Answer: The concentration of [tex]COF_2[/tex] remains at equilibrium is 0.40 M.
Explanation:-
Initial concentration of [tex]COF_2[/tex] = 2 M
The given balanced equilibrium reaction is,
            [tex]2COF_2(g)\rightleftharpoons CO_2(g)+CF_4(g[/tex]
Initial conc. Â Â Â Â Â Â Â Â Â Â Â 2 M Â Â Â Â Â Â 0 Â Â Â Â Â Â 0 Â Â Â Â Â Â
At eqm. conc. Â Â Â Â Â (2-2x) M Â Â Â Â Â Â x M Â Â Â Â x M
The expression for equilibrium constant for this reaction will be,
[tex]K_c=\frac{[CO_2]\times [CF_4]}{[COF_2]^2}[/tex]
Now put all the given values in this expression, we get :
[tex]4.20=\frac{(x)\times (x)}{(2-2x)^2}[/tex]
By solving the term 'x', we get :
x = Â 0.80 M
Thus, the concentrations of [tex]COF_2[/tex] at equilibrium is : (2-2x) = 2-2(0.80)=0.40 M
The concentration of [tex]COF_2[/tex] remains at equilibrium is 0.40 M.