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March 27, 2017

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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 - ,

    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 - ,

    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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