The data given in the table below is for the following reaction (run at 25°C): 2 NO+ O → 2 NO, Initial Concentration (M) Experiment Initial rate NO Mol/(L - s) 1 0.0010 0.0010 7 x 10 2. 0.0010 0.0020 14 x 10 3. 0.0010 0.0030 21 x 10+ 4. 0.0020 0.0030 84 x 10* 0.0030 0.0030 189 x 10 1. Determine the orders of the reaction with respect to NO and O, (show all work): (2-2) 14810-Ms [00010] Lo.002] IK [0.0010]"[0.001o NO 2 od ocder 2= (2)°=70-1 (3-4) 84x6-6 Mi- klo002a". [o.0030] k [a001o". [a0080] 40(2)2m-2 02= 44 order 71
The data given in the table below is for the following reaction (run at 25°C): 2 NO+ O → 2 NO, Initial Concentration (M) Experiment Initial rate NO Mol/(L - s) 1 0.0010 0.0010 7 x 10 2. 0.0010 0.0020 14 x 10 3. 0.0010 0.0030 21 x 10+ 4. 0.0020 0.0030 84 x 10* 0.0030 0.0030 189 x 10 1. Determine the orders of the reaction with respect to NO and O, (show all work): (2-2) 14810-Ms [00010] Lo.002] IK [0.0010]"[0.001o NO 2 od ocder 2= (2)°=70-1 (3-4) 84x6-6 Mi- klo002a". [o.0030] k [a001o". [a0080] 40(2)2m-2 02= 44 order 71
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 7.61P
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How do I find the overall order and the rate constant?
![The data given in the table below is for the following reaction (run at 25°C):
2 NO+ O
→ 2 NO,
Initial Concentration (M)
Experiment
Initial rate
NO
Mol/(L - s)
1
0.0010
0.0010
7 x 10
2.
0.0010
0.0020
14 x 10
3.
0.0010
0.0030
21 x 10+
4.
0.0020
0.0030
84 x 10*
0.0030
0.0030
189 x 10
1. Determine the orders of the reaction with respect to NO and O, (show all work):
(2-2)
14810-Ms [00010] Lo.002]
IK [0.0010]"[0.001o
NO 2 od ocder
2= (2)°=70-1
(3-4)
84x6-6 Mi- klo002a". [o.0030]
k [a001o". [a0080]
40(2)2m-2
02= 44 order
71](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc247cb8d-78ad-4f89-a50f-ab06364ea5f4%2F827a405d-e127-4fb7-8a65-895fbff4c25d%2F1hczqt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The data given in the table below is for the following reaction (run at 25°C):
2 NO+ O
→ 2 NO,
Initial Concentration (M)
Experiment
Initial rate
NO
Mol/(L - s)
1
0.0010
0.0010
7 x 10
2.
0.0010
0.0020
14 x 10
3.
0.0010
0.0030
21 x 10+
4.
0.0020
0.0030
84 x 10*
0.0030
0.0030
189 x 10
1. Determine the orders of the reaction with respect to NO and O, (show all work):
(2-2)
14810-Ms [00010] Lo.002]
IK [0.0010]"[0.001o
NO 2 od ocder
2= (2)°=70-1
(3-4)
84x6-6 Mi- klo002a". [o.0030]
k [a001o". [a0080]
40(2)2m-2
02= 44 order
71
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