Question

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

2COF2(g)⇌CO2(g)+CF4(g),    Kc=5.90

If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium?

Part B:

Consider the reaction

CO(g)+NH3(g)⇌HCONH2(g),    Kc=0.860

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?

Answer 1

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

5.9 = (1*x)^2/(2-2*x)^2

sqrt(5.9) = (1*x)/(2-2*x)

2.429 = (1*x)/(2-2*x)

4.858-4.858*x = 1*x

4.858-5.858*x = 0

x = 0.829

At equilibrium:

[COF2] = 2.0-2x = 2.0-2*0.829 = 0.342 M

Answer: 0.342 M

B)

ICE Table:

[CO] [NH3] [HCONH2]

initial 1.0 2.6 0

change -1x -1x +1x

equilibrium 1.0-1x 2.6-1x +1x

Equilibrium constant expression is

kc = [HCONH2]/[CO]*[NH3]

0.86 = (1*x)/((1-1*x)(2.6-1*x))

0.86 = (1*x)/(2.6-3.6*x + 1*x^2)

2.236-3.096*x + 0.86*x^2 = 1*x

2.236-4.096*x + 0.86*x^2 = 0

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

a = 0.86

b = -4.096

c = 2.236

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 = 9.085

roots are :

x = 4.134 and x = 0.629

x can't be 4.134 as this will make the concentration negative.so,

x = 0.629

At equilibrium:

[HCONH2] = x = 0.629 M

Answer: 0.629 M