Problem 23 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Consider the following equilibrium in a closed container $$ \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}) $$ At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statements holds true regarding the equilibrium constant \(\left(K_{p}\right)\) and degree of dissociation \((\alpha)\) ? (a) neither \(K_{p}\) nor \(\alpha\) changes (b) both \(K_{p}\) and \(\alpha\) change (c) \(K_{p}\) changes, but \(\alpha\) does not change (d) \(K_{p}\) does not change, but \(\alpha\) changes

Step-by-Step Solution

Answer

(d) \(K_p\) does not change, but \(\alpha\) changes.

Step 1: Understanding the Equilibrium Constant

The equilibrium constant, \(K_p\), is dependent on temperature and not on the changes like pressure or volume caused by altering the container's size, as long as the temperature remains constant. Thus, changes in the volume do not affect the value of \(K_p\).

Step 2: Assessing the Degree of Dissociation

The degree of dissociation, \(\alpha\), is the fraction of the original \(N_2O_4\) that dissociates into \(NO_2\). When the volume is halved, the pressure increases, because pressure is inversely proportional to volume (Boyle's Law). For a reaction where moles of gas increase, high pressure typically shifts the equilibrium towards fewer gas molecules to reduce pressure, according to Le Chatelier's principle. Thus, \(\alpha\) decreases.

Step 3: Conclusion on the Given Options

Based on the explanations above, \(K_p\) remains unchanged despite the decrease in volume, and \(\alpha\) changes because the system responds to the pressure change. Therefore, the correct statement is option (d): \(K_p\) does not change, but \(\alpha\) changes.

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K_p \) when dealing with partial pressures of gases, is a crucial concept that helps us understand chemical equilibria. It is defined as the ratio of the concentrations of the products to the reactants, each raised to the power of their coefficients in the balanced chemical equation, at equilibrium.

One important thing to remember is that \( K_p \) is solely dependent on temperature. This means that changes in pressure or volume, as long as the temperature remains constant, do not affect \( K_p \). Thus, even if the volume of the reaction container is halved, which increases pressure, the value of \( K_p \) stays the same. This is because the equilibrium constant reflects the position of equilibrium which is fundamentally set by temperature.

For example, in the reaction \(\text{N}_2\text{O}_4 (g) \rightleftharpoons 2 \text{NO}_2 (g)\), the equilibrium constant \( K_p \) will only change if the temperature changes, not due to physical changes in the system like volume or pressure.

Degree of Dissociation

The degree of dissociation, represented by \( \alpha \), measures the extent to which a chemical species separates into other chemicals in a reaction. It is essentially the fraction of the initial amount of a reactant that dissociates at equilibrium.

When the volume of a reaction container is halved, Boyle's Law tells us that pressure increases. In the reaction \(\text{N}_2\text{O}_4 (g) \rightleftharpoons 2 \text{NO}_2 (g)\), decreasing volume increases pressure, which impacts \( \alpha \). According to Le Chatelier's principle, the system will attempt to counteract this pressure change by favoring the side with fewer moles of gas, in this case shifting towards \( \text{N}_2\text{O}_4 \) and thus decreasing \( \alpha \).

Understanding how \( \alpha \) behaves in response to volume changes is crucial. It doesn't remain constant as equilibrium tries to re-adjust to maintain pressure balance.

Le Chatelier's Principle

Le Chatelier's principle is a fundamental concept that predicts the behavior of a reaction at equilibrium when it undergoes a change in concentration, temperature, or pressure. Simply put, if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the disturbance.

In the case of our reaction, \(\text{N}_2\text{O}_4 (g) \rightleftharpoons 2 \text{NO}_2 (g)\), if the volume is decreased causing an increase in pressure, Le Chatelier's principle suggests that the equilibrium position shifts towards the side with fewer moles of gas. Since \(\text{N}_2\text{O}_4\) has fewer moles compared to \(2 \text{NO}_2\), the system shifts left, reducing the degree of dissociation. This counteraction helps lower the increased pressure.

Le Chatelier's principle thus helps us predict and comprehend how chemical equilibria respond to external stresses, making it an invaluable tool in the study of chemistry.