Question

for the reaction pcl5(g) ------ pcl3(g)+cl2 (g)

K = 0.050 at 150 C. A quantity of pcl5 is placed in a 5 liter container at 150 C when equilibrium is established, the container holds 0.50 moles of cl2. How many moles of pcl5 were originally placed in the container?

Answer 1

Given equilibrium moles of Cl2 is =0.50 mol

Volume of the container is = 5 L

So concentration of Cl₂ at equilibrium is = number of moles / volume

                                                                = 0.50 mol / 5 L

                                                               = 0.10 M

                             PCl₅(g)    PCl₃(g)    + Cl₂ (g)

initial conc              c                    0                 0

change                  -a                  +a               +a

Equb conc            c-a                  a                 a

Given Equb concentration of Cl₂ is [Cl₂] = a = 0.10 M

So the equb concentration of PCl₃ is [PCl₃]= a = 0.10 M

   The Equb concentration of PCl₅ is [PCl₅]= c-a = c-0.10

Equilibrium constant ,    K = ([PCl₃][Cl₂])/[PCl₅]

                                0.050 = ( 0.10 x 0.10 ) / (c-0.10)

                               c-0.10 = 0.2

                                      c = 0.2+0.10

                                         = 0.30 M

So initial concentration of PCl5 is = 0.30 M

Therefore initial moles of PCl₅ is = Molarity x Volume in L

                                                     = 0.30 M x 5 L

                                                     = 1.50 moles

Therefore 1.50 moles of PCl₅ were originally placed in the container