Posted by Writeacher on Thursday, January 15, 2009 at 8:43pm.
For the following reaction at 600. K, the equilibrium constant, Kp, is 11.5.
PCl5(g) = PCl3(g) + Cl2(g)
Suppose that 2.510 g of PCl5 is placed in an evacuated 470. mL bulb, which is then heated to 600. K.
(a) What would be the pressure of PCl5 if it did not dissociate?
___atm
(b) What is the partial pressure of PCl5 at equilibrium?
___ atm
(c) What is the total pressure in the bulb at equilibrium?
___atm
(d) What is the degree of dissociation of PCl5 at equilibrium?
___%
i got a to be 1.263 but i cnat find b or c and i didn't get to d yet
im having a lot of trouble! help!!

Chem/math question  from Sam  bobpursley, Thursday, January 15, 2009 at 9:04pm
I didn't check a)
But to get b,c you have to write the equilibrium expression
and solve for the moles of the products at equialbrium. I assume you know how to do that.
Then, you can calculate b, c given the moles of each product, and the moles left of the reactant.

Chem/math question  from Sam  DrBob222, Thursday, January 15, 2009 at 9:14pm
Your answer to a is correct IF the unit is atmospheres.
b. Set up at ICE chart.
Initial: pressure PCl5 = 1.263 atm
partial p PCl3 = 0
partial p Cl2 = 0
change: partial p PCl3 = +y
partial p Cl2 = y
partial pressure PCl5 = y
equilibrium pressures:
add the columns to obtain
partial p PCl3 = y
partial p Cl2 = y
partial pressure PCl5 = 1.263y
Now plug all that into the equilibrium constant expression; Kp = 
and solve for y, the only unknown in the equation. I expect you will get a quadratic equation. That will give you y and 1.263y (1.263y will be the partial pressure of PCl5).
c. Add partial pressure PCl5 + partial pressure PCl3 + partial pressure Cl2.
d. The degree of dissociation is
partial p PCl5 at equilibrium/partial p PCl5 initially. By the way, if you then multiply that by 100 you will have percent dissociation. The degree of dissociation IS NOT expressed in percent terms.
part c.
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