Question

Suppose a 500. ml flask is filled with 0.90 mol of NO,, 0.10 mol of CO and 0.50 mol of NO. The following reaction becomes possible: NO2(g) +CO(g) = NO(g) + CO2(g) The equilibrium constant K for this reaction is 0.172 at the temperature of the flask. Calculate the equilibrium molarity of CO. Round your answer to two decimal places. IM xs ?

Answer 1

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

= 0.90 mol / 0.500 L

= 1.8 M

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

= 0.10 mol / 0.500 L

= 0.20 M

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

= 0.50 mol / 0.500 L

= 1.0 M

ICE Table:

[NO2] [CO] [NO] [C02] 1.8 0.2 1.0 initial change equilibrium - 1x - 1x +1x +1x 1.8–1x 0.2-1x 1.0+1x +1x

Equilibrium constant expression is

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

0.172 = (1 + 1*x)(1*x)/((1.8-1*x)(0.2-1*x))

0.172 = (1*x + 1*x^2)/(0.36-2*x + 1*x^2)

6.192*10^-2-0.344*x + 0.172*x^2 = 1*x + 1*x^2

6.192*10^-2-1.344*x-0.828*x^2 = 0

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

a = -0.828

b = -1.344

c = 6.192*10^-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 = 2.011

roots are :

x = -1.668 and x = 4.483*10^-2

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

x = 4.483*10^-2

At equilibrium:

[CO] = 0.2-1x = 0.2-1*0.04483 = 0.155 M

Answer: 0.155 M