PCl3(g) + Cl2(g) <---> PCl5(g) Kc = 0.18 In an initial mixture of 0.300M PCl3 (g), 0.400M Cl2 (g), what are the equilibrium concentrations of all gases?

The equilibrium concentrations are #[PCl_3] = 0.280M#, #[Cl_2] = 0.380M#, #[PCl_5] = 0.02M#.

One can approach this problem by using the ICE method (more here: http://en.wikipedia.org/wiki/RICE_chart); even before doing any calculations, the value of the equilibrium constant can give us an idea on how the result will look like; since #K_c<1#, the reaction will favor the reactans, so we'd expect bigger equilibrium concentrations for #PCl_3# and #Cl_2#, than for #PCl_5#.

..#PCl_3(g) + Cl_2(g) rightleftharpoons PCl_5(g)#
I:. 0.300.............0.400..............0
C:... -x....................-x...................+x
E:..(0.300-x).....(0.400-x)...........x

So, #K_c = ([PCl_5])/([PCl_3]*[Cl_2]) = x/((0.300 - x) * (0.400-x)) = 0.18#

The equation #0.18x^2 - 1.126x + 0.0216 = 0# produces two values for #x#, #x_1# = #6.23# and #x_2# = #0.020#; since concentrations cannot be negative, the correct value will be #x = 0.020# (#x = 6.23# would have produced negative concentrations for #PCl_3# and #Cl_2#).

Therefore, the equilibrium concentrations for all the gases are

#[PCl_3] = 0.300 - 0.020 = 0.28M#
#[Cl_2] = 0.400 - 0.020 = 0.38M#
#[PCl_5] = 0.020 M#

Notice that the initial predictions on the relative concentration values is correct, the concentrations of the reactans are bigger than the concentration of the product.