Consider the table of initial rates for the reaction: [NH4¹+]o, mol/L 0.100 0.100 0.200 Experiment 1 2 3 What is the rate law for this reaction? NH4¹++NO₂¹-N₂ + 2H₂O. [NO₂¹]o, mol/L 0.0050 0.010 0.010 Initial Rate, mol/(Ls) 1.35 x 10-7 2.70 x 10-7 1.10 x 10-6
Consider the table of initial rates for the reaction: [NH4¹+]o, mol/L 0.100 0.100 0.200 Experiment 1 2 3 What is the rate law for this reaction? NH4¹++NO₂¹-N₂ + 2H₂O. [NO₂¹]o, mol/L 0.0050 0.010 0.010 Initial Rate, mol/(Ls) 1.35 x 10-7 2.70 x 10-7 1.10 x 10-6
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
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![Consider the table of initial rates for the reaction:
[NH4¹+]o, mol/L
0.100
0.100
0.200
Experiment
1
2
3
What is the rate law for this reaction?
A. rate = k [NH4¹+] [NO₂¹]¹
rate = k [NH4¹+]²[NO₂¹-]¹
B.
NH4¹+ + NO₂¹ → N₂ + 2H₂O.
[NO₂¹-]o, mol/L
0.0050
0.010
0.010
Initial Rate, mol/(Ls)
1.35 x 10-7
2.70 x 10-7
1.10 x 10-6
C. rate = k [NH4¹+]¹[NO₂¹-1²
D. rate = k [NH4¹+]¹[NO₂¹-]¹](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F032f9f2f-4a86-4f7c-abbd-a161821f35fd%2F8bf5ef80-c41c-409a-a439-9146bd2ce137%2Fpqgdkdt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Consider the table of initial rates for the reaction:
[NH4¹+]o, mol/L
0.100
0.100
0.200
Experiment
1
2
3
What is the rate law for this reaction?
A. rate = k [NH4¹+] [NO₂¹]¹
rate = k [NH4¹+]²[NO₂¹-]¹
B.
NH4¹+ + NO₂¹ → N₂ + 2H₂O.
[NO₂¹-]o, mol/L
0.0050
0.010
0.010
Initial Rate, mol/(Ls)
1.35 x 10-7
2.70 x 10-7
1.10 x 10-6
C. rate = k [NH4¹+]¹[NO₂¹-1²
D. rate = k [NH4¹+]¹[NO₂¹-]¹
![Cindy decomposed a metal hydroxide via the reaction: Mt(OH)3 MtO + H₂O, and then collected
concentration and time data. After analyzing on her TI84 the following data were obtained.
[Mt(OH)3] vs. time
1n[Mt(OH)3] vs. time
a = -1.72 x 10-5
b = 0.450
r² = 0.917
What is the order with respect to Mt(OH)3?
B. 1
A. 0
a = -5.80 x 10-5
b = -0.770
r² = 0.999
C. 2
[Mt(OH)3]¹ vs. time
a = 2.10 x 10-4
b = 2.00
r² = 0.981
D. 3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F032f9f2f-4a86-4f7c-abbd-a161821f35fd%2F8bf5ef80-c41c-409a-a439-9146bd2ce137%2F8ayjmt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Cindy decomposed a metal hydroxide via the reaction: Mt(OH)3 MtO + H₂O, and then collected
concentration and time data. After analyzing on her TI84 the following data were obtained.
[Mt(OH)3] vs. time
1n[Mt(OH)3] vs. time
a = -1.72 x 10-5
b = 0.450
r² = 0.917
What is the order with respect to Mt(OH)3?
B. 1
A. 0
a = -5.80 x 10-5
b = -0.770
r² = 0.999
C. 2
[Mt(OH)3]¹ vs. time
a = 2.10 x 10-4
b = 2.00
r² = 0.981
D. 3
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