Part A Carbonyl fluoride, COF2, is an important intermediate used in the product

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=8.20

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=0.790

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?

Explanation / Answer

Part A

2 COF2 <=> CO2 + CF4
2.00 -2x <=> x & x

K = [CO2] [CF4] / [COF2)^2

8.20 = [x] [x] / [2 - 2x]^2

8.20 [2 - 2x]^2 = [x] [x]

8.20 [4 - 8x +4x2] = x2

32.8 - 65.6x + 32.8x2 = x2

32.8 - 65.6x + 31.8x2 = 0

x = 0.851

what concentration of COF2 remains at equilibrium:
2.00 - 2x =

2.00 - 1.702 = 0.298 Molar

Part B

Kc = [HCONH2] / [CO][NH3]
let x = moles/L of HCONH2 formed
[CO] = 1.00 M - x
[NH3] = 2.00 M - x
[HCONH2] = x
0.790 = x / (1.00-x)(2.00-x) = x / (2.00 - 3.00 x + x^2)
0.790x^2 - 3.37x + 1.58 = 0
x = 0.536
HCONH2 = 0.536 M

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.