# 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?

Jul 12, 2014

The equilibrium concentration of ${\text{H}}_{2}$ is 0.012 mol/L.

#### Explanation:

Step 1. Calculate the initial concentrations

["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.

$\textcolor{w h i t e}{m m m m m m m} {\text{2H"_2"S"color(white)(m) ⇌color(white)(m) "2H"_2 + "S}}_{2}$
$\text{I/mol·L"^"-1":color(white)(m) "0.096 67} \textcolor{w h i t e}{m m m m} 0 \textcolor{w h i t e}{m m} 0$
$\text{C/mol·L"^"-1":color(white)(mm) "-"2xcolor(white)(mmmm) "+2"xcolor(white)(m) "+} x$
$\text{E/mol·L"^"-1": "0.096 67 - 2} x \textcolor{w h i t e}{l m m} 2 x \textcolor{w h i t e}{m l} x$

Write the ${K}_{\text{c}}$ expression and solve for $x$.

${K}_{\text{c" = (["H"_2]^2["S"_2])/["H"_2"S"]^2 = 9.30 × 10^"-8}}$

((2x)^2×x)/(("0.096 67" – 2x)^2) = 9.30 × 10^"-8"

0.09667/(9.30 ×10^"-8") = 1 × 10^"-6" > 400. ∴x ≪ "0.099 67"

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

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

x^3 = 2.173 × 10^"-10"

x = 6.012 × 10^"-4"

["H"_2] = 2xcolor(white)(l) "mol/L" = 2 × 6.012 × 10^"-4"color(white)(l) "mol/L" = "0.0012 mol/L"