1) 1/2 N2 (g) 6.96 x 10-16 2) 2 NO2 (g) 4.5 x 10-13 + 1/2 02 (g) NO 2 NO ← (g) Keg (g) + O2 (g) 3) 2 NO2 (g) N2₂04 (9) 6.7 Kea = = Keq = Net reaction: N₂ (g) + 2 O₂ (g) N₂O4 (9) ←
1) 1/2 N2 (g) 6.96 x 10-16 2) 2 NO2 (g) 4.5 x 10-13 + 1/2 02 (g) NO 2 NO ← (g) Keg (g) + O2 (g) 3) 2 NO2 (g) N2₂04 (9) 6.7 Kea = = Keq = Net reaction: N₂ (g) + 2 O₂ (g) N₂O4 (9) ←
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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
Transcribed Image Text:What would be the equilibrium constant of the net reaction
given the following reactions? When submitting answers in
scientific notation, please use E to represent "x10". An
example would be 4.1 x 10-2, this would be counted correct
as 4.1E-2.
1) 1/2 N2 (g) + 1/2 O2 (g) -
6.96 x 10-16
2) 2 NO2 (g).
4.5 x 10-13
3) 2 NO2 (g)
6.7
2 NO
NO (g)
N₂04 (g)
(g) Keq =
+ O2 (g)
(g)
Keq =
=
Keq=
Net reaction: N₂ (g) + 2 O₂ (g) N₂O4 (9)
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