The equilibrium constant KP for the following reaction is 4 atm^2 at 300 K. AB(s) ---> A(g) + B(g)

Question

What is the equilibrium pressure of A(g) and B(g) above AB(s) at 300 K? What is KC at 300K?

Answer 1

Kp = pA*pB = 4.
I would let pA = p which means pB = p and solve for p.
Kp = Kc(RT)^{delta n}

Answer 2

Could you please explain what you wrote above? Like can you replace all the constants with the numbers given so I get what you are saying...Thanks! Also how will I find the equilibrium pressure and Kc?

Answer 3

And how do I solve for p? I'm confused :(

Answer 4

Kp = 4 = partial pressure of A x partial pressure of B
4=p*p
4 = p^2
p=2 atm
partial pressure A = 2 atm at eqilibrium.
partial pressure of B = 2 atm at equilibrium.
Kp = Kc(RT)^{delta n}
You know Kp and you know R and T. You can calculate delta n (it's mols reactants - moles products = 2-0=2)
Note that Kp = pA*pB and does NOT (repeat NOT) include AB since it is a solid.

Answer 5

So is this right:

4=Kc*((0.08206)(300K)^2)= .006600

Also what is the answer to this part of the question:

What is the equilibrium pressure of A(g) and B(g) above AB(s) at 300 K?

Answer 6

Kc is right but you worked it wrong.
It's 4 = Kc*(0.08206*300)^2
What do you not understand about partial pressure? Do you know what that means? I have posted the equilibrium partial pressures for both A and B in all of my responses but you continue to ask what is the answer.

Answer 7

Ohh I'm sorry I didn't see that! Sorry for the mishap :( I really appreciate your help!

Answer 8

You're welcome.

Answer 9

Hey, I just wanted to help and point out that delta n = moles of gaseous products - moles of gaseous reactants according to my chem lecture.