$H_{2}$ $H_2$ (g) + $F_{2}$ $F_2$ (g) $⇌ 2 H F$ $\rightleftharpoons2HF$ (g), find final values of ${[H_{2}]}_{e q}$ $[H_2]_(eq)$ , ${[F_{2}]}_{e q}$ $[F_2]_(eq)$ , and ${[H F]}_{e q}$ $[HF]_(eq)$ ?

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$H_{2}$ $H_2$ (g) + $F_{2}$ $F_2$ (g) $⇌ 2 H F$ $\rightleftharpoons2HF$ (g), find final values of ${[H_{2}]}_{e q}$ $[H_2]_(eq)$ , ${[F_{2}]}_{e q}$ $[F_2]_(eq)$ , and ${[H F]}_{e q}$ $[HF]_(eq)$ ?

$k = 1.15 \times 10^{2}$ $k=1.15xx10^2$
If $3.000$ $3.000$ mol of $H_{2}$ $H_2$ , $F - 2$ $F-2$ and $H F$ $HF$ are added to a $1.500$ $1.500$ L flask...

(find final values)

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Answer 1 · 2018-06-26T21:48:31

${[H_{2}]}_{e q} = {[F_{2}]}_{e q} = 0.471 M$ $["H"_2]_(eq) = ["F"_2]_(eq) = "0.471 M"$
${[HF]}_{e q} = 5.06 M$ $["HF"]_(eq) = "5.06 M"$

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

The reaction was:

$H_{2} (g) + F_{2} (g) ⇌ 2 HF (g)$ $"H"_2(g) + "F"_2(g) rightleftharpoons 2"HF"(g)$

In a $1.500 L$ $"1.500 L"$ flask, the volume is shared, so the initial concentrations are:

${[H_{2}]}_{i} = \frac{3.000 mols}{1.500 L} = 2.000 M$ $["H"_2]_i = "3.000 mols"/"1.500 L" = "2.000 M"$

${[F_{2}]}_{i} = \frac{{3.000 mols F}_{2}}{1.500 L} = 2.000 M$ $["F"_2]_i = "3.000 mols F"_2/"1.500 L" = "2.000 M"$

${[HF]}_{i} = \frac{3.000 mols HF}{1.500 L} = 2.000 M$ $["HF"]_i = "3.000 mols HF"/"1.500 L" = "2.000 M"$

So, the ICE table uses concentrations to give:

$H_{2} (g) + F_{2} (g) ⇌ 2 HF (g)$ $"H"_2(g) " "+" " "F"_2(g) " "rightleftharpoons" " 2"HF"(g)$

$I 2.000 2.000 2.000$ $"I"" "2.000" "" "" "2.000" "" "" "2.000$
$C - x - x + 2 x$ $"C"" "-x" "" "" "-x" "" "" "+2x$
$E 2.000 - x 2.000 - x 2.000 + 2 x$ $"E"" "2.000-x" "2.000-x" "2.000+2x$

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

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

$Q_{c} = \frac{{(2.000 M HF)}^{2}}{{2.000 M H}_{2} \cdot {2.000 M F}_{2}} = 1$ $Q_c = ("2.000 M HF")^2/("2.000 M H"_2 cdot "2.000 M F"_2) = 1$

(Since $Q_{c} < K_{c}$ $Q_c < K_c$ , the reaction wants to shift towards the products, so our assumption is good.)

Therefore, we can set up the $K_{c}$ $K_c$ expression:

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

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.

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

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

$10.72 (2.000 - x) = 2.000 + 2 x$ $10.72(2.000 - x) = 2.000 + 2x$

$21.45 - 10.72 x = 2.000 + 2 x$ $21.45 - 10.72x = 2.000 + 2x$

$19.45 = 12.72 x$ $19.45 = 12.72x$

$x = 1.53 M$ $x = "1.53 M"$

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

Anyways, we now get the equilibrium concentrations to be:

${[H_{2}]}_{e q} = {[F_{2}]}_{e q} = 2.000 - x = 0.471 M$ $color(blue)(["H"_2]_(eq) = ["F"_2]_(eq) = 2.000 - x = "0.471 M")$

${[HF]}_{e q} = 2.000 + 2 x = 5.06 M$ $color(blue)(["HF"]_(eq) = 2.000 + 2x = "5.06 M")$

And just to check,

$K_{c} = \frac{{5.06}^{2}}{(0.471) (0.471)} = 115.41 \approx 115$ $K_c = (5.06^2)/((0.471)(0.471)) = 115.41 ~~ 115$ $\sqrt{}$ $color(blue)(sqrt"")$

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$H_{2}$ $H_2$ (g) + $F_{2}$ $F_2$ (g) $⇌ 2 H F$ $\rightleftharpoons2HF$ (g), find final values of ${[H_{2}]}_{e q}$ $[H_2]_(eq)$ , ${[F_{2}]}_{e q}$ $[F_2]_(eq)$ , and ${[H F]}_{e q}$ $[HF]_(eq)$ ?

$k = 1.15 \times 10^{2}$ $k=1.15xx10^2$
If $3.000$ $3.000$ mol of $H_{2}$ $H_2$ , $F - 2$ $F-2$ and $H F$ $HF$ are added to a $1.500$ $1.500$ L flask...

(find final values)

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$H_{2}$ $H_2$ (g) + $F_{2}$ $F_2$ (g) $⇌ 2 H F$ $\rightleftharpoons2HF$ (g), find final values of ${[H_{2}]}_{e q}$ $[H_2]_(eq)$ , ${[F_{2}]}_{e q}$ $[F_2]_(eq)$ , and ${[H F]}_{e q}$ $[HF]_(eq)$ ?

$k = 1.15 \times 10^{2}$ $k=1.15xx10^2$
If $3.000$ $3.000$ mol of $H_{2}$ $H_2$ , $F - 2$ $F-2$ and $H F$ $HF$ are added to a $1.500$ $1.500$ L flask...

(find final values)