# 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?

##### 1 Answer

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

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

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

Since we are at a high temperature, we can assume the gases are all ideal (full transfer of kinetic energy in collisions, minimal intermolecular forces, etc).

(The exact temperature doesn't really matter though. If it is high, we have ideal gases. That's the point.)

We are given:

#"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

Since the container is shared, the volumes are all equal. Construct an **ICE table** (initial, change, equilibrium) using molarity:

#"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

Since we have **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

We are fortunate, since this is a perfect square.

#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.

So:

#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 means 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 = (["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 is what we started with.