at the very end of the 2nd image. can you elaborate as to how Kc1 was achieved in 1/Kc1 ? I understand how to get the 1, but i dont understand why there is a Kc1 under the 1. The same thing applies to the Kc2 to the right of this. can you elaborate more on this please? Example: Given the equilibrium constants for the reactions2 NO(g) ⇒ N₂(g) + O₂(g)2 NO(g) + O₂(g) ⇒ 2 NO₂(g)Kc1 = 2.3 x 1024Kc₂ = 6.4 x 10⁹find the numerical value for Kc3, the equilibrium constant for thereaction=N₂(g) + 2 O₂(g) ⇒ 2 NO₂(g) Kc3 ? 2 NO(g) ⇒ N₂(g) + O₂(g)2 NO(g) + O₂(g) ⇒ 2 NO₂(g)N₂(g) + 2 O₂(g) ⇒ 2 NO₂(g)KC3[NO₂]² [NO]²[N₂][0₂]² [N₂][0₂]=—Kc1 = [N₂][O₂] = 2.3 x 10²4[NO]²Kc2KC3[NO]²[0₂]—12.3 x 1024[NO₂]²[NO]²[0₂][NO₂] = 1[NO₂]²[N₂][0₂]²= 6.4 x 10⁹KclKC2(6.4 x 10⁹) = 2.8 x 10-¹5

Answered: Image /qna-images/answer/5ca043ce-64d4-45a9-8522-4db56a88424b.jpg

Answered: Example: Given the equilibrium…

Chemistry

9th Edition

ISBN:9781133611097

Author:Steven S. Zumdahl

Publisher:Steven S. Zumdahl

Chapter13: Chemical Equilibrium

Section: Chapter Questions

Problem 13Q: Suppose a reaction has the equilibrium constant K = 1.3 108. What does the magnitude of this...

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at the very end of the 2nd image. can you elaborate as to how Kc1 was achieved in 1/Kc1 ? I understand how to get the 1, but i dont understand why there is a Kc1 under the 1. The same thing applies to the Kc2 to the right of this. can you elaborate more on this please?

Example: Given the equilibrium constants for the reactions
2 NO(g) ⇒ N₂(g) + O₂(g)
2 NO(g) + O₂(g) ⇒ 2 NO₂(g)
Kc1 = 2.3 x 1024
Kc₂ = 6.4 x 10⁹
find the numerical value for Kc3, the equilibrium constant for the
reaction
=
N₂(g) + 2 O₂(g) ⇒ 2 NO₂(g) Kc3 ?

2 NO(g) ⇒ N₂(g) + O₂(g)
2 NO(g) + O₂(g) ⇒ 2 NO₂(g)
N₂(g) + 2 O₂(g) ⇒ 2 NO₂(g)
KC3
[NO₂]² [NO]²
[N₂][0₂]² [N₂][0₂]
=
—
Kc1 = [N₂][O₂] = 2.3 x 10²4
[NO]²
Kc2
KC3
[NO]²[0₂]
—
1
2.3 x 1024
[NO₂]²
[NO]²[0₂]
[NO₂] = 1
[NO₂]²
[N₂][0₂]²
= 6.4 x 10⁹
Kcl
KC2
(6.4 x 10⁹) = 2.8 x 10-¹5

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But why is KC1 inverted? I still dont understand why. why isn't it just Kc2 x Kc1.

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