In: Chemistry
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 0.121 moles/liter in a sealed container and no product was present initially?
Br2(g) + F2(g) ⇔ 2 BrF(g)
I 0.121 0.121 0
C -x -x 2x
E 0.121-x 0.121-x 2x
Kc = [BrF]2/[Br2][F2]
54.7 = (2x)2 /(0.121-x)(0.121-x)
54.7 = (2x/0.121-x)2
7.34 = 2x/0.121-x
2x = 7.34(0.121-x)
x = 0.095
[F2] = 0.121-x = 0.121-0.095 = 0.026M
[Br2] = 0.121-x = 0.121-0.095 = 0.026M
[BrF] = 2x = 2*0.095 = 0.19M