Question

Phosphorus pentachloride decomposes according to the chemical equation PC15(g) PC13(g) + Cl2 (g) Kc = 1.80 at 250° C A 0.475 mol sample of PCI, (g) is injected into an empty 4.90 L reaction vessel held at 250° C. Calculate the concentrations of PCI, (g) and PCI,(g) at equilibrium. [PCls] [PCl3]

Answer 1

Initial concentration of PCl5 = mol of PCl5 / volume in L

= 0.475 mol / 4.90 L

= 2.33 M

ICE Table:

[PCl5] [PCl3] [Cl2]

initial 2.33 0 0

change -1x +1x +1x

equilibrium 2.33-1x +1x +1x

Equilibrium constant expression is

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

1.8 = (1*x)(1*x)/((2.33-1*x))

1.8 = (1*x^2)/(2.33-1*x)

4.194-1.8*x = 1*x^2

4.194-1.8*x-1*x^2 = 0

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

a = -1

b = -1.8

c = 4.194

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 = 20.02

roots are :

x = -3.137 and x = 1.337

since x can't be negative, the possible value of x is

x = 1.337

At equilibrium:

[PCl5] = 2.33-1x = 2.33-1* 1.337 = 0.993 M

[PCl3] = +1x = +1* 1.337 = 1.337 M

[Cl2] = +1x = +1* 1.337 = 1.337 M

Answer:

[PCl5] = 2.30.993 M

[PCl3] = 1.34 M