Question

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?

Answer 1

The equilibrium concentration of `"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.

`color(white)(mmmmmmm)"2H"_2"S"color(white)(m) ⇌color(white)(m) "2H"_2 + "S"_2`
`"I/mol·L"^"-1":color(white)(m) "0.096 67" color(white)(mmmm)0 color(white)(mm)0`
`"C/mol·L"^"-1":color(white)(mm) "-"2xcolor(white)(mmmm) "+2"xcolor(white)(m) "+"x`
`"E/mol·L"^"-1": "0.096 67 - 2"xcolor(white)(lmm) 2xcolor(white)(ml) x`

Write the `K_"c"` expression and solve for `x`.

`K_"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"`

∴ `(4x^3)/"0.096 67"^2 = 9.30 × 10^"-8"`

`4x^3 = "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"`