Equilibrium calculations

Question

The equilibrium constant, K_c, for the following reaction is 83.3 at 500 K.

PCl₃(g) + Cl₂(g) PCl₅(g)

Calculate the equilibrium concentrations of reactant and products when 0.282 moles of PCl₃ and 0.282 moles of Cl₂ are introduced into a 1.00 L vessel at 500 K.

[PCl3]	=	M
[Cl2]	=	M
[PCl5]	=	M

Answer 1

Set up an ICE table

PCl₃(g) + Cl₂(g) <==> PCl₅(g)

0.282........0.282..............0.........I

-x..............-x...................+x........C

0.282-x....0.282-x............x.........E

Keq = [PCl₅] / [PCl₃][Cl₂]

83.3 = x /(0.282-x)(0.282-x)

83.3 = x / x2 - 0.564x + 0.0795

I'm no math guru, but solving for x, I get

x = 0.223 and if this is correct, then @ equilibrium...

[PCl₃] = 0.282 - 0.223 = 0.059 M

[Cl₂] = 0.282 - 0.223 = 0.059 M

[PCl₅] = 0.223 M