Carbonyl fluoride. COF_2 is an important intermediate used in the production of
ID: 497038 • Letter: C
Question
Carbonyl fluoride. COF_2 is an important intermediate used in the production of fluorine-containing compounds. For instance, it is used to make the refrigerant carbon tetrafluoride, CF_4 via the reaction 2COF_2(g) CO_2(g) + CF_4(g), K_c = 4.20 If only COF_2 is present initially at a concentration of 2.00 M. what concentration of COF_2 remains at equilibrium? Express your answer with the appropriate units. Consider the reaction CO(g) + NH_3 (g) HCONH_2 (g), K_c = 0.890 If a reaction vessel initially contains only CO and NH_3 at concentrations of 1.00 M and 2.00 M, respectively, what will the concentration of HCONH_2 be at equilibrium? Express your answer with the appropriate units.Explanation / Answer
A)
2COF2 <----> CO2 + CF4
2.00 0 0 (initial)
2.00-2x x x (at equilibrium)
Kc = [CO2][CF4]/[COF2]^2
4.20 = x^2/(2-2x)^2
4.20*(4+4x^2-8x) =x^2
16.8 + 16.8*x^2 - 33.6 x = x^2
15.8*x^2 - 33.6*x + 16.8 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 15.8
b = -33.6
c = 16.8
Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a
b^2-4*a*c = 67.2
roots are :
x = 1.3 and x = 0.8
x can't be 1.3 as it will make [COF2] negative
so, x = 0.8
[COF2] = 2.0 - 2x = 2.0 - 2*0.8 = 0.4 M
Answer: 0.4 M
B)
CO + NH3 <----> HCONH2
1 2 0 (INITIAL)
1-x 2-x x (AT EQUILIBRIUM)
Kc = [HCONH2]/[CO][NH3]
0.890 = x / (1-x)(2-x)
0.890 = x / (2 -3x +x^2)
0.890*x^2 - 2.67*x + 1.78 = x
0.890*x^2 - 3.67*x + 1.78 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 0.89
b = -3.67
c = 1.78
Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a
b^2-4*a*c = 7.13
roots are :
x = 3.56 and x = 0.56
x can't be 3.56
so,
x = 0.56
[HCONH2]=x = 0.56 M