For each of the questions, four choices have been provided. Select the correct alternative. In which among the following reactions, the formation of product is favoured by decreasing the temperature or volume? (a) \(2 \mathrm{SO}_{3(\mathrm{~g})} \rightleftharpoons 2 \mathrm{SO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}-\mathrm{q}\) (b) \(\mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NO}_{(\mathrm{g})}-\mathrm{q}\) (c) \(4 \mathrm{NH}_{3(\mathrm{~g})}+5 \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons 4 \mathrm{NO}_{(\mathrm{g})}+6 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}, \Delta \mathrm{H}\) \(=\) -ve (d) \(2 \mathrm{NO}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NO}_{2(\mathrm{~g})}, \Delta \mathrm{H}=-\mathrm{ve}\)

Le Chatelier's Principle states that when a system at equilibrium is subjected to a change in temperature, pressure, or concentration of reactants or products, the system will adjust itself to re-establish equilibrium. For temperature changes, if the reaction is exothermic (ΔH < 0), decreasing the temperature will shift the equilibrium towards the formation of products. If the reaction is endothermic (ΔH > 0), decreasing the temperature will shift the equilibrium towards the formation of reactants.

For volume changes, if the volume of the system is decreased, the equilibrium will shift towards the side with the fewer moles of gas. If the volume is increased, the equilibrium will shift towards the side with more moles of gas.

(a) The given reaction is exothermic (ΔH < 0) and has more moles of gas on the left side. Therefore, decreasing the temperature will shift the equilibrium towards the formation of products, while decreasing the volume will shift the equilibrium towards the formation of reactants. This option does not satisfy both conditions. (b) The given reaction is endothermic (ΔH > 0) and has an equal number of moles of gas on both sides. Decreasing the temperature will shift the equilibrium towards the formation of reactants, while changing the volume will not affect the equilibrium. This option does not satisfy both conditions. (c) The given reaction is exothermic (ΔH < 0) and has more moles of gas on the left side. Therefore, decreasing the temperature will shift the equilibrium towards the formation of products, while decreasing the volume will shift the equilibrium towards the formation of products as well. This option satisfies both conditions. (d) The given reaction is exothermic (ΔH < 0) and has the same number of moles of gas on both sides. Decreasing the temperature will shift the equilibrium towards the formation of products, while changing the volume will not affect the equilibrium. This option does not satisfy both conditions.

Based on the analysis in Step 3, the correct alternative is (c) \(4 \mathrm{NH}_{3(\mathrm{~g})}+5 \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons 4 \mathrm{NO}_{(\mathrm{g})}+6 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}, \Delta \mathrm{H}=\) -ve as both decreasing the temperature and volume favor the formation of products in this reaction.