# #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 **initial concentrations** are:

#["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"")#