can anyone help with this equilibrium problem?

Question

for the following reaction at a certain temperature
H2 (g) + F2 (g) -> 2HF (g)
(NOTE: !!!!!!!! it's at equilibrium !!!!!!!!!!!!!)



it is found that the equilibrium concentrations in a 5.00 liter rigid container are [H2] =0.05 M, [F2]=0.01M, and [HF] =0.4M. if 0.2 mol of F2 is added to this equilibrium mixture, calculate the concentrations of all gases once equilibrium is reestablished.

Answer 1

Be careful.
First of all you are given concentrations, thus you find with the equilibrium data Kc and not Kp

H2 + F2 <=> 2HF

Kc= [HF]^2 / [H2][F2]= (0.4^2)/(0.05*0.01) = 320

Now you add 0.2 mol F2. BUT the volume is 5 L, thus you add 0.2/5= 0.04 mole/L F2.
So now you shift the equilibrium to the right having 0.01+0.04=0.05 M initial concentration for F2.

.. .. .. .. .. .. H2 + F2 <=> 2HF
Initial .. .. 0.05 .. 0.05 .. .. 0.4
React .. .. .. x .. .. x
Produce .. .. .. .. .. .. .. .. .. 2x
At Equil. 0.05-x.. 0.05-x .. 0.4+2x

Kc=[HF]^2 / [H2][F2]= (0.4+2x)^2 / (0.05-x)^2 =>
Kc= [(0.4+2x) / (0.05-x)]^2=>
squareroot (Kc) = (0.4+2x)/(0.05-x) =>
17.889*(0.05-x)=0.4+2x =>
x=(17.889*0.05-0.4)/19.889 =0.0248=0.025
Thus [H2]=[F2]= 0.05-0.025=0.025 M and
[HF]= 0.4+2*0.025= 0.45 M

Answer 2

Work out Kp first.

Then work out the moles of each, then the change, and then go back to molarities.

The number of moles of F2 initially present is 0.05, H2 = 0.01, and HF = 2. You raise the F2 to (0.05 + 0.2) = 0.25.
So F2 will now be 0.25 - x, H2 will be 0.01 - x and HF will be 2 + 2x. Now divide each by 5, and put the values in the Kp expression and solve the quadratic.