Question

The equilibrium constant, K_p, for the following reaction is 2.01 at 500 K:

PCl₃(g) + Cl₂(g) <----->PCl₅(g)

Calculate the equilibrium partial pressures of all species when PCl₃ and Cl₂, each at an intitial partial pressure of 1.56 atm, are introduced into an evacuated vessel at 500 K.

PPCl3	=	atm
PCl2	=	atm
PPCl5	=	atm

Answer 1

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

1.56           1.56           0       (initial)

1.56-x       1.56-x       x       (at equilibrium)

Kp = p(PCl5) / (p(PCl3)p(Cl2))

2.01 = x / (1.56-x)^2

2.01*(1.56-x)^2 = x

2.01*(2.434 + x^2 - 3.12*x) = x

4.892 + 2.01*x^2 - 6.27*x = x

2.01*x^2 - 7.27*x + 4.892 = 0

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

a = 2.01

b = -7.27

c = 4.892

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 = 13.52

roots are :

x = 2.723 and x = 0.894

x can't be 2.723 as this will make the concentration negative.so,

x = 0.894

So, At equilibrium,

p(PCl3) = 1.56-x = 1.56 - 0.894 = 0.666 atm

p(Cl2) = 1.56-x = 1.56 - 0.894 = 0.666 atm

p(PCl5) = x = 0.894 atm