Question

Suppose a 500. mL flask is filled with 1.6 mol of O, and 0.70 mol of NO. The following reaction becomes possible: N2(g)02g)2NO (g) The equilibrium constant K for this reaction is 9.43 at the temperature of the flask Calculate the equilibrium molarity of NO. Round your answer to two decimal places. Ом X

Answer 1

Initial concentration of O2 = mol of O2 / volume in L
= 1.6 mol / 0.500 L
= 3.2 M

Initial concentration of NO = mol of NO / volume in L
= 0.70 mol / 0.500 L
= 1.4 M

ICE Table:

   

[02] [N2] [NO] initial 3.2 1.4 change +1x +1x -2x equilibrium 3.2+1x +1x 1.4-2x

Equilibrium constant expression is
Kc = [NO]^2/[O2]*[N2]
9.43 = (1.96-5.6*x + 4*x^2)/((3.2 + 1*x)(1*x))
9.43 = (1.96-5.6*x + 4*x^2)/(3.2*x + 1*x^2)
30.18*x + 9.43*x^2 = 1.96-5.6*x + 4*x^2
-1.96 + 35.78*x + 5.43*x^2 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 5.43
b = 35.78
c = -1.96

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 = 1.322*10^3

roots are :
x = 5.434*10^-2 and x = -6.643

since x can't be negative, the possible value of x is
x = 5.434*10^-2

At equilibrium:
[NO] = 1.4-2x = 1.4-2*0.05434 = 1.29 M

Answer:   1.29 M