A reaction mixture initially contains 2.8 M $H_{2} O$ $H_2O$ and 2.6 M $S O_{2}$ $SO_2$ . How do you determine the equilibrium concentration of $H_{2} S$ $H_2S$ if Kc for the reaction at this temperature is $1.3 \times 10^{- 6}$ $1.3 xx 10^-6$ ?

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A reaction mixture initially contains 2.8 M $H_{2} O$ $H_2O$ and 2.6 M $S O_{2}$ $SO_2$ . How do you determine the equilibrium concentration of $H_{2} S$ $H_2S$ if Kc for the reaction at this temperature is $1.3 \times 10^{- 6}$ $1.3 xx 10^-6$ ?

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Answer 1 · 2016-07-22T01:51:02

$[H_{2} S] = 0.12 M$ $["H"_2"S"] = "0.12 M"$

Explanation:

The first thing to do here is write the equilibrium reaction

$2 {SO}_{2 (g)} + 2 H_{2} O_{(g)} ⇌ 2 H_{2} S_{(g)} + 3 O_{2 (g)}$ $2"SO"_ (2(g)) + 2"H"_ 2"O"_ ((g)) rightleftharpoons 2"H"_ 2"S"_ ((g)) + 3"O"_ (2(g))$

Now, you know that at a certain temperature, the equilibrium constant for this reaction is equal to

$K_{c} = 1.3 \cdot 10^{- 6}$ $K_c = 1.3 * 10^(-6)$

Right from the start, you can tell just by looking at the value of $K_{c}$ $K_c$ that the equilibrium concentration of hydrogen sulfide, $H_{2} S$ $"H"_2"S"$ , will be lower than the equilibrium concentration of the two reactants.

This is the case because you have $K_{c} < 1$ $K_c < 1$ , which means that at this temperature you can expect the equilibrium mixture to contain more reactants than products.

The next thing to do here is use an ICE table to find the equilibrium concentration of hydrogen sulfide

$2 {SO}_{2 (g)} + 2 H_{2} O_{(g)} ⇌ 2 H_{2} S_{(g)} + 3 O_{2 (g)}$ $" "2"SO"_ (2(g)) " "+" " 2"H"_ 2"O"_ ((g)) rightleftharpoons 2"H"_ 2"S"_ ((g)) " "+" " 3"O"_ (2(g))$

$I a a a a a 2.6 a a a a a a a a a a a a 2.8 a a a a a a a a 0 a a a a a a a a a a a 0$ $color(purple)("I")color(white)(aaaaacolor(black)(2.6)aaaaaaaaaaaacolor(black)(2.8)aaaaaaaacolor(black)(0)aaaaaaaaaaacolor(black)(0)$
$C a a a (- 2 x) a a a a a a a a (- 2 x) a a a a (+ 2 x) a a a a a a a (+ 3 x)$ $color(purple)("C")color(white)(aaacolor(black)((-2x))aaaaaaaacolor(black)((-2x))aaaacolor(black)((+2x))aaaaaaacolor(black)((+3x))$
$E a a a 2.6 - 2 x a a a a a a a 2.8 - 2 x a a a a a a 2 x a a a a a a a a a a 3 x$ $color(purple)("E")color(white)(aaacolor(black)(2.6-2x)aaaaaaacolor(black)(2.8-2x)aaaaaacolor(black)(2x)aaaaaaaaaacolor(black)(3x)$

By definition, the equilibrium constant for the reaction will be

$K_{c} = \frac{{[H_{2} S]}_{2}^{2} \cdot {[O_{2}]}_{2}^{3}}{{[{SO}_{2}]}_{2}^{2} \cdot {[H_{2} O]}_{2}^{2}}$ $K_c = (["H"_2"S"]^2 * ["O"_2]^3)/(["SO"_2]^2 * ["H"_2"O"]^2)$

In your case, this expression is equivalent to

$K_{c} = \frac{{(2 x)}^{2} \cdot {(3 x)}^{3}}{{(2.6 - 2 x)}^{2} \cdot {(2.8 - 2 x)}^{2}}$ $K_c = ( (2x)^2 * (3x)^3)/( (2.6 - 2x)^2 * (2.8 - 2x)^2)$

$K_{c} = \frac{4 x^{2} \cdot 27 x^{3}}{{(2.6 - 2 x)}^{2} \cdot {(2.8 - 2 x)}^{2}}$ $K_c = (4x^2 * 27x^3)/( (2.6 - 2x)^2 * (2.8 - 2x)^2)$

$K_{c} = \frac{108 x^{5}}{{(2.6 - 2 x)}^{2} \cdot {(2.8 - 2 x)}^{2}} = 1.3 \cdot 10^{- 6}$ $K_c = (108x^5)/( (2.6 - 2x)^2 * (2.8 - 2x)^2) = 1.3 * 10^(-6)$

Now, because the value of $K_{c}$ $K_c$ is so small compared with the initial concentrations of water vapor and sulfur dioxide, you can use the approximations

$2.8 - 2 x \approx 2.8$ $2.8 - 2x ~~ 2.8" "$ and $2.6 - 2 x \approx 2.6$ $" "2.6 - 2x ~~ 2.6$

This will give you

$1.3 \cdot 10^{- 6} = \frac{108 x^{5}}{{2.6}^{2} \cdot {2.8}^{2}}$ $1.3 * 10^(-6) = (108x^5)/(2.6^2 * 2.8^2)$

which allows you to calculate $x$ $x$ by

$x = \sqrt[5]{\frac{1.3 \cdot {2.6}^{2} \cdot {2.8}^{2} \cdot 10^{- 3}}{108}} = 0.05767$ $x = root(5)( (1.3 * 2.6^2 * 2.8^2 * 10^(-3))/108) = 0.05767$

Keep in mind that because the equilibrium concentration of hydrogen sulfide is $2 x$ $2x$ , you will have

$[H_{2} S] = 2 \times 0.05767 M = 0.11534 M$ $["H"_2"S"] = 2 xx "0.05767 M" = "0.11534 M"$

Rounded to two sig figs, the number of sig figs you have for the initial concentrations of sulfur dioxide and water vapor, the answer will be

$["H"_2"S"] = color(green)(|bar(ul(color(white)(a/a)color(black)("0.12 M")color(white)(a/a)|)))$

As predicted, the equilibrium concentration of hydrogen sulfide is lower than the equilibrium concentrations of the two reactants, which are

$["SO"_2] = 2.6 - 2 * "0.05767 M" = "2.5 M"$

$["H"_2"O"] = 2.8 - 2 * "0.05767 M" = "2.7 M"$

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A reaction mixture initially contains 2.8 M $H_{2} O$ $H_2O$ and 2.6 M $S O_{2}$ $SO_2$ . How do you determine the equilibrium concentration of $H_{2} S$ $H_2S$ if Kc for the reaction at this temperature is $1.3 \times 10^{- 6}$ $1.3 xx 10^-6$ ?

1 Answer

Explanation:

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A reaction mixture initially contains 2.8 M $H_{2} O$ $H_2O$ and 2.6 M $S O_{2}$ $SO_2$ . How do you determine the equilibrium concentration of $H_{2} S$ $H_2S$ if Kc for the reaction at this temperature is $1.3 \times 10^{- 6}$ $1.3 xx 10^-6$ ?