Question

Suppose a 500 ml flask is filled with 1.7 mol of CO, 0.40 mol of H.O and 1.9 mol of CO,The following reaction becomes possible: CO(e)+H,0(B) - CO (8)+H.) The equilibrium constant K for this reaction is 0.305 at the temperature of the flask. Calculate the equilibrium molarity of CO. Round your answer to two decimal places

Answer 1

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

= 1.7 mol / 0.500 L

= 3.4 M

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

= 0.40 mol / 0.500 L

= 0.80 M

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

= 1.9 mol / 0.500 L

= 3.8 M

ICE Table:

[CO] [H20] [002] [H2] initial 3.4 0.8 3.8 change -1x -1x +1x +1x equilibrium 3.4–1x 0.8-1x 3.8+1x +1x

Equilibrium constant expression is

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

0.305 = (3.8 + 1*x)(1*x)/((3.4-1*x)(0.8-1*x))

0.305 = (3.8*x + 1*x^2)/(2.72-4.2*x + 1*x^2)

0.8296-1.281*x + 0.305*x^2 = 3.8*x + 1*x^2

0.8296-5.081*x-0.695*x^2 = 0

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

a = -0.695

b = -5.081

c = 0.8296

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 = 28.12

roots are :

x = -7.471 and x = 0.1598

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

x = 0.1598

At equilibrium:

[CO] = 3.4-1x = 3.4-1*0.1598 = 3.24 M

Answer: 3.24 M