[HBr] = moles/L = 3.20 mol/12.0 L = 0.267 M
[H2] = 1.50 mol/12.0 L = 0.125 M
                            Â
2HBr       <====> 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
-2x                        Â
+x         +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
-----------------------------------------------------------------------
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
