The reaction which proceeds towards completion in the forward direction is — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Criterion for a Reaction to Proceed Towards Completion A chemical reaction proceeds almost to completion in the forward direction if its equilibrium constant (\(K_{\text{eq}}\)) is exceptionally large (\(K_{\text{eq}} \gg 1\)) and its standard Gibbs free energy change (\(\Delta G^\circ\)) is highly negative (\(\Delta G^\circ \ll 0\)). This is typically observed in: 1. Acid-base neutralization reactions involving strong acids or bases, driven by the formation of highly stable, weakly dissociating water molecules (\(\ce{H2O}\)). 2. Reactions that form highly insoluble precipitates or highly stable complexes. 3. Strongly favored redox processes with a large positive cell potential (\(E^\circ_{\text{cell}} > 0\)). Step 2 - Analyze Option (A) The given reaction is: \[\ce{Fe2O3(s) + 6HCl(aq) <=> 2FeCl3(aq) + 3H2O(l)}\] * **Nature of Reactants:** Iron(III) oxide (\(\ce{Fe2O3}\)) is a basic metal oxide, and hydrochloric acid (\(\ce{HCl}\)) is a very strong mineral acid. * **Reaction Type:** This is an acid-base neutralization reaction. The oxide anions (\(\ce{O^{2-}}\)) from the basic oxide lattice react with the hydronium ions (\(\ce{H^+}\)) from the strong acid to form extremely stable covalent water molecules: \[\ce{O^{2-} + 2H+ -> H2O(l)}\] * **Thermodynamic Driving Force:** Because water is an extremely weak electrolyte with an exceptionally low autoionization constant (\(K_w = 1.0 \times 10^{-14}\) at \(25^\circ\text{C}\)), the reverse reaction (hydrolysis to regenerate a strong acid and basic metal oxide) is practically non-existent. * **Conclusion:** This neutralization is highly exothermic (\(\Delta H^\circ \ll 0\)) and highly spontaneous (\(\Delta G^\circ \ll 0\)). Consequently, the equilibrium constant for this reaction is extraordinarily large, and the reaction proceeds almost to \(100\%\) completion in the forward direction. Step 3 - Analyze Option (B) The given reaction is: \[\ce{SnCl4 + Hg2Cl2 <=> SnCl2 + 2HgCl2}\] * **Nature of Reaction:** This represents a redox process where tin(IV) (\(\ce{Sn^4+}\)) acts as an oxidizing agent to oxidize mercury(I) (\(\ce{Hg2^2+}\)) to mercury(II) (\(\ce{Hg^2+}\)), while being reduced to tin(II) (\(\ce{Sn^2+}\)). * Since the standard reduction potential of the mercury(II)/mercury(I) couple (\(+0.92\text{ V}\)) is much higher than that of tin(IV)/tin(II) (\(+0.15\text{ V}\)), the reverse reaction is highly spontaneous. * **Conclusion:** The forward reaction has a very small equilibrium constant (\(K_{\text{eq}} \ll 1\)) and does not proceed towards completion. Step 4 - Analyze Option (C) The given reaction is: \[\ce{NH3(aq) + H2O(l) + NaCl(aq) <=> NH4Cl(aq) + NaOH(aq)}\] * **Net Ionic Equation:** After canceling spectator ions, this reduces to the basic ionization of ammonia: \[\ce{NH3(aq) + H2O(l) <=> NH4+(aq) + OH-(aq)}\] * **Equilibrium Constant:** \(K_b \approx 1.8 \times 10^{-5}\) * **Conclusion:** Since \(K_b \ll 1\), the equilibrium lies heavily to the left. The forward reaction does not proceed towards completion. Step 5 - Analyze Option (D) The given reaction is: \[\ce{2CuI(s) + I2(s) + 4K+(aq) <=> 2Cu^{2+}(aq) + 4KI(aq)}\] * In aqueous chemistry, \(\ce{Cu^{2+}}\) ions spontaneously oxidize \(\ce{I^-}\) ions to \(\ce{I2}\) while forming the highly insoluble \(\ce{CuI}\) precipitate (\(K_{sp} \approx 1 \times 10^{-12}\)): \[\ce{2Cu^{2+}(aq) + 4I-(aq) -> 2CuI(s) + I2(s)}\] * **Conclusion:** The backward reaction proceeds quantitatively to completion. The forward reaction has a negligible equilibrium constant. Step 6 - Final Conclusion Only reaction (A) represents a neutralization process between a basic oxide and a strong acid, which is highly thermodynamically favored in the forward direction. \[\boxed{\text{A}}\]