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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=8.90 if only cof2 is present initially at a concentration of 2.00 m, what concentration of cof2 remains at equilibrium?

Respuesta :

Dejanras
Answer is:concentration of COF₂ remains at equilibrium is 0,503 M.
Chemical reaction: 2COF₂ ⇄ CO₂ + CF₄.
c₀(COF₂) = 2,00 M.
Kc = 8,90.
[CO₂] = [CF₄] = x; equilibrium concentration.
From chemical reaction: n(COF₂) : n(CO₂) = 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.
[COF₂] = 2,00 M - 1,497 M = 0,503 M.

baraltoa

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 =    x * x /( 2-x)  = 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.

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