Question

The equilibrium constant for the reaction of fluorine gas with bromine gas at 300 K is 54.7 and the reaction is: Br2(g) + F2(g) ⇔ 2 BrF(g) What is the equilibrium concentration of fluorine if the initial concentrations of bromine and fluorine were 0.111 moles/liter in a sealed container and no product was present initially?

Answer 1

The correct answer is 0.024 M

Step-by-step 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=
`([BrF ]^(2) )/([ F_(2) ][Br_(2)  ])`

Kc=
`((2x)^(2) )/((0.111-x)(0.111-x))`

54.7=
`(4x^(2) )/((0.111-x)^(2) )`

We can take the square root of each side of the equation and we obtain:

7.395=
`(2x)/((0.111-x))`

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.