Question

Phosphorus pentachloride decomposes according to the chemical equation. PCl5(g) ightleftharpoons PCI3(g) + Cl2(g) Kc = 1.80 at 250 �C A 0.318 mol sample of PCl5(g) is injected into an empty 3.35 L reaction vessel held at 250 �C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium.

Answer 1

initial concentration of PCl5 = mol of PCl5 / molar mass of PCl5

= 0.318 mol / 3.35 L

= 0.0949 M

Let's prepare the ICE table

[PCl5] [PCl3] [Cl2]

initial 0.0949 0 0

change -1x +1x +1x

equilibrium 0.0949-1x +1x +1x

Equilibrium constant expression is

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

1.8 = (1*x)(1*x)/((0.0949-1*x))

1.8 = (1*x^2)/(0.0949-1*x)

0.17082-1.8*x = 1*x^2

0.17082-1.8*x-1*x^2 = 0

Let's solve this quadratic equation

Comparing it with general form: (ax^2+bx+c=0)

a = -1

b = -1.8

c = 0.1708

solution of quadratic equation is found by below formula

x = {-b + √(b^2-4*a*c)}/2a

x = {-b - √(b^2-4*a*c)}/2a

b^2-4*a*c = 3.923

putting value of d, solution can be written as:

x = {1.8 + √(3.923)}/-2

x = {1.8 - √(3.923)}/-2

solutions are :

x = -1.89 and x = 9.036*10^-2

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

x = 9.036*10^-2

At equilibrium:

[PCl5] = 0.0949 - x = 0.0949- 0.09036 = 0.00454 M

[PCl3] = x = 0.09036 M

[PCl5] = 0.00454 M

[PCl3] = 0.0904 M