Problem 40 - A Complete Resource Book in Chemistry for JEE Main

Question

In a reaction \(\mathrm{A}_{2}(\mathrm{~g})+4 \mathrm{~B}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}_{4}(\mathrm{~g}) ; \Delta \mathrm{H}<0\) The formation of \(\mathrm{AB}_{4}(\mathrm{~g})\) will be favoured by (a) low temperature and high pressure (b) high temperature and high pressure (c) low temperature and low pressure (d) high temperature and low pressure

Step-by-Step Solution

Answer

The formation of \(\mathrm{AB}_4\) will be favored by low temperature and high pressure (option a).

Step 1: Understand the Reaction Type

The given reaction is an exothermic reaction because \( \Delta H < 0 \). This means heat is released when the reaction occurs.

Step 2: Analyze Effects of Temperature

For exothermic reactions, lowering the temperature shifts the equilibrium to the right, favoring the formation of products. This is because the system will try to absorb heat to counter the effect of lowering the temperature.

Step 3: Analyze Effects of Pressure

The balanced chemical equation shows a decrease in total moles of gas: from 5 moles (\(\mathrm{A}_2 + 4\mathrm{B}_2\)) to 2 moles (\(2\mathrm{AB}_4\)). According to Le Chatelier's Principle, increasing the pressure favors the side of the reaction with fewer moles of gas. Therefore, high pressure will favor the formation of \(\mathrm{AB}_4\).

Step 4: Combine Temperature and Pressure Effects

To favor the formation of \(\mathrm{AB}_4\), we need conditions that favor the right side of the equilibrium. Thus, low temperature and high pressure will both contribute to driving the equilibrium towards forming more \(\mathrm{AB}_4\).

Key Concepts

Exothermic Reactions

In chemistry, whether a reaction is exothermic or endothermic is determined by the sign of its enthalpy change, denoted as \( \Delta H \). If \( \Delta H \) is negative, like in our given reaction, it indicates that the reaction is exothermic. This means that the reaction releases heat to the surroundings. Exothermic reactions are quite common and include processes like burning fuels, and many synthesis reactions. A simple way to remember is that exothermic means "giving off heat." This makes these reactions energetically favorable at lower temperatures because, according to Le Chatelier's Principle, reducing the temperature will cause the system to shift in a direction that compensates for that change—in this case, moving toward the exothermic direction.However, remember that while exothermic reactions release energy, not all the energy necessarily translates into a high reactivity unless conditions like temperature and pressure are optimally adjusted.

Effects of Temperature on Equilibrium

Temperature plays a critical role in chemical equilibria, especially in exothermic reactions. According to Le Chatelier's Principle, a system at equilibrium will shift to counteract imposed changes. When you decrease the temperature for an exothermic reaction, the system will shift towards the product side, where heat is released, to counterbalance the temperature drop.For the reaction \( \mathrm{A}_{2} + 4\mathrm{B}_{2} \rightleftharpoons 2 \mathrm{AB}_{4} \), lowering the temperature would favor the formation of \( \mathrm{AB}_{4} \). To sum up:

• In exothermic reactions, low temperatures promote product formation.
• The system shifts towards the exothermic side to "create" more heat.

Be mindful, however, that extremely low temperatures might slow down the reaction rate, even if they favor product formation. Therefore, a balance must be struck between favorable equilibrium conditions and reaction rates.

Effects of Pressure on Equilibrium

Pressure adjustments can significantly influence reactions involving gaseous reactants and products. Le Chatelier's Principle states that if a change is made to a system at equilibrium, the system will move in such a way as to counteract the change. When pressure is increased, the equilibrium will shift towards the side of the reaction with fewer moles of gas. In our specific reaction \( \mathrm{A}_{2} + 4\mathrm{B}_{2} \rightleftharpoons 2 \mathrm{AB}_{4} \), the left side has 5 moles of gas, while the right side has only 2 moles. Therefore, increasing the pressure will shift the equilibrium to the right, favoring the formation of \( \mathrm{AB}_{4} \).Here's how it works:

• High pressure favors the side with fewer gas molecules.
• This effect encourages the formation of products with lower total volume.

Applying high pressure to this reaction, therefore, supports more efficient product formation, contributing further to achieving optimal conditions alongside a low temperature.