More chem help PLEASE!?

Question

I have no idea....

For the following reaction at a certain temperature, it is found that the equilibrium concentrations in a 5.00 L rigid container are [H2] = 0.0500 M, [F2] = 0.0100 M, and [HF] = 0.400 M. If 0.330 mol of F2 is added to this equilibrium mixture, calculate the concentrations of all gases once equilibrium is reestablished.

H2(g) + F2(g) 2 HF(g)

[H2]=
[F2]=
[HF]=

Answer 1

For gas phase reactions,
H2(g) + F2(g) ⇌ 2HF(g)
The equilibrium constant K can be calculated as:
K = [HF]^2/{ [H2]*[F2] } = 0.4^2/(0.05*0.01) = 320

After the addition of 0.330 mole of F2, the F2 concentration becomes (before reaction):
[F2] = 0.330mol /5.00L + 0.0100 M = 0.0760M
Assume an additional X(M) of [F2] will be converted to HF at the new equilibrium. Thus the new concentration of HF is: [HF] = (0.400+2X) M.
...........H2(g) + F2(g) ⇌ 2HF(g), K = 320
Initial: 0.0500..0.0760...0.400
Final: 0.05-X...0.07-X...0.4+2*X
K = 320 = (0.4+2*X)^2 /((0.05-X)*(0.07-X))
or: (0.4+2*X)^2 = 320*(0.05-X)*(0.07-X)
0.16+1.6*X+4*X^2 = 320*(0.0035-0.12*X+X^2)
316*X^2-40*X+0.96 = 0
The solutions to this quadratic equation are:
X1=0.0944 (to be omitted. Do you see why?)
X2 = 0.0322 (M)
So finally:
[H2] = 0.0178M
[F2] = 0.0378M
[HF] = 0.4644M