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 0.131 moles/liter in a sealed container and no product was present initially?

Answer 1

                          Br₂(g) + F₂(g) ⇔ 2 BrF(g)

initial conc          0.131     0.131           0

change                -a           -a              +2a

Equb conc      0.131-a    0.131-a           2a

Equilibrium constant , K = [BrF]² / ( [Br₂][F₂])

                             54.7 = (2a)² / [(0.131-a)(0.131-a)]

                2a / (0.131-a) = 7.39

solving we get a = 0.103 M

So the equilibrium concentration of fluorine = 0.131-a = 0.131 - 0.103 = 0.028 M