Question

Suppose a 500. ml flask is filled with 1.3 mol of NO, and 1.5 mol of NO. The following reaction becomes possible: NO2(g) +NO(g) + 2NO2(g) The equilibrium constant K for this reaction is 9.06 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places. i M x 6 ?

Answer 1

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

= 1.3 mol / 0.500 L

= 2.6 M

Initial concentration of NO = mol of NO / volume in L

= 1.5 mol / 0.500 L

= 3.0 M

ICE Table:

(NO3] [NO] [NO2) initial 2.6 3.0 change - 1x - 1x +2x equilibrium 2.6–1x 3.0-1x +2x

Equilibrium constant expression is

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

9.06 = (4*x^2)/((2.6-1*x)(3-1*x))

9.06 = (4*x^2)/(7.8-5.6*x + 1*x^2)

70.67-50.74*x + 9.06*x^2 = 4*x^2

70.67-50.74*x + 5.06*x^2 = 0

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

a = 5.06

b = -50.74

c = 70.67

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 = 1.144*10^3

roots are :

x = 8.355 and x = 1.671

x can't be 8.355 as this will make the concentration negative.so,

x = 1.671

At equilibrium:

[NO2] = +2x = +2*1.671 = 3.34 M

Answer: 3.34 M