The equilibrium constant Kc for the following reaction is 4.59 Ã— 10-7 at 730oC. 2HBr(g) â‡Œ...

3.6 Ratings (362 Votes)

[HBr] = moles/L = 3.20 mol/12.0 L = 0.267 M

[H2] = 1.50 mol/12.0 L = 0.125 M

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 2HBrÂ Â Â Â Â Â Â <====>Â Â H2Â Â Â Â Â Â + Br2

initial (M)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.267Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.125Â Â Â Â Â Â Â Â 0

change(M)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â -2xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +xÂ Â Â Â Â Â Â Â Â +x

equilibrium(M)Â Â Â Â Â (0.267-2x)Â Â Â Â Â Â Â Â Â Â Â Â (0.125 +x)Â Â Â Â x

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Kc = [H2][Br2]/[HBr]^2

4.59 x 10^-7 = (x)(0.125 + x)/(0.267 - 2x)^2

4.59 x 10^-7 = 0.125x + x^2/(0.0713 - 1.068x + 4x^2)

1.84 x 10^-6x^2 - 4.902 x 10^-7x + 3.27 x 10^-8 = 0.125x + x^2

x^2 + 0.125x + 4.902 x 10^-7 = 0

x = 0.125 M

So the concentration of [Br2] = 0.125 M

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