Question

Suppose a 250. ml flask is filled with 0.50 mol of CO, 0.60 mol of NO and 2.0 mol of CO2. The following reaction becomes possible: NO2(g) + CO(g) = NO(g) + CO2(g) The equilibrium constant K for this reaction is 3.29 at the temperature of the flask. Calculate the equilibrium molarity of CO. Round your answer to two decimal places. IM x 6 ?

Answer 1

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

= 0.50 mol / 0.250 L

= 2.0 M

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

= 0.60 mol / 0.250 L

= 2.4 M

Initial concentration of CO2 = mol of CO2 / volume in L

= 2.0 mol / 0.250 L

= 8.0 M

ICE Table:

[CO] (NO2) [NO] [002] initial 2.0 2.4 8.0 change +1x +1x - 1x -1x equilibrium 2.0+1x +1x 2.4-1x 8.0-1x

Equilibrium constant expression is

Kc = [NO]*[CO2]/[CO]*[NO2]

3.29 = (2.4-1*x)(8-1*x)/((2 + 1*x)(1*x))

3.29 = (19.2-10.4*x + 1*x^2)/(2*x + 1*x^2)

6.58*x + 3.29*x^2 = 19.2-10.4*x + 1*x^2

-19.2 + 16.98*x + 2.29*x^2 = 0

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

a = 2.29

b = 16.98

c = -19.2

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 = 4.642*10^2

roots are :

x = 0.9968 and x = -8.412

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

x = 0.9968

At equilibrium:

[CO] = 2.0+1x = 2.0+1*0.9968 = 3.00 M

Answer: 3.00 M