If #K_c# at a certain temperature for twice the formation reaction of #"HF"(g)# in a sealed rigid #"10-L"# container is #1.0 xx 10^2#, and one starts with #"1.00 mol"# of each reactant, what are the concentrations at equilibrium for each species?

Question

Jeremiah Adams · Accepted Answer

#["H"_2(g)]_(eq) = "0.017 M"#

#["F"_2(g)]_(eq) = "0.017 M"#

#["HF"(g)]_(eq) = "0.17 M"#

We can assume that all of the gases are perfect because of the high temperature (minimal intermolecular forces, complete kinetic energy transfer in collisions, etc.).

(The important thing to remember is that ideal gases exist at high temperatures; the precise temperature is not very important.)

We receive:

#"H"_2(g) + "F"_2(g) rightleftharpoons 2"HF"(g)#

#K_c = 1.0 xx 10^2#

We don't know what units are involved, but I assume it is #"mol/L"# since we are given #"1.00 mols"# of both reactants for initial mols and a vessel volume of #"10.0 L"#. This would give a molar concentration, #"mol"/"L"#, or #"M"#.

The volumes are all equal because the container is shared. Using molarity, create an ICE table (initial, change, equilibrium):

#"H"_2(g) " "+" " "F"_2(g) " "rightleftharpoons" " color(red)(2)"HF"(g)#

#"I"" "0.100" "" "" "0.100" "" "" "" "0.000#

#"C"" "-x" "" "" "-x" "" "" "" "+color(red)(2)x#

#"E"" "(0.100-x)" "(0.100-x)" "color(red)(2)x#

Remember that the mols in the reaction correspond to the multiple of molarity gained going towards equilibrium, so we have #+2x# for #"HF"#, and not just #+x#.

Since we have #K_c#, we just need to write its mass action expression:

#K_c = (["HF"]^color(red)(2))/(["H"_2]["F"_2])#

#= (color(red)(2)x)^color(red)(2)/((0.100-x)(0.100-x)) = (color(red)(2)x)^color(red)(2)/(0.100-x)^2 = 1.0 xx 10^2#

The #color(red)(2)# coefficient for #"HF"# again shows up in the exponent of the equilibrium constant, just as it does in the change in mols as #+color(red)(2)x#.

We are fortunate, since this is a perfect square. #1.0 xx 10^2 = 100#, so:

#sqrt100 = sqrt(K_c) = sqrt((color(red)(2)x)^color(red)(2)/(0.100-x)^2)#

#= (color(red)(2)x)/(0.100-x) = 10#

(we ignore the negative root since #K_c > 0#, always.

Thus:

#10(0.100-x) = 2x#

#=> 1.00 - 10x = 2x#

#=> 1.00 = 12x#

#=> x = 1.00/12 = 5/6 cdot 1/10 = 0.08bar(3)# #"mols/L"#

This implies that:

#color(blue)(["H"_2(g)]_(eq)) = 0.100 - 0.08bar(3) = color(blue)("0.017 M")#

#color(blue)(["F"_2(g)]_(eq)) = 0.100 - 0.08bar(3) = color(blue)("0.017 M")#

#color(blue)(["HF"(g)]_(eq)) = color(red)(2) xx 0.08bar(3) = color(blue)("0.17 M")#

This should make sense, since #K_c# was #100#, and indeed,

#K_c = (["HF"]_(eq)^color(red)(2))/(["H"_2]_(eq)["F"_2]_(eq)) = ("0.17 M")^2/(("0.017 M")("0.017 M"))#

#= cancel(("0.17 M")("0.17 M"))/(0.1cdotcancel("0.17 M")cdot0.1cdotcancel("0.17 M"))#

#= 1/(0.01)#

#= 100 = 1.0 xx 10^2#,

which was our initial point.

Luna Chen · Answer

undefined $$ [HF]_{\text{eq}} = \sqrt{K_c} = \sqrt{1.0 \times 10^2} = 10 \, \text{M} $$
$$ [H_2]_{\text{eq}} = [F_2]_{\text{eq}} = \frac{x}{V} = \frac{x}{10} \, \text{M} $$
$$ K_c = \frac{[HF]_{\text{eq}}^2}{[H_2]_{\text{eq}}[F_2]_{\text{eq}}} $$
$$ 1.0 \times 10^2 = \frac{(10)^2}{\left(\frac{x}{10}\right)^2} $$
$$ x = 10 \, \text{M} $$
$$ [H_2]_{\text{eq}} = [F_2]_{\text{eq}} = \frac{10}{10} = 1 \, \text{M} $$
$$ [HF]_{\text{eq}} = 10 \, \text{M} $$
$$ \text{At equilibrium: } [HF] = 10 \, \text{M}, \, [H_2] = 1 \, \text{M}, \, [F_2] = 1 \, \text{M} $$