Part A. Carbonyl fluoride, COF2, is an important intermediate used in the produc
ID: 1008194 • Letter: P
Question
Part A. 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=7.30 If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium? Part B Consider the reaction CO(g)+NH3(g)HCONH2(g), Kc=0.660 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
Part A. 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=7.30 If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium?
2COF2(g)CO2(g)+CF4(g), Kc=7.30
Kc = [CO2][CF4] / [COF2]²
7.30 = x² / (2.00 - x)²
Taking square root of both side:
sqrt(7.30) = x / (2.00 - x)
2.70 = x/ 2.00-x
5.40 – 2.70 x= x
5.40 = 3.70 x
X = 5.40/3.70
X= 1.459 or 1.46
[COF2] = 2.00 - 1.46 M = 0.54 M
Part B Consider the reaction CO(g)+NH3(g)HCONH2(g), Kc=0.660 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?
Kc = [HCONH2]/[CO][NH3]
0.660 = x/(1.00-x)(2.00-x),
0.660 = x/2.00-1.00 x -2.00 x –x^2
0.660 = x/2.00-3.00 x –x^2
1.32 -1.98 x – 0.660 x^2= x
0.66 x^2 +2.98 x -1.32=0
X= 0.406