The equilibrium constant, K c, is 0.022 at 25 °C for the reaction below. What is the concentration of PCl 5 at equilibrium if a reaction is initiated with 0.80 mole of PCl 5 in a 1.00-liter container?
PCl5(g) --> PCl3(g) + Cl2(g)
0.80 mole PCl5 in 1 L = 0.80 M
..................PCl5(g) --> PCl3(g) + Cl2(g)
I..................0.80...............0..............0
C..................-x..................x..............x
E.................0.80-x.............x..............x
Kc = 0.022 = (PCl3)(Cl2)/(PCl5)
Substitute the E line into the Kc expression and solve for x, then evaluate 0.80-x and that's the answer to the problem.