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The correct equilibrium constant expression for this equation is;
[tex]\rm K_{eq}=\dfrac{(C0_2)(CF_4)}{(COF_2)^2}[/tex]
Given
Consider the equation below.
[tex]\rm {2COF_2(g)}\rightleftharpoons C0_2(g)+CF_4(g)[/tex]
In an equilibrium constant expression, you do not include the solid substances; only gases and dissolved substances.
It is the quotient of the product of the concentrations of the species on the right-hand side of the equilibrium equation, each raised to its corresponding coefficient, and the product of the concentrations of the species on the left-hand side, each raised to its corresponding coefficient.
Therefore,
The correct equilibrium constant expression for this equation is;
[tex]\rm K_{eq}=\dfrac{(C0_2)(CF_4)}{(COF_2)^2}[/tex]
Hence, the correct equilibrium constant expression for this equation is;
[tex]\rm K_{eq}=\dfrac{(C0_2)(CF_4)}{(COF_2)^2}[/tex].
To know more about Equilibrium constant click the link is given below.
https://brainly.com/question/19240570
Answer:
b then c
Step-by-step explanation:
look at the image
