balanced chemical equation and the molecularity of each step, and write the rate law for the fifth step. * 2NO(g) 2H,(g) (1) (2) (3) N,O2(g) + H(g) (4) HO(g) + H(g) (5) H(g) + N,0(g) N,O2(g) 4H(g) N,0(g) + HO(g) H,O(g) HO(g) + N,(g) O 2NO(g) + 2H2(g) -> N2(g) + 2H20(g); all are bimolecular; rate5 = k5 [H][N20] O NO(g) + H2(g) --> N2(g) + 2H20(g); all are bimolecular; rate5 = k5 [N20] 2NO(g) + 2H2(g) -> N2(g) + 2H20(g); step 2 is unimolecular while the others are bimolecular, rate5 = k5 [H][N20] NO(g) + H2(g) --> N2(g) + 2H20(g); step 2 is unimolecular while the others are bimolecular, rate5 = k5 [H][N20] %3!
balanced chemical equation and the molecularity of each step, and write the rate law for the fifth step. * 2NO(g) 2H,(g) (1) (2) (3) N,O2(g) + H(g) (4) HO(g) + H(g) (5) H(g) + N,0(g) N,O2(g) 4H(g) N,0(g) + HO(g) H,O(g) HO(g) + N,(g) O 2NO(g) + 2H2(g) -> N2(g) + 2H20(g); all are bimolecular; rate5 = k5 [H][N20] O NO(g) + H2(g) --> N2(g) + 2H20(g); all are bimolecular; rate5 = k5 [N20] 2NO(g) + 2H2(g) -> N2(g) + 2H20(g); step 2 is unimolecular while the others are bimolecular, rate5 = k5 [H][N20] NO(g) + H2(g) --> N2(g) + 2H20(g); step 2 is unimolecular while the others are bimolecular, rate5 = k5 [H][N20] %3!
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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Question
![Below are the elementary steps are proposed for a mechanism, what is the
balanced chemical equation and the molecularity of each step, and write the rate
law for the fifth step. *
2NO(g)
2H,(8)
(1)
(2)
(3) N2O2(g) + H(g)
(4) HO(g) + H(g)
(5) H(g) + N,O(g) –→ HO(g) + N2(g)
N,O2(g)
4H(g)
N,0(g) + HO(g)
H,O(g)
2NO(g) + 2H2(g) -> N2(g) + 2H20(g); all are bimolecular; rate5 = k5 [H][N20]
O NO(g) + H2(g) --> N2(g) + 2H20(g); all are bimolecular; rate5 = k5 [N20]
2NO(g) + 2H2(g) -> N2(g) + 2H20(g); step 2 is unimolecular while the others are
bimolecular; rate5 = k5 [H][N20]
NO(g) + H2(g) --> N2(g) + 2H20(g); step 2 is unimolecular while the others are
bimolecular; rate5 = k5 [H][N20]
%3D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F39436fd9-6408-4e00-b831-60d6b2073658%2F91430fac-4396-468d-a0dd-26b6dab07c78%2Fh0i6b9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Below are the elementary steps are proposed for a mechanism, what is the
balanced chemical equation and the molecularity of each step, and write the rate
law for the fifth step. *
2NO(g)
2H,(8)
(1)
(2)
(3) N2O2(g) + H(g)
(4) HO(g) + H(g)
(5) H(g) + N,O(g) –→ HO(g) + N2(g)
N,O2(g)
4H(g)
N,0(g) + HO(g)
H,O(g)
2NO(g) + 2H2(g) -> N2(g) + 2H20(g); all are bimolecular; rate5 = k5 [H][N20]
O NO(g) + H2(g) --> N2(g) + 2H20(g); all are bimolecular; rate5 = k5 [N20]
2NO(g) + 2H2(g) -> N2(g) + 2H20(g); step 2 is unimolecular while the others are
bimolecular; rate5 = k5 [H][N20]
NO(g) + H2(g) --> N2(g) + 2H20(g); step 2 is unimolecular while the others are
bimolecular; rate5 = k5 [H][N20]
%3D
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