Carbonyl fluoride, COF2, is an important intermediate used in the production of
ID: 539966 • 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=7.20 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=0.830
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
First, let us define the equilibrium constant for any species:
The equilibrium constant will relate product and reactants distribution. It is similar to a ratio
The equilibrium is given by
rReactants -> pProducts
Keq = [products]^p / [reactants]^r
For a specific case:
aA + bB = cC + dD
Keq = [C]^c * [D]^d / ([A]^a * [B]^b)
Q1.
2COF2(g)CO2(g)+CF4(g) Kc = 7.20
initially
[COF2] = 2
[CO2] = 0
[CF4] = 0
in equilbirium
[COF2] = 2 -2x
[CO2] = 0 + x
[CF4] = 0 + x
K = [CO2][CF4]/([COF2]^2)
7.20 = x*x/(2 -2x)^x
sqrt(7.20) = x / (2-2x)
2-2x = 1/2.6832 x
(1/2.6832+2)x = 2
x = 2/ (1/2.6832 +2)
x = 0.843
[COF2] = 2 -2x = 2-2*0.843 = 0.314 M
Q2.
K = [HCONH2] /([CO][NH3])
initially
[CO] = 1
[NH3] = 2
[HCONH2] = 0
in equilbirum
[CO] = 1 - x
[NH3] = 2 - x
[HCONH2] = 0+x
substitute
K = [HCONH2] /([CO][NH3])
0.83 = (x) / (1-x)(2-x)
1-3x +x^2 = 1.205x
x^2 - 4.205x +1 = 0
x = 0.253
[HCONH2] = 0+x = 0.253 M