Answer:
0.396 M
Explanation:
Let's consider the following reaction.
2 COF₂(g) ⇌ CO₂(g) + CF₄(g)
We can find the concentrations at equilibrium using an ICE Chart.
2 COF₂(g) ⇌ CO₂(g) + CF₄(g)
I 2.00 0 0
C -2x +x +x
E 2.00-2x x x
The concentration equilibrium constant (Kc) is:
[tex]Kc=4.10=\frac{[CO_{2}][CF_{4}]}{[COF_{2}]^{2} } =\frac{x^{2} }{(2.00-2x)^{2} } =(\frac{x}{2.00-2x} )^{2} \\\sqrt{4.10} = \frac{x}{2.00-2x}\\4.05-4.05x=x\\x=0.802[/tex]
The concentration of COF₂ at equilibrium is:
[COF₂] = 2.00-2x = 2.00-2(0.802) = 0.396 M