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)?
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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