Question

Suppose a 500 ml flask is filled with 1.6 mol of H, and 1.7 mol of HCl. The following reaction becomes possible: H2(g) +C12(g) + 2HCI(g) The equilibrium constant K for this reaction is 8.78 at the temperature of the flask. Calculate the equilibrium molarity of Cl,. Round your answer to two decimal places. OM x 6 ?

Answer 1

Initial concentration of H2 = mol of H2 / volume in L

= 1.6 mol / 0.500 L

= 3.2 M

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

= 1.7 mol / 0.500 L

= 3.4 M

ICE Table:

[H2] [C12] [HCL] initial 3.2 3.4 change +1x +1x -2x equilibrium 3.2+1x +1x 3.4-2x

Equilibrium constant expression is

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

8.78 = (11.56-13.6*x + 4*x^2)/((3.2 + 1*x)(1*x))

8.78 = (11.56-13.6*x + 4*x^2)/(3.2*x + 1*x^2)

28.1*x + 8.78*x^2 = 11.56-13.6*x + 4*x^2

-11.56 + 41.7*x + 4.78*x^2 = 0

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

a = 4.78

b = 41.7

c = -11.56

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 = 1.96*10^3

roots are :

x = 0.269 and x = -8.992

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

x = 0.269

At equilibrium:

[Cl2] = +1x = +1*0.2689 = 0.269 M

Answer: 0.27 M