Question

A 11.1 g sample of PCl5 is added to a sealed 1.75 L flask and the reaction is allowed to come to equilibrium at a constant temperature. At equilibrium, 49.6% of the PCl5 remains. What is the equilibrium constant, Kc, for the reaction? Equation: PCl5 (g)<----> PCl3(g) + Cl2(g)

Answer 1

1st find the initial concentration of PCl5

Molar mass of PCl5,

MM = 1*MM(P) + 5*MM(Cl)

= 1*30.97 + 5*35.45

= 208.22 g/mol

mass(PCl5)= 11.1 g

number of mol of PCl5,

n = mass of PCl5/molar mass of PCl5

=(11.1 g)/(208.22 g/mol)

= 5.331*10^-2 mol

volume , V = 1.75 L

Molarity,

M = number of mol / volume in L

= 5.331*10^-2/1.75

= 0.0305 M

PCl5 (g)   <—>    PCl3 (g)   +   Cl2 (g)

0.0305       0           0       (initial)

0.0305-x       x           x       (at equilibrium)

At equilibrium, 49.6 % of PCl5 remains

So, percent decomposed = 100 - 49.6 - 50.4 %

% decomposed = x*100/initial concnetration

50.4 = x*100/0.0305

x = 0.0154

Now use:

Kc = [PCl3][Cl2]/[PCl5]

= x*x / (0.0305-x)

= (0.0154*0.0154)/(0.0305-0.0154)

= 0.0157

= 1.57*10^-2

Answer: 1.57*10^-2