Question

Phosphorus pentachloride decomposes according to the chemical equation

PCl5(g)↽−−⇀PCl3(g)+Cl2(g)?c=1.80 at 250 ∘CPCl5(g)↽−−⇀PCl3(g)+Cl2(g)Kc=1.80 at 250 ∘C

A 0.252 mol0.252 mol sample of PCl5(g)PCl5(g) is injected into an empty 3.40 L3.40 L reaction vessel held at 250 ∘C.250 ∘C.

Calculate the concentrations of PCl5(g)PCl5(g) and PCl3(g)PCl3(g) at equilibrium.

Answer 1

Initial concentration of PCl5 = mol of PCl5 / volume in L
= 0.252 mol / 3.40 L
= 0.0741 M

ICE Table:

  

[PC15] [PC13] [C12] initial 0.0741 change -1x +1x +1x equilibrium 0.0741-1x +1x +1x

Equilibrium constant expression is
Kc = [PCl3]*[Cl2]/[PCl5]
1.8 = (1*x)(1*x)/((7.41*10^-2-1*x))
1.8 = (1*x^2)/(7.41*10^-2-1*x)
0.1334-1.8*x = 1*x^2
0.1334-1.8*x-1*x^2 = 0
This is quadratic equation (ax^2+bx+c=0)
a = -1
b = -1.8
c = 0.1334

Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 3.774

roots are :
x = -1.871 and x = 7.128*10^-2

since x can't be negative, the possible value of x is
x = 7.128*10^-2

At equilibrium:
[PCl5] = 0.0741-1x = 0.0741-1*0.07128 = 0.00282 M
[PCl3] = +1x = +1*0.07128 = 0.07128 M

Answer:
[PCl5] = 0.00282 M
[PCl3] = 0.0713 M