Problem 73 - A Complete Resource Book in Chemistry for JEE Main

Question

The oxidation of \(\mathrm{SO}_{2}\) by \(\mathrm{O}_{2}\) to \(\mathrm{SO}_{3}\) is an exothermic reaction. They yield of \(\mathrm{SO}_{3}\) will be maximum if (a) temperature is reduced and pressure is increased (b) temperature is increased and pressure is kept constant (c) both temperature and pressure are reduced (d) both temperature and pressure are increased

Step-by-Step Solution

Answer

(a) temperature is reduced and pressure is increased

Step 1: Analyze the Reaction

The chemical equation for the oxidation of \( \mathrm{SO}_{2} \) by \( \mathrm{O}_{2} \) to form \( \mathrm{SO}_{3} \) is \( 2\mathrm{SO}_{2} + \mathrm{O}_{2} \rightarrow 2\mathrm{SO}_{3} \). This is an exothermic reaction, meaning that it releases heat as it proceeds.

Step 2: Apply Le Chatelier's Principle for Temperature

Le Chatelier's principle states that if a system in equilibrium is disturbed, it will adjust in a way that counteracts the disturbance. Since the reaction is exothermic, decreasing the temperature will favor the forward reaction, increasing the yield of \( \mathrm{SO}_{3} \). Thus, reducing the temperature increases the yield.

Step 3: Apply Le Chatelier's Principle for Pressure

Le Chatelier's principle also indicates that increasing pressure in a system will favor the side of the reaction with fewer moles of gas. In this reaction, 3 moles of gas (2 \( \mathrm{SO}_{2} \) and 1 \( \mathrm{O}_{2} \)) react to form 2 moles of \( \mathrm{SO}_{3} \). Hence, increasing the pressure will shift the equilibrium toward the production of more \( \mathrm{SO}_{3} \).

Step 4: Choose the Optimal Conditions

From Steps 2 and 3, reducing the temperature and increasing the pressure will both favor the forward reaction, maximizing the yield of \( \mathrm{SO}_{3} \). Therefore, the optimal conditions are to decrease the temperature while increasing the pressure.

Key Concepts

Exothermic Reaction

An exothermic reaction is a type of chemical reaction that releases energy in the form of heat. Imagine a reaction where, instead of needing a heater to keep things warm, the reaction itself heats up the surroundings. This is what happens when an exothermic reaction occurs.
When sulfur dioxide (\(\mathrm{SO}_2\)) reacts with oxygen (\(\mathrm{O}_2\)) to form sulfur trioxide (\(\mathrm{SO}_3\)), it is an exothermic reaction. The release of heat means that energy is leaving the system. Typically, this released energy manifests as a rise in temperature of the surroundings.
This characteristic of exothermic reactions is crucial when applying Le Chatelier's Principle. If the temperature of an exothermic reaction is decreased, the equilibrium will shift to produce more products (in this case, \(\mathrm{SO}_3\)) to counteract the change by "releasing" more heat. This is because the system "wants" to stay in balance, and by generating more heat through the forward reaction, it compensates for the decrease in temperature.

Chemical Equilibrium

Chemical equilibrium is a fascinating concept in chemistry where reactions reach a state of balance. Even though the reactions continue to happen, the amounts of reactants and products stay constant over time. This means that the forward reaction (reactants to products) and the reverse reaction (products to reactants) occur at the same rate. It's like a seesaw perfectly balanced with equal weight on both sides.
When we talk about the oxidation of \(\mathrm{SO}_2\) to \(\mathrm{SO}_3\), reaching equilibrium is like hitting a pause where both the formation of \(\mathrm{SO}_3\) and the breakdown back to \(\mathrm{SO}_2\) and \(\mathrm{O}_2\) happen simultaneously and continuously.
By applying Le Chatelier's Principle, it's possible to manipulate the conditions (such as temperature and pressure) to shift the equilibrium, thus favoring either the production of more reactants or products. For this exothermic reaction, decreasing the temperature and increasing the pressure will shift the equilibrium towards more \(\mathrm{SO}_3\), maximizing its yield.

Oxidation of SO2

The oxidation of sulfur dioxide (\(\mathrm{SO}_2\)) to sulfur trioxide (\(\mathrm{SO}_3\)) is a key industrial process, primarily because \(\mathrm{SO}_3\) is used in the production of sulfuric acid, one of the most widely used chemicals worldwide. This reaction is represented by the equation: \[2\mathrm{SO}_2 + \mathrm{O}_2 \rightarrow 2\mathrm{SO}_3\] In this reaction, \(\mathrm{SO}_2\) undergoes oxidation as it gains oxygen to form \(\mathrm{SO}_3\).

To ensure this reaction produces a maximum yield of \(\mathrm{SO}_3\), we can apply Le Chatelier's Principle. An understanding of the role of pressure is essential here. Since the reaction reduces the total number of gas molecules from three to two, increasing the pressure favors the forward reaction.
In practical terms, maximizing the yield is important for efficiency and cost-effectiveness in chemical manufacturing. Manufacturers strive to optimize conditions to favor the production of \(\mathrm{SO}_3\) while ensuring that the reaction remains safe and sustainable.