Please answer! Part A: Carbonyl fluoride, COF2, is an important intermediate use

Question

Please answer!

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

Part B:

Consider the reaction

CO(g)+NH3(g)?HCONH2(g),    Kc=0.770

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?

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

Explanation / Answer

a)

Chemical Equation --------- 2COF2(g) =========> CO2(g) -------- + --------- CF4(g)

Intial Concentration ----------- 2 ------------------------------ 0 ------------------------------- 0

Change in Conc. ----------- -- - 2x --------------------------- x ------------------------------- x

Equilibrium Conc. ----------- 2 - 2x ------------------------- x ------------------------------- x


Kc = [CO2][CF4] / [COF2]^2

Kc = (x)(x) / (2 - 2x)^2

Kc = x^2 / (2 - 2x)^2 = 4.50 (solve for x algebraically)

x = [CO2] = [CF2] = 0.8093

[COF2] = 2 - 2x = 2 - (2)(0.8093)

[COF2] = 0.3814 (ANSWER)

b)

Chemical Reaction ----------- CO(g) -------- + -------- NH3(g) ======> HCONH2(g)

Initial Conc. ---------------------- 1 ----------------------------- 2 ------------------------- 0

Change in Conc. --------------- - x -------------------------- - x ------------------------ x

Equilibrium Conc. ----------- 1 - x ------------------------- 2 - x ---------------------- x


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

Kc = x / (1 - x)(2 - x) = 0.850 (simplify and solve for x algebraically)

x = [HCONH2] = 0.5518 (ANSWER)

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