At 22 °C, K_p= 0.070 for the equilibrium:
NH₄HS (s) NH₃(g) + H₂S (g)
A sample of solid NH₄HS is placed in a closed vessel and allowed to equilibrate. Calculate the equilibrium partial pressure (atm) of ammonia, assuming that some solid NH₄HS remains.

Question

At 22 °C, K_p= 0.070 for the equilibrium:
NH₄HS (s) At 22 °C, Kp = 0.070 for the equilibrium: NH4HS (s) NH3 (g) + H2S (g) A sample of solid NH4HS is placed in a closed vessel and allowed to equilibrate. Calculate the equilibrium partial pressure (atm) of ammonia, assuming that some solid NH4HS remains. A) 0.070 B) 0.26 C) 3.8 D) 0.52 E) 4.9 × 10- 3

At 22 °C, K<sub>p </sub>= 0.070 for the equilibrium: NH<sub>4</sub>HS (s) NH<sub>3 </sub>(g) + H<sub>2</sub>S (g) A sample of solid NH<sub>4</sub>HS is placed in a closed vessel and allowed to equilibrate. Calculate the equilibrium partial pressure (atm) of ammonia, assuming that some solid NH<sub>4</sub>HS remains. A) 0.070 B) 0.26 C) 3.8 D) 0.52 E) 4.9 × 10<sup>-</sup><sup> </sup><sup>3</sup>

NH₃(g) + H₂S (g)
A sample of solid NH₄HS is placed in a closed vessel and allowed to equilibrate. Calculate the equilibrium partial pressure (atm) of ammonia, assuming that some solid NH₄HS remains.

Quizplus · Accepted Answer

The equilibrium constant is given by: K_p= (P_₃ * P_H₂S) / P_NH₄HS Since NH₄HS is a solid, its partial pressure is not included in the equilibrium expression. Therefore, we only need to consider the gases: ₃ and H₂S. Let x be the equilibrium partial pressure of ₃ and H₂S. Then, according to the stoichiometry of the balanced equation, the equilibrium partial pressure of NH₄HS is (0.070 - x). Substituting these values into the equilibrium expression and solving for x, we get: 0.070 = (x^2) / (0.070 - x) 0.070 - x = (x^2) / 0.070 x^2 = 0.0049 - 0.07x x^2 + 0.07x - 0.0049 = 0 Solving this quadratic equation, we get: x = 0.256 atm or x = -0.326 atm Since partial pressures cannot be negative, we reject the negative solution and conclude that the equilibrium partial pressure (atm) of ₃ and H₂S is 0.256 atm.