# 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?

Jun 7, 2017

["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:

$\text{H"_2(g) + "F"_2(g) rightleftharpoons 2"HF} \left(g\right)$

${K}_{c} = 1.0 \times {10}^{2}$

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

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

$\text{H"_2(g) " "+" " "F"_2(g) " "rightleftharpoons" " color(red)(2)"HF} \left(g\right)$

$\text{I"" "0.100" "" "" "0.100" "" "" "" } 0.000$

$\text{C"" "-x" "" "" "-x" "" "" "" } + \textcolor{red}{2} x$

$\text{E"" "(0.100-x)" "(0.100-x)" } \textcolor{red}{2} x$

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

Since we have ${K}_{c}$, we just need to write its mass action expression:

${K}_{c} = \left(\left[{\text{HF"]^color(red)(2))/(["H"_2]["F}}_{2}\right]\right)$

$= {\left(\textcolor{red}{2} x\right)}^{\textcolor{red}{2}} / \left(\left(0.100 - x\right) \left(0.100 - x\right)\right) = {\left(\textcolor{red}{2} x\right)}^{\textcolor{red}{2}} / {\left(0.100 - x\right)}^{2} = 1.0 \times {10}^{2}$

The $\textcolor{red}{2}$ coefficient for $\text{HF}$ again shows up in the exponent of the equilibrium constant, just as it does in the change in mols as $+ \textcolor{red}{2} x$.

We are fortunate, since this is a perfect square. $1.0 \times {10}^{2} = 100$, so:

$\sqrt{100} = \sqrt{{K}_{c}} = \sqrt{{\left(\textcolor{red}{2} x\right)}^{\textcolor{red}{2}} / {\left(0.100 - x\right)}^{2}}$

$= \frac{\textcolor{red}{2} x}{0.100 - x} = 10$

(we ignore the negative root since ${K}_{c} > 0$, always.

So:

$10 \left(0.100 - x\right) = 2 x$

$\implies 1.00 - 10 x = 2 x$

$\implies 1.00 = 12 x$

$\implies x = \frac{1.00}{12} = \frac{5}{6} \cdot \frac{1}{10} = 0.08 \overline{3}$ $\text{mols/L}$

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 ${K}_{c}$ was $100$, and indeed,

${K}_{c} = \left(\left[\text{HF"]_(eq)^color(red)(2))/(["H"_2]_(eq)["F"_2]_(eq)) = ("0.17 M")^2/(("0.017 M")("0.017 M}\right)\right)$

$= \cancel{\left(\text{0.17 M")("0.17 M"))/(0.1cdotcancel("0.17 M")cdot0.1cdotcancel("0.17 M}\right)}$

$= \frac{1}{0.01}$

$= 100 = 1.0 \times {10}^{2}$,

which is what we started with.