Question

Objective Knowledge Check Suppose a 250. mL flask is filled with 0.20 mol of H20, 0.10 mol of CO2 and 1.5 mol of Hy. The following reaction becomes possible: COG) + H2O@eCO3%)+H,®) The equilibrium constant for this reaction is 4.23 at the temperature of the flask. Calculate the equilibrium molarity of CO. Round your answer to two decimal places Ом $ ?

Answer 1

Initial concentration of H2O = mol of H2O / volume in L

= 0.20 mol / 0.250 L

= 0.80 M

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

= 0.10 mol / 0.250 L

= 0.40 M

Initial concentration of H2 = mol of H2 / volume in L

= 1.5 mol / 0.250 L

= 6.0 M

ICE Table:

[H20] [CO] [C02] [H2] initial 0.8 0 0.4 6.0 change +1x +1x -1x -1x equilibrium 0.8+1x 0.4-1x 6.0-1x +1x

Equilibrium constant expression is

Kc = [CO2]*[H2]/[H2O]*[CO]

4.23 = (0.4-1*x)(6-1*x)/((0.8 + 1*x)(1*x))

4.23 = (2.4-6.4*x + 1*x^2)/(0.8*x + 1*x^2)

3.384*x + 4.23*x^2 = 2.4-6.4*x + 1*x^2

-2.4 + 9.784*x + 3.23*x^2 = 0

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

a = 3.23

b = 9.784

c = -2.4

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 = 1.267*10^2

roots are :

x = 0.2281 and x = -3.257

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

x = 0.2281

At equilibrium:

[CO] = +1x = +1*0.2281 = 0.228 M

Answer: 0.23 M