Nitrogen dioxide decomposes according to the reaction given below where K p = 4.

Question

Nitrogen dioxide decomposes according to the reaction given below where Kp = 4.48 ? 10-13 at a certain temperature.

If 0.59 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)?
PNO
atm
PO2
atm

Nitrogen dioxide decomposes according to the reaction given below where Kp = 4.48 ? 10^-13 at a certain temperature. 2 NO2(g) arrow 2 NO(g) + O2(g) If 0.59 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)? PNO atm PO2 atm

Explanation / Answer

Answer:  Kp = P,NO^2 x P,O2 / P, NO2^2

I calculate for .27 atm pressure
2 NO2(g) <---> 2 NO(g) + O2(g)
Initial 0.27 0 0
Change -2x +2x x
Equib 0.27 - 2x 2x x

Solve Quad.

4.48x10^(-13) = 2x^(2) / (0.27-2x)

2x^(2) + 8.96x10^(-13)x - 1.21x10^(-13) = 0

x = 2.37x10^(-7) or same as negative( use positive)

P, NO = 2x = 2x2.37x10^(-7) = 4.74x10^(-7) atm

P, O2 = x = 2.37x10^(-7) atm

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