Question

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Hydrogen sulfide decomposes according to the following reaction, for which

K_c = 9.30 × 10⁻⁸ at 700°C:

2 H₂S(g) ⇌ 2 H₂(g) + S₂(g)

If 0.55 mol of H₂S is placed in a 3.0−L container, what is the equilibrium concentration of H₂(g) at 700°C?

? M

Answer 1

The equilibrium concentration of H₂ is 0.012 mol/L.

Calculate the initial concentrations

[H₂S]₀ = 0.29mol3.0L = 0.096 67 mol/L (2 significant figures + 2 guard digits)

Write the balanced equation and set up an ICE table.

2H₂S ⇌ 2H₂ + S₂
I/mol·L⁻¹: 0.096 67; 0; 0
C/mol·L⁻¹: -2x; +2x; +x
E/mol·L⁻¹: 0.096 67 - 2x; 2x; x

Write the Kc expression and solve for x

Kc=[H₂]2[S₂][H₂S]2 = 9.30 × 10⁻⁸

(2x)2×x(0.09667–2x)2 = 9.30 × 10⁻⁸

Because Kc is so small, we can assume that 2x is negligible compared to 0.09667. Then we have

4x30.096672 = 9.30 × 10⁻⁸

4x3 = 0.096 67² × 9.30 × 10⁻⁸ = 8.691 × 10⁻¹⁰

x3 = 2.173 × 10⁻¹⁰

x = 6.012 × 10⁻⁴

1

[H₂] = 2x mol/L= 2 × 6.012 × 10⁻⁴ mol/L = 0.001 20 mol/L