Question

Phosphorous pentachloride decomposes according to the reaction

PCL₅ (g) ---> PCL₃ +CL₂

A 13.9 g sample of PCl5 is added to a sealed 1.50 L flask and the reaction is allowed to come to equilibrium at a constant temperature. At equilibrium, 39.8% of the PCl5 remains. What is the equilibrium constant, Kc, for the reaction?

Answer 1

molar mass of PCl5 = 208.2 g/mol
number of mole of PCl5 = (given mass of PCl5)/(molar mass of PCl5)
= (13.9/208.2) mole
= 0.0668 mole

molarity = (number of mole of PCl5)/(volume of solution in L)
= 0.0668/1.50
= 0.0445 M

At equilibrium, 39.8% of the PCl5 remains means (100 - 39.8)% decompose
that is x = 60.2 % of 0.0445 M

x = 0.02679
if we assume intial concentration of PCl5 be c
c = 0.0445
reaction
PCL5 <---> PCL3 + CL2
c 0 0 intial
c-x x x at equlibrium

Kc = ([PCl3]*[Cl2])/[PCl5]
= (x)^2/(c-x)
= (x^2)/(c-x)

= (0.02679)^2 / ( 0.0445-0.02679 )

=7.18*10^-4 / 0.01771

=0.0405
Answer : 0.0405