
Respuesta :
Answer is:concentration of COF₂ remains at equilibrium is 0,503 M.
Chemical reaction: 2COF₂ ⇄ CO₂ + CF₄.
câ‚€(COFâ‚‚) = 2,00 M.
Kc = 8,90.
[COâ‚‚] = [CFâ‚„] = x; equilibrium concentration.
From chemical reaction: n(COFâ‚‚) : n(COâ‚‚) = 2 : 1.
[COF₂] = 2 · [CO₂] = 2x
Kc = [CO₂] · [CF₄] / [COF₂]².
Kc = x² / (2-x)².
8,9 = x² / (2-x)², square root formula.
2,98 = x / 2-x.
5,96 - 2,98x = x.
x = 1,497 M.
[COFâ‚‚] = 2,00 M - 1,497 M = 0,503 M.
Chemical reaction: 2COF₂ ⇄ CO₂ + CF₄.
câ‚€(COFâ‚‚) = 2,00 M.
Kc = 8,90.
[COâ‚‚] = [CFâ‚„] = x; equilibrium concentration.
From chemical reaction: n(COFâ‚‚) : n(COâ‚‚) = 2 : 1.
[COF₂] = 2 · [CO₂] = 2x
Kc = [CO₂] · [CF₄] / [COF₂]².
Kc = x² / (2-x)².
8,9 = x² / (2-x)², square root formula.
2,98 = x / 2-x.
5,96 - 2,98x = x.
x = 1,497 M.
[COFâ‚‚] = 2,00 M - 1,497 M = 0,503 M.
Answer:
0.32 m
Explanation:
           2 COF2  ========= CO2 + CF4 ( all gases)
initially          2             0      0
equilibrium      2-x           x      x
Then the equilibrium constant given by the coefficients in the balanced equation is:
K = Â Â x * x /( 2-x) Â = 8.90
Then  x^2/(2-x) = 8.90
It gets a bit messy since we get to solve  a  quadratic equation :
    x^2 = 17.80 - 8.90 x
    x^2 + 8.90 x - 17.80 = 0
After solving we get that x is equal to 1.68
Therefore the amount of COF2 at equilibrium is ( 2 - 1.68 ) m or
0.32 m
Note: We can check answer by plugging the results in the equilibrium equation.