The equilibrium constant, Kc, for the following reaction is 1.29×10-2 at 600 K.
COCl2(g) ----> CO(g) + Cl2(g)
Calculate the equilibrium concentrations of reactant and products when 0.362 moles of COCl2(g) are introduced into a 1.00 L vessel at 600 K.
[COCl2] = _____ M
[CO] = _____ M
[Cl2] = ____ M
ICE Table:
Equilibrium constant expression is
Kc = [CO]*[Cl2]/[COCl2]
0.0129 = (1*x)(1*x)/((0.362-1*x))
0.0129 = (1*x^2)/(0.362-1*x)
4.67*10^-3-1.29*10^-2*x = 1*x^2
4.67*10^-3-1.29*10^-2*x-1*x^2 = 0
This is quadratic equation (ax^2+bx+c=0)
a = -1
b = -1.29*10^-2
c = 4.67*10^-3
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 = 1.885*10^-2
roots are :
x = -7.509*10^-2 and x = 6.219*10^-2
since x can't be negative, the possible value of x is
x = 6.219*10^-2
At equilibrium:
[COCl2] = 0.362-1x = 0.362-1*0.06219 = 0.29981 M
[CO] = +1x = +1*0.06219 = 0.06219 M
[Cl2] = +1x = +1*0.06219 = 0.06219 M
Answer:
[COCl2] = 0.300 M
[CO] = 0.0622 M
[Cl2] = 0.0622 M
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