In the gaseous phase PCl5 dissociates according to the following equilibrium: PCl5⇌PCl3+Cl2 and Kc=0.8 at 613K .Determine the equilibrium composition when 0.12 moles/L of the three compounds are mixed at 613K.

Answer 1

First you must determine which way the reaction will shift to react equilibrium. That is done with Qc.
Qc = (PCl3)(Cl2)/(PCl5) = 0.12*0.12/0.12 = 0.12= Qc and Kc = 0.8 so Qc is too small which means products are too low and reactant is too high so shift must be from left to right.
......PCl5 ==> PCl3 + Cl2
I....0.12......0.12...0.12
C.....-x.........x......x
E..0.12-x....0.12+x...0.12+x

Substitute the E line into Kc expression and solve for x, then evaluate each component according to the E line. You will need to solve the quadratic; I get approx 0.07 M

Answer 2

Using your numbers, looks like Kc < Qc. Rx shifts from product side to reactant side. Shouldn't the PCl5 increase and PCl3 & Cl2 decrease?

Answer 3

No. Qc = 0.12; Kc = 0.8
Qc < Kc (or in your language Kc > Qc). Therefore, to make Qc = Kc requires products to increase and reactant to decrease. If you try shifting the reaction the other way you get a negative number for x and that can't be. Try working the problem and if x is 0.0732 then PCl5 = 0.12-0.0732 =0.0468. PCl3 = Cl2 = 0.12 + 0.0732 = 0.1932. Then (0.1932)^2/(0.0486) = 0.798 which rounds to 0.8. QED