A. Carbonyl fluoride, COF2, is an important intermediate used in the production

Question

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=4.90

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

B.

Consider the reaction

CO(g)+NH3(g)HCONH2(g),    Kc=0.890

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.

Explanation / Answer

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

4.9 = (1*x)^2/(2-2*x)^2

sqrt(4.9) = (1*x)/(2-2*x)

2.2136 = (1*x)/(2-2*x)

4.42719-4.42719*x = 1*x

4.42719-5.42719*x = 0

x = 0.81574

At equilibrium:

[COF2] = 2.0-2x = 2.0-2*0.81574 = 0.36851 M

[CO2] = +1x = +1*0.81574 = 0.81574 M

[CF4] = +1x = +1*0.81574 = 0.81574 M

Answer:

[COF2] = 0.369 M

B)

ICE Table:

[CO] [NH3] [HCONH2]

initial 1.0 2.0 0

change -1x -1x +1x

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

Equilibrium constant expression is

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

0.89 = (1*x)/((1-1*x)(2-1*x))

0.89 = (1*x)/(2-3*x + 1*x^2)

1.78-2.67*x + 0.89*x^2 = 1*x

1.78-3.67*x + 0.89*x^2 = 0

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

a = 0.89

b = -3.67

c = 1.78

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 = 7.132

roots are :

x = 3.562 and x = 0.5615

x can't be 3.562 as this will make the concentration negative.so,

x = 0.5615

At equilibrium:

[HCONH2] = x = 0.5615 M

Answer: 0.5615 M

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