Question

9. The equilibrium constant(K.) for the gaseous dissociation of Iodine molecules to lodine atoms (shown below) 12(8) 21(8) is 3.76 x 10-3 at 1000K. If 0.15 mole of 12(g) is placed in a 12.3-L flask at 1000 K, what are the concentrations of 12(g) and I(g) when the reaction comes to equilibrium?

Answer 1

Initial concentration of I2 = mol of I2 / volume in L

= 0.15 mol / 12.3 L

= 0.0122 M

ICE Table:

[12] [1] initial 0.0122 change - 1x +2x equilibrium 0.0122-1x +2x

Equilibrium constant expression is

Kc = [I]^2/[I2]

0.00376 = (4*x^2)/((1.22*10^-2-1*x))

4.587*10^-5-3.76*10^-3*x = 4*x^2

4.587*10^-5-3.76*10^-3*x-4*x^2 = 0

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

a = -4

b = -3.76*10^-3

c = 4.587*10^-5

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 = 7.481*10^-4

roots are :

x = -3.889*10^-3 and x = 2.949*10^-3

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

x = 2.949*10^-3

At equilibrium:

[I2] = 0.0122-1x = 0.0122-1*0.002949 = 0.00925 M

[I] = +2x = +2*0.002949 = 0.00590 M

Answer:

[I2] = 0.00925 M

[I] = 0.00590 M