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?
(b) What is the partial pressure of PCl5 at equilibrium?
(c) What is the total pressure in the bulb at equilibrium?
(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
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.
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