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Chapter 4: Problem 15

In which case, a spontaneous reaction is impossible at any temperature? (a) \(\Delta H+v e, \Delta S+v e\) (b) \(\Delta H+\mathrm{ve}, \Delta S\)-ve (c) \(\Delta H-v e, \Delta S-\) ve (d) In all cases

Short Answer

Expert verified
Option (b) is impossible for a spontaneous reaction at any temperature.

Step by step solution

01

Understanding the Sign Conventions

For a chemical reaction to be spontaneous, the Gibbs free energy change, denoted as \( \Delta G \), should be negative. The relationship is given by \( \Delta G = \Delta H - T\Delta S \). Here, \( \Delta H \) represents the change in enthalpy and \( \Delta S \) denotes the change in entropy. The sign of \( T \) is always positive since temperature (\( T \)) is in Kelvin, and cannot be negative.
02

Analyzing Option (a)

In option (a), both \( \Delta H \) and \( \Delta S \) are positive (+ve). This implies that \( \Delta G = \Delta H - T\Delta S \). For spontaneity, \( \Delta G \) should be negative. This can happen at high temperatures because the \( T\Delta S \) term will dominate and exceed \( \Delta H \). So, a spontaneous reaction is possible at high temperatures.
03

Analyzing Option (b)

In option (b), \( \Delta H \) is positive (+ve) and \( \Delta S \) is negative (-ve). Thus, \( \Delta G = \Delta H - T\Delta S \) remains positive because both terms \( \Delta H \) and \( -T\Delta S \) contribute positively to \( \Delta G \). Consequently, the reaction cannot be spontaneous at any temperature because \( \Delta G \) is never negative.
04

Analyzing Option (c)

In option (c), \( \Delta H \) is negative (-ve) and \( \Delta S \) is also negative (-ve). Here, \( \Delta G = \Delta H - T\Delta S \). A lower temperature may allow \( \Delta H \) to be more negative than \( -T\Delta S \), making \( \Delta G \) negative and the reaction spontaneous. Hence, a spontaneous reaction is possible at low temperatures.
05

Summarizing All Options

From the analysis: Option (a) can be spontaneous at high temperatures, option (c) at low temperatures, and option (b) cannot be spontaneous at any temperature because \( \Delta G \) is always positive due to both contributions being positive.

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Key Concepts

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

Enthalpy
Enthalpy, denoted as \( \Delta H \), is a measure of the total energy content of a system. It accounts for both the energy needed to break bonds and the energy released when new bonds form.
  • When \( \Delta H \) is negative, the process is exothermic, meaning it releases heat.
  • Conversely, a positive \( \Delta H \) indicates an endothermic process, where the system absorbs heat.

Understanding enthalpy is essential in determining whether a reaction requires or releases energy. In the context of spontaneous reactions, the sign of \( \Delta H \) can significantly influence the Gibbs free energy, \( \Delta G \).
Together with temperature and entropy, enthalpy helps predict whether a reaction is likely to occur on its own.
Entropy
Entropy, represented by \( \Delta S \), measures the disorder or randomness of a system. It is a crucial factor in thermodynamics and plays a vital role in the spontaneity of reactions.
  • A positive \( \Delta S \) implies an increase in disorder, which is often favored in natural processes.
  • A negative \( \Delta S \) suggests a more ordered system, which may require energy input.

Entropy contributes to the Gibbs free energy equation through the term \( T \Delta S \). Depending on the sign of \( \Delta S \), this term can either help or hinder a reaction's spontaneity.
Think of entropy as the universe's tendency toward chaos, where reactions that increase entropy are often spontaneous, especially at high temperatures.
Spontaneous Reaction
A spontaneous reaction is one that occurs without needing external energy. The Gibbs free energy change, \( \Delta G \), is the primary indicator of whether a reaction is spontaneous.
  • If \( \Delta G \) is negative, the reaction is spontaneous.
  • If \( \Delta G \) is positive, the reaction is non-spontaneous unless conditions change.

The formula \( \Delta G = \Delta H - T \Delta S \) combines enthalpy \( \Delta H \), entropy \( \Delta S \), and temperature \( T \) to determine \( \Delta G \).
  • At higher temperatures, the \( T \Delta S \) term becomes more significant, which can shift \( \Delta G \) to become negative if \( \Delta S \) is positive.
  • At lower temperatures, \( \Delta H \) tends to dominate the equation.
Thus, spontaneous reactions depend heavily on the balance between enthalpy and entropy, and the conditions under which they occur. Understanding these elements is key to predicting reaction behaviors in chemistry.

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Most popular questions from this chapter

The work done by a system is \(8 \mathrm{~J}\), when \(40 \mathrm{~J}\) heat is supplied to it. The change in internal energy of the system during the process: (a) \(32 \mathrm{~J}\) (b) \(40 \mathrm{~J}\) (c) \(36 \mathrm{~J}\) (d) \(44 \mathrm{~J}\)

If an endothermic reaction is non-spontaneous at freezing point of water and becomes feasible at its boiling point, then (a) \(\Delta H\) is \(-v e, \Delta S\) is \(+\) ve (b) \(\Delta H\) and \(\Delta S\) both are +ve (c) \(\Delta H\) and \(\triangle S\) both are -ve (d) \(\Delta H\) is \(+\mathrm{ve}, \Delta S\) is \(-\mathrm{ve}\)

Heat of neutralization of an acid with a base is \(13.7 \mathrm{kcal}\) when (a) both acid and base are weak (b) acid is weak and base is strong (c) acid is strong and base is weak (d) both acid and base are strong

An example of extensive property is (a) temperature (b) internal energy (c) viscosity (d) surface tension

Given that bond energies of \(\mathrm{H}-\mathrm{H}\) and \(\mathrm{Cl}-\mathrm{Cl}\) are \(430 \mathrm{~kJ} / \mathrm{mol}\) and \(240 \mathrm{~kJ} / \mathrm{mol}\), respectively. \(\Delta H_{f}\) for \(\mathrm{HCl}\) is \(-90 \mathrm{~kJ} / \mathrm{mol}\). Bond enthalpy of \(\mathrm{HCl}\) is (a) \(380 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(425 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(245 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(290 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
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