The equilibrium constant is 1.48 for the reaction below at 800.0 K. What would be the equilibrium constant with respect to pressure, Kp, at this temperature? 2 NO2 (g) rit 2 NO + O2 (g) (g)
The equilibrium constant is 1.48 for the reaction below at 800.0 K. What would be the equilibrium constant with respect to pressure, Kp, at this temperature? 2 NO2 (g) rit 2 NO + O2 (g) (g)
Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter14: Chemical Equilibrium
Section: Chapter Questions
Problem 65P
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Question
![The equilibrium constant is 1.48 for the reaction below at
800.0 K. What would be the equilibrium constant with
respect to pressure, Kp, at this temperature?
2 NO2 (g)
دام
2 NO (g)
+ O2 (g)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe241491e-60fc-432d-9a68-dcbb1fe61326%2Fda00a275-28e4-4493-8188-6a0d528d106e%2F5hzw1ng_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The equilibrium constant is 1.48 for the reaction below at
800.0 K. What would be the equilibrium constant with
respect to pressure, Kp, at this temperature?
2 NO2 (g)
دام
2 NO (g)
+ O2 (g)
Expert Solution
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Step 1
Given reaction:-
2 NO2(g) <==> 2NO(g) + O2(g)
Here,
∆n = (Gaseous product)-(Gaseous reactant)
∆n = (2+1) - (2)
∆n = 1
=> Given Kc = 1.48
Temperature (T) = 800 K
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