Carbonyl fluoride, COF2, is an important intermediate used in the production of
ID: 497286 • Letter: C
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=4.30
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.
Part B
Consider the reaction
CO(g)+NH3(g)HCONH2(g), Kc=0.820
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?
[COF2] =Explanation / Answer
2COF2(g)CO2(g)+CF4(g), Kc=4.30
K = [CO2][CF4]/([CF2]2)
initially:
[COF2] = 2
[CF4] = 0
[CO2] = 0
in equilibirum
[COF2] = 2-2x
[CF4] = 0+x
[CO2] = 0+x
substitute in K
K = [CO2][CF4]/([CF2]2)
4.3 = (x*x)/(2-2x)^2
sqrt(4.3) = x/(2-2x)
2.0736 = x/(2-2x)
2.0736*2 - 2*2.0736x = x
(2*2.0736+1)x = 2.0736*2
x = (2.0736*2 ) / ((2*2.0736+1)) = 0.80571
[COF2] = 2-2x = 2-2*0.80571 = 0.38858
[CF4] = 0+x = 0.80571
[CO2] = 0+x = 0.80571
proof
Q = (0.80571*0.80571) / 0.38858^2= 4.299, which is pretty near to 4.30
B)
for
CO(g)+NH3(g)HCONH2(g), Kc=0.820
initially
[HCONH2] = 0
[CO] = 1
[NH3] = 2
in equilibrium
[HCONH2] = 0 +x
[CO] = 1 - x
[NH3] = 2 - x
substitute in K
K = HCONH2 /(CO)(NH3)
0.82 = (x)/((1 - x)(2 - x))
0.82 (2-3x+x^2) = x
0.82 *2 -3*0.82 x + 0.82 x^2 = x
0.82 x^2 + ( -3*0.82 -1)x + 0.82 *2= 0
0.82 x^2 -3.46x +1.64 = 0
solve for x
0.544
[HCONH2] = 0 +x = 0.544 M