Hydrogen sulphide decomposes according to the following reaction: 2 H ₂ S(g) 2 H ₂ (g) + S ₂ (g) If the percentage dissociation of H ₂ S is 1 % at an equilibrium total pressure of 16 bar , the equilibrium constant K_p for the reaction is x 10⁻⁶ . The value of x is _____ (Nearest integer).

Step-by-step solution

Let the initial moles of H ₂ S be 1 . At equilibrium, moles of H ₂ S = 1 - , moles of H ₂ = , moles of S ₂ = 2 . Total moles at equilibrium = 1 - + + 2 = 1 + 2 1 (since is very small). Partial pressures at equilibrium: P_ H ₂ S P P_ H ₂ P P_ S ₂ 2 P The equilibrium constant K_p is given by: K_p = (P_ H ₂ )^2 (P_ S ₂ ) (P_ H ₂ S )^2 K_p = ( P)^2 ( 2 P ) P^2 = ^3 P 2 Given = 1 % = 0.01 and P = 16 bar : K_p = (0.01)^3 16 2 = 8 10⁻⁶ Thus, x = 8 . Answer: 8