Question

Suppose a 500 ml flask is filled with 1.2 mol of NO, and 0.40 mol of NO the following reaction becomes possible: NO,() +NO(g) - 2NO(g) The equilibrium constant K for this reaction is 7.29 at the temperature of the flask Calculate the equilibrium molarity of NO. Round your answer to two decimal places

Answer 1

Initial concentration of NO3 = mol of NO3 / volume in L

= 1.2 mol / 0.500 L

= 2.4 M

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

= 0.40 mol / 0.500 L

= 0.80 M

ICE Table:

INO] [N03] [N02] initial 2.4 0.8 change +1x -2x +1x equilibrium 2.4+1x +1x 0.8-2x

Equilibrium constant expression is

Kc = [NO2]^2/[NO3]*[NO]

7.29 = (0.64-3.2*x + 4*x^2)/((2.4 + 1*x)(1*x))

7.29 = (0.64-3.2*x + 4*x^2)/(2.4*x + 1*x^2)

17.5*x + 7.29*x^2 = 0.64-3.2*x + 4*x^2

-0.64 + 20.7*x + 3.29*x^2 = 0

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

a = 3.29

b = 20.7

c = -0.64

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 = 4.367*10^2

roots are :

x = 3.077*10^-2 and x = -6.321

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

x = 3.077*10^-2

At equilibrium:

[NO] = +1x = +1*0.03077 = 0.0308 M

Answer: 0.0308 M