Abs vs Time Trial 1 Ln(Abs) vs Time Trial 1 1/Abs vs Time Trial 1 0.35 0.00 9.00 20 40 60 80 100 120 140 160 180 8.00 y= 0.0297x + 2.4624 R=0.989 0.3 y=-0.0012x + 0.3231 -a.s0 7.00 R= 0.9771 0.25 y=0.0059x - 1.0624 6.00 0.2 -1.00 R0.9993 5.00 0.15 4.00 . .. -1.50 3.00 0.1 2.00 0.05 -2.00 1.00 0.00 20 40 60 80 100 120 140 160 180 -2.50 20 40 60 80 100 120 140 160 180

Chemistry for Engineering Students
3rd Edition
ISBN:9781285199023
Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter11: Chemical Kinetics
Section: Chapter Questions
Problem 11.4PAE: What are the steps in the Chapman cycle? Explain the importance.
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I need help finding the rate law for each graph. The first graph is in order zero, the middle is in order first and the last is in order second. 

Abs vs Time Trial 1
Ln(Abs) vs Time Trial 1
1/Abs vs Time Trial 1
0.35
0.00
9.00
20
40
60
80
100
120
140
160
180
8.00
y = 0.0297x + 2.4624
R2 = 0.989
0.3
y = -0.0012x + 0.3231
R2 = 0.9771
-0.50
7.00
0.25
y = -0.0059x - 1.0624
R2 = 0.9993
6.00
0.2
-1.00
5.00
...
4.00
0.15
-1.50
3.00
....
0.1
2.00
0.05
-2.00
1.00
0.00
40
60
80
100
120
140
160
180
-2.50
20
40
60
80
100
120
140
160
180
20
Transcribed Image Text:Abs vs Time Trial 1 Ln(Abs) vs Time Trial 1 1/Abs vs Time Trial 1 0.35 0.00 9.00 20 40 60 80 100 120 140 160 180 8.00 y = 0.0297x + 2.4624 R2 = 0.989 0.3 y = -0.0012x + 0.3231 R2 = 0.9771 -0.50 7.00 0.25 y = -0.0059x - 1.0624 R2 = 0.9993 6.00 0.2 -1.00 5.00 ... 4.00 0.15 -1.50 3.00 .... 0.1 2.00 0.05 -2.00 1.00 0.00 40 60 80 100 120 140 160 180 -2.50 20 40 60 80 100 120 140 160 180 20
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