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 = 1.263-y

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 1.263-y (1.263-y 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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