Question

Tall- 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(9) = CO2(g) + CF4(9), K= 7.80 at 850 K If only COF2 is present initially at a pressure of 260 kPa, what is the partial pressure of COF2 at equilibrium? You can convert the units of pressure from kPa to bar, using the relation 1 kPa = 0.01 bar. Express your answer with the appropriate units. ► View Available Hint(s) T: Å Pcor, = | Value O ? Units PCOF, = You have already submitted this answer. Enter a new answer. No credit lost. Try again. Submit Previous Answers

Answer 1

Part A

Answer

0.394 bar

Explanation

2COF2(g) <------> CO2(g) + CF4(g)

K = P_CO2 × P_CF4/(P_COF2)² = 7.80

Initial partial pressure

P_COF2 = 2.6 bar

P_CO2 = 0

P_CF4 = 0

Change in partial pressure

P_COF2 = -2x

P_CO2 = +x

P_CF4 = + x

Equilibrium partial pressure

P_COF2 = 2.6 - 2x

P_CO2 = x

P_CF4 = x

so,

x²/(2.6 - 2x)² = 7.80

x / (2.6 - 2x) = 2.793

x = 7.2618 - 5.586x

6.586x = 7.2618

x = 1.103 bar

Therefore, at equilibrium

P_COCl2 = 2.6 - (2×1.103) = 0.394 bar

Part B

Answer

1.624 bar

Explanation

CO(g) + NH3(g) ------> HCONH2(g)

K = P_HCONH2/ ( P_CO × P_NH3) = 2.70

Initial partial pressure

P_CO= 2.40

P_NH3 = 2.40

P_HCONH2 = 0

change in partial pressure

P_CO = - x

P_NH3 = - x

P_HCONH2 = +x

Equilibrium concentration

P_CO = 2.40 - x

P_NH3 = 2.40 - x

P_HCONH2 = x

so,

x/( 2.40 -x)² = 2.70

solving for x

x = 1.624

Therefore, at equilibrium

P_HCONH2 = 1.624 bar