The equilibrium constant of the reaction h2 + Cl2= 2HCl at 100 C is 50. Of a one liter flask containing 1 mole of H2 is connected to a two liter flask containing two moles of Cl2, how many moles of HCl will be formed at 100 oC
The general expression of equilibrium constant is Keq = [concentration of products]^n / [concentration of reactants]^m, where n and m are the stoichiometric coefficients for that species. In this problem,
H2 + Cl2 -> 2 HCl
the Keq = [HCl]^2 / [H2][Cl2]
We square the concentration of HCl since it has a stoichiometric coefficient of 2 in the reaction. Now, we know that Keq = 50, so we need to find the concentration of H2 and the concentration of Cl2. We'll need molarity for this, which is (moles / liters).
The total volume in this reaction is 3 liters (one from the flask with H2, two from the flask with Cl2). The molarity of H2 will just be its moles divided by 3 liters, or 1/3 moles per liter. Cl2 will be 2/3 moles per liter. Now our expression reads:
50 = [HCl]^2 / (1/3)(2/3)
Rearranging for HCl gives: [HCl]^2 = 11.11
Square root both sides and [HCl] = 3.33 moles per liter
This is not our answer; it's a concentration that tells us how many moles we have per liter in this reaction. We have 3 liters, so 3.33 moles per liter * 3 liters = about 10 moles.