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

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= $\frac{[BrF ]^{2} }{[ F_{2} ][Br_{2} ]}$

Kc= $\frac{(2x)^{2} }{(0.111-x)(0.111-x)}$

54.7= $\frac{4x^{2} }{(0.111-x)^{2} }$

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

7.395= $\frac{2x}{(0.111-x)}$

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.