Question

Suppose a 500. mL flask is filled with 1.5 mol of CO, 1.2 mol of NO and 1.0 mol of CO,. The following reaction becomes possible: NO2(g) +CO(g) = NO(g) + CO2(g) The equilibrium constant K for this reaction is 6.78 at the temperature of the flask. Calculate the equilibrium molarity of NO2. Round your answer to two decimal places. Пм x 6 ?

Answer 1

Initial concentration of CO = mol of CO / volume in L
= 1.5 mol / 0.500 L
= 3.0 M

Initial concentration of NO = mol of NO / volume in L
= 1.2 mol / 0.500 L
= 2.4 M

Initial concentration of CO2 = mol of CO2 / volume in L
= 1.0 mol / 0.500 L
= 2.0 M

ICE Table:

[CO] [NO2] [NO] [002] initial 3.0 2.4 2.0 change +1x +1x -1x - 1x equilibrium 3.0+1x +1x 2.4-1x 2.0-1x

Equilibrium constant expression is
Kc = [NO]*[CO2]/[CO]*[NO2]
6.78 = (2.4-1*x)(2-1*x)/((3 + 1*x)(1*x))
6.78 = (4.8-4.4*x + 1*x^2)/(3*x + 1*x^2)
20.34*x + 6.78*x^2 = 4.8-4.4*x + 1*x^2
-4.8 + 24.74*x + 5.78*x^2 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 5.78
b = 24.74
c = -4.8

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 = 7.23*10^2

roots are :
x = 0.1859 and x = -4.466

since x can't be negative, the possible value of x is
x = 0.1859

At equilibrium:
[NO2] = +1x = +1*0.1859 = 0.186 M

Answer: 0.19 M