Question

1. The equilibrium constant, Kc, for the reaction PC13(g) + Cl2(g) = Pcl3(g) equals 49 at 230°C. If 0.400 mol each of phosphorus trichloride and chlorine are added to a 4.0-L reaction vessel, what is the equilibrium composition of the mixture at 230°C?

Answer 1

[PCl3] = [Cl2] = number of mol / volume

= 0.400 mol / 4.0 L

= 0.10 M

ICE Table:

[PC13] [012] [PC15] initial 0.1 0.1 -1x - 1x +1x change equilibrium 0.1-1x 0.1-1x +1x

Equilibrium constant expression is

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

49.0 = (1*x)/((0.1-1*x)(0.1-1*x))

49.0 = (1*x)/(1*10^-2-0.2*x + 1*x^2)

0.49-9.8*x + 49*x^2 = 1*x

0.49-10.8*x + 49*x^2 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 49

b = -10.8

c = 0.49

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 = 20.6

roots are :

x = 0.1565 and x = 6.389*10^-2

x can't be 0.1565 as this will make the concentration negative.so,

x = 6.389*10^-2

At equilibrium:

[PCl3] = 0.1-1x = 0.1-1*0.06389 = 0.03611 M

[Cl2] = 0.1-1x = 0.1-1*0.06389 = 0.03611 M

[PCl5] = +1x = +1*0.06389 = 0.06389 M

Answer:

[PCl3] = 0.0361 M

[Cl2] = 0.0361 M

[PCl5] =0.0639 M