chemistry

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Phosphorus pentachloride decomposes according to the chemical equation
PCl5(g) -----> PCl3 (g)+ Cl2(g)
Kc= 1.80 at 250 degrees C

A 0.463 mol sample of PCl5(g) is injected into an empty 4.80 L reaction vessel held at 250 °C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium.


PLEASE HELP!!!! Can you confirm the correct answer as well. I have one more attempt. Thanks

  • chemistry -

    M PCl5 = 0.463 mols/4.80L = 0.09646 which is more places than allowed but I like to keep one extra place, go through the calculations and round in the last step.
    ..........PCl5 ==>PCl3 + Cl2
    initial.0.09646......0......0
    change...-x.........x.......x
    equil..0.09646-x.....x......x

    Kc = 1.80 = (PCl3)(Cl2)/(PCl5)
    Substitute from the ICE chart into the Kc expression and solve for x, then for 0.09646

  • chemistry -

    What answers did u get for PCl5 and PCl3...I want to verify them please.

  • chemistry -

    I didn't work it. Give me your answers and I will check it.

  • chemistry -

    I got x=0.091766....now what do i do? I'm confused. Also can check if this is right for x?

  • chemistry -

    I obtained 0.0918M for x and that is (PCl3). The problem also asked for (PCl5) and that is 0.09646-x = ? and round to 3 s.f. I obtained 0.00466 for that.

  • chemistry -

    yes it is correct! thank you :) I really appreciate it!

  • chemistry -

    How do you solve for x with the equation 1.8 = x^2 / .0964 - x? i don't understand what to do isolate x

  • chemistry -

    I want to know that too ^
    That always stumps me on the hw!

  • chemistry -

    you cannot isolate X by its self without using the quadratic equation:

    x = -B +or- square-root(B^2 - 4AC)
    ------------------------------
    2A

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