Question

At 1,600 K, the equilibrium constant for the reaction below is Kc = 0.215. Br2(g) ⇌ 2 Br(g) What is the percent dissociation of Br2 at 1,600 K, if the initial concentration of Br2 is 2.25 mol L−1 ?

Answer 1

1. ICE Table:

   [Br2] [Br]

   initial 2.25

   change -1x +2x

   equilibrium 2.25-1x +2x

   Equilibrium constant expression is

   kc = [Br]^2/[Br2]

   0.215 = (4*x^2)/((2.25-1*x))

   0.4838-0.215*x = 4*x^2

   0.4838-0.215*x-4*x^2 = 0

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

   a = -4

   b = -0.215

   c = 0.4838

   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 = 7.786

   roots are :

   x = -0.3757 and x = 0.3219

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

   x = 0.3219

   % dissociation = x*100 / initial concentration

   = 0.3219*100 / 2.25

   = 14.3 %

   Answer: 14.3 %