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1 N2(g) + O2(g) =NO2(g) Kp = 1.0 × 10 1 NOC(g) + O2(g) = NO2CI(g) Kp = 1.1 × 10² 2 1 NO2(g) + Cl2(g) = NO2CI(g) Kp 0.3

1 N2(g) + O2(g) =NO2(g) Kp = 1.0 × 10 1 NOC(g) + O2(g) = NO2CI(g) Kp = 1.1 × 10² 2 1 NO2(g) + Cl2(g) = NO2CI(g) Kp 0.3

Living By Chemistry: First Edition Textbook
Living By Chemistry: First Edition Textbook
1st Edition
ISBN:9781559539418
Author:Angelica Stacy
Publisher:Angelica Stacy
ChapterU6: Showtime: Reversible Reactions And Chemical Equilibrium
SectionU6.2: How Backward: Reversible Reactions
Problem 1E
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From the following equation with their corresponding equilibrium constants at 298 K,find the Kc of the reaction given below.

N2(9) + O2(g) + Cl2(g) 큰 NOCI2 2 NOC K = ? + 2(g)
Transcribed Image Text:N2(9) + O2(g) + Cl2(g) 큰 NOCI2 2 NOC K = ? + 2(g)
N2(g) + O2(g) = NO2(g) Kp = 1.0 × 10–9 NOCI(g) + O2(g) = NO2CI(g) 2 Kp = 1.1 × 10² NO2(g) + Cl2(g) 2 NO2CI(g) Kp = 0.3
Transcribed Image Text:N2(g) + O2(g) = NO2(g) Kp = 1.0 × 10–9 NOCI(g) + O2(g) = NO2CI(g) 2 Kp = 1.1 × 10² NO2(g) + Cl2(g) 2 NO2CI(g) Kp = 0.3
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