What is the concentration when equilibrium is reached?

The balanced equation is

#"2HI(g)" ⇌ "H"_2(g) + "I"_2(g)#

Now, we can set up an ICE table.

#color(white)(mmmmmmm)"2HI(g)" ⇌ color(white)(l)"H"_2(g) + color(white)(l)"I"_2(g)#
#"I/mol·L"^"-1": color(white)(mm)0.068 color(white)(mmmll)0color(white)(mmmll)0#
#"C/mol·L"^"-1": color(white)(mm)"-2"xcolor(white)(mmmll)"+"xcolor(white)(mmll)"+"x#
#"E/mol·L"^"-1": color(white)(m)"0.068-2"xcolor(white)(mmll)xcolor(white)(mmmll)x#

#K_text(c) = (["H"_2]["I"_2])/(["HI"]^2) = x^2/("0.0068-2"x)^2 = 0.0180#

#x/("0.068-2"x) = 0.134#

#x = 0.134("0.068-2"x) =0.00912 - 0.268x#

#1.268x = "0.009 12"#

#x = "0.007 19"#

#"[HI]" = ("0.068-2"x)color(white)(l) "mol/L" = "(0.068 -2×0.007 19) mol/L" = "(0.068-0.0144) mol/L" = "0.054 mol/L"#

Check:

#K = "0.007 19"^2/0.054^2 = 0.018"#

It checks!

#x = "0.00719 M"#. I'll leave it to you to figure out how this leads to the equilibrium concentrations by reading below.

Here we have nonzero amounts of #"HI"(g)# initially, so it must react (through dissociation) to form #"H"_2(g)# and #"I"_2(g)#.

#2"HI"(g) rightleftharpoons "H"_2(g) + "I"_2(g)#

#"I"" "0.068" "" "" "0" "" "0#
#"C"" "-2x" "" "+x" "+x#
#"E"" "0.068-2x" "x" "" "x#

The equilibrium expression is then:

#K_c = 0.0180 = x^2/(0.068 - 2x)^2#

This is a perfect square, so...

#sqrt(0.0180) = x/(0.068 - 2x)#

We take the positive square root, as it must lead to some #K_c' = sqrt(K_c)#, for the same reaction with half the coefficients.

#sqrt(0.0180)(0.068) - 2sqrt(0.0180)x = x#

#x = (sqrt(0.0180)(0.068))/(1 + 2sqrt(0.0180))#

The physical solution is then #x = "0.00719 M"#. Therefore, the equilibrium concentrations are:

#x = color(blue)([H_2] = [I_2] = "0.00719 M")#

#0.068 - 2x = color(blue)([HI] = "0.0536 M")#

And this can be verified:

#K_c = ([H_2][I_2])/([HI]^2) = (0.00719)^2/(0.0536)^2 ~~ 0.0180# #color(blue)(sqrt"")#