+Concentration effects on reaction rates Run [1] [S2O8²-] time (mol/L) (mol/L) (sec) la 0.08 0.04 1/time (sec-1) 41.0 0.0244 temp log[I] log log(time) (°C) [S2O82-] 23.3 -1.10 -1.40 1.61 lb 0.08 0.04 34.0 0.0294 22.9 -1.10 -1.40 1.53 lc 0.08 0.04 56.0 0.0179 21.6 -1.10 -1.40 1.75 1 avg 0.08 0.04 43.7 0.0239 22.6 -1.10 -1.40 1.63 2 0.06 0.04 54.7 0.0183 23.0 -1.22 -1.40 1.74 3 0.04 0.04 80.0 0.0125 22.8 -1.40 -1.40 1.90 4 0.02 0.04 179.0 0.0127 21.6 -1.70 -1.40 1.90 5 0.08 0.03 51.0 0.0196 22.4 -1.10 -1.52 1.70 6 0.08 0.02 95.0 0.0105 23.4 -1.10 -1.70 1.98 7 0.08 0.01 197.0 0.0051 23.4 -1.10 -2.00 2.30 8 0.08 0.04 16.1 0.0621 22.6 -1.10 -1.40 1.20 hanisms of Reactions-1 Saved V This question is not optional a. Assuming the rate expression for this system is in the form, rate - k[I][S208²], it should be possible to determine the values of x and y from the data obtained. Show why a plot of log(time) vs log[I] for runs 1-4 will give a slope of -x and log(time) vs. log[S208] for runs 1, 5, 6, and 7 will give a slope of-y. Perform the graphing exercises listed above and determine the values of x and y. Insert your graphs from Excel here (they can both be on the same graph). Persulfate Log [1] mol/L Log (Time) vs Log [1] -1.8 -16 y=-1.7303x + 1.7465 R² = 0.766 -1.4 -12 -1 -0.8 -0.6 -0.4 -0.2 16 1.65 1.7 1.75 1.8 1.85 19 1.95 0 Log (Time) -25 Log (Persulfate) vs Log Time -2 -15 -1 -0.5 y=-0.8502x-0.0374 R=0.9882 0.5 1 15 Time N 2,5
+Concentration effects on reaction rates Run [1] [S2O8²-] time (mol/L) (mol/L) (sec) la 0.08 0.04 1/time (sec-1) 41.0 0.0244 temp log[I] log log(time) (°C) [S2O82-] 23.3 -1.10 -1.40 1.61 lb 0.08 0.04 34.0 0.0294 22.9 -1.10 -1.40 1.53 lc 0.08 0.04 56.0 0.0179 21.6 -1.10 -1.40 1.75 1 avg 0.08 0.04 43.7 0.0239 22.6 -1.10 -1.40 1.63 2 0.06 0.04 54.7 0.0183 23.0 -1.22 -1.40 1.74 3 0.04 0.04 80.0 0.0125 22.8 -1.40 -1.40 1.90 4 0.02 0.04 179.0 0.0127 21.6 -1.70 -1.40 1.90 5 0.08 0.03 51.0 0.0196 22.4 -1.10 -1.52 1.70 6 0.08 0.02 95.0 0.0105 23.4 -1.10 -1.70 1.98 7 0.08 0.01 197.0 0.0051 23.4 -1.10 -2.00 2.30 8 0.08 0.04 16.1 0.0621 22.6 -1.10 -1.40 1.20 hanisms of Reactions-1 Saved V This question is not optional a. Assuming the rate expression for this system is in the form, rate - k[I][S208²], it should be possible to determine the values of x and y from the data obtained. Show why a plot of log(time) vs log[I] for runs 1-4 will give a slope of -x and log(time) vs. log[S208] for runs 1, 5, 6, and 7 will give a slope of-y. Perform the graphing exercises listed above and determine the values of x and y. Insert your graphs from Excel here (they can both be on the same graph). Persulfate Log [1] mol/L Log (Time) vs Log [1] -1.8 -16 y=-1.7303x + 1.7465 R² = 0.766 -1.4 -12 -1 -0.8 -0.6 -0.4 -0.2 16 1.65 1.7 1.75 1.8 1.85 19 1.95 0 Log (Time) -25 Log (Persulfate) vs Log Time -2 -15 -1 -0.5 y=-0.8502x-0.0374 R=0.9882 0.5 1 15 Time N 2,5
Introduction to General, Organic and Biochemistry
11th Edition
ISBN:9781285869759
Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Chapter7: Reaction Rates And Chemical Equilibrium
Section: Chapter Questions
Problem 7.56P
Related questions
Question
I would like my graphs checked please. Do they look right? Do I have iodine and persulfate on the right axis ?
![+Concentration effects on reaction rates
Run
[1]
[S2O8²-]
time
(mol/L)
(mol/L)
(sec)
la
0.08
0.04
1/time
(sec-1)
41.0 0.0244
temp log[I] log log(time)
(°C)
[S2O82-]
23.3 -1.10
-1.40
1.61
lb
0.08
0.04
34.0
0.0294
22.9 -1.10
-1.40
1.53
lc
0.08
0.04
56.0
0.0179
21.6
-1.10
-1.40
1.75
1 avg
0.08
0.04
43.7
0.0239
22.6
-1.10
-1.40
1.63
2
0.06
0.04
54.7
0.0183
23.0
-1.22
-1.40
1.74
3
0.04
0.04
80.0
0.0125
22.8
-1.40
-1.40
1.90
4
0.02
0.04
179.0
0.0127
21.6
-1.70
-1.40
1.90
5
0.08
0.03
51.0
0.0196
22.4
-1.10
-1.52
1.70
6
0.08
0.02
95.0
0.0105
23.4
-1.10
-1.70
1.98
7
0.08
0.01
197.0
0.0051
23.4
-1.10
-2.00
2.30
8
0.08
0.04
16.1
0.0621
22.6
-1.10
-1.40
1.20](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F98788fff-5d55-4f08-bd33-832b3c76eea7%2Fa40a116b-bd3c-4ee8-a427-f32328a29cae%2Fdb8sfes_processed.jpeg&w=3840&q=75)
Transcribed Image Text:+Concentration effects on reaction rates
Run
[1]
[S2O8²-]
time
(mol/L)
(mol/L)
(sec)
la
0.08
0.04
1/time
(sec-1)
41.0 0.0244
temp log[I] log log(time)
(°C)
[S2O82-]
23.3 -1.10
-1.40
1.61
lb
0.08
0.04
34.0
0.0294
22.9 -1.10
-1.40
1.53
lc
0.08
0.04
56.0
0.0179
21.6
-1.10
-1.40
1.75
1 avg
0.08
0.04
43.7
0.0239
22.6
-1.10
-1.40
1.63
2
0.06
0.04
54.7
0.0183
23.0
-1.22
-1.40
1.74
3
0.04
0.04
80.0
0.0125
22.8
-1.40
-1.40
1.90
4
0.02
0.04
179.0
0.0127
21.6
-1.70
-1.40
1.90
5
0.08
0.03
51.0
0.0196
22.4
-1.10
-1.52
1.70
6
0.08
0.02
95.0
0.0105
23.4
-1.10
-1.70
1.98
7
0.08
0.01
197.0
0.0051
23.4
-1.10
-2.00
2.30
8
0.08
0.04
16.1
0.0621
22.6
-1.10
-1.40
1.20
![hanisms of Reactions-1 Saved V
This question is not optional
a. Assuming the rate expression for this system is in the form, rate - k[I][S208²],
it should be possible to determine the values of x and y from the data obtained.
Show why a plot of log(time) vs log[I] for runs 1-4 will give a slope of -x and
log(time) vs. log[S208] for runs 1, 5, 6, and 7 will give a slope of-y. Perform
the graphing exercises listed above and determine the values of x and y. Insert
your graphs from Excel here (they can both be on the same graph).
Persulfate
Log [1] mol/L
Log (Time) vs Log [1]
-1.8
-16
y=-1.7303x + 1.7465
R² = 0.766
-1.4
-12
-1
-0.8
-0.6
-0.4
-0.2
16
1.65
1.7
1.75
1.8
1.85
19
1.95
0
Log (Time)
-25
Log (Persulfate) vs Log Time
-2
-15
-1
-0.5
y=-0.8502x-0.0374
R=0.9882
0.5
1
15
Time
N
2,5](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F98788fff-5d55-4f08-bd33-832b3c76eea7%2Fa40a116b-bd3c-4ee8-a427-f32328a29cae%2Fnnm3x3k_processed.jpeg&w=3840&q=75)
Transcribed Image Text:hanisms of Reactions-1 Saved V
This question is not optional
a. Assuming the rate expression for this system is in the form, rate - k[I][S208²],
it should be possible to determine the values of x and y from the data obtained.
Show why a plot of log(time) vs log[I] for runs 1-4 will give a slope of -x and
log(time) vs. log[S208] for runs 1, 5, 6, and 7 will give a slope of-y. Perform
the graphing exercises listed above and determine the values of x and y. Insert
your graphs from Excel here (they can both be on the same graph).
Persulfate
Log [1] mol/L
Log (Time) vs Log [1]
-1.8
-16
y=-1.7303x + 1.7465
R² = 0.766
-1.4
-12
-1
-0.8
-0.6
-0.4
-0.2
16
1.65
1.7
1.75
1.8
1.85
19
1.95
0
Log (Time)
-25
Log (Persulfate) vs Log Time
-2
-15
-1
-0.5
y=-0.8502x-0.0374
R=0.9882
0.5
1
15
Time
N
2,5
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