Carbonyl fluoride, COF_2, is an important intermediate used in the production of
ID: 937710 • 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 = 4.60 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 = 0.790 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
KC = 4.60
Kc = [CO2][CF4]/[COF2]^2
in the equilbirium:
[CO2]= 0 + x
[CF4]= 0 +x
[COF2]= 2-x
substitute
Kc = [CO2][CF4]/[COF2]^2
4.6 = (x*x)/(2-x)^2
solve for x
sqrt(4.6) = x/(2-x)
2.1447*(2-x) = x
4.2894 - 2.1447x = x
3.1447x = 4.2894
x= 4.2894 /3.1447 = 1.364
substitute
[CO2]= 0 + x = 1.364
[CF4]= 0 +x = 1.364
[COF2]= 2-x 2-1.364 = 0.636
[COF2]= 0.636
B)
Kc = 0.79
Kc = [HCONH2] /[CO][NH3]
in equilbrium:
[HCONH2]= 0+x
[CO] = 1 -x
[NH3] = 2-x
substitute
Kc = [HCONH2] /[CO][NH3]
0.79= x /(2-x)(1 -x)
(2-3x +x^2)= x/0.79
1.26582x = (2-3x +x^2)
x^2 -4.2658 +2= 0
x = 0.536 and 3.732 (last one cant be since then we hav enegative concnetrations)
then
[HCONH2]= 0+x = 0.536
[CO] = 1 -x = 1-0.536 = 0.464
[NH3] = 2-x = 2-0.536 = 1.464