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?

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.

the correct option is c.

isotopes refers to different forms of the same element with the same atomic number but different mass number as a result of difference in their neutron numbers. isotopes have different relative atomic mass but share the same chemical properties.

the earth would become to overheated and everything alive on it will eventually die from overheating. plants will eventually die as well, water will evaporate, and it will be nothing but a hot planet without anything living on it.