# H_2 (g) + F_2 (g) \rightleftharpoons2HF (g), find final values of [H_2]_(eq), [F_2]_(eq), and [HF]_(eq)?

## $k = 1.15 \times {10}^{2}$ If $3.000$ mol of ${H}_{2}$, $F - 2$ and $H F$ are added to a $1.500$ L flask... (find final values)

Jun 26, 2018

["H"_2]_(eq) = ["F"_2]_(eq) = "0.471 M"
["HF"]_(eq) = "5.06 M"

I assume you have given the equilibrium constant ${K}_{c}$ (not $k$, which is a rate constant). Also, we used ${K}_{c}$ because you gave concentrations.

The reaction was:

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

In a $\text{1.500 L}$ flask, the volume is shared, so the 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:

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

$\text{I"" "2.000" "" "" "2.000" "" "" } 2.000$
$\text{C"" "-x" "" "" "-x" "" "" } + 2 x$
$\text{E"" "2.000-x" "2.000-x" } 2.000 + 2 x$

...remembering that coefficients go into the change in concentration.

Here we assumed that the reaction goes forward. If you calculate ${Q}_{c}$, you would find that here it equals $1$:

${Q}_{c} = \left({\text{2.000 M HF")^2/("2.000 M H"_2 cdot "2.000 M F}}_{2}\right) = 1$

(Since ${Q}_{c} < {K}_{c}$, the reaction wants to shift towards the products, so our assumption is good.)

Therefore, we can set up the ${K}_{c}$ expression:

$1.15 \times {10}^{2} = {\left(2.000 + 2 x\right)}^{2} / \left(\left(2.000 - x\right) \left(2.000 - x\right)\right)$

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.

$\sqrt{1.15 \times {10}^{2}} = 10.72 = \frac{2.000 + 2 x}{2.000 - x}$

We only take ${K}_{c} > 0$ because equilibrium constants can never be negative. Proceed to solve for $x$:

$10.72 \left(2.000 - x\right) = 2.000 + 2 x$

$21.45 - 10.72 x = 2.000 + 2 x$

$19.45 = 12.72 x$

$x = \text{1.53 M}$

Note that we lost a solution of $x = \text{2.69 M}$, but that's OK because it is nonphysical; $2.000 - x$ would have given a negative concentration for the reactants.

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,

${K}_{c} = \frac{{5.06}^{2}}{\left(0.471\right) \left(0.471\right)} = 115.41 \approx 115$ color(blue)(sqrt"")