2 H2S(g) ⇄ 2 H2(g) + S2(g) Kc = 9.3× 10-8 at 400ºC 0.25 moles of H2S are placed in a 3.0 L container and the system is allowed to reach equilibrium. Calculate the concentration of H2 at equilibrium.

The concentration of H₂ at equilibrium = 1.09 x 10⁻³ M

Explanation:

                     2H₂S(g)   ⇄   2H₂(g)  +  S₂(g)

Initial             0.0833

Equilibrium   0.0833-x           x             x/2

Since = 9.3x10⁻⁸ which is very small

So, 0.0833 - x = 0.083

= 9.3x10⁻⁸ = (x² x) / 2(0.0833)²

     x³ = 0.6458 x 10⁻⁹ x 2

     x  =  1.09 x 10⁻³

[H₂] = 1.09 x 10⁻³ M

[H2] = 0.001086 M

Explanation:

Step 1: Data given

Kc = 9.3 * 10^-8

Temperature = 400 °C = 673 K

Number of moles H2S = 0.25 moles

Volume = 3.0 L

Step 2: The balanced equation

2 H2S(g) ⇄ 2 H2(g) + S2(g)

Step 3: The initial concentrations

[H2S] = 0.25 moles / 3.0 L = 0.083 M

[H2] = 0M

[S2] = 0M

Step 4: The concentrations at the equilibrium

[H2S] = (0.083 -2X)M

[H2] = 2XM

[S2] = X M

Kc = [S2][H2]² /[H2S]²

Kc = x * (2X)² / ( 0.083 - X)²

9.3 * 10^-8 = 4X³ / (0.083 - X)²

9.3 * 10^-8 = 4X³ / 0.083 ²

4X³ = 6.41*10^-10

X = 0.000543

[H2S] = (0.083 -2X) = 0.081914 M

[H2] = 2XM =0.001086 M

[S2] = X M = 0.000543 M

c) should be the answer.