[Chocotaco]

Just have a few questions on a couple of problems, hopefully if I do this one right the other one would follow ok

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 K_p 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


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

Am I going in the right direction or is it completely wrong? I'm not too sure if it is. Any help would be appreciated.

[rabolisk]

You're going in the right direction.

[Chocotaco]

Ok alright that's great to hear

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 K_p would be 1.32? Would that be correct or missing a part?

Thanks

[rabolisk]

Sorry I misread the question. At equilibrium the partial pressure of AB is 0.28 atm. What you have tabled is if the initial partial pressure is 0.28 atm, which is unsolvable.

[Chocotaco]

Although I'm not quite sure what you mean exactly, but given the problem has no answer?

[rabolisk]

The problem is solvable. Note that K_p is the ratio of equilibrium partial pressures. You are given the equilibrium partial pressure of one of the products.

You interpreted the problem wrong. Your calculations would be correct if you were given the initial partial pressure of AB. But this would not be solvable. Basically I mistakenly said you were doing the right thing when you were not. My apologies.

[Chocotaco]

I appreciate for helping out but I don't know what else to do. Maybe I'm over thinking but I tried something else and doubt it was even right. Basically did this:

      A3B5 <--> A2B4  +  AB

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

x2/3.42-x= Kp

[rabolisk]

That is correct, except you're given one more information.