When NH₄HS(s) is vaporized in an empty vessel and maintained at 20°C, the equilibrium is established — Chemical Equilibrium Chemistry Question
Question
When NH₄HS(s) is vaporized in an empty vessel and maintained at 20°C, the equilibrium is established: NH₄HS(s) ⇌ NH₃(g) + H₂S(g) and the total pressure of gases at equilibrium is 0.4 atm. When NH₄HS(s) is vaporized in the presence of NH₃(g) in the same vessel at 20°C, the partial pressure of NH₃(g) at equilibrium is 0.5 atm. Which of the following is/are correct statement(s)?
💡 Solution & Explanation
From the first equilibrium, P_total = 0.4 atm, so P_NH₃ = 0.2 atm and P_H₂S = 0.2 atm. K_p = P_NH₃ × P_H₂S = 0.2 × 0.2 = 0.04 atm². In the second equilibrium, P_NH₃ = 0.5 atm. Using K_p, 0.04 = 0.5 × P_H₂S => P_H₂S = 0.08 atm. (Option A is correct). Since H₂S comes only from dissociation, 0.08 atm of NH₃ was also formed. Initial NH₃ = 0.5 - 0.08 = 0.42 atm. (Option B is correct). The dissociation pressure of H₂S dropped from 0.2 atm to 0.08 atm, so the extent of dissociation decreased. (Option C is correct). Mole fraction of H₂S = 0.08 / (0.5 + 0.08) = 0.08 / 0.58 ≠ 0.5. (Option D is incorrect). Therefore, correct answer is A,B,C.