Question

Be sure to answer all parts. The dissociation of molecular lodir Is represented as 13C) = 2(g) brium constant for the reaction is 3.80 X 10. Suppose you start with 0.0454 mol of 1 in a 2.33-L flask the concentrations of the gases at equilibrium? Wha the equilibrium concentration of 1,? What is the equilibrium concentration of I?

Answer 1

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

= 0.0454 mol / 2.33 L

= 0.0195 M

ICE Table:

[I2] [I] initial 0.0195 change -1x +2x equilibrium 0.0195-1x +2x

Equilibrium constant expression is

Kc = [I]^2/[I2]

3.8*10^-5 = (4*x^2)/((1.95*10^-2-1*x))

7.41*10^-7 - 3.8*10^-5*x = 4*x^2

7.41*10^-7 - 3.8*10^-5*x - 4*x^2 = 0

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

a = -4

b = -3.8*10^-5

c = 7.41*10^-7

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 = 1.186*10^-5

roots are :

x = -4.352*10^-4 and x = 4.257*10^-4

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

x = 4.257*10^-4

At equilibrium:

[I2] = 0.0195-1x = 0.0195-1*0.0004257 = 0.01907 M

[I] = +2x = +2*0.0004257 = 0.0008514 M

Answer:

[I2] = 0.0191 M

[I] = 8.514*10^-4 M