Hydrogen sulfide decomposes according to the following reaction, for which Kc=9.30E-8 at 700 degrees Celsius. 2 H2S(g) --> 2 H2(g) + S2(g) If 0.29 moles of H2S is placed in a 3.0-L container, What is the equilibrium concentration of H2(g) at 700 degrees Celsius?

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Hydrogen sulfide decomposes according to the following reaction, for which Kc=9.30E-8 at 700 degrees Celsius. 2 H2S(g) --> 2 H2(g) + S2(g) If 0.29 moles of H2S is placed in a 3.0-L container, What is the equilibrium concentration of H2(g) at 700 degrees Celsius?

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Answer 1 · 2014-07-12T19:37:41

The equilibrium concentration of $H_{2}$ $"H"_2$ is 0.012 mol/L.

Explanation:

Step 1. Calculate the initial concentrations

${[H_{2} S]}_{0} = \frac{0.29 mol}{3.0 L} = 0.0967 mol/L$ $["H"_2"S"]_0 = "0.29 mol"/"3.0 L" = "0.0967 mol/L"$

Step 2. Write the balanced equation and set up an ICE table.

$m m m m m m m {2H}_{2} S m ⇌ m {2H}_{2} + S_{2}$ $color(white)(mmmmmmm)"2H"_2"S"color(white)(m) ⇌color(white)(m) "2H"_2 + "S"_2$
${I/mol\cdotL}^{-1} : m 0.096 67 m m m m 0 m m 0$ $"I/mol·L"^"-1":color(white)(m) "0.096 67" color(white)(mmmm)0 color(white)(mm)0$
${C/mol\cdotL}^{-1} : m m - 2 x m m m m +2 x m + x$ $"C/mol·L"^"-1":color(white)(mm) "-"2xcolor(white)(mmmm) "+2"xcolor(white)(m) "+"x$
${E/mol\cdotL}^{-1} : 0.096 67 - 2 x l m m 2 x m l x$ $"E/mol·L"^"-1": "0.096 67 - 2"xcolor(white)(lmm) 2xcolor(white)(ml) x$

Write the $K_{c}$ $K_"c"$ expression and solve for $x$ $x$ .

$K_{c} = \frac{{[H_{2}]}_{2}^{2} [S_{2}]}{{[H_{2} S]}_{2}^{2}} = 9.30 \times 10^{-8}$ $K_"c" = (["H"_2]^2["S"_2])/["H"_2"S"]^2 = 9.30 × 10^"-8"$

$\frac{{(2 x)}^{2} \times x}{{(0.096 67 - 2 x)}^{2}} = 9.30 \times 10^{-8}$ $((2x)^2×x)/(("0.096 67" – 2x)^2) = 9.30 × 10^"-8"$

$\frac{0.09667}{9.30 \times 10^{-8}} = 1 \times 10^{-6} > 400$ $0.09667/(9.30 ×10^"-8") = 1 × 10^"-6" > 400$ . ∴ $x ≪ 0.099 67$ $x ≪ "0.099 67"$

∴ $\frac{4 x^{3}}{{0.096 67}^{2}} = 9.30 \times 10^{-8}$ $(4x^3)/"0.096 67"^2 = 9.30 × 10^"-8"$

$4 x^{3} = {0.096 67}^{2} \times 9.30 \times 10^{-8} = 8.691 \times 10^{-10}$ $4x^3 = "0.096 67"^2 × 9.30 × 10^"-8" = 8.691 × 10^"-10"$

$x^{3} = 2.173 \times 10^{-10}$ $x^3 = 2.173 × 10^"-10"$

$x = 6.012 \times 10^{-4}$ $x = 6.012 × 10^"-4"$

$[H_{2}] = 2 x l mol/L = 2 \times 6.012 \times 10^{-4} l mol/L = 0.0012 mol/L$ $["H"_2] = 2xcolor(white)(l) "mol/L" = 2 × 6.012 × 10^"-4"color(white)(l) "mol/L" = "0.0012 mol/L"$

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Hydrogen sulfide decomposes according to the following reaction, for which Kc=9.30E-8 at 700 degrees Celsius. 2 H2S(g) --> 2 H2(g) + S2(g) If 0.29 moles of H2S is placed in a 3.0-L container, What is the equilibrium concentration of H2(g) at 700 degrees Celsius?

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