For the reaction: NO2(g) → 1/2O2(g) + NO (g) The following data were obtained: [NO₂] (M) Natural Log [NO2] (M) 0.01 -4.605170186 100 Time (s) 0 50 100 300 200 0.00481 -5.337058195 0.00787 -4.844697217 127.064803 0.00649 -5.037492748 Concentration NO2 0.012 0.0038 -5.572754212 263.1578947 Determine the rate law of this reaction and the value of k, the rate constant. The graphs of the data appear below. 0.01 0.008 0.006 0.004 0.002 0 0 1/[NO2] 100 154.0832049 200 Time (s) 207.9002079 y=-1.97E-05x +9.15E-03 R² = 9.27E-01 1/[NO2] 300 250 200 150 8 100 50 。 300 0 50 400 100 Natural Log [NO2] 4.5 -4.7 -5.1 -5.3 -5.5 -5.7 0 y=0.5431x +99.833 R² = 1 150 200 Time (s) 100 250 200 Time (s) 300 350 y = -0.0032x-4.6679 R² = 0.9824 300 400
For the reaction: NO2(g) → 1/2O2(g) + NO (g) The following data were obtained: [NO₂] (M) Natural Log [NO2] (M) 0.01 -4.605170186 100 Time (s) 0 50 100 300 200 0.00481 -5.337058195 0.00787 -4.844697217 127.064803 0.00649 -5.037492748 Concentration NO2 0.012 0.0038 -5.572754212 263.1578947 Determine the rate law of this reaction and the value of k, the rate constant. The graphs of the data appear below. 0.01 0.008 0.006 0.004 0.002 0 0 1/[NO2] 100 154.0832049 200 Time (s) 207.9002079 y=-1.97E-05x +9.15E-03 R² = 9.27E-01 1/[NO2] 300 250 200 150 8 100 50 。 300 0 50 400 100 Natural Log [NO2] 4.5 -4.7 -5.1 -5.3 -5.5 -5.7 0 y=0.5431x +99.833 R² = 1 150 200 Time (s) 100 250 200 Time (s) 300 350 y = -0.0032x-4.6679 R² = 0.9824 300 400
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter11: Chemical Kinetics: Rates Of Reactions
Section: Chapter Questions
Problem 114QRT
Related questions
Question
![For the reaction:
NO2(g) → 1/2 O2(g) + NO (g)
The following data were obtained:
Time (s)
0
50
100
[NO₂]
(M)
0.01
Concentration NO2
0.00787 -4.844697217 127.064803
200 0.00481 -5.337058195
0.00649 -5.037492748 154.0832049
300 0.0038 -5.572754212
0.012
0.01
0.008
Natural Log
[NO2] (M)
Determine the rate law of this reaction and the value of k, the rate constant. The graphs of the data appear below.
0.006
0.004
-4.605170186 100
0.002
0
0
1/[NO2]
100
200
Time (s)
207.9002079
1/[NO2]
263.1578947
y = -1.97E-05x + 9.15E-03
R² = 9.27E-01
300
250
200
150
100
50
0
300
0
50
400
100
Natural Log [NO2]
150
-4.5
-4.7
-4.9
-5.1
-5.3
-5.5
-5.7
0
200
Time (s)
y = 0.5431x + 99.833
R² = 1
100
250
300
200
Time (s)
350
y = -0.0032x - 4.6679
R² = 0.9824
300
400](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5c433f37-c484-4136-9ada-41e91862a98c%2Fc31c568f-4f33-4a69-845c-326bc7e534bc%2F8iitzzi_processed.png&w=3840&q=75)
Transcribed Image Text:For the reaction:
NO2(g) → 1/2 O2(g) + NO (g)
The following data were obtained:
Time (s)
0
50
100
[NO₂]
(M)
0.01
Concentration NO2
0.00787 -4.844697217 127.064803
200 0.00481 -5.337058195
0.00649 -5.037492748 154.0832049
300 0.0038 -5.572754212
0.012
0.01
0.008
Natural Log
[NO2] (M)
Determine the rate law of this reaction and the value of k, the rate constant. The graphs of the data appear below.
0.006
0.004
-4.605170186 100
0.002
0
0
1/[NO2]
100
200
Time (s)
207.9002079
1/[NO2]
263.1578947
y = -1.97E-05x + 9.15E-03
R² = 9.27E-01
300
250
200
150
100
50
0
300
0
50
400
100
Natural Log [NO2]
150
-4.5
-4.7
-4.9
-5.1
-5.3
-5.5
-5.7
0
200
Time (s)
y = 0.5431x + 99.833
R² = 1
100
250
300
200
Time (s)
350
y = -0.0032x - 4.6679
R² = 0.9824
300
400
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