Answer:
The concentration of H₂ at equilibrium = 1.09 x 10⁻³ M
Explanation:
[tex][H_{25}]_{initial} = \frac{0.25}{3} = 0.0833[/tex]
2H₂S(g) ⇄ 2H₂(g) + S₂(g)
Initial 0.0833
Equilibrium 0.0833-x x x/2
Since [tex]K_c[/tex] = 9.3x10⁻⁸ which is very small
So, 0.0833 - x = 0.083
[tex]K_c[/tex] = 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
Answer:
[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
