15.00 g of solid ammonium hydrogen sulfide is introduced into a 500.-mL flask at 25^$\circ$C, the flask is sealed, and the system is allowed to reach equilibrium. What is the partial pressure of ammonia in this flask if K_p = 0.108 at 25^$\circ$C for NH₄HS(s) ⇌ NH₃(g) + H₂S (g)

Question

Quizplus · Accepted Answer

The equilibrium reaction is $NH_4HS(s) \rightleftharpoons NH_3(g) + H_2S(g)$. Given $K_p = 0.108$ at $25^\circ C$, and since the stoichiometry of $NH_3$ and $H_2S$ is 1:1, their partial pressures at equilibrium will be equal. The total pressure $P_{total}$ is the sum of the partial pressures of $NH_3$ and $H_2S$, so $P_{total} = P_{NH_3} + P_{H_2S} = 2P_{NH_3}$ because $P_{NH_3} = P_{H_2S}$. Given $K_p = P_{NH_3} \times P_{H_2S} = (P_{NH_3})^2$, solving for $P_{NH_3}$ gives $P_{NH_3} = \sqrt{K_p} = \sqrt{0.108} \approx 0.329$ atm.

1500 G of Solid Ammonium Hydrogen Sulfide Is Introduced into a 500.-ML