session.masteringchemistry.com 6 of9 Part A Carbonyl fluoride, COF2, is an impor
ID: 594133 • Letter: S
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session.masteringchemistry.com 6 of9 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, CF via the reaction 2COF2(g) CO2(8)+CF (g), Ke 4.50 If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF remains at equilibrium? Express your answer with the appropriate units. > View Available Hint(s) ? icor,1 = 10.38 |m Submit Incorrect: One attempt remaining: Try Again Part B Consider the reaction C0(g):, NH, (g) ist HCONH2 (g), Ke= 0.610Explanation / Answer
A)
ICE Table:
[COF2] [CO2] [F2]
initial 2.0 0 0
change -2x +1x +1x
equilibrium 2.0-2x +1x +1x
Equilibrium constant expression is
Kc = [CO2]*[F2]/[COF2]^2
4.5 = (1*x)^2/(2-2*x)^2
sqrt(4.5) = (1*x)/(2-2*x)
2.121 = (1*x)/(2-2*x)
4.24264 - 4.24264*x = 1*x
4.24264 - 5.24264*x = 0
x = 0.809
At equilibrium:
[COF2] = 2.0-2x = 2.0-2*0.809 = 0.381 M
Answer: 0.381 M
B)
ICE Table:
[CO] [NH3] [HCONH2]
initial 1.0 2.0 0
change -1x -1x +1x
equilibrium 1.0-1x 2.0-1x +1x
Equilibrium constant expression is
Kc = [HCONH2]/[CO]*[NH3]
0.61 = (1*x)/((1-1*x)(2-1*x))
0.61 = (1*x)/(2-3*x + 1*x^2)
1.22-1.83*x + 0.61*x^2 = 1*x
1.22-2.83*x + 0.61*x^2 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 0.61
b = -2.83
c = 1.22
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 = 5.032
roots are :
x = 4.158 and x = 0.481
x can't be 4.158 as this will make the concentration negative.so,
x = 0.481
At equilibrium:
[HCONH2] = +1x = +1*0.48096 = 0.481 M
Answer: 0.481 M
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