How Do You Calculate Kc and Kp for a Reversible Gas Reaction?

[asing]

Homework Statement

When 1.000 mol PCl5 is intorduced into a 5.000L container @500K, 78.50% of PCl5 dissociates to give equilibrium mixture PCl5, PCl3, Cl2 : PCl5--->PCl3+Cl2 (this eq. is reversible)

(a) calculate Kc and Kp
(b) if the initial concentrations are [PCl5]=.500M, [PCl3]=.150M, and [Cl2=.600], in which direction does the rx proceed to reach equilibrium? what are the concentrations when the mixture reaches equilibrium?

Homework Equations

Kc=[Products]/[reactants]
Kp=Kc(RT)^n

The Attempt at a Solution

I attempted to find Kc of the initial problem, by getting the molarity (.2) of the reactant. I then multiplied by .785 to get the percent to put in the numerator, but am not getting the right answer in the book. I know in part b i must use the chart to convert to quadratic equation, but am having trouble with the first part (surprisingly, as its usually the second part which gives me trouble). I have tried this either which way and still can not get the right answer. Any input would be great. Thanks!

 

[Borek]

Not seeing all the numbers it is hard to say - you can be missing something, it can be just a math error. Please show your complete work, step by step.

 

[asing]

Yeah turns out I had the right idea, but didn't do it right. I went and saw my professor and he walked me through it. I had assumed that the .2M stayed the same for the PCl5, however, when the equation went to equilibrium, I had forgotten to use the 21.5% that was left, as the PCl5.

.2M *.785 gives .157, then .2-.157 gives .043, which is what I should have used for the PCl5 @ equilibrium, rather than the .2. Thanks for the input. Glad to know that people actually respond and are willing to help out.