8 7.22 x 10
-5
1.250 13.4 METHOD OF INITIAL RATES (Part B of Lab) 1.
In order to determine the kinetics of the reaction via the Method of Initial Rates, what data will you plot for each of your eight runs? Concentration of A vs. Time
2.
You will take the first 10 seconds worth of data on each these plots and perform linear fits to these 10 data points. What valuable information can be obtained from this line? How does this relate to the kinetics of the reaction? Slope = ∆𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪
∆𝑻𝑻𝑪𝑪𝑻𝑻𝑪𝑪
This slope is used as the initial rate.
3.
The table below provides examples of linear curve fit data for each of the eight sample runs. Use this information to determine the initial rate for each run. Record these rates in the table below. Run [A] at t = 0 sec (M) [B] (M) Temp (°C) Equation of Line (t = 0s to t = 10s) Rate (M/s) 1 7.24 x 10
-5
1.250 23.4 y
= (4.32 x 10
-7
)
x + 7.24 x 10
-5
4.32 x 10
-7
2 7.22 x 10
-5
1.250 23.3 y
= (4.31 x 10
-7
)
x + 7.22 x 10
-5
4.31 x 10
-7
3 3.61 x 10
-5
1.250 23.4 y
= (1.09 x 10
-7
)
x + 3.61 x 10
-5
1.09 x 10
-7
4 3.63 x 10
-5
1.250 23.4 y
= (1.08 x 10
-7
)
x + 3.63 x 10
-5
1.08 x 10
-7
5 7.25 x 10
-5
0.625 23.2 y
= (2.14 x 10
-7
)
x + 7.25 x 10
-5
2.14 x 10
-7
6 7.24 x 10
-5
0.625 23.5 y
= (2.17 x 10
-7
)
x + 7.24 x 10
-5
2.17 x 10
-7
7 7.23 x 10
-5
1.250 13.5 y
= (2.33 x 10
-7
)
x + 7.23 x 10
-5
2.33 x 10
-7
8 7.22 x 10
-5
1.250 13.4 y
= (2.35 x 10
-7
)
x + 7.22 x 10
-5
2.35 x 10
-7
Document ID:
104 Pre-Lab, Kinetics of Erioglaucine #2, KEY Date of last revision:
2/9/24 Page 2 of 5
4.
From the initial rates you listed in the table, determine: a.
Order of reaction in A Second
b.
Order of reaction in B First
c.
Rate law Rate = k[A]
2
[B]
d.
Rate constant at room temperature Using data from Run 1, k = 65.9 M
-2
s
-1
k =
Rate
[A]
2
[B]
= 4.32 × 10
−7
M/s
(7.24 × 10
−5
M)
2
(1.250 M)
=
𝟔𝟔𝟔𝟔
.
𝟗𝟗
𝐌𝐌
−𝟐𝟐
𝐬𝐬
−𝟏𝟏
Determination of Activation Energy (Parts A and B of Lab) Kinetics studies provide the opportunity to determine activation energies for reactions through the use of the Arrhenius equation. By determining the value of k
at two different temperatures, the Arrhenius equation can be arranged into this combined equation: 𝐥𝐥𝐥𝐥 �
𝒌𝒌
𝟐𝟐
𝒌𝒌
𝟏𝟏
�
=
𝑬𝑬
𝑪𝑪
𝐑𝐑
�
𝟏𝟏
𝑻𝑻
𝟏𝟏
−
𝟏𝟏
𝑻𝑻
𝟐𝟐
�
where: k
1
is the rate constant at temperature T
1 k
2
is the rate constant at temperature T
2 E
a
is the activation energy in J/mol R is the universal gas constant (8.314 J/mol K) For each of the methods of analysis you use in this lab, you will be asked to determine the activation energy. Refer to the equation above when you are asked to complete this calculation. 5. Using the data from the table provided in Question 3 above, find the activation energy for this reaction. k
2
=
Rate
[A]
2
[B]
= 2.33 × 10
−7
M/s
(7.23 × 10
−5
M)
2
(1.250 M)
= 35.7 M
−2
s
−1
𝑬𝑬
𝑪𝑪
= R × �
ln
�
𝑘𝑘
2
𝑘𝑘
1
�
�
1
𝑇𝑇
1
−
1
𝑇𝑇
2
�
�
= (8.314 J mol
−1
K
−1
)
�
ln
�
35.7 M
−2
s
−1
65.9 M
−2
s
−1
�
�
1
296.5 K
−
1
286.6 K
�
�
= 43,745
J
mol
=
𝟒𝟒𝟒𝟒
.
𝟕𝟕
𝐤𝐤𝐤𝐤
/
𝐦𝐦𝐦𝐦𝐥𝐥
Document ID:
104 Pre-Lab, Kinetics of Erioglaucine #2, KEY Date of last revision:
2/9/24 Page 3 of 5
ANALYSIS BY INTEGRATED RATES (Part B of Lab)
In order to determine the kinetics of the reaction via the Method of Integrated Rates, you will plot three different types of plots and observe which one produces a linear relationship. 6.
Use the information organizer below to summarize what these three different types of plots and the information each provides. Plotted Data If Plot is Linear: Order of Reactant Slope [A]
t
vs. t
Zero
-
k
obs
ln[A]
t
vs. t
First
-
k
obs
1/[A]
t
vs. t
Second
k
obs
7. For the hypothetical reaction between A and B, the plot of 1/[A]
t
vs. t
for Run 1 is linear with a curve-
fit equation of y = 82.375
x + 13,812. The same plot for Run 5 is y = 41.095
x
+ 13,835. Based on this information, determine: a.
Order of reaction in A Second Plot of
1/[A]
t
vs. t is LINEAR
b.
The value of k
obs
for Run 1 82.4 M
-1
s
-1
Slope = k
obs
c.
The value of k
obs
for Run 5 41.1 M
-1
s
-1
Slope = k
obs
8. Determine the order of the reaction in B. Clearly show your reasoning/work. (Simple investigation or using the equation listed as Equation (8) in the Background section of this lab manual entry). 𝑏𝑏
= ln(
𝑘𝑘
𝑜𝑜𝑜𝑜𝑜𝑜2
)
−
ln (
𝑘𝑘
𝑜𝑜𝑜𝑜𝑜𝑜1
)
ln[
𝐵𝐵
]
2
−
ln[
𝐵𝐵
]
1
𝑏𝑏
= ln(
𝑘𝑘
𝑜𝑜𝑜𝑜𝑜𝑜2
)
−
ln (
𝑘𝑘
𝑜𝑜𝑜𝑜𝑜𝑜1
)
ln[
𝐵𝐵
]
2
−
ln[
𝐵𝐵
]
1
= 𝐥𝐥𝐥𝐥
(
𝟒𝟒𝟏𝟏
.
𝟏𝟏
)
− 𝐥𝐥𝐥𝐥
(
𝟖𝟖𝟐𝟐
.
𝟒𝟒
)
𝐥𝐥𝐥𝐥
(
𝟎𝟎
.
𝟔𝟔𝟐𝟐𝟔𝟔
𝐌𝐌
)
− 𝐥𝐥𝐥𝐥
(
𝟏𝟏
.
𝟐𝟐𝟔𝟔𝟎𝟎
𝐌𝐌
)
=
𝟏𝟏
𝑭𝑭𝑪𝑪𝑪𝑪𝑭𝑭𝑪𝑪
𝑪𝑪𝑪𝑪𝒐𝒐𝑪𝑪𝑪𝑪
𝑪𝑪𝑪𝑪
𝑩𝑩
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filler
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FeCl3
- Drop of H₂O
- Grinding
- rt
OH
1.13909
1, 11199
1.13579
Chemicals Required:
2.88069
2.8843
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Iron(III) chloride - 1.10 g
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-sumins
2.88 242
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P
Procedure:
2.8
0.1082
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