Question

A 3.07 mol sample of NO, (g) is added to a 4.25 L vessel and heated to 100°C. N,0.9) = 2 NO (9) K = 0.360 at 100°C Calculate the concentrations of NO, (g) and N, O,() at equilibrium. [NO, 1 = 0.427 [N,9,1 = 0.508

Answer 1

Initial concentration of NO2 = mol of NO2 / volume in L

= 3.07 mol / 4.25 L

= 0.722 M

ICE Table:

[N204] [NO2] initial 0.722 change +1x -2x equilibrium +1x 0.722-2x

Equilibrium constant expression is

Kc = [NO2]^2/[N2O4]

0.36 = (0.5213-2.888*x + 4*x^2)/((1*x))

0.36*x = 0.5213-2.888*x + 4*x^2

-0.5213 + 3.248*x-4*x^2 = 0

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

a = -4

b = 3.248

c = -0.5213

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 = 2.209

roots are :

x = 0.2202 and x = 0.5918

x can't be 0.5918 as this will make the concentration negative.so,

x = 0.2202

At equilibrium:

[N2O4] = +1x = +1*0.2202 = 0.220 M

[NO2] = 0.722-2x = 0.722-2*0.2202 = 0.282 M

Answer:

[N2O4] = 0.220 M

[NO2] = 0.282 M