equilibrium: N2(g) + O2(g) 2NO(g) At a certain temperature the equilibrium const

Question

equilibrium: N2(g) + O2(g) 2NO(g)
At a certain temperature the equilibrium constant for the reaction is 0.0255. What is the partial pressure of NO gas at equilibrium if the initial pressure of all the gases (both reactants and products) is 0.300 atm?

Explanation / Answer

N2(g) + O2(g) ------>> 2NO(g) initially 0.3 atm 0.3 atm 0 atm at eqm 0.3 - y 0.3 - y 2y K = [NO]^2/([N2][O2]) 0.0255 = (2y)^2/(0.3-y)^2 taking square root of both the sides, 0.1597 = 2y/(0.3-y) 0.0048 - 0.1597y = 2y y = 0.0022 the partial pressure of NO gas at equilibrium = 2y = 0.0044 atm

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