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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) \(\Delta \mathrm{H}=+\mathrm{ve}, \mathrm{T} \Delta \mathrm{S}=+\mathrm{ve}, \mathrm{T} \Delta \mathrm{S}>\Delta \mathrm{H}\)

Short Answer

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

Step by step solution

01

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.
02

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.
03

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.
04

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.
05

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.
06

Conclusion of Favorable Conditions

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Classify each of the following processes as spontaneous or non-spontaneous. I. \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}), \mathrm{T}=25^{\circ} \mathrm{C}\) vessel open to atomsphere with \(50 \%\) relative humidity. II. \(\mathrm{H}_{2} \mathrm{O}(\mathrm{s}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}), \mathrm{T}=25^{\circ} \mathrm{C}, \mathrm{P}=1 \mathrm{~atm}\) (a) I and II are both non-spontaneous (b) I and II are both spontaneous (c) I is non-spontaneous and II is spontaneous (d) I is spontaneous and II is non-spontaneous

In thermodynamics, a process is called reversible when (a) the surroundings are always in equilibrium with the system (b) there is no boundary between system and sur roundings (c) the surroundings and system change into each other (d) the system changes into the surroundings sponta neously

The incorrect statement(s) among the following is/ are (a) For a system undergoing a cyclic change, \(\oint \frac{\mathrm{fq}}{\mathrm{T}}>0\). (b) A real crystal has lower entropy than ideal crystal. (c) Pressure is an extensive property. (d) A reversible process is always dynamic in nature.

Which are the intensive properties? (a) Volume (b) Enthalpy (c) Temperature (d) Refractive index

Identify the correct statement regarding entropy. (a) at absolute zero temperature, entropy of a perfectly crystalline substance is +ve. (b) at absolute zero temperature, entropy of a perfectly crystalline substance is taken to be zero (c) at absolute zero temperature, entropy of all crystalline substances in taken to be 0 . (d) at \(0^{\circ} \mathrm{C}\), entropy of a perfectly crystalline substance is taken to be 0 .

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