In the system, LaCl3(s) + (g) + heat ⇌ LaClO(s) + 2(g), equilibrium is established. More water vapou — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Write the heterogeneous equilibrium equation \[\ce{LaCl3(s) + H2O(g) + heat <=> LaClO(s) + 2HCl(g)}\] Step 2 - Write the Kp expression Since \ce{LaCl3(s)} and \ce{LaClO(s)} are pure solids (activity = 1), they are excluded from Kp: \[K_p = \frac{p_{\ce{HCl}}^2}{p_{\ce{H2O}}}\] Step 3 - Relate initial and new equilibrium states At initial equilibrium: \[K_p = \frac{p_{\ce{HCl}}^2}{p_{\ce{H2O}}}\] At new equilibrium (T constant, so Kp unchanged), with \(p'_{\ce{H2O}} = 2p_{\ce{H2O}}\): \[K_p = \frac{(p'_{\ce{HCl}})^2}{p'_{\ce{H2O}}} = \frac{(p'_{\ce{HCl}})^2}{2p_{\ce{H2O}}}\] Step 4 - Solve for the factor of change in HCl pressure Equating both Kp expressions: \[\frac{p_{\ce{HCl}}^2}{p_{\ce{H2O}}} = \frac{(p'_{\ce{HCl}})^2}{2p_{\ce{H2O}}}\] \[(p'_{\ce{HCl}})^2 = 2 \cdot p_{\ce{HCl}}^2\] \[p'_{\ce{HCl}} = \boxed{\sqrt{2} \cdot p_{\ce{HCl}}}\] Step 5 - Evaluate all options - **Option (A) 2 times**: Incorrect. Would apply if HCl had stoichiometric coefficient 1 (not 2). - **Option (B) √2 times**: Correct. HCl is squared in Kp; doubling P_H2O doubles P_HCl², so P_HCl increases by √2. - **Option (C) 1/√2 times**: Incorrect. Adding water vapour shifts equilibrium forward (Le Chatelier), so P_HCl increases, not decreases. - **Option (D) 4 times**: Incorrect. Mathematical error — applies if P_H2O increase directly scaled P_HCl.