For the equilibrium br2(g)⇌2Br(g), Kc=0.00104 at 1285 degrees C. What will be the equilibrium concentration of both species if 1.00 mole Br2 is put into 1.00 L at 1285 degrees C?
Br2 <>>> 2Br
Kc= [Br]^2/(1-2Br)
solve for [Br], looks like a quadratic equation
1.04e-3= x^2/(1-x)
where x will be conc Br, and 1-2x will be conc of Br2