posted by Jerry on .
For the following reaction at a certain temperature:
2( ) 2( ) ( ) 2 H F HF g g g +
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 0.200 mol F2 is added to this equilibrium mixture,
calculate the concentrations of all gases once equilibrium is reestablished.
I don't know what the gibberish you wrote stands for. If I ignore that, we can go to the problem which is
......H2 + F2 ==> 2HF--is that right?
Ka = (H2)(F2)/(HF^2)
Plug in the E concns and solve for Ka. Then 0.2 mol F added which is 0.2 mol/5 L = 0.04M so initial for F2 = 0.01+0.04 0.05M
Substitute E line into Ka and solve x then evaluate HF, H2 and F2.