Thursday

July 24, 2014

July 24, 2014

Posted by **Writeacher** on Thursday, January 15, 2009 at 8:43pm.

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 didnt get to d yet

im having a lot of trouble! help!!

- Chem/math question - from Sam -
**bobpursley**, Thursday, January 15, 2009 at 9:04pmI 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 9:14pmYour 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.

**Related Questions**

Chemistry - For the following reaction at 600. K, the equilibrium constant, Kp ...

chem sorry the last one did not have the question - A sample of PCl5 weighing 2....

Chem II - PCl5 dissociates according to the reaction: PCl5(g) ↔ PCl3(g) + ...

chemistry - how would you begin this question? An empty 15.0 L cylinder, .500 ...

Chem 2 - Phosphorus pentachloride decomposes according to the chemical equation ...

chem - A sample of PCl5 weighing 2.69 grams is placed in a 1.000 liter flask and...

CHEM - A 0.229 mol sample of PCL5 is injected into an empty 3.20 L reaction ...

Chemistry - When a sample of PCl5(g) (0.02087 mol/L) is placed in 83.00 L ...

Chemistry - Phosphorus pentachloride decomposes according to the chemical ...

Chem II - The equilibrium constant, Kp for the reaction PCl5 <---> PCl3 + ...