chemistry help? very lost! equilibrium concentration?

Question

At a particular temperature, K = 1.00×102 for the reaction:


H2(g) + F2(g) ⇔ 2HF(g)


In an experiment, at this temperature, 1.30 mol of H2 and 1.30 mol of F2 are introduced into a 1.01-L flask and allowed to react. At equilibrium, all species remain in the gas phase.
What is the equilibrium concentration of H2, HF, F2?

Answer 1

OK, the equilibrium equation is K = [HF]^2/[H2][F2]. You have to square the numerator since you get two moles of HF on the right side.

Since you have a 1.01 L flask (an odd volume, to be sure, in this case) the initial concentration of each of the two reactants is 1.30/1.01 = 1.29 M. Now if x is the concentration of HF after the reaction, then you used x/2 of each of the reactants, and the final concentration of the reactants would be 1.29 - x/2. So the equation that you need to solve is:

x^2/((1.29-x/2)*(1.29-x/2)) = 1.00 x 10^2.

You can solve this using the quadratic formula.

This will give you the final concentration of HF, and by subtracting x/2 from 1.29, you'll get the final concentration of each of the two reactants.

Answer 2

Never fear, whatisname is here.
Set up your equilibrium expression in terms of moles. [HF]^2/[H+][F-] = 100

When the reaction starts, you have 1.3 moles of each reactant and no product. At equilbrium, you lose x moles of each reactant and form 2x moles of product. Then
[2x]^2/[1.3-x]^2 = 100
4x^2= 100 * [1.69-2.6x+x^2]
Solve for x. Discard the weird root.
NOTE: solving (2x)^2/[1.3-x]^2 = 100 by Trial and Error is an honorable alternative. x should be between 1 and 1.1.