Problem 124 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following conditions are favourable for the feasibility of a reaction ? (a) \(\Delta H=-v e, T \Delta S=+v e\) (b) \(\Delta \mathrm{H}=-\mathrm{ve}, \mathrm{T} \Delta \mathrm{S}=-\mathrm{ve}, \mathrm{T} \Delta \mathrm{S}<\Delta \mathrm{H}\) (c) \(\Delta \mathrm{H}=+\mathrm{ve}, \mathrm{T} \Delta \mathrm{S}=+\mathrm{ve}, \mathrm{T} \Delta \mathrm{S}<\Delta \mathrm{H}\) (d) \(\Delta \mathrm{H}=+\mathrm{ve}, \mathrm{T} \Delta \mathrm{S}=+\mathrm{ve}, \mathrm{T} \Delta \mathrm{S}>\Delta \mathrm{H}\)

Step-by-Step Solution

Answer

Conditions (a), (b), and (d) are favorable for the reaction.

Step 1: Understand Spontaneity Criteria

A reaction is favorable when the Gibbs free energy change \( \Delta G \) is negative. The Gibbs free energy is given by \( \Delta G = \Delta H - T \Delta S \), where \( \Delta H \) is the enthalpy change and \( T \Delta S \) is the product of temperature and the entropy change.

Step 2: Analyze Option (a)

For option (a), \( \Delta H = -ve \) (negative) and \( T \Delta S = +ve \) (positive). This results in \( \Delta G = \Delta H - T \Delta S \). Since \( \Delta H \) is negative and \( T \Delta S \) is positive, the overall \( \Delta G \) will definitely be negative, making the reaction favorable.

Step 3: Analyze Option (b)

In option (b), both \( \Delta H = -ve \) and \( T \Delta S = -ve \) with \( |T \Delta S| < |\Delta H| \). Since both terms are negative, \( \Delta G = \Delta H - T \Delta S = negative - (negative) = negative + positive \). Here, since \(|T \Delta S| < |\Delta H|\), \( \Delta G \) remains negative. Thus, this reaction is also favorable.

Step 4: Analyze Option (c)

For option (c), \( \Delta H = +ve \), \( T \Delta S = +ve \), and \( T \Delta S < \Delta H \). This yields \( \Delta G = \Delta H - T \Delta S = positive - positive \). Since \( T \Delta S < \Delta H \), \( \Delta G \) will be positive, making the reaction unfavorable.

Step 5: Analyze Option (d)

In option (d), \( \Delta H = +ve \), \( T \Delta S = +ve \), and \( T \Delta S > \Delta H \). Here, \( \Delta G = \Delta H - T \Delta S = positive - larger positive \). Because \( T \Delta S > \Delta H \), \( \Delta G \) becomes negative, thus the reaction is favorable.

Step 6: Conclusion of Favorable Conditions

Based on the above analysis, options (a), (b), and (d) lead to negative \( \Delta G \) values, indicating favorable reaction conditions.

Key Concepts

Gibbs Free Energy

Gibbs Free Energy, denoted as \( \Delta G \), is a fundamental concept in thermodynamics that predicts the direction of chemical reactions. This value helps us determine whether a reaction can occur spontaneously without external influence. To find \( \Delta G \), use the equation:

• \( \Delta G = \Delta H - T \Delta S \)

Here, \( \Delta H \) represents enthalpy change and \( T \Delta S \) embodies temperature multiplied by the change in entropy. A key point to remember is that for a reaction to be spontaneous, \( \Delta G \) should be negative.
Detecting a negative \( \Delta G \) implies the reaction can proceed on its own under given conditions, releasing free energy that could potentially do work. Hence, analyzing \( \Delta G \) helps chemists and scientists determine the feasibility and directionality of reactions. Understanding \( \Delta G \) is essential for studying reaction mechanisms and for applications in fields like biochemistry, materials science, and engineering.

Enthalpy

Enthalpy, symbolized as \( \Delta H \), represents the total heat content of a system. It quantifies the heat absorbed or released during a reaction at constant pressure.
For a better grasp, imagine \( \Delta H \) as the heat 'currency' exchanged during chemical processes. If \( \Delta H \) is negative, the reaction is exothermic, meaning it releases heat to the surroundings, often making \( \Delta G \) negative, a favorable condition for reaction spontaneity.
However, if \( \Delta H \) is positive, the reaction is endothermic, absorbing heat and possibly rendering the reaction unfavorable.

• Exothermic Reaction: \( \Delta H < 0 \)
• Endothermic Reaction: \( \Delta H > 0 \)

Understanding how enthalpy interacts with entropy and temperature in the equation for Gibbs Free Energy helps predict the energetic feasibility of reactions.

Entropy

Entropy, denoted as \( \Delta S \), is a measure of disorder or randomness within a system. In the universe of chemistry, systems tend toward maximum entropy. The change in entropy represents the difference in randomness from reactants to products.
A positive \( \Delta S \) indicates an increase in disorder, often aiding in lowering \( \Delta G \) for spontaneity. Conversely, a negative \( \Delta S \) reflects a more ordered system, making ascertaining spontaneity more challenging.

• \( \Delta S > 0 \): Increased disorder, potentially favorable.
• \( \Delta S < 0 \): Decreased disorder, possibly unfavorable.

When calculating \( \Delta G \), the contribution of \( T \Delta S \) can significantly impact whether a reaction proceeds. Distinguishing how entropy impacts balance in the equation for Gibbs Free Energy is vital for understanding reaction dynamics in physical and chemical processes.

Spontaneity of Reactions

The spontaneity of reactions is a key concept in thermodynamics, assessing whether reactions will proceed without external input. If \( \Delta G \) is negative, the reaction is spontaneous under the given conditions, indicating an energetically favorable transition.
Several factors influence spontaneity:

• Enthalpy Change \( (\Delta H) \): Determines heat flow—negative \( \Delta H \) favors spontaneity.
• Entropy Change \( (\Delta S) \): Entropy increase drives reactions towards spontaneity.
• Temperature: Affects the magnitude of \( T \Delta S \), shifting balance in the \( \Delta G \) equation.

To summarize, conditions where \( \Delta H \) is negative and \( T \Delta S \) is positive are generally favorable. However, even when both values are positive, a high enough \( T \Delta S \) can surpass \( \Delta H \), resulting in negative \( \Delta G \). By recognizing these conditions, one can predict whether a reaction will occur spontaneously or require external energy. Analyzing \( \Delta G \) aids in applications from industrial synthesis to biological metabolism.