log [S208^2-] (mol/L) Log [1] mol/L a. Assuming the rate expression for this system is in the form, rate - k[I]*[S₂Os²], 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[S2082] for runs 1, 5, 6, and 7 will give a slope of-y. b. Perform the graphing exercises listed above and determine the values of x and y. Insert your graphs from Excel here. Log (Time) vs Log [1] 0 -0.2 1/6 1.65 1.7 1.75 1.8 1.85 1.9 1.95 -0.4 -0.6 -0.8 -1 -1.2 -1.4 -1.6 -1.8 0 0 -0.5 y=-1.7303x+1.7465 R²=0.766 Log (Time) Log (time) vs. log [S208^2-] 0.5 y=-0.8502x - 0.0374 R² = 0.9882 -1 -1.5 -2 -2.5 log (Time) 1.5 2 2.5 -1.4 -1.52 -1.7 -2
log [S208^2-] (mol/L) Log [1] mol/L a. Assuming the rate expression for this system is in the form, rate - k[I]*[S₂Os²], 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[S2082] for runs 1, 5, 6, and 7 will give a slope of-y. b. Perform the graphing exercises listed above and determine the values of x and y. Insert your graphs from Excel here. Log (Time) vs Log [1] 0 -0.2 1/6 1.65 1.7 1.75 1.8 1.85 1.9 1.95 -0.4 -0.6 -0.8 -1 -1.2 -1.4 -1.6 -1.8 0 0 -0.5 y=-1.7303x+1.7465 R²=0.766 Log (Time) Log (time) vs. log [S208^2-] 0.5 y=-0.8502x - 0.0374 R² = 0.9882 -1 -1.5 -2 -2.5 log (Time) 1.5 2 2.5 -1.4 -1.52 -1.7 -2
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter12: Chemical Kinetics
Section: Chapter Questions
Problem 1ALQ: Define stability from both a kinetic and thermodynamic perspective. Give examples to show the...
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![log [S208^2-] (mol/L)
Log [1] mol/L
a. Assuming the rate expression for this system is in the form, rate - k[I]*[S₂Os²],
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[S2082] for runs 1, 5, 6, and 7 will give a slope of-y.
b. Perform the graphing exercises listed above and determine the values of x and y.
Insert your graphs from Excel here.
Log (Time) vs Log [1]
0
-0.2 1/6
1.65
1.7
1.75
1.8
1.85
1.9
1.95
-0.4
-0.6
-0.8
-1
-1.2
-1.4
-1.6
-1.8
0
0
-0.5
y=-1.7303x+1.7465
R²=0.766
Log (Time)
Log (time) vs. log [S208^2-]
0.5
y=-0.8502x - 0.0374
R² = 0.9882
-1
-1.5
-2
-2.5
log (Time)
1.5
2
2.5
-1.4
-1.52
-1.7
-2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F98788fff-5d55-4f08-bd33-832b3c76eea7%2F2a4b6132-ae24-47cb-8ea5-9f3eceac21f3%2F7vf8eul_processed.jpeg&w=3840&q=75)
Transcribed Image Text:log [S208^2-] (mol/L)
Log [1] mol/L
a. Assuming the rate expression for this system is in the form, rate - k[I]*[S₂Os²],
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[S2082] for runs 1, 5, 6, and 7 will give a slope of-y.
b. Perform the graphing exercises listed above and determine the values of x and y.
Insert your graphs from Excel here.
Log (Time) vs Log [1]
0
-0.2 1/6
1.65
1.7
1.75
1.8
1.85
1.9
1.95
-0.4
-0.6
-0.8
-1
-1.2
-1.4
-1.6
-1.8
0
0
-0.5
y=-1.7303x+1.7465
R²=0.766
Log (Time)
Log (time) vs. log [S208^2-]
0.5
y=-0.8502x - 0.0374
R² = 0.9882
-1
-1.5
-2
-2.5
log (Time)
1.5
2
2.5
-1.4
-1.52
-1.7
-2
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