Question

Suppose a 250. mL flask is filled with 1.2 mol of Cl₂, 1.8 mol of CHCl₃ and 0.90 mol of HCl. The following reaction becomes possible:


Cl₂(g)+CHCl₃(g) ⇌ HCl(g)+CCl₄(g)


The equilibrium constant K for this reaction is 0.855 at the temperature of the flask.

Calculate the equilibrium molarity of HCl. Round your answer to two decimal places.


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Answer #1

Initial concentration of Cl2 = mol of Cl2 / volume in L
= 1.2 mol / 0.250 L
= 4.8 M

Initial concentration of CHCl3 = mol of CHCl3 / volume in L
= 1.8 mol / 0.250 L
= 7.2 M

Initial concentration of HCl = mol of HCl / volume in L
= 0.90 mol / 0.250 L
= 3.6 M


ICE Table:

  [C12] [CHC13] [HCl] [CC14] initial 4.8 7.2 3.6 change - 1x - 1x +1x +1x equilibrium 4.8-1x 7.2–1x 3.6+1x +1x       

Equilibrium constant expression is
Kc = [HCl]*[CCl4]/[Cl2]*[CHCl3]
0.855 = (3.6 + 1*x)(1*x)/((4.8-1*x)(7.2-1*x))
0.855 = (3.6*x + 1*x^2)/(34.56-12*x + 1*x^2)
29.55-10.26*x + 0.855*x^2 = 3.6*x + 1*x^2
29.55-13.86*x-0.145*x^2 = 0
This is quadratic equation (ax^2+bx+c=0)
a = -0.145
b = -13.86
c = 29.55

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 = 2.092*10^2

roots are :
x = -97.67 and x = 2.086

since x can't be negative, the possible value of x is
x = 2.086

At equilibrium:
[HCl] = 3.6+1x = 3.6+1*2.086 = 5.69 M

Answer: 5.69 M

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