Answer:
The correct answer is 0.024 M
Explanation:
First we use an ICE table:
Br₂(g) + F₂(g) ⇔ 2 BrF(g)
I 0.111 M 0.111 M 0
C -x -x 2 x
E 0.111 -x 0.111-x 2x
Then, we replace the concentrations of reactants and products in the Kc expression as follows:
Kc=
Kc=
54.7=
We can take the square root of each side of the equation and we obtain:
7.395=
0.111(7.395) - 7.395x= 2x
0.82 - 7.395x= 2x
0.82= 2x + 7.395x
⇒ x= 0.087
From the x value we can obtain the concentrations in the equilibrium:
[F₂]= [Br₂]= 0.111 -x= 0.111 - 0.087= 0.024 M
[BrF]= 2x= 2 x (0.087)= 0.174 M
So, the concentration of fluorine (F₂) at equilibrium is 0.024 M.
