Chemistry: Equilibrium Math Question 7

Question

Consider the equilibrium between hydrogen sulfide and its gaseous elements:

2 H₂S(g) ↔ 2 H₂(g) + S₂(g) K_P = 2.023E-6 at 908.0 K

if a vessel is initially charged with 8.35 bar of pure H₂S(g), what is the final equilibrium pressure of H₂(g)?

Answer 1

2 H₂S(g) ===> 2 H₂(g) + S₂(g)

8.35..................0.................0..........Initial

-2x..................+2x.............+x..........Change

8.35-2x............2x...............x...........Equilibrium

K = (H₂)²(S₂)/(H₂S)²

2.023x10^-6 = (2x)²(x)/(8.35-2x)²

2.023x10^-6 = 4x³/4x² - 33.4x + 69.72

Using an online calculator for cubic equations, x = 0.0326

Final equilibrium pressure of H2(g) = 2 x 0.0326 = 0.0652 bar