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Question

The equilibirium constant for the reaction H2(g) + Br2(g) <--------> 2HBr(g) is 2.18 x 10^6 at 730 degrees C. Starting with 3.20 moles of HBr in a 12.0L vessel, cacculate the concentrations of H2, Br2, and HBr at equilibrium.

Answer 1

Note that our reaction is running in reverse.
Co of HBr(g) = 3.2/12= 0.275 mol//l
Let 2x be the amount of HBr(g) decomposed.
Ce of HBr = 0.275-2x
From the equilibrium reaction.
Ce of H2 = x
Ce of Br2 = x
So we have: (0.275-2x)/x^2 = 2.18x106
Calculate it out.