Chem. Equilibrium question

I kinda want to know if I did it right, hopefully I did.

After being introduced into a container at an initial pressure of 3.42 atm A3B5 decomposes into A2B4 and AB. At equilibrium the partial pressure of AB is 0.28 atm.
What is Kp for this equilibrium?

So I'll do a table chart for it

      A3B5 <--> A2B4  +  AB

Initial   3.42  |  0   |  0
Change     -x   |  +x  | +x
Equi      3.42-x|  x   | .28+x


Kp= (x)(.28+x)/3.42-x

So...


Kp= (x)(.28+x)/3.42-x=

x2+1.28x-3.42=0

Using the quadratic formula, I would get x= 1.32, -2.6. The first value 1.32 would make sense...so Kp would be 1.32? Would that be correct or missing a part?

Thanks

i don't get it. at equilibrium the partial pressure of AB is 0.28atm, right? but at initial state the partial pressure was 0, so there was an increase of 0.28 atm, right?

it should actually be:

Initial   3.42  |  0   |  0
Change     -x   |  +x  | +x
Equi      3.42-x|  x   | .28

meaning x=.28atm and Kp=(.28)^2/(3.42-.28)

Kp=0.025, that is

CAN SOMEONE TELL ME WHO IS RIGHT AND WHO IS WRONG

1.   Do the ICE chart
2.   Relate the definition of the equlibrium constant ([products]/[reactants] with the value of the equilibrium constant for this problem.
3.   Solve equation.

(Yes throw out the negative value if you obtain one).

http://www.anontalk.se/

I WAS RIGHT, OP WAS WRONG! BOOYAH!