Question

In: Chemistry

The equilibrium constant, Kp, for the following reaction is 2.01 at 500 K: PCl3(g) + Cl2(g)...

The equilibrium constant, Kp, 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

Solutions

Expert Solution

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


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