When (s) is vaporized in an empty vessel and maintained at 20°C, equilibrium (s) ⇌ (g) + (g) establi — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Analyze the First Experiment to Find the Equilibrium Constant (\(K_p\)) The heterogeneous decomposition equilibrium: \[\ce{NH4HS(s) <=> NH3(g) + H2S(g)}\] Since \(\ce{NH4HS(s)}\) is a pure solid, it does not appear in the equilibrium expression. In the first experiment (empty vessel), the gases are produced in a 1:1 ratio, so: \[p_{\ce{NH3}} = p_{\ce{H2S}} = p\] Given total pressure = 0.4 atm: \[2p = 0.4\text{ atm} \implies p = 0.2\text{ atm}\] \[K_p = p_{\ce{NH3}} \cdot p_{\ce{H2S}} = 0.2 \times 0.2 = 0.04\text{ atm}^2\] Temperature is constant at 20°C, so \(K_p = 0.04\text{ atm}^2\) for both experiments. Step 2 - Analyze the Second Equilibrium with Added Ammonia Let initial pressure of added \(\ce{NH3}\) = \(P_0\). Let \(p'\) = partial pressure of \(\ce{H2S}\) formed by dissociation. By stoichiometry, both \(\ce{NH3}\) and \(\ce{H2S}\) pressures increase by \(p'\): \[\begin{array}{lccccc} \text{Species} & \ce{NH4HS(s)} & \ce{<=>} & \ce{NH3(g)} & + & \ce{H2S(g)} \ \hline \text{Initial (atm)} & \text{solid} & & P_0 & & 0 \ \text{Change (atm)} & - & & +p' & & +p' \ \text{Equilibrium (atm)} & \text{solid} & & P_0 + p' & & p' \ \end{array}\] Given: \(p_{\ce{NH3}} = P_0 + p' = 0.5\text{ atm}\) Using \(K_p = 0.04\text{ atm}^2\): \[0.5 \times p' = 0.04 \implies p' = \frac{0.04}{0.5} = \boxed{0.08\text{ atm}}\] So the final partial pressure of \(\ce{H2S}\) = 0.08 atm. **Option (A) is correct.** Step 3 - Determine the Initial Pressure of Ammonia \[P_0 = 0.5 - p' = 0.5 - 0.08 = \boxed{0.42\text{ atm}}\] Initial pressure of \(\ce{NH3(g)}\) was 0.42 atm. **Option (B) is correct.** Step 4 - Evaluate the Change in Extent of Dissociation * First experiment: extent of dissociation = \(p = 0.2\text{ atm}\) * Second experiment: extent of dissociation = \(p' = 0.08\text{ atm}\) Since \(0.08 < 0.2\), the extent of dissociation decreased. The added \(\ce{NH3(g)}\) acts as a common product (Le Chatelier's principle), shifting equilibrium backward. **Option (C) is correct.** Step 5 - Evaluate the Mole Fraction of H2S at Second Equilibrium \[P_{\text{total, 2}} = 0.5 + 0.08 = 0.58\text{ atm}\] \[x_{\ce{H2S}} = \frac{0.08}{0.58} \approx 0.138 \neq 0.5\] **Option (D) is incorrect.** Step 6 - Summary * **Option (A)**: Correct — \(p_{\ce{H2S}} = 0.08\text{ atm}\) * **Option (B)**: Correct — Initial \(p_{\ce{NH3}} = 0.42\text{ atm}\) * **Option (C)**: Correct — Extent of dissociation decreased from 0.2 to 0.08 atm * **Option (D)**: Incorrect — Mole fraction of \(\ce{H2S}\) ≈ 0.138, not 0.5