Question

For the following reaction K_c = 2.20 ✕ 10² at 74°C.

CO(g) + Cl₂(g) ↔ COCl₂(g)

Find the equilibrium concentrations of all chemical species starting with [CO] = 0.179 M and [Cl₂] = 0.287 M.
[CO] = M
[Cl₂] = M
[COCl₂] = M

Answer 1

                    [CO]                [Cl2]               [COCl2]           

initial             0.179               0.287               0                 

change              -1x                 -1x                 +1x               

equilibrium         0.179-1x            0.287-1x            +1x               

Equilibrium constant expression is
Kc = [COCl2]/[CO]*[Cl2]
2.2 = (1*x)/((0.179-1*x)(0.287-1*x))
2.2 = (1*x)/(0.05137-0.466*x1*x^2)
0.11302-1.0252*x2.2*x^2 = 1*x
0.11302-2.0252*x2.2*x^2 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 2.2
b = -2.025
c = 0.113

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 = 3.107

roots are :
x = 0.8609 and x = 5.968*10^-2

x can't be 0.8609 as this will make the concentration negative.so,
x = 5.968*10^-2

At equilibrium:
[CO] = 0.179-1x = 0.179-1*0.05968 = 0.119 M
[Cl2] = 0.287-1x = 0.287-1*0.05968 = 0.227 M
[COCl2] = 0+1x = 0+1*0.05968 = 0.0597 M