Question

The equilibrium constant, Kc, for the following reaction is 77.5 at 600 K. CO(g) + Cl2(g) COCl2(g) Calculate the equilibrium concentrations of reactant and products when 0.576 moles of CO and 0.576 moles of Cl2 are introduced into a 1.00 L vessel at 600 K.

[CO] = M

[Cl2] = M

[COCl2] = M

Answer 1

Since volume is 1L, concentration will be same as number of mol

CO(g) + Cl2(g) <—> COCl2(g)

0.576   0.576       0       (initial)

0.576-x   0.576-x       x       (at equilibrium)

Kc = [COCl2]/[CO][Cl2]

77.5 = x / (0.576-x)^2

77.5* (0.576-x)^2 = x

77.5*( 0.3318 - 1.152*x + x^2) = x

25.71 - 89.28*x + 77.5*x^2 = x

77.5*x^2 - 90.28*x + 25.71 = 0

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

a = 77.5

b = -90.28

c = 25.71

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 = 1.804*10^2

roots are :

x = 0.669 and x = 0.496

x can't be 0.669 as this will make the concentration negative.so,

x = 0.496

At equilibrium:

[Cl2] = 0.576 - x = 0.576 - 0.496 = 0.008 M

[CO] = 0.576 - x = 0.576 - 0.496 = 0.008 M

[COCl2] = x = 0.496 M