Question

Suppose a 500. mL flask is filled with 1.0 mol of CHCl, 1.6 mol of Hcl and 0.30 mol of CCI,. The following reaction becomes p Cl(g)+CHCI (g) HCI (s)+CC () The equilibrium constant K for this reaction is 0.671 at the temperature of the flask. Calculate the equilibrium molarity of Cl,. Round your answer to two decimal places. ?

Answer 1

Initial concentration of CHCl3 = mol of CHCl3 / volume in L

= 1.0 mol / 0.500 L

= 2.0 M

Initial concentration of HCl = 1.6 mol / 0.500 L

= 3.2 M

Initial concentration of CCl4 = 0.30 mol / 0.500 L

= 0.60 M

ICE Table:

[СHC13] [Cl2] CC14] initial 2.0 0 3.2 0.6 change -2x +1x +1x -1x equilibrium 2.0+1x 3.2-2х 0.6-1x +1x

Equilibrium constant expression is

Kc = [HCl]^2*[CCl4]/[CHCl3]*[Cl2]

0.671 = (3.2-2*x)(0.6-1*x)/((2 + 1*x)(1*x))

0.671 = (1.92-4.4*x + 2*x^2)/(2*x + 1*x^2)

1.342*x + 0.671*x^2 = 1.92-4.4*x + 2*x^2

-1.92 + 5.742*x-1.329*x^2 = 0

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

a = -1.329

b = 5.742

c = -1.92

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 = 22.76

roots are :

x = 0.3653 and x = 3.955

x can't be 3.955 as this will make the concentration negative.so,

x = 0.3653

At equilibrium:

[Cl2] = x = 0.3653 M

Answer: 0.37 M