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=4.60

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.
Hints

[COF2] =

Part B

Consider the reaction

CO(g)+NH3(g)⇌HCONH2(g),    Kc=0.760

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

Ans. Stoichiometry of balanced reaction – 1 mol COF₂ dissociates into 0.5 mol each of CO₂ and CF.

Create an ICE table as shown in figure-

Equilibrium constant, Kc = [CO₂] [CF] / [COF₂]²

            Or, 4.60 = (0.5X) (0.5X) / (2.0- X)²

            Or, 4.60 (X² – 4X + 4) = 0.25X²

            Or, 4.60X² – 18.4X + 18.4 – 0.25X² = 0

            Or, 4.35X² – 18.4X + 18.4 = 0

Solving the quadratic equation, we get following two roots-

            X1 = 2.607                 ; X2 = 1.622

Since, equilibrium consecration can’t be greater than 2.00, reject X1.

Hence, X = 1.622 M.

Now,

            [COF₂] ay equilibrium = 2.0 M – X = (2.0 – 1.622) M = 0.378 M

# Part B: Stoichiometry of balanced reaction – 1 mol each of CO and NH3 reacts to form 1 mol HCONH2.

Create an ICE table as shown in figure-

Equilibrium constant, Kc = [HCONH2] / ([CO] [NH3])

            Or, 0.760 = X / [(1.00 - X) (2.00 -X)]

            Or, 0.760 (X² – 3.0X + 2.0) = X

            Or, 0.760X² – 3.0X + 2.0 –X = 0

            Or, 0.760X² – 4.0X + 2.0 = 0

Solving the quadratic equation, we get following two roots-

            X1 = 4.70                   ; X2 = 0.559

Since, equilibrium consecration can’t be greater than 1.00, reject X1.

Hence, X = 0.559 M.

Now,

            [HCONH2] ay equilibrium = X M = 0.559 M