For the reaction below at a certain temperature, it is found that the equilibrium concentrations in a 4.87-L rigid container are

Question

[H2] = 0.0496 M
[F2] = 0.0116 M
[HF] = 0.429 M.
H2(g) + F2(g) <==> 2 HF(g)
If 0.185 mol of F2 is added to this equilibrium mixture, calculate the concentrations of all gases once equilibrium is reestablished.

Answer 1

K = (HF)^2/(H2)(F2).
Substitute equilibrium concns and calculate K.
Then
0.185 mol F2/4.87 L = 0.0380 M added. The equilibrium concns become the initial values for the next calculation.
.......H2 + F2 ==> 2HF
I..0.0496..0.0116..0.429
add........0.0380........
C.....-x...-x........+2x
E..0.0496-x..0.0496-x..0.429+2x

Substitute the E line into a new Ka expression and solve for x, then evaluate the E line for final concentrations.