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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=4.90 If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium?

Answer :

Answer : The concentration of [tex]COF_2[/tex] remains at equilibrium will be, 0.37 M

Explanation :  Given,

Equilibrium constant = 4.90

Initial concentration of [tex]COF_2[/tex] = 2.00 M

The balanced equilibrium reaction is,

                       [tex]2COF_2(g)\rightleftharpoons CO_2(g)+CF_4(g)[/tex]

Initial conc.    2 M                 0             0

At eqm.         (2-2x) M          x M         x M

The expression of equilibrium constant for the reaction will be:

[tex]K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}[/tex]

Now put all the values in this expression, we get :

[tex]4.90=\frac{(x)\times (x)}{(2-2x)^2}[/tex]

By solving the term 'x' by quadratic equation, we get two value of 'x'.

[tex]x=1.291M\text{ and }0.815M[/tex]

Now put the values of 'x' in concentration of [tex]COF_2[/tex] remains at equilibrium.

Concentration of [tex]COF_2[/tex] remains at equilibrium = [tex](2-2x)M=[2-2(1.219)]M=-0.582M[/tex]

Concentration of [tex]COF_2[/tex] remains at equilibrium = [tex](2-2x)M=[2-2(0.815)]M=0.37M[/tex]

From this we conclude that, the amount of substance can not be negative at equilibrium. So, the value of 'x' which is equal to 1.291 M is not considered.

Therefore, the concentration of [tex]COF_2[/tex] remains at equilibrium will be, 0.37 M

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