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

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Chemistry

The equilibrium constant for the reaction of fluorine gas withbromine 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 theinitial concentrations of bromine and fluorine were 0.121moles/liter in a sealed container and no product was presentinitially?

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4.1 Ratings (823 Votes)

           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
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