Part A Carbonyl fluoride COF2 is an important intermediate u

Part A Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it is used to make the refrigerant carbon tetrafluoride, CF4 via the reaction The concentrations of reactants and products for a chemical reaction can be calculated if the equilibrium constant for the reaction and the starting concentrations of reactants and/or products are known. 2COFolg) CO2(g) +CF4(g), K.-5.50 If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium? Express your answer with the appropriate units View Available Hint(s) COP.I-Value Units Submit Part B Consider the reaction CO(g) + NH3 (g) HCONH2(g), = 0.720 If a reaction vessel initially contains only CO and NH3 at concentrations of 1.00 M and 2.00 M, respectively, what will the concentration of HCONH2 be at equilibrium? Express your answer with the appropriate units View Available Hint(s)

Solution

A)
ICE Table:

                    [COF2]              [CO2]               [CF4]             

initial             2.0                 0                   0                 

change              -2x                 +1x                 +1x               

equilibrium         2.0-2x              +1x                 +1x               

Equilibrium constant expression is
Kc = [CO2]*[CF4]/[COF2]^2
5.5 = (1*x)^2/(2-2*x)^2
sqrt(5.5) = (1*x)/(2-2*x)
2.345208 = (1*x)/(2-2*x)
4.69042-4.69042*x = 1*x
4.69042-5.69042*x = 0
x = 0.82427

At equilibrium:
[COF2] = 2.0-2x = 2.0-2*0.82427 = 0.35147 M

Answer: 0.351 M

B)
ICE Table:

                    [CO]                [NH3]               [HCCONH2]         

initial             1.0                 2.0                 0                 

change              -1x                 -1x                 +1x               

equilibrium         1.0-1x              2.0-1x              +1x               

Equilibrium constant expression is
Kc = [HCCONH2]/[CO]*[NH3]
0.720 = (1*x)/((1-1*x)(2-1*x))
0.720 = (1*x)/(2-3*x + 1*x^2)
1.44-2.16*x + 0.72*x^2 = 1*x
1.44-3.16*x + 0.72*x^2 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 0.72
b = -3.16
c = 1.44

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 = 5.838

roots are :
x = 3.872 and x = 0.5165

x can\'t be 3.872 as this will make the concentration negative.so,
x = 0.5165

At equilibrium:
[HCCONH2] = x = 0.5165 M

Answer: 0.5165 M