Other way of solving the following problem:

Suppose 6.000 mol of F₂ and 3.000 mol of H₂ are mixed in a 3.000 L container to make HF . The equilibrium constant at a certain temperature is 1.15x10 ².

Calculate the equilibrium concentrations, given:

H_2(g) + F_2(g) 2 HF_(g)

In such problems, it is very important to distinguish between

(1) the number of moles initially found in the reaction vessel

(2) the number of moles that actually react or form

(3) the number of moles found at equilibrium, which is what we use in the expression after dividing by litres.

	1 H2(g)	1 F2(g)	2 HF(g)
Initial # of moles	3.000	6.000	0 (since it is not mentioned,we have to assume there was none introduced.
Moles reacting/forming	x, since we don’t know how much reacts	x, since the ratio of H2(g) to F2(g) in the equation is 1: 1	2x , since 2 moles of HF form for every 1 mole of reacting H2.
# of moles at equilibrium	3-x remain	6-x remain	0 + 2x now exist.

We write an equilibrium law expression, and then substitute the equilibrium number of moles into that expression. Don’t forget to divide by the number of litres, which in this example is 3.0 L.

K = [HF]²/([H₂][F₂]) = =115

Now the “9” cancels, which could lead to the false impression that the 3 L was a useless insertion, but the convenient canceling does NOT always happen.

4x² =115(3-x)(6-x)

115(18 - 9x + x²) = 4x²

_{x = 6.419, 2.905}

If we had only 3.00 moles of hydrogen initially, only 2.90 is the sensible answer.

Now we can wrap up the problem and calculate the equilibrium concentrations:

[H₂ ]= (3-2.905)/3 = 0.033M

[F₂] = 1.03 M

[HF] = 1.93 M