Carbonyl fluoride, COF 2 , is an important intermediate used in the production o
ID: 915266 • 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 =9.00 If only COF 2 is present initially at a concentration of 2.00 M , what concentration of COF 2 remains at equilibrium? B. CO(g)+NH 3 (g)HCONH 2 (g), K c =0.600 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?
Explanation / Answer
K ) 9
COF2 = 2
CO2 = 0
CF4 = 0
In equilibrium
COF2 = 2 - 2x
CO2 = 0 + x
CF4 = 0 + x
Substitute in K
K = [CO2][CF4] / [COF2]^2
9 = (x*x) /(2-2x)^2
sqr(9) = x/(2-2x)
3(2-2x) = x
6 - 6x = x
6 = 7x
x = 6/7= 0.857
[COF2] = 2 - 2x = 2-2*0.857 = 0.286
B)
B. CO(g)+NH 3 (g)HCONH 2 (g), K c =0.600 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?
K = [HCONH2]/ ([CO][NH3])
K = 0.6
initially
[CO] = 1
[NH3] = 2
[HCONH ] = 0
in euqilibrium
[CO] = 1 -x
[NH3] = 2 -x
[HCONH ] = 0 +x
subtitute
K = [HCONH2]/ ([CO][NH3])
0.6 = x/(1-x)(2-x)
1-3x + x^2 = 1.666x
x^2 -4.666x +1 = 0
x = 0.225
In equlibrium
[HCONH ] = 0 +x = 0.225