At a given temperature, the equilibrium constant (Kp) for the decomposition of dinitrogen tetraoxide to nitrogen dioxide is 0.154. If 0.198 atm N2O4 is sealed in a flask, what partial pressure (in atm) of NO2 will exist at equilibrium? N2O4(g) ⇄ 2NO2(g)
At a given temperature, the equilibrium constant (Kp) for the decomposition of dinitrogen tetraoxide to nitrogen dioxide is 0.154. If 0.198 atm N2O4 is sealed in a flask, what partial pressure (in atm) of NO2 will exist at equilibrium? N2O4(g) ⇄ 2NO2(g)
General Chemistry - Standalone book (MindTap Course List)
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter14: Chemical Equilibirum
Section: Chapter Questions
Problem 14.62QP: The following reaction has an equilibrium constant Kc equal to 3.59 at 900C....
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At a given temperature, the equilibrium constant (Kp) for the decomposition of dinitrogen tetraoxide to nitrogen dioxide is 0.154. If 0.198 atm N2O4 is sealed in a flask, what partial pressure (in atm) of NO2 will exist at equilibrium?
N2O4(g) ⇄ 2NO2(g)
(no scientific notation)
Expert Solution
Introduction
In equilibrium concepts, it is quite common to come across various equilibrium constants. The equilibrium constant (K) becomes quite helpful while expressing the relationship between products and reactants for a particular reaction (under equilibrium conditions).
Given data:
The equilibrium constant Kp is 0.154.
Initial pressure of N2O4 is 0.198 atm.
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