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=6.70 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=0.780

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

A)
reaction taking place is
2COF2 <--> CO2 + CF4
2 0 0 initaly
2-2x x x at equlibrium

Kc = ([CO2]*[CF4])/[COF2]^2
6.7 = x^2/(2-2x)^2
6.7 = (x/2-2x)^2
2.6 = x/(2-2x)
2.6*(2-2x) = x
5.2-(5.2)x = x
(6.2)x = 5.2
x = 0.89

equlibrium concentration of COF2 = 2=2X
= 2-2*(0.89)
= 2-1.78
= 0.22 M
Answer : 022 M

B)
reaction taking placeis
CO + NH3 <--> HCONH2
1 2 0 initialy
1-x 2-x x at equlibrium
Kc = [HCONH2]/([CO]*[NH3])
0.78 = X/{(1-X)*(2-x)}
0.78*(2-3x+x^2) = x
x is very small so, we can neglect x^2 term
0.78*(2-3x) = x
1.56-(2.34)x = x
(3.34)x = 1.56
x = 0.47

equlibrium cocentration of HCONH2 = x
= 0.47 M
Answer : 0.47 M