For the equilibrium: NH2COONH4(s) ⇌ 2(g) + (g), Kp = 27×2^(√2) at 800 K and equilibrium pressure is — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the reaction components and their phases We are given the heterogeneous decomposition reaction of solid ammonium carbamate (\ce{NH2COONH4}): \[\ce{NH2COONH4(s) <=> 2NH3(g) + CO2(g)}\] In this heterogeneous equilibrium, the reactant is a pure solid and the activity (active mass) of a pure solid is constant and taken as unity (\(1\)): \[a_{\ce{NH2COONH4(s)}} = 1\] Therefore, the solid reactant does not appear in the expression for the equilibrium constant \(K_p\), which depends solely on the partial pressures of the gaseous products: \[K_p = (p_{\ce{NH3}})^2 \cdot p_{\ce{CO2}}\] Step 2 - Express the partial pressures of the gases Let \(P_{ ext{total}}\) represent the total pressure of the gaseous mixture at equilibrium. According to the stoichiometry, for every 1 mole of solid reactant that decomposes: * \(2 ext{ moles}\) of \ce{NH3(g)} * \(1 ext{ mole}\) of \ce{CO2(g)} Mole ratio: \(n_{\ce{NH3}} : n_{\ce{CO2}} = 2 : 1\) Total gaseous moles per stoichiometric unit = \(2 + 1 = 3\). Using Dalton's Law of Partial Pressures: \[p_{\ce{NH3}} = rac{2}{3} P_{ ext{total}}\] \[p_{\ce{CO2}} = rac{1}{3} P_{ ext{total}}\] Step 3 - Express \(K_p\) in terms of the total pressure \[K_p = \left(rac{2}{3} P_{ ext{total}} ight)^2 \left(rac{1}{3} P_{ ext{total}} ight) = rac{4}{27} P_{ ext{total}}^3\] Given \(P_{ ext{total}} = 22 ext{ atm}\) at \(800 ext{ K}\): \[K_p = rac{4 imes 22^3}{27} ext{ atm}^3\] Step 4 - Calculate \(\lambda\) \[\lambda = P_{ ext{total}} - 1 = 22 - 1 = oxed{21}\] Step 5 - Explain each option * **Option (A) 21**: Correct. \(\lambda = P_{ ext{total}} - 1 = 22 - 1 = 21\). * **Option (B) 22**: Incorrect. This is \(P_{ ext{total}}\) itself, not \(\lambda\). * **Option (C) 11**: Incorrect. Half of total pressure — algebraic error. * **Option (D) 12**: Incorrect. Arithmetic error.