What is the partial pressure of ammonia at equilibrium?

[ThatDude]

Homework Statement

Ammonia gas decomposes to nitrogen and hydrogen spontaneously with an equilibrium constant, Kp = 5.9 x 10⁹. Calculate the partial pressure of ammonia at equilibrium given that the partial pressure of nitrogen and hydrogen are both 2.7 atm initially.

[Hint: If you need to use anything more than a second order equation you are on the wrong track]

2.The attempt at a solution

I tried constructing an ICE table, but it ends up being a 3rd degree equation.

 

[Borek]

Kp is so high you can safely assume (almost) everything decomposed.

 

[abitslow]

Sorry, this problem makes no sense. Ptot = Pn+Ph+Pa at all times, right? You are given that Ph=Pn=2.7 initially. What does this mean? That Pa = 0?, = 0.00001?, = 2.7?, = 100000? There is NO way to determine what the total pressure is at ANY time, unless you know something more than what is given. So, I suppose who ever wrote this meant AT EQUILIBRIUM rather than "initially". Then its simple.
Oh, I should also note (although you need to ignore this and pretend the ideal gas law is perfect) that the IDL is generally correct to between 0.5% and 7% with some outliers at 15% depending on how low the temperature is and how high the pressure is (and how close you are to any critical point). Why is the relevant? Well we know for sure that the pressure is about 5.4 atm (even if we assume Pa is 0, initially (which is another way to solve it, and is probably as valid as assuming the author screwed up by using "initially" - for brownie points solve it both ways. :) )
Anyway, 0.5% of 5 atm is 0.025 atm - meaning that the calculated Pa isn't going to be anywhere CLOSE to the actual Pa. a=ammonia, h = hydrogen, and n = nitrogen - I assume you get that.