Problem 22 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The Haber's process for the formation of \(\mathrm{NH}_{3}\) at \(298 \mathrm{~K}\) is \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} ; \Delta H=-46.0 \mathrm{~kJ} ;\) Which of the following is the correct statement (a) The condition for equilibrium is $$ G_{\mathrm{N}_{2}}+3 G_{\mathrm{H}_{2}}=2 G_{\mathrm{NH}_{3}} $$ where \(G\) is Gibb's free energy per mole of the gaseous species measured at that partial pressure. (b) On adding \(\mathrm{N}_{2}\), the equilibrium will shift to forward direction because according to \(\mathrm{II}^{\text {nd }}\) law of thermodynamics, the entropy must increase in the direction of spontaneous reaction (c) The catalyst will increase the rate of forward reaction by 2 times and that of backward reaction by \(1.5\) times (d) None of these

Step-by-Step Solution

Answer

Statement (a) is correct.

Step 1: Understanding Equilibrium Condition

For a reaction at equilibrium, the Gibbs free energy change, \( \Delta G \), is zero. Therefore, the condition for equilibrium is that the Gibbs free energies of the reactants and products are equal in a way that satisfies the stoichiometry of the reaction. For the reaction \( \mathrm{N}_{2} + 3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} \), the correct relation should be \( G_{\mathrm{N}_{2}} + 3 G_{\mathrm{H}_{2}} = 2 G_{\mathrm{NH}_{3}} \). This matches statement (a).

Step 2: Analyze Effect of Adding N2

Le Chatelier's principle states that if a system at equilibrium experiences a change in concentration, temperature, or pressure, the system will adjust to counteract the change. Adding \( \mathrm{N}_{2} \) would shift the equilibrium to the right (forward direction) to produce more \( \mathrm{NH}_{3} \). However, the explanation based on entropy increase does not fully justify this shift, making statement (b) partially incorrect.

Step 3: Catalyst's Effect on Reaction Rate

A catalyst speeds up both the forward and backward reactions equally by lowering the activation energy, maintaining the equilibrium constant unchanged. Statement (c) suggests unequal increase rates (2 times for forward and 1.5 times for backward), which is incorrect.

Step 4: Conclusion on Statements

Having analyzed each statement, only statement (a) accurately describes the condition of equilibrium using Gibbs free energy. Statements (b) and (c) are incorrect based on thermodynamic principles and the role of a catalyst.

Key Concepts

Gibbs Free Energy

Gibbs free energy is a vital concept when studying the Haber's process and other chemical reactions. It represents the maximum reversible work a thermodynamic system can perform at constant temperature and pressure. In the context of the Haber's process, the reaction involved is \[ \mathrm{N}_{2} + 3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}. \]For this reaction to be at equilibrium, the change in Gibbs free energy \( \Delta G \) must be zero. This means the Gibbs free energy of the reactants and products aligns with the stoichiometry of the reaction. That's why the equilibrium condition is represented by the equation:

• \( G_{\mathrm{N}_{2}} + 3 G_{\mathrm{H}_{2}} = 2 G_{\mathrm{NH}_{3}} \)

In simpler terms, the energy of the reactants equals the energy of the products when the system is at equilibrium. To check if a reaction will occur spontaneously, we look at the change in Gibbs free energy. If it's negative, the process is spontaneous. For a complete understanding, remember that the reaction is only at equilibrium when \( \Delta G = 0 \).
This is pivotal in confirming the first statement in the exercise solution that correctly reflects the equilibrium using Gibbs free energy.

Le Chatelier's Principle

Le Chatelier's principle is a fundamental principle in chemistry that helps predict how changes affect a system at equilibrium. When we apply this principle to the Haber's process, we can comprehend how the system reacts to changes. For example, adding more \( \mathrm{N}_{2} \) to the reaction causes the equilibrium to shift to the right, resulting in more \( \mathrm{NH}_{3} \) being produced. This happens to counterbalance the change introduced, essentially trying to "use up" the added \( \mathrm{N}_{2} \).
However, the explanation in the exercise relying solely on the increase in entropy is inadequate. While increasing entropy can make some reactions more favorable, it's not the main reason for the adjustment here. Instead, Le Chatelier's principle helps explain the reaction's response more effectively. According to this principle, the system takes actions to counteract the change made, aligning the reaction back to its optimal state.

Catalyst Effect

The role of a catalyst in the Haber's process is to speed up the reaction without a change in the equilibrium position. Catalysts achieve this by lowering the activation energy required for the reaction. Importantly, they affect both the forward and backward reactions equally.
The step-by-step solution correctly dismisses the erroneous statement that suggested a catalyst changes the rates inequitably—doubling the forward reaction and increasing the backward reaction by only 1.5 times. This is incorrect as the catalyst's function ensures it does not alter the equilibrium constant or the ratio of forward to backward rates. By decreasing the activation energy, a catalyst accelerates the rates at exactly the same proportional rate, maintaining the equilibrium conditions. Hence, its essential purpose is to allow the system to reach equilibrium faster without altering the ratios of stimulated reactions.