for the reaction PCl>5 = PCl> + Cl>2 K>c = 33.3 at 760 degrees C. In a container at equilibrium there are

Question

1.29x10^-3 mol/L of PCl>3 and 1.87x10^-1 mol/L of Cl>2. Calculate the equilibrium molar concentration of PCl>5 in the vessel.

Answer 1

K>c=(1.29*10^3)* (1.87 * 10^-1)/equilibrium molar concenteration of PCl5

Answer 2

Balance the equation first:
PCl5 <-> PCl3 + Cl2

then, put it in equilibrium expression:
[PCl3][Cl2]/[PCl5] = Kc = 33.3

Also, mol/L is the meaning of molarity. 1.29x10^-3M is the concentration of PCl3 in the reaction vessel, and 1.87x10^-1M is the concentration of Cl2 in the reaction vessel.

So you have:
[1.29x10^-3M][1.87x10^-1M]/ [PCl5] = Kc = 33.3

solving for PCl5:
[1.29x10^-3M][1.87x10^-1M]/ [PCl5] = Kc = 33.3
1/[PCl5] = 33.3/[1.29x10^-3M] [1.87x10^-1M]
[PCl5] = [1.29x10^-3M] [1.87x10^-1M]/33.3
[PCl5] = 7.244x10^-6M