Problem 15 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following equilibrium, in gaseous phase, would be unaffected by an increase in pressure? (a) \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) (b) \(\mathrm{N}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}\) (d) \(\mathrm{CO}_{2}+{ }^{1} / 2 \mathrm{O}_{2} \rightleftharpoons \mathrm{CO}_{2}\)

Step-by-Step Solution

Answer

Reaction (b) \(\text{N}_2 + \text{O}_2 \rightleftharpoons 2\text{NO}\) is unaffected by pressure change.

Step 1: Understanding Pressure Effect

An increase in pressure usually affects equilibria by shifting the reaction towards the side with fewer moles of gas, according to Le Chatelier's Principle. This is because increasing pressure affects the volume and the system attempts to counteract the change by shifting equilibrium towards the side with less volume (fewer moles).

Step 2: Analyze Reaction (a)

For the equilibrium \( ext{N}_2 + 3 ext{H}_2 \rightleftharpoons 2 ext{NH}_3\), the left side has 4 moles of gas (1 N₂ + 3 H₂) and the right side has 2 moles of NH₃. Increasing pressure will shift the equilibrium towards "less" moles side, which is the NH₃ side.

Step 3: Analyze Reaction (b)

The reaction \( ext{N}_2 + ext{O}_2 \rightleftharpoons 2 ext{NO}\) has 2 moles of gas on both sides. Since the number of moles of gas is equal on both sides, an increase in pressure will not affect the equilibrium position.

Step 4: Analyze Reaction (c)

In the reaction \( ext{N}_2 ext{O}_4 \rightleftharpoons 2 ext{NO}_2\), there is 1 mole of gas on the left and 2 moles of gas on the right. An increase in pressure would shift the equilibrium towards the left, where there is less moles of gas.

Step 5: Analyze Reaction (d)

For the equilibrium \( ext{CO}_2 + rac{1}{2} ext{O}_2 \rightleftharpoons ext{CO}_2\), one side has 1.5 moles (1 CO₂ + 0.5 O₂) and the other side has 1 mole of CO₂. A pressure increase will shift the equilibrium towards the CO₂ side.

Step 6: Conclusion

Since Reaction (b) has an equal number of moles of gas on both sides, an increase in pressure will not affect its equilibrium position.

Key Concepts

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental concept in chemistry that helps us understand how systems at equilibrium respond to external changes, such as pressure, temperature, or concentration shifts.
This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change and restore equilibrium.
It is like a balance that tips in response to stress, finding a new point of balance.
When applied to pressure changes, Le Chatelier's Principle predicts that an increase in pressure will drive the equilibrium toward the side of the reaction with fewer gas moles.

• This is because increasing pressure effectively reduces the volume available to gaseous molecules.
• The system responds by favoring the formation of fewer gas molecules to reduce pressure.

Understanding this principle allows chemists to predict the behavior of chemical reactions under different conditions and is particularly useful in industrial applications to optimize reactions for desired products.

Pressure Effects on Equilibrium

The effect of pressure on equilibrium is closely related to the number of gas moles in a chemical reaction.
When the pressure of a system at equilibrium increases, the system will shift to reduce this pressure.
This usually involves moving the equilibrium position toward the side of the reaction that has fewer gas molecules.
Let's look at some examples to understand this better.

• For a reaction such as \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\), increasing the pressure shifts the equilibrium toward the production of \(\mathrm{NH}_{3}\), as there are fewer moles (2 moles) on that side than on the reactant side (4 moles).
• Conversely, if there are equal numbers of gas moles on both sides of the reaction, as in \(\mathrm{N}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}\), then changing pressure will not affect the equilibrium.

This concept is crucial when manipulating variables in processes like the Haber process for ammonia production, where maximizing product yield is important.

Reaction Moles Analysis

Analyzing the moles in a reaction is crucial in predicting how pressure changes will affect equilibrium.
By counting the moles of gaseous reactants and products on both sides of the equation, one can determine the direction in which the equilibrium will shift upon a pressure change.
This is because the number of moles directly influences the gas volume, connecting to how the system compensates for pressure changes.

• For instance, in a reaction like \(\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}\), the equilibrium will shift to the left, toward \(\mathrm{N}_{2} \mathrm{O}_{4}\), when pressure is increased since one side has fewer moles (1 vs 2 moles).
• Similarly, counting moles in \(\mathrm{CO}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{CO}_{2}\) reveals a shift toward the side with fewer moles of gas, favoring \(\mathrm{CO}_{2}\).

By mastering reaction moles analysis, students gain valuable insights into how chemical equilibria will adapt under different pressures, which is not only helpful in academics but also in practical industrial chemistry.