Chemistry
posted by Elizabeth on .
For the reaction below 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 0.340 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)

You have two problems here.
First, write Kc expression, substitute eq concns, and calculate Kc.
Then you can address the second half of the problem.Note:I have omitted all of th zeros to save space but you should include them.
..............H2+ F2 ==> 2HF
initial....0.05M.0.01M...0.400
add..............(0.340 mol/5L = ?)
change........x..x......+2x
eqil.....0.05x..0.078x...0.400+2x
Substitute the equil concns into the Kc expression and solve for x. Then plug x into the eq conditions of the ICE chart above and solve for the individual concns of each element/compound.