Equilirum question?

Question

Consider the following equilibrium at 298 K:
H2(g) + Cl2(g) « 2HCl(g)

Analysis of the initial contents of a 14.0L flask reveals that there are 3.6 moles of H2, 3.6 moles of Cl2, and no HCl. The system is then allowed to reach equilibrium; the equilibrium constant Kc is 3.3E-2. Calculate the number of moles of HCl present at equilibrium.

Answer 1

Equilibrium expression for this rxn is Kc = [HCl]^2 / ([H2][Cl2])
Kc is the appropriate rather than Kp b/c rxn is at constant volume.
Initial amounts: H2 = 3.6mol, Cl2 = 3.6mol, HCl = 0mol
Equil amounts: H2 = 3.6-x, Cl2 = 3.6-x, HCl = 2x (1:2 reaction stoichiometry)
To get concentrations (for plugging into Kc expression) divide each amount by 14L, however factor/divisor eventually cancels out b/c it is present on top and bottom of Kc expression (squared in both cases)...

3.3E-2 = (2x/14)^2 / {[(3.6-x)/14] [(3.6-x)/14]}

This rearranges (after simplification) to 4x^2 = 0.42768-0.2376x+0.033x^2.
The solution is x=0.30 mol, but we want 2x (HCl = 2x), so HCl = 0.60 mol (rounded to 2 sig figs)

Answer 2

You need to recognize that if x moles of H2 reacts with x moles of Cl2, you will form 2x moles of HCl.

So, at equilibrium you will be left with 3.6-x moles of both H2 and Cl2, and will have formed 2x moles of HCl.

Plug those numbers of moles into the equation for Kc, and solve for x. You'll have to deal with a quadratic in this problem, because I don't think you should ignore x compared to 3.6 moles.

Once you find x, 2x will be the number of moles of HCl present at equilibrium.