[lilcybergimp]

I need help on a simple problem. The problem states, A mixture of 0.500 mol H₂ and 0.500 mol I₂ was placed in a 1.00 L stainless-steel flask at 430 °C. The equilibrium constant K_c for the reaction H₂ (g) + I₂ (g) <--> 2HI (g) is 54.3 at this temperature. Calculate the concentrations of H₂, I₂ and HI at equilibrium.

All I need to know is how to find the molarity. Since I already know that the equalibrium equation is:
K_c = Products/Reactants, where product is (product's Molarity/ 1M) and reactants is (reactant's Molarity/ 1M). I know molarity is Moles of solute/Liters of solution. I am having trouble finding the Liters of solution.

If anyone can help that would be great. Thanks

[Borek]

There is no solution, this all happens in gaseous phase. Still, molarity is n/V.

Your equilibrium equation doesn't look correct to me. Do you know what reaction quotient is?

[lilcybergimp]

I am looking in the book, and from what I can see is that this equation would be set up like this:
 K_c = [HI]² / [H₂] [I₂]. The problem has given me the moles for H₂ and I₂, so wouldn't I need to convert this to molarity? And when I get the moarity for H₂ and I₂, I can solve for the molarity of HI.

[lilcybergimp]

I also have a follow up question for this same equation [ H₂ (g) + I₂ (g) <-->
2HI (g)]. I want to know how you would find the equalibrium concentration, when given the initial concentration. For example, the initial concentrations of H₂, I₂ and
HI were 0.00623 M, 0.00414 M and 0.0224 M; how would I go about finding the equalibrium concentration.

[xiankai]

here molarity = moles, because the volume is 1.00L and hence can be effectively left out of the equation. since we are dealing with gases, the 'volume' of the solution can be taken to be the volume of the container, which is 1.00L.

as for your second question, think about where the reaction will head (excess reagents, but on which side?) and then form an algebraic equation with 'x' as the amount that will react/be produced. fit it into the equilibrium constant and work out the math.