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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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lilyrose lilyrose
wrote...
11 years ago
The equilibrium constant for dissociation of dinitrogen tetroxide
 N2O4(g) ? 2 NO2(g)
 is KP = 0.308763 at 298.0 K
 If the total equilibrium pressure at 298.0 K is 0.45 bar, what is the partial pressure of NO2(g)?
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lilysketcher Author
wrote...
#1
11 years ago
[P(NO2)]^2 / [P(N2O4)] = Kp

P(NO2) + P(N2O4) = P(total)

P(N2O4) = P(total) - P(NO2)

[P(NO2)]^2 / [P(total) - P(NO2)] = Kp

P(NO2) = x

x^2 + Kp(x) - (P(total)*Kp) = 0

Kp = 0.308763
P(total) = 0.45

solve for x
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