Question

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=4.30

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.

[COF2] =

Part B

Consider the reaction

CO(g)+NH3(g)⇌HCONH2(g),    Kc=0.820

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

2COF2(g)⇌CO2(g)+CF4(g),    Kc=4.30

K = [CO2][CF4]/([CF2]2)

initially:

[COF2] = 2

[CF4] = 0

[CO2] = 0

in equilibirum

[COF2] = 2-2x

[CF4] = 0+x

[CO2] = 0+x

substitute in K

K = [CO2][CF4]/([CF2]2)

4.3 = (x*x)/(2-2x)^2

sqrt(4.3) = x/(2-2x)

2.0736 = x/(2-2x)

2.0736*2 - 2*2.0736x = x

(2*2.0736+1)x = 2.0736*2

x = (2.0736*2 ) / ((2*2.0736+1)) = 0.80571

[COF2] = 2-2x = 2-2*0.80571 = 0.38858

[CF4] = 0+x = 0.80571

[CO2] = 0+x = 0.80571

proof

Q = (0.80571*0.80571) / 0.38858^2= 4.299, which is pretty near to 4.30

B)

for

CO(g)+NH3(g)⇌HCONH2(g),    Kc=0.820

initially

[HCONH2] = 0

[CO] = 1

[NH3] = 2

in equilibrium

[HCONH2] = 0 +x

[CO] = 1 - x

[NH3] = 2 - x

substitute in K

K = HCONH2 /(CO)(NH3)

0.82 = (x)/((1 - x)(2 - x))

0.82 (2-3x+x^2) = x

0.82 *2 -3*0.82 x + 0.82 x^2 = x

0.82 x^2 + ( -3*0.82 -1)x + 0.82 *2= 0

0.82 x^2 -3.46x +1.64 = 0

solve for x

0.544

[HCONH2] = 0 +x = 0.544 M