Problem 27 - A Complete Resource Book in Chemistry for JEE Main

Question

For equilibrium reaction \(2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})+14.6 \mathrm{~J}\), increase in tem- perature would (a) favour the formation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) (b) stop reaction (c) favour the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{4}\) (d) no alter the equilibrium

Step-by-Step Solution

Answer

Increasing temperature favors the decomposition of \( \text{N}_2\text{O}_4 \).

Step 1: Analyze the Reaction Equation

The given reaction is \( 2 \text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g) + 14.6 \text{ J} \). Since the reaction releases heat, it is an exothermic reaction.

Step 2: Apply Le Chatelier's Principle

According to Le Chatelier's principle, if a system in equilibrium undergoes a change in temperature, the system will adjust to counteract the change. For an exothermic reaction, an increase in temperature shifts the equilibrium towards the reactants to absorb the added heat.

Step 3: Determine Temperature Effect on Equilibrium

Given that increasing the temperature favors the reaction that absorbs heat (endothermic direction), the equilibrium will shift towards the decomposition of \( \text{N}_2\text{O}_4 \) into \( \text{NO}_2 \). Therefore, increasing the temperature will favor the reverse reaction.

Key Concepts

Understanding Equilibrium Reactions

Equilibrium reactions occur when the rate of the forward reaction is equal to the rate of the reverse reaction. This balance results in stable concentrations of reactants and products over time. When a reaction reaches equilibrium, it doesn't mean the reactions have stopped; rather, they continue to occur but at equal rates. In our given reaction, \(2 \text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g) + 14.6 \text{ J}\), both the formation and decomposition processes occur simultaneously.

• The forward reaction converts \(\text{NO}_2\) to \(\text{N}_2\text{O}_4\) and releases energy, since it is exothermic.
• The reverse reaction breaks \(\text{N}_2\text{O}_4\) back into \(\text{NO}_2\), requiring energy input (endothermic).

Knowing how equilibrium works helps in predicting how changes in conditions, like temperature, can shift the balance between reactants and products.

Exothermic Reactions and Their Characteristics

Exothermic reactions release energy, usually in the form of heat, as they proceed. This energy release is part of what drives the reaction forward. In the equation \(2 \text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g) + 14.6 \text{ J}\), the formation of \(\text{N}_2\text{O}_4\) from \(\text{NO}_2\) releases 14.6 Joules of energy, hence it is classified as an exothermic process.
Exothermic reactions have key characteristics:

• They often result in a temperature increase in their surroundings.
• The products are typically lower in energy compared to the reactants.
• They can reach equilibrium where energy release and absorption (reactant formation) balance.

Understanding these characteristics helps in predicting reaction shifts when external conditions like temperature change.

The Effect of Temperature on Equilibrium

Temperature can significantly influence the direction of an equilibrium reaction. According to Le Chatelier's Principle, a change in temperature will cause the system to adjust in such a way as to oppose that change. For an exothermic reaction, like the conversion of \(\text{NO}_2\) to \(\text{N}_2\text{O}_4\), increasing the temperature adds heat to the system.

• The system will seek to counteract this by favoring the endothermic direction, which absorbs heat.
• In this case, the equilibrium shifts towards the decomposition of \(\text{N}_2\text{O}_4\), forming more \(\text{NO}_2\).
• This shift is a direct response to the added heat, aiming to restore balance.

Recognizing how temperature affects equilibrium reactions allows us to manipulate conditions to obtain desired concentrations of reactants or products.