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Nitrogen Dioxide decomposes according to the reaction 2 NO2 (g) <----> 2 NO(g) + O2 (g)...

Nitrogen Dioxide decomposes according to the reaction 2 NO2 (g) <----> 2 NO(g) + O2 (g) Where Kp= 4.48 x 10-13 at a certain temperature. If 0.08 atm of NO2 is added to a container and allowed to come to equilibrium, what are the equilibrium partial pressures of NO(g) and O2(g)?

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Answer #1

           2 NO2 (g) <----> 2 NO(g) + O2 (g)

initial     0.08 atm           0 atm       0 atm

change      -2x               +2x          x

eequil     0.08-2x             2x           x


Kp = pO2*pNO^2/pNO2^2

(4.48*10^-13) = x*(2x)^2/(0.08-2x)^2

approximately, x = 8.95*10^-6

partial pressure of NO = 2X = 2*8.95*10^-6 = 1.79*10^-5 atm

partial pressure of O2 = x = 8.95*10^-6 atm

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