posted by Mina
At 1285°C, the equilibrium constant for the reaction
Br2 (gas) <---> 2Br (gas) is Keq= 1.04 X 10^ -3
A 0.200-L vessel containing an equilibrium mixture of gases has 0.245 g. Br2 (gas) in it. What is the mass of Br (gas) in the vessel? What are the concentrations of Br2 and Br?
0.245 g Br2. 0.245g/159.808 = 0.00153 moles Br2. M = 0.00153 moles/0.200 L = 0.00767 M.
Br2(g) ==> 2Br(g)
Keq = (Br)^2/(Br2) = 1.04E-3
You know (Br2) and you know Keq, solve for (Br) which will be in moles/L. To find g Br, convert (Br) in mols/L to moles in 0.200 L, then to grams (g = moles x molar mass). Post your work if you get stuck.
I'm having some difficulty getting the answer...I completely follow your logic and your numbers make sense, but I end up getting 1.129 as my final answer while the back of the book says it should be 0.00767....