The equilibrium constant for the reaction of fluorine gas with bromine gas at 300 K is...

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?

Expert Solution

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

queen_honey_blossom answered 1 year ago

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