Question

Suppose a 500. ml flask is filled with 1.2 mol of Cl, and 0.80 mol of HCl. The following reaction becomes possible: H2(g) + Cl2(g) = 2HCl (g) The equilibrium constant K for this reaction is 0.419 at the temperature of the flask. Calculate the equilibrium molarity of Cl. Round your answer to two decimal places. xs ?

Answer 1

Initial concentration of Cl2 = mol of Cl2 / volume in L

= 1.2 mol / 0.500 L

= 2.4 M

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

= 0.80 mol / 0.500 L

= 1.6 M

ICE Table:

[C12] [H2] [HCl) initial 2.4 1.6 +1x +1x -2x change equilibrium 2.4+1x +1x 1.6-2x

Equilibrium constant expression is

Kc = [HCl]^2/[Cl2]*[H2]

0.419 = (2.56-6.4*x + 4*x^2)/((2.4 + 1*x)(1*x))

0.419 = (2.56-6.4*x + 4*x^2)/(2.4*x + 1*x^2)

1.006*x + 0.419*x^2 = 2.56-6.4*x + 4*x^2

-2.56 + 7.406*x-3.581*x^2 = 0

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

a = -3.581

b = 7.406

c = -2.56

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 = 18.17

roots are :

x = 0.4388 and x = 1.629

x can't be 1.629 as this will make the concentration negative.so,

x = 0.4388

At equilibrium:

[Cl2] = 2.4+1x = 2.4+1*0.4388 = 2.84 M

Answer: 2.84 M