0.1 mol of PCl5 is vaporised in a  litre vessel at 260 C. Calculate the concentration of Cl2 at equilibrium. If the equilibrium constant for the dissosciation of PCl5 is 0.0414

For the reaction 

 PCl₅ (g) ⇔ PCl₃ (g) + Cl₂ (g)

 K_c = [PCl₃][Cl₂] / [PCl₅]

We are given that the volume of the container is one litre. The initial molar concentration of PCl₅ is 

 = number of moles of PCl₅ / Volume of solution

 = 0.1 / 1 = 0.1 molar

Suppose change in molar concentration of PCl₅ is (-y) mol L^-1, then 

Now substituting the value of equilibrium concentrations in the expression for equilibrium constant, we get

 K_c = [y][y] / [0.1 -y]

or y² + 0.0414y - 0.00414 = 0

 Solving for y , we get 2 solutions. They are 

 y = -0.883 and y = 0.0414 moles

since concentration of must be positive, therefore 0.0414 mol L^-1 is chosen as the appropriate concentration. Therefore equilibrium concentration of PCl₅ is = (0.1- 0.0414) mol L^-1 = 0.0586 mol L^-1

[PCl₃] = [Cl₂] = 0.0414 mol L^-1