#H_2# (g) + #F_2# (g) #\rightleftharpoons2HF# (g), find final values of #[H_2]_(eq)#, #[F_2]_(eq)#, and #[HF]_(eq)#?
#k=1.15xx10^2#
If #3.000# mol of #H_2# , #F-2# and #HF# are added to a #1.500# L flask...
(find final values)
If
(find final values)
1 Answer
#["H"_2]_(eq) = ["F"_2]_(eq) = "0.471 M"#
#["HF"]_(eq) = "5.06 M"#
I assume you have given the equilibrium constant
The reaction was:
#"H"_2(g) + "F"_2(g) rightleftharpoons 2"HF"(g)#
In a
#["H"_2]_i = "3.000 mols"/"1.500 L" = "2.000 M"#
#["F"_2]_i = "3.000 mols F"_2/"1.500 L" = "2.000 M"#
#["HF"]_i = "3.000 mols HF"/"1.500 L" = "2.000 M"#
So, the ICE table uses concentrations to give:
#"H"_2(g) " "+" " "F"_2(g) " "rightleftharpoons" " 2"HF"(g)#
#"I"" "2.000" "" "" "2.000" "" "" "2.000#
#"C"" "-x" "" "" "-x" "" "" "+2x#
#"E"" "2.000-x" "2.000-x" "2.000+2x# ...remembering that coefficients go into the change in concentration.
Here we assumed that the reaction goes forward. If you calculate
#Q_c = ("2.000 M HF")^2/("2.000 M H"_2 cdot "2.000 M F"_2) = 1#
(Since
Therefore, we can set up the
#1.15 xx 10^2 = (2.000 + 2x)^2/((2.000 - x)(2.000 - x))#
This is nice in that it is a perfect square. That is one situation where no approximations need to be made and it is quite solvable without the quadratic formula.
#sqrt(1.15 xx 10^2) = 10.72 = (2.000 + 2x)/(2.000 - x)#
We only take
#10.72(2.000 - x) = 2.000 + 2x#
#21.45 - 10.72x = 2.000 + 2x#
#19.45 = 12.72x#
#x = "1.53 M"#
Note that we lost a solution of
Anyways, we now get the equilibrium concentrations to be:
#color(blue)(["H"_2]_(eq) = ["F"_2]_(eq) = 2.000 - x = "0.471 M")#
#color(blue)(["HF"]_(eq) = 2.000 + 2x = "5.06 M")#
And just to check,
#K_c = (5.06^2)/((0.471)(0.471)) = 115.41 ~~ 115# #color(blue)(sqrt"")#