Question

In an experiment 4.0 mole of N204 was introduced to a previously evacuated 5.0L vessel and was allowed to react until equilibrium was reached. Show the steps to calculate the concentration of all species For thu reaction at equilibrium " O 2N022 N2 0419)

Answer 1

Initial concentration of N2O4 = mol of N2O4 / volume in L

= 4.0 mol / 5.0 L

= 0.80 M

ICE Table: [N02 [N204] initial 0.8 change +2x -1x equilibrium +2x 0.8-1x

Equilibrium constant expression is

Kc = [N2O4]/[NO2]^2

70.0 = (0.8-1*x)/((4*x^2))

2.8*10^2*x^2 = 0.8-1*x

-0.8 + 1*x + 2.8*10^2*x^2 = 0

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

a = 2.8*10^2

b = 1

c = -0.8

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 = 8.97*10^2

roots are :

x = 5.17*10^-2 and x = -5.527*10^-2

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

x = 5.17*10^-2

At equilibrium:

[NO2] = +2x = +2*0.0517 = 0.10339 M

[N2O4] = 0.8-1x = 0.8-1*0.0517 = 0.7483 M

Answer:

[NO2] = 0.10 M

[N2O4] = 0.75 M